notes for the course analisi di fourier - unimi.itthe fourier transform 11 2.3. the schwartz space...

NOTES FOR THE COURSE ANALISI DI FOURIER

MARCO M. PELOSO

Contents

1. Fourier series 11.1. The torus 11.2. Norm convergence of Fourier series 52. The Fourier transform 72.1. Convolutions 72.2. The Fourier transform 112.3. The Schwartz space 152.4. The space of tempered distributions 193. The Marcinkiewicz interpolation theorem and the Calderon–Zygmund decomposition 283.1. The Marcinkiewicz interpolation theorem 283.2. The Calderon–Zygmund decomposition of an L1-function 314. The Hilbert transform 334.1. The Lp-boundedness of the Hilbert transform 364.2. Further properties of the Hilbert transform 395. Singular integrals 415.1. The Calderon–Zygmund theorem 415.2. Homogeneous distributions 435.3. The Riesz transforms 465.4. Solution of the Laplace equation 475.5. The heat operator 515.6. The Riesz potentials 525.7. †Construction of Calderon–Zygmund kernels 565.8. †Vector-valued singular integrals 626. Fourier multipliers 676.1. The space Mp of Lp-bounded Fourier multipliers 676.2. Multiple Fourier series and convergence in the Lp-norm 716.3. The Sobolev spaces Hs 746.4. †The Mihlin–Hormander multiplier theorem 786.5. †Proof of Thm. 6.24 807. Partial Differential Operators 827.1. The Sobolev spaces Hs 827.2. Partial differential operators 858. †Littlewood–Paley theory 908.1. An application 908.2. The Littlewood–Paley theorem 938.3. The Marcinkiewicz multiplier theorem 969. Symbols and pseudodifferential operators 989.1. The kernel of a pseudodifferential operator 1019.2. Sobolev continuity of a pseudodifferential operator 104

Appunti per il corso Analisi di Fourier per i Corsi di Laurea in Matematica dell’Universita di Milano, a.a.2017/18.

A † indicates the parts that are not required for the exam.– June 25, 2018.

9.3. Asymptotic expansion of symbols 1059.4. Adjoints and products of pseudodifferential operators 1089.5. Local regularity of elliptic operators 11210. Sum of squares of vector fields 115Appendix A. Some results on Lp spaces 121Appendix B. Some exercises 125References 127

HARMONIC ANALYSIS 1

1. Fourier series

In this brief first section we recall the main facts about classical Fourier series on the torus,in part as a historical introduction, in part to motivate some of the results and techniques thatwe will devolop later in the notes.

1.1. The torus. We will denote the additive group of real numbers by R and by Z the subgroupconsisting of the integers.

We define the torus T to be the group which is the quotient T = R/2πZ. There is an obviousidentification between functions on T and 2π-periodic functions on R.

This also allows an implicit introduction of notions such as continuity, differentiability, etc.for functions on T. The Lebesgue measure on T, also, can be defined by means of the aboveidentification: a function f is integrable on T if the corresponding 2π-periodic function, whichwe denote again by f , is integrable on [0, 2π) and we set∫

Tf(t) dt =

∫ 2π

0f(x) dx .

A fundamental property of the measure dt on T is that it is translation invariant, that is, forall t0 ∈ R ∫

Tf(t− t0) dt =

∫Tf(t) dt ,

as it is easily checked, using the periodicity of f in R. (Here, of course, we have defined thefunction t 7→ f(t− t0) via the translated of the periodic function x 7→ f(x) on the real line.)

We recall that the unit circle ∂D = ζ ∈ C : ζ = eiθ inherits the multiplicative structurefrom C and it is homeomorphic (as a topological group) to the torus T via the mapping t 7→ eit,thus defining on T the structure of a topological group.

With an abuse of notation, we are going to use the same letter to denote functions on ∂Dand T, that is, we write

f(t) = f(eit)

for t ∈ R and f 2π-periodic.

We denote by Lp(T), 1 ≤ p < ∞ the Lebesgue space of measure (equivalence classes of)functions such that

‖f‖Lp(T) :=( 1

2π

∫T|f(t)|p dt

)1/p

is finite, and by L∞(T) the space of essentially bounded functions on T. Notice that, since thetotal mass of T is finite, for 1 ≤ p ≤ q ≤ ∞ we have the (continuous) embedding

Lq(T) ⊂ Lp(T) .

We have the following fundamental definition of Fourier coefficient.

Definition 1.1. Let f ∈ L1(T) (or, as an integrable function on the unit circle ∂D). We definethe n-th Fourier coefficient of the restriction of f

f(n) =1

2π

∫ 2π

0f(eiθ)e−inθ dθ .

We turn to some elementary properties of Fourier coefficients.

2 M. M. PELOSO

Proposition 1.2. Let f, g ∈ L1(T), then

(i) (f + g) (n) = f(n) + g(n).

(ii) For any complex number α, (αf) (n) = αf(n).

(iii) If f is the complex conjugate of f , then (f) (n) = f(−n).

(iv) Writing fτ (t) = f(t− τ), for t, τ ∈ T, we have (fτ ) (n) = f(n)e−inτ .

(v) |f(n)| ≤ ‖f‖L1(T).

Proof. The proof is elementary and left as an exercise. 2

Given f, g ∈ L1(T), the convolution on T is defined as

f ∗ g(t) =1

2π

∫ 2π

0f(t− τ)g(τ) dτ . (1.1)

Proposition 1.3. For f, g ∈ L1(T), f ∗ g ∈ L1(T) and

(i) ‖f ∗ g‖L1 ≤ ‖f‖L1‖g‖L1;

(ii) (f ∗ g) (n) = f(n)g(n).

Proof. Exercise. 2

Definition 1.4. A function of the form

p(t) =

N∑k=−N

akeikt ,

for a positive integer N , is called a trigonometric polynomial.

It is immediate to observe that a trigonometric polynomial p is smooth, hence continuous andin L1 and that, if p is as above, p(k) = ak, for all k (observing that ak = 0 for |k| > N).

Therefore, if f ∈ L1 and p is as above,

(f ∗ p) (k) = akf(k) . (1.2)

Definition 1.5. A family of 2π-periodic functions ΦNN=1,2,... is called a summability kernelif they satisfy the following properties:

(1)1

2π

∫ 2π

0ΦN (eiθ) dθ = 1 for N = 1, 2, . . . ;

(2) ‖ΦN‖L1 ≤ C with C independent of N ;

(3) for every δ > 0, limN→+∞

∫δ≤|θ|≤π

|ΦN (eiθ)| dθ = 0.

We can give an analogous definition of summability kernel using a countinous parameter t > 0(and typically letting t→ 0+).

We recall (see [Ka] e.g.) that if ΦNN>0 is a summability kernel on T and g ∈ Lp(T),1 ≤ p < ∞, then ΦN ∗ g → g in the Lp-norm, and that if g ∈ C(T), then ΦN ∗ g → g in thesup-norm.

HARMONIC ANALYSIS 3

Two of the most useful summability kernels, and also best known, are Fejer’s kernel (whichwe denote by KN) defined by

KN (t) =N∑

j=−N

(1− |j|

N + 1

)eijt , (1.3)

and the Poisson kernel, given by

Pr(t) =+∞∑j=−∞

r|n|eijt = 1 + 2+∞∑j=1

rj cos(jt) =1− r2

1− 2r cos t+ r2. (1.4)

The fact that KN satisfies condition (1) in Def. 1.5 above is obvious; that KN satisfies (2)and (3) is clear from the following result.

Lemma 1.6. We have that

KN (t) =1

N + 1

(sin(N+1

2 t)

sin t2

)2

.

Proof. Recall that

sin2( t

2

)=

1

2(1− cos t) = −1

4e−it +

1

2− 1

4eit .

A direct computation of the coefficients in the product shows that(− 1

4e−it +

1

2− 1

4eit) N∑j=−N

(1− |j|

N + 1

)eijt =

1

N + 1

(− 1

4e−i(N+1)t +

1

2− 1

4ei(N+1)t

).

This proves the lemma. 2

We now have some consequences of the fact that KN is a summability kernel.

Corollary 1.7. The trigonometric polynomials are dense in Lp(T), 1 ≤ p <∞ and in C(T) inthe uniform norm.

Moreover, if f ∈ L1(T) and f(n) = 0 for all n, then f = 0.

Proof. If we set σN (f) = KN ∗ f , by (1.2) it follows that σN (f) is a trigonometric polynomial.The two statements now follow from the properties of summability kernels. 2

The terms σN (f) are classically called the Cesaro means, that the corollary says that theyconverge in norm to f .

Corollary 1.8. (The Riemann-Lebesgue Lemma) Let f ∈ L1(T), then

limn→+∞

f(n) = 0 .

Proof. Let ε > 0 and let P be a trigonometric polynomial on T such that ‖f − P‖L1 < ε. If |n|is greater than degree of P , then

|f(n)| = |(f − P )(n)| ≤ ‖f − P‖L1 < ε .

The conclusion follows. 2

4 M. M. PELOSO

Definition 1.9. Given an L1 function f , the formal series

+∞∑k=−∞

f(k)eikt

is called the Fourier series of f .

Perhaps one of the main questions in harmonic analysis is to establish under which conditionthe Fourier series of f converges back to f . We observe that the partial sum

Σn(f)(t) =

n∑k=−n

f(k)eikt =( n∑k=−n

eikt)∗ f(t) ,

is then given by the convolution with the kernel Dn

Dn(t) =n∑

k=−neikt =

sin((n+ 1

2)t)

sin t2

, (1.5)

called the Dirichlet kernel. The last equality follows since(eit2 − e−i

t2

)( n∑k=−n

eikt)

=

n∑k=−n

ei(k+ 12

)t − e−i(t−12

)t = ei(n+ 12

)t − e−i(n+ 12

)t .

It is important to notice that the Dirichlet kernel is not a summability kernel, since it satisfies(1), but not (2-3) in Def. 1.5. For, in particular

‖Dn‖L1(T) =1

2π

∫ π

−π

∣∣∣sin ((n+ 12)t)

sin t2

∣∣∣ dt ≥ 1

2π

∫ δ

−δ

∣∣∣sin ((n+ 12)t)

sin t2

∣∣∣ dt ≥ Cn . (1.6)

Hence, the question of the norm convergence of the Fourier series of f to f iteself in theLp-norm, is a much harder question.

In some respects the greatest success in representing functions by means of their Fourier serieshappens for square summable functions. The reason is that L2(T) is a Hilbert space1, its innerproduct being defined by

〈f, g〉 =1

2π

∫Tf(t)g(t) dt , (1.7)

Lemma 1.10. The set of exponentials eintn∈Z form a complete orthonormal system for L2(T).

Proof. The orthogonality and the fact that are of norm 1 is a straightforward calculation, thecompleteness follows Corollary 1.7. The details are left as an exercise. 2

We denote by `2 the Hilbert space of square summable sequences ak, that is,

`2 =ak : ak ∈ C, k ∈ Z and

+∞∑k=−∞

|ak|2 <∞. (1.8)

From the abstract theory of Hilbert spaces and the lemma it now follows the following theorem.

Theorem 1.11. Let f ∈ L2(T). Then

1For a quick tailored introduction to the theory of Hilbert spaces and convergence of Fourier series see, e.g.[Ka].

HARMONIC ANALYSIS 5

(i)+∞∑

k=−∞|f(k)|2 =

1

2π

∫ 2π

0|f(t)|2 dt;

(ii) f = limN→+∞

N∑k=−N

f(k)eikt in the L2(T) norm;

(iii) for any ak ∈ `2, there exists a unique f ∈ L2(T) such that ak = f(k);(iv) let f, g ∈ L2(T), then

1

2π

∫ 2π

0f(t)g(t) dt =

+∞∑k=−∞

f(k)g(k) .

1.2. Norm convergence of Fourier series. We have the following fundamental definition.On the other hand, we have

Definition 1.12. Given f ∈ Lp(T), 1 ≤ p <∞, or f ∈ C(T), we say that the Fourier series of

f ,∑

j∈Z f(j)eijt converges in norm to f if we set Σn(f) =∑n

j=−n f(j)eijt we have

‖Σn(f)− f‖ → 0 as n→ +∞ ,

where ‖ · ‖ denotes the appropriate norm. In this case, we say that the given space admitsconvergence of the Fourier series.

The question of the norm convergence of Fourier series while is tightly connected to theboundedness of the mapping of real harmonic function to its harmonic conjugate, as we willsoon see. Suppose u is a real harmonic function on the unit disk. Then we know that u admitsa harmonic conjugate v, that is, a real harmonic function such that u+ iv is holomorphic in D.The function v is uniquely determined modulo constants, so we require v to vanish at the originz = 0, i.e. we set v(0) = 0.

We observe that if g is trigonometric polynomial of degree m, then Σn(g) = g for n ≥ m;hence we trivially have norm convergence in any norm for all trigonometric polynomials.

We reduce the problem of the convergence of Fourier series to a single estimate.

Theorem 1.13. With the notation above, let B be any of the Banach spaces Lp(T), 1 ≤ p <∞,or C(T). Then B admits norm convergence of Fourier series if and only if there exists a constantM > 0 such that for any positive integer n

‖Σn‖ ≤M ,

where ‖Σn‖ denotes the operator norm of Σn on B.

Proof. If Σn(f) converges in norm for any f ∈ B, then ‖Σn(f)‖ is a bounded sequence. By theuniform boundedness principle ([Ru]), the operator norms ‖Σn‖ are uniformly bounded.

Conversely, suppose that ‖Σn‖ ≤M , for some M > 0 and all n. Given f ∈ B, fix ε > 0. Letg be a trigonometric polynomial such that ‖f − g‖B < ε. Then,

‖Σn(f)− f‖B ≤ ‖Σn(f)− Σn(g)‖B + ‖Σn(g)− g‖B + ‖g − f‖B≤Mε+ ε ,

for n sufficiently large. This proves the result.

We begin by observing that Fourier series do not converge in L1(T)-norm, nor in uniformnorm.

6 M. M. PELOSO

Theorem 1.14. Let B be either L1(T) or C(T). Then B does not admit norm convergence ofFourier series.

Proof. By the previous theorem, it suffices to estimates the operators norms ‖Σn‖ as n →+∞, when B = L1(T) and B = C(T). Since Σn(f) = Dn ∗ f , we have that ‖Σn(f)‖B ≤‖Dn‖L1(T)‖f‖B, so that ‖Σn‖B ≤ ‖Dn‖L1(T). On the other hand, let B = L1(T) first. Recallthat Fejer’s kernel (1.3) KN is such that ‖KN‖L1(bT ) = 1, and KN ∗ f → f in the norm of B.Then,

‖Σn‖ ≥ ‖Σn(KN )‖L1(T) = ‖KN ∗Dn‖L1(T) → ‖Dn‖L1(T) as N → +∞.Therefore, when B = L1(T), ‖Σn‖ = ‖Dn‖L1(T) ≥ Cn, see (1.6). Now let B = C(T). We wishto show also in this case ‖Σn‖ → +∞ as n→ +∞. Let gn ∈ C(T) be given by sgn(Dn) wheresgn(Dn) is continuous, and we may assume that the measure of the set where gn 6= sgn(Dn) isless than ε > 0. Then,

Σ(gn)(0) =1

2π

∫ 2π

0gn(t)Dn(t) dt ≥ ‖Dn‖L1(T) − ε ≥ Cn− ε→ +∞ .

This completes the proof. We now relate the problem of finding a uniform estimate for the norms ‖Σn‖ to the (harmonic)

conjugation.

Definition 1.15. Given f ∈ C∞(T), we define the conjugate Fourier series the series∑j∈Z−i sgn(j)f(j)eijt = Q(f)(eit) .

We say that the Banach space B admits conjugation if Q extends to a bounded operator on B.

We also define the Szego projection. Given f ∈ L2(T), we define

Sf(t) =+∞∑n=0

f(n)eint . (1.9)

Theorem 1.16. Let B any of the Banach spaces Lp(T), 1 ≤ p < ∞, or C(T). Then thefollowing are equivalent:

(i) B admits convergence of Fourier series;(ii) Q is bounded on B;(iii) the Szego projection S is bounded on B.

Proof. Since 12(I + iQ) = S − 1

2M , where Mg = g(0), (ii) and (iii) are clearly equivalent.Next, we notice that

Σ]2n(f) :=

2n∑j=0

f(j)eijt = eintΣn

(e−intf

).

Thus, the operator norms of Σ]2n are uniformly bounded if and only if the operator norms of Σn

are uniformly bounded. The equivalence of (i) and (iii) now follows from Thm. 1.13. Thus, we are left with determining whether, say, the operator Q is bounded on Lp(T), 1 <

p < ∞. To study this boundedness, we need the theory of singular integrals, and in particularof the analysis of the so-called Hilbert transform. We will address these and further issues in allthe remaining parts of these notes.

HARMONIC ANALYSIS 7

2. The Fourier transform

We begin by introducing some notation and basic definitions and recalling properties of a fewfunction spaces.

We denote partial derivatives by ∂xj . We denote by N the set of non-negative integers. Ifα is a multi-index, i.e. α ∈ Nn, we write ∂αx = ∂α1

x1 · · · ∂αnxn , |α| =

∑nj=1 α1 + · · · + αn, and

α! =∏nj=1 α1 · · ·αn.

Given a function f on Rn we denote by τy the translation by y, that is,

τyf(x) = f(x− y) . (2.1)

We recall that when g is a continuous function, its support supp g is the smallest closed setoutside of which it vanishes, that is, it is the closure of g−1

(c0

). We denote by Cc = Cc(Rn)

the space of continuous function with compact support and by C0 = C0(Rn) the continuousfunctions that vanish at infinity, that is, lim|x|→+∞ g(x) = 0.

The following result is elementary and its proof is left as an exercise (see Exercise 2 inAppendix B).

Lemma 2.1. Endowed with the uniform norm, the space C0(Rn) is a Banach space and Cc(Rn)

is a dense subspace. Functions in C0 are uniformly continuous.

2.1. Convolutions. If f and g are measurable functions on Rn, their convolution f ∗ g is thefunction defined as

f ∗ g(x) =

∫f(x− y)g(y) dy

for all x for which the integral exists.The question of measurability of the convolution deserves a discussion. For f Lebesgue

measurable, the function F (x, y) = f(x− y) turns out to be measurable on Rn ×Rn. One cansee this fact by writing F = f g where g(x, y) = x − y and show that g−1(E) of a Lebesguemeasurable set is still a Lebesgue measurable set.2

We now set some notation that will be use sistematically throughout these notes.Given f on Rn and t > 0 we define

ft(x) = t−nf(x/t) and f t(x) = f(tx) . (2.2)

Notice that, if f ∈ L1 then∫ft(x) dx =

∫t−nf(x/t) dx =

∫f(y) dy

and that, if the integral is well defined (for instance if f, g ∈ L2)∫ft(x)g(x) dx =

∫f(x)gt(x) dx ,

as a simple change of variables shows.

2In general, if f and g are Lebesgue measurable functions, say f, g : R→ R, then the composition f g, evenwhen defined, may not be measurable. One way to remedy this, is to redefine f and g on sets of measure 0 andtake them to be Borel measurable. A function f is said to be Borel measurable if f−1(U) of any open set U is aBorel set, that is countable union and/or intersection of open and closed sets.

8 M. M. PELOSO

The next result is elementary, however we need to define the support of a function f that islocally integrable, but defined only a.e. When f is only a measurable function we define

supp f = c(∪ A : A open and f = 0 a.e. on A

)). (2.3)

Thus, supp f is a closed set and outside of which f = 0 a.e. For consistency, we need to checkthat if f is also continuous, the above definition coincides with the one given for continuousfunctions. To do so, we only need to check that if E is a closed set outside of which f vanishes,then E contains supp f as defined (2.3). But this is clear passing to the complementary sets(and reversing the inclusion).

Proposition 2.2. Assume that the integrals below exist. Then

(i) f ∗ g = g ∗ f ;(ii) f ∗ (g ∗ h) = (f ∗ g) ∗ h;(iii) τy(f ∗ g) = (τyf) ∗ g = f ∗ (τyg);(iv) supp(f ∗ g) ⊆ supp f + supp g = x ∈ Rn : x = y + z, y ∈ supp f, z ∈ supp g.

From Fubini’s theorem it is clear that the convolution of two L1-functions is still in L1. Butmore can be said. Now we prove the following.

Theorem 2.3. The following properties hold.

(i) (Young’s inequality) If f ∈ L1 and g ∈ Lp, with 1 ≤ p ≤ ∞ then f ∗ g exists for almostevery x, f ∗ g ∈ Lp and

‖f ∗ g‖p ≤ ‖f‖1‖g‖p .(ii) (Generalized Young’s inequality) Let 1 ≤ p, q, r ≤ ∞ be such that 1/p + 1/q = 1 + 1/r.

Let f ∈ Lp, q ∈ Lq, then f ∗ g ∈ Lr and

‖f ∗ g‖r ≤ ‖f‖p‖g‖q .

(iii) If f ∈ Lp and g ∈ Lp′, where 1p + 1

p′ = 1, 1 ≤ p <∞, then f ∗ g ∈ C0 and

‖f ∗ g‖∞ ≤ ‖f‖p‖g‖p′ .

(iv) If f ∈ L1 and g ∈ Ck and ∂αx g ∈ L∞ for |α| ≤ k, then, for |α| ≤ k, ∂αx (f ∗ g) ∈ C0 and

∂αx (f ∗ g) = f ∗ ∂αx g .

Proof. Notice that (i) follows from (ii) by taking p = 1 and q = r in (ii). Now we prove (ii).Notice that 1/p′ + 1/q′ + 1/r = 1. Then, by the generalized Holder’s inequality3

|f ∗ g(x)| ≤∫|f(x− y)| |g(y)| dy

=

∫|f(x− y)|p/r|f(x− y)|1−p/r|g(y)|q/r|g(y)|1−q/r dy

≤(∫|f(x− y)|p|g(y)|q dy

)1/r(∫|f(x− y)|q′(1−p/r) dy

)1/q′(∫|g(y)|p′(1−q/r) dy

)1/p′

=

(∫|f(x− y)|p|g(y)|q dy

)1/r

‖f‖1−p/rp ‖g‖1−q/rq ,

3See Exercise 4.

HARMONIC ANALYSIS 9

where we have used the fact that 1/q′ = 1/p − 1/r so that q′(1 − p/r) = p and analogously1/p′ = 1/q − 1/r and p′(1− q/r) = q. Then,∫

|f ∗ g(x)|r dx ≤∫ (∫

|f(x− y)|p|g(y)|q dy)‖f‖r−pp ‖g‖r−qq dx

= ‖f‖r−pp ‖g‖r−qq

∫ (|g(y)|q

∫|f(x− y)|p dx

)dy

= ‖f‖rp‖g‖rqfrom which the conclusion follows.

For part (iii) see we use differentiation under the integral sign, see Exercise 5. Let |α| = 1, thatis, α = ej . Then, ∂xjg is bounded and |f(y)∂xjg(x− y)| ≤ C|f(y)|, where f ∈ L1, indepently ofx. Then we can pass the derivative under the integral sign

∂xj

(∫f(y)g(x− y) dy

)=

∫f(y)∂xjg(x− y) dy ,

and finally proceed by induction. 2

Theorem 2.4. Let ϕ ∈ L1 and let∫ϕ(x) dx = a. Then,

(i) If f ∈ Lp, 1 ≤ p <∞, then f ∗ ϕt → af in the Lp-norm, as t→ 0.(ii) If f is bounded and uniformly continuous, then ϕt ∗ f → af uniformly as t→ 0.

Proof. We have

f ∗ ϕt(x)− af(x) =

∫ (f(x− y)− f(x)

)ϕt(y) dy

=

∫ (f(x− tz)− f(x)

)ϕ(z) dz

=

∫ ((τtzf)(x)− f(x)

)ϕ(z) dz .

Now we apply Minkowski’s integral inequality4

‖f ∗ ϕt − af‖p =∥∥∫ ((τtzf)− f

)(·)ϕ(z) dz

∥∥p≤∫ ∥∥(τtzf)− f

∥∥p|ϕ(z)| dz .

But ‖(τtzf)−f‖p ≤ 2‖f‖p and tends to 0 as t→ 0+ for each z (see Exercise I.6). Thus, assertion(i) follows from the dominated convergence theorem.

If f is bounded and uniformly continuous, then it is still true that ‖(τtzf)−f‖∞ ≤ 2‖f‖∞ andtends to 0 as t→ 0+ for each z. Thus, assertion (ii) also follows from the dominated convergencetheorem. 2

Definition 2.5. We denote by D (or C∞c ) the space of C∞-functions with compact support,and more generally by C∞c (Ω) the space of C∞-functions having compact support contained inΩ.

The next result is elementary, however, we invite the readers to convince themselves of itsvalidity.

4See Appendix, Theorem A.8.

10 M. M. PELOSO

Lemma 2.6. The space D is non-empty; for instance the function

ϕ(x) =

e− 1

1−|x|2 if |x| < 1

0 if |x| ≥ 1

is an element of D.

By translation and dilation, it is easy to produce an element ψ of D with suppψ ⊆ B(x0, δ);it suffices to take ϕ((x− x0)/δ).

Proposition 2.7. (C∞-Urysohn’s Lemma) Let K be a compact set in Rn, U an open set thatcontains K. Then there exists f ∈ D such that f ≥ 0, f = 1 on K and f vanishes outside U .

Proof. Let δ = dist(K, cU) and let V = y : dist(y,K) < δ/3. Let ψ ∈ D,∫ψ = 1,

suppψ ⊆ x : |x| ≤ δ/3. Notice that ψ(x) = cϕ(3x/δ), with ϕ as in the previous lemma,satisfies the requirements for a suitable constant c. Then we set f = χV ∗ ψ. We have that,

supp f ⊆ V + suppψ ⊆ dist(x,K) < 2δ/3 ⊆ U ,

Moreover, 0 ≤ f ≤ 1, and notice that for x ∈ K, y ∈ V , |x− y| < δ/3 and

f(x) =

∫χV (y)ψ(x− y) dy =

∫V ∩y: |x−y|<δ/3

ψ(x− y) dy

=

∫y: |x−y|<δ/3

ψ(x− y) dy = 1 .


Corollary 2.8. The space D is dense in Lp, 1 ≤ p <∞.

Proof. We have shown that, if f ∈ Lp, 1 ≤ p < ∞, and ϕ ∈ D,∫ϕ = 1, then f ∗ ϕt → f in

the Lp-norm. Let f ∈ Lp and let ε > 0 be fixed. Let K be a compact set chosen in such a waythat ‖fχ cK‖p < ε/2. Next, let t0 be such that then ‖fχK − (fχK) ∗ ϕt‖p < ε/2 for 0 < t < t0.Notice that (fχK) ∗ ϕt ∈ D for all t > 0. Then

‖f − (fχK) ∗ ϕt‖p < ‖f − fχK‖p + ‖fχK − (fχK) ∗ ϕt‖p < ε . 2

We conclude this section with a result (Prop. 2.10) that we will not use, but that is quiteuseful in a variety of problems.

Lemma 2.9. Let E be a compact set in Rn and let Uj, j = 1 . . . , N be a cover of E. Then,there exist ϕ1, . . . , ϕN ∈ D such that suppϕj ⊆ Uj for all j,

∑j ϕj(x) ≤ 1 for all x, and∑

j ϕj(x) = 1 for all x ∈ E.

Proof. We first observe that there exist compact sets Ej such that Ej ⊂ Uj and E = ∪Nj=1Kj .

Indeed, for each x ∈ E there exists a ball B(x, rx) such that B(x, rx) ⊆ Uj for some j = 1, . . . , N .Since E is compact, we can select B(x1, rx1), . . . , B(xM , rxM ) so that E ⊆ ∪Mk=1B(xk, rxk). Now,we let

Ej = ∪B(xk, rxk) : B(xk, rxk) ⊆ Uj

.

Next, Prop. 2.7 guarantees the existence of ψj ∈ D with support in Uj and equal to 1 on Ej .Then we define ϕj = ψj/(

∑j ψj).

HARMONIC ANALYSIS 11

It is easy to see that ϕj ∈ D, 0 ≤ ϕj ≤ 1 and that if x ∈ E, then x ∈ Ej for at least one j, sothat ϕj(x) 6= 0 (in fact ϕj(x) = 1) for at least one j. Therefore, if x ∈ E,

ϕ(x) =

N∑j=1

ϕj(x) =

N∑j=1

ψj(x)∑Nk=1 ψk(x)

= 1 .

Proposition 2.10. (Partition of unity) Let Uk be a locally finite open cover5 of Rn. Then,there exists ϕk ⊆ D such that

• suppϕk ⊆ Uk for all k;

• 0 ≤ ϕk ≤ 1 for all k;

•∑

k ϕk(x) = 1 for all x ∈ Rn.

Proof. Let Q be the closed unit cube Q = x : 0 ≤ xj ≤ 1, j = 1, . . . , n and B the ball centeredat the center of Q and radius 1, so that Q ⊂ B. For κ ∈ Zn we set Qκ = Q+κ and Vκ = B+κ.Then, Rn = ∪κ∈ZQκ, Vκ is open containing Qκ and each x ∈ Rn belongs to at most Cn of theVκ’s, with Cn independent of κ.

For each κ ∈ Zn, we can find a finite sub-collection U (κ)j j=1,...,Nκ of Uk that covers Qκ.

For each κ ∈ Zn fixed, we now apply Lemma 2.9 in the case E = Qκ and finite open cover

U (κ)j , j = 1, . . . , Nκ and obtain ψκ,j, j = 1, . . . , Nκ, such that ψκ,j ∈ D, suppψκ,j ⊂ U

(κ)j ,

and 0 ≤ ψκ,j ≤ 1.Finally we set

ϕκ,j(x) =ψκ,j(x)∑

κ∈Zn∑Nκ

j=1 ψκ,j(x).

It is easy now to convince oneself that the functions ϕκ,j , κ ∈ Zn, j = 1, . . . , Nκ satisfy theconclusions in the statement.

2.2. The Fourier transform. We now come to the main object of our study. For f ∈ L1(Rn)we define the Fourier transform of f as

f(ξ) =

∫f(x)e−2πixξ dx .

We will also denote it by Ff , so that to emphasyse the action of the Fourier transform as amapping.

It is clear that f is well defined for f ∈ L1 and that ‖f‖∞ ≤ ‖f‖1. Moreover, f is continuous,as an immediate application of the dominated convergence theorem. Thus,

F : L1(Rn)→ C(Rn) ∩ L∞(Rn) (2.4)

is bounded.We now see the first elementary properties of the Fourier transform.

Proposition 2.11. Let f, g ∈ L1. Then, the following hold.

(i) For any y, η ∈ Rn, F(τyf)(ξ) = e−2πiyξ f(ξ) and τηf(ξ) = F(e2πixηf)(ξ).

(ii) F(f ∗ g) = f g.

(iii) If xαf ∈ L1 for |α| ≤ k, then f ∈ Ck and ∂αξ f(ξ) = F((−2πix)αf

)(ξ).

5A collection Uk of sets is called a locally finite cover of X if (i) ∪kUk ⊇ X, and (ii) for each x ∈ X thereexist only finitely many Uk1 , . . . , Ukm that contains x.

12 M. M. PELOSO

(iv) Suppose f ∈ Ck, ∂αx f ∈ L1 for |α| ≤ k, then F(∂αx f)(ξ) = (2πiξ)αf(ξ).

Proof. (i) is elementary:

(τyf ) (ξ) =

∫f(x− y)e−2πixξ dx =

∫f(x)e−2πi(x+y)ξ dx

= e−2πiyξ f(ξ) ,

and analogously for the other part of the statement.(ii) is also elementary. By Fubini’s theorem,

(f ∗ g) (ξ) =

∫∫f(x− y)g(y) dye−2πixξ dx

=

∫∫f(x− y)e−2πi(x−y)ξ dxg(y)e−2πiyξ dy

= f(ξ)g(ξ) .

(iii) We argue as in the proof of Thm. 2.2 (iii). We begin with the case |α| = 1 and observethat

|∂ξj(f(x)e−2πixξ

)| ≤ C|xjf(x)| ∈ L1(Rn) .

Thus, we can pass the differentiation under the integral sign and then proceed by induction.In order to prove (iv), we first assume that f is such that ∂αx f ∈ C0 for |α| ≤ k (in particular

if f also has compact support). Under this assumption, assume that |α| = 1 first. Then, since∂xjf vanishes at ∞, by integration by parts we have∫

∂xjf(x)e−2πixξ dx = −∫f(x)(−2πiξj)e

−2πixξ dx

= 2πiξj f(ξ) .

The case when |α| > 1 follows by induction. The general case now follows by approximatingf ∈ L1 such that ∂αx f ∈ L1 when |α| ≤ k, as follows. Let η ∈ C∞c , 0 ≤ η ≤ 1, η = 1 when|x| ≤ 1, and η = 0 when |x| ≥ 2. Clearly ηεf ∈ L1 having compact support, so the first part ofthe argument applies. Then, it is easy to see that ∂αx

(ηεf)→ ∂αx f in L1 as ε → 0+, for every

|α| ≤ k.6 Since F : L1 → C ∩ L∞ is bounded (recall (2.4)) we have that

(2πiξ)α(ηεf) (ξ)→ (2πiξ)αf(ξ) ,

but also

(2πiξ)α(ηεf) (ξ) = F(∂αx (ηεf)

)(ξ)→ F(∂αx f)(ξ)

uniformly. The conclusion thus follows. 2

Corollary 2.12. (Riemann–Lebesgue Lemma) The Fourier transform maps L1 continuouslyinto C0(Rn):

F : L1(Rn)→ C0(Rn) .

6See also Exercise 8.


Proof. We only need to show that for f ∈ L1, f vanishes at ∞ (that is lim|x|→+∞ ∂αx f(x) = 0).

For f ∈ L1 ∩ C1c we have that

|ξ||f(ξ)| ≤n∑j=1

∣∣ξj ∫ f(x)e−2πixξ dx∣∣

≤ c0 ,

i.e. |ξ|f is bounded, that is f ∈ C0. Finally, the result follows by the density of L1 ∩ C1c in L1.

For, given g ∈ L1 and ε > 0, let fn ∈ L1 ∩ C1c such that ‖fn − g‖∞ ≤ ‖fn − g‖1 → 0. Thus,

fn ∈ C0 and converges uniformly to g. The conlusion now follows. 2

We now compute a fundamental Fourier transform.

Lemma 2.13. Let a > 0 and f(x) = e−aπ|x|2. Then

f(ξ) =1

an/2e−π|ξ|

2/a .

Proof. We begin with the case n = 1. By the previous proposition, we can differentiate underthe integral sign and get

(f)′(ξ) =(−2πixe−aπx

2)(ξ) =

(i

a

)((e−aπx

2)′)(ξ)

=

(i

a

)(2πiξ)f(ξ)

=

(−2π

a

)ξf(ξ) .

Therefore, the function eπξ2/af(ξ) satisfies the differential equation

d

dξeπξ

2/af(ξ) = 0 ,

i.e. it is a constant. In order to compute the constant, we select ξ = 0 and recall that∫e−t

2dt =√

π, so that

f(0) =

∫e−aπx

2dx =

1√a.

The n-dimensional case follows from an application of Fubini’s theorem:

f(ξ) =

∫. . .

∫e−2πixξe−aπ|x|

2dx1 · · · dxn

=n∏j=1

∫e−2πixjξje−aπ|xj |

2dxj

=

n∏j=1

(1√ae−πξ

2j /a

)=

1

an/2e−π|ξ|

2/a . 2

14 M. M. PELOSO

Remark 2.14. When we consider the two types of dilations defined in 2.2 we easily see havethat

ft(ξ) = f(tξ) = (f)t(ξ) and f t(ξ) = t−nf(ξ/t) = (f)t(ξ) , (2.5)

(Exercise).

We now prove

Theorem 2.15. (Inversion Theorem) Let f ∈ L1 be such that f ∈ L1. Then

f(x) =

∫e2πiξxf(ξ) dξ .

Proof. Let t > 0 and set

ψ(ξ) = e2πixξ−πt|ξ|2 .

Then,

(Fψ)(y) = τx(Fe−πt2|ξ|2)(y) = t−nτx(e−π|y|

2/t2)

= ϕt(x− y) ,

where ϕ(x) = e−π|x|2.

Next notice that, if f, ϕ ∈ L1, then∫f(x)ϕ(x) dx =

∫∫f(x)ϕ(ξ)e−2πixξ dxdξ =

∫f(ξ)ϕ(ξ) dξ . (2.6)

Therefore, ∫e2πixξ−πt|ξ|2 f(ξ) dξ =

∫ψ(ξ)f(ξ) dξ =

∫φ(y)f(y) dy

=

∫ϕt(x− y)f(y) dy = f ∗ ϕt(x) .

Since∫ϕ(x) dx =

∫e−π|x|

2dx = 1, we have that f ∗ ϕt → f in the L1 norm as t → 0. On the

left hand side of the equalities above we can use the dominated convergence theorem to show

that∫e2πixξ−πt|ξ|2 f(ξ) dξ →

∫e2πixξ f(ξ) dξ as t→ 0. 2

As an immediate consequence of the Inversion Theorem we have the following.

Corollary 2.16. If f ∈ L1 and f = 0, then f = 0.

We now extend the definition of the Fourier transform to functions in L2.

Theorem 2.17. (Plancherel Theorem) Let f ∈ L1 ∩ L2. Then f ∈ L2 and F|L1∩L2 extends

uniquely to a unitary isomorphism of L2.

Proof. Consider the space H = f ∈ L1 : f ∈ L1. We use the facts that f ∈ L1 implies f ∈ L∞and that L1 ∩L∞ ⊂ L2 to see that S ⊂ H ⊂ L2. Since S is dense in L2, also H is. Notice that,for g ∈ H, by the inversion theorem, setting h = g

h(ξ) =

∫e−2πixξ g(x) dx =

∫e2πixξ g(x) dx = g(ξ) .

Therefore, ∫fg dx =

∫fh, dx =

∫fh dξ =

∫f g dξ .

Thus, F restricted to H preserves the L2-scalar product, so that, by taking f = g we see that‖f‖2 = ‖f‖2. Since F(H) = H and H is dense in L2, F extends by continuity to a uniquemapping that is still an isometry. Thus, its image is closed and contains the dense subspace H,that is, F is a surjective isomorphism of L2.

Finally, one has to show that such an extension coincides with F on L1 ∩ L2. Suppose

f ∈ L1 ∩ L2 and let ϕ(x) = e−π|x|2. Consider f ∗ ϕt. Then f ∗ ϕt ∈ L1 ∩ L2, F(f ∗ ϕt) =

fF(e−πt|·|2) ∈ L1 (since f is bounded). Therefore, f ∗ϕt ∈ H and f ∗ϕt → f both in L1 and L2

norms, and F(f ∗ ϕt)→ f both uniformly and in L2 norm. 2

2.3. The Schwartz space. We recall a few basic facts about seminormed linear spaces.7

Let V be a complex linear space. A function % : V → [0,+∞) is called a seminorm if

(a) %(λv) = |λ|v for all v ∈ V and λ ∈ C;(b) %(v1 + v2) ≤ %(v1) + %(v2) for all v1, v2 ∈ V.

Notice that if we also require that %(v) = 0 implies v = 0, then % would be a norm.Let V be a linear space on which there exists a family %αα∈A of seminorms with the following

property: For each pair of points v1, v2 in V there exists %α such that %α(v1) 6= %α(v2). In thiscase we say that the family of seminorms separates the points in V.

A linear space V on which there exists a family %αα∈A of seminorms that separates thepoints is called a seminormed space.

On a seminormed space V with seminorms %αα∈A we define a topology by defining a systemof open sets

Ux,α,ε =y ∈ V : %α(x− y) < ε

.

Let V be a (complex) seminormed space and suppose that adimits a countable family P ofseminorms %k. Let τP be the topology defined above, which also is the coarsest topology thatmake continuous the identity

i : (V, τP)→ (V, %k)for all k ∈ N.

We say that P is separating if for each v ∈ V there exists k ∈ N such that %k(v) > 0. We willalways assume that P is separating.

Proposition 2.18. With the above notation, the topology τP enjoys the following properties.

(i) The space (V, τP) is a locally convex Hausdorff space.(ii) The finite intersections of the sets

Bk,n = v : %k(v) < 1/n

form a fundamental system of neighborhoods of 0.(iii) The topology τP is induced by the metric distance

dP(v, w) =∑k

1

2k%k(v − w)

1 + %k(v − w).

7For proofs and more on this topic, see G. B. Folland, Real Analysis, Modern Techniques and Their Applications,2 Ed..

16 M. M. PELOSO

(iv) A sequence vn in V converges to v in the τP topology if and only if it converges to vin all the seminorms %k, that is if

limn→+∞

%k(v − vn) = 0

for all k ∈ N.(v) A linear functional T on V, that is, a linear operator from V to C, is continuous in the

τP topology if and only if there exist seminorms %k1 , . . . , %kN and constant C > 0 suchthat

|T (v)| ≤ CN∑j=1

%kj (v)

for all v ∈ V.

The distance dP is invariant, that is

dP(v + z, w + z) = dP(v, w)

for all v, w, z ∈ V.

We also recall that if T a linear operator between two linear normed spaces X ,Y, then T iscontinuous if and only if it is bounded, that is there exists a constant C > 0 such that

‖Tx‖Y ≤ C‖x‖X ,

for all x ∈ X . The proof of this fact is simple. If T linear is bounded, then it is Lipschitz, withLipschitz constant less or equal to C:

‖Tx1 − Tx2‖Y = ‖T (x1 − x2)‖Y ≤ C‖x1 − x2‖X ,

which of course implies continuity. Conversely, if it continuous, the inverse image of the ball ofradius r > 0 is contained in the ball of radius R > 0, so that, for some R′ < R, and all x with‖x‖X ≤ R′,

‖Tx‖Y ≤ r .But then, for all x ∈ X ,

‖Tx‖Y =

∥∥∥∥‖x‖XR′ T( R′x

‖x‖X

)∥∥∥∥Y≤ ‖x‖X

R′r .

As a consequence of the results discussed so far, we have the following proposition involvingthe continuity (or boundedness) of a linear operator acting between seminormed spaces.

Proposition 2.19. Let V,W be seminormed linear spaces admitting family of seminorms %α,σβ, resp. Let T : V → W be a linear mapping. Then T is continuous if and only if, for eachseminorm σβ on W there exist seminorms %α1 , . . . , %αN and a constant C = Cβ,α1,...,αN

> 0 suchthat for all v ∈ V we have

σβ(Tv) ≤ CN∑j=1

%αj (v) .

A space of function of fundamental importance is the Schwartz space S(Rn) (often denotedas S). We denote by %(α,β) the seminorm

%(α,β)(f) = supx∈Rn

|xα∂βf(x)|

and define

S =f ∈ C∞(Rn) : %(α,β)(f) <∞, for all multi-indices α, β

.

Notice that for every positive integer N there exists a constant C = CN > 0 such that for allx ∈ Rn we have

1

C

∑|α|≤N

|xα| ≤ (1 + |x|)N ≤ C∑|α|≤N

|xα| . (2.7)

(We leave the simple proof as an exercise.)Observe that if ϕ ∈ S, for every positive integer N

(1 + |x|)N |ϕ(x)| ≤ C∑|α|≤N

%α,0(ϕ) ≤ CN ,

so that

|ϕ(x)| ≤ CN(1 + |x|)N

.

Thus, Schwartz functions are in C0, hence uniformly continuous. Moreover, they decay fasterthan the reciprical of any polynomial. This fact easily implies that S is contained in Lp, for allp, 1 ≤ p ≤ ∞, and in particular that for 1 ≤ p <∞,

‖ϕ‖pp =

∫(1 + |x|)Np|ϕ(x)|p(1 + |x|)−Np dx

≤ C∑

|α|≤[Np]+1

%α,0(ϕ)(1 + |x|)−Np dx ,

which is finite if N is chosen so that Np− (n− 1) > 1.

It is clear that S contains D, so it is not empty. However, such inclusion is proper since e−|x|2,

1/ coshx are examples of funtions in S that do not have compact support.

Lemma 2.20. Let p be a polinomial of degree d and γ be a multi-index. Then, the mapping

T : S 3 ϕ→ ∂γx(pϕ) ∈ S

is bounded (i.e. continuous).

Proof. For a given seminorm %(α,β) and any ϕ ∈ S we have

%(α,β)

(∂γxϕ

)= %(α,β+γ)(ϕ) ,

so ϕ 7→ ∂γxϕ is a continuous mapping of S into itself.On the other hand, if p(x) =

∑|α′|≤d aα′x

α′ , then

∂βx (pϕ)

=∑

|α′|≤d, |β′|≤|β|

bα′xα′∂β

′x (ϕ)

18 M. M. PELOSO

for some coefficients bα′ , so that

%(α,β)(pϕ) ≤ C∑

|α′|≤d, |β′|≤|β|

%(α+α′,β′β)(pϕ) .

Thus, also ϕ → pϕ is a continuous mapping of S into itself, and the conclusion follows ascomposition of continuous functions.

Proposition 2.21. The space S is a complete metric space (i.e. a Frechet space) in the topologydefined by the seminorms %(α,β). Moreover, C∞0 is dense in S.

Proof. Using (iii) of the previous Prop. 2.18, we only need to prove that S is complete in thegiven topology.

Let fk be a Cauchy sequence in S, then %(α,β)(fj − fk)→ 0 as j, k → +∞ for all (α, β) . In

particular, ∂βfk converges uniformly to a function gβ for all β. Denote by ej the j-th elementof the canonical basis in Rn. Notice that

g0(x+ tej)− g0(x) = limk→+∞

fk(x+ tej)− fk(x)

= limk→+∞

∫ t

0∂xjfk(x+ sej) ds

=

∫ t

0gej (x+ sej) ds .

Therefore, ∂xjg0 = gej , and by induction on |β| we obtain that gβ = ∂βg0. Finally,

|xα||∂βxfk(x)− ∂βxg0(x)| = limj→+∞

|xα||∂βxfk(x)− ∂βxfj(x)|

≤ limj→+∞

%(α,β)(fk − fj)

≤ ε ,

for j, k ≥ k0. Then %(α,β)(fk − g0)→ 0 as k → +∞.In order to show that C∞0 is dense in S, observe that, if η ∈ C∞c with η(0) = 1 and ϕ ∈ S,

then, for every ε > 0, ηεϕ is in C∞c , where ηε(x) = η(εx). Now it is easy to check that ηεϕ→ ϕin the S-topology, as ε→ 0. For, ηεϕ− ϕ is identically 0 for |ξ| ≤ 1/ε, so that

%α,β(ηεϕ− ϕ) = sup|x|≥1/ε

∣∣xα∂βx (ηεϕ− ϕ)(x)∣∣

≤ sup|x|≥1/ε

(∣∣xα∂βx (ηεϕ)(x)∣∣+∣∣xα∂βxϕ(x)

∣∣)≤ Cε

∑|γ|≤|β|

sup|x|≥1/ε

∣∣xα∂γxϕ(x)∣∣ .

Using the fact that xα∂γxϕ ∈ C0 for all α and γ, for a fixed ε, we can find ε0 such that the righthand side above is less or equal to ε. The conclusion now follows. 2

A consequence of Prop. 2.11 and the Inversion Theorem is the following.

Proposition 2.22. The Fourier transform F : S → S is a continuous bijection, with continuousinverse.

We remark that, as a consequence of these facts, we obtain a simple proof that the convolutionof two Schwartz functions is still a Schwartz function. For, ϕ ∗ ψ = F−1(ϕψ), where ϕψ ∈ S.

Proof. In order to prove the continuity of F , and therefore the one of F−1, we need to showthat for every semi-norm %α,β, there exist multi-indices α1, β1, . . . , αN , βN and C > 0 such thatfor every f ∈ S,

%(α,β)

(F(f)

)≤ C

N∑j=1

%(αj ,βj)(f) .

But,

ξα∂βξ (f) = cβξαF(xβf) = cα,βF

(∂αx (xβf)

).

%(α,β)

(F(f)

)= sup

ξ∈Rn|ξα∂βξ (f)(ξ)|

= cα,β supξ∈Rn

∣∣F(∂αx (xβf))(ξ)∣∣

≤ C‖∂αx (xβf)‖1≤ C

∑|α′|≤N

%(α′,0)

(∂αx (xβf)

).

It follows that F is bounded from S into itself. The Inversion theorem, Thm. 2.15, shows thatϕ 7→ ϕ(−·) is its inverse. So, F is one to one, onto and its inverse is also continuous. 2

We conclude this part by observing that, using the bounds in (2.7), if we set

%(N,β)(f) = supx∈Rn

(1 + |x|)N |∂βxf(x)| ,

for all non-negative intergers N and multi-indices β, we obtain a countable family of seminorms%N,β that defines a topology in S equivalent to the given one.

2.4. The space of tempered distributions. We define the space S ′ of tempered distributionsas the dual space of S. We endow S ′ with the weak-∗ topology, that is, the weakest topologythat makes the elements of S continuous on S ′, that is, the functionals

S ′ 3 u 7→ u(ϕ)

for ϕS fixed. A ngbh basis for the weak-∗ topology is given by the sets

Iϕ1,...,ϕN ;ε(u) =v ∈ S ′ : |u(ϕj)− v(ϕj)| < ε , j = 1, . . . , N

, (2.8)

where ϕ1, . . . , ϕN are in S and ε > 0.

There are many noticeable examples of tempered distributions.

Examples 2.23. (i) Functions in S, in any Lp class, 1 ≤ p ≤ ∞ give rise to bounded linearfunctionals on S, by setting:

Lf (ϕ) =

∫fϕ dx ϕ ∈ S, f ∈ Lp (or S) .

20 M. M. PELOSO

For,

|Lf (ϕ)| ≤∫|fϕ| dx ≤ ‖f‖Lp‖ϕ‖Lp′

= ‖f‖Lp(∫

Rn

(1 + |x|)−Np′(1 + |x|)Np′ |ϕ(x)|p′ dx)1/p′

≤ C‖f‖Lp supx(1 + |x|)N |ϕ(x)|

≤ C∑|α|≤N

%(α,0)(ϕ) ,

dove N e scelto ≥ n+ 1.(ii) A function f is called tempered if there exists N > 0 such that (1 + |x|)−Nf ∈ L1. Then,

the formula

Lf (ϕ) =

∫ϕf dx

defines a tempered distribution, since

|∫ϕf dx| ≤ sup

x(1 + |x|)N |ϕ(x)|

∫(1 + |x|)−N |f | dx ≤ C

∑|α|≤N

%(α,0)(ϕ) .

Analogously, a Borel measure µ is called tempered if there exists N > 0 such that∫

(1 +

|x|)−Nd|µ| <∞. Then, the pairing

Lµ(ϕ) =

∫ϕd|µ|

defines a tempered distribution. Then, Dirac deltas are examples of tempered distributions.Notice however, not every C∞ function defines a tempered distribution; for instance one with

exponential growth.

Remark 2.24. On S ′ we can introduce a few operations, besides the ones that define the vectorspace structure.

(I) Differentiation. For u ∈ S ′ and α a multi-index we set

∂αu(ψ) = u((−1)|α|∂αψ

).

Notice that, if u ∈ S ⊂ S ′, then ∂αu ∈ S and therefore it defines again an element of S ′ by theintegral pairing, and integrating by parts the boundary terms equal 0 (since both u and ψ areSchwartz functions)

(∂αu)(ψ) =

∫(∂αu)ψ dx = −

∫(∂α−e1u)(∂x−1ψ) dx

= (−1)|α|∫u∂αψ dx .

Hence, the definition of derivative of a distribution coincides with the classical definition if thedistribution admits classical derivatives.

(II) Multiplication by a smooth function of polinomial growth. Given f ∈ C∞, we say that fis of moderate growth if for each multi-index α there exist an integer N and a positive constantC such that

|∂αx f(x)| ≤ C(1 + |x|)N .Clearly such a function defines an element of S ′, and if ϕ ∈ S, then fϕ ∈ S. Moreover, if u ∈ S ′,we can define another element fu of S ′ by setting

(fu)(ϕ) = u(fϕ) .

(III) Fourier transform. Given u ∈ S ′ we define the tempered distribution u by setting

u(ψ) = u(ψ) ,

for any ψ ∈ S. Observe that, using identity (2.6), this defintion extends the definition of theFourier transform given on S; that is, if u above is in fact an element of S, then the equalityabove is justified by (2.6).

Proposition 2.25. The Fourier transform F is a surjective topological isomorphism of S ′ ontoitself.

Proof. Clearly F maps S ′ into itself, and using the fact that F is a bijection on S, it is easy tosee that it is a bijection on S ′ too. For, if u(ϕ) = 0 for all ϕ, then u(ϕ) = 0 for all ϕ ∈ S, thatis, u(ψ) = 0 for all ψ ∈ S, i.e. u = 0. In order to see that F is onto, given u ∈ S ′, define v ∈ S ′by setting v(ψ) = u(F−1ψ). Then, v is well defined and

u(ϕ) = u(F−1ϕ

)= v(ϕ) = v(ϕ) ,

for all ϕ ∈ S. Then, u = Fv, that is, F is onto. Notice also that we have shown that the inverseFourier transform on tempered distributions is given by

F−1u = u F−1 .

The fact that F and F−1 are continuous on S ′ follows from the description of the weak-∗topology: a ngbh basis of u ∈ S ′ is given by the sets

Iϕ1,...,ϕN ;ε(u) =v ∈ S ′ : |u(ϕj)− v(ϕj)| < ε , j = 1, . . . , N

,

where ϕ1, . . . , ϕN are in S and ε > 0. Let v ∈ Iϕ1,...,ϕN ;ε(u). Then,∣∣u(ϕj)− v(ϕj)∣∣ =

∣∣u(ϕj)− v(ϕj)∣∣ < ε ,

if v ∈ Iϕ1,...,ϕN ;ε(u), where Iϕ1,...,ϕN ;ε(u) has the obvious meaning. Therefore,

F(Iϕ1,...,ϕN ;ε(u)

)⊂∣∣u(ϕj)− v(ϕj)

∣∣and this shows that F : S ′ → S ′ is continuous and we are done. 2

(IV) Convolution with a Schwartz function ϕ. We first define the operator ˇ on functions bysetting ϕ(x) = ϕ(−x). Then, for ϕ ∈ S and u ∈ S ′ we set8

(u ∗ ϕ)(x) = u(τxϕ). (2.9)

Proposition 2.26. Let ϕ,ψ ∈ S and u ∈ S ′. Then

8We adopt the convention that τxϕ = ϕ(· − x) = ϕ(x− ·).

22 M. M. PELOSO

(i) u ∗ ϕ ∈ C∞ and for every multi-index α

∂α(u ∗ ϕ) = (∂αu) ∗ ϕ = u ∗ (∂αϕ);

(ii) u ∗ ϕ is of moderate growth, so that u ∗ ϕ ∈ S ′;(iii) we have ∫

Rn

u(τxϕ)ψ(x) dx = u

(∫Rn

(τxϕ)(y)ψ(x) dx

);

(iv) the action of u ∗ ϕ on S is given by the formula

(u ∗ ϕ)(ψ) = u(ϕ ∗ ψ) ;

(v) F(u ∗ ϕ) = uϕ and F(uϕ) = u ∗ ϕ;

(vi) (u ∗ ϕ) ∗ ψ = u ∗ (ϕ ∗ ψ).

Notice that the pairing in (iv) extends the case when also u ∈ S that we obtain by switchingthe integration order:

(u ∗ ϕ)(ψ) =

∫ (∫u(y)ϕ(x− y) dy

)ψ(x) dx

=

∫u(y)

(∫ϕ(x− y)ψ(x) dx

)dy

=

∫u(y)(ϕ ∗ ψ)(y) dy .

The same comments applies to the identity in (iv): if u is also in S, then the convolutions aregiven by absolutely convergent integrals and we have

(u ∗ ϕ) ∗ ψ(x) =

∫ (∫u(z)ϕ(y − z) dz

)ψ(x− y) dy =

∫u(z)

(∫ϕ(y − z)ψ(x− y) dy

)dz

=

∫u(z)

(∫ϕ(y′)ψ(x− z − y′) dy′

)dz =

∫u(z)(ϕ ∗ ψ)(x− z) dz

= u ∗ (ϕ ∗ ψ)(x) .

Proof. (i) We first claim that

ϕ(t) :=ϕ(·+ te1)− ϕ

t=

1

t

(τ−te1ϕ− ϕ

)→ ∂x1ϕ (2.10)

as t→ 0 in S.Assume the validity of the claim for the moment. Using the fact that the convolution com-

mutes with translations, it follows that

(u ∗ ϕ)(x+ te1)− (u ∗ ϕ)(x)

t=

1

t

[u(τx+te1ϕ)− u(τxϕ)

]= u

(1

t

[τte1(τxϕ)− τxϕ

])→ u

(− ∂y1(τxϕ)

)= −u

(τx∂y1(ϕ)

)= u

(τx(∂y1ϕ)

)= (u ∗ ∂y1ϕ)(x) ,

as t→ 0, by definition (2.9).

Iterating this argument we obtain that u ∗ ϕ is C∞ and

∂α(u ∗ ϕ) = u ∗ (∂αϕ) ,

for all multi-indices α. Since

u ∗ (∂αϕ)(x) = u(τx(∂αϕ)

)= (−1)|α|u

(∂α(τxϕ)

)= (∂αu)

(τxϕ)

= (∂αu) ∗ ϕ(x) ,

also the first equality in (i) follows, modulo equality (2.10).In order to prove (2.10) it suffices to prove that

F(1

t

(τ−te1ϕ− ϕ

)− ∂x1ϕ

)→ 0

in S, that is, (t−1(e2πitξ1 − 1

)− 2πiξ1

)ϕ→ 0

in S, as t→ 0. It this is easy to check that the function m(ξ) = t−1(e2πitξ1 − 1

)− 2πiξ1 is C∞,

of moderate growth together with all its derivatives, uniformly so in t, and they all tend to 0uniformly as t→ 0+; hence the conclusion. This proves (i).

(ii) In ordet to show that u∗ϕ is of moderate growth we need to show that for each multi-indexα there exist an integer N and a positive constant C such that

|∂αx (u ∗ ϕ)(x)| ≤ C(1 + |x|)N .

Since ∂αx (u ∗ ϕ) = u ∗ (∂αxϕ), and ∂αxϕ is again a Schwartz function, it suffices to prove the caseα = 0. Since u ∈ S ′, given ϕ ∈ S there exist semi-norms %(α1β1), . . . , %(αN ,βN ) and C > 0 suchthat

|u(τxϕ)| ≤ CN∑j=1

%(αjβj)(τxϕ) .

Now, for any semi-norm %(α,β),

%(α,β)

(τxϕ)

= supy∈Rn

∣∣yα∂βy (τxϕ(y))∣∣ = sup

y∈Rn

∣∣yα∂βy (ϕ(x− y))∣∣

= supy∈Rn

∣∣(x+ y)α∂βyϕ(y)∣∣ ,

which is clearly bounded by a polynomial in x.9

(iii) We have the following

Claim. Let η ∈ C∞0 . Then, the Riemann sums of the integral∫Rn(τxϕ)(y)η(x) dx converge in

the topology of S to ϕ ∗ η(y).

For simplicity of notation assume n = 1, the general case is similar. If η ∈ C∞0 , supp η ⊆ I,denoting by x0 < x1 < · · · < xN a partition of the interval I, ∆j = xj − xj−1, xj an arbitrary

9The reader should check that expicitly this polynomial is a constant times (1 + |x|)k∑|α′|≤|α| %(α′,β)(ϕ).

24 M. M. PELOSO

point in Ij = (xj−1, xj), we have∫Iϕ(x− y)η(x) dx = lim

N→+∞

N∑j=1

∆jϕ(xj − y)η(xj)

and we want to show that the sequence

fN (y) :=N∑j=1

∆jϕ(xj − y)η(xj)→∫Iϕ(x− y)η(x) dx (2.11)

in S as N → +∞. In fact,∫Iϕ(x− y)η(x) dx− fN (y) =

N∑j=1

∫Ij

(ϕ(x− y)η(x)− ϕ(xj − y)η(xj)

)dx

By the uniform continuity of the integrand, as a function of x, uniformly in y, given ε > 0 thereexists δ > 0 such that ∣∣ϕ(x− y)η(x)− ϕ(xj − y)η(xj)

∣∣ < ε

when ∆j < δ, for all j = 1, . . . , N .Therefore, choosing a partition of I such that ∆j < δ for all j, that is N ≥ |I|/δ, we have∣∣∣ ∫

Iϕ(x− y)η(x) dx− fN (y)

∣∣∣ ≤ ε N∑j=1

∫Ij

= ε|I| .

This shows that

%(0,0)

(∫Iϕ(x− ·)η(x) dx− fN

)→ 0

as N → +∞. The cases of the seminorms %(α,β) for all other α and β follow from this one.In the case |β| > 0 we have

∂βy


)= (−1)|β|

(∫I∂βyϕ(x− ·)η(x) dx− fβ;N

),

where fβ;N is defined as fN in (2.11) with ϕ replaced by ∂βyϕ. Hence,

%(0,β)


)≤ C%(0,0)

(∫I∂βyϕ(x− ·)η(x) dx− fβ;N

)→ 0

as N → +∞,In the case |α| > 0,

yα(∫

Iϕ(x− ·)η(x) dx− fN

)(y)

=

∫I(y − x+ x)αϕ(x− y)η(x) dx−

N∑j=1

∆j(y − xj + xj)αϕ(xj − y)η(xj)

=∑

α′+β′=α

cα′,β′(∫

I(x− y)α

′ϕ(x− y)xβ

′η(x) dx−

N∑j=1

∆j(y − xj)α′ϕ(xj − y)xβ

′

j η(xj)).


Therefore,

%(α,0)


)≤ C

∑α′+β′=α

%(0,0)

(∫Iϕα′(x− ·)ηβ′(x) dx− fα′,β′;N

)→ 0

as N → +∞, where ϕα′ = xα′ϕ, ηβ′ = xβ

′η and fα′,β′;N is defined as fN , with ϕ and η replaced

by ϕα′ , ηβ′ , resp.The claim, at least in the case n = 1, now follows by combining the above cases. We leave

the simple details to the reader.Therefore, when η ∈ C∞0 (I),

u(∫

R

(τxϕ)(y)η(x) dx

)= u

(∫Iϕ(x− y)η(x) dx

)= u

(lim

N→+∞

N∑j=1

∆jϕ(xj − y)η(xj))

= limN→+∞

N∑j=1

∆ju(ϕ(xj − ·)

)η(xj)

=

∫Iu(ϕ(x− ·)

)η(x) dx

=

∫Ru(τxϕ)η(x) dx .

Thus,

u(∫ (

τxϕ)(·)η(x) dx

)=

∫u(τxϕ)η(x) dx (2.12)

when η ∈ C∞0 (I).The case n > 1 can be proved with an analogous argmunent, which is just more involved

notationally (because of the expression of the Riemann sums in several variables and the decom-position of the expressions of the form (y − x+ x)α). Again, we leave the details to the reader,and therefore we consider that (2.12) (has been proved and) it is valid in the case of Rn.

Next, for a generic ψ ∈ S, let ηk ∈ C∞0 , ηk → ψ in S, then, as k → +∞,

∫ (

τxϕ)(y)ηk(x) dx→

∫ (τxϕ)(y)ψ(x) dx in S;

∫u(τxϕ)ηk(x) dx→

∫u(τxϕ)ψ(x) dx .

These conditions imply that

u(∫ (

τxϕ)(·)ψ(x) dx

)= lim

k→+∞u(∫ (

τxϕ)(·)ηk(x) dx

)= lim

k→+∞

∫u(τxϕ)ηk(x) dx

=

∫u(τxϕ)ψ(x) dx .

This proves (iii).

26 M. M. PELOSO

In order to prove the identity (iv), notice that, by (iii)(u ∗ ϕ

)(ψ) =

∫Rn

u(τxϕ)ψ(x) dx

= u(∫

Rn

(τxϕ)(y)ψ(x) dx

)= u

(∫Rn

ϕ(x− y)ψ(x) dx)

= u(ϕ ∗ ψ) .

(v) For all ψ ∈ S, since u ∗ ϕ ∈ S ′, using (iv) we have

F(u ∗ ϕ)(ψ) = (u ∗ ϕ)(ψ) = u(ϕ ∗ ψ) = u

(F(ϕψ)

)= u(ϕψ) = (uϕ)(ψ) .

This shows that F(u ∗ ϕ) = uϕ Next, we wish to show that F(uϕ) = u ∗ ϕ. For we have ψ ∈ S,

F(uϕ)(ψ) = (uϕ)(ψ) = u(ϕψ) = u(F(F−1ϕ ∗ ψ)

)= u

(F−1ϕ ∗ ψ

)=(u ∗ (F−1ϕ)

)(ψ) =

(u ∗ ϕ)

)(ψ) .

This implies F(uϕ) = u ∗ ϕ, as we wished to show.Finally (vi) follows at one from (v) by checking the equality of their Fourier transforms:

F((u ∗ ϕ) ∗ ψ

)= F

((u ∗ ϕ)

)(ψ) = (uϕ)(ψ) = F

(u ∗ (ϕ ∗ ψ)

).

(V) We conclude this part with a simple observation. It follows from (i) and (iv) that if u ∈ S ′

(∂αu) (ϕ) = (∂αu)(ϕ) = (−1)|α|u(∂αϕ

)= u

(F((2πiξ)αϕ

))= (2πiξ)αu(ϕ) .

A consequence of these facts is that S is dense in S ′.

Proposition 2.27. If ϕ ∈ S,∫ϕ = 1 and u ∈ S ′, then u ∗ ϕt → u in S ′, as t→ 0. Moreover,

C∞0 , hence S, is dense in S ′ in its topology.

Proof. Given ψ ∈ S, we have

(u ∗ ϕt)(ψ) = u(ϕt ∗ ψ)→ u(ψ)

as t→ 0, since ϕt ∗ ψ → ψ in S and∫ϕ = 1.

Next, arguing as in the proof of Prop. 2.21, let η ∈ C∞0 , 0 ≤ η ≤ 1, η(x) = 1 if |x| ≤ 1 and

η(x) = 0 if |x| ≥ 2. Since u ∗ ϕt ∈ C∞ (and it is of moderate growth), ηt′(u ∗ ϕt) ∈ C∞0 . For

ψ ∈ S we have,(ηt′(u ∗ ϕt)

)(ψ) = (u ∗ ϕt)(ηt

′ψ) = (u ∗ ϕt)(ψ) + (u ∗ ϕt)(ηt

′ψ − ψ)

= u(ψ) + (u ∗ ϕt − u)(ψ) + u(ηt′ψ − ψ) + (u ∗ ϕt − u)(ηt

′ψ − ψ) .

Given ε > 0 we can choose t′ > 0 small enough so that |u(ηt′ψ − ψ)| < ε, and then t > 0 small

enough so that |(u ∗ ϕt − u)(ηt′ψ − ψ)| < ε and also |(u ∗ ϕt − u)(ψ)| < ε. Hence,∣∣(ηt′(u ∗ ϕt)− u)(ψ)

∣∣ < 3ε .

The conclusion now follows. 2

The next result, that we state without proof (for which we refer to [StWe, Thm. 3.16]) is anexample of the significance of the space of tempered distributions.

Theorem 2.28. Let T : Lp(Rn) → Lq(Rn) be a bounded linear operator, 1 ≤ p, q ≤ ∞, thatcommutes with translations (that is, T (τxf) = τx(Tf)).

Then, there exists a tempered distribution K such that, for f ∈ S, Tf = f ∗K.

28 M. M. PELOSO

3. The Marcinkiewicz interpolation theorem and the Calderon–Zygmunddecomposition

3.1. The Marcinkiewicz interpolation theorem. We recall that, given a measurable func-tion f on Rn, its distribution function αf is the function, defined for λ ≥ 0

αf (λ) =∣∣x ∈ Rn : |f(x)| > λ

∣∣ . (3.1)

Using Fubini’s theorem, it is easy to see that for p > 0

‖f‖pLp = p

∫ +∞

0λp−1αf (λ) dλ . (3.2)

We also observed that

αf1+f2(λ) ≤ αf1(λ/2) + αf2(λ/2) ,

since

x ∈ R : |(f1 + f2)(x)| > λ ⊆ x ∈ R : |f1(x)| > λ/2 ∪ x ∈ R : |f2(x)| > λ/2 .

An operator T defined on measurable functions on Rn is said to be sublinear if

- |T (f0 + f1)(x)| ≤ |T (f0)(x)|+ |T (f1)(x)|;- |T (λf)(x)| = |λ||Tf(x)|;

for all x ∈ Rn, λ ∈ C. We remark that, if T is a linear operator, |T | is sublinear. Moreover,a typical example of an operator that is genuinely sublinear is the Hardy–Littlewood maximalfunction Mf(x) = supB(x,r),r>0

1|B|∫B |f(x)|dx.

Definition 3.1. Given a sublinear operator T defined on measurable functions on Rn is ofweak-type (p, q) if there exists a constant C > 0 such that

αTf (λ) =∣∣x ∈ Rn : |Tf(x)| > λ

∣∣ ≤ (Cλ‖f‖Lp

)q,

while it is said to be strong-type (p, q) if

T : Lp → Lq

is bounded.

It is easy to see that if T is of strong-type (p, q), then it is also of weak-type (p, q). For,

αTf (λ) =

∫x∈Rn: |Tf(x)|>λ

dx

≤∫x∈Rn: |Tf(x)|>λ

(|Tf(x)|

λ

)qdx

≤ 1

λq‖Tf‖qLq

≤(Cλ‖f‖Lp

)q.

Theorem 3.2. (Marcinkiewicz interpolation theorem) Let T be a sublinear operator de-fined on Lp0(Rn) + Lp1(Rn), 1 ≤ pj , qj ≤ +∞, j = 0, 1. Suppose that T is of weak-type (p0, q0)and of weak-type (p1, q1). Then, T is of strong-type (p, q), where p and q are given by therelations

1

p=

θ

p0+

1− θp1

, and1

q=

θ

q0+

1− θq1

,

and 0 < θ < 1.

Proof. We will prove the theorem only in the case q0 = p0 and q1 = p1. Then it suffices toassume that p0 0 be given. For a constant c > 0 to be selected later, we decompose f asf = f0 + f1, where

f0 = fχx:|f(x)|>cλ, and f1 = fχx:|f(x)|≤cλ .

Notice that f0 ∈ Lp0 and f1 ∈ Lp1 , since p0 cλ(x)

< (cλ)p0−p|f(x)|pχx:|f(x)|>cλ(x) , (3.3)

so that f0 ∈ Lp0 . Analogous reasoning shows that f1 ∈ Lp1 :

|f1(x)|p1 ≤ (cλ)p1−p|f(x)|pχx:|f(x)|≤cλ(x) . (3.4)

Then,

|T (f0 + f1)(x)| ≤ |T (f0)(x)|+ |T (f1)(x)|and

αT (f0+f1)(λ) ≤ αTf0(λ/2) + αTf1(λ/2) .

We distinguish two different cases. First assume that p1 < +∞. Then,

‖Tf‖pLp = p

∫ +∞

0λp−1αTf (λ) dλ

≤ p∫ +∞

0λp−1αTf0(λ/2) dλ+ p

∫ +∞

0λp−1αTf1(λ/2) dλ

≤ p∫ +∞

0λp−1

(2A0

λ‖f0‖Lp0

)p0dλ+ p

∫ +∞

0λp−1

(2A1

λ‖f1‖Lp1

)p1dλ

≤ p(2A0)p0∫ +∞

0λp−p0−1

∫x:|f(x)|>cλ

|f(x)|p0 dx dλ

+ p(2A1)p1∫ +∞

0λp−p1−1

∫x:|f(x)|≤cλ

|f(x)|p1 dx dλ

≤ p(2A0)p0∫Rn

|f(x)|p0∫ |f(x)|/c

0λp−p0−1 dλ dx

+ p(2A1)p1∫Rn

|f(x)|p1∫ +∞

|f(x)|/cλp−p1−1 dλ dx

= p

((2A0)p0

(p− p0)(cp−p0)+

(2A1)p0

(p1 − p)(cp−p1)

)‖f‖pLp .

30 M. M. PELOSO

In the second case, that is, p1 = +∞, then T : L∞ → L∞ is bounded. We may choosec = 1/(2A1), where A1 is the (L∞, L∞)-norm of T , so that αTf1(λ/2) = 0.10 Then, we have that

‖Tf‖pLp ≤ p∫ +∞

0λp−1

(2A0

λ‖f0‖Lp0

)p0dλ

= p(2A0)p0∫Rn

|f(x)|p0∫ |f(x)|/c

0λp−p0−1 dλ dx

=p

p− p0(2A0)p0(2A1)p−p1‖f‖pLp .

This proves the theorem. 2

As an application, we show that the Hardy–Littlewood maximal operator M is bounded onLp, 1 0

1

|B(x, r)|

∫B(x,r)

|f(y)| dy .

We need a lemma. This type of result is called a covering lemma.

Lemma 3.3. Let B1, B2, . . . , Bk be a finite collection of open balls in Rn. Then there existsa sub-collection Bj1 , . . . , Bj` of pairwise disjoint balls such that

∑i=1

|Bji | ≥ 3−n∣∣ k⋃j=1

Bj∣∣ .

Proof. We letBj1 to be a ball of largest volume among the balls in the collection B1, B2, . . . , Bk.Next, we select Bj2 among the remaining balls that are disjoint from Bj1 to be of largest volume,and we proceed in the same fashion. Since we have a finite number of balls, this process willterminate after ` steps.

Thus, we have constructed a sub-collection Bj1 , . . . , Bj` of pairwise disjoint balls. It Bmwas not selected, that is, m 6∈ j1, . . . , j`, this means that Bm intersects one of the ballsBj1 , . . . , Bj`, say Bji , with |Bji | ≥ |Bm|. Then, 3Bji ⊇ Bm, where 3B denotes the ball withthe same center as B, with radius 3 times as large. The conclusion now follows.

Theorem 3.4. The operator M is weak-type (1, 1) and bounded on Lp, 1 0, then we wish to show that there exists C > 0 such that∣∣x ∈ Rn : |Mf(x)| > λ∣∣ ≤ C

λ‖f‖L1 .

Let E be a compact subset of Ωλ := x ∈ Rn : |Mf(x)| > λ. For any x ∈ Ωλ there is an openball Bx = B(x, rx) such that x ∈ Bx and

1

|Bx|

∫Bx

|f(y)| dy > λ .

10For, |Tf1(x)| ≤ A1‖f1‖L∞ ≤ A1cλ ≤ λ/2, so that x : |Tf1(x)| > λ/2 = ∅.

Since E is compact, finitely many such balls B1, B2, . . . , Bk cover E. Then, we can applyLemma 3.3 to obtain a sub-collection Bj1 , . . . , Bj` of disjoint balls such that | ∪kj=1 Bj | ≤3n∑`

i=1 |Bji |. Therefore, for every compact E ⊆ Ωλ we have

|E| ≤∣∣ k⋃j=1

Bj∣∣ ≤ 3n

∑i=1

|Bji |

≤ 3n∑i=1

1

λ

∫Bji

|f(y)| dy

≤ C

λ‖f‖L1 .

This proves the theorem.

3.2. The Calderon–Zygmund decomposition of an L1-function. We now introduce afundamental decomposition of the whole space Rn, the dyadic decomposition. We define theunit cube to be the set [0, 1)n and the set Q0 to be the family of sets obtained by translatingby mej , where m ∈ Z, and for j = 1, . . . , n, ej is an element of the canonical basis, and all theirpossible combinations, that is,

Q0 =Q = [0, 1)n +

n∑j=1

mjej ,mj ∈ Z.

Analogously, for k ∈ Z we define the family Qk as

Qk =Q = [0, 2−k)n +

n∑j=1

2−kmjej ,mj ∈ Z.

The union of the famlies Qk, k ∈ Z, is called the family of dyadic cubes in Rn.It is immediate to see that these sets are cubes with sides parallel to the axes, that a cube in

Qk have vertices at adjacent points in the lattices (2−kZ)n. Moreover, the following propertiesare also easily checked.

(1) For each k fixed, the cubes in Qk are (mutually) disjoint and their union is all of Rn.(2) Given any two dyadic cubes, they are either disjoint, or one is contained in the other

one.(3) Given j, k ∈ Z, with j < k, then each cube in Qk is contained in a unique cube in Qj

and contains 2k+1 cubes in Qk+1.

The proof of Thm. 4.6 uses an important and far-reaching result, called the Calderon–Zygmund decomposition, that here we state and prove in the setting of Rn.

Theorem 3.5. Let f ∈ L1(Rn) and non-negative. Given λ > 0, there exists a sequence Qjofdisjoint dyadic cubes such that

(i) f(x) ≤ λ for almost all x 63 ∪jQj;

(ii)∣∣ ∪j Qj∣∣ ≤ 1

λ‖f‖L1;

(iii) λ <1

|Qj |

∫Qj

f(x) dx ≤ 2nλ.

32 M. M. PELOSO

Proof. Let λ > 0 be fixed. Since f ∈ L1, there exists k0 ∈ Z so that

1

|Q|

∫Qf dx ≤ λ

for all Q ∈ Qk0 .Now we consider the cubes of the next generation, that are obtained from the cubes in Qk0

by bisecting each cube of the collection Qk0 . We say that a cube Q ∈ Qk0+1 belongs to thesequence we week if

1

|Q|

∫Qf dx > λ

while (by construction),

1

|Q′|

∫Q′f dx ≤ λ for (the unique) Q′ ∈ Qk0 such that Q′ ⊇ Qk0 .

If a cube is not chosen, that is, 1|Q|∫Q f dx ≤ λ, we iterate this process and we bisect it by

considering the dydiac cubes of the next generation that are contained in it. Then, we say thata cube Qj ∈ Qj , j > k0 is in the sequence if

1

|Qj |

∫Qj

f dx > λ ,

while,1

|Q′|

∫Q′f dx ≤ λ for Q′ ∈ Qj−1 such that Q′ ⊇ Qj .

We need to check that the sequence Qj so constructed satisfies the required conditions. Bydefinition, if Qj ∈ Qj is a cube in the sequence and Q ∈ Qj−1 contains Qj then

λ <1

|Qj |

∫Qj

f(x) dx ≤ 1

2−n|Q|

∫Qf(x) dx ≤ 2nλ ,

so that (iii) is satisfied.Next,

| ∪j Qj | =∑j

|Qj | <1

λ

∑j

∫Qj

f(x) dx

≤ 1

λ‖f‖L1 ,

so that (ii) also holds.Finally, (i) follows from the Lebesgue differentiation theorem. If x ∈ c(∪jQj), then for all

dyadic cubes Q containing x then1

|Q|

∫Qf dx ≤ λ .

Letting the side length of the cube containing x tend to 0, by the Lebesgue differentiationtheorem we obtain

f(x) = lim`(Q)→0

1

|Q|

∫Qf dx ≤ λ ,

for a.a. x ∈ c(∪jQj). This proves the theorem.

4. The Hilbert transform

Throughout this section we assume that n = 1 and consider the real line R. We are goingto introduce and study perhaps the most important example of an integral operator that isLp-bounded. By this we mean that there exists a (measurable) kernel K(x, y) defined a.e. onR×R and such that for the integral operator

Tf(x) =

∫Rf(y)K(x, y) dy

defined on a suitable dense class of functions of Lp(R) there exists a constant A = Ap dependingon p, but not on the given function f , such that

‖Tf‖Lp(R) ≤ A‖f‖Lp(R) .

Next we introduce a temperated distribution of fundamental importance.

Definition 4.1. We define the principal value of 1/x as the tempered distribution

p.v.1

x(ϕ) = lim

ε→0

∫|x|≥ε

ϕ(x)

xdx .

Definition 4.2. We define the Poisson kernel Py(x), defined for x ∈ R and y > 0 as the kernel

Py(x) =1

π

y

y2 + x2.

We remark that the kernel Py(x) can be viewed as the “dilated” of the L1 function 1π

11+x2

, so

that Py(x) is a so-called summability kernel.For a given function f ∈ Lp(R), 1 ≤ p <∞, we define the Poisson integral of f the function

Py ∗ f(x) =1

π

∫Rf(x− t)Py(t) dt =

1

π

∫Rf(x− t) y

y2 + t2dt .

Definition 4.3. We are also goint to consider the family of tempered distributions Qy, withy > 0, given by

Qy(x) =1

π

x

y2 + x2.

The kernel Qy(x) is called the conjugate Poisson kernel.For a function f ∈ Lp(R), 1 ≤ p <∞, we define the conjugate Poisson integral of f to be the

function

Qy ∗ f(x) =1

π

∫Rf(x− t)Qy(t) dt =

1

π

∫Rf(x− t) t

y2 + t2dt .

We remark that Qy is not an L1 function, so that Qy(t) is not a summability kernel (andalso that here we are using an abuse of notation, since the subindex t does not represent thedilation defined in (2.2)). Nonetheless, the convolution Qy ∗ f is well defined and continuous(since instead Qy ∈ Lq, for 1 < q ≤ ∞).

Remark 4.4. The three objects above arose naturally when trying to study the regularity ofthe mapping that associates to a given real harmonic function f on the upper half-plane (or onthe unit disk) its harmonic conjugate.

34 M. M. PELOSO

To briefly illustrate this circle of ideas, suppose that f is a real function that is in Lp(T),where T is the unit circle, that is, the boundary of the unit disk in the complex plane (and Tis identified with the one-dimensional torus, see [Ka]).

Consider its Poisson integral P(f), a real harmonic function on the unit disk, that we still

denote by f . Then f admits a harmonic conjugate function f . It can be proved that f definesa boundary function (still denoted by f).

It is important to notice that the Cauchy integral of f

C(f)(z) =1

2πi

∫γ

f(ζ)

ζ − zdζ

where the integral is the complex line integral over the boundary of the unit disk, produces aholomorphic function F . Recall that the Poisson kernel (on the unit disk) satisfies the identity

P = C + C − 1 ,

that is, P = Re (2C−1). Thus, the real part of 2C(F )−f(0) coincides with f , and its imaginary

part with f .Hence, the function f can be obtained as the integral of f against the imaginary part of

the Cauchy kernel, kernel that is called the conjugate Poisson kernel Q. Moreover, it can beproved that, as the variable z approaces the boundary, f(z) = Q(f)(z) tends to a function on

the boundary, that we still denote by f .The correspondence of the boundary functions f 7→ f is the “Hilbert transform”.

We collect in the next statement the definition and main properties of the Hilbert transform.

Theorem 4.5. The following properties hold:

(i)∣∣p.v.

1

x(ϕ)∣∣ ≤ C(‖ϕ′‖∞ + ‖xϕ‖∞

)= C

(%(0,1)(ϕ

′) + %(1,0)(ϕ));

(ii) in S ′, limt→0

Qt =1

πp.v.

1

x;

(iii) F( 1

πp.v.

1

x

)(ξ) = −i sgn(ξ).

The following expressions are equivalent and define the Hilbert transform Hf of a functionf ∈ S:

(i’) Hf(x) =( 1

πp.v.

1

x

)∗ f(x);

(ii’) Hf(x) = limt→0

Qt ∗ f(x);

(iii’) F(Hf)(ξ) = −i sgn(ξ)f(ξ).

Proof. Using the fact that∫ε≤|x|≤1 1/x dx = 0 we write

p.v.1

x(ϕ) = lim

ε→0

∫ε≤|x|≤1

ϕ(x)

xdx+

∫1≤|x|

ϕ(x)

xdx

= limε→0

∫ε≤|x|≤1

ϕ(x)− ϕ(0)

xdx+

∫1≤|x|

ϕ(x)

xdx

=

∫|x|≤1

ϕ(x)− ϕ(0)

xdx+

∫1≤|x|

ϕ(x)

xdx .

Now (i) follows using the mean value theorem.

In order to prove (ii) consider the functions u(ε)(x) = 1xχ|x|≥ε. These are locally integrable

and bounded, hence they define tempered distributions. By definition of the principal valuedistribution we have that

limε→0

u(ε) = p.v.1

x.

Therefore, we wish to prove that

limt→0

(Qt −

1

πu(t)

)= 0

in S ′.Let ϕ ∈ S be fixed. We have(

πQt − u(t)

)(ϕ) =

∫R

xϕ(x)

t2 + x2dx−

∫|x|≥t

ϕ(x)

xdx

=

∫|x|<t

xϕ(x)

t2 + x2dx−

∫|x|≥t

( x

t2 + x2− 1

x

)ϕ(x) dx

=

∫|x|<1

xϕ(tx)

1 + x2dx−

∫|x|≥1

ϕ(tx)

x(1 + x2)dx .

Now we can pass to the limit as t→ 0, by applying the dominated convergence theorem, bymajorizing the integrands using the fact that |vp(tx)| ≤ ‖ϕ‖L∞ . The limit equals ϕ(0) times∫

|x|<1

x

1 + x2dx−

∫|x|≥1

1

x(1 + x2)dx = 0 ,

since the functions under the integral signs are odd and the region of integration symmetric withrespect to the origin.

Next we compute the Fourier transform of 1πp.v. 1

x , and we do this using (ii) and the continuityof F on S ′; that is, we use the equality

F( 1

πp.v.

1

x

)= lim

t→0FQt .

We observe that11

F−1(− i sgn(ξ)e−2πt|ξ|

)(x) =

∫R−i sgn(ξ)e−2πt|ξ|e2πixξ dξ

= i

∫ 0

−∞e2π(t+ix)ξ dξ − i

∫ +∞

0e2π(−t+ix)ξ dξ

=i

2π(t+ ix)+

i

2π(−t+ ix)

=1

π

x

t2 + x2= Qt(x) .

11The Fourier transform of Qt can be computed directly using the calculus of indefinite integrals in complexanalysis that relies on the residue theorem. Here we use the inverse Fourier transform to avoid using complexanalysis.

36 M. M. PELOSO

Therefore, we obtain that

F( 1

πp.v.

1

x

)= lim

t→0FQt = lim

t→0−i sgn(ξ)e−2πt|ξ|

= −i sgn(ξ) ,

as we wish to show.12

We define the Hilbert transform Hf of a Schwartz function f as

Hf(x) =( 1

πp.v.

1

x

)∗ f(x) ,

and observe that

Hf(x) = limε→0

u(ε) ∗ f(x) = limε→0

1

π

∫Rf(x− y)

1

yχ|y|≥ε(y) dy

= limε→0

1

π

∫|y|≥ε

f(x− y)

ydy . (4.1)

Finally, (i’)-(iii’) follow from the first part and Remark 2.24. 2

4.1. The Lp-boundedness of the Hilbert transform. The regularity result for the Hilberttransform is the following.

Theorem 4.6. The Hilbert transform H, initially defined on Schwartz functions, can be extentedto an operator, still denoted by H, which is weak-type (1, 1) and strong-type (p, p) when 1 0 such that∣∣∣x ∈ R : |Hf(x)| > λ

∣∣∣ ≤ C

λ‖f‖L1(R) ,

and

‖Hf‖Lp(R) ≤ C‖f‖Lp(R)

when 1 < p <∞.

Proof. We first observe that H is bounded on L2, since by Plancherel’s theorem

‖Hf‖L2 = ‖(Hf ) ‖L2 = ‖ sgn(ξ)f‖L2 = ‖f‖L2 .

12Notice however that the limits above are in the S ′-sense, that is, for every ϕ ∈ S,

limt→0

(− i sgn(ξ)e−2πt|ξ|)(ϕ) = lim

t→0

∫R

−i sgn(ξ)e−2πt|ξ|ϕ(ξ) dξ =

∫R

−i sgn(ξ)ϕ(ξ) dξ =(− i sgn(ξ)

)(ϕ).

Next, suppose that we have shown that H is weak-type (1, 1). Then, by Marcinkiewicz itfollows that H is strong -type (p, p), for 1 < p ≤ 2. If 2 < p <∞ we use duality:

‖Hf‖Lp = sup‖g‖

Lp′≤1

∣∣∣ ∫RHf(x)g(x) dx

∣∣∣≤ sup‖g‖

Lp′≤1

∣∣∣ ∫Rf(x)Hg(x) dx

∣∣∣≤ sup‖g‖

Lp′≤1‖f(x)‖Lp‖Hg‖Lp′

≤ C‖f‖Lp ,

that is, H is bounded on Lp also when 2 0 and

we form the Calderon–Zygmund decomposition at height λ. We obtain a sequence of disjointinterval Ij such that:

(i) f(x) ≤ λ for almost all x 63 ∪jIj ;

(ii)∣∣ ∪j Ij∣∣ ≤ 1

λ‖f‖L1 ;

(iii) λ <1

|Ij |

∫Ij

f(x) dx ≤ 2λ.

Now we decompose f as sum f = g + b, where

g(x) =

f(x) if x 6∈ ∪jIj1

|Ij |

∫Ij

f(x) dx if x ∈ Ij ,

and

b(x) =∑j

bj(x) =∑j

(f(x)− 1

|Ij |

∫Ij

f)χIj (x) .

(The decomposition f = g + b is what is called the Calderon–Zygmund decomposition of thenon-negative L1 function f .)

We observe that 0 ≤ g(x) ≤ λ, if x 6∈ ∪jIj , while 0 ≤ g(x) = 1|Ij |∫Ijf(x) ≤ 2λ if x ∈ Ij .

Hence, g(x) ≤ 2λ, and also g ∈ L1 since it is obviously integrable on c(∪jIj) and on each Ij(that are disjoint)

∫Ijg(x)dx =

∫Ijf(x)dx <∞. Notice in particular that

∫R g =

∫R f . On the

other hand, b has mean equal to 0, since each bj does.We now proceed. Since |Hf(x)| ≤ |Hg(x)|+ |Hb(x)|, we have

∣∣x : |Hf(x)| > λ∣∣ ≤ ∣∣x : |Hg(x)| > λ/2

∣∣+∣∣x : |Hb(x)| > λ/2

∣∣ .

38 M. M. PELOSO

Now, on the “good function” g we use the L2-boundedness of the Hilbert transform to obtain∣∣x : |Hg(x)| > λ/2∣∣ =

∫x:|Hg(x)|>λ/2

dx ≤∫x:|Hg(x)|>λ/2

|Hg(x)|2

(λ/2)2dx

≤ 4

λ2

∫R|Hg(x)|2 dx =

4

λ2

∫Rg(x)2 dx

≤ 8

λ

∫Rg(x) dx =

8

λ

∫Rf(x) dx .

Next, let I∗j denote the interval with the same center cj as Ij with twice the length. Then weestimate,∣∣x : |Hb(x)| > λ/2

∣∣ =∣∣x ∈ (∪jI∗j ) : |Hb(x)| > λ/2

∣∣+∣∣x ∈ c(∪jI∗j ) : |Hb(x)| > λ/2

∣∣≤ | ∪j I∗j |+

∣∣x ∈ c(∪jI∗j ) : |Hb(x)| > λ/2∣∣

≤ 2

λ‖f‖L1 +

2

λ

∫R\(∪jI∗j )

|Hb(x)| dx .

Since |Hb(x)| ≤∑

j |Hbj(x)| a.e., it will suffice to prove that∑j

∫R\I∗j

|Hbj(x)| dx ≤ C‖f‖L1 . (4.2)

Now, bj 6∈ S, nonetheless, for x 6∈ I∗j we have the formula

Hbj(x) =

∫Ij

bj(y)

x− ydy .

Recall that bj has mean 0 and that we denote by cj the center of the interval Ij . Notice thaty ∈ Ij implies that |y − cj | ≤ |Ij |/2 and that x 6∈ I∗j and y ∈ Ij imply that |x− y| > |x− cj |/2.Then, ∫

R\I∗j|Hbj(x)| dx =

∫R\I∗j

∣∣∣ ∫Ij

bj(y)

x− ydy∣∣∣ dx

=

∫R\I∗j

∣∣∣ ∫Ij

bj(y)

(1

x− y− 1

x− cj

)dy∣∣∣ dx

≤∫Ij

|bj(y)|∫R\I∗j

|y − cj ||x− y| |x− cj |

dx dy

≤∫Ij

|bj(y)|∫R\I∗j

|Ij ||x− cj |2

dx dy

= 2

∫Ij

|bj(y)| dy .

Therefore, ∑j

∫R\I∗j

|Hbj(x)| dx ≤ 2∑j

∫Ij

|bj(y)| dy ≤ 4‖f‖L1 ,

which gives (5.33). This concludes the proof.

4.2. Further properties of the Hilbert transform. We now make a few observations.(1) We have shown that, when f ∈ S, then Hf satisfies a weak-type (1, 1) bound and a

strong-type (p, p) bound. We now extend the definition of Hf to all of Lp and all of L1, withthe same bounds.

If f ∈ Lp, 1 < p <∞, then there exists a sequence fn of Schwartz functions converging inLp to f :

limn→+∞

‖fn − f‖Lp = 0

Then, in order to define Hf , notice that ‖Hfn −Hfm‖Lp ≤ C‖fn − fm‖Lp , so that Hfn is aCauchy sequence in Lp and converges to g ∈ Lp. We set then Hf = g. It is immediate to seethat this definition is well-given, in the sense that does not depend on the choice of the sequencefn and that H then satisfies the bound ‖Hf‖Lp ≤ C‖f‖Lp , with the same constant C as inThm. 4.6.

(2) If f ∈ L1, then there exists a sequence fn of Schwartz functions converging in L1 to f .The weak (1, 1) inequality gives that, for all ε > 0 fixed,

limn,m,→+∞

∣∣x ∈ R :∣∣(Hfn −Hfm)(x)

∣∣ > ε∣∣ ≤ C

ε‖fn − fm‖L1 = 0 .

Then, Hfn is a Cauchy sequence in measure, so it converges to a measurable function g a.e..We set, Hf = g. It is easy to chech that H satisfies the same weak-type (1, 1) bound on all ofL1.

(3) When p = 1, the strong-type inequality fails. For instance, if we take f = χ[0,1], then we

have13

Hf(x) = limε→0

1

π

∫|y|≥ε

f(y)

x− ydy = lim

ε→0

1

π

∫ 1

ε

1

x− ydy

= limε→0

1

π

∫ x−ε

x−1

1

tdt

=1

πlog

|x||x− 1|

.

It is clear that Hf 6∈ L1, while f ∈ L1.

(4) Having defined the Hilbert transform of a function f ∈ Lp, for 1 ≤ p <∞ as limit in norm(when 1 < p < ∞) or in measure (when p = 1), we now wish to defined it pointwise as well.In this part we will omit the proofs and we refer the reader to [Du]. Consider the truncatedintegrals

Hεf(x) =1

π

∫|y|≥ε

f(x− y)

ydy .

Proposition 4.7. Let 1 ε is in Lq for all 1 < q ≤ ∞. Then, the function

1

π

∫1

yχ|y|>ε(y)f(x− y) dy = Hεf(x) ,

13Here we use the equatity at the end of the proof of Thm. 4.5.

40 M. M. PELOSO

is well defined for all f ∈ Lp, 1 ≤ p <∞.Now,

F(1

yχ|y|>ε

)(ξ) = lim

N→+∞

∫ε<|y|<N

e−2πiyξ

ydy

= limN→+∞

∫ε<|y|<N

−i sin(2πyξ)

ydy

= −2i sgn(ξ) limN→+∞

∫ 2πN |ξ|

−2πε|ξ|

sinx

xdx .

Therefore,∣∣F( 1

yχ|y|>ε)(ξ)∣∣ ≤ 2π, for all ε > 0. This implies that Hε is of strong-type (2, 2),

with constants uniform in ε.The weak-type (1, 1) boundedness also follows with uniform constant, so the strong-type (p, p)

follows, with uniform constant, by interpolation and duality.We consider now the case 1 < p < ∞; the case p = 1 begin analogous. If f ∈ Lp, then let

fn be in S converging to f in Lp. Then, using the uniform bound for the Lp-boundedness ofHε (and the fact that fn ∈ S), we have

Hf = limn→+∞

Hfn = limn→+∞

limε→0

Hεfn = limε→0

limn→+∞

Hεfn = limε→0

Hεf .

This proves the lemma. We conclude this section by stating the following result concerning the pointwise definition

of the Hilbert transform.

Theorem 4.8. Let 1 ≤ p <∞, and f ∈ Lp. Then

Hf(x) = limε→0

Hεf(x) a.e.x ∈ R .

We only mention that, similarly to the case of the Hardy–Littlewood maximal function andthe Lebesgue differentiation theorem, the proof relies on the boundedness of a maximal function.

SetH∗f(x) = sup

ε>0|Hεf(x)| .

Then the following result holds true.

Theorem 4.9. The maximal Hilbert transform H∗ is weak-type (1, 1) and strong-type (p, p) for1 < p <∞.

5. Singular integrals

5.1. The Calderon–Zygmund theorem. Main result of this section is the following theorem.Given a function K that is locally integrable in Rn \ 0, and it is also a tempered distribution,it makes sense to consider the convolution K ∗ f with a Schwartz function f . Therefore, wedefine the operator

Tf(x) = K ∗ f(x) ,

initially defined on Schwartz functions.The next result generalizes the theorem on the bounededness of the Hilbert transform.

Theorem 5.1. Let K be a tempered distribution that coincides with a locally integrable functionon Rn \ 0. Suppose that

|K(ξ)| ≤ A , (5.1)

and furthermore ∫|x|>2|y|

|K(x− y)−K(x)| dx ≤ B . (5.2)

Then, the operator T is bounded from Lp into itself, i.e., it is of strong-type (p, p) for 1 λ

∣∣ ≤ C

λ‖f‖L1 .

Definition 5.2. A tempered distribution K that satisfies the conditions in the theorem, thatis, that coincides with a locally integrable function on Rn \ 0 and satisfies (5.1) and (5.2) iscalled a Calderon–Zygmund convolution kernel.

Corollary 5.3. Let K ∈ C1(Rn \ 0) be a tempered distribution that coincides with a locallyintegrable function on Rn \ 0. Suppose that K sastisfies (5.1) and also

|∇K(x)| ≤ B′

|x|n+1. (5.3)

Then, for the operator T the same conclusions hold as in the previous theorem.

Proof. In order to prove the corollary, assuming the validity of the theorem, it suffices to showthat (5.3) implies (5.2). But this follows from the mean value theorem. Let σ(x, y) denote thesegment having endpoints in x − y and x. Then, using the fact that |x| > 2|y|, it holds that

|x− y| ≥ |x| − |x|2 = |x|2 , so that, if z ∈ σ(x, y) we have

|z| ≥ min|x− y|, |x| ≥ |x|2.

Therefore,

|K(x− y)−K(x)| ≤ |y| supz∈σ(x,y)

|∇K(z)| ≤ B′|y| supz∈σ(x,y)

1

|z|n+1

≤ 2n+1B′|y||x|n+1

.

42 M. M. PELOSO

Hence,14 ∫|x|>2|y|

|K(x− y)−K(x)| dx ≤ 2n+1B′|y|∫|x|>2|y|

1

|x|n+1dx

= 2n+1B′ωn|y|∫ +∞

2|y|r−2 dr

= B .

Thus, (5.3) implies (5.2) and the conclusion follows from Thm. 5.1.

Proof of Thm. 5.1. The proof goes along the same lines of the proof of Thm. 4.6.By the assumption (5.1) and the Plancherel theorem it immediately follows that T is bounded

on L2:

‖Tf‖2L2 =

∫Rn

|F(K ∗ f)(ξ)|2 dξ =

∫Rn

|K(ξ)f(ξ)|2 dξ ≤ A2‖f‖2L2 .

We now observe that the adjoint operator T ∗ has kernel K∗(x) = K(−x), so that it alsosatisfies hypotheses (5.1) and (5.2). Thus, if we prove the weak-(1, 1) inequality for T , byinterpolation it would follow that T is Lp-bounded for 1 0 be fixed. We decompose f according to Thm. 3.5, as in the proof of Thm. 4.6.We repeat it for simplicity: we a sequence of disjoint cubes Qj such that:

(i) f(x) ≤ λ for almost all x 63 ∪jQj ;

(ii)∣∣ ∪j Qj∣∣ ≤ 1

λ‖f‖L1 ;

(iii) λ <1

|Qj |

∫Qj

f(x) dx ≤ 2nλ.

Now we decompose f as sum f = g + b, where

g(x) =

f(x) if x 6∈ ∪jQj1

|Qj |

∫Qj

f(x) dx if x ∈ Qj ,

and

b(x) =∑j

bj(x) =∑j

(f(x)− 1

|Qj |

∫Qj

f)χQj (x) .

Arguing as in the proof of Thm. 4.6, we have∣∣x : |Tf(x)| > λ∣∣ ≤ ∣∣x : |Tg(x)| > λ/2

∣∣+∣∣x : |Tb(x)| > λ/2

∣∣ ,and then ∣∣x : |Tg(x)| > λ/2

∣∣ ≤ C

λ

∫Rf(x) dx .

14Here, and in the remaining of the notes, we denote by ωn the volume of the unit sphere Sn−1 in Rn.15This boundedness is equivalent to the Lp-boundedness of T ∗ for 1 < p ≤ 2.


Next, ∣∣x : |Tb(x)| > λ/2∣∣ ≤ C

λ‖f‖L1 +

2

λ

∫R\(∪jQ∗j )

|Tb(x)| dx .

where Q∗j denote the cube with the same center cj as Qj with side length 2√n-times longer.

Then we reduce ourselves to show that∑j

∫R\Q∗j

|Tbj(x)| dx ≤ C‖f‖L1 ,

which, in turn, is implied by the inequality∫R\Q∗j

|Tbj(x)| dx ≤ C‖bj‖L1 . (5.4)

We are going to use the Hormander condition (5.2) and the fact that bj has integral equal to0. For x 6∈ Q∗j ,

Tbj(x) =

∫Qj

K(x− y)bj(y) dy =

∫Qj

[K(x− y)−K(x− cj)

]bj(y) dy .

Therefore, ∫Rn\Q∗j

|Tbj(x)| dx ≤∫Qj

|bj(y)|∫Rn\Q∗j

∣∣K(x− y)−K(x− cj)∣∣ dx dy

≤ B∫Qj

|bj(y)| dy ,

since

Rn \Q∗j ⊆x ∈ Rn : |x− cj | > 2|y − cj |

,

so that ∫Rn\Q∗j

∣∣K(x− y)−K(x− cj)∣∣ dx ≤ ∫

|x−cj |>2|y−cj |

∣∣K(x− y)−K(x− cj)∣∣ dx

=

∫|x′|>2|y−cj |

∣∣K(x′ − (y − cj))−K(x′)∣∣ dx′

≤ B .

Thus, (5.4) follows and we are done.

5.2. Homogeneous distributions. In order to describe some classical and fundamental exam-ples of singular integrals in Rn, examples that generalize the case of the Hilbert transform in thecase of the real line, we need a preliminary discussion of the so-called homogenous distributions.

We recall that a function f is said to be homogenous of degree a if for all λ > 0 and all x ∈ Rn

f(λx) = λaf(x) .

44 M. M. PELOSO

Given another function ϕ, we see that∫Rn

f(x)ϕλ(x) dx = λ−n∫Rn

f(x)ϕ(x/λ) dx =

∫Rn

f(λx′)ϕ(x′) dx′

= λa∫Rn

f(x′)ϕ(x′) dx′ .

Therefore, we may extend the definition of homogeneity to tempered distributions as follows.

Definition 5.4. Given a tempered distribution u, we say that u is homogenous of degree a > 0if for all ϕ ∈ S and λ > 0 we have

u(ϕλ) = λau(ϕ) .

Example 5.5. It is immediate to see that the distribution 1πp.v. 1

x on S(R) is homogeous ofdegree −1, while m(ξ) = −i sgn(ξ) is a homogeneous (function) of degree 0.

We now wish to generalize the tempered distribution 1πp.v. 1

x and the Hilbert transform tohigher dimensions. A typical situation will be the following.

Let Ω(x) be a function in Rn \ 0 homogeneous of degree 0, that is, such that Ω(λx) = Ω(x)for all λ > 0 and x ∈ Rn \ 0. Notice that Ω, as all homogeneous functions, is uniquelydetermined by its values on the unit sphere Sn−1. Suppose that Ω ∈ L1(Sn−1) and∫

Sn−1

Ω(x′) dσ(x′) = 0 ,

where dσ denotes the Lebesgue surface measure on Sn−1. We define the tempered distribution

u(ϕ) = p.v.

∫Rn

Ω(x)

|x|nϕ(x) dx

= p.v.

∫ +∞

0

∫Sn−1

Ω(x′)ϕ(rx′) dσ(x′)r−1 dr

=

∫ +∞

0

∫Sn−1

Ω(x′)(ϕ(rx′)− ϕ(0)

)dσ(x′)r−1 dr ,

where we have used the fact that Ω has zero integral on the unit sphere. Now, using the meanvalue theorem, as in the case of 1

πp.v. 1x , it is easy to see that the last integral above converges

absolutely, for ϕ ∈ S(Rn).

Lemma 5.6. If u ∈ S ′ is a tempered distribution homogeneous of degree a, then its Fouriertransform is a tempered distribution homogeneous of degree −n− a.

Proof. Recall the identities (2.5). For ϕ ∈ S and λ > 0 we have

u(ϕλ) = u(ϕλ)

= u((ϕ)λ)

= λ−nu((ϕ)λ−1

)= λ−n−au

(ϕ)

= λ−n−au(ϕ) ,

as we wished to show. 2

Lemma 5.7. Let 0 < a < n. Then the function |x|−a is locally integrable and defines a tempereddistribution. The Fourier transform, as an element of S ′ satisfies the equality(

|x|−a)

(ξ) = cn,a|ξ|a−n ,

where

cn,a =πa−n/2Γ

(n−a

2

)Γ(a2

) . (5.5)

Proof. We recall that the Fourier transform of a radial function is also radial.16 Suppose firstthat n/2 < a < n, so that

|x|−a = |x|−aχ|x|≤1 + |x|−aχ|x|>1 = f1(x) + f2(x) ,

where f1 ∈ L1 and f2 ∈ L2. Then we can compute the Fourier transform of |x|−a as a function.Its Fourier transform is a radial function, and by the previous lemma, homogeneous of degree−n+ a. Hence, it is a constant multiple of |ξ|a−n, i.e.,(

|x|−a)

(ξ) = cn,a|ξ|a−n .

We now compute the constant cn,a. Using the Parseval formula (2.6) and the identity(e−π|x|

2 )(ξ) = e−π|ξ|

2of Lemma 2.13 we have∫

Rn

e−π|x|2 |x|−a dx = cn,a

∫Rn

e−π|ξ|2 |ξ|a−n dξ . (5.6)

Now, setting πr2 = s we see that∫ +∞

0e−πr

2rb dr =

1

2√π

∫ +∞

0e−s( sπ

)b/2 1√sds

=1

2π(1+b)/2

∫ +∞

0e−ss(b−1)/2 ds

=1

2π(1+b)/2Γ((1 + b)/2

).

From (5.6) it then follows

cn,a =1

2π(n−a)/2Γ((n− a)/2

)[ 1

2πa/2Γ(a/2)]−1

=πa−n/2Γ

(n−a

2

)Γ(a2

) .

This proves the proposition in the case n/2 < a < n.In the case 0 < a < n/2 we use the first part, since in this case n/2 < n − a < n, and the

inversion formula, valid also in S ′. The case a = n/2 follows by passing to the limit, and usingthe continuity of the Fourier transform in S ′. 2

16This is immediate to check, using the fact that a funciton f is radial if and only if f(Ox) = f(x) for allx ∈ Rn and all orthogonal transformations O.

46 M. M. PELOSO

5.3. The Riesz transforms. We now define the main generalizations of the Hilbert transform,the Riesz transforms Rj , j = 1, . . . , n.

Given f ∈ S, we set

Rjf(x) = cn p.v.

∫Rn

yj|y|n+1

f(x− y) dy , (5.7)

where the constant cn is given by17

cn =Γ(n+1

2

)π(n+1)/2

.

Notice that for n = 1 we have c1 = 1/π and we recover the definition of the Hilbert transform.

Lemma 5.8. For f ∈ L2(Rn) and j = 1, . . . , n we have that

(Rjf ) (ξ) = −i ξj|ξ|f(ξ) .

Hence,n∑j=1

R2j = −I .

Proof. We wish to show that

F(cn p.v.

yj|y|n+1

)(ξ) = −i ξj

|ξ|,

as tempered distributions.We claim that, again as tempered distributions,

∂xj |x|−n+1 = (1− n) p.v.xj|x|n+1

, (5.8)

Assuming the claim we finish the proof. Using Lemma 5.7 and Remark 2.24 (V) we have

F(

p.v.xj|x|n+1

)(ξ) =

1

1− nF(∂xj |x|−n+1

)(ξ)

=2πiξj1− n

F(|x|−n+1

)(ξ)

=2πiξj1− n

πn2−1Γ(1

2)

Γ(n−12 )

|ξ|−1

= −i π(n+1)/2

Γ((n+ 1)/2

) ξj|ξ|

= − i

cn

ξj|ξ|

,

as we wish to prove.

17We recall that the volume of the unit ball Bn and of the unit sphere Sn−1 in Rn are, resp., |Bn| =

πn/2/Γ((n/2) + 1

)and σ(Sn−1) = 2πn/2/Γ(n/2), resp.

It only remains to prove the claim. For ϕ ∈ S we have

(1− n) p.v.xj|x|n+1

(ϕ) = limε→0+

(1− n)

∫Rn

xj

(ε2 + |x|2)(n+1)/2ϕ(x) dx

= limε→0+

∫Rn

∂xj1

(ε2 + |x|2)(n−1)/2ϕ(x) dx

= − limε→0+

∫Rn

1

(ε2 + |x|2)(n−1)/2∂xjϕ(x) dx

= −∫Rn

1

|x|n−1∂xjϕ(x) dx

=(∂xj |x|1−n

)(ϕ) .

This proves the claims and therefore the lemma. 2

Theorem 5.9. The Riesz transforms Rj, j = 1, . . . , n, are of weak-type (1, 1) and of strong-type(p, p), for 1 < p <∞.

Proof. It suffices to show that the the tempered distributions Kj(x) = cn p.v.(xj/|x|n+1

)satisfy

the hypotheses of Thm. 5.1, for j = 1, . . . , n, resp.Clearly Kj coincides with a locally integrable function on Rn \ 0. Moreover, by the lemma,

Kj(ξ) = −i ξj|ξ|

,

which is clearly bounded.Finally, on Rn \ 0

|∇Kj(x)| ≤ cn1

|x|n+1+ cn

n+ 1

2

|xj ||x||x|n+3

≤ C 1

|x|n+1.

Thus, (5.2) is satisfied and we are done. 2

5.4. Solution of the Laplace equation. In this section we consider the Laplace operator

∆ =n∑j=1

∂2xj , (5.9)

and prove a regularity result for the solution of the equation

∆g = f .

Given a partial differential operator P with constant coefficients, we say that a tempereddistribution E is a fundamental solution for P if PE = δ0, the Dirac delta at the origin, that is,if

P (f ∗ E) = f ∗ PE = f ∗ δ0 = f ,

for all f ∈ S. Notice in particular that, for a given f ∈ S, the tempered distribution f ∗E solvesthe equation Pg = f .

We now compute the fundamental solution for ∆.

48 M. M. PELOSO

Proposition 5.10. Let ωn = 2πn/2/Γ(n/2

)denote the volume of the unit sphere in Rn and

define

E(x) =1

(2− n)ωn

1

|x|n−2for n > 2 ,

and

E(x) =1

2πlog |x| for n = 2 .

Then E is a fundamental solution for ∆.

Proof. We consider a family of regularized tempered distributions, namely Eε, given by

Eε(x) =1

(2− n)ωn

1

(ε2 + |x|2)(n−2)/2for n > 2 ,

and

Eε(x) =1

4πlog(ε2 + |x|2) for n = 2 .

It is easy to see that Eε → E pointwise as ε → 0, and that there exists g ∈ L1loc such that

|Eε| ≤ g for all ε ≤ 1, in both cases n = 2 and n > 2. (It suffices to take |E| when n > 2 and∣∣ log |x|∣∣+ 1 when n = 2.) Hence, Eε → E in S ′ and therefore, also ∆Eε → ∆E in S ′ .

Let ϕ ∈ S, we wish to show that ∆Eε(ϕ)→ ϕ(0) as ε→ 0. We provide the details for n > 2.We have

∂xjEε(x) =

1

(2− n)ωn· (2− n)xj

1

(|x|2 + ε2)n/2

∂2xjE

ε(x) =1

ωn

(1

(|x|2 + ε2)n/2−

nx2j

(|x|2 + ε2)(n+2)/2

),

so that

∆Eε(x) =1

ωn

n∑j=1

(1

(|x|2 + ε2)n/2−

nx2j

(|x|2 + ε2)(n+2)/2

)

=n

ωn

ε2

(|x|2 + ε2)(n+2)/2.

Therefore,∆Eε(x) = ψε(x) ,

where, as usual, ψε(x) = ε−nψ(x/ε) and

ψ = ∆E1 =n

ωn

1

(|x|2 + 1)(n+2)/2.

Next we use the fact that ψ(x) = ψ(−x) to see that

∆Eε(ϕ) =

∫∆Eε(−x)ϕ(x) dx = ψε ∗ ϕ(0)→ (

∫ψ)ϕ(0) ,

by the properities of the summability kernels.Thus, we only need to compute

∫ψ. But, by passing to polar coordinates and making the

change of variables s = r2/(1 + r2), so that ds = 2(1 + r2)−2rdr, and

rn−1

(1 + r2)(n+2)/2=( r2

1 + r2

)(n−2)/2 r

(1 + r2)2.

We then find ∫ψ(x) dx =

n

ωn

∫1

(|x|2 + 1)(n+2)/2dx

= n

∫ +∞

0

1

(1 + r2)(n+2)/2rn−1 dr

=n

2

∫ 1

0s(n−2)/2 ds = 1 .

This completes the proof (at least in the case n > 2, the case n = 2 being left to the reader). 2

We now have the regularity result for the solutions of the Laplace equation.

Theorem 5.11. Let f ∈ S be given. Define the operator

Tf(x) = ∂2xjxk

f ∗ E = f ∗ ∂2xjxk

E ,

j, k = 1, . . . , n. Then, T extends to an operator that is of weak-type (1, 1) and of strong-type(p, p), for 1 < p <∞.

Remark 5.12. Notice that the result shows that, for a given f ∈ Lp, it is possible to define thesolution u = f ∗ E of the equation ∆u = f is such a way that the map T (f) = ∂2

xjxk(f ∗ E) is

bounded on Lp.On the other hand, the map I defined as f 7→ u = f ∗ E is bounded, when n ≥ 3,

I : Lp → Lq

as a consequence of Thm. 5.19 of the next section, where

1

q=

1

p− 2

n.

Proof of Thm. 5.11. It suffices to show that the kernel K = ∂2xjxk

E satisfies the hypotheses of

Thm. 5.1. It is clear that, by Lemma 5.7,

F(∂2xjxk

E)(ξ) = −(2π)2ξjξkE(ξ)

= Cnξjξk|ξ|−2 ,

so that (5.1) is satisfied.On the other hand, also the condition (5.3) is also easily seen to be satisfied: if x 6= 0,∣∣∇(∂2

xjxkE)(x)∣∣ ≤ C 1

|x|n+1,

The result now follows from 5.1. 2

We now consider the Dirichlet problem for the Laplacian on the half-space∂2t f(x, t) + ∆u(x, t) = 0 for (x, t) ∈ Rn+1, t > 0

u(x, 0) = f(x) for x ∈ Rn .(5.10)

Here ∆ = ∆x =∑n

j=1 ∂2xJ

, so that ∂2t + ∆x is the Laplacian in Rn+1. Moreover,

Rn+1+ :=

(x, t) ∈ Rn ×R : t > 0

.

50 M. M. PELOSO

We assume that the boundary data g ∈ Lp(Rn) and we look for a solution f defined on Rn+1

such that f(·, t) ∈ S ′(Rn) for all fixed t > 0. In this case, we can compute the Fourier transformw.r.t. the variable x ∈ Rn. We have that

F((∂2t + ∆x)f(·, t)

)(ξ) = ∂2

t f(ξ, t) + F(∆x)f(·, t)

)(ξ) = ∂2

t f(ξ, t)− |2πξ|2f(ξ, t) .

For the time being we consider ξ as a fixed parameter and write ϕ = f(t, ·). Then the Dirichletproblem (5.10) has been transformed into the Cauchy initial value problem

ϕ′′(t)− a2ϕ(t) = 0

ϕ(0) = b ,

where a = |2πξ| and b = g(ξ). Since the characteristic equation has roots λ± = ±|2πξ|, thesolutions of the ODE are

ϕ(t) = c+e|2πξ|t + c−e

−|2πξ|t .

Then, the solution of (5.10) has to satisfy

F−1(c+e|2πξ|t + c0e

−|2πξ|t)(x)

and the only possibility for the (inverse) Fourier transform to make sense is that c+e|2πξ|t +

c0e−|2πξ|t ∈ S ′(Rn). This forces c+ = 0 so that the solution of the Cauchy problem satifsies the

initial condition

ϕ(0) = c− = b = g(ξ) .

Therefore,

f(x, t) = F−1(ge−|2π·|t

)(x)

=(g ∗ F−1(e−|2π·|t)

)(x) .

Thus, we need to compute F−1(e−|2π·|t))(x). Observe that by Remark 2.14 we have that

F−1(e−|2π·|t))(x) = Pt(x), so it suffices to determine Ψ such that P (x) = e−|2πξ.

We define the Poisson kernel for the half-space

Pt(x) = cnt(

|x|2 + t2)n+1

2

(5.11)

where ωn =Γ(n+1

2)

πn+12

. We have the following

Proposition 5.13. The function P (x) = ωn1(

1+|x|2)n+1

2∈ L1(Rn) and

P (ξ) = e−|2πξ| .

For the proof, see Exercise 9, Appendix B.18

18Check the constants cn and ωn.


5.5. The heat operator. We now turn our attention to the heat operator in Rn+1 ∂t−∆. Webegin by considering the Cauchy problem in the upper-half space Rn× (0,∞), where we denotethe variables (x, t) ∈ Rn ×R, called the Cauchy problem for the heat equation,

∂tu(x, t)−∆u(x, t) = 0 for (x, t) ∈ Rn+1, t > 0

u(x, 0) = f(x) for x ∈ Rn .(5.12)

We assume that f ∈ S ′ and that the solution u is a tempered distribution in x, for any tfixed. Then, by taking the Fourier transform in the x-variable, we obtain the ordinary differentialequation Cauchy problem (for any ξ ∈ Rn fixed)

∂tu(ξ, t) + 4π2|ξ|2u(ξ, t) = 0 for t > 0

u(ξ, 0) = f(ξ) .

We can easily solve this Cauchy problem and find that the solution is

u(ξ, t) = f(ξ)e−4π2|ξ|2t for t > 0 .

Therefore, by undoing the Fourier transform we obtain

u(x, t) = f ∗K√t(x) ,

where

K√t(x) = K(x, t) = F−1(e−4π2|ξ|2t) =

1

(4πt)n/2e−|x|

2/4t ,

by Prop. 2.13. Notice that

K√t(x) = t−n/2K1(x/t1/2) ,

so that∫K√t dx =

∫K1 dx = 1, for all t > 0.19

We now compute a fundamental solution for the heat operator. We define K on all of Rn+1

by setting

K(x, t) =

(4πt)−n/2e−|x|

2/4t for t > 0

0 for t ≤ 0 .(5.13)

Since∫Rn K(x, t) dx = 1 for t > 0, K(x, t) is integrable on every region of the form Rn×(−∞, T ].

Proposition 5.14. The function defined by (5.13) is a fundamental solution for the heat oper-ator.

Proof. Let ε > 0 be given. Define K(ε)(x, t) = K(x, t) for t > ε and K(ε)(x, t) = 0 for t ≤ ε.

Then K(ε) → K in the sense of distributions in Rn+1 as ε→ 0. Then, we wish to show that

〈(∂tK(ε) −∆xK(ε)), ϕ〉 = 〈K(ε), (−∂t −∆x)(ϕ)〉 → ϕ(0, 0)

as ε→ 0, for every ϕ ∈ D(Rn+1).

19The notation K√t is not standard, as this kernel is classically defined as Kt. We have elected to use the

former notation for consistency.

52 M. M. PELOSO

Now, by integrating by parts we find that

〈K(ε), (−∂t −∆x)(ϕ)〉 =

∫ +∞

ε

∫Rn

K(x, t)(−∂t −∆x)(ϕ)(x, t) dxdt

=

∫ +∞

ε

∫Rn

−∆xK(x, t)ϕ(x, t) dxdt+

∫Rn

∫ +∞

ε∂tK(x, t)ϕ)(x, t) dtdx

+

∫Rn

K(x, ε)ϕ(x, ε) dx

=

∫ +∞

ε

∫Rn

(∂t −∆x)K(x, t)ϕ(x, t) dxdt+

∫Rn

K(x, ε)ϕ(x, ε) dx

= 0 +

∫Rn

K(−x, ε)ϕ(x, ε) dx

since K is even in x and it is annihilated by the heat operator on Rn × (ε,+∞). But, the righthand side above is (the convolution being in Rn)∫

Rn

K(−x, ε)ϕ(x, ε) dx = K(ε) ∗ ϕ(·, ε)(0)

= K(ε) ∗ ϕ(·, 0)(0) +[K(ε) ∗ ϕ(·, ε)−K(ε) ∗ ϕ(·, 0)

](0) .

The former term tends to ϕ(0, 0) as ε→ 0, while the latter one is bounded by

supx∈Rn

|ϕ(x, ε)− ϕ(x, 0)|‖K(ε)‖1 = supx∈Rn

|ϕ(x, ε)− ϕ(x, 0)|

which tends to 0 as ε→ 0. 2

We conclude this part by solving the problem (5.12).

Theorem 5.15. Let 1 ≤ p < +∞ and g ∈ Lp(Rn) be given. Then the boundary value problem(5.12) admits a solution f(x, t) such that f(·, t)→ g in Lp(Rn) as t→ 0+.

Proof. Setting f = g ∗K√t we see that f solves the equation in Rn+1+ and f(·, t) = g ∗K√t → g

in Lp(Rn) as t→ 0+.

5.6. The Riesz potentials. We now discuss some operators Iα, called the Riesz potentials, thatarise in a natural way in connection with the Laplacian. The operators Iα are integral operatorswhose kernel is homogenous of degree −(n − α), then locally integrable when 0 < α < n, caseto which we restrict ourselves.

Definition 5.16. Let 0 < α < n and define the Riesz potential of order α the integral operator

Iα(f)(x) =1

(2π)α1

cn,n−α

∫Rn

1

|x− y|n−αf(y) dy ,

where cn,n−α is the constant defined in (5.5). (It will be sometimes convenient to set a = n−α.Notice that 0 < a < n exactly when 0 < α < n.)

We remark that the Laplacian satisfies the identity

(−∆f ) (ξ) = (2π|ξ|)2f(ξ) ,

for f ∈ S. By Lemma 5.7, it follows that

(Iαf ) (ξ) =1

(2π)α1

cn,n−α

(|x|−(n−α) ∗ f

)(ξ)

= (2π|ξ|)−αf(ξ) .

Hence, it is natural to define the fractional powers of −∆ by setting((−∆)−α/2f

)(ξ) = (2π|ξ|)−αf(ξ) , (5.14)

that is, by setting

(−∆)−α/2 = Iα .

We now study the (Lp, Lq)-boundedness of the operators Iα. We begin with a simple resultthat will give us a necessary condition on p, q and α. We set Dλf(x) = fλ(x) = f(λx), forλ > 0.

Lemma 5.17. For λ > 0 we have the follwing identities:

(i) Dλ−1IαDλ = λ−αIα;

(ii) ‖Dλf‖Lp = λ−n/p‖f‖Lp;

(iii) ‖Dλ−1Iα(f)‖Lq = λn/q‖Iα(f)‖Lq .

Proof. These are elementary. For a given function f , sufficiently regular, we have

Dλ−1IαDλ(f)(x) =(2π)α

cn,n−α

∫Rn

1

|λ−1x− y|n−αf(λy) dy

=(2π)α

cn,n−α

1

λn

∫Rn

1

|λ−1(x− z)|n−αf(z) dz

= λ−αIα(f)(x) ,

as we claimed.(ii) and (iii) are simply a change of variables. 2

Lemma 5.18. Let 0 < α < n, 1 < p, q < ∞ be given. Then, if Iα : Lp(Rn) → Lq(Rn) isbounded, then necessarily

1

q=

1

p− α

n.

Proof. If Iα is bounded, there exists a constant C > 0 such that

‖Iα(f)‖Lq ≤ C‖f‖Lp

for all f ∈ Lp. Then, for all λ > 0 it must hold that

‖Iα(Dλf)‖Lq ≤ C‖Dλf‖Lp . (5.15)

Now, by Lemma 5.17,

‖Iα(Dλf)‖Lq = ‖DλDλ−1Iα(Dλf)‖Lq = λ−n/qλ−α‖Iα(f)‖Lq ,

Then, (5.15) is equivalent to

‖Iα(f)‖Lq ≤ Cλnq

+α−np ‖f‖Lp .

54 M. M. PELOSO

Since this must hold for all λ > 0, if the exponent of λ in the inquality above is 6= 0, by lettingλ→ 0, or λ→ +∞ we get a contradiction. The conclusion now follows. 2

We now prove the positive result, valid in the case 1 < p, q <∞.

Theorem 5.19. Let 0 < α < n, 1 ≤ p, q <∞ satisfy

1

q=

1

p− α

n, (5.16)

and consider the fractional integral Iα. For f ∈ Lp, the integral defining Iα(f) converges abso-lutely. Moreover,

(i) if 1 < p, q <∞Iα : Lp(Rn)→ Lq(Rn)

is strong-type (p, q);(ii) if p = 1, then Iα is weak-type (1, q).

Proof. Clearly, it suffices to consider the convolution operator T : f 7→ K ∗ f , where K(x) =|x|−n+α and prove (i) and (ii) for this operator.

By the general form of the Marcinkiewicz interpolation theorem, if we show that T is weak-type (p1, q1) and also weak-type (1, q0), both pairs satisfying the condition (5.16), then it followsthat it is strong-type (p, q), where

1

p=

θ

p1+ 1− θ , and

1

q=

θ

q1+

1− θq0

.

Notice that1

q= θ( 1

p1− α

n

)+ (1− θ)

(1− α

n

)=

θ

p1+ 1− θ − α

n=

1

p− α

n.

We write K = K1 +K∞, where

K1(x) = K(x)χ|x|≤R , and K∞(x) = K(x)χ|x|>R ,

and where R > 0 is a positive constant to be selected later.We begin by observing that K1 ∈ L1(Rn), since∫

Rn

|K1(x)| dx =

∫|x|≤R

|x|−n+α dx = ωn

∫ R

0rα−1 dr

=ωnαRα . (5.17)

On the other hand, K∞ ∈ Lp′, where p′ is the exponent conjugate to p. For,∫

Rn

|K∞(x)|p′ dx =

∫|x|>R

|x|(−n+α)p′ dx

= ωn

∫ +∞

Rrn(1−p′)+αp′−1 dr

=ωn

n(p′ − 1)− αp′Rn(1−p′)+αp′ , (5.18)

and the integral is convergent since n(1− p′) +αp′ < 0, that is, −np +α < 0, which is equivalent

to q <∞.Now, let f ∈ Lp, 1 ≤ p < ∞. Then, the integral defining K ∗ f converges absolutely since

K1 ∈ L1 and K∞ ∈ Lp′. (Make sure you agree with this assertion.) This proves the first part of

the statement.In order to prove (i) and (ii), it suffices to show that, for 1 ≤ p < q < ∞ such that 1/q =

1/p− α/n, T is weak-type (p, q), that is,∣∣x : |K ∗ f(x)| > λ∣∣ ≤ (C ‖f‖Lp

λ

)q. (5.19)

Notice that we may assume that ‖f‖Lp = 1 (as it is easy to check, and we invite you to doso).

Then we have,∣∣x : |K ∗ f(x)| > λ∣∣ ≤ ∣∣x : |K1 ∗ f(x)| > λ/2

∣∣+∣∣x : |K∞ ∗ f(x)| > λ/2

∣∣ .Now, using (5.17) and the assumption ‖f‖Lp = 1, we have∣∣x : |K1 ∗ f(x)| > λ/2

∣∣ =

∫x:|K1∗f(x)|>λ/2

dx ≤∫ ( |K1 ∗ f(x)|

λ/2

)pdx

=2p‖K1 ∗ f‖pLp

λp≤

2p‖K1‖pL1‖f‖pLpλp

≤ C(Rαλ

)p. (5.20)

Next, using (5.18) we see that

‖K∞ ∗ f‖L∞ ≤ ‖K∞‖Lp′‖f‖Lp = ‖K∞‖Lp′ ≤ CRα−n/p . (5.21)

We now choose R such that

CRα−n/p =λ

2, that is, R = C ′λ1/(α−n/p) = C ′λ−q/n .

With choice of R, by (5.21) it follows that∣∣x : |K∞ ∗ f(x)| > λ/2∣∣ = 0 .

Therefore, by (5.21) (recalling that ‖f‖Lp = 1),∣∣x : |K ∗ f(x)| > λ∣∣ ≤ C(Rα

λ

)p≤ C

(λ−q(α/n+1/q)

)p= Cλ−q = C

(‖f‖Lpλ

)q.

This concludes the proof of the theorem. 2

56 M. M. PELOSO

5.7. †Construction of Calderon–Zygmund kernels. In order to prove the Mihlin-Hormandermultiplier theorem, Theorem 6.24 that follows, we describe a method, that is quite standard, toconstruct Calderon–Zygmund convolution kernels, as defined in Def. 5.2.

Definition 5.20. An Lp-function f is said to satisfy the Lp-Lipschitz condition of order α,α ∈ (0, 1), if there exists a constant C > 0 such that(∫

Rn

|f(x− h)− f(x)|p dx)1/p

≤ C|h|α

for every h ∈ Rn. In this case we set

‖f‖Λpα := ‖f‖Lp(Rn) + suph6=0|h|−α

(∫Rn

|f(x− h)− f(x)|p dx)1/p

.

Remark 5.21. The following facts are easy to prove.20

(1) An Lp-function f is in Λpα if and only if for every λ > 0 we have

sup0<|h|<λ

|h|−α(∫

Rn

|f(x− h)− f(x)|p dx)1/p

<∞ ;

(2) let ϕ ∈ C∞0 be identically 1 in a nghb of the origin and set f = |x|α−n/pϕ, then f ∈ Λpα.

Lemma 5.22. Let f ∈ Λ1α. Then, there exists C > 0 such that

|f(ξ)| ≤ C(1 + |ξ|)−α‖f‖Λ1α.

Proof. Notice that

−f(ξ) = e−πi∫Rn

f(x)e−2πixξ dx =

∫Rn

f(x)e−2πi(x+ξ/(2|ξ|2))ξ dx

=

∫Rn

f(x− ξ/(2|ξ|2)

)e−2πixξ dx .

Therefore,

|f(ξ)| = 12

∣∣∣ ∫Rn

[f(x)− f

(x− ξ/(2|ξ|2)

)]e−2πixξ dx

∣∣∣≤ 1

2

∫Rn

∣∣∣f(x)− f(x− ξ/(2|ξ|2)

)∣∣∣ dx≤ C|ξ|−α‖f‖Λ1

α.

On the other hand|f(ξ)| ≤ ‖f‖L1 ≤ ‖f‖Λ1

α,

so that(1 + |ξ|)α|f(ξ)| ≤ C(1 + |ξ|α)|f(ξ)| ≤ C‖f‖Λ1

α.


We will need a smooth decomposition of the function identically 1.

Lemma 5.23. There exists a function ϕ ∈ C∞0 such that

20Exercise.


(i)∑j∈Z

ϕ(2−jξ) = 1 for all ξ 6= 0 .

Moreover, there exists another function ϕ0 ∈ C∞0 such that

(ii) ϕ0(ξ) +

+∞∑j=1

ϕ(2−jξ) = 1 for all ξ .

Proof. Let ϕ0 ∈ C∞0 (Rn) such that ϕ ≥ 0, ϕ0(ξ) = 1 if |ξ| ≤ 1 and ϕ0(ξ) = 0 if |ξ| ≥ 2. (Suchfunction exists by the C∞-Urysohn’s lemma, Lemma 2.7). We set

ϕ(ξ) = ϕ0(ξ)− ϕ0(2ξ) .

Notice that ϕ(ξ) = 0 if |ξ| ≤ 1/2 or |ξ| ≥ 2, that is,

suppϕ ⊆ ξ : 1/2 ≤ |ξ| ≤ 2 .By induction, it is elementary to prove that, for ` ≥ 1 and all ξ,

ϕ0(ξ) +∑j=1

ϕ(2−jξ) = ϕ0(2−`ξ) . (5.22)

Letting `→ +∞ we obtain (ii).Next, again by induction on `′, with `′ ≤ 0, it is easy to show that∑

j=`′

ϕ(2−jξ) = ϕ0(2−`ξ)− ϕ0(2−`′+1ξ) . (5.23)

Letting ` → +∞ and `′ → −∞ we obtain (i). (Notice that ϕ0(2−`ξ) − ϕ0(2−`′+1ξ) → 1 only

when ξ 6= 0.) 2

Corollary 5.24. Let ϕ0, ϕ be as in Lemma 5.23, and set

ψ(0) = Fϕ0 , , ψ(j) = F(ϕ(2−j ·)

)=(Fϕ)

2−j, for j = 1, 2, . . . (5.24)

where(Fϕ)

2−j(ξ) = 2jn(Fϕ)(2jξ) denotes the dilation defined in (2.2). Then the following

properties hold:

(i)∑N

j=0 ψ(j) → δ0 in S ′ as N → +∞;

(ii) ‖ψ(j)‖L1 ≤ C, ‖ψ(j)‖L∞ ≤ C2nj, for all j;

(iii) for all j,∫ψ(j) = 0.

Proof. It is immediate to see that the sum in Lemma 5.23 converges in S ′. Then

F(ϕ0 +

N∑j=1

ϕ(2−j ·))

=

N∑j=0

ψ(j) → δ0

as N → +∞ in S ′.Next,

‖ψ(j)‖L1 = ‖(Fϕ)2−j‖L1 = ‖Fϕ‖L1 ≤ ‖Fϕ0‖L1 + ‖F(ϕ0(2·)‖L1 ≤ C .Moreover,

‖ψ(j)‖L∞ ≤ ‖Fψ(j)‖L1 = ‖ϕ(2−j ·)‖L1 = 2jn‖ϕ‖L1 = C2jn .

58 M. M. PELOSO

Finally, ∫ψ(j)(x) dx = F−1(ψ(j))(0) = ϕ2−j (0) = ϕ(0) = 0 .

This concludes the proof. 2

Lemma 5.25. Let α ∈ (0, 1) and let p < (n + α)/n. Then, if f ∈ Λ1α we have f ∈ Lp and

‖f‖Lp ≤ C‖f‖Λ1α

.

Proof. Let f ∈ Λ1α, and let ψ(j) be as in the previous lemma.

Using (iii), observe that for j ≥ 1,

‖f ∗ ψ(j)‖L1 =

∫Rn

∣∣∣ ∫Rn

(f(x− y)− f(x)

)ψ(j)(y) dy

∣∣∣ dx≤∫Rn

∫Rn

∣∣f(x− y)− f(x)∣∣ dx|ψ(j)(y)| dy

≤ ‖f‖Λ1α

∫Rn

|y|α |ψ(j)(y)| dy

= ‖f‖Λ1α

∫Rn

|y|α |2jnFϕ(2jy)| dy

≤ 2−jα‖f‖Λ1α

∫Rn

|z|α |Fϕ(z)| dz

≤ C2−jα‖f‖Λ1α.

Moreover, by (i)

‖f ∗ ψ(j)‖L∞ ≤ ‖f‖L1 ‖ψ(j)‖L∞ ≤ C2nj‖f‖L1 ≤ C2nj‖f‖Λ1α.

Hence,

‖f ∗ ψ(j)‖pLp ≤ ‖f ∗ ψ(j)‖

p−1L∞

∫Rn

|f ∗ ψ(j)(x)|p dx ≤ C2nj(p−1)2−jα‖f‖pΛ1α.

Therefore,

+∞∑j=0

‖f ∗ ψ(j)‖Lp = limN→+∞

N∑j=0

‖f ∗ ψ(j)‖Lp

≤ C‖f‖Λ1α

limN→+∞

N∑j=0

2j[n(p−1)−α]/p

≤ C‖f‖Λ1α,

if n(p− 1)− α < 0, that is, p < (n+ α)/n.Consequentely,

∑+∞j=0 f ∗ψ(j) convereges in the Lp-norm, and since it converges in S ′ to f (at

least when f ∈ S), we obtain that

‖f‖Lp ≤+∞∑j=0

‖f ∗ ψ(j)‖Lp

≤ C‖f‖Λ1α,

when p < (n+ α)/n. 2

We now obtain the reward for our work in this setting with the following result.

Theorem 5.26. Let k(j) ∈ L1(Rn), j ∈ Z, be functions such that there exist constants ε > 0,α ∈ (0, 1) and C > 0 such that for all j ∈ Z,

(1)

∫Rn

(1 + |x|)ε|k(j)(x)| dx ≤ C;

(2)

∫Rn

k(j)(x) dx = 0;

(3) ‖k(j)‖Λ1α≤ C.

Then the series∑

j∈Z(k(j)

)2j

converges in S ′ to a Calderon–Zygmund kernel.

Proof. We wish to show that∑

j∈Z(k(j)

)2j

converges in S ′ to a distribution K that coincides on

Rn \ 0 with a locally integrable function and that K satisfies the conditions (5.2) and (5.1).

Step 1. We begin by showing that the series of the Fourier transforms of the(k(j)

)2j

convergesabsolutely, that is, ∑

j∈Z

∣∣F((k(j))2j)(ξ)∣∣ ≤ C . (5.25)

We split the sum in two, the first one being for j such that |2jξ| ≥ 1. Using Lemma 5.22 and(3) we have21

∑j: |2jξ|≥1

∣∣F((k(j)

)2j

)(ξ)∣∣ =

∑j: |2jξ|≥1

∣∣F(k(j)

)(2jξ)

∣∣≤ C

∑j: |2jξ|≥1

(1 + |2jξ|)−α‖k(j)‖Λ1α

≤ C∑

j: |2jξ|≥1

(1 + |2jξ|)−α

≤ C∑

j: |2jξ|≥1

(2j |ξ|)−α

= C|ξ|−α∑

j: 2j |ξ|≥1

2−jα

≤ C . (5.26)

21 Here we also use the elementary estimates∑j≥j0 q

j ' qj0 if 0 < q < 1, and∑j≤j0 q

j ' qj0 if q > 1.

60 M. M. PELOSO

When |2jξ| < 1 we use (1), (2) and the estimate |e−2πit − 1| ≤ C|t|ε for |t| < 1 (and footnote21 again). We have∑

j: |2jξ|<1

∣∣F((k(j)

)2j

)(ξ)∣∣ =

∑j: |2jξ|<1

∣∣∣ ∫Rn

k(j)(x)(e−2πi2jξx − 1

)dx∣∣∣

≤ C∑

j: |2jξ|<1

∫Rn

|k(j)(x)| |2jξx|ε dx

≤ C|ξ|ε∑

j: |2jξ|<1

2εj

≤ C . (5.27)

Estimates (5.26) and (5.27) now give (5.25). We set

u(ξ) =∑j∈ZF((k(j)

)2j

)(ξ) ∈ L∞(Rn) .

Since, for each f ∈ S, by the dominated convergence theorem,∫Rn

u(ξ)f(ξ) dξ =∑j∈Z

∫Rn

F((k(j)

)2j

)(ξ)f(ξ) dξ

so that∑

j∈ZF((k(j)

)2j

converges in S ′. By continuity of the (inverse) Fourier transform in S ′,∑j∈Z

(k(j)

)2j

= K (5.28)

for some K ∈ S ′. This completes Step 1.

Step 2. We wish to prove that the distribution K defined in (5.28) is a Calderon–Zygmundkernel.

We first show that K coincides with a locally integrable function on Rn \ 0, by showingthat

∑j∈Z

(k(j)

)2j

converges in L1(E), for every compact E ⊂ Rn.

It suffices to consider compact sets E of the form x : 2m ≤ |x| ≤ 2m+1, for some m ∈ Z.22

Notice that, with an obvious change of variables,∫x: 2m≤|x|≤2m+1

|(k(j)

)2j

(x)| dx =

∫x: 2m−j≤|x|≤2m−j+1

|k(j)(x)| dx .

Assume first that j ≤ m. Then using (1) we have,∫x: 2m−j≤|x|≤2m−j+1

|k(j)(x)| dx ≤ 2(j−m)ε

∫x: 2m−j≤|x|≤2m−j+1

|x|ε |k(j)(x)| dx

≤ C2(j−m)ε .

22Make sure you agree with this reduction.

On the other hand, if j > m we use (3) and Lemma 5.25 we have∫x: 2m−j≤|x|≤2m−j+1

|k(j)(x)| dx ≤(∫|x|≤2m−j+1

|k(j)(x)|p dx)1/p∣∣|x| ≤ 2m−j+1

∣∣1/p′≤ ‖k(j)‖Lp2n(m−j+1)/p′

≤ C2n(m−j+1)/p′ ,

for a suitable p (and p′).Therefore, ∑

j∈Z‖k(j)‖L1(E) ≤

∑j≤m

C2(j−m)ε +∑j>m

C2n(m−j+1)/p′ ≤ C ,

since both series converge.

We now show that K satisfies the two conditions (5.1) and (5.2). But, by construction and(5.25)

|K(ξ)| ≤∑j∈Z

∣∣F((k(j))2j)(ξ)∣∣ ≤ C ,

so that (5.1) holds.Finally, to prove that K satisfies Hormander’s condition, let h ∈ Rn \ 0 and assume that

2m < 2|h| ≤ 2m+1. Then we have∫|x|>2|h|

|K(x− h)−K(x)| dx ≤∑j∈Z

∫|x|>2m

|(k(j)

)2j

(x− h)−(k(j)

)2j

(x)| dx

=∑j∈Z

∫|x|>2m−j

|k(j)(x− 2−jh)− k(j)(x)| dx

=∑j<m

∫|x|>2m−j

|k(j)(x− 2−jh)− k(j)(x)| dx

+∑j≥m

∫|x|>2m−j

|k(j)(x− 2−jh)− k(j)(x)| dx

≤ 2∑j<m

∫|x|>2m−j−1

|k(j)(x)| dx+∑j≥m|2−jh|α‖k(j)‖Λ1

α

≤ 2∑j<m

2(j−m)ε

∫|x|>2m−j−1

|x|ε |k(j)(x)| dx+ C∑j≥m|2−jh|α

≤ C∑j<m

2(j−m)ε + C∑j≥m|2−jh|α

≤ C ,

where we have used conditions (1) and (3) in the hypotheses. This proves Step 2 and hence thetheorem. 2

62 M. M. PELOSO

5.8. †Vector-valued singular integrals. We conclude this part by extending Thm. 5.1 tothe case of vector-valued functions. This extension will be used in Section 8, when developingthe Littlewood–Paley theory.

Although this part of the theory works in the case of functions taking values in a separableBanach space, we restrict our attention to the case of functions with values in a separable HilbertspaceH.

We say that a function f from Rn to H is measurable if for every v ∈ H, the mapping

Rn 3 x 7→ 〈f(x), v〉H ∈ C

is measurable.For 1 ≤ p ≤ ∞ we define the space Lp(Rn,H) to be the space of measurable functions

f : Rn → H such that (∫Rn

‖f(x)‖pH dx)1/p

<∞ .

Then, Lp(Rn) = Lp(Rn,C), but we will still denote it by Lp for simplicity.A typical example of Lp(Rn,H) is a function of the form f(x) = fs(x)v, where fs is a scalar-

valued function in Lp(Rn) and v is a fixed element of H. The subset

Lp ⊗H =f : f =

∑Nj=1fjvj where fj ∈ Lp, vj ∈ H

,

of finite linear combination of functions of the form above, is dense in Lp(Rn,H). (The proofof this fact is simple, for this and other facts about this topic, see [Ru].)

Given f =∑N

j=1 fjvj ∈ L1 ⊗H, we define its integral to be the element of H given by∫Rn

f(x) dx =N∑j=1

(∫Rn

fj(x) dx)vj .

This map f 7→∫f(x)dx extends to all of L1(Rn,H) by density. Then, for f ∈ L1(Rn,H),∫

f(x)dx is the unique element of H such that⟨∫f(x)dx, v

⟩H

=

∫Rn

〈f(x), v〉H dx (5.29)

for all v ∈ H.Similarly reasoning works also in the case of f ∈ Lp(Rn,H) and g ∈ Lp′(Rn,H): notice that

by applying the standard Holder’s inequality it follows that∫Rn

∣∣⟨f(x), g(x)⟩H∣∣ dx ≤ (∫

Rn

‖g(x)‖p′

H dx)1/p′(∫

Rn

‖f(x)‖pH dx)1/p

= ‖g‖Lp′ (Rn,H)‖f‖Lp(Rn,H) . (5.30)

Therefore∫Rn〈f(x), g(x)

⟩Hdx converges and it turns out that

‖g‖Lp′ (Rn,H) = sup∣∣∣ ∫

Rn

〈f(x), g(x)〉 dx∣∣∣ , ‖f‖Lp(Rn,H) = 1

.

This implies that Lp′(Rn,H) ⊆

(Lp(Rn,H)

)∗. In fact equality holds (in the Hilbert case), that

is, (Lp(Rn,H)

)∗ ≡ Lp′(Rn,H) .

We conclude this preliminary discussion introducing the singular integral in the vector-valuecase.

Let H1,H2 be Hilbert spaces and denote by L(H1,H2) the space of bounded linear operators

from H1 to H2. Suppose f : Rn → H1 and ~K : Rn \0 → L(H1,H2) are measurable functions.We may consider the integral operator

~Tf(x) =

∫Rn\0

K(y)f(x− y) dy (5.31)

Here clearly the expression ~K(y)f(x − y) can only be interpreted as the element ~K(y) of

L(H1,H2) acting on f(x− y) ∈ H1; hence, it ~Tf(x) is well defined, it is an element of H2.

Theorem 5.27. Suppose ~T is a bounded linear operator from Lr(Rn,H1) to Lr(Rn,H2) forsome r, 1 < r <∞, defined by the integral operator with kernel K as in (5.31). Further, assumethat K satisfies the (vector-valued) Hormander condition∫

|x−y|>2|y|‖ ~K(x− y)− ~K(x)‖L(H1,H2) dx ≤ B . (5.32)

Then ~T is bounded from Lp(Rn,H1) to Lp(Rn,H2) when 1 λ

∣∣ ≤ C

λ‖f‖L1(Rn,H1) .

Proof. For simplicity, we are going to drop the “ ~ ”-notation.The proof does not follow from the scalar case, but, it follows from the same proof. We begin

by observing that the adjoint of T has kernel K∗(−x), since

〈Tf, g〉 =

∫Rn

⟨∫Rn

K(x− y)f(y)dy, g(x)⟩H2

dx

=

∫Rn

∫Rn

〈K(x− y)f(y), g(x)〉H2 dydx

=

∫Rn

∫Rn

〈f(y), K∗(x− y)g(x)〉H2 dxdy

=

∫Rn

⟨f(y),

∫Rn

K∗(x− y)g(x)dx⟩H1

dy .

If we show that T is weak-type (1, 1), by Marcinkiewicz interpolation theorem it would followthat T is of strong-type (p, p) for 1 < p ≤ r. The operator T ∗ satisfies the same assumption as

T , with the assumption that T ∗ is a bounded linear operator from Lr′(Rn,H2) to Lr

′(Rn,H1),

so that it is also of strong-type (p, p) for 1 < p ≤ r′. Thus, by duality, T is of strong-type (p, p)for r < p <∞, so it follows that T is of strong-type (p, p) for 1 < p <∞.

Thus, (as in the scalar case) it suffices to show that T is weak-type (1, 1).We now proceed as in the proof of Thm. 5.1. Hence, we only sketch the argument, indicating

the main differences.

The main point is to generalize the Calderon–Zygmund decomposition of an L1-function. Inthe proof of Thm. 4.6, as well as in Thm. 3.5, we assumed that f is non-negative (since wewrote an arbitrary complex-valued function as f = (Re f)+ − (Re f)− + i

[(Im f)+ − (Im f)−

]).

We provide the details for sake of completeness.

64 M. M. PELOSO

Given λ > 0, by applying the Calderon–Zygmund decomposition to the function ‖f(x)‖H1 wefind that there exists a sequence of disjoint cubes Qj such that:

(i) 0 ≤ ‖f(x)‖H1 ≤ λ for almost all x 6∈ ∪jQj ;

(ii)∣∣ ∪j Qj∣∣ ≤ 1

λ‖f‖L1(Rn,H1);

(iii) λ <1

|Qj |

∫Qj

‖f(x)‖H1 dx ≤ 2nλ.

Now we decompose f as the sum f = g + b, where

g(x) = f(x)χ c∪jQj(x) +∑j

( 1

|Qj |

∫Qj

f(y) dy)χQj (x) ,

and

b(x) =∑j

bj(x) =∑j

(f(x)− 1

|Qj |

∫Qj

f(y) dy)χQj (x) .

Notice that, as f(x), g(x) and b(x) are vectors in H1, for a.a. x ∈ Rn, and that 1|Qj |

∫Qjf(y)dy

is also a well-defined element of H1, since f ∈ L1(Rn,H1) and by the identity (5.29).Notice that, similarly to the proof of Thm.’s 4.6 and 5.1, we have that

‖g(x)‖H1 ≤ ‖f(x)‖H1χ c∪jQj(x) +∑j

∥∥∥ 1

|Qj |

∫Qj

f(y) dy∥∥∥H1

χQj (x)

≤ λχ c∪jQj(x) +∑j

( 1

|Qj |

∫Qj

‖f(y)‖H1 dy)χQj (x)

≤ 2nλ ;

and also that ∫Rn

‖g(x)‖H1 dx ≤∫Rn

‖f(x)‖H1 dx .

(This last inequality follows since g(x) = f(x) on c(∪jQj), while∫Qjg(x)dx =

∫Qjf(x)dx for

all Qj .)

On the other hand,∫Rn b(x)dx = 0H1 , the zero element in H1.

We now proceed and in the proof of Thm. 5.1. Since

‖Tf(x)‖H2 ≤ ‖Tg(x)‖H2 + ‖Tb(x)‖H2 ,

we have ∣∣x : ‖Tf(x)‖H2 > λ∣∣ ≤ ∣∣x : ‖Tg(x)‖H2 > λ/2

∣∣+∣∣x : ‖Tb(x)‖H2 > λ/2

∣∣ .


Now, using the boundedness of T : Lr(Rn,H1)→ Lr(Rn,H2),∣∣x : ‖Tg(x)‖H2 > λ/2∣∣ =

∫x:‖Tg(x)‖H2

>λ/2dx ≤

∫x:‖Tg(x)‖H2

>λ/2

‖Tg(x)‖rH2

(λ/2)rdx

≤ C

λr

∫Rn

‖Tg(x)‖rH2dx

≤ C

λr

∫Rn

‖g(x)‖rH1dx ≤ C

λ

∫Rn

‖g(x)‖H1 dx

=C

λ

∫Rn

‖f(x)‖H1 dx .

Next, let Q∗j denote the cube with the same center cj as Qj with side length√

2n times longer.Then we estimate,∣∣x : ‖Tb(x)‖H2 > λ/2

∣∣ =∣∣x ∈ (∪jQ∗j ) : ‖Tb(x)‖H2 > λ/2

∣∣+∣∣x ∈ c(∪jQ∗j ) : ‖Tb(x)‖H2 > λ/2

∣∣≤ | ∪j Q∗j |+

∣∣x ∈ c(∪jQ∗j ) : ‖Tb(x)‖H2 > λ/2∣∣

≤ 2

λ‖f‖L1(Rn,H1) +

2

λ

∫Rn\(∪jQ∗j )

‖Tb(x)‖H2 dx .

Since ‖Tb(x)‖H2 ≤∑

j ‖Tbj(x)‖H2 a.e., it will suffice to prove that∑j

∫R\Q∗j

‖Tbj(x)‖H2 dx ≤ C‖f‖L1(Rn,H1) . (5.33)

Notice that, since∫Qjbj(y)dy = 0H1 , also∫

Qj

K(x− cj)bj(y)dy = K(x− cj)∫Qj

bj(y)dy = 0H2 ,

so that∫R\Q∗j

‖Tbj(x)‖H2 dx =

∫R\Q∗j

∥∥∥∫Qj

K(x− y)bj(y) dy∥∥∥H2

dx

=

∫R\Q∗j

∥∥∥∫Qj

(K(x− y)bj(y)−K(x− cj)bj(y)

)dy∥∥∥H2

dx

≤∫R\Q∗j

∫Qj

∥∥(K(x− y)−K(x− cj))bj(y)

∥∥H2dy dx

≤∫R\Q∗j

∫Qj

∥∥K(x− y)−K(x− cj)∥∥L(H1,H2)

‖bj(y)‖H1 dy dx

≤∫Qj

‖bj(y)‖H1

∫R\Q∗j

∥∥K(x− y)−K(x− cj)∥∥L(H1,H2)

dx dy

≤ B∫Qj

‖bj(y)‖H1 dy .

66 M. M. PELOSO

Therefore, ∑j

∫R\Q∗j

‖Tbj(x)‖H2 dx ≤ B∑j

∫Qj

‖bj(y)‖H1 dy ≤ 2B‖f‖L1(Rn,H1) ,

estimates that completes the proof. 2


6. Fourier multipliers

In this chapter we can to the analysis of the so-called Fourier multipliers, that is, of theoperators, initially defined on Schwartz functions, of the form

Tm(f)(x) = F−1(mf)(x) ,

where m ∈ S ′(Rn). We will often identify the operator Tm with the distribution m and callthem indifferently a Fourier multiplier.

Observe that m ∈ S ′ implies that m = K for some K ∈ S ′, so that

Tm(f) = K ∗ f .

Therefore, the singular integrals (of convolution type) that we have seen in the previous sections,are particular cases of Fourier multipliers.

Notice that f ∈ S, so that mf ∈ S ′ and it makes sense to define its inverse Fourier transform.We mention, without proving, the following result, due to Hormander, that shows that every

bounded linear operator between Lp-spaces that commutes with translations is given by theconvolution with a tempered distribution K; hence it is a Fourier multiplier.

Recall that, if a ∈ Rn we define [τaf ](x) = f(x+ a). An operator T is said to commute withtranslations if

τa[T (f)

](x) = T

([τaf ]

)(x) ,

for all a ∈ Rn, and a.a. x.

Theorem 6.1. Let 1 ≤ p, q ≤ ∞. Suppose that T : Lp(Rn) → Lq(Rn) is a bounded operatorthat commutes with translations. Then there exists a unique u ∈ S ′ such that

T (f) = u ∗ f

for all f ∈ S.

For the proof we refer to [StWe], Thm. 3.16.We will first show a few basic properties of these operators, and then describe some sufficient

conditions on the function m to ensure that Tm is a bounded operator on Lp.

6.1. The space Mp of Lp-bounded Fourier multipliers. We define the space

Mp =m ∈ S ′ : Tm : Lp → Lp is bounded

,

and we set

‖m‖Mp = ‖Tm‖(Lp,Lp) ≡ sup‖f‖Lp=1

‖Tm(f)‖Lp .

By Plancherel’s theorem we know that if m is bounded then Tm is bounded on L2. Theconverse of this assertion is also true, as consequence of the following result.

Lemma 6.2. If Tm is bounded on Lp then m is bounded. In particular, Tm is bounded on L2 isand only if m ∈ L∞ and, in this case, ‖Tm‖(L2,L2) = ‖m‖L∞. Finally, Mp =Mp′, if p and p′

are conjugate exponents.

68 M. M. PELOSO

Proof. If m in not bounded, then it is easy to see that Tm cannot be bounded on L2 (exercise).Now, it is easy to see that ‖Tm‖(L2,L2) = ‖m‖L∞ . For, by Plancherel theorem,

‖Tmf‖L2 = ‖mf‖L2 ≤ ‖m‖L∞‖f‖L2 = ‖m‖L∞‖f‖L2 ,

so that ‖Tm‖(L2,L2) ≤ ‖m‖L∞ .Next, if ε > 0 is fixed, let

Eε =x : |x| ≤ 1/ε and |m(x)| > ‖m‖L∞ − ε

.

Then Eε has positive and finite measure, and let f ∈ L2 be such that f = χEε . Hence,

‖Tmf‖2L2 = ‖mf‖2L2 =

∫|m(ξ)f(ξ)|2 dξ

> (‖m‖L∞ − ε)2

∫|f(ξ)|2 dξ

= (‖m‖L∞ − ε)2‖f‖2L2 .

It follows that ‖Tm‖(L2,L2) ≥ ‖m‖L∞ , and then equality holds.Next we show that if Tm is bounded on Lp, 1 ≤ p < ∞, then m is bounded. For, notice

that the adjoint operator T ∗m is given by convolution with K. Thus, T ∗m is also bounded on Lp.

Since it is bounded on Lp′, by Marcinkiewicz interpolation theorem23 is also bounded on Lr for

r between p and p′; hence in particular on L2. This implies that m ∈ L∞ and we are done. 2

Proposition 6.3. For 1 < p < q < 2 we have the inclusion

M1 ⊂Mp ⊂Mq ⊂M2 (6.1)

and moreover,

‖m‖L∞ ≤ ‖m‖Mq ≤ ‖m‖Mp .

Proof. This follows easily from the previous lemma and the Marcinkiewicz theorem, if m ∈M1,then m is also in M2 and therefore in Mr for all 1 < r < 2. The same argument proves theother inclusions in (6.1).

Using again the Riesz–Thorin theorem, for 1 < p < 2 we have that

‖Tm‖(L2,L2) ≤ ‖Tm‖1/2(Lp,Lp)‖Tm‖

1/2

(Lp′ ,Lp′ )= ‖Tm‖(Lp,Lp) ,

since1

2=

1/2

p+

1/2

p′.

A similar argument shows that, for 1 < p < q < 2, ‖Tm‖(Lq ,Lq) ≤ ‖Tm‖(Lp,Lp) and we are done.2

Proposition 6.4. For 1 ≤ p < ∞ the space Mp is a Banach space with respect to the norm‖ · ‖Mp. Moreover, Mp is closed under pointwise multiplication and it is a Banach algebra.24

23Or, more simply by the Riesz-Thorin Theorem, [Ka] or [So].24An algebra A which is a Banach space w.r.t. the norm ‖·‖A is called a Banach algebra if the product satisfies

the inequality ‖xy‖A ≤ ‖x‖A‖y‖A for all x, y ∈ A.

Proof. It suffices to consider the case 1 ≤ p ≤ 2.It is obvious that Mp is a linear space and that the equality ‖m‖Mp = ‖Tm‖(Lp,Lp) defines a

norm on Mp.Next, if m1,m2 ∈Mp, then

(Tm1m2f ) (ξ) = m1(ξ)m2(ξ)f(ξ) =(Tm1(Tm2)f

)(ξ) ,

so that Tm1m2 = Tm1Tm2 . Moreover,

‖Tm1m2f‖Lp ≤ ‖Tm1‖(Lp,Lp)‖Tm2f‖Lp≤ ‖Tm1‖(Lp,Lp)‖Tm2‖(Lp,Lp)‖f‖Lp ,

that is,

‖m1m2‖Mp = ‖Tm1m2‖(Lp,Lp) ≤ ‖Tm1‖(Lp,Lp)‖Tm2‖(Lp,Lp) = ‖m1‖Mp‖m2‖Mp .

This proves that Mp is a Banach algebra w.r.t. the pointwise product.

We now show that Mp is complete. Let mj be a Cauchy sequence in Mp. By Prop. 6.3 itfollows that mj is a Cauchy sequence in L∞, so it converges, in the L∞-norm, hence a.e., toa function m. We wish to show that mj → m in Mp.

Let f ∈ S be fixed. Then, by Lebesgue’s dominated convergence theorem we have that

Tmj (f)(x) =

∫Rn

f(ξ)mj(ξ)e2πixξ dξ →

∫Rn

f(ξ)m(ξ)e2πixξ dξ = Tm(f)(x)

a.e. By Fatou’s lemma it follows that∫Rn

|Tm(f)(x)|p dx ≤ lim infj→+∞

∫Rn

|Tmj (f)(x)|p dx

≤ lim infj→+∞

‖mj‖Mp‖f‖Lp ,

that is,

‖m‖Mp ≤ lim infj→+∞

‖mj‖Mp ≤ Cp

since mj is a Cauchy sequence, hence the sequence of the norms is bounded.From the last inequality it also follows that

‖m−mk‖Mp ≤ lim infj→+∞

‖mj −mk‖Mp < ε

if k ≥ k0, since mj is a Cauchy sequence. This proves the proposition. 2

Examples 6.5.(1) The first example we consider is m(ξ) = −i sgn(ξ). Then Tm is the Hilbert transform,

and we know that Tm is weak-type (1, 1) and bounded on Lp for 1 < p <∞.

(2) Consider the interval (a, b), and assume for the moment that it is bounded. Notice that

χ(a,b)(ξ) =1

2

[sgn(ξ − a)− sgn(ξ − b)

].

Let a ∈ R and observe that the operator (called the modulation operator)

Maf(x) = e2πiaxf(x)

70 M. M. PELOSO

is bounded on Lp for all p, and also that (Maf ) (ξ) = f(ξ− a) = (τaf)(ξ). Then notice that, forf ∈ S(R),

χ(a,b)(ξ)f(ξ) =1

2

[sgn(ξ − a)− sgn(ξ − b)

]f(ξ)

=1

2

[τa(

sgn(ξ)τ−a(f))(ξ)− τb

(sgn(ξ)τ−b(f)

)(ξ)]

=i

2F([MaHM−a −MbHM−b

](f))

(ξ) , (6.2)

where H denotes the Hilbert transform. (This last equality can be easily be checked by com-

puting the Fourier transform on the right hand side, recalling that (Hf ) (ξ) = −i sgn(ξ)f(ξ).)Therefore,

F−1(χ(a,b)

)∗ f =

i

2

[MaHM−a −MbHM−b

](f) ,

so that

Tχ(a,b)(f) =

i

2

[MaHM−a −MbHM−b

](f) . (6.3)

Next, suppose that the interval is unbounded. Let (a, b) = (a,+∞), the other case beginanalogous. Then

χ(a,+∞)(ξ) =1

2

[1 + sgn(ξ − a)

],

so that, arguing as in (6.2),

χ(a,+∞)(ξ)f(ξ) =i

2F([− iI +MaHM−a

](f))

(ξ) ,

i.e.

Tχ(a,+∞)(f) =

i

2

[MaHM−a − iI

](f) . (6.4)

Therefore, it follows from (6.3) and (6.4) that, for all a, b ∈ R ∪ ±∞, m = χ(a,b) is inMp(R) for 1 1. The Hilbert transform can be extended to define a family of Lp-bounded Fourier

multipliers in Rn. For j = 1, . . . , n define mj(ξ) = −i sgn(ξj) and hence Tmj (f) = F−1(mj f).

It is easy to see that

Tmjf(x) =1

πp.v.

∫ +∞

−∞

1

tf(x1, . . . , xj − t, . . . , xn) dt .

Now, taking j = n for simplicity, using the boundedness of the Hilbert transfom in the lastvariable we have

‖Tmnf‖pLp(Rn) =

∫Rn−1

∫R|Tmnf(x1, . . . , xn−1, xn)|p dxn dx1 · · · dxn−1

≤ C∫Rn−1

∫R|f(x1, . . . , xn−1, xn)|p dxn dx1 · · · dxn−1

= C‖f‖pLp(Rn) .

It also follows that, if a ∈ Rn, b ∈ R and

E =ξ ∈ Rn : a · ξ > b

,

then m = χE ∈Mp, 1 < p <∞. (Exercise.)

(4) Let m = χ|ξ|<1 be the characteristic function of the unit ball in Rn. If n = 1 it followsfrom the previous discussion that m ∈ Mp for all p, 1 1 that χ|ξ|<1 ∈Mp if and only if p = 2.

We conclude this section by stating the characterization ofM1, a significant result, for whoseproof we refer to [StWe] or [Gr]. We recall that a finite complex-valued Borel measure µ definesa tempered distribution by setting µ(f) =

∫f(x)dµ(x).

Theorem 6.6. A measurable function m ∈ M1 if and only if m is the Fourier transform of afinite complex-valued Borel measure µ.

6.2. Multiple Fourier series and convergence in the Lp-norm. In this section we brieflyillustrate an application of the theory of Fourier multipliers.

We define the n-dimensional torus to be the quotient group Tn = Rn/Zn. It is easy toconvince oneself that Tn can be identified with the compact set [0, 2π]× · · · × [0, 2π] ≡ [0, 2π]n.

Given an integrable function f on Tn we define its (multiple) Fourier series to be the expression∑k∈Zn

f(k)e2πik·t , (6.5)

where t ∈ Tn and

f(k) =

∫Tnf(t)e−2πik·t dt .

The main question is in what sense the expression in (6.5) represents f , that is, if it convergesand if it does converge, whether it converges to f .

By elementary theory of Hilbert spaces applied to the case of H = L2(Tn) it follows at oncethat, if f ∈ L2(Tn) then the series in (6.5) converges (unconditionally) to f in the L2(Tn)-norm.

If p 6= 2 and we are still interested in the norm convergence of the series in (6.5), when n > 1we need to specify what we mean by the summation, since Zn lacks of a natural order. Therefore,we proceed as follows.

Definition 6.7. We call convex body an open convex set C ⊂ Rn containing the origin. Forλ > 0 we denote by λC the dilated of C by λ, that is the set y = λx : x ∈ C, λ > 0, and define

SλC (f) =∑

k∈Zn∩λCf(k)e2πik·t . (6.6)

We say that the Fourier series converge in Lp(Tn), 1 ≤ p < ∞ in the summation methodgiven by the convex body C, or more briefly, in the C-sense, if

limλ→+∞

∥∥∥SλC (f)− f∥∥∥Lp(Tn)

= 0 ,

for every f ∈ Lp(Tn) ∩ C(Tn).

Our goal now is to relate the norm convergence of Fourier series to the boundedness of asingle operator, as we have done in the case of the Fourier series on the torus. Namely, we have

Theorem 6.8. Let 1 < p < +∞ and let a convex body C in Rn be given. Then the followingconditions are equivalent.

72 M. M. PELOSO

(i) The space Lp(Tn), 1 ≤ p < ∞, admits norm convergence for the Fourier series in thesummation method given by the convex body C.

(ii) There exists C > 0 such that

supλ>0‖SλC‖(Lp(Tn),Lp(Tn)) ≤ C .

(where ‖SλC‖(Lp(Tn),Lp(Tn)) denotes the norm of SλC as an operatort on Lp(Tn)).(iii) The function χC ∈Mp(R

n).

We introduce the notion of Fourier multipliers on Tn.

Definition 6.9. Let ω : Zn → C be a complex-valued function on Zn. We say that ω isa bounded Fourier multiplier on Lp(Tn) if the operator Tω initially defined on trigonometricpolynomials as

Tω(f)(t) =∑k∈Zn

ω(k)f(k)e2πik·t

extends to a bounded operator Tω : Lp(Tn)→ Lp(Tn).

The following result is not difficult to prove, but for lack of adequate time we refer to [So],Ch. IV for its proof.

Theorem 6.10. The following implications hold.

(1) If m ∈Mp(Rn) and m is continuous on Zn, then m := m|Zn ∈Mp(T

n), and ‖m‖Mp(Tn) ≤‖m‖Mp(Rn).

(2) If m is a function defined a.e. in Rn and everywhere on Zn and it is such that mε ∈Mp(T

n) for every ε > 0 and

supε>0‖Tmε‖Mp(Tn) ≤ C ,

for some C > 0, then m ∈Mp(Rn) and ‖m ∈Mp(R

n)‖Mp(Rn) ≤ C.

We recall that mε(x) = m(εx). We are now ready to prove Thm. 6.8.

Proof of Thm. 6.8. We first observe that same proof of Thm. 1.13 shows that (i) is equivalentto (ii).

Next, we notice that since C is convex and 0 ∈ C, λ1 < λ2 implies that λ1C ( λ2C. Inparticular, the summation method given by C and any of its dilated coincides. Moreover, ∂(C) ⊆(1 + ε)C \ (1− ε)C so that

|∂(λC)| ≤(

(1 + ε)n − (1− ε)n)|C| ≤ εC|C|

for every ε > 0. Thus, |∂(C)| = 0. Thus, χλC is a.e. continuous in Rn, and clearly definedeverywhere on Zn Moreover, for k ∈ Zn there exists just one value λ > 0 such that k ∈ ∂(λC);call such value λk. Thus, we may select an increasing sequence λ1 < λ2 < · · · , λm → +∞, suchthat no point of Zn lies on ∂(λmC) for every m, in such a way that so that χλmC is continuouson Zn, for every m, and moreover, for each λ > 0, there exists m such that

SλC = SλmC .

Now we observe that, χλC(k) = χ1/λC (k) so that for any trigonometric polynomial f ,

SλCf(t) =∑

k∈Zn∩λCf(k)e2πik·t =

∑k∈Zn

χλC(k)f(k)e2πik·t

=∑k∈Zn

χ1/λC (k)f(k)e2πik·t

= Tχ1/λCf(t) .

Suppose then that (ii) holds. This gives that Tχ1/λC

is a bounded operator on Lp(Tn), with

uniformly bounded norms, that is, χ1/λC ∈Mp(T

n) and

‖χ1/λC ‖Mp(Tn) ≤ C .

Thm. 6.10 (2) gives that χC ∈Mp(Rn).

Suppose that (i) holds. Having selected the increasing sequence λm as above, χλmC iscontinuous on Zn, then χλmC|Zn ∈Mp(T

n), by Thm. 6.10 (1). Therefore,

supλ>0‖SλC‖(Lp(Tn),Lp(Tn)) = sup

m∈N‖SλmC ‖(Lp(Tn),Lp(Tn)) = sup

m∈N‖T

χ1/λmC‖(Lp(Tn),Lp(Tn))

= supm∈N

‖χ1/λmC |Zn

‖Mp(Tn)

≤ supm∈N

‖χ1/λmC ‖Mp(Rn)

= ‖χC‖Mp(Rn) .

In order to prove the last equality, it suffices to notice the norm ‖ · ‖Mp(Rn) is dilation invariant,that is,

‖mε‖Mp(Rn) = ‖m‖Mp(Rn) .

for every ε > 0. We leave the elementary verification to the reader. This completes the proof.

A subset S of Rn is called a half-space if there exist a ∈ Rn and b ∈ R such that

S =x ∈ Rn : a · x > b

.

A bounded subset P ⊂ Rn is called a convex polyhedron if it is convex, contains the orgin,and it is the intersection of a finite number of half-spaces.

The most typical examples of convex bodies are the unit ball B, which has smooth boundaryand nowhere vanishing curvature, and a convex polyhedron, that instead has only Lipschitzboundary a.e. flat.

Theorem 6.11. Let P be a convex polyhedron. Then χP ∈ Mp, 1 1 we easily reduce ourselves to the boundedness of the Riesz transforms. For, if

P ∩Nj=1 Sj , where Sj are half-spaces, then χP =∏Nj=1 χSj . Hence,

TχP = T∏Nj=1 χSj

= TχS1 · · ·TχSN .

74 M. M. PELOSO

Thus, it suffices to show that each TχSj is bounded on Lp(Rn), 1 0

,

and this follows from Example 6.5 (3). 2

The next result came as a big surprise. In the early ’70 C. Fefferman proved that the charac-teristic function of the unit ball in Rn with n > 1 is not a bounded Fourier multiplier in Lp(Rn),unless p = 2.

Theorem 6.12. Let m = χB be the characteristic function of the unit ball B in Rn, wheren > 1. Then m ∈Mp if and only if p = 2.

The proof of this theorem is quite elaborated and for the time being we refer to [St2] Ch. XSection 2.5, or [So] Ch. 4, Section 3. (The latter proof is self-contained and it is essentially theoriginal proof by C. Fefferman, while the latter one shows the deep connection of this problemwith other important concepts and open problems in modern harmonic analysis.)

6.3. The Sobolev spaces Hs. We have remarked that the characteristic function of the unitball in Rn, with n > 1, defines a Fourier multiplier that is bounded on Lp only for p = 2. It isclear that, besides the boundedness, the function m needs to posses some regularity. In orderto measure regularity we introduce a family of spaces, called the Sobolev spaces, whose elementshave derivatives (in some weak sense) in L2.

We begin our brief introduction with the case of a non-negative integer k. We define Hk =Hk(Rn) to be the space of L2 functions f such that the distributional derivatives ∂αf ∈ L2 forall α with |α| ≤ k. On Hk we can define an inner product that makes Hk into a Hilbert space:

〈f |g〉k =∑|α|≤k

∫(∂αf)(∂αg) .25

We wish to extend this definition to non-integral values of k and also to have a more “efficient”way to represent the inner product. We observe that, by Plancherel theorem, f ∈ Hk if andonly if ξαf ∈ L2 for all α, |α| ≤ k. We claim that there exist positive constants c1, c2 such that,for all ξ ∈ Rn,

c1

(1 + |ξ|2)k/2 ≤

∑|α|≤k

|ξα| ≤ c2

(1 + |ξ|2)k/2 .

For, if α is a multi-index with |α| ≤ k, if |ξ| ≥ 1 then

|ξα| ≤ |ξ|k ≤ (1 + |ξ|2)k/2 ,

while, if |ξ| ≤ 1,

|ξα| ≤ 1 ≤ (1 + |ξ|2)k/2 .

On the other hand, since |ξ|k and∑k

j=1 |ξj |k are both homogeneous of degree k non-vanishingfor ξ 6= 0, we have

(1 + |ξ|2)k/2 ≤ c0(1 + |ξ|k) ≤ c0

(1 + c′0

k∑j=1

|ξj |k)≤ C

∑|α|≤k

|ξα| .

25The fact that this inner product defines a Hilbert space, hence in particular complete, is left as an Exercise.

By the claim, it follows that f ∈ Hk if and only if (1 + |ξ|2)k/2f ∈ L2. Moreover,

c1

∫ (1 + |ξ|2)k|f(ξ)|2 dξ ≤

∑|α|≤k

∫|ξα|2|f(ξ)|2 dξ

≤ c∑|α|≤k

∫|∂αf(x)|2 dx

≤ c′2∫ (

1 + |ξ|2)k|f(ξ)|2 dξ ,

so that the two norms(∑|α|≤k

∫|∂αf(x)|2 dx

)1/2

and

(∫ (1 + |ξ|2)k|f(ξ)|2 dξ

)1/2

are equivalent, and the latter one is defined for all k real, not just non-negative integer.

Definition 6.13. We define the Bessel potential of order s ∈ R of a distribution f ∈ S ′ as

Λsf = F−1

((1 + |ξ|2

)s/2f

).

Notice that Λsu is well defined for any u ∈ S ′, since, for any Schwartz function ψ, if ϕ = F−1ψ,

(Λsu)(ψ) = (Λsu)(ϕ) =((1 + |ξ|2)s/2u

)(ϕ)

=

∫(1 + |ξ|2)s/2u(ξ)ϕ(ξ) dξ = u

((1 + |ξ|2)s/2ϕ

)and (1 + |ξ|2)s/2ϕ(ξ) is again a Schwartz function.

Definition 6.14. Let s ∈ R. We define the Sobolev space Hs of order s ∈ R as the space ofdistributions f ∈ S ′ such that Λsf ∈ L2, that is

Hs =f ∈ S ′ :

∫|f(ξ)|2(1 + |ξ|2)s dξ <∞

.

We endow Hs with the inner product

〈f |g〉s =

∫(Λsf)(Λsg) ,

which also gives the norm

‖f‖Hs =

(∫|f(ξ)|2(1 + |ξ|2)s dξ

)1/2

.

Proposition 6.15. The following properties are elementary.

(i) The Fourier transform F is a unitary isomorphism from Hs to L2((1 + |ξ|2)sdξ

).

(ii) The Schwartz space S is dense in Hs for all s ∈ R.(iii) If s > t then ‖ · ‖Ht ≤ ‖ · ‖Hs so that Hs is a subspace of Ht and it is also a dense

subspace.(iv) The Bessel potential Λs is a unitary isomorphism from Ht to Ht−s for all s, t ∈ R. Its

inverse is Λ−s.

76 M. M. PELOSO

(v) H0 = L2, so that Hs ⊂ L2 for all s > 0. For s < 0 the elements of Hs may not befunctions.

(vi) The operators ∂α are continuous from Hs to Hs−|α|, for all s and α.

Proposition 6.16. For s ∈ R, the L2 inner product 〈·|·〉 induces a unitary isomorphism betweenH−s and the dual space (Hs)∗ of Hs.

Proof. Let f, ϕ ∈ S. Then, by Plancherel’s theorem,∣∣〈ϕ|f〉∣∣ =∣∣〈ϕ|f〉∣∣ ≤ ∫ (1 + |ξ|2)−s/2|f(ξ)|(1 + |ξ|2)s/2|ϕ(ξ)| dξ

≤(∫

(1 + |ξ|2)−s|f(ξ)|2 dξ)1/2(∫

(1 + |ξ|2)s|ϕ(ξ)|2 dξ)1/2

= ‖f‖H−s‖ϕ‖Hs .

Therefore, the linear functional ϕ 7→ 〈ϕ|f〉 extends to all of Hs with norm ≤ ‖f‖H−s . But,we actually have equality of norms, since if g = (Λ−2sf),

〈f |g〉 = 〈f |g〉 =

∫(1 + |ξ|2)−s|f(ξ)|2 dξ = ‖f‖2H−s

= ‖f‖H−s‖g‖Hs .

Notice that the pairing 〈ϕ|f〉 is well defined for all f ∈ H−s and ϕ ∈ Hs.Finally, let L ∈

(Hs)∗

. Then L F−1 is a bounded linear functional on L2((1 + |ξ|2)sdξ

), so

that there exists g ∈ L2((1 + |ξ|2)sdξ

)such that

L(ϕ) = L F−1(ϕ) =

∫ϕ(ξ)g(ξ)(1 + |ξ|2)s dξ

=

∫ϕ(ξ)FΛ2s

((F−1g)(ξ)

)dξ

= 〈ϕ|Λ2s(F−1g)〉= 〈ϕ|f〉 .

But, f = Λ2s(F−1g) ∈ H−s (since F−1g ∈ Hs). This proves the proposition. 2

We conclude this part with the Sobolev immersion theorem. We set26

Ck0 = f ∈ Ck(Rn) : ∂αf ∈ C0, for |α| ≤ k .

Lemma 6.17. Let f be such that f ∈ Hs, where s > n/2. Then f ∈ L1 and for all ε < s− n/2we have ∫

Rn

|f(x)|(1 + |x|)ε dx ≤ Cε‖f‖Hs .

26We recall that we denote by C0 the space of continuous functions that vanishes at infinity.

Proof. Using the Cauchy-Schwarz inequality we have∫|f(x)|(1 + |x|)ε dx ≤ 2ε

∫|f(x)|(1 + |x|2)ε/2 dx

≤ 2ε(∫

|f(x)|2(1 + |x|2)s dx

)1/2(∫(1 + |x|2)ε−s dx

)1/2

(6.7)

≤ Cε‖f‖Hs ,

where Cε = 2ε( ∫

(1 + |x|2)ε−sdx)1/2

is finite since ε < s− n/2. 2

Theorem 6.18. (Sobolev Embedding Theorem) Let t > k + (n/2). Then, Ht embedds continu-ously in Ck0 .

Proof. Let g ∈ Ht and |α| ≤ k. By the previous lemma, F−1(∂αg

)∈ L1, since ∂αg ∈ Hs, where

s = t− |α| > n/2, and∫Rn

|F−1(∂αx g

)(x)|(1 + |x|)ε dx ≤ Cε‖∂αg‖Hs ≤ Cε‖g‖Ht .

By the Riemann-Lebesgue lemma, ∂αx g ∈ C0, for |α| ≤ k. 2

For sake of clarity, we isolate the result that we need in the remaining of this part.

Corollary 6.19. Let s > n/2. Then, if f ∈ Hs it follows that f is continuous and

‖f‖L∞ ≤ C‖f‖Hs .

We conclude this part with a lemma that we are going to need later on.

Lemma 6.20. Let f ∈ Hs and ϕ ∈ S. Then fϕ ∈ Hs and

‖fϕ‖Hs ≤ Cϕ‖f‖Hs .

Proof. We begin with the preliminary observation that

1 + |ξ + ξ′|2 ≤ 1 + 2(|ξ|2 + |ξ′|2) ≤ 2(1 + |ξ|2)(1 + |ξ′|2) . (6.8)

Next, since (fϕ) = f ∗ ϕ, using the Cauchy-Schwarz intequality, we have∫Rn

|(fϕ)(ξ)|2(1 + |ξ|2)s dξ =

∫Rn

∣∣∣ ∫Rn

f(ξ − ξ′)ϕ(ξ′) dξ′∣∣∣2(1 + |ξ|2)s dξ

≤ ‖ϕ‖L1

∫Rn

∫Rn

|f(ξ − ξ′)|2|ϕ(ξ′)| dξ′ (1 + |ξ|2)s dξ

≤ ‖ϕ‖L1

∫Rn

|ϕ(ξ′)|∫Rn

|f(ξ)|2 (1 + |ξ + ξ′|2)s dξ dξ′

≤ 2s‖ϕ‖L1

∫Rn

|ϕ(ξ′)|(1 + |ξ′|2)s∫Rn

|f(ξ)|2 (1 + |ξ|2)s dξ dξ′

≤ Cϕ‖f‖2Hs ,

where we have used the fact that ϕ ∈ S. This proves the lemma. 2

78 M. M. PELOSO

6.4. †The Mihlin–Hormander multiplier theorem. In this section we prove the Mihlin–Hormander multiplier theorem. The Mihlin–Hormander condition that guarantees that a func-tion gives rise to a Fourier multiplier that is bounded on Lp(Rn) has the following features:

(i) it is invariant under the dilations m 7→ m(r·) for r > 0, in the sense that m satisfies thiscondition then also m(r·) does;

(ii) the operator Tm : f 7→ (F−1m) ∗ f is an operator whose kernel K = (F−1m) is aCalderon–Zygmund convolution kernel, hence it is of weak-type (1, 1) and thereforebounded on Lp(Rn), 1 < p < ∞;– in order to show this we are going to appeal toThm. 5.26.

We are now ready to define the space of Mihlin-Hormander multipliers.

Definition 6.21. Let 0 < a0 < a 0‖m(r·)ψ‖Hs , (6.9)

for s > n/2. We denote the space of such multipliers by MH s.

Remark 6.22.(1) Although we are not going to prove this assertion, and the one below, it is important to

notice that the above definition is independent on the choice of ψ and that, a different ψ justgives rise to an equivalent norm.

(2) The norm ‖ · ‖MH s is invariant under dilation, in the sense that if m ∈ MH s and mr isgiven by mr(ξ) = m(rξ), where r > 0, then mr ∈ MH s and ‖mr‖MH s = ‖m‖MH s .27

Lemma 6.23. For s > n/2, if m ∈ MH s, then m is bounded and

‖m‖L∞ ≤ C‖m‖MH s .

Proof. By assumption, the functions m(r·)ψ are in Hs with s > n/2 and norms uniformlybounded. By Cor. 6.19, it follows that ‖m(r·)ψ‖L∞ = ‖mψ(r−1·)‖L∞ ≤ C‖m‖MH s , with Cindependent of r > 0.

Hence,

supξ|m(ξ)|2 = sup

jsup

2−j≤|ξ|≤2−j+1

|m(ξ)|2

= supj

sup2−j≤|ξ|≤2−j+1

∑k

|m(ξ)ψ(2−kξ)|2

= supj

sup2−j≤|ξ|≤2−j+1

∑k∈j−1,...,j+2

|m(ξ)ψ(2−kξ)|2

≤ 4 supk

supξ∈Rn

|m(ξ)ψ(2−kξ)|2

≤ C‖m‖MH s .


27Exercise: Prove (1) and (2).

Theorem 6.24. (Mihlin–Hormander) Let m ∈ MH s, with s > n/2. Then, the multiplieroperator Tm is bounded on Lp, 1 bn/2c+ 1, be such that

supr>0

r|α|(r−n

∫r/2<|ξ|<2r

|∂αm(ξ)|2 dξ)1/2

<∞ , (6.10)

for all |α| ≤ k. Then, m ∈Mp, for 1 0

r|α|(r−n

∫r/2<|ξ|<2r

|∂αm(ξ)|2 dξ)1/2

= supr>0

(∫1/2<|ζ|<2

r2|α||(∂αm)(rζ)|2 dζ)1/2

= supr>0

(∫1/2<|ζ|<2

|(∂αm(r·)

)(ζ)|2 dζ

)1/2

.

This quantity is, by the assumption (6.10), finite. We prove that (6.9) is also satisfied. For,if |α| ≤ k, by the Leibnitz rule, using the facts that suppψ ⊆ 1/2 ≤ |ξ| ≤ 2 and that|∂βψ(ξ)| ≤ C, we see that∥∥∂α(m(r·)ψ

)∥∥L2 =

∥∥∑β≤α

cα,β∂βm(r·)∂α−βψ

∥∥L2

≤ C∑β≤α

(∫Rn

∣∣(∂βξm(r·))(ξ)∂α−βψ(ξ)

∣∣2 dξ)1/2

≤ C∑|β|≤k

(∫1/2<|ξ|<2

|(∂βm(r·)

)(ξ)|2 dξ

)1/2

,

where the constant C is independent of r > 0.Therefore, m(r·)ψ ∈ Hk with norm uniformly bounded in r > 0, that is, m ∈ MH k, with

k > bn/2c+ 1. Applying Thm. 6.24 we obtain that m ∈Mp, for 1 < p <∞.

Finally, if m satisfies (6.11), then

r|α|(r−n

∫r/2<|ξ|<2r

|∂αm(ξ)|2 dξ)1/2

≤ Cr|α|(r−n

∫r/2<|ξ|<2r

|ξ|−2|α| dξ

)1/2

≤ C ,

that is, (6.10) is also satisfied, and the corollary is proven. 2

80 M. M. PELOSO

6.5. †Proof of Thm. 6.24. Let m ∈ MH s with s > n/2 and let ψ be as in Lemma 5.23 (i),that is, ψ ∈ C∞0 and such that

∑j∈Z ψ(2jξ) = 1 for all ξ 6= 0.

We set mj(ξ) = m(2−jξ)ψ(ξ), so that∑j∈Z

(mj)2j (ξ) =

∑j∈Z

mj(2jξ) = m(ξ)

for all ξ 6= 0.Morevore, we define

K = F−1m, and k(j) = F−1mj . (6.12)

Notice that then we have

FK = m =∑j∈Z

(mj)2j =

∑j∈Z

(Fk(j)

)2j=∑j∈ZF((k(j)

)2j

)= F

(∑j∈Z

(k(j)

)2j

),

that is,

K =∑j∈Z

(k(j)

)2j,

where the convergence is in S ′.We wish to prove that the k(j)’s satisfy the conditions (1)-(3) in Thm. 5.26.

We begin with (1). Using Lemma 6.17 we have∫(1 + |x|)ε|k(j)(x)| dx =

∫(1 + |x|)ε|F−1mj(x)| dx

≤ Cε‖mj‖Hs = Cε‖m(2−j ·)ψ‖Hs

≤ Cε‖m‖MH s .

Next, ∫k(j)(x) dx = Fk(j)(0) = mj(0) = 0 ,

since ψ(0) = 0; fact that proves (2).

Finally we prove (3). It suffices to show that28

sup0<|h|<1

|h|−α∫|k(j)(x− h)− k(j)(x)| dx <∞ .

28See footnote 20.

Since s > n/2, using the Cauchy-Schwarz inequality we have that for any α ∈ (0, 1) and0 < |h| < 1

|h|−α∫Rn

|k(j)(x− h)− k(j)(x)| dx

≤ |h|−α(∫

Rn

|k(j)(x− h)− k(j)(x)|2(1 + |x|2)s dx)1/2

= |h|−α(∫

Rn

∣∣∣ ∫ 1

0h · ∇k(j)(x− th) dt

∣∣∣2(1 + |x|2)s dx)1/2

≤ |h|1−α(∫

Rn

∫ 1

0

∣∣∇k(j)(x− th)∣∣2 dt(1 + |x|2)s dx

)1/2

≤ C(∫

Rn

∣∣∇k(j)(x)∣∣2 ∫ 1

0(1 + |x+ th|2)s dt dx

)1/2

≤ C(∫

Rn

∣∣∇k(j)(x)∣∣2(1 + |x|2)s dx

)1/2

≤ Cn∑i=1

‖F−1(∂ξik(j))‖Hs ,

where we have used an estimate as in (6.8).Thus, we only need to prove that

n∑i=1

‖F−1(∂ξik(j))‖Hs ≤ C‖m‖MH s . (6.13)

Notice that

F−1(∂ξik(j)) = 2πiξiF(k(j))(ξ) = 2πiξimj(ξ) = 2πiξim(2jξ)ψ(ξ)

=(2πiξiη(ξ)

)m(2jξ)ψ(ξ) ,

where η ∈ C∞0 and it is identically 1 on the support of ψ. Applying Lemma 6.20 we have thenn∑i=1

‖F−1(∂ξik(j))‖Hs ≤ Cn∑i=1

‖(ξiη)m(2j ·)ψ‖Hs

≤ C‖m(2j ·)ψ‖Hs

≤ C‖m‖MH s ,

thus proving (10.4) and therefore the theorem. 2

82 M. M. PELOSO

7. Partial Differential Operators

In this section we apply the Fourier analysis techniques developed so far, to the study ofpartial differential operators and equations. We begin by introducing a scale of spaces, calledSobolev spaces and denoted by Hs, that measure smoothness of functions, and that are suitablydefined using the Fourier transform.

7.1. The Sobolev spaces Hs. Let k be a non-negative integer. We define Hk = Hk(Rn) tobe the space of L2 functions f such that the distributional derivatives ∂αf ∈ L2 for all α with|α| ≤ k. On Hk we can define an inner product that makes Hk into a Hilbert space:

〈f |g〉k =∑|α|≤k

∫(∂αf)(∂αg) .29

We wish to extend this definition to non-integral values of k and also to have a more “efficient”way to represent the inner product. We observe that, by Plancherel theorem, f ∈ Hk if andonly if ξαf ∈ L2 for all α, |α| ≤ k. We claim that there exist positive constants c1, c2 such that,for all ξ ∈ Rn,

c1

(1 + |ξ|2)k/2 ≤

∑|α|≤k

|ξα| ≤ c2

(1 + |ξ|2)k/2 .

For, if α is a multi-index with |α| ≤ k, if |ξ| ≥ 1 then

|ξα| ≤ |ξ|k ≤ (1 + |ξ|2)k/2 ,

while, if |ξ| ≤ 1,

|ξα| ≤ 1 ≤ (1 + |ξ|2)k/2 .

On the other hand, since |ξ|k and∑k

j=1 |ξj |k are both homogeneous of degree k non-vanishingfor ξ 6= 0, we have

(1 + |ξ|2)k/2 ≤ c0(1 + |ξ|k) ≤ c0

(1 + c′0

k∑j=1

|ξj |k)≤ C

∑|α|≤k

|ξα| .

By the claim, it follows that f ∈ Hk if and only if (1 + |ξ|2)k/2f ∈ L2. Moreover,

c1

∫ (1 + |ξ|2)k|f(ξ)|2 dξ ≤

∑|α|≤k

∫|ξα|2|f(ξ)|2 dξ

≤ c∑|α|≤k

∫|∂αf(x)|2 dx

≤ c′2∫ (

1 + |ξ|2)k|f(ξ)|2 dξ ,

so that the two norms(∑|α|≤k

∫|∂αf(x)|2 dx

)1/2

and

(∫ (1 + |ξ|2)k|f(ξ)|2 dξ

)1/2

29We will keep the notation 〈·, ·〉 to indicate the bilinear pairing between a test function and a distribution,while 〈·|·〉 will denote a hermitian inner product.

are equivalent, and the latter one is defined for all k real, not just non-negative integer.

Definition 7.1. We define the Bessel potential of order s ∈ R of a distribution f ∈ S ′ as

Λsf = F−1

((1 + |ξ|2

)s/2f

).

Notice that Λsf is well defined for any f ∈ S ′, since, for any Schwartz function ψ, if ϕ = F−1ψ

〈Λsf, ψ〉 = 〈Λsf, ϕ〉 = 〈(1 + |ξ|2)s/2f , ϕ〉

=

∫(1 + |ξ|2)s/2f(ξ)ϕ(ξ) dξ = 〈f , (1 + |ξ|2)s/2ϕ〉

and (1 + |ξ|2)s/2ϕ(ξ) is again a Schwartz function.

Definition 7.2. We define the Sobolev space Hs of order s ∈ R as the space of distributionsf ∈ S ′ such that Λsf ∈ L2, that is

Hs =f ∈ S ′ :

∫|f(ξ)|2(1 + |ξ|2)s dξ <∞

.

we endow Hs with the inner product

〈f |g〉s =

∫(Λsf)(Λsg)

which also gives the norm

‖f‖Hs =

(∫|f(ξ)|2(1 + |ξ|2)s dξ

)1/2

.

Proposition 7.3. The following properties are elementary.

(i) The Fourier transform F is a unitary isomorphism from Hs to L2((1 + |ξ|2)sdξ

).

(ii) The Schwartz space S is dense in Hs for all s ∈ R.(iii) If s > t then ‖ · ‖Ht ≤ ‖ · ‖Hs so that Hs is a subspace of Ht and it is also a dense

subspace.(iv) The Bessel potential Λs is a unitary isomorphism from Ht to Ht−s for all s, t ∈ R. Its

inverse is Λ−s.(v) H0 = L2, so that Hs ⊂ L2 for all s > 0. For s < 0 the elements of Hs may not be

functions.(vi) The operators ∂α are continuous from Hs to Hs−|α|, for all s and α.

Proposition 7.4. For s ∈ R, the L2 inner product 〈·|·〉 induces a unitary isomorphism betweenH−s and the dual space (Hs)∗ of Hs.

Proof. Let f, ϕ ∈ S. Then, by Plancherel,∣∣〈ϕ|f〉∣∣ =∣∣〈ϕ|f〉∣∣ ≤ ∫ (1 + |ξ|2)−s/2|f(ξ)|(1 + |ξ|2)s/2|ϕ(ξ)| dξ

≤(∫

(1 + |ξ|2)−s|f(ξ)|2 dξ)1/2(∫

(1 + |ξ|2)s|ϕ(ξ)|2 dξ)1/2

= ‖f‖H−s‖ϕ‖Hs .

84 M. M. PELOSO

Therefore, the linear functional ϕ 7→ 〈ϕ|f〉 extends to all of Hs with norm ≤ ‖f‖H−s . But,we actually have equality of norms, since if g = (Λ−2sf),

〈f |g〉 = 〈f |g〉 =

∫(1 + |ξ|2)−s|f(ξ)|2 dξ = ‖f‖2H−s

= ‖f‖H−s‖g‖Hs .

Notice that the pairing 〈ϕ|f〉 is well defined for all f ∈ H−s and ϕ ∈ Hs.Finally, let L ∈

(Hs)∗

. Then L F−1 is a bounded linear functional on L2((1 + |ξ|2)sdξ

), so

that there exists g ∈ L2((1 + |ξ|2)sdξ

)such that

L(ϕ) = L F−1(ϕ) =

∫ϕ(ξ)g(ξ)(1 + |ξ|2)s dξ

=

∫ϕ(ξ)FΛ2s

((F−1g)(ξ)

)dξ

= 〈ϕ|Λ2s(F−1g)〉= 〈ϕ|f〉 .

But, f = Λ2s(F−1g) ∈ H−s (since F−1g ∈ Hs). This proves the proposition. 2

We conclude this part with the Sobolev immersion theorems. We set

Ck(0) = f ∈ Ck(Rn) : ∂αf ∈ C(0), for |α| ≤ k .

Theorem 7.5. (Sobolev Embedding Theorem) Let s > k + (n/2). Then, Hs embedds continu-ously in Ck(0).

Proof. Let f ∈ Hs. Then,∫|(∂αf) (ξ)| dξ = (2π)|α|

∫|ξαf(ξ)| dξ ≤

∫(1 + |ξ|2)k/2 dξ

≤ (2π)|α|(∫|f(ξ)|2(1 + |ξ|2)s dξ

)1/2(∫(1 + |ξ|2)k−s dξ

)1/2

= (2π)|α|(∫

(1 + |ξ|2)k−s dξ

)1/2

‖f‖Hs .

Now,∫

(1 + |ξ|2)k−s dξ is finite exactly when 2(k − s) < −n, that is s > k + (n/2).This argument shows that, when s > k + (n/2), (∂αf) ∈ L1. By the Fourier inversion

theorem, and the Reimann-Lebesgue theorem, it follows that (∂αf) ∈ C(0). 2

This result can be made more precise, taking into account also fractional regularity. For0 < α < 1 we define the Lipschitz space

Λα(Rn) =f ∈ C ∩ L∞(Rn) : sup

x 6=y|f(x)− f(y)|/|x− y|α <∞

.

Then we define

Ck,α(0) (Rn) =f ∈ Ck(0)(R

n) : ∂βf ∈ Λα for all β, |β| = k.

Then we have

Theorem 7.6. (Sobolev Embedding Theorem, second version) Let 0 < α < 1 and s = (n/2)+α.Then, Hs embedds continuously in Λα. If s = (n/2) + k + α. Then, Hs embedds continuously

in Ck,α(0) (Rn).

Proof. Exercise (or see [H], Theorem 4.5.13). 2

7.2. Partial differential operators. A linear partial differential operator with smooth coeffi-cients30 is an operator of the form

Pf(x) =∑|α|≤m

aα(x)∂αx f(x) .

The smooth functions aα are called the coefficients of the operator P and m is called the orderof the operator. If the coefficients aα are all constants, then

(Pf) (ξ) =∑|α|≤m

aα(2πiξ)αf(ξ) .

Thus, in order to obtain a cleaner expression, we introduce the differential operator

D = (2πi)−1∂ so that Dα = (2πi)−|α|∂α . (7.1)

In this way, if P =∑|α|≤m aαD

α, then

(Pf) (ξ) =∑|α|≤m

aα(ξ)αf(ξ) .

We will often denote an operator P =∑|α|≤m aαD

α as p(D), where p is the polynomial

of degree m∑|α|≤m aαξ

α. The principal sympbol of P is defined to the top-order part of the

polynomial p:

pm(ξ) =∑|α|=m

aαξα .

We now give two important definitions.

Definition 7.7. Let P =∑|α|≤m aα(x)Dα be a PDO with smooth coefficients on an open set

Ω. We say that P is hypoellptic on Ω if for every open set Ω′ ⊆ Ω and u ∈ D′ such that Pu isC∞ on Ω′, then u is also C∞ on Ω′.

Definition 7.8. Let P =∑|α|≤m aα(x)Dα be a PDO with smooth coefficients on an open set

Ω. We say that P is locally solvable in a point x0 ∈ Ω if there exists a neighborhood of x0 suchthat for every f ∈ D there exists u ∈ D′ such that

Pu = f on U .

Given a PDO P with constant coefficients, we say that a distribution E is a fundamentalsolution for P if PE = δ0, the Dirac delta at the origin, that is if

〈f, PE〉 = 〈 tPf,E〉 = f(0)

for all f ∈ D.

Lemma 7.9. Let P be a PDO with constant coefficients. Then, the following are equivalents.

30Sometime it will be written PDO for short.

86 M. M. PELOSO

(i) P is locally solvable at a point x0 ∈ Rn;(ii) P is locally solvable at any point x0 ∈ Rn;(iii) P admits a fundamental solution.

Proof. It is a simple matter to see that a PDO P =∑|α|≤m aα(x)Dα has constant coefficients

if and only if it is invariant under translations. For, let y ∈ Rn, then

P (τyf)(x) =∑|α|≤m

aα(x)Dα(τyf)(x)

=∑|α|≤m

aα(x)(Dαf)(x− y)

= τy

( ∑|α|≤m

(τ−yaα)(Dαf)

)(x)

and the above equals τy(Pf) if and only if, τ−yaα = aα. Now, it is easy to see that (i) and (ii)are equivalent.

Next, if (iii) holds, let E be a fundamental solution for P . Given x0 ∈ Rn fix a neighborhoodU of x0 and a distribution u that we assume to have compact support (otherwise, replace u byu′ = uϕ, where ϕ ∈ D, ϕ = 1 on U). Let f = ψ(u ∗ E), where ψ ∈ D, ψ = 1 on U . Then

Pf = P(ψ(u ∗ E)

)= ψP (u ∗ E) + v

= ψ(u ∗ δ) + v

= ψu+ v .

Notice that the distribution v is obtained when some of the derivatives falls on ψ. Since ψ = 1on U , the support of v does not intersect U . Therefore, on U , Pf = ψu = u; that is P is locallysolvable at x0.

Finally, the proof that (ii) implies (iii) (i.e. the existence of a fundamental solution) requiresfurther notion of functional analysis and we refer to [H] for a proof. 2

We now show that all PDO’s with constant coefficients admits a fundamental solution, i.e.are locally solvable.

Theorem 7.10. (Malgrange-Ehrenpreis) Let P =∑|α|≤m aαD

α be a constant coefficient PDO.

Then P admits a fundmental solution.

Let f ∈ D be given. Let P = p(D). If a tempered distribution u solves the equation Pu = f ,then

p(ξ)u(ξ) = f(ξ) ,

We would like to define

u(x) =

∫e2πixξ f(ξ)

p(ξ)dξ ,

the problem being the zeros of p on the real space Rn. The proof is based on a deformation ofthe integral to a path in Cn in order to avoid the zeros of p. We begin with a reduction and acouple of lemmas.

By a rotation we may assume that the polynomial p has the leading term that does not vanishat the point (0, . . . , 0, 1). By multiplying by a non-zero constant, we may assume that

p(ξ) = ξmn + lower order terms in ξn .


Therefore, writing ξ = (ξ′, ξn) with ξ′ ∈ Rn−1, we can find λj(ξ′), j = 1, 2, . . . ,m such that

p(ξ′, ξn) =

m∏j=1

(ξn − λj(ξ′)

).

Lemma 7.11. There exists a measurable function φ : Rn−1 → [−m,m] such that

infξ′∈Rn−1

minj=1,2,...,m

|φ(ξ′)− λj(ξ′)|

≥ 1 .

Proof. Let ξ′ be fixed. The interval [−m − 1,m + 1] can be divided into the m + 1 intervals[−m − 1,−m + 1], . . . , [m − 1,m + 1]. At least one of them does contain any of the valuesImλj(ξ), j = 1, 2, . . . ,m. Define φ(ξ′) to be the middle point of that interval. More precisely,define

Uk =ξ′ ∈ Rn−1 : Imλj(ξ

′) 6∈ [−m− 1 + 2k,−m+ 1 + 2k), j = 1, 2, . . . ,m.

The sets Uk cover Rn−1 and they are Borel sets since the functions Imλj are continuous. Thus,set

φ(ξ′) = −m+ 2k for Uk \ ∪k−1k=0Ul

for 0 ≤ k ≤ m. 2

Lemma 7.12. Let h be a polynomial of one complex variable of degree m, with leading coefficient1. Suppose h(0) 6= 0 and let λ1, . . . , λm be its zeros. Then

|h(0)| ≥ (d/2)m

where d = minj |λj |.

Proof. We can write h(z) =∏mj=1(z − λj) so that∣∣∣∣h(z)

h(0)

∣∣∣∣ =m∏j=1

∣∣∣∣1− z

λj

∣∣∣∣ ≤ 2m

when |z| ≤ d. Notice that h(m)(z) is constant and equals m!. By Cauchy integral formula forthe derivatives

m! = h(m)(0) =

∣∣∣∣ ∫ m!

2πi

∫|z|=d

h(z)

zm+1dz

∣∣∣∣≤ m!2m|h(0)| 1

2π

∫ 2π

0

1

dm+1ddθ

=m!2m

dm|h(0)| .

This proves the desired result. 2

Proof of Theorem 7.10. With φ as in the previous lemma set

u(x) =

∫Rn−1

∫Im ξn=φ(ξ′)

e2πixξ f(ξ)

p(ξ)dξndξ .

88 M. M. PELOSO

We apply Lemma 7.12 to the polynomial h(z) = p(ξ′, ξn + z). Notice that h(z) = 0 whenz = λj(ξ

′)− ξn, j = 1, . . . ,m. Then

|p(ξ)| = |h(0)| ≥(|λj(ξ′)− ξn|

2

)m≥(∣∣Im (λj(ξ′)− ξn)∣∣

2

)m=

(∣∣λj(ξ′)− φ(ξ′)∣∣

2

)m≥ 2−m

when Im ξn = φ(ξ′). By the Paley-Wiener Theorem we know that f is an entire function thatdecays rapidly when |Re ξ| → +∞, as long as Im ξ remains bounded. This shows that theintegral converges absolutely.

The same argument shows that we can differentiate under the integral sign as many times aswe like. Therefore, u ∈ C∞.

Now,

Pu(x) =

∫Rn−1

∫Im ξn=φ(ξ′)

e2πixξ f(ξ) dξndξ′ .

We use Cauchy integral theorem to deform the contour of integration to bring back the inte-gration in ξn on the real axis. For, let ΓR be the contour in the complex ξn-plane given by therectangle of vertices ξn = R, ξn = R + iImφ(ξ′), ξn = −R + iImφ(ξ′), and ξn = −R, with this

orientation. Now,∫

ΓRe2πixξ f(ξ) dξn = 0, by Cauchy theorem. If we show that the integrals over

the two vertical sides tend to 0 as R→ +∞, we would obtain that∫Rn−1

∫Im ξn=φ(ξ′)

e2πixξ f(ξ) dξndξ′ =

∫Rn−1

∫ +∞

−∞e2πixξ f(ξ) dξndξ

′ .

By Fourier inversion theorem the result now would follow. Thus, consider the integral over theright vertical segment

∣∣ ∫ Imφ(ξ′)

0e2πi(x′ξ′+xn(R+is)f(ξ′, R+ is)ds

∣∣ ≤ ∫ Imφ(ξ′)

0e−2πxns|f(ξ′, R+ is)|ds

≤ me2π|xn|mcN (1 +R)−N

which tends to 0 as R→ +∞.Thus, we have shown that given f ∈ D there exists u ∈ C∞ such that Pu = f . We finally

construct a fundamental solution for P .Define E by setting

〈E, f〉 =

∫Rn−1

∫Im ξn=φ(ξ′)

f(−ξ)p(ξ)

dξndξ′ .

As before, we see that the integral converges absolutely. Moreover,

〈PE, f〉 = 〈E, tPf〉


and notice that tP is the operator with symbol p(−ξ). Therefore, deforming the integral in ξnas before,

〈PE, f〉 = 〈E, tPf〉 =

∫Rn−1

∫Im ξn=φ(ξ′)

( tPf) (−ξ)p(ξ)

dξndξ′

=

∫Rn−1

∫Im ξn=φ(ξ′)

f(−ξ) dξndξ′

=

∫Rn

f(ξ)dξ

= f(0) = 〈δ0, f〉 .This finishes the proof of the theorem. 2

We conclude this section by characterizing the hypoelliptic PDO with constant coefficientsby the regularity of their fundamental solutions.

Proposition 7.13. Let P be a PDO with constant coefficients. Then, the following are equiva-lent.

(i) P admits a fundamental solution that is in C∞(Rn \ 0);(ii) every fundamental solution of P is in C∞(Rn \ 0);

(iii) P is hypoelliptic.

Proof. Suppose (iii) holds and let E be any fundamental solution. Then, from the equationPE = δ0 we deduce that E ∈ C∞(Rn \ 0) since δ0 is. Thus, (ii) holds. (ii) implies (i) trivially.In order to show that (i) implies (iii) we need the following fact: If u and v are distributions (i.e.elements of D′) with f ∈ C∞ \0 and v with compact support, then sing supp (u ∗ v) ⊆ supp v.

Assuming the claim for now, we finish the proof. Let K be a fundamental solution of P whichis C∞(Rn \ 0). Let g be a distribution such that Pg is C∞ on an open set Ω. Let x ∈ Ω andlet ϕ ∈ C∞0 (B(x, ε)) be ≡ 1 on B(x, ε/2). Then, P (ϕg) = ϕPg + v, where v ≡ 0 on B(x, ε/2)and outside B(x, ε). We write

K ∗ P (ϕg) = K ∗ ϕPg +K ∗ v .Now, ϕPg is a C∞0 function, so that K ∗ϕPg is C∞. Also, K ∗v is C∞ on B(x, ε/2) by the claim.Therefore, K ∗ P (ϕg) is C∞ on B(x, ε/2). But,

K ∗ P (ϕg) = PK ∗ (ϕg) = ϕg .

Therefore ϕg is C∞ on B(x, ε/2), but ϕ ≡ 1 on B(x, ε/2), so that ϕg = g on that ball. Therefore,g ∈ C∞ on B(x, ε/2)

Finally, we prove the claim. Let x 6∈ supp v. Let ε > 0 small so that B(x, ε)∩ supp v = ∅. Letψ ∈ C∞0 (B(0, ε/2)), ψ = 1 on B(0, ε/4). Then

u ∗ v = (ψu) ∗ v + [(1− ψ)u] ∗ v .But, (1− ψ)u ∈ C∞, so that [(1− ψ)u] ∗ v is C∞ in Rn. Moreover,

supp((ψu) ∗ v

)⊆ suppψ + supp g

which does not intersect B(x, ε/2).Hence, on B(x, ε/2) u ∗ v = [(1− ψ)u] ∗ v and we are done. 2

90 M. M. PELOSO

8. †Littlewood–Paley theory

In order to present the Littlewood–Paley theory we need the extension of the singular integralsto the case of vector-valued functions developed in Subsection 5.8.

8.1. An application. A simple application of the theory of vector-valued singular integrals is

the following result. In this theorem the vector-valued kernel ~K is a constant sequence. In this

case it is easy to prove the boundedness of the corresponding vector-valued singular integral ~T .

Theorem 8.1. Let T be a convolution operator which is bounded L2(Rn) and whose integralkernel satisfy the Hormander condition (5.2). Let 1 < p, r < ∞. Then we have the strong(p, p)-bound ∥∥∥(∑

j

|Tfj |r)1/r∥∥∥

p≤ Cp,r

∥∥∥(∑j

|fj |r)1/r∥∥∥

p,

and the weak (1, 1)-bound∣∣∣x ∈ Rn :(∑

j

|Tfj |r)1/r

> λ∣∣∣ ≤ Cr

λ

∥∥∥(∑j

|fj |r)1/r∥∥∥

L1.

Proof. We have stated the theorem in the case of the `r-norm, 1 < r <∞. Since we stated andproved Thm. 5.27 in the case of functions taking values in a Hilbert space (rather than in aseparable, reflexive, Banach space), here we consider only the case r = 2.

We only have to check that the hypotheses of Thm. 5.27 are satisfied.In the current situation we consider functions f : Rn → `2, that is, f(x) = fj(x), where fj

are scalar valued functions. Then, we have that

~T (f) = ~T(fj

)=K ∗ fj

.

Then ~T (f) = ~K ∗ f , where

~K(x) : `2 → `2

aj 7→ K(x)aj .

Hence, ~T : L2(Rn, `2) → L2(Rn, `2) is clearly bounded since, (as in the discussion prior to thetheorem) if f ∈ L2(Rn, `2),

‖~T (f)‖2L2(Rn,`2) =

∫Rn

∑j

|K ∗ fj(x)|2 dx =

∫Rn

∑j

|K(ξ)|2|fj(ξ)|2 dξ

≤ A‖f‖2L2(Rn,`2) .

Next, ∫|x|>2|y|

‖K(x− y)−K(x)‖L(`2,`2) dx =

∫|x|>2|y|

|K(x− y)−K(x)| dx ≤ B .

The result now follows from Thm. 5.27.

In order to illustrate the Littlewood–Paley theorem we consider a collection Ej of mutuallydisjoint measurable sets in Rn and the operator S initially defined on Schwartz functions,

Sf = F−1(∑

j

(χEj f)). (8.1)

Then, it is immediate to see that S is bounded on L2, since, by Plancherel’s theorem

‖Sf‖2L2 = ‖∑j

(χEj f)‖2L2 =

∫Rn

∣∣∑j

(χEj f)(ξ)∣∣2 dξ =

∑j

∫Ej

|f(ξ)|2 dξ = ‖f‖2L2 . (8.2)

Starting from this simple observation we now state the following result. Notice that in thistheorem, we restrict ourselves to the 1-dimensional case.

Theorem 8.2. Let Ej = (−2j+1,−2j ] ∪ [2j , 2j+1) ⊂ R and let Sj be defined as(Sjf

)(ξ) = χEj (ξ)f(ξ) .

Then, for 1 0 such that, for all f ∈ Lp(R)

1

C‖f‖Lp ≤

∥∥∥(∑j

|Sjf |2)1/2∥∥∥

Lp≤ C‖f‖Lp .

We will prove this theorem as a consequence of the Littlewood–Paley theorem, Thm. 8.5 inthe next section, that holds true also in Rn. In that theorem we will deal with an operator asthe one in (8.1). Notice that ∑

j

χEj (ξ) =∑j

χE0(2−jξ) = 1

for all ξ ∈ R \ 0. We will need a smooth decomposition of the function identically 1, and wethen recall Lemma 5.23: There exists a function ϕ ∈ C∞0 such that

(i)∑j∈Z

ϕ(2−jξ) = 1 for all ξ 6= 0 .

Moreover, there exists another function ϕ0 ∈ C∞0 such that

(ii) ϕ0(ξ) +

+∞∑j=1

ϕ(2−jξ) = 1 for all ξ .

A consequence of the lemma is the following.

Corollary 8.3. There exists ψ ∈ S(Rn) such that suppψ ⊆ ξ : 1/2 ≤ |ξ| ≤ 2 and∑j∈Z|ψ(2−jξ)|2 = 1 for ξ 6= 0 . (8.3)

Proof. With ϕ as in the Lemma, we define Φ by setting (Φ)2 = ϕ so that(F(Φ2−j )

)2(ξ) =

(Φ(2−jξ)

)2= ϕ(2−jξ) ,

and therefore, ∑j∈Z|F(Φ2−j )(ξ)|2 =

∑j∈Z

ϕ(2−jξ) = 1

for ξ 6= 0.

Then, it suffices to set ψ = Φ. 2

The next result is the first part of the Littlewood–Paley Thm. In a sense, it is the simplest,and least interesting, part. It shows that a certain decomposition on the Fourier transform sideof a function gives rise to an operator that is bounded in Lp. It is more interesting to show that,

92 M. M. PELOSO

under an additional hypothesis, such an operator is also bounded from below– as we shall seein Thm. 8.5.

Theorem 8.4. (Littlewood–Paley Thm., part 1) Let Φ ∈ S(Rn) be such that

∑j∈Z| Φ(2−jξ)|2 ≤ C

for all ξ 6= 0, and define ∆jf = f ∗ Φ2−j . Then, for 1 0 such that

∥∥∥(∑j∈Z|∆jf |2

)1/2∥∥∥Lp≤ C‖f‖Lp . (8.4)

Proof. If we set

~T (f) = ∆jfj∈Z

if suffices to show that ~T satisfies the hypotheses of Thm. 5.27, when

~T : Lp(Rn)→ Lp(Rn, `2) .

As we have seen already, ~T is bounded when p = 2. For,

‖~Tf‖2L2(Rn,`2) =

∫Rn

‖~T (f)(x)‖2`2 dx =

∫Rn

∑j∈Z|∆j(f)(x)|2 dx

=∑j∈Z

∫Rn

|Φ(2−jξ)|2|f(ξ)|2 dξ

≤ C∫Rn

|f(ξ)|2 dξ

= C‖f‖2L2 .

Thus, it will suffice to show that

‖∇ ~K(x)‖L(C,`2) ≤C

|x|n+1.

Now,

∇ ~K(x) =∇Φ2−j (x)

=∇(2njΦ(2jx)

)=

2(n+1)j(∇Φ)(2jx)

.

Then, using the fact that Φ is a Schwartz function, for any integer N > 0 there exists a constantCN > 0 such that

‖∇ ~K(x)‖`2 =(∑j∈Z

∣∣2(n+1)j(∇Φ)(2jx)

∣∣2)1/2

≤∑j∈Z

2(n+1)j∣∣(∇Φ

)(2jx)

∣∣≤ CN

∑j∈Z

2(n+1)j min(1, |2jx|−N

)≤ CN

(∑j≤j0

2(n+1)j +∑j>j0

2(n+1)j |2jx|−N)

= CN

(∑j≤j0

2(n+1)j +1

|x|N∑j>j0

2(n+1−N)j),

with j0 to be selected. Choosing j0 so that 2j0 ' |x|−1, and N > n+ 1 we have31,

‖∇ ~K(x)‖`2 ≤ CN(

2(n+1)j0 +2(n+1−N)j0

|x|N)≤ C

|x|n+1,

as we wished to show. 2

8.2. The Littlewood–Paley theorem.

Theorem 8.5. (Littlewood–Paley theorem) Let ψ ∈ S(Rn) be as in (8.3). Let Φ = F−1ψand set

∆jf = f ∗ Φ2−j .

Then, for 1 0 such that

1

C‖f‖Lp ≤

∥∥∥(∑j

|∆jf |2)1/2∥∥∥

Lp≤ C‖f‖Lp .

Proof. The bound from above follows directly from Thm. 8.4, where we use only the estimate∑j∈Z |ψ(2−jξ)|2 ≤ C.For the bound from below, we notice that

‖f‖L2 =(∫

Rn

|f(ξ)|2∑j∈Z|ψ(2−jξ)|2 dξ

)1/2

=(∫

Rn

∑j∈Z|f(ξ)ψ(2−jξ)|2 dξ

)1/2

=∥∥∥(∑

j

|∆jf |2)1/2∥∥∥

L2.

31Recall footnote 21.

94 M. M. PELOSO

Having the identity ‖f‖L2 = ‖~T (f)‖L2(Rn,`2), we can polarized it32, to obtain that for all f, g ∈L2(Rn) ∫

fg =

∫ ∑j

∆jf∆jg .

Therefore,

‖f‖Lp = sup‖g‖

Lp′=1

∣∣ ∫ f(x)g(x) dx∣∣ = sup

‖g‖Lp′=1

∣∣∣ ∫ ∑j

∆jf(x)∆jg(x) dx∣∣∣

≤ sup‖g‖

Lp′=1

∫ (∑j

|∆jf(x)|2)1/2(∑

j

|∆jg(ξ)|2)1/2

dx

≤ sup‖g‖

Lp′=1

(∫ (∑j

|∆jf(x)|2)p/2

dx)1/p(∫ (∑

j

|∆jg(ξ)|2)p′/2

dx)1/p′

≤ C(∫ (∑

j

|∆jf(x)|2)p/2

dx)1/p

, (8.5)

where we have used the estimate from above. This proves the theorem. 2

We are now ready to prove Thm. 8.2. We recall that in this situation the space dimension isn = 1.

Proof of Thm. 8.2. Let ψ ∈ C∞0 (R), suppψ ⊆ ξ : 1/2 ≤ |ξ| ≤ 4 and such that ψ(ξ) = 1 if1 ≤ |ξ| ≤ 2, that is, if ξ ∈ E0. Therefore,

ψ2−j (ξ)χEj (ξ) = ψ(2−jξ)χE0(2−jξ) = χE0(2−jξ) = χEj (ξ)

for all ξ. This implies that, if we define ∆jf by setting(∆jf

)(ξ) = ψ(2−jξ)f(ξ) ,

then

∆jSjf = Sjf .

Of course we can write ∆jf = Φ2−j ∗ f , where Φ = ψ.Notice that for any given ξ, since suppψ ⊆ ξ : 1/4 ≤ |ξ| ≤ 4, there at most 3 indices

j1, j2, , j3 such that ψj(ξ) 6= 0. Therefore,∑j

|ψ(2−jξ)|2 ≤ C ,

for all ξ 6= 0.We now make the following claim: For 1 0 such that for all

g = gj ∈ Lp(Rn, `2)∥∥∥(∑j

|Sjgj |2)1/2∥∥∥

Lp≤ C

∥∥∥(∑j

|gj |2)1/2∥∥∥

Lp= C‖g‖Lp(Rn,`2) . (8.6)

32Recall the identity 4〈f, g〉H = ‖f + g‖2H − ‖f − g‖2H + i‖f + ig‖2H − i‖f − ig‖2H valid on any Hilbert space,

and notice that in our case 〈f, g〉H =∫ ∑

j ∆jf∆jg.


Assuming the claim we obtain the estimate from above, since∥∥∥(∑j

|Sjf |2)1/2∥∥∥

Lp=∥∥∥(∑

j

|Sj∆jf |2)1/2∥∥∥

Lp

≤ C∥∥∥(∑

j

|∆jf |2)1/2∥∥∥

Lp

≤ C‖f‖Lp ,

by Thm. 8.4.The estimate from below follows by polarizing the identity (8.2) and using duality, as in the

proof of Thm. 8.5, inequality (8.5), by replacing ∆j with Sj .Thus, we only need to prove the claim.

We now consider the vector-valued operator ~S : Lp(Rn, `2)→ Lp(Rn, `2) defined as

~S(g) =Sj(gj)

≡Kj ∗ gj

where Kj = F−1(χEj ). For this operator we wish to prove the bound (8.6).

Instead of proving that the operator ~S and its kernel satisfy the hypotheses of Thm. 5.27, weproceed in a direct way.

Recall Example (6.5) (2). By (6.3) above we have that

Sj(gj)(x) = F−1(χ(−2j+1,−2j ] + χ[2j ,2j+1)

)∗ gj(x)

=i

2

[M−2j+1HM2j+1 −M−2jHM2j +M2jHM−2j −M2j+1HM−2j+1

](gj) ,

so that∣∣Sj(gj)(x)∣∣ ≤ ∣∣H(M2j+1gj

)(x)∣∣+∣∣H(M2jgj

)(x)∣∣+∣∣H(M−2jgj

)(x)∣∣+∣∣H(M−2j+1gj

)(x)∣∣ .

Therefore, in order to prove the claim it suffices to show that each of the four operators on theright hand side above satisfies (8.6), that is,∥∥∥(∑

j

|H(Majgj

)|2)1/2∥∥∥

Lp≤ C

∥∥∥(∑j

|gj |2)1/2∥∥∥

Lp,

where aj is a sequence of real numbers.Applying Thm. 8.1 we have that∥∥∥(∑

j

|H(Majgj

)|2)1/2∥∥∥

Lp≤ C

∥∥∥(∑j

|Majgj |2)1/2∥∥∥

Lp

= C∥∥∥(∑

j

|gj |2)1/2∥∥∥

Lp.

This proves the claim and therefore the theorem. 2

We conclude this part with an extension of Thm. 8.2 to n-dimensions.

96 M. M. PELOSO

The dyadic cubes in Rn can be seen as the cartesian product of dyadic intervals of the samesize; e.g. χ[0,1)n(x) = χ[0,1)(x1) · · ·χ[0,1)(xn). A dyadic rectangle R in Rn is the cartesian productof dyadic intervals in Ij1 , . . . , Ijn of possibly different scales:

χR(x) = χIj1 (x1) · · ·χIjn (xn) .

We consider the operator (S1j1f)

(ξ) = χIj1 (ξ1)f(ξ) ,

and more generally,(S~jf

)(ξ) =

(S1j1 · · ·S

njnf)

(ξ) = χIj1 (ξ1) · · ·χIjn (ξn)f(ξ) .

Corollary 8.6. Let f ∈ Lp(Rn), 1 0 suchthat

1

Cp‖f‖Lp ≤

∥∥∥( ∑~j ∈Zn

|S~jf |2)1/2∥∥∥

Lp≤ Cp‖f‖Lp ,

where Ij1, . . . , Ijn are decompositions of the real line R into dyadic intervals of (possibly

different) scales 2k1 , . . . , 2kn, resp.

Proof. Notice that, if we set R~j , the collection R~j~j∈bZn is a decomposition of Rn into mutually

disjoint dyadic rectangles.The estimate from above follows at once from Thm. 8.4. The estimate from below follows

instead by applying Thm. 8.2 to the operators Skjk , k = 1, 2, . . . , n.

8.3. The Marcinkiewicz multiplier theorem. We conclude with another multiplier theorem,the Marcinkiewicz multiplier theorem, which is tightly connected with the theory of productspaces.

The simplest form of the Mihlin-Hormander condition is given by the inequality (6.11) fork ≥ bn/2c+ 1. We now consider the condition∣∣∂αm(ξ)

∣∣ ≤ C|ξ1|−α1 · · · |ξn|−αn . (8.7)

Notice that when n = 1 it coincides with the Mihlin-Hormander condition. But, when n > 1the two conditions are different. In the theorem we give a more general condition for the Lp-boundedness of the multiplier operator. It is easy to see that the condition is satisfied if (8.7)holds, for k ≥ n.

Definition 8.7. Let s = (s1, . . . , sn) ∈ Rn. We define the product Sobolev space Hs as the spaceof tempered distributions f such that∫

Rn

(1 + |ξ1|2)s1 · · · (1 + |ξn|2)sn |f(ξ)|2 dξ1 · · · dξn <∞ .

Similarly to Lemma 6.17 one can easily prove the following.

Lemma 8.8. Let g ∈ S ′ be such that g ∈ Hs with si > 1/2 for i = 1, . . . , n. Then g ∈ L1(Rn)and for 0 < ε < si − 1/2 for all i∫

Rn

(1 + |x1|2)ε · · · (1 + |xn|2)ε|g(x)| dx ≤ Cε‖g‖Hs .

Definition 8.9. For i = 1, . . . , n, let ηi ∈ C∞(R) be a function as in the hypothesis of Lemma5.23 and set η(ξ) = η1(ξ1) · · · ηn(ξn). For r = (r1, . . . , rn) ∈ (R+)n and ξ = (ξ1, . . . , ξn) ∈ Rn wealso set rξ = (r1ξ1, . . . , rnξn).

A tempered distribution m is said to be a Marcinkiewicz multiplier if

supr∈(R+)n

‖m(r·)η‖Hs =: ‖m‖Ms <∞ .

Theorem 8.10. (Marcinkiewicz multiplier theorem) Let m be a Marcinkiewicz multiplierin Hs, with si > 1/2 for i = 1, . . . , n. Then m is a bounded Fourier multiplier on Lp for1 0 such that for all0 < k ≤ n

supIj1 ,...,Ijk

∫Ij1×···×Ijk

∣∣∂kξj1 ···ξjkm(ξ)∣∣ dξj1 · · · dξjk ≤ B , (8.8)

where Iji, i = 1, . . . , k, are dyadic intervals in R.

98 M. M. PELOSO

9. Symbols and pseudodifferential operators

In this section we introduce the notion of pseudodifferential operator. This turns out to beof fundamental importance in the analysis of differential operators and their inverses. We beginwith the definition of symbol.

Definition 9.1. Let X,Y be open sets in Rn. For m a real number, we denote by Sm(X × Y )the set of all a ∈ C∞(X × Y ×Rn) such that, for all KX ,KY compact sets contained in X andY respectively, and all multi-indices α, β there exists a constant C = CKX ,KY ,α,β such that∣∣∂βx,y∂αξ a(x, y, ξ)

∣∣ ≤ C(1 + |ξ|)m−|α| for (x, y, ξ) ∈ KX ×KY ×Rn .

More generally, for 0 ≤ %, δ ≤ 1, one defines the class of symbols Sm%,δ(X × Y ) the set of all

a ∈ C∞(X×Y ×Rn) such that, for all KX ,KY compact sets contained in X and Y respectively,and all multi-indices α, β there exists a constant C > 0 such that∣∣∂βx,y∂αξ a(x, y, ξ)

∣∣ ≤ C(1 + |ξ|)m−%|α|+δ|β| for (x, y, ξ) ∈ KX ×KY ×Rn .

But, in these notes we restrict ourselves to the classical, simpler case % = 1 and δ = 0.

It is immediate to see that, if m is a positive integer and aα(x, y) are smooth functionsof x and y, and a(x, y, ξ) :=

∑|α|≤m aα(x, y)ξα (that is, a is a polynomial in ξ), then a ∈

Sm(Rn ×Rn ×Rn).

We now consider symbols of (only apparent) simpler nature.

Definition 9.2. For m a real number, we denote by Sm the set of all a ∈ C∞(Rn ×Rn) suchthat for all K compact in Rn and all multi-indices α, β there exists a constant C = Cα,β,K > 0such that ∣∣∂βx∂αξ a(x, ξ)

∣∣ ≤ C(1 + |ξ|)m−|α| for (x, ξ) ∈ K ×Rn .

We now define a pseudodiffential operator P with symbol a.

Definition 9.3. Let a ∈ Sm. We define the pseudodifferential operator Pa with symbol a theoperator, initially defined on Schwartz functions, as

Pau(x) =

∫a(x, ξ)e2πix·ξu(ξ) dξ .

Notice that, since u ∈ S and for fixed x, |a(x, ξ)| ≤ C(1 + |ξ|)m, the integral above convergesabsolutely.

Moreover, if, for a ∈ Sm, we consider the application defined on D

Pa : u 7→x

a(x, y, ξ)e2πi(x−y)·ξu(y) dydξ

the integral converges absolutely, for any fixed x, if m < −n, since then |a(x, y, ξ)u(y)| ≤C(1 + |ξ|)−n−ε|u(y)|, which is in L1(Rn ×Rn). For generic m ∈ R, the double integral

xe2πi(x−y)·ξa(x, y, ξ)u(y) dydξ

does not converge absolutely and it must be interpreted as an iterated integral in the followingway. Observe that, for all positive integers N ,

e2πi(x−y)ξ =(I −∆y)

N

(1 + 4π2|ξ|2)N(e2πi(x−y)ξ

),

so that, integrating by parts in the inner integral in the y-variable we obtain thatx

e2πi(x−y)·ξa(x, y, ξ)u(y) dydξ

=x (I −∆y)

N

(1 + 4π2|ξ|2)N(e2πi(x−y)ξ

)a(x, y, ξ)u(y) dydξ

=x e2πi(x−y)ξ

(1 + 4π2|ξ|2)N(I −∆y)

N(a(x, y, ξ)u(y)

)dydξ . (9.1)

For N > 0 large enough, the integral above now converges absolutely.

Definition 9.4. Let a ∈ Sm. We define the pseudodiffential operator Pa with symbol a theoperator, initially defined on test functions (i.e. smooth functions with compact support), byequation (9.1) above.

Notice that if the symbol a ≡ a does not depend on y, then

Pau(x) =x

a(x, ξ)e2πi(x−y)·ξu(y) dydξ =

∫a(x, ξ)e2πix·ξu(ξ) dξ = Pau(x) ,

that is, the two definitions coincide in this case.A few examples are now in order. Let P be the partial differential operator with smooth

coefficients of order m, given by Pf(x) =∑|α|≤m aα(x)Dα

xf(x). By the Fourier inversion

theorem

Pf(x) =∑|α|≤m

aα(x)

∫e2πix·ξξαf(ξ) dξ

=

∫e2πix·ξ

( ∑|α|≤m

aα(x)ξα)f(ξ) dξ

=

∫e2πix·ξa(x, ξ)f(ξ) dξ

where a(x, ξ) :=∑|α|≤m aα(x)ξα. Thus, P = Pa is a pseudo differential operator of order m

(that is the definitions of orders coincide).If the operator P has constant coefficients aα, then P = Pa where the symbol a(x, ξ) :=∑|α|≤m aαξ

α = a(ξ) is just a polynomial in ξ.

Definition 9.5. For m ∈ R we denote by Ψm the set of pseudodifferential operators withsymbol a ∈ Sm.

Example 9.6. For each m ∈ R, the function σm(ξ) = (1 + |ξ|2)m/2 defines and element ofSm(Rn), so that Λm ∈ Ψm. Moreover, if P is a PDO (with C∞ coefficients) of order k, thenP ∈ Ψk.

If a ∈ Sm, then ∂αx a ∈ Sm while ∂αξ a ∈ Sm−|α| for any multiindex α.

The next result is rather obvious.

Lemma 9.7. If m1 < m2 are real numbers, then Sm1 ⊆ Sm2. If a1 ∈ Sm1 and a2 ∈ Sm2, thena1a2 ∈ Sm1+m2 and a1 + a2 ∈ Smax(m1,m2).

100 M. M. PELOSO

It is natural to then set

S∞ = ∪m∈RSm , Ψ∞ = ∪m∈RΨm and S−∞ = ∩m∈RSm , Ψ−∞ = ∩m∈RΨm .

We mention that in general, if b is a measurable function, then the Fourier multiplier operatorMb is defined as

Mbf(x) = F−1(b(ξ)f(ξ)

)(x) .

Before presenting the next example, we prove the following simple fact.

Lemma 9.8. Let Mb be a Fourier multiplier operator as above. Then, the following conditionsare equivalent:

(i) b is bounded, i.e. b ∈ L∞(Rn);

(ii) Mb : L2(Rn)→ L2(Rn) is bounded;

(iii) for all s ∈ R, Mb : Hs(Rn)→ Hs(Rn) is bounded.

Proof. By Plancherel’s theorem it is immediate to see that∫|Mb(f)(x)|2 dx =

∫|b(ξ)f(ξ)|2 dξ ≤ ‖b‖2L∞‖f‖2L2 ,

that is (i) implies (ii). It is also easy to see that if g 7→ bg is bounded on L2, then b must bebounded (Exercise V.2). By Plancherel again it follows then that (ii) implies (i).

Finally, if (i) holds, then

‖Mb(f)‖2s =

∫|b(ξ)f(ξ)|2(1 + |ξ|2)s dξ

≤ ‖b‖2L∞∫|f(ξ)|2(1 + |ξ|2)s dξ

= C‖f‖2s .

Hence, (i) implies (iii). And, if (iii) holds, also (ii) holds, so does (i) 2

Notice that this result holds true even without the assuption that b ∈ C∞, and thus withoutany assumption on the behaviour of the derivatives of b.

Corollary 9.9. If m ≤ 0 and b = b(ξ) ∈ Sm then Pb : Hs → Hs for all s ∈ R. If ψ ∈ S thenthe mappings u 7→ ψu and u 7→ ψ ∗ u, initially defined on S, are in Ψ0.

We wish to check that a pseudodifferential operator Pa : S → D′ boundedly. Let then ϕ ∈ Sand ψ ∈ D and consider 〈Paϕ,ψ〉. Let K = suppψ, then we have

|〈Paϕ,ψ〉| =∣∣x e2πixξa(x, ξ)ϕ(ξ) dξ ψ(x) dx

∣∣≤∫|ϕ(ξ)|

∫|a(x, ξ)| |ψ(x)| dx dξ

≤ CK∫|ϕ(ξ)|(1 + |ξ|)m

∫|ψ(x)| dx dξ

≤ C∑|α|≤N

‖ϕ‖(α,n+1)‖ψ‖(0,n+1) , (9.2)

for some N large enough.


More is true. Let E ′ denote the space of distributions having compact support. That is,u ∈ D′ (or in S ′) is in E ′ if there exists a compact set K such that 〈ψ, u〉 = 0 for all ψ ∈ D suchthat suppψ ∩K = ∅.

Then, if a ∈ Sm, then Pa : E ′ → D′ continuously. Now we have a definition that will be usedin what follows.

Definition 9.10. A linear operator T : E ′ → C∞ is called a smoothing operator.

9.1. The kernel of a pseudodifferential operator. If T is an integral operator initiallydefined on S having values in S ′, (or initially defined on D having values in D′)

Tϕ(x) =

∫Rn

K(x, y)ϕ(y) dy ,

then, for ψ also in S we have that, if the double integral makes sense,

〈Tϕ, ψ〉 =xK(x, y)ϕ(y)ψ(x) dydx ,

or, in general,

〈Tϕ, ψ〉 = 〈K, ϕ⊗ ψ〉 .

The distribution K on Rn × Rn is called the distributional kernel of the operator T . In thissetting, recall also the so-called Schwartz kernel theorem, Theorem ??

We wish to calculate the distributional kernel of a pseudodifferential operator. Let a ∈ Sm,ϕ ∈ S and ψ ∈ D, then, using (9.2), we have

〈Paϕ,ψ〉 =x

e2πixξa(x, ξ)ϕ(ξ)ψ(x) dξdx .

Now, let χ ∈ C∞0 , χ = 1 in a neighborhood of the origin. Then set aε(x, ξ) = χε(ξ)a(x, ξ), sothat aε has compact support in ξ, while tends to a as ε→ 0. Then,

〈Paϕ,ψ〉 = limε→0

ye−2πi(y−x)ξaε(x, ξ)ϕ(y)ψ(x) dydξdx

= limε→0

ye−2πi(y−x)ξχ(εξ)a(x, ξ) dξ ϕ(y)ψ(x) dydx

= limε→0

x (χε ∗ F2a(x, ·)

)(y − x)ϕ(y)ψ(x) dydx

=xF2a(x, y − x)ϕ(y)ψ(x) dydx ,

where F2a(x, ·) denotes the Fourier transform of a(x, ξ) in the second variable. (Notice that thisis well defined, since if a ∈ Sm, a(x, ·) is C∞ and of moderate growth.)

Theorem 9.11. Let a ∈ Sm and let K(x, y) = F2a(x, y − x) be its distributional kernel. Then,

(i) if |α| > m+n+k, the distribution uα = zαF2a(x, z) is in fact a function in Ck(Rn×Rn)and its derivatives of order ≤ k are bounded in K ×Rn, for all compact K ⊆ Rn;

(ii) if |α| > m + n + k, the distribution (x − y)αF2a(x, y − x) is in fact a function inCk(Rn×Rn), in particular K ∈ C∞(Rn×Rn\4R2n), where 4R2n = (x, y) ∈ Rn×Rn :x = y.

102 M. M. PELOSO

Proof. By the discussion above, it suffices to prove (i). Now, the distribution uα is the Fouriertransform in the second variable of Dα

ξ a(x, ξ). If |α| > m + n + k, for x varying in a compact

set K, |Dαξ a(x, ξ)| ≤ C(1 + |ξ|)m−|α|, which is integrable. Thus we can apply Proposition 2.11

and the result follows. 2

Corollary 9.12. If P ∈ Ψ−∞, then P is a smoothing operator.

Proof. By the previous theorem it follows that the distributional kernel of P is in fact C∞(Rn×Rn). Then, for u a distribution with compact support we have

Pu(x) = 〈K(x, ·), u〉 .

By (i) in the Theorem, we can differentiate in x under the pairing of duality (which can bethought to be an integral, by approximating u with elements of D) to obtain that

DαxPu(x) = 〈Dα

xK(x, ·), u〉 .

This proves the result. 2

Definition 9.13. We say that a subset Ω of Rn×Rn is proper if for every compact set K ⊂ Rn

π−1x (K) ∩Ω and π−1

y (K) ∩Ω are compact sets. Here we denote by πx and πy the projections ofRn ×Rn onto its first and second component, respectively.

A pseudodifferential operator Pa is said to be properly supported if its distributional kernelK(x, y) is properly supported.

An important fact, that we leave without proof (see [Fo2]) is that the composition of twoproperly supported operators is still properly supported.

Proposition 9.14. Let a = a(x, y, ξ) ∈ Sm. Then, the pseudodifferential operator Pa can bewritten as sum of a properly supported and of a smoothing pseudodifferential operators.

Proof. Let χ be a test function, supported on the unit ball and identically 1 on the ball of radius1/2. Write the symbol pseudodifferential operator as sum

a(x, y, ξ) = χ(x− y)a(x, y, ξ) +(1− χ(x− y)

)a(x, y, ξ) ≡ a1(x, yξ) + a∞(x, y, ξ) .

This gives the desired decomposition. 2

Notice that if a(x, y, ξ) is supported in a neigborhood of the diagonal 4R2n , then, for x andξ fixed the function y 7→ a(x, y, ξ) is C∞ with compact support. Then, Paf is well defined forany C∞ function f . We will make use of this remark in the next result.

The notion of properly supported pseudodifferential operators allows us to make a reduction.This is a somewhat technical result, which is however of some significance.

Theorem 9.15. Let a ∈ Sm and Pa properly supported. Set

b(x, ξ) = e−2πixξPa(e2πi(·)ξ)(x) .

Then b ∈ Sm and Pa = Pb modulo Ψ−∞.

Proof. We begin by arguing as in the discussion preceeding Theorem 9.11. Let χ ∈ C∞0 , χ = 1in a neighborhood of the origin. Then set aε(x, y, η) = χε(η)a(x, y, η), so that aε has compactsupport in η, while tends to a as ε → 0 (as a tempered distribution in η, depending smoothly


in x, y). Recall also that, for fixed x, a(x, y, η), and so also aε(x, y, η), has compact support iny. Let Paε be the corresponding operator. Then, for a given ϕ ∈ S,

Paεϕ(x) =x

e2πi(x−y)ηaε(x, y, η)ϕ(y) dydη

=x

e2πi(x−y)ηaε(x, y, η)

∫e2πiyξϕ(ξ) dξ dydη

=

∫e2πixξϕ(ξ)

(e−2πixξ

xe2πi(x−y)ηaε(x, y, η)e2πiyξ dydη

)dξ

=

∫e2πixξϕ(ξ)

(e−2πixξPaε

(e2πi(·)ξ)(x)

)dξ .

Passing to the limit as ε→ 0+, we obtain that

Paϕ(x) = limε→0+

Paεϕ(x)

= limε→0+

∫e2πixξϕ(ξ)

(e−2πixξPaε

(e2πi(·)ξ)(x)

)dξ

=

∫e2πixξϕ(ξ)

(e−2πixξPa

(e2πi(·)ξ)(x)

)dξ

=

∫e2πixξϕ(ξ)b(x, ξ) dξ ,

where Pau must be interpreted as in Definition 9.4.Thus, we only need to show that

b(x, ξ) = e−2πixξPa(e2πi(·)ξ)(x)

= e−2πixξx

e2πi(x−y)ηa(x, y, η)e2πiyξ dydη

is an element of Sm (where, again, the last double integral above is an iterated integral thatmust be must be interpreted in the sense of Definition 9.4).

Now we set σ(x, y, η) = a(x, x+ y, η), so that

b(x, ξ) =x

e2πi(x−y)(η−ξ)a(x, y, η) dydη

=x

e2πi(x−y)(η−ξ)σ(x, y − x, η) dydη

=x

e−2πiz(η−ξ)σ(x, z, η) dydη

=

∫F2σ(x, η − ξ, η) dη .

Hence,

b(x, ξ) =

∫F2σ(x, η, η + ξ) dη . (9.3)

For x varying in a fixed compact set K, y 7→ σ(x, ·, η) is C∞ with compact support so thatfor any N > 0, ∣∣∂βx∂αζ F2σ(x, η, ζ)

∣∣ ≤ CK,α,β,N (1 + |ζ|)m−|α|(1 + |η|)−N . (9.4)

Next we apply the simple fact that

(1 + |x|2)s(1 + |y|2)−s ≤ 2|s|(1 + |x− y|2)|s| (9.5)

104 M. M. PELOSO

for all x, y ∈ Rn and s ∈ R, that follows from the triangle inequality

(1 + |x|2) ≤ 2(1 + |x− y|2)(1 + |y|2) .

Then, by applying (9.5) with x = ξ + η and y = ξ to (9.4), we obtain∣∣∂βx∂αξ F2σ(x, η, η + ξ)∣∣ ≤ CK,α,β,N (1 + |ξ + η|)m−|α|(1 + |η|)−N

≤ CK,α,β,N (1 + |ξ|)m−|α|(1 + |η|)∣∣m−|α|∣∣−N .

Inserting this estimate in (9.3) we obtain∣∣∂βx∂αξ b(x, ξ)∣∣ ≤ CK,α,β,N ∫ (1 + |ξ|)m−|α|(1 + |η|)∣∣m−|α|∣∣−N dη

≤ CK,α,β(1 + |ξ|)m−|α| ,for all x ∈ K and ξ ∈ Rn. This shows that b ∈ Sm. 2

9.2. Sobolev continuity of a pseudodifferential operator. We now wish to study themapping properties of pseudodifferential operators. The nature of the definition of symboloffers no control in the behaviour of the symbol itself, and consequentely of the action of apseudodifferential operator on a test function (say) Paϕ(x) when the output variable x tends toinfinity.

For this reason we need to restrict our attention to the local behaviour of Paϕ. We then havethe following definition.

Definition 9.16. Let s ∈ R. We set

Hsloc =

f ∈ S ′ : ϕf ∈ Hs for all ϕ ∈ C∞0

.

The next result shows that a pseudodifferential operator of order m maps Hs into Hs−mloc

continuously, for all s,m ∈ R.

Theorem 9.17. Let P be a pseudodifferential operator of order m. Then, for all s ∈ R,Pa : Hs → Hs−m

loc is continuous, that is, for each χ ∈ C∞0 there exists a constant C = Cχ,s,m > 0such that

‖χPaϕ‖Hs−m ≤ C‖ϕ‖Hs .

Proof. Since we wish to show that Pa(ϕ) is in Hsloc, we may just assume that a(x, ξ) has compact

support in the x-variable– it suffices to multiply a by the cut-off function χ– and prove that

Pa : Hs → Hs−m

for all s ∈ R. Thus, we assume that a has compact support in x throughout the rest of theproof.

We begin by making a reduction. We may assume that Pa is of order 0. In fact, supposea ∈ Sm and define b(x, ξ) = (1 + |ξ|2)−m/2a(x, ξ). Then b ∈ S0 as it is immediate to check (andstill has compact support in x). Moreover,

Pa(ϕ)(x) =

∫e2πix·ξb(x, ξ)(1 + |ξ|2)m/2ϕ(ξ) dξ = Pb(Λmϕ)(x) .

If we had shown the Hs-boundedness of operators of order 0, then it would follow that

‖Pa(ϕ)‖Hs−m = ‖Pb(Λmϕ)‖Hs−m ≤ c‖Λmϕ‖Hs−m ≤ c‖ϕ‖Hs .


Thus, we assume that a ∈ S0. Since a has compact support in x, we may compute its Fouriertransform in x. We set

F1a(t, ξ) =

∫e−2πit·xa(x, ξ) dx ,

so that

a(x, ξ) =

∫e2πit·xF1a(t, ξ) dt .

An integration by parts now shows that for all multi-indices α

(2πit)αF1a(t, ξ) =

∫ [∂αx a(x, ξ)

]e−2πit·x dx .

Therefore,

supξ|F1a(t, ξ)| ≤ CN (1 + |t|)−N (9.6)

for arbitrary N ≥ 0.Notice that for ϕ ∈ S

F(Paϕ)(ξ) =

∫e−2πixξ

∫e2πixηa(x, η)ϕ(η) dηdx

=

∫F1a(ξ − η, η)ϕ(η) dη .

Finally, let s ∈ R. Then, using the estimate (9.6) above, Holder inequality, (9.4) and choosingN large enough, we have

‖Pa(ϕ)‖2Hs =

∫ ∣∣ ∫ F1a(ξ − η, η)ϕ(η) dη∣∣2(1 + |ξ|2)s dξ

≤ CN∫ (∫

(1 + |ξ − η|2)−N |ϕ(η)| dη)2

(1 + |ξ|2)s dξ

≤ CN∫|ϕ(η)|2

∫(1 + |ξ − η|2)−N (1 + |ξ|2)s dξdη

≤ CN∫|ϕ(η)|2(1 + |η|2)s dη

∫(1 + |ξ − η|2)−N+|s| dξdη

≤ C‖ϕ‖2Hs ,

by choosing N > |s|+ n, as we may. Here we have used (9.5) with x = ξ and y = η.This concludes the proof. 2

9.3. Asymptotic expansion of symbols. We now present the notion of symptotic expansionof our symbols.

Definition 9.18. Let aj ∈ Smj , mj ∈ R, j = 0, 1, 2, . . . . We say that a symbol a has theasymptotic expansion

∑j aj , and we write

a ∼∑j

aj ,

106 M. M. PELOSO

if for every M > 0 there exists NM > 0 such that

a−NM∑j=0

aj ∈ S−M .

Proposition 9.19. Let aj ∈ Smj , j = 0, 1, 2, . . . , where mj −∞. Then there exists a symbola ∈ Sm0, unique modulo symbols in S−∞ (and we will usually say, with an abuse of language,modulo smoothing operators), such that

a ∼+∞∑j=0

aj .

Proof. Let θ be a C∞ function that is identically 0 when |ξ| ≤ 1 and identically 1 when |ξ| ≥ 2.Let Ωj be an increasing sequence of bounded open set tending to the whole of Rn, say Ωj =B(0, Rj), with Rj +∞.

Let t0 = 1 < t1 < t2 < · · · be an increasing sequence of positive numbers tending to +∞, tobe selected later. We set

a(x, ξ) =+∞∑j=0

θ(ξ/tj)aj(x, ξ) . (9.7)

Notice that θ(ξ/tj) = 0 when |ξ| ≤ tj , so that, for each (x, ξ) fixed, only finitely many terms inthe series are non-zero. Therefore, a(x, ξ) is a well-defined C∞ function.

Now we claim that we can find a sequence tj in such a way that∣∣∂βx∂αξ (θ(ξ/tj)aj(x, ξ))∣∣ ≤ 2−j(1 + |ξ|)mj−1−|α| (9.8)

for |x| ≤ Rj and |α|+ |β| ≤ j.Assuming the claim for the moment, we show that the function a defined in (9.7) has the

required properties. Having fixed a compact set K ⊂ Rn, let k be such that K ⊂ B(0, Rk) andlet |α| + |β| ≤ k. We use the facts that, for j ≤ k aj ∈ Smj , and for j > k the claim above, toobtain that

|∂βx∂αξ a(x, ξ)| ≤k∑j=0

∣∣∂βx∂αξ (θ(ξ/tj)aj(x, ξ))∣∣++∞∑j=k+1

∣∣∂βx∂αξ (θ(ξ/tj)aj(x, ξ))∣∣≤

k∑j=0

Cj(1 + |ξ|)mj−|α| ++∞∑j=k

2−j(1 + |ξ|)mj−1−|α|

≤ Ck(1 + |ξ|)m0−|α| + C(1 + |ξ|)mk−1−|α| (9.9)

≤ C(1 + |ξ|)m0−|α| .

Therefore, a ∈ Sm0 . We now need to show that a admits the right asymptotic expansion. Wecan write

a(x, ξ)−k−1∑j=0

aj(x, ξ) =(a(x, ξ)−

k−1∑j=0

θ(ξ/tj)aj(x, ξ))

+(k−1∑j=0

(θ(ξ/tj)− 1

)aj(x, ξ)

)=: b(x, ξ) + r(x, ξ) .

By the previous argument (9.9) we see that

b(x, ξ) = a(x, ξ)−k−1∑j=0

θ(ξ/tj)aj(x, ξ) =+∞∑j=k

θ(ξ/tj)aj(x, ξ)

is a well-defined C∞ function and it satisfied the differential inequalities

|∂βx∂αξ b(x, ξ)| ≤ C(1 + |ξ|)mk−|α| ,

that is, b ∈ Smk . On the other hand, notice that

supp(1− θ(ξ/tj)) ⊆ ξ : |ξ| ≤ 2tj ⊆ ξ : |ξ| ≤ 2tk (9.10)

for j = 0, . . . , k − 1. Then r(x, ξ) has compact support in ξ, that is, r ∈ S−∞. These two factsimply that

a(x, ξ)−k−1∑j=0

aj(x, ξ) ∈ Smk ,

as we wished to show.In order to show that this asymptotic expansion is unique (that is, that a is unique modulo

smoothing terms), let b(x, ξ) be another symbol in Sm0 such that b ∼∑

j aj . Then, givenM > 0, let k be such that mk < −M so that

a(x, ξ)− b(x, ξ) =(a(x, ξ)−

k−1∑j=0

aj(x, ξ))−(b(x, ξ)−

k−1∑j=0

aj(x, ξ))∈ Smk .

Hence, a − b ∈ Smk for all k, that is, a − b ∈ S−∞, i.e. a = b modulo S−∞, as we wished toshow.

It only remains to prove the claim (9.8). We begin by observing that (9.10) also gives thatθ(ξ/tj) = 1 for all ξ with |ξ| ≥ 2tj and all j. Then, if |α| > 0, ∂αξ θ(ξ/tj) has compact support

for all j, and, in fact, |ξ| ≤ 2tj for ξ ∈ supp ∂αξ θ(ξ/tj), for all j. Hence,

1

t|α|j

≤ C(1 + |ξ|)−|α|

for ξ ∈ supp ∂αξ θ(ξ/tj) and for all j, where the constant C is independent of j. Thus,

|∂αξ θ(ξ/tj)| =1

t|α|j

|(∂αξ θ)(ξ/tj)|

≤ C(1 + |ξ|)−|α| supξ|(∂αξ θ)(ξ)|

≤ Cα(1 + |ξ|)−|α| ,

where the constant Cα is again independent of j. Notice that the above estimate holds true alsowhen α = 0.

108 M. M. PELOSO

Therefore33, for |α|+ |β| ≤ j and x ∈ B(0, Rj),∣∣∂βx∂αξ (θ(ξ/tj)aj(x, ξ))∣∣ =∣∣∣∑γ≤α

(αγ

)∂γξ θ(ξ/tj)∂

βx∂

α−γξ aj(x, ξ)

∣∣∣≤∑γ≤α

(αγ

)Cγ(1 + |ξ|)−|γ|Cα,γ(1 + |ξ|)mj−(|α|−|γ|)

≤ Cj(1 + |ξ|)mj−|α| ,

for some constant Cj = maxCα,β : |α| + |β| ≤ j, |x| ≤ Rj , |α| + |β| ≤ j. Recall that θ(ξ/tj)is supported in |ξ| ≥ tj and that mj −∞. Inductively, having chosen tj−1 we select tj insuch a way that

|ξ| ≥ tj implies Cj(1 + |ξ|)mj ≤ 2−j(1 + |ξ|)mj−1 .

This establish the claim and prove the proposition. 2

9.4. Adjoints and products of pseudodifferential operators. In order to compute theadjoint of a pseudodifferential operator we need to rely on the more general definition of symbola(x, y, ξ) given in Definition 9.1.

Theorem 9.20. Let a ∈ Sm and Pa properly supported. Let

b(x, ξ) = e−2πixξPa(e2πi(·)ξ)(x)

be as in Theorem 9.15. Then b admits asymptotic expansion

b(x, ξ) ∼∞∑j=0

bj(x, ξ) ,

where

bj(x, ξ) =∑|α|=j

1

α!∂αξ D

αy a(x, y, ξ)∣∣

y=x

.

Proof. From Theorem 9.15 we know that b is well defined and is in Sm. We need to show thatb has the prescribed asymptotic expansion.

We use the notation from the proof of Theorem 9.15. From equation (9.3) we know that

b(x, ξ) =

∫F2σ(x, η, η + ξ) dη ,

and from (9.4) that∣∣∂βx∂αξ F2σ(x, η, ξ)∣∣ ≤ CK,α,β,N (1 + |ξ|)m−|α|(1 + |η|)−N ,

for x ∈ K, K compact, and any N > 0.Observe that this inequality implies that, for |α| = j,∣∣∣∣∂βx∂α′ξ ∫ ∂αξ F2σ(x, η, ξ)ηα dη

∣∣∣∣ ≤ CK,α,α′,β,N (1 + |ξ|)m−j−|α′|∫

(1 + |η|)−N |η|j dη

≤ C(1 + |ξ|)m−j−|α′| , (9.11)

33We use the product formula for higher order partial derivatives.

that is, ∫∂αξ F2σ(x, η, ξ)ηα dη ∈ Sm−j for |α| = j .

Next, we expand the function ξ 7→ F2σ(x, η, ξ):∣∣∣∣F2σ(x, η, ξ + η)−∑|α|<k

∂αξ F2σ(x, η, ξ)

α!ηα∣∣∣∣

≤ ck sup|α|=k, 0≤t≤1

|η|k∣∣∣∂αξ F2σ(x, η, ξ + tη)

∣∣∣≤ CK,k,N sup

0≤t≤1|η|k(1 + |ξ + tη|)m−k(1 + |η|)−N .

If |η| < 12 |ξ| (taking N = k) we have∣∣∣∣F2σ(x, η, ξ + η)−

∑|α|<k


α!ηα∣∣∣∣ ≤ C ′(1 + |ξ|)m−k ,

and if |η| ≥ 12 |ξ| (taking N large) we have∣∣∣∣F2σ(x, η, ξ + η)−

∑|α|<k


α!ηα∣∣∣∣ ≤ C”(1 + |η|)−n−k .

Integrating, we have∣∣∣∣ ∫ F2σ(x, η, ξ + η) dη −∑|α|<k

1

α!

∫∂αξ F2σ(x, η, ξ)ηα dη

∣∣∣∣=

∣∣∣∣b(x, ξ)− ∑|α|<k

1

α!


∣∣∣∣≤ C ′

∫|η|< 1

2|ξ|

(1 + |ξ|)m−k dη + C”

∫|η|≥ 1

2|ξ|

(1 + |η|)−n−k dη

≤ Ck(

(1 + |ξ|)m+n−k + (1 + |ξ|)−k)

≤ Ck(1 + |ξ|)mk .

Notice that mk −∞, as k → +∞.A repeatition of the same argument shows that, for all multi-indices α, β and K compact,

there exists C = Cα,β,K such that, for all x ∈ K, ξ ∈ Rn,∣∣∣∣∂βx∂αξ b(x, ξ)− ∑|α|<k

1

α!∂βx∂

αξ


∣∣∣∣ ≤ C(1 + |ξ|)mk−|α| .

Therefore, since mk −∞, given any M > 0, we can find k = k(M) in such a way thatmk < −M and

b(x, ξ)−k−1∑j=0

∑|α|=j

1

α!

∫∂αξ F2σ(x, η, ξ)ηα dη ∈ S−M .

110 M. M. PELOSO

Thus,

b(x, ξ) ∼+∞∑j=0

∑|α|=j

1

α!

∫∂αξ F2σ(x, η, ξ)ηα dη .

Finally, notice that∫∂αξ F2σ(x, η, ξ)ηα dη =

[Dαy

∫e2πiyη∂αξ F2σ(x, η, ξ) dη

]y=0

=[Dαy ∂

αξ σ(x, y, ξ)

]y=0

=[Dαy ∂

αξ a(x, y, ξ)

]y=x

.


Definition 9.21. Let a ∈ Sm, Pa be the corresponding pseudodifferential operator. Assumethat Pa is properly supported. Then, we define the transpose tPa and the adjoint P ∗a resp., as

〈Paϕ,ψ〉 = 〈ϕ, tPaψ〉 , and 〈Paϕ|ψ〉 = 〈ϕ|P ∗aψ〉 , resp. .

Lemma 9.22. Let a ∈ Sm. Then the transpose tPa of Pa is the pseudodifferential operatorPσ where σ(x, y, ξ) = a(y,−ξ), while the adjoint P ∗a is the pseudodifferential operator Pτ where

τ(x, y, ξ) = a(y, ξ).

Proof. We only present the case of tPa since the other case is essentially identical.For ϕ,ψ ∈ D (recall that Paϕ is only C∞, so we need at least ψ to be smooth and with

compact support to define the pairing and have convergent integrals), we write

〈Paϕ,ψ〉 =x

e2πixξa(x, ξ)ϕ(ξ)ψ(x) dξdx

=

∫ϕ(ξ)g(ξ) dξ

=

∫ϕ(y)g(y) dy = 〈ϕ, g〉 ,

where g is given by

g(ξ) =

∫e2πixξa(x, ξ)ψ(x) dx

and so g, and therefore g too, is a Schwartz function. For, g is clearly C∞. Moreover, for everyα,

ξαg(ξ) =

∫ (Dαxe

2πixξ)a(x, ξ)ψ(x) dx

= (−1)|α|∫e2πixξDα

x

(a(x, ξ)ψ(x)

)dx ,

and this last integral is bounded since ψ has compact support. Similar reasoning shows that allthe derivatives of g decay rapidly.

Therefore,

tPaψ(y) = g(y) =x

e2πi(x−y)ξa(x, ξ)ψ(x) dxdξ

=x

e2πi(y−x)ξa(x,−ξ)ψ(x) dxdξ ,


and the lemma is proved. 2

Theorem 9.23. Let a ∈ Sm and suppose Pa is properly supported. Denote by aP ′ and aP ∗ thesymbols of tPa and P ∗a resp. Then, aP ′ , aP ∗ ∈ Sm and we have

aP ′(x, ξ) ∼∑|α|≥0

(−1)|α|

α!Dαx∂

αξ a(x,−ξ)

aP ∗(x, ξ) ∼∑|α|≥0

1

α!Dαx∂

αξ a(x, ξ) .

Proof. Now this follows at once. Let’s consider the case of a tP . By the lemma we know thattPa = Pσ, where σ(x, y, ξ) = a(y,−ξ). Since Pa is properly supported, so is Pσ. By Theorems9.20 and 9.15 we then have that, modulo smoothing operators, Pσ = Pa tP , where

aP ′(x, ξ) ∼∑|α|≥0

(−1)|α|

α!Dαx∂

αξ a(x,−ξ) .


Theorem 9.24. Let a ∈ Sm, b ∈ Sm′. Then PaPb ∈ Ψm+m′ and its symbol σ = σPaPb has

asymptotic expansion

σ(x, ξ) ∼∑|α|≥0

1

α!∂αξ a(x, ξ)Dα

x b(x, ξ) .

Proof. Since P = (P ′)′, we can write

Pϕ(x) =x

e2πi(x−y)ξaP ′(y,−ξ)ϕ(y) dydξ ,

so that

Pϕ(ξ) =

∫e−2πiyξaP ′(y,−ξ)ϕ(y) dyd .

Therefore,

PbPaϕ(x) =

∫e2πixξb(x, ξ)Pϕ(ξ) dξ =

xe2πi(x−y)ξb(x, ξ)aP ′(y,−ξ)ϕ(y) dydξ .

This means that PbPa = Pτ , where

τ(x, y, ξ) = b(x, ξ)aP ′(y,−ξ) .Clearly τ ∈ Sm+m′ . We obtain the asymptotic expansion by applying Theorem 9.15. Modulo

smoothing operators, Pτ = Pσ, where σ has asymptotic expansion

σ(x, ξ) ∼∑|α|≥0

1

α!∂αξ D

αx

[b(x, ξ)aP ′(y,−ξ)

]∣∣x=y

.

The fact that∑|α|≥0

1

α!∂αξ D

αx

[b(x, ξ)aP ′(y,−ξ)

]∣∣x=y

∼∑|α|≥0

1

α!∂αξ a(x, ξ)Dα

x b(x, ξ)

is a (fairly) elementary verification that is left to the reader. 2

112 M. M. PELOSO

Corollary 9.25. Let Pa, Pb be properly supported, with a ∈ Sm, b ∈ Sm′. Then, if σ denotesthe symbol of PaPb we have that

σ(x, ξ) = a(x, ξ)b(x, ξ) modSm+m′−1 .

9.5. Local regularity of elliptic operators. We now introduce the classical notion of anelliptic partial differential operator.

Definition 9.26. Let P =∑|α|≤m aα(x)Dα be a PDO with smooth coefficients. We say that

P is elliptic at x0 if pm(x0, ξ) 6= 0 for all ξ ∈ Rn \ 0, that is if∑|α|=m

aα(x0)ξα 6= 0, for all ξ ∈ Rn \ 0 .

Notice that, in this case there exists a constant C0 > 0 such that∣∣∣ ∑|α|=m

aα(x0)ξα∣∣∣ ≥ C0|ξ|m ,

(where C0 can be taken to be the minimum of the left hand side on the sphere |ξ| = 1).The operator P is said to be elliptic on an open set Ω if it is elliptic at every point x ∈ Ω.

In analogy with the classical definition, we introduce the notion of elliptic symbols.

Definition 9.27. Let a ∈ Sm, where m is a real number, and let Pa be the correspondingpseudodifferential operator. We say that P is elliptic if, for every compact set K ⊂ Rn thereexist positive constants C1, C2 such that

|a(x, ξ)| ≥ C1|ξ|m ,

for all x ∈ K and ξ ∈ Rn with |ξ| ≥ C2.

Now a technical lemma.

Lemma 9.28. Let a ∈ Sm be elliptic. Then, there exists χ ∈ C∞(Rn ×Rn) such that, for allcompact set K there exist constant c, C > 0 such that

(i) χ(x, ξ) = 1 for |ξ| ≥ C;

(ii) |a(x, ξ)| ≥ c|ξ|m when χ(x, ξ) 6= 0.

Proof. Exercise V.6. 2

Definition 9.29. Let P ∈ Ψ∞ be a pseudodifferential operator. We say that a properlysupported pseudodifferential operator Q is a parametrix for P if,

PQ− I, QP − I ∈ Ψ−∞ .

Theorem 9.30. If P ∈ Ψm is elliptic, then P has a parametrix Q ∈ Ψ−m.Moreover, if P is properly supported, u ∈ E ′ and Pu ∈ Hs

loc, then u ∈ Hm+sloc . In particular, if

Pu ∈ C∞(Ω) for some open set Ω, then u ∈ C∞(Ω).

Proof. Let P = Pa where a ∈ Sm is elliptic, and let χ be as in Lemma 9.28. We set

b0(x, ξ) =

χ(x,ξ)a(x,ξ) if χ(x, ξ) 6= 0

0 if χ(x, ξ) = 0 .

It is clear that b0 is well defined. Indeed, having fixed a compact set K ⊂ Rn, letting x vary

in K we have that |a(x, ξ)| ≥ cK |ξ|m when χ(x, ξ) 6= 0. Moreover, b0(x, ξ) =[a(x, ξ)

]−1when

|ξ| ≥ CK , since χ(x, ξ) = 1 when |ξ| ≥ CK .For the time being, we work on a fixed compact set K ⊂ Rn and let x ∈ K. All the constants

that appear will depend on K and the equality holds for x ∈ K, even if not explicitely indicated.To complete the proof it will then suffice to repeat the argument for compact sets Kj Rn.

Now we wish to estimate∣∣∂βx∂αξ b0(x, ξ)

∣∣ as |ξ| → +∞. Thus, it suffices to compute the above

derivatives for |ξ| ≥ CK .Notice that, avoiding to write the arguments of the various functions, ∂ξja

−1 = −a−2∂ξja,

∂2ξjξk

a−1 = a−3[2∂ξja∂ξka+ a∂2

ξjξka].

Inductively, one can easily show that

∂αξ[a(x, ξ)

]−1=[a(x, ξ)

]−1−|α|A(x, ξ) ,

where A is a finite sum of the form

∂γ1ξ a(x, ξ) · · · ∂γkξ a(x, ξ) ,

where |γ1| + · · · + |γk| = |α|. Notice that necessarly k ≤ |α|. Analogous formula holds for thepartial derivatives in x.

We now restrict ourselves to the case m ≥ 0. The case m < 0 can be obtained along the samelines. We leave the easy checking to the reader.

Then, using the fact that∣∣[a(x, ξ)]−1

∣∣ ≤ C(1+ |ξ|)−m (or ≤ C(1+ |ξ|)−|m| in the general case)for |ξ| ≥ CK and the above remark, we have∣∣∂βx∂αξ b0(x, ξ)

∣∣ =∣∣∣∂βx∂αξ [a(x, ξ)

]−1∣∣∣

≤ C(1 + |ξ|)−m(1+|α|)∑

|γ1|+···+|γk|=|α|

(1 + |ξ|)m−|γ1| · · · (1 + |ξ|)m−|γk|

≤ C(1 + |ξ|)−m(1+|α|)(1 + |ξ|)m|α|−|α|

≤ C(1 + |ξ|)−m−|α| .

This shows that b0 ∈ S−m.Moreover,

b0(x, ξ)a(x, ξ) =

χ(x, ξ) if χ(x, ξ) 6= 0

0 if χ(x, ξ) = 0

= 1− r0(x, ξ) ,

where r0 ∈ S−∞ as it is easy to check, using the fact that supp r0 ⊂ ∪KRnK × |ξ| ≤ CK.Now, it follows from Theorem 9.24 that the symbol of the operator Pb0Pa, denote it by σPb0Pa ,

satisfiesσPb0Pa(x, ξ) = b0(x, ξ)a(x, ξ)− r′1(x, ξ) ,

where r′1 ∈ S−1, that is, σPb0Pa = 1− r1 with r1 ∈ S−1.

114 M. M. PELOSO

Next we set

b1(x, ξ) =

χ(x,ξ)r1(x,ξ)

a(x,ξ) if χ(x, ξ) 6= 0

0 if χ(x, ξ) = 0 .

Then an argument similar as the one above shows that b1(x, ξ) ∈ S−m−1 and that

b1(x, ξ)a(x, ξ) = (1− χ(x, ξ))r1(x, ξ) = r1(x, ξ)− r′2(x, ξ) ,

where r′2 ∈ S−2.Again, denote by σ(Pb0+Pb1 )Pa the symbol of (Pb0 + Pb1)Pa. By Theorem 9.24 we have that

σ(Pb0+Pb1 )Pa(x, ξ) =(1− r1(x, ξ)

)+(r1(, ξ)− r2(x, ξ)

)= 1− r2(x, ξ) ,

where r2 ∈ S−2.Inductively, having picked k, one constructs bj with j ≥ 2 with bj ∈ Sm−j for j < k in such a

way thatσ(Pb0+Pb1+···+Pbj )Pa(x, ξ) = 1− rj+1

where rj+1 ∈ S−j−1, for all j < k.Then, by the asymptotic expansion of symbols Theorem 9.15 we obtain there exists a properly

supported b such that b ∼∑

j bj , b ∈ S−m and such that, for all k,

σPbPa = σ(Pb0+Pb1+···+Pbk )Pa(x, ξ) = 1− rk+1 ∈ S−k−1 .

This proves that PbPa − I ∈ Ψ−∞,With the same procedure, we can construct b′ ∈ Sm such that PaPb′ = I modulo Ψ−∞.But then also,

PaPb − I = (PaPb′ − I)− PaPb′ + PaPbPaPb′ + PaPb − PaPbPaPb′= (PaPb′ − I) + Pa(PbPa − I)Pb′ − PaPb(PaPb′ − I)

∈ Ψ−∞ .

Then Pb is a parametrix for Pa. This proves the first part of the theorem.The second and third part follow immediately from Theorem 9.17 (Exercise V.7). 2

As a corollary we also obtain the following classical result.The elliptic partial differential operators enjoy of the so-called local regularity property. More

precisely, they satisfy the following apriori local estimate, which follows from Theorem 9.30.

Theorem 9.31. Let Ω be a bounded open set and let P =∑|α|≤m aα(x)Dα be a PDO with

smooth coefficients, elliptic in a neighborhood of Ω. Then, for every s ∈ R there exists aconstant C = Cs > 0 such that for all u ∈ C∞c (Ω) we have

‖u‖Hs ≤ Cs(‖Pu‖Hs−m + ‖u‖Hs−1

).

Proof. Exercise V.8. 2


10. Sum of squares of vector fields

We now turn our attention to operators with variable coefficients that are not elliptic, butshare some regularity properties with them.

Finally, we study the famous Hormander’s theorem on the hypoellipticity of sums of squaresof vector fields.

We fix an open set Ω in Rn. A vector field (with smooth coefficients) is a first order PDOwith no constant term

X =

n∑j=1

aj(x)∂xj .

We will (typically) assume that the coefficients aj are real-valued. (In case we are interested incomplex-valued coefficients, we will refer to them as complex-vector fields.)

We consider the PDO P defined as

P =

k∑j=1

X2j +X0 + b(x) , (10.1)

where X0, . . . , Xk are (real-valued) vector fields and b is a smooth real-valued function.

Definition 10.1. The commutator of two vector fieldsX =∑n

j=1 aj(x)∂xj and Y =∑n

j=1 bj(x)∂xjis the vector field, denoted by [X,Y ], defined by the action on a smooth function f

[X,Y ](f) = (XY − Y X)(f) .

It is easy to check that indeed (XY − Y X) is a PDO of first order, with no constant termand that

[X,Y ](f)(x) = (XY − Y X)(f)(x) = X

( n∑j=1

bj(x)∂xjf(x)

)− Y

( n∑j=1

aj(x)∂xjf(x)

)

=

n∑j,k=1

ak(x)(∂xkbj(x)

)∂xjf(x)−

n∑j,k=1

bk(x)(∂xkaj(x)

)∂xjf(x)

=

n∑j=1

( n∑k=1

ak(x)∂xkbj(x)− bk(x)∂xkaj(x)

)∂xjf(x) ,

since the second order terms cancel out.

Definition 10.2. We denote by X1 the collection of the Xj ’s, j = 0, 1, . . . , k and inductively, byXm the set of Xm−1 and the collection of the commutators [X,Y ], where X ∈ X1 and Y ∈ Xm−1.

We say that the operator P as in (10.1) is of finite type of order m in Ω if Xm spans the wholetangent space of Rn at every point in Ω.

Theorem 10.3. (Hormander) Let P be the operator in (10.1). Suppose that P is of finitetype at every point in Ω. Then P is hypoelliptic in Ω.

This result follows by the next, more precise, theorem.

116 M. M. PELOSO

Theorem 10.4. (Hormander) Let P be the operator in (10.1). Suppose that Xm spans thetangent space at every point in Ω. Then for every s ∈ R and N > 0 there exists a constant Cs,Nsuch that

‖u‖Hs+ε ≤ Cs,N(‖Pu‖Hs + ‖u‖H−N

),

for every u ∈ C∞c (Ω), where ε = 21−m.

Typical examples of operators to which these theorems apply are the heat operator P =∑nj=1 ∂

2xj − ∂t and the Grushin operator in R2, P = ∂2

x + x2∂2y . Another classical example is

the so-called sub-Laplacian L on the Heisenberg group. The Heisenberg group H is identifiedwith Rn ×Rn ×R, with coordinates (x, y, t), and the operator L = −1

2

∑nj=1X

2j + Y 2

j , where

Xj = ∂xj −y2∂t, Xj = ∂yj + x

2∂t, and T = ∂t. Then, it is easy to see that [Xj , Yj ] = −T , so thatthe Hormander condition is satisfied with m = 2.

We write Xj =∑n

`=1 aj`∂x` , and by shrinking the domain Ω, we may assume that b and

aj,` ∈ C∞(Ω) for all j, `.

Lemma 10.5. Let P be defined as in (10.1). Then there exists C > 0 such that

k∑j=1

‖Xju‖2 ≤ C(|(Pu|u)|+ ‖u‖2

),

for all u ∈ C∞c (Ω).

Proof. We denote by X∗j the adjoint of Xj (w.r.t. the L2-inner product). It is easy to check that

(recall that the Xj are real-valued) X∗j = −Xj + hj , where hj = −∑n

`=1 ∂x`aj`. Integrating byparts we see that

−(X2j u|u) = −(Xju|X∗j u) = (Xju|Xju)− (Xju|hju)

≤ ‖Xju‖2 + C‖Xju‖ ‖u‖= ‖Xju‖2 +O

(‖Xju‖ ‖u‖

),

while

(X0u|u) = −(u|X0u) +O(‖u‖2) ,

so that

Re (X0u|u) = O(‖u‖2) .

Therefore,

k∑j=1

‖Xju‖2 = −k∑j=1

(X2j u|u) +O

(‖Xju‖ ‖u‖

)= −Re (Pu|u) +O

( k∑j=1

‖Xju‖ ‖u‖+ ‖u‖2). (10.2)


Using the estimate ab ≤ ε2

2 a2 + 1

2ε2b2, valid for all a, b ∈ R, ε > 0, we see that

O( k∑j=1

‖Xju‖ ‖u‖)≤ C

k∑j=1

‖Xju‖ ‖u‖

≤ C

2

(ε2

k∑j=1

‖Xju‖2 +1

ε2‖u‖2

)

=Cε2

2

k∑j=1

‖Xju‖2 + C ′‖u‖2 .

By choosing ε > 0 suitably small, we can take part of the right hand side of (10.2) on the leftand obtain the result. 2

Lemma 10.6. Let L be a PDO operator with smooth coefficients of order `, s ∈ R. Then theoperators LΛs and ΛsL are psuedodifferential operators of order ` + s, while the commutator[Λs, L] is a psuedodifferential operator of order `+ s− 1, in the sense that

LΛs, ΛsL : Htloc → H

t−(`+s)loc ,

while[Λs, L] : Ht

loc → Ht−(`+s−1)loc

for all t ∈ R.

Proof. This follows at once from the results in Section 5. 2

Theorem 10.7. If Xm spans the tangent space at every point in Ω, then there exist ε > 0 andC > 0 such that

‖u‖2Hε ≤ C( k∑j=0

‖Xju‖2 + ‖u‖2),

for all u ∈ C∞c (Ω). Here we can take ε = 21−m.

Proof. We denote by Zm a generic element of Xm. Then,

‖u‖2Hε ≤ C( k∑j=0

‖Dju‖2Hε−1 + ‖u‖2)≤ C

( ∑Zm∈Xm

‖Zmu‖2Hε−1 + ‖u‖2),

where the last sum is over finitely many terms. Therefore, it suffices to show that, for eachZm ∈ Xm

‖Zmu‖2Hε−1 ≤ C( k∑j=0

‖Xju‖2 + ‖u‖2),

for some ε > 0.If m = 1 we can take ε = 1. Let then m ≥ 2 and assume the estimate valid for m − 1 and

ε = 21−(m−1). We may assume that Zm = XZm−1 − Zm−1X, with X ∈ X1. Then,

‖Zmu‖2Hε−1 = (Zmu|Λ2ε−2Zmu)

= (XZm−1u|Λ2ε−2Zmu)− (Zm−1Xu|Λ2ε−2Zmu) , (10.3)

118 M. M. PELOSO

and we estimate these last two terms separately.By Lemma 10.6, using the fact that 2ε− 1 ≤ 0, we have

|(XZm−1u|Λ2ε−2Zmu)| = |(Zm−1u|XΛ2ε−2Zmu)|+O(|(Zm−1u|Λ2ε−2Zmu)|

)= |(Zm−1u|XΛ2ε−2Zmu)|+O

(|(Λ2ε−1Zm−1u|Λ−1Zmu)|

)= |(Zm−1u|XΛ2ε−2Zmu)|+O

(‖Zm−1u‖H2ε−1‖u‖

)= |(Zm−1u|Λ2ε−2ZmXu)|+O

(‖Zm−1u‖H2ε−1‖u‖

)≤ C

(‖Xu‖2 + ‖Zm−1u‖2H2ε−1 + ‖u‖2

).

Next,

|(Zm−1Xu|Λ2ε−2Zmu)| = |(Xu|Zm−1Λ2ε−2Zmu)|+O(‖u‖‖Xu‖

)≤ C

(‖Xu‖2 + ‖Zm−1u‖2H2ε−1 + ‖u‖2

).

Substituting into (10.3) and using the induction hypothesis, by taking ε = 21−m, we obtain

‖Zmu‖2Hε−1 ≤ C( k∑j=0

‖Xju‖2 + ‖Zm−1u‖2H2ε−1 + ‖u‖2)

≤ C( k∑j=0

‖Xju‖2 + ‖Zm−1u‖2H21−(m−1)−1

+ ‖u‖2)

≤ C( k∑j=0

(‖Xju‖2 + ‖u‖2

).


We now prove

Theorem 10.8. If Xm spans the tangent space at every point in Ω, then there exist ε > 0 andC > 0 such that

‖u‖2Hε ≤ C(‖Pu‖2 + ‖u‖2

),

for all u ∈ C∞c (Ω). Here we can take ε = 21−m.

Proof. The proof is quite simple in the case that P =∑k

j=1X2j , i.e. there is no first order term

X0 and no zero order term b in P .Under these assumptions it suffices to combine Lemma 10.5 and Theorem 10.7:

‖u‖2Hε ≤ C( k∑j=0

(‖Xju‖2 + ‖u‖2

)≤ C

(|(Pu|u)|+ ‖u‖2

)≤ C

(‖Pu‖2 + ‖u‖2

).

This proves the theorem in the special case P =∑k

j=1X2j .

For the general case we refer to [?] Thm. 8.2.6. 2

We now “boot-strap” the argument of the previous theorem to prove


Theorem 10.9. Let Xm span the tangent space at every point in Ω. Then, for every s ∈ R andN > 0 there exists C = Cs,m,N > 0 such that

‖u‖2Hs+ε ≤ C(‖Pu‖2Hs + ‖u‖2H−N

),

for all u ∈ C∞c (Ω). Here, again, ε = 21−m.

Proof. We wish to apply Thm. 10.8 to Λsu. We assume that the estimates in Thm. 10.8 holdsfor functions u in C∞c (Ω′), with Ω′ open, Ω′ ⊆ Ω.

Let ϕ,ϕ1 be smooth cut-off functions supported in Ω, ϕ = 1 on Ω′ and ϕ1 = 1 on suppϕ.We now claim that for each s ∈ R and N > 0 there exists C > 0 such that for all v ∈ S

‖(1− ϕ1)Λs(ϕv)‖ ≤ C‖v‖H−N .

This depends on the fact that the distribuitions 1 − ϕ1 and ϕv have disjoint supports, so that(1− ϕ1)Λs(ϕv) ∈ Ψ−∞.

Next, we assume the claim. Notice that u = ϕu (since (1− ϕ)u = 0). We then have

‖u‖Hs+ε = ‖Λsu‖Hε

≤ ‖ϕ1Λsu‖Hε + ‖(1− ϕ1)Λs(ϕu)‖Hε

≤ C(‖P (ϕ1Λsu)‖+ ‖ϕ1Λsu‖+ ‖u‖H−N

)≤ C

(‖[P,ϕ1Λs]u)‖+ ‖Pu‖Hs + ‖u‖Hs + ‖u‖H−N

). (10.4)

Notice that [P,ϕ1Λs] is a pseudodifferential operator that can be written as

[P,ϕ1Λs] =

k∑j=1

QsjXj +Qsk+1 (10.5)

where we denote by Qs, Qsj etc. a generic element in Ψs. For, by Thm. 9.24 we have

[P,ϕ1Λs] =[ k∑j=1

X2j +X0 + b, ϕ1Λs

]

=k∑j=1

[X2j , ϕ1Λs

]+Qs ,

where Qs ∈ Ψs with the convention fixed above. Now,[X2j , ϕ1Λs

]= X2

j ϕ1Λs − ϕ1ΛsX2j = X2

j ϕ1Λs −Xjϕ1ΛsXj +Xjϕ1ΛsXj − ϕ1ΛsX2j

= Xj [Xj , ϕ1Λs] + [Xj , ϕ1Λs]Xj = [Xj , ϕ1Λs]Xj +[Xj , [Xj , ϕ1Λs]

]+ [Xj , ϕ1Λs]Xj

= QsjXj +Q′sj .

This proves (10.5). As a consequence,

‖[P,ϕ1Λs]u‖ ≤ C( k∑j=1

‖Xju‖Hs + ‖u‖Hs

). (10.6)

120 M. M. PELOSO

Next, applying Lemma 10.5 and using (10.6) we have,

‖Xju‖2Hs = ‖ΛsXju‖2 ≤ C(‖ϕΛsXju‖2 + ‖(1− ϕ)Λsϕ1Xju‖2

)≤ C

(‖XjϕΛsu‖2 + ‖[Xj , ϕΛsu]‖2 + ‖u‖2H−N

)≤ C

(∣∣(P (ϕΛsu)|ϕΛsu)∣∣+ ‖u‖2Hs + ‖u‖2H−N

)≤ C

(∣∣(ϕΛsP (u)|ϕΛsu)∣∣+

∣∣([P,ϕΛs](u)|ϕΛsu)∣∣+ ‖u‖2Hs + ‖u‖2H−N

)≤ C

(‖Pu‖Hs‖u‖Hs +

( k∑j=1

‖Xj‖Hs + ‖u‖Hs

)‖u‖Hs + ‖u‖2Hs + ‖u‖2H−N

)

≤ C( k∑j=1

‖Xj‖Hs

)‖u‖Hs + C

(‖Pu‖2Hs + ‖u‖2Hs + ‖u‖2H−N

). (10.7)

Therefore, using the “small constant-big constant” argument we obtain,

k∑j=1

‖Xju‖2Hs ≤ C( k∑j=1

‖Xj‖Hs

)‖u‖Hs + C

(‖Pu‖2Hs + ‖u‖2Hs + ‖u‖2H−N

)≤ Cε

k∑j=1

‖Xj‖2Hs + C(‖Pu‖2Hs + ‖u‖2Hs + ‖u‖2H−N

),

so that,k∑j=1

‖Xju‖Hs ≤ C(‖Pu‖Hs + ‖u‖Hs + ‖u‖H−N

),

for some suitable large constant C > 0. Inserting the above estimate and the one from (10.6)into (10.4) we see that

‖u‖Hs+ε ≤ C(‖Pu‖Hs + ‖u‖Hs + ‖u‖H−N

). (10.8)

In order to conclude the proof it suffices to show that for any δ > 0 there exists a constantC = Cε,s,N,δ such that

‖u‖Hs ≤ δ‖u‖Hs+ε + C‖u‖H−N ,

and then choose δ > 0 sufficiently small in order to bring the term δ‖u‖Hs on the left hand sideof (10.8). This concludes the proof of the theorem. 2

Appendix A. Some results on Lp spaces

We begin by recalling a few basic facts about the Lp spaces in Rn.For proofs and further details of the facts and ideas presented here, we refer to [WZ], or also

[Fo1] and [Ru], for example.

We denote by Lp = Lp(Rn) the space of Lebesgue measurable functions, which are p-integrablefunctions, if 0 ≤ p <∞, and for p =∞ essentially bounded. These norms are

‖f‖p =

(∫Rn

|f(x)|p dx)1/p

‖f‖∞ = infα≥0

∣∣x ∈ Rn : |f(x)| > α∣∣ = 0 .

It is clear that Lp are vector spaces, since

|f + g|p ≤ 2p[|f |p + |g|p

].

In order to prove that ‖ · ‖p is indeed a norm we need to prove the triangle inequality. Webegin by showing that when 0 |A|1/p + |B|1/p = ‖χA‖p + ‖χB‖p .

We now prove some basic inequalities concerning the Lp-norms. For p ≥ 1 we denote by p′ itsconjugate exponent (i.e. 1/p+ 1/p′ = 1, which gives p′ = p/(p− 1) if p > 1, p = 1 and p′ =∞being conjugate exponents).

Theorem A.1. (Holder’s inequatlity) Let 1 < p <∞, and let p′ be its conjugate exponent. If fand g are measurable functions, we have

‖fg‖1 ≤ ‖f‖p‖g‖p′ .

Equality holds if and only if α|f |p = β|g|p′ for some constants α and β.

Proof. The result is obvious if either ‖f‖p = 0 or ‖g‖p′ = 0. In fact, in this case either f = 0a.e. or g = 0 a.e., which gives fg = 0 a.e.

The result is also clear if either ‖f‖p =∞ or ‖g‖p′ =∞.Next we claim that for a, b ≥ 0 and 0 < λ < 1 then

aλb1−λ ≤ λa+ (1− λ)b . (A.1)

For, suppose b = 0, then the inequality holds. If b 6= 0, setting t = a/b, inequality (A.1) becomes

tλ ≤ λt+ (1− λ) .

Let ϕ(t) = tλ − λt, we obtain ϕ′(t) = λ(t(λ−1) − 1

), which is > 0 for 0 < t < 1 and < 0 when

t > 1. Thus, the maximum is attained when t = 1 with value 1 − λ. This proves (A.1) andequality holds only if t = 1, i.e. a = b.

Now suppose that both ‖f‖p and ‖g‖p′ are different from 0. We use the claim above with

a =(|f(x)|/‖f‖p

)p, b =

(|g(x)|/‖g‖p′

)p′and λ = 1/p .

We obtain|f(x)||g(x)|‖f‖p‖g‖p′

≤ 1

p

|f(x)|‖f‖p

+1

p′|g(x)|‖g‖p′

.

122 M. M. PELOSO

Integrating both sides of the inequality gives

1

‖f‖p‖g‖p′

∫|f(x)||g(x)| dx ≤ 1

p+

1

p′= 1 .

This proves the inequality. Finally equality holds if and only if a = b, i.e.(|f(x)|/‖f‖p

)p=(|g(x)|/‖g‖p′

)p′,

that is α|f |p = β|g|p′ . 2

Theorem A.2. (Minkowski’s inequatlity) Let 1 ≤ p ≤ ∞, and let p′ be its conjugate exponent.If f and g are measurable functions, we have

‖f + g‖p ≤ ‖f‖p + ‖g‖p′ .

Proof. The result is clear if p = 1 or p = +∞, and also if f + g = 0.Suppose that 1 < p <∞, and f + g 6= 0 a.e. Then

|f + g|p ≤(|f |+ |g|

)|f + g|p−1

and using Holder’s inequality with conjugate exponents p and p′ we obtain∫|f + g|p dx ≤ ‖f‖p

∥∥|f + g|p−1∥∥p′

+ ‖g‖p∥∥|f + g|p−1

∥∥p′

=(‖f‖p + ‖g‖p

)∥∥|f + g|p−1∥∥p′

=(‖f‖p + ‖g‖p

)(∫|f + g|p dx

)1/p′

.

Since 1/p′ = 1− 1/p it follows that

‖f + g‖p ≤ ‖f‖p + ‖g‖p′ . 2

Theorem A.3. For 1 ≤ p ≤ +∞, Lp is a Banach space.

Proof. We need to prove the completeness. We begin with the case p = +∞. Let fk be aCauchy sequence in L∞. Then

|fk(x)− fm(x)| ≤ ‖fk − fm‖∞for all x outside a set Ek,m of measure 0. If E = ∪∞k,m=1Ek,m, then |E| = 0 and

|fk(x)− fm(x)| ≤ ‖fk − fm‖∞for all x 6∈ E.

Thus, fk converges uniformly to a bounded limit f outside a set of measure 0, i.e. ‖fk −f‖∞ → 0.

Now let 1 ≤ p <∞. We use the following characterization of Banach spaces: A normed spaceX is complete if and only if every series converging absolutely it converges in norm (that is, if∑

k ‖xk‖X <∞ then there exists x ∈ X such that limN→+∞ ‖∑N

k=1 xk − x‖X = 0).

Let fk be a sequence of functions in Lp such that∑

k ‖fk‖Lp = C <∞. Let gN =∑N

k=1 |fk|and g =

∑∞k=1 |fk|. By the monotone convergence theorem

‖g‖pp = limN→+∞

∫|gN (x)|p dx ≤ Cp .

Then g ∈ Lp, and in particular g(x) <∞ a.e. This implies that the series∑

k fk converges a.e.to a function f . For such a function, |f | ≤ g, so that f ∈ Lp. Finally,

|f −N∑k=1

fk|p ≤ (g + g)p = 2pgp ,

so that by the dominated convergence theorem

limN→+∞

∥∥f − N∑k=1

fk∥∥pp

= limN→+∞

∫ ∣∣f(x)−N∑k=1

fk(x)∣∣p dx = 0 ,

that is∑

k fk converges to f in the Lp-norm. 2

Proposition A.4. For 1 ≤ p <∞, the set of simple functions f : f =∑N

k=1 αkχEk , with |Ek|<∞, for all k is dense in Lp.

Proof. We first show that if f is measurable, then there exists a sequence of simple functionssm such that |s1| ≤ |s2| ≤ · · · ≤ |f |, sm → f pointwise, and sm → f uniformly on any set wheref is bounded. By writing f = f+ − f− we may assume that f ≥ 0.

Let m be a positive integer and let 0 ≤ k ≤ 22m − 1. It suffices to define

Ekm = f−1((k2−m, (k + 1)2−m]

), and Fm = f−1

((2m,+∞]

),

and set

sm =22m−1∑k=0

k2−mχEkm + 2mχFm .

It is not difficult to check that 0 ≤ sm ≤ sm+1 ≤ f for all m = 1, 2, . . . and that 0 ≤ f−sm ≤ 2−m

on the set where f ≤ 2m. The claim now follows.

Now, let f ∈ Lp, and let sm be a sequence of simple measurable functions convergingpointwise a.e. to f , |sm| ≤ f . Then, sm ∈ Lp for all m and |f − sm|p ≤ 2p|f |p, which is afunction in L1. Then, by the dominated converge theorem, ‖f − sm‖p → 0 as m→ +∞. 2

Proposition A.5. For 1 ≤ p < ∞ the set of continuous functions with compact support Cc isdense in Lp.

Proof. Since the simple functions are dense in Lp, it suffices to show that for any measurable setE, with |E| <∞, we can approximate χE in the Lp-norm with compactly suported, continuousfunctions. Given any measurable set E, |E| < ∞, we can find a (so-called Borel set) E′, withE′ ⊆ E, |E \ E′| = 0, such that for any ε > 0 there exist an open set U and a compact set Ksuch that

K ⊆ E′ ⊆ U , |U \K| < ε .

Now, by Urysohn’s lemma, we can find f ∈ Cc such that χK ≤ f ≤ χU . Therefore,

‖χE − f‖p = ‖χE′ − f‖p ≤ |U \K|1/p ≤ ε1/p . 2

Next we wish to describe the dual space of Lp, when 1 ≤ p <∞.

Theorem A.6. (Reverse Holder’s inequality) Let g be a measurable function, 1 ≤ p <∞. Then

‖g‖p′ = sup∣∣∫ fg dx∣∣ : ‖f‖p = 1

.

124 M. M. PELOSO

Proof. Set M(g) = sup∣∣ ∫ fg dx∣∣ : ‖f‖p = 1

. From Holder’s inequality we know that, if

g ∈ Lp′ , ∣∣ ∫ fg dx∣∣ ≤ ‖f‖p‖g‖p′

that is M(g) ≤ ‖g‖p′ . The statement holds true if g = 0 a.e. If g 6= 0 and p′ <∞, select

f = sgn g(|g|/‖g‖p′)p′−1

where(sgn g

)(x) = g(x)/|g(x)| if g(x) 6= 0 and equals 0 if g(x) = 0. Then,

‖f‖pp =1

‖g‖p(p′−1)

p′

∫|g(x)|p(p′−1) dx

= 1

while

M(g) ≥∣∣ ∫ fg dx

∣∣ =1

‖g‖p(p′−1)

p′

∫|g(x)|p′ dx

‖g‖p′ .

Finally, if p′ =∞, given any ε > 0, let E be a set such that

E ⊆ x : |g(x)| > ‖g‖∞ − ε, 0 < |E| <∞ .

Select

f =1

|E|sgn gχE .

Then

M(g) ≥∣∣ ∫ fg dx

∣∣ =1

|E|

∫E|g(x)| dx ≥ ‖g‖∞ − ε .


Theorem A.7. (Duality of the Lp spaces) For 1 ≤ p < ∞ the dual of Lp is Lp′, with equatity

of norms.

Proof. By Holder’s inequality is clear that any f ∈ Lp′ gives rise to a bounded linear functionalLf on Lp, 1 ≤ p <∞. Conversely, let L be a bounded linear functional on Lp, 1 ≤ p <∞. We

wish to find an f ∈ Lp′ such that

L(g) =

∫gf dx ,

that is, L = Lf . The construction of such an f goes a little beyond the scope of these notes andlectures. Therefore, we refer to [Fo1], or [WZ] for a proof. 2

We conclude this section with a few further inequalities.

Theorem A.8. (Minkowski’s integral inequality) Let 1 ≤ p ≤ ∞, f a measurable functiondefined on the product space Rn1 × Rn2. Suppose that f(·, y) ∈ Lp for a.e. y and that thefunction y 7→ ‖f(·, y)‖p ∈ L1. Then the function f(x, ·) ∈ L1 for a.e. x, the function x 7→∫f(x, y) dy ∈ Lp, and ∥∥∫ f(·, y) dy

∥∥p≤∫‖f(·, y)‖p dy .

Proof. When p = 1 this is just Fubini’s theorem on iterated integrals. Let 1 < p <∞, then weuse the duality between Lp spaces. Let p′ be the conjugate exponent of p, and g ∈ Lp′ . We havethen ∣∣ ∫ (∫ f(x, y) dy

)g(x) dx

∣∣ ≤x|f(x, y)||g(x)| dxdy

≤∫ (∫

|f(x, y)|p dx)1/p

‖g‖p′ dy

= ‖g‖p′∫‖f(·, y)‖p dy .

By taking the supremum over g ∈ Lp′ , ‖g‖p′ = 1, the statement follows. 2

Theorem A.9. (Chebyshev’s inequality) Let 1 ≤ p <∞. Then, for any λ > 0∣∣x : |f(x)| > λ∣∣ ≤ (‖f‖p

λ

)p.

Proof. Define Eλ = x : |f(x)| > λ. We have

‖f‖pp =

∫|f |p dx ≥

∫Eλ

|f |p dx ≥ λp∫Eλ

dx = λp|Eλ| .

This gives the desired inequality. 2

Appendix B. Some exercises

1. Evaluate the following limits, justifying your answers:

limk→+∞

∫ k

0

(1− x

k

)kex/2 dx lim

k→+∞

∫ k

0

(1 +

x

k

)ke−2x dx .

2. Prove Lemma 2.1.

3. (Measure of the ball and sphere in Rn.) Recall that for a, b > 0,∫Rn

e−a|x|2dx =

(π/a

)n/2, and

∫ 1

0(1− s)a−1sb−1 ds =

Γ(a)Γ(b)

Γ(a+ b),

(where the Euler’s Gamma function is defined as Γ(x) =∫∞

0 tx−1e−t dt).Use these facts and integration in polar coordinates to compute the surface measure of the

unit sphere Sn−1 in Rn and the volume of the ball of radius r, Br:

σ(Sn−1) =2πn/2

Γ(n2 ), |Br| =

πn/2

Γ(n2 + 1).

4. (Generalized Holder’s inequality.) Show that if fj ∈ Lpj , j = 1, . . . ,m and∑m

j=1 1/pj = 1

then∏mj=1 fj ∈ L1 and ∫

|m∏j=1

fj(x)| dx ≤m∏j=1

‖fj‖pj .

126 M. M. PELOSO

5. (Differentiation under the integral sign.) Let f : Rn × [a, b] be measurable and denote by(x, y) the variables in Rn ×R. Suppose ∂yf exists and suppose that there exists g ∈ L1(Rn)such that |∂yf(x, y)| ≤ |g(x)| for all (x, y) ∈ Rn × [a, b]. Show that

F (y) =

∫Rn

f(x, y) dx

is differentiable and that F ′(y) =∫Rn ∂yf(x, y) dx. Use this result to prove Theorem 2.3 (iii).

6. Show that the translation is a continuous operation in Lp when 1 ≤ p < ∞, but it is notcontinuous for p =∞.

7. Let f, g ∈ L2. Show that both sides of the equality F(f ∗ g) = f g are well defined, and provesuch an equality.

8. Let 1 ≤ p < ∞, f ∈ Lp such that ∂αx f ∈ Lp for |α| ≤ N . Show that there exists a sequenceof functions ϕk in C∞0 such that ∂αxϕk → ∂αx f in Lp, for all |α| ≤ N .

9. Prove that F(ϕ)(x) = e−2π|ξ|, where ϕ(x) = ωn(1+|x|2)−(n+1)/2 and ωn = Γ((n+1)/2

)π−(n+1)/2,

by proving: (i) for β ≥ 0,

e−β =1

π

∫ +∞

−∞

e−iβt

1 + t2dt ;

(ii) for β ≥ 0,

e−β =

∫ +∞

0

e−se−β2/4s

√πs

ds ;

(iii) set β = 2π|ξ| in (ii) and use Lemma 2.13.

10. Let G be an invertible linear transfomation of Rn and let g = f G, for f ∈ L1. Express gin terms of f (that is, prove that

g(ξ) =1

|detG|f(tG−1(ξ)

). )

11. Let δ denote the Dirac delta at the origin, that is the distribution 〈δ, ϕ〉 = ϕ(0). Prove that

∂αδ(ξ) = (2πiξ)α and that (xα) = (−2πi)−|α|∂αδ.

12. On R1, let Φ be the distribution defined as

〈Φ, ϕ〉 = limε→0

∫|x|≥ε

1

πxφ(x) dx .

Show that the equality above defines indeed a distribution and show that the limit equals∫|x|≤1

ϕ(x)− ϕ(0)

πxdx+

∫|x|>1

ϕ(x)

πxdx .

Moreover, prove that Φ = −isgn ξ. Call g = Φ. Prove that f 7→ f ∗ g extends to a unitaryoperator on L2 (called the Hilbert transform).


References

[Du] J. Duoandikoetxea, Fourier Analysis, Graduate Studies in Mathematics 29, American MathematicalSociety, Rhode Island 2001.

[Fo1] G. B. Folland, Real Analysis, Modern Techniques and Their Applications. Second Edition, Wyley & Sons,New York 1999.

[Fo2] G. B. Folland, Introduction to Partial Differential Equations. Second Edition, Princeton University Press,Princeton 1995.

[Gr] L. Grafakos, Classical Fourier Analysis, GTM 249, Springer-Verlag Ed., New York 2008.[H] L. Hormander, Anaysis of Partial Diffrential Operators, vol 1, Distribution theory and Fourier analysis,

Springer-Verlag, Berlino 2003.[Ka] Y. Katznelson, An Introduction to Harmonic Analysis, Dover, New York 1968.[Ru] W. Rudin, Functional Analysis, 2nd Ed., Mc Graw Hill, Toronto 1991.[So] P.M. Soardi, Serie di Fourier in piu variabili, Quaderni dell’Unione Matematica Italiana 26 , Pitagora

Ed., Bologna 1984.[St1] E. M. Stein, Singular Integrals and Differentiability Properties of Functions, Princeton Univ. Press, Prince-

ton 1970.[St2] E. M. Stein, Harmonic Analysis, Real-variable Methods, Orthogonality, and Oscillatory Integrals, Prince-

ton Univ. Press, Princeton 1993.[StWe] E. M. Stein, G. Weiss, Introduction to Fourier Analysis on Euclidean Spaces, Princeton Univ. Press,

Princeton 1971.[WZ] R. Wheeden, A. Zygmund, Measure and Integral, An Introduction to Real Analysis, Marcel Dekker, New

York 1977.

Dipartimento di Matematica, Universita degli Studi di Milano, Via C. Saldini 50, 20133 Milano,Italy

E-mail address: [email protected]

URL: http://wwww.mat.unimi.it/~peloso/

notes for the course analisi di fourier - unimi.itthe fourier transform 11 2.3. the schwartz space...

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