MATH 247B - HARMONIC ANALYSIS

Joseph Breen

Notes based on lectures from Math 247B at UCLA, Spring 2017, taught by Monica Visan.Last updated: May 31, 2017

Contents

1 4/3 — 4/7: Restriction of the Fourier Transform 21.1 Formulating the restriction conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Tomas-Stein for the sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Tomas-Stein for the parabaloid and Strichartz inequalities . . . . . . . . . . . . . . . . . . . . 7

2 4/10 — 4/14: Harmonic Analysis on the Torus 132.1 Basic properties of the Fourier transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Bourgain’s dispersive estimate on the torus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 4/17 — 5/1: Rearrangement Inequalities 233.1 Definitions and basic estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2 The one-dimensional Riesz rearrangement inequality . . . . . . . . . . . . . . . . . . . . . . . 293.3 Steiner symmetrization; the multi-dimensional Riesz rearrangement inequality . . . . . . . . 323.4 The Polya-Szego inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.5 The hydrogen atom energy functional . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4 5/3 — 5/8: Compactness in Lp Spaces 434.1 The Riesz compactness theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 434.2 The Strauss lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

5 5/8 — 5/19: Optimal Constants I: Gagliardo-Nirenberg 505.1 Motivation: the focusing cubic NLS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.2 The sharp Gagliardo-Nirenberg inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.3 Concentration compactness for Gagliardo-Nirenberg . . . . . . . . . . . . . . . . . . . . . . . 57

5.3.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575.3.2 The concentration compactness principle . . . . . . . . . . . . . . . . . . . . . . . . . . 59

6 5/19 — 5/26: Optimal Constants II: Sobolev Embedding 656.1 Motivation: the focusing energy-critical NLS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 656.2 Motivation: existence of bound states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 686.3 The concentration compactness principle for Sobolev embedding . . . . . . . . . . . . . . . . 72

1

1 4/3 — 4/7: Restriction of the Fourier Transform

1.1 Formulating the restriction conjecture

Fix d ≥ 2. If f ∈ L1(Rd), then f ∈ C0(Rd). In particular, it makes sense to restrict f to a set of measure 0 inRd. On the other hand, if f ∈ L2(Rd), we only know that f ∈ L2(Rd), and thus restricting to measure 0 setsis not well-defined.

We can similarly pose the question of restriction if we know that f ∈ Lp(Rd) for 1 < p < 2. Ideally,we would like a critical value of p between 1 and 2 for which restriction is well-defined for p smaller thanthis critical value, but this is unfortunately not the case. It turns out that there are functions that belong toLp(Rd) for all p > 1 but for which f blows up on a hyperplane.

Example 1.1. Let ψ : Rd−1 → C be a smooth cutoff function. Define f : Rd → C by

f(x1, . . . , xd) =ψ(x2, . . . , xd)

1 + |x1|.

Observe that ∫Rd|f(x)|p dx =

∫Rd−1

ψ(x2, . . . , xd) dx2 . . . dxd ·∫

1

(1 + |x1|)pdx1 <∞

for any p > 1, hence f ∈ Lp(Rd) for all p > 1. Computing the Fourier transform of f yields

f(ξ) = (2π)−d2

∫e−ix·ξ

ψ(x2, . . . , xd)

1 + |x1|dx

= (2π)−12 ψ(ξ2, . . . , ξd)

∫e−ix1ξ1

1 + |x1|dx1.

On the hyperplane {ξ1 = 0}, the integral∫e−ix1ξ1

1+|x1| dx1 =∫

11+|x1| dx1 is infinite, hence the restriction of f to

this set is not well-defined.

Stein discovered that if a surface S of measure 0 has “sufficient curvature,” then one may restrict f toS when f ∈ Lp(Rd) for certain values of p. Thus, we can informally state the question of restriction of theFourier transform as follows: for which zero-measure sets S and which 1 < p < 2 can one meaningfully fto S if f ∈ Lp(Rd)?

There are three surfaces which have received a lot of attention:

- The sphere: Ssphere ={ξ ∈ Rd : |ξ| = 1

}.

- The parabaloid: Sparab ={ξ ∈ Rd : ξd = 1

2 |ξ|2}

.

- The cone: Scone ={ξ ∈ Rd : ξd = |ξ|

}.

Here we are using the notation ξ = (x1, . . . , xd−1).These three surfaces are endowed with the following canonical measures:

- The sphere is given the usual surface measure dσ.

- The parabaloid is given the measure dσ defined via the integrals∫Sparab

f(ξ) dσ(ξ) :=

∫Rd−1

f

(ξ,

1

2|ξ|2)dξ.

2

- The cone is given the measure dσ defined via the integrals∫Scone

f(ξ) dσ(ξ) :=

∫Rd−1

f(ξ, |ξ|

) dξ|ξ|.

With these surfaces and measures we give a quantitative formulation of the restriction problem: we seekan estimate of the form ∥∥∥f |S∥∥∥

Lq(S,dσ).p,q,S ‖f‖Lp(Rd) .

An estimate of the form above we shall refer to as R(p→ q). We will focus on the case S = Ssphere.First, we make a few remarks.

1. If p = 1, then R(p→ q) holds for all 1 ≤ q ≤ ∞. Indeed, by Holder’s inequality,∥∥∥f |S∥∥∥Lq(S,dσ)

. σ(S)1q

∥∥∥f∥∥∥L∞(S,dσ)

. ‖f‖L1(Rd)

because S has finite measure.

2. If p = 2, then R(p→ q) fails for every 1 ≤ q ≤ ∞ by our discussion from above.

3. If R(p→ q) holds, then R(p→ q) holds for all p ≤ p and q ≤ q. Indeed, by Holder’s inequality,∥∥∥f |S∥∥∥Lq(S,dσ)

. σ(S)1q−

1q

∥∥∥f |S∥∥∥Lq(S,dσ)

.

Let ϕ be a function such that ϕ is compactly supported and ϕ = 1 on a ball containing S. Then∥∥∥f |S∥∥∥Lq(S,dσ)

.∥∥∥f |S ·ϕ∥∥∥

Lq(S,dσ).∥∥∥f ∗ ϕ∥∥∥

Lq(S,dσ). ‖f ∗ ϕ‖Lp(Rd)

where the last inequality is becauseR(p→ q) holds. Next, we apply Young’s inequality with r chosenso that 1 + 1

p = 1p + 1

r . Note that, for such a choice of r, we have 1r = 1 + p−p

pp ≤ 1 so that such a choiceis possible. Then ∥∥∥f |S∥∥∥

Lq(S,dσ). ‖f‖Lp(Rd) ‖ϕ‖Lr(Rd) . ‖f‖Lp(Rd)

as desired.

Because of this fact, in restriction problems we adopt a philosophy of maximizing p and q for whichR(p→ q) holds.

We also remark that it is sufficient to consider Schwartz functions, then invoke density to extend toLebesgue spaces. Also, in modern papers dealing with the restriction problem, the formulation is oftenpresented using the adjoint of the expression R(p → q). In particular, let R : Lp(Rd) → Lq(S, dσ) be therestriction operator given by Rf = f |S . We wish to compute the adjoint R∗ : Lq

′(S, dσ) → Lp

′(Rd). For a

Schwartz function f , we have

〈Rf, g〉L2(S,dσ) =

∫S

(Rf)(ξ)g(ξ) dσ(ξ) = (2π)−d2

∫S

∫Rde−ix·ξf(x) dx g(ξ) dσ(ξ)

=

∫Rdf(x)(2π)−

d2

∫S

eix·ξg(ξ) dσ(ξ) dx

=

∫Rdf(x) (g dσ)

ˇdx.

This implies that R∗g = (g dσ)ˇ. Thus, R(p → q) holds if and only if R∗(q′ → p′) holds, i.e., if there is anestimate of the form ∥∥(g dσ)

∥∥Lp′ (Rd)

.p,q,S ‖g‖Lq′ (S,dσ) .

We wish to find necessary conditions for R∗(q′ → p′) to hold. Consider the following two examples:

3

1. Suppose that g ≡ 1. Recall from a computation done in 247A that |(dσ)ˇ| . 〈x〉− d−12 . Thus, (dσ)ˇ∈ Lp′

precisely when d−12 p′ > d, which is equivalent to p < 2d

d+1 . Thus, in this case, R∗(q′ → p′) holds when

p <2d

d+ 1.

2. (The Knapp example) Let κ ⊆ S be a cap on the sphere centered at the point ξ0 = (0, . . . , 0,−1) ofhorizontal radius 1/R for someR >> 1. Near ξ0, we can parametrize the cap κ and Taylor expand via

ξd = −√

1− |ξ|2 = −(

1− |ξ|2

2+O(|ξ|4)

)= −1 +O(R−2)

because |ξ| ≤ 1/R on κ. So the ξd-height of the cap is on the order of R−2, i.e., the cap κ is containedin a cylinder D of radius 1/R and thickness ∼ 1/R2. Let T denote the “dual” cylinder to D, i.e., thecylinder centered at 0 with radius R and height ∼ R2.

Let g = χκ. Then‖g‖Lq′ (S,dσ) = σ(κ)

1q′ . (R−1)

d−1q′ .

Also,

(g dσ)ˇ(x) = (2π)−d2

∫κ

eix·ξ dσ(ξ) = (2π)−d2 e−ixd

∫κ

eix(ξ−ξ0) dσ(ξ).

Note that, for ξ ∈ κ,

|(ξ − ξ0)i| .

{1/R 1 ≤ i ≤ d− 1

1/R2 i = d

and for x ∈ T ,

|xi| .

{R 1 ≤ i ≤ d− 1

R2 i = d.

Thus, for most x ∈ T , we have|(g dσ)ˇ(x)| & σ(κ) ∼ R−(d−1).

So ∥∥(g dσ)∥∥Lp′ (Rd)

&∥∥(g dσ)

∥∥Lp′ (T )

& R−(d−1)|T |1p′ ∼ R−(d−1)R

d−1+2p′ = R−(d−1)R

d+1p′ .

Therefore, if R∗(q′ → p′) holds, then R−(d−1)Rd+1p′ . R

− d−1q′ . Letting R→∞ gives the condition

d+ 1

p′≤ d− 1

q.

Thus, by considering these two examples, we see that two necessary scaling conditions for R∗(q′ → p′)

to hold are 1 ≤ 2dd+1 and d+1

p′ ≤d−1q . We might wonder whether such conditions are also sufficient. Formally,

we pose the following conjecture.

Conjecture 1.1 (Restriction cojecture for the sphere). Let d ≥ 2, and let S = Sd−1 ⊆ Rd be the unit sphereendowed with the canonical surface measure. Then∥∥∥f |S∥∥∥

Lq(S,dσ).p,q ‖f‖Lp(Rd) ,

or equivalently ∥∥(g dσ)∥∥Lp′ (Rd)

.p,q ‖g‖Lq′ (S,dσ) ,

whenever 1 ≤ p ≤ 2dd+1 and 1 ≤ q ≤ d−1

d+1p′.

4

Zygmund proved this conjecture for d = 2; for all other d, this remains an open problem.

1.2 Tomas-Stein for the sphere

We have the following special case of the restriction conjecture, due to Tomas and Stein.

Theorem 1.1 (Tomas-Stein). Let d ≥ 2, and let S = Sd−1 ⊆ Rd be the unit sphere endowed with thecanonical surface measure. Then ∥∥∥f |S∥∥∥

L2(S,dσ).p ‖f‖Lp(Rd)

whenever 1 ≤ p ≤ 2(d+1)d+3 .

Note that, when q = 2, the scaling conditions in the restriction conjecture become 1 ≤ p ≤ 2dd+1 and

p′ ≥ 2(d+1)d−1 , which is equivalent to p ≤ 2(d+1)

d−3 . Note that, for d > 1, 2(d+1)d+3 < 2d

d+1 , so the Tomas-Steintheorem is indeed the full restriction conjecture for the sphere in the special case q = 2.

Proof. Attempt 1: Our first approach relies on using the decay condition |σ(x)| . 〈x〉− d−12 . We wish to

show that the operator R : Lp(Rd) → L2(S, dσ) is bounded, which is equivalent to showing boundednessof R∗ : L2(S, dσ)→ Lp

′(Rd), which is equivalent to showing boundedness of R∗R : Lp(Rd)→ Lp

′(Rd).

Let f be a Schwartz function. By definition, we have

(R∗Rf)(x) = (Rf dσ)ˇ(x) = (2π)−d2

∫S

eix·ξ(Rf)(ξ) dσ(ξ) = (2π)−d2

∫S

eix·ξ f(ξ) dσ(ξ)

= (2π)−d2 (2π)−

d2

∫S

eix·ξ∫Rde−iy·ξf(y) dy dσ(ξ)

= (2π)−d2

∫Rdf(y)σ(x− y) dy.

This implies that R∗Rf = f ∗ σ. Next, because |σ(x)| . 〈x〉− d−12 , we have σ ∈ L

2dd−1 ,∞(Rd). Therefore, by

the Hardy-Littlewood-Sobolev inequality,

‖R∗Rf‖Lp′ (Rd) . ‖f‖Lp(Rd) ‖σ‖L 2dd−1

,∞(Rd)

. ‖f‖Lp

provided that 1 + 1p′ = 1

p + d−12d . This is equivalent to 2

p = d−12d , hence p′ = 4d

d−1 , hence p = 4d3d+1 . Thus,

R(p → 2) holds for all 1 ≤ p ≤ 4d3d+1 . But a quick calculation shows that 4d

3d+1 <2(d+1)d+3 , so this proof does

not give the full scaling range from the statement of the theorem.

Attempt 2: Next, we use both the decay of |σ(x)| and the oscillation. We have

ϕ(x) +∑N>1

ψN (x) = 1

where ϕ and ψN are the usual Littlewood projections, and N is a dyadic number. Thus, we have thedecomposition

f ∗ σ = f ∗ (ϕσ) +∑N>1

f ∗ (ψN σ).

By the triangle inequality,

‖R∗Rf‖Lp′ (Rd) ≤ ‖f ∗ (ϕσ)‖Lp′ (Rd) +∑N>1

‖f ∗ (ψN σ)‖Lp′ (Rd) .

5

Consider the first term. Note that 1 + 1p′ = 1

p + 1r when r = p′

2 . Because p < 2, p′ > 2, hence r = p′

2 is asensible choice. Thus, by Young’s convolution inequality, we have

‖f ∗ (ϕσ)‖Lp′ (Rd) . ‖f‖Lp(Rd) ‖ϕσ‖Lp′2 (Rd)

.

Because ϕ is a bounded function with compact support, ‖ϕσ‖Lp′2 (Rd)

< ∞. Thus, ‖f ∗ (ϕσ)‖Lp′ (Rd) .

‖f‖Lp(Rd).Next, we consider the sum. Our goal is to find an estimate of the form

‖f ∗ (ψN σ)‖Lp′ (Rd) . N−ε(p) ‖f‖Lp(Rd)

for some ε(p) > 0, in order for the dyadic sum to be finite. Our strategy is to use real interpolation betweena L1 → L∞ and L2 → L2 type estimate.

Note that, because ψN is supported on an annulus of radius ∼ N and because of the decay of |σ|, wehave

‖f ∗ (ψN σ)‖L∞(Rd) . ‖f‖L1(Rd) ‖ψN σ‖L∞(Rd) . N−d−1

2 ‖f‖L1(Rd) .

In the L2 case, we use Plancharel to get the estimate

‖f ∗ (ψN σ)‖L2(Rd) =∥∥∥f · (ψN σ)

∥∥∥L2(Rd)

≤∥∥∥f∥∥∥

L2(Rd)

∥∥∥ψN ∗ dσ∥∥∥L∞(Rd)

= ‖f‖L2(Rd)

∥∥∥ψN ∗ dσ∥∥∥L∞(Rd)

.

We have(ψN ∗ dσ)(x) =

∫S

ψN (x− y)dσ(y) = Nd

∫S

ψ(N(x− y)) dσ(y).

So for any m, since ψ is Schwartz,

|(ψN ∗ dσ)(x)| . Nd

∫S

1

〈N(x− y)〉mdσ(y)

= Nd

∫|x−y|≤1/N ; y∈S

dσ(y) +Nd∑M≤N

∫|x−y|∼1/M ; y∈S

(M

N

)mdσ(y).

The first integral∫|x−y|≤1/N ; y∈S dσ(y) is the measure of a cap on the sphere of radius . 1/N , hence∫

|x−y|≤1/N ; y∈S dσ(y) . (1/N)d−1. Similarly,∫|x−y|∼1/M ; y∈S dσ(y) is the measure of two caps on the sphere

with radius . 1/M , so in total we have

|(ψN ∗ dσ)(x)| . Nd

(1

N

)d−1

+Nd∑M≤N

(M

N

)m(1

M

)d−1

= N +N∑M≤N

(M

N

)m−(d−1)

.

The sum in question is finite if m > d− 1. For such a choice of m, we then have |(ψN ∗ dσ)(x)| . N , hence∥∥∥ψN ∗ dσ∥∥∥L∞(Rd)

. N . Therefore, ‖f ∗ (ψN σ)‖L2(Rd) . N ‖f‖L2(Rd).

Now we interpolate between the estimates ‖f ∗ (ψN σ)‖L∞(Rd) . N−d−1

2 ‖f‖L1(Rd) and ‖f ∗ (ψN σ)‖L2(Rd) .

N ‖f‖L2(Rd). Note that1

p′=

2/p′

2+

1− 2/p′

∞.

Thus,

‖f ∗ (ψN σ)‖Lp′ (Rd) . ‖f ∗ (ψN σ)‖2p′

L2(Rd)‖f ∗ (ψN σ)‖

1− 2p′

L∞(Rd). N

2p′N

− d−12

(1− 2

p′

)‖f‖

2p′

L2(Rd)‖f‖

1− 2p′

L1(Rd)

. N− d−1

2 + d+1p′ ‖f‖Lp(Rd) .

6

Set ε(p) = −d−12 + d+1

p′ . Then ε(p) > 0 precisely when p′ > 2(d+1)d−1 , or equivalently p < 2(d+1)

d+3 . Thus, we

have proven the desired estimate in all cases save for the endpoint p = 2(d+1)d+3 .

Attempt 3: The above argument, which proves the theorem save for the endpoint case, is due to Tomas.The loss of the endpoint is essentially due to the use of the triangle inequality in the dyadic sum. To recoverthe endpoint case, Stein used complex interpolation with estimates of the form∥∥∥∥∥∑

N>1

Nd−1

2 +itf ∗ (ψN σ)

∥∥∥∥∥L∞(Rd)

. ‖f‖L1(Rd)∥∥∥∥∥∑N>1

N−1+itf ∗ (ψN σ)

∥∥∥∥∥L2(Rd)

. ‖f‖L2(Rd) .

We will not consider the details of the argument here; for reference, see Tao’s Math 254 notes.

1.3 Tomas-Stein for the parabaloid and Strichartz inequalities

Finally, we consider the Tomas-Stein result for the paraboloid. For convenience which will be explainedlater, we increase the dimension from d to d+ 1 and seek an estimate of the form∥∥(g dσ)

∥∥L

2(d+2)d (Rd+1)

. ‖g‖L2(Sparab,dσ)

where dσ is now the canonical measure on the paraboloid.The reason for the dimension increase is that we will connect the problem of Fourier restriction to es-

timates on dispersive partial differential equations in space and time. In particular, we will show that theTomas-Stein estimate for the paraboloid is equivalent to a certain symmetric Strichartz estimate.

Recall from Math 247A in discussing the linear Schrodinger equation, we defined the free propagatoroperator eit

∆2 by

eit∆2 u = F−1

ξ

(e−it

|ξ|22 u0(ξ)

)where u0(x) = u(0, x). Because eit

∆2 is a unitary operator, we have∥∥∥eit∆

2 f∥∥∥L2x

= ‖f‖L2x

by Placharel. We also proved the following dispersive estimate:∥∥∥eit∆2 f∥∥∥L∞x

. |t|− d2 ‖f‖L1x.

Using the Marcinkiewicz interpolation theorem, we obtain the generalized dispersive estimate∥∥∥eit∆2 f∥∥∥Lpx

. |t|−d(12−

1p ) ‖f‖

Lp′x

(1)

for 2 ≤ p ≤ ∞. With this estimate we derive the following inequality.

Theorem 1.2 (Strichartz inequality). Let f ∈ S(Rt × Rdx). Then1∥∥∥eit∆2 f∥∥∥LqtL

rx

. ‖f‖L2x

for 2 ≤ q, r ≤ ∞ satisfying 2q + d

r = d2 and (d, q, r) 6= (2, 2,∞).

1The notation ‖·‖LqtLrx means the Lrx norm followed by the Lq

t norm.

7

Proof. We first demonstrate the scaling relation, assuming the inequality. For λ > 0, define fλ(x) =

λ−d2 f(xλ

). Note that ‖fλ‖L2

x= ‖f‖L2

x. Then(eit

∆2 fλ

)(x) = λ−

d2

(ei

tλ2

∆2 fλ

)(xλ

).

Thus, ∥∥∥it∆2 fλ

∥∥∥LqtL

rx

= λ−d2

∥∥∥(ei tλ2∆2 fλ

)(xλ

)∥∥∥LqtL

rx

= λ−d2 + 2

q+ dr

∥∥∥eit∆2 f∥∥∥LqtL

rx

. λ−d2 + 2

q+ dr ‖fλ‖L2

x.

Thus in order for this to hold for any λ, we need −d2 + 2q + d

r = 0.

Next, we prove the inequality. Define the operator T (t) by T (t)f = eit∆2 f . We want to show that

T : L2(Rd)→ LqtLrx(Rt ×Rdx) is bounded. This is equivalent to showing that T (t)T ∗(t) : Lq

′

t Lr′

x (Rt ×Rdx)→LqtL

rx(Rt × Rdx) is bounded.

Observe thatT ∗(t)F =

∫Re−is

∆2 F (s) ds.

Thus,

T (t)T ∗(t)F =

∫Rei(t−s)

∆2 F (s) ds.

So, by the usual integral triangle equality followed by the generalized dispersive estimate, we have

‖T (t)T ∗(t)F‖LqtLrx ≤∥∥∥∥∫

R

∥∥∥ei(t−s) ∆2 F (s)

∥∥∥Lrx

ds

∥∥∥∥Lqt

.

∥∥∥∥∫R|t− s|−d(

12−

1r ) ‖F (s)‖Lr′x ds

∥∥∥∥Lqt

.

Observe that the integral inside the t-norm is a convolution between the function t 7→ |t|−d(12−

1r ) and

s 7→ ‖F (s)‖Lr′x . Using Hardy-Littlewood-Sobolev with the scaling relation 1 + 1q = 1

q′ + 1q/2 then gives

‖T (t)T ∗(t)F‖LqtLrx . ‖F‖Lq′t L

r′x

∥∥∥|t|−d( 12−

1r )∥∥∥Lq2,∞

t

.

The quantity∥∥∥|t|−d( 1

2−1r )∥∥∥Lq2,∞

t

is finite, and hence T (t)T ∗(t) is bounded, provided d(

12 −

1r

)< 1 and

d(

12 −

1r

)q2 = 1. The latter is true when d

2 −dr = 2

q which is the desired scaling relation.Note, however, that d

(12 −

1r

)< 1 coupled with the scaling relation d

(12 −

1r

)q2 = 1 implies that q > 2,

so we have not proven the endpoint case (q, r) =(

2, 2dd−2

). This case, which is more difficult, was proven

by Keel and Tao.

When q = r, the above inequality is known as a symmetric Strichartz inequality. In this case, the scalingrelation gives

2

q+d

q=d

2⇒ q =

2(d+ 2)

d.

Because 2(d+2)d > 2, we have thus proven:

Theorem 1.3 (Symmetric Strichartz inequality). Let f ∈ S(Rt × Rdx). Then∥∥∥eit∆2 f∥∥∥L

2(d+2)d

t,x

. ‖f‖L2x.

The connection between the symmetric Strichartz inequality and the Fourier restriction problem as al-luded to above is the following result.

8

Lemma 1.4. The Tomas-Stein theorem for the paraboloid is equivalent to the symmetric Strichartz inequal-ity; that is, the estimate ∥∥(g dσ)

∥∥L

2(d+2)d (Rd+1)

. ‖g‖L2(Sparab,dσ)

is equivalent to the symmetric Strichartz estimate.

Proof. Let u0 ∈ S(Rd) and ϕ ∈ S(R×Rd). We adopt the notation ϕ = Ft,xϕ to denote the space-time Fouriertransform of ϕ. Let ψ = F−1

x ϕ.We wish to calculate the space-time Fourier transform of eit

∆2 u0, in the distributional sense. We have⟨

eit∆2 u0, ϕ

⟩t,x

=⟨eit

∆2 u0, ϕ

⟩t,x

= (2π)−d2

⟨eit

∆2 u0,

∫Rde−ix·ξψ(t, ξ) dξ

⟩t,x

= (2π)−d2

∫Rd

∫R

∫Rd

(eit

∆2 u0

)(x)e−ix·ξψ(t, ξ) dx dt dξ

=

∫Rd

∫Re−it

|ξ|22 u0(ξ)ψ(t, ξ) dt dξ

= (2π)12

∫Rdu0(ξ) [Ftψ(ξ)]

(|ξ|2

2

)dξ.

Observe that [Ftψ(ξ)](|ξ|22

)= ϕ

(|ξ|22 , ξ

). Thus,⟨

eit∆2 u0, ϕ

⟩t,x

= (2π)12 〈u0 dσparab, ϕ〉t,x .

So, in the distributional sense, and up to a factor of (2π)12 ,

eit∆2 u0 = u0 dσparab ⇒ eit

∆2 u0 = Ft,x (u0 dσparab) .

Let g(ξ, |ξ|

2

2

)= u0(ξ). Then

‖g‖L2(Sparab,dσ) = ‖u0‖L2 = ‖u0‖L2

and (g dσparab)ˇ

= eit∆2 u0. Thus, the symmetric Strichartz estimate

∥∥∥eit∆2 u0

∥∥∥L

2(d+2)d

t,x

. ‖u0‖L2 yields the

Tomas-Stein estimate ∥∥(g dσ)∥∥L

2(d+2)d (Rd+1)

. ‖g‖L2(Sparab,dσ) .

The converse direction is essentially the same direction.

With this lemma, we have therefore demonstrated that the Tomas-Stein result holds for the paraboloid.However, it is of interest to prove that the Tomas-Stein estimate on the sphere actually implies the Tomas-Stein estimate on the paraboloid, without appealing directly to Strichartz estimates.

Proposition 1.1. The Tomas-Stein estimate on the sphere implies the Tomas-Stein estimate on the paraboloid.

Before proving the proposition, we compute explicitly the surface area measure on the sphere.

Lemma 1.5. The canonical surface measure on the sphere Sd is given explicitly by

dσSd =1√

1− |ξ|2dξ.

9

Proof. Let φ : Rd → Rd+1 be a coordinate parametrization of the sphere given by φ(ξ) = (ξ,√

1− |ξ|2).Then

dσSd =√

det [(∇φ)T (∇φ)] dξ.

Thus, it remains to compute√

det [(∇φ)T (∇φ)].We have

∇φ(ξ) =

1 0 · · · 0

0 1 · · · 0...

......

...0 0 · · · 1−ξ1√1−|ξ|2

−ξ2√1−|ξ|2

· · · −ξd√1−|ξ|2

.

Thus,

(∇φ(ξ))T (∇φ(ξ)) =

[δij +

ξiξj1− |ξ|2

]1≤i,j≤d

= I +|ξ|2

1− |ξ|2Pξ

where Pξ =[ξiξj|ξ|2

]is the projection matrix onto the ξ direction.

Consequently,

(∇φ(ξ))T (∇φ(ξ))ξ =

(I +

|ξ|2

1− |ξ|2Pξ

)ξ = ξ +

|ξ|2

1− |ξ|2ξ =

(1

1− |ξ|2

)ξ

so that (∇φ)T (∇φ) has an eigenvector ξ with eigenvalue 11−|ξ|2 . Next, observe that if x is a vector orthogonal

to ξ, then (I +

|ξ|2

1− |ξ|2Pξ

)x = x

so that (∇φ)T (∇φ) has a d− 1-dimensional eigenspace corresponding to the eigenvalue 1. Thus,√det [(∇φ)T (∇φ)] =

1√1− |ξ|2

as desired.

With this, we now prove Proposition 1.1.

Proof. Assume the Tomas-Stein estimate on the sphere, that is,∥∥(g dσSd)∥∥L

2(d+2)d (Rd+1)

. ‖g‖L2(Sd,dσSd

) .

Because the Tomas-Stein estimate on the paraboloid is equivalent to the symmetric Strichartz estimate, wewant to derive an estimate of the form∥∥∥eit∆

2 u0

∥∥∥L

2(d+2)d

t,x (R×Rd). ‖u0‖L2

x

where we can take u0 ∈ C∞c (Rd) by density.Towards this goal, fix λ >> 1. Define a function gλ on the sphere Sd as follows: gλ(ξ,

√1− |ξ|2) = λ

d2 u0(λξ)

√1− |ξ|2

gλ(ξ,−√

1− |ξ|2) = 0.

10

Because u0 ∈ C∞c (Rd), for λ large, gλ is supported on a small ξ-region, i.e., a small cap around the northpole. Using the surface area lemma from above, we have

‖gλ‖2L2(Sd,dσSd

) = λd∫|u0(λξ)|2(1− |ξ|2)

1√1− |ξ|2

dξ

= λd∫|u0(λξ)|2

√1− |ξ|2 dξ

=

∫|u0(ξ)|2

√1− |ξ|

2

λdξ.

Thus, for very large λ we have ‖gλ‖2L2(Sd,dσSd

) ∼ ‖u0‖L2ξ

= ‖u0‖L2x.

Next, we compute (gλ dσSd)ˇ.

(gλ dσSd)ˇ(t, x) = (2π)−d+1

2

∫eitω+ix·ξλ

d2 u0(λξ)

√1− |ξ|2 1√

1− |ξ|2dξ δ(ω =

√1− |ξ|2)

= (2π)−d+1

2

∫eit√

1−|ξ|2+ix·ξλd2 u0(λξ) dξ

= (2π)−d+1

2 λ−d2

∫eit√

1− |ξ|2

λ2 +i xλ ·ξu0(ξ) dξ.

By Taylor expansion, we have√

1− |ξ|2

λ2 = 1− |ξ|2

2λ2 +O(|ξ|4λ4

). Also note that

(2π)−d+1

2 λ−d2

∫eit−

it|ξ|2

2λ2 +i xλ ·ξu0(ξ) dξ = (2π)−d+1

2 λ−d2 eit ·

[ei

tλ2

∆2 u0

] (xλ

).

Thus, we can write

(gλ dσSd)ˇ(t, x) = (2π)−d+1

2 λ−d2 eit ·

[ei

tλ2

∆2 u0

] (xλ

)+ (2π)−

d+12 λ−

d2

∫eit−

it|ξ|2

2λ2 +i xλ ·ξu0(ξ)

[eit

(√1− |ξ|

2

λ2 −1+|ξ|2

2λ2

)− 1

]dξ.

By the Taylor expansion comment and by the usual arc-length type estimate, for λ large we have∣∣∣∣∣eit(√

1− |ξ|2

λ2 −1+|ξ|2

2λ2

)− 1

∣∣∣∣∣ . |t| |ξ|4λ4.

Consequently,∣∣∣∣∣(2π)−d+1

2 λ−d2

∫eit−

it|ξ|2

2λ2 +i xλ ·ξu0(ξ)

[eit

(√1− |ξ|

2

λ2 −1+|ξ|2

2λ2

)− 1

]dξ

∣∣∣∣∣ . λ−d2−4|t|

∫|u0(ξ)||ξ|4 dξ

. λ−d2−4|t|

where the last inequality follows from the fact that u0 ∈ C∞c (Rd). Thus,∥∥(gλ dσSd)∥∥L

2(d+2)d (Rd+1)

&∥∥(gλ dσSd)

∥∥L

2(d+2)d (|t|≤λ2(1+ε),|x|≤λ1+ε)

& λ−d2

∥∥∥[ei tλ2∆2 u0

] (xλ

)∥∥∥L

2(d+2)d (|t|≤λ2(1+ε),|x|≤λ1+ε)

− λ− d2−4λ2(1+ε)λ(2(1+ε)+d(1+ε)) d2(d+2)

11

where the final power of λ comes from considering the volume of the set. Changing variables and simpli-fying the power of λ in the error term then gives∥∥(gλ dσSd)

∥∥L

2(d+2)d (Rd+1)

& λ−d2 λ(2+d) d

2(d+2)

∥∥∥eit∆2 u0

∥∥∥L

2(d+2)d (|t|≤λ2ε,|x|≤|λ|ε)

− λ−2+2ε+ dε2 .

Thus, ∥∥∥eit∆2 u0

∥∥∥L

2(d+2)d

t,x (|t|≤λ2ε,|x|≤|λ|ε)+O

(λ−2+ d+4

2 ε). ‖u0‖L2

x.

As λ→∞, for ε << 1 we then get ∥∥∥eit∆2 u0

∥∥∥L

2(d+2)d

t,x (Rd+1). ‖u0‖L2

x

as desired.

12

2 4/10 — 4/14: Harmonic Analysis on the Torus

2.1 Basic properties of the Fourier transform

In the interest of studying dispersive partial differential equations on the torus, we discuss the Fouriertransform in this context. Let Td := Rd/Zd.

Definition 2.1. For a C∞-function f : Td → C, we define its Fourier transform f : Zd → C by

f(k) =

∫Tde−2πik·xf(x) dx.

Then f(k) = 〈f, ek〉where ek(x) = e2πik·x.

Observe that the characters ek are orthonormal, since

〈ek, el〉 =

∫Tde2πi(k−l)·x = δk=l.

We quickly establish other familiar facts about the Fourier transform.

Proposition 2.1 (Bessel’s Inequality). For f ∈ C∞(Td),∑k∈Zd

| 〈f, ek〉 |2 ≤ ‖f‖2L2(Td) .

Proof. Let S ⊆ Zd be a finite set. Then

0 ≤

∥∥∥∥∥f −∑k∈S

〈f, ek〉 ek

∥∥∥∥∥2

L2(Td)

=

⟨f −

∑k∈S

〈f, ek〉 ek, f −∑l∈S

〈f, el〉 el

⟩

= ‖f‖2L2(Td) − 2∑k∈S

〈f, ek〉 〈ek, f〉+

⟨∑k∈S

〈f, ek〉 ek,∑l∈S

〈f, el〉 el

⟩= ‖f‖2L2(Td) − 2

∑k∈S

| 〈f, ek〉 |2 +∑| 〈f, ek〉 |2

where the last equality follows from orthogonality of the ek’s. Simplifying and rearranging gives∑k∈S

| 〈f, ek〉 |2 ≤ ‖f‖2L2(Rd)

for every finite set S, hence the desired inequality.

This fact shows that if f ∈ C∞(Td), then∑k∈Zd 〈f, ek〉 ek ∈ L2(Td). Unsurprisingly, they are in fact the

same function.

Proposition 2.2 (Fourier inversion). If f ∈ C∞(Td), then

f =∑k∈Zd

〈f, ek〉 ek =∑k∈Zd

f(k)e2πik·x.

13

Proof. Suppose for the sake of contradiction that f 6=∑k∈Zd 〈f, ek〉 ek. Then by Stone-Weierstrass, there is

a trigonometric polynomial g such that⟨f −

∑k∈Zd

〈f, ek〉 ek, g

⟩6= 0.

But observe that, for any character el,⟨f −

∑k∈Zd

〈f, ek〉 ek, el

⟩= 〈f, el〉 − 〈f, el〉 = 0.

This is a contradiction.

2.2 Bourgain’s dispersive estimate on the torus

We now consider the linear Schrodinger equation on the torus. Initially, we consider initial data which is acharacter: {

i∂tu = ∆2

u(0) = ek. (2)

Taking the spacial Fourier transform of this equation yields{i∂tu(t, n) = −(2π)2 |n|2

2 u(t, n)

u(0, n) = δk=n

(3)

and solving this ODE in time givesu(t, n) = e2π2it|n|2δk=n.

Inverting the Fourier transform, it follows that a solution to (2) takes the form

u(t, x) =∑n∈Zd

e2πin·xe2π2it|n|2δk=n = e2πik·x+2π2it|k|2 .

We adopt the notation eit∆2 ek(x) = e2πik·x+2π2it|k|2 . For a general smooth function f , we have

eit∆2 f(x) = eit

∆2

∑k∈Zd

〈f, ek〉 ek =∑k∈Zd

〈f, ek〉 e2πik·x+2π2it|k|2 .

When we studied the linear Schrodinger equation in Rd, we proved certain dispersive estimates about howsolutions disperse over time. On the torus, there is no space to have dispersive decay over time.

In 1993, Bourgain proved the following dispersive estimate for the linear Schrodinger equation on theone-dimensional torus.

Theorem 2.1 (Bourgain 1993). Let ϕ ∈ C∞c (R) with suppϕ ⊆ [−2, 2], ϕ ≡ 1 on [−1, 1], and |ϕ′′| . 1. FixN ≥ 1. Let K(t, x) =

∑n∈Z ϕ

(nN

)e2πinx+2πin2t. If 0 ≤ a ≤ q ≤ N , (a, q) = 1, and |t− a/q| < 1

Nq , then

|K(t, x)| . N

√q

(1 +N

∣∣∣t− aq

∣∣∣ 12

) .

By convention, we have (0, q) = q. Before proving the theorem, we make a few remarks. Observethat K(t, x) = ei

t2π

∆2 P≤N where P≤N is a Littlewood-Paley projection operator. Also, we note that the

hypothesis of the theorem is not very restrictive, due to Dirichlet’s lemma.

14

Lemma 2.2 (Dirichlet). Fix N ≥ 2 and t ∈ [0, 1). There exists 1 ≤ q ≤ N and 0 ≤ a ≤ q such that (a, q) = 1

and∣∣∣t− a

q

∣∣∣ < 1Nq

Proof. If t = 0, then taking a = 0 and q = 1 gives the desired result.Thus, we may assume that t ∈ (0, 1). Let {α} denote the fractional part of a positive real number α.

Consider the set{ {tn} : 0 ≤ n ≤ N } .

By the pigeonhole principle, there exists 0 ≤ m < n ≤ N such that |{tn} − {tm}| < 1N . Thus,

|tn− btnc − (tm− btmc)| < 1

N⇒ |t(n−m)− (btnc − btmc)| < 1

N.

Let q = n−m and b = btnc − btmc. Then 1 ≤ q ≤ N , b ≥ 0, and |tq − b| < 1N .

If b ≤ q, then let a = b(q,b) and q = q

(q,b) . Then (a, q) = 1, 1 ≤ q ≤ N , 0 ≤ a ≤ q, and

|tq − b| < 1

N⇒

∣∣∣∣t− b

q

∣∣∣∣ < 1

Nq⇒

∣∣∣∣t− a

q

∣∣∣∣ < 1

Nq≤ 1

Nq

as desired.If b > q, then tq < q < b, so

|tq − b| < 1

N⇒ |tq − q| < 1

N.

Thus, set a = q = 1. Then ∣∣∣∣t− a

q

∣∣∣∣ = |t− 1| < 1

Nq≤ 1

N=

1

Nq

as desired.

Next, we remark the the estimate in Bourgain’s result cannot be improved. In particular, when t = aq

with (a, q) = 1, 0 ≤ a ≤ q ≤ N for odd q, the estimate is sharp. To see this, we rely on the following fact.

Lemma 2.3. Let q be odd, (a, q) = 1, and b ∈ Z. Then∣∣∣∣∣q−1∑n=0

e2πin bq+2πin2 aq

∣∣∣∣∣ =√q.

Proof. We compute: ∣∣∣∣∣q−1∑n=0

e2πin bq+2πin2 aq

∣∣∣∣∣2

=

(q−1∑n=0

e2πin bq+2πin2 aq

)q−1∑m=0

e2πim bq+2πim2 a

q

=

q−1∑m=0

q−1∑n=0

e2πin bq+2πin2 aq e2πim b

q+2πim2 aq

=

q−1∑m=0

q−1∑n=0

e2πi(n−m) bq+2πi(n2−m2) aq .

15

For a fixed m, we make the change of variables n = m + l. By periodicity of the exponential, we can stillsum from l = 0 to l = q − 1. Thus, since n2 −m2 = (m+ l)2 −m2 = 2lm+ l2, we have∣∣∣∣∣

q−1∑n=0

e2πin bq+2πin2 aq

∣∣∣∣∣2

=

q−1∑m=0

q−1∑l=0

e2πil bq+2πi(2lm+l2) aq

=

q−1∑l=0

e2πil bq+2πil2 aq

q−1∑m=0

e2πim 2laq .

Summing the inner geometric sum, we have

q−1∑m=0

e2πim 2laq =

q 2la = 0 mod q

1−e2πi2laqq

1−e2πi2laq

2la 6= 0 mod q=

{q 2la = 0 mod q

0 2la 6= 0 mod q.

So the only terms which contribute to the outer sum are the ones in which 2la = 0 mod q. Since q is oddand 2l is even, 2la = 0 mod q if and only if l = 0. Thus, the only remaining term of the sum si l = 0, hence∣∣∣∣∣

q−1∑n=0

e2πin bq+2πin2 aq

∣∣∣∣∣2

=

q−1∑m=0

e2πim = q

and we are done.

Before demonstrating that the Bourgain estimate is sharp in the indicated case, we present one morestandard lemma.

Lemma 2.4 (Poisson summation). Let ϕ ∈ S(R). Then∑n∈Z

ϕ(x+ n) =∑n∈Z

ϕ(n)e2πinx

for all x ∈ R. In particular,∑n∈Z ϕ(n) =

∑n∈Z ϕ(n).

Proof. Define F1(x) =∑n∈Z ϕ(x + n) and F2(x) =

∑n∈Z ϕ(n)e2πinx. Note that both F1 and F2 are 1-

periodic in x. Also, since ϕ is Schwartz, for every x, each sum converges absolutely in n and the sumconverges uniformly in x on compact sets. Therefore, F1 and F2 are both continuous functions on T. Assuch, to give the desired equality it suffices to prove that F1 and F2 have the same Fourier coefficients. Thedefinition of F2 immediately gives F2(k) = ϕ(k) for all k ∈ Z.

Next, we compute:

F1(k) =∑n∈Z

∫ 1

0

ϕ(x+ n)e−2πikx dx =∑n∈Z

∫ n+1

n

ϕ(y)e−2πik(y−n) dy =∑n∈Z

∫ n+1

n

ϕ(y)e−2πiky dy

=

∫Rϕ(y)e−2πiky dy

= ϕ(k).

The in particular statement follows from considering x = 0.

Next, define a (tempered) distribution on R by S ′(R) 3 f :=∑n∈Z δn, where δn(ϕ) := ϕ(n) for a

Schwartz function ϕ. We wish to compute the Fourier transform of f (as a distribution). Let ϕ ∈ S(R). By

16

definition, we have ⟨f , ϕ

⟩= 〈f, ϕ〉 =

∑n∈Z〈δn, ϕ〉 =

∑n∈Z

ϕ(n)

=∑n∈Z

ϕ(n)

where the final equality follows from Poisson summation. Therefore, f =∑n∈Z δn in the distributional

sense.Following this, we wish to compute the time evolution e−i

t2π∆Rf . As before, let ϕ be Schwartz. Then⟨

e−it

2π∆Rf, ϕ⟩

=⟨f, ei

t2π∆Rϕ

⟩=∑n∈Z

(ei

t2π∆Rϕ

)(n) =

∑n∈Z

(ei

t2π∆Rϕ

)(n)

=∑n∈Z

e2πitn2

ϕ(n)

=

∫R

(∑n∈Z

e2πitn2−2πinx

)ϕ(x) dx.

Therefore, as a distribution,e−i

t2π∆Rf =

∑n∈Z

e2πitn2+2πinx.

Observe that this is also a distribution on T, being 1-periodic in x. In S ′(T), we have

e−it

2π∆Rf = e−it

2π∆Tδ0.

Now, let q be odd, 0 ≤ a ≤ q with (a, q) = 1, and set t = a/q. Then

e−iaq

δR2π f =

∑n∈Z

e2πi aq n2+2πinx.

For each n, write n = mq + r where 0 ≤ r ≤ q − 1. Then

e−iaq

∆R2π f =

∑m∈Z

q−1∑r=0

e2πi(mq+r)xe2πi aq (mq+r)2

=

q−1∑r=0

e2πirx+2πi aq r2 ∑m∈Z

e2πimqx+2πim2q2 aq+2πi(2mqr) aq

=

q−1∑r=0

e2πirx+2πi aq r2 ∑m∈Z

e2πimqx.

We claim that, as distributions,∑m∈Z e

2πimqx =∑n∈Z

1q δ0

(x− n

q

). To see this, let ϕ be Schwartz and note

that (∑n∈Z

1

qδ0

(x− n

q

))(ϕ) =

∑n∈Z

1

qϕ

(n

q

).

Let ψ(x) = ϕ(xq

). Then ψ(n) = ϕ

(nq

)and ψ(n) = qϕ(nq) and so by Poisson summation we have(∑

n∈Z

1

qδ0

(x− n

q

))(ϕ) =

∑n∈Z

ϕ(nq).

On the other hand, (∑m∈Z

e2πimqx

)(ϕ) =

∑m∈Z

ϕ(mq).

17

Thus, ∑m∈Z

e2πimqx =∑n∈Z

1

qδ0

(x− n

q

).

So

e−iaq

∆R2π f =

q−1∑r=0

e2πirx+2πi aq r2 ∑n∈Z

1

qδ0

(x− n

q

)=∑n∈Z

1

qδ0

(x− n

q

) q−1∑r=0

e2πir nq +2πi aq r2

.

By the lemma from above,q−1∑r=0

e2πir nq +2πi aq r2

=√qeiθn

for some phase θn depending on n. Thus,

e−iaq

∆R2π f =

∑n∈Z

1√qeiθnδ0

(x− n

q

)and as a distribution in S ′(T),

e−iaq

∆T2π δ0 =

q−1∑n=0

1√qeiθnδn/q.

This shows that refocusing occurs at times a/q for (a, q) = 1.Next, by commutativity of Fourier multipliers, we have

e−iaq

∆T2π P≤Nδ0 =

q−1∑n=0

1√qeiθnP≤Nδn/q =

q−1∑n=0

1√qNϕ

(N

(x− n

q

)).

Taking absolute values and considering the support of ϕ (?) gives∣∣∣∣K (aq , x)∣∣∣∣ . N

√q.

Hence, Bourgain’s estimate is sharp.We require one last proposition before proving the main result.

Proposition 2.3. Fix N ≥ 2. Let ϕ ∈ C∞c with suppϕ ⊆ [−2, 2]. Then∣∣∣∣∣∑n∈Z

ϕ( nN

)e2πinα

∣∣∣∣∣ . min

{N,

1

N dist2(α,Z)

}.

N

1 +N2 dist2(α,Z)

for any α ∈ R.

Proof. The latter inequality is immediate. To show the first inequality. It suffices to show that the sum inquestion is bounded by both terms in the minimum term.

The first term is straightforward. By the compact support of ϕ, we have∣∣∣∣∣∑n∈Z

ϕ( nN

)e2πinx

∣∣∣∣∣ ≤∑n∈Z

∣∣∣ϕ( nN

)∣∣∣ . N.

18

To bound the sum by the second term, we would like to sum integration / summation by parts. Note that

2e2πinα − e2πi(n+1)α − e2πi(n−1)α = e2πinα[2− e2πiα − e−2πiα

]= e2πinα

[−(eπiα − e−πiα

)2]= 4e2πinα

(eπiα − e−πiα

2i

)2

= 4e2πinα sin2(πα).

Thus,

e2πinα =2e2πinα − e2πi(n+1)α − e2πi(n−1)α

4 sin2(πα).

Substituting this into the sum and changing variables to factor out a common e2πinα gives∑n∈Z

ϕ( nN

)e2πinα =

1

4 sin2(πα)

∑n∈Z

e2πinα

(2ϕ( nN

)− ϕ

(n− 1

N

)− ϕ

(n+ 1

N

)).

The quantity in the parentheses is the different quotient for the second derivative of ϕ. Thus,∣∣∣∣2ϕ( nN )− ϕ(n− 1

N

)− ϕ

(n+ 1

N

)∣∣∣∣ . 1

N2.

This, together with the triangle inequality and the compact support of ϕ, gives,∣∣∣∣∣∑n∈Z

ϕ( nN

)e2πinα

∣∣∣∣∣ . 1

sin2(πα)·N · 1

N2.

Observe that sin2(πα) ∼ dist(α,Z). This gives the desired result.

Finally, we prove Theorem 2.1.

Proof. As in the statement of the proof, let N ≥ 1, 1 ≤ q ≤ N , 0 ≤ a ≤ q with (a, q) = 1,∣∣∣t− a

q

∣∣∣ < 1Nq , and

defineK(t, x) =

∑n∈Z

ϕ( nN

)e2πinx+2πin2t.

We compute:

|K(t, x)|2 =

(∑n∈Z

ϕ( nN

)e2πinx+2πin2t

)(∑m∈Z

ϕ(mN

)e2πimx+2πim2t

)=∑n,m∈Z

ϕ( nN

)ϕ(mN

)e2πi(n−m)x+2πi(n−m)(n+m)t

=∑

{|k|≤4N}

∑{l=kmod 2}

Ψ

(k

N,l

N

)e2πil(x+kt)

where in the last equality we have made the change of variables l = n − m and l = n + m, definedΨ(kN ,

lN

):= ϕ

(nN

)ϕ(mN

), observed that l and k are both either even or odd, and used the fact that suppϕ ⊆

[−2, 2] to restrict the range of k = n+m.If k is even, the inner sum becomes∑

m∈ZΨ

(k

N,

2m

N

)e2πim(2x+2kt).

19

For a fixed k, the function Ψ(kN ,

2mN

)satisfies the hypothesis of Proposition 2.3. Thus,∣∣∣∣∣∑

m∈ZΨ

(k

N,

2m

N

)e2πim(2x+2kt)

∣∣∣∣∣ . N

1 +N2 dist2(2x+ 2kt,Z).

Similarly, if k is odd, then the inner sum becomes

∑m∈Z

Ψ

(k

N,

2m+ 1

N

)e2πi(x+kt)e2πim(2x+2kt)

and thus∣∣∣∣∣∑m∈Z

Ψ

(k

N,

2m+ 1

N

)e2πi(x+kt)e2πim(2x+2kt)

∣∣∣∣∣ =

∣∣∣∣∣e2πi(x+kt)∑m∈Z

Ψ

(k

N,

2m+ 1

N

)e2πim(2x+2kt)

∣∣∣∣∣.

N

1 +N2 dist2(2x+ 2kt,Z).

So we have|K(t, x)|2 .

∑|k|≤4N

N

1 +N2 dist2(2x+ 2kt,Z).

Next, observe that for a real number y,

dist(2y,Z) = 2 dist

(y,

1

2Z)

= 2 min

{dist (y,Z) ,dist

(y +

1

2,Z)}

.

Hence, it suffices to consider just one of the cases in the minimum, and consequently to prove the theoremit suffices to bound the sum ∑

|k|≤4N

N

1 +N2 dist2(x+ kt,Z).

By the hypotheses of the theorem, we can write t = aq + τ for some |τ | < 1

Nq . Also, write k = mq + s wherer ∈ {0, . . . , q − 1}. Observe that for the given range of k, |m| ≤ 5N

q . We have

dist(x+ kt,Z) = dist

(x+ (mq + r)

(a

q+ τ

),Z)

) = dist

(x+

ar

q+ (mq + r)τ,Z

).

We make two remarks.

1. Note that |(mq + r)τ | < 4N · 1Nq = 4

q .

2. Since (a, q) = 1 and necessarily (r, q) = 1, it follows that (ar, q) = 1. Thus, the map r 7→ arq induces a

bijection between the sets {0, . . . , q − 1} and{

0, 1q , . . . ,

q−1q

}.

Continuing,

∑|k|≤4N

N

1 +N2 dist2(x+ kt,Z)=

∑|m|≤ 5N

q

q−1∑r=0

N

1 +N2 dist2(x+ ar

q + (mq + r)τ,Z) .

We split the inner sum over r into two: a sum over bad residues, and a sum over good residues. A badresidue r∗ is one such that

dist

(x+

ar∗

q+ (mq + r∗)τ,Z

)<

16

q.

20

All other residues are good. Note that∣∣∣∣{ r : dist

(x+

ar

q+ (mq + r)τ,Z

)<

16

q

}∣∣∣∣ ≤ ∣∣∣∣{ r : dist

(x+

ar

q,Z)<

20

q

}∣∣∣∣≤∣∣∣∣{ s ∈ {0, . . . , q − 1} : dist

(s

q,Z− x

)<

20

q

}∣∣∣∣ .The first inequality follows from remark 1, and the second inequality follows from remark 2.

Because dist(sq ,Z− x

), for q sufficiently large there are at most 1 + 19 + 19 = 39 bad residues.

We first sum the good residues. If r is good, then

dist

(x+

ar

q+ (mq + r)τ,Z

)≥ 16

q

and thus, since (mq + r)τ ≤ 4q , dist

(x+ ar

q ,Z)≥ 12

q . So

∑|m|≤ 5N

q

∑{r:good}

N

1 +N2 dist2(x+ ar

q + (mq + r)τ,Z) .

∑|m|≤ 5N

q

∑{r:good}

N

1 +N2 dist2(x+ ar

q ,Z)

.∑|m|≤ 5N

q

∞∑d=1

N

1 +N2 d2

q2

.N

q

∞∑d=1

q2

Nd2

. q.

We will see that this bound is acceptable.It remains to sum the bad residues. In order for a residue r∗ to be bad, there must exists k∗ = m∗q + r∗

such that

dist(x+ k∗t,Z) = dist

(x+

ar∗

q+ (m∗q + r∗)τ,Z

)<

16

q.

If k = mq + r∗, then

dist(x+ kt,Z) = dist(x+ k∗t+ (k − k∗)t,Z)

= dist

(x+

ar∗

q+ k∗τ + (m−m∗)q

(a

q+ τ

),Z)

= dist (x∗ + (m−m∗)qτ,Z)

where x∗ = x+ ar∗

q + k∗τ . We have

|(m−m∗)qτ | ≤ 10N

q· q · 1

Nq≤ 10

q.

The worst case scenario in this estimate is when x∗ ∈ Z. Thus, using the change of variables d = m −m∗,we have

∑|m|≤ 5N

q

∑{r:bad}

N

1 +N2 dist2(x+ ar

q + (mq + r)τ,Z) .

∑{r∗:bad}

10Nq∑

d=0

N

1 +N2d2q2τ2

.

10Nq∑

d=0

N

1 +N2d2q2τ2

21

because there are only 39 bad residues.Thus, in total, we have shown that

|K(t, x)|2 . q +∑

0≤d.Nq

N

1 +N2d2q2τ2.

First, consider the sum on the right. While summing over d, it is either the case that 1 is the dominant termin the denominator, or N2d2q2τ2 is the dominant term. Furthermore, we consider the cases τ & 1

N2 andτ << 1

N2 separately. Thus,

∑0≤d.N

q

N

1 +N2d2q2τ2.

{ ∑0≤d≤ 1

NqτN +

∑1

Nqτ<d.Nq

1N2d2q2τ2 τ & 1

N2∑0≤d.N

qN τ << 1

N2

.

{1qτ + 1

Nq2τ2Nqτ τ & 1N2

N2

q τ << 1N2

.

{1qτ τ & 1

N2

N2

q τ << 1N2

.

Note that if τ & 1N2 , then 1

qτ . N2

q . Thus,{1qτ τ & 1

N2

N2

q τ << 1N2

. min

{1

qτ,N2

q

}.N2

qmin

{1

τN2, 1

}.N2

q· 1

1 + τN2.

So we have|K(t, x)| . √q +

N√q(1 +N

√τ)

where we have invoked equivalence of finite dimensional norms in taking the square root. Since the termon the right is exactly the term we wanted, it remains to show that

√q is an acceptable bound. This follows

fromq(1 +N

√τ) ≤ q + qN

1√Nq≤ N +N . N

so that√q .

N√q(1 +N

√τ).

22

3 4/17 — 5/1: Rearrangement Inequalities

Our next topic of study is the theory of rearrangements. Rearrangement inequalities have applications toproblems involving the minimization of energy functionals and other variational problems, and can be usedin showing the existence and uniqueness of minimizers. There are also connections to classical problemssuch as the isoperimetric inequality.

3.1 Definitions and basic estimates

We begin by recalling the layer-cake decomposition of a Borel measurable function f : Rd → [0,∞); namely,the identity

f(x) =

∫ ∞0

χ{f>λ}(x) dλ.

This equality follows from the observation that

χ{f>λ}(x) =

{1 f(x) > λ

0 f(x) ≤ λ=

{1 λ < f(x)

0 λ ≥ f(x)= χ[0,f(x))(λ)

and so ∫ ∞0

χ{f>λ}(x) dλ =

∫ ∞0

χ[0,f(x))(λ) dλ =

∫ f(x)

0

dλ = f(x).

This decomposition will frequently be used in the proofs of the inequalities to come. With this in mind, webegin discussing rearrangements.

Definition 3.1.

1. If A ⊆ Rd is Borel measurable, its symmetric rearrangement is the set

A∗ ={x ∈ Rd : |x| < r

}where r is chosen so that |A∗| = |A|.

2. Let f : Rd → C be a Borel measurable function which vanishes at infinity, that is, |{|f | > λ}| <∞ forall λ > 0. The symmetric decreasing rearrangement of f is given by

f∗(x) =

∫ ∞0

χ{|f |>λ}∗(x) dλ.

The intuition and geometry behind the rearrangement of a set A is clear: A∗ is simply the open ballcentered at the origin with the same measure as A. To give intuition for the symmetric rearrangement of afunction, we compute a simple example. Suppose that f : R→ C is given by f(x) = χ[−2,−1](x) + χ[1,2](x).Fix λ > 0. Then

{|f | > λ}∗ =

{([−2,−1] ∪ [1, 2])∗ 0 < λ < 1

∅∗ λ ≥ 1=

{(−1, 1) 0 < λ < 1

∅ λ ≥ 1.

Therefore,

f∗(x) =

∫ 1

0

χ{|f |>λ}∗(x) dλ =

∫ 1

0

χ(−1,1)(x) dλ =

{1 x ∈ (−1, 1)

0 x /∈ (−1, 1)= χ(−1,1)(x).

23

For a slightly more interesting example, suppose that g : R→ C is given by g(x) = χ[−2,−1](x) + 2χ[1,2](x).Fix λ > 0. Then

{|g| > λ}∗ =

([−2,−1] ∪ [1, 2])∗ 0 < λ < 1

([1, 2])∗ 1 ≤ λ < 2

∅∗ λ ≥ 2

=

(−1, 1) 0 < λ < 1(− 1

2 ,12

)1 ≤ λ < 2

∅ λ ≥ 2

.

Therefore,

g∗(x) =

∫ 1

0

χ(−1,1)(x) dλ+

∫ 2

1

χ(− 12 ,

12 )(x) dλ = χ(−1,1)(x) + χ(− 1

2 ,12 )(x).

Roughly, the symmetric decreasing rearrangement of a function takes the mass under the graph of thefunction and rearranges it to return a function which is radially symmetric and decreasing with the sametotal mass.

These examples suggest some basic properties of symmetric decreasing rearrangements.

Proposition 3.1. If f : Rd → C is a Borel measurable function which vanishes at infinity, then f∗ is nonneg-ative, radially symmetric, radially non-increasing, and lower-semicontinuous.

Proof. Nonnegativity is clear. Radial symmetry of f∗ follows from the fact that the set {|f | > λ}∗ is radiallysymmetric. The fact that f∗ is radially decreasing follows from the observation that if λ > µ, then {|f | >λ}∗ ⊆ {|f | > µ}∗.

To prove lower-semicontinuity, recall that f∗ is lower-semicontinuous if and only if {f∗ > λ} is openfor all λ. We have{

x ∈ Rd : f∗(x) > λ}

=

{x ∈ Rd :

∫ ∞0

χ{|f |>t}∗(x) dt > λ

}=⋃t>λ

{|f | > t}∗.

Thus, {f∗ > λ} is open.

Next, observe that the rearrangement function satisfies the following monotonicity property.

Proposition 3.2. If f, g : Rd → C are Borel measurable functions which vanishes at infinity, and if |f | ≤ |g|,then f∗ ≤ g∗.

Proof. If |f | ≤ |g|, then {|f | > λ} ⊆ {|g| > λ}. Thus, {|f | > λ}∗ ⊆ {|g| > λ}∗, and the claim follows.

Next, we make an observation that will be used frequently. By definition,

f∗(x) =

∫ ∞0

χ{|f |>λ}∗(x) dλ.

On the other hand, we have the layer-cake decomposition of f∗, which gives

f∗(x) =

∫ ∞0

χ{f∗>λ}(x) dλ.

This suggests the equality {|f | > λ}∗ = {f∗ > λ}. Indeed, this is the case.

Proposition 3.3. If f : Rd → C is a Borel measurable function which vanishes at infinity, then

{|f | > λ}∗ = {f∗ > λ}.

24

Proof. Note that

|{|f | > t}∗| = |{|f | > t}| =∫Rdχ{|f |>t}(x) dx.

As t→ λ from above, by the dominated convergence theorem, it then follows that

|{|f | > t}∗| →∫Rdχ{|f |>λ}(x) dx = |{|f | > λ}| = |{|f | > λ}∗|.

Because {f∗ > λ} =⋃t>λ{|f | > t}∗, we have thus shown |{f∗ > λ}| = |{|f | > λ}∗|. Because f∗ is radially

symmetric and lower-semicontinuous, {f∗ > λ} is an open ball centered at the origin. By definition of a setrearrangement, {|f | > λ}∗ is also an open ball centered at the origin. As two open balls of equal measurewith same center, {|f | > λ}∗ = {f∗ > λ}.

Corollary 3.0.1. If f : Rd → C is a Borel measurable function which vanishes at infinity, then

‖f‖Lp(Rd) = ‖f∗‖Lp(Rd)

for 1 ≤ p ≤ ∞.

Proof. This follows from

‖f‖pLp(Rd) = p

∫ ∞0

λp−1|{|f | > λ}| dλ

when 1 ≤ p <∞. The case p =∞ is similar.

In preparation for our first major result, we make a few remarks. We say that a function f : Rd → [0,∞)

is strictly symmetric decreasing if f is radial and if |x| > |y| implies that f(x) < f(y). Note that this impliesf(x) > 0 for all x ∈ Rd. Also, f∗ = f .

The following inequality is usually attributed to Hardy and Littlewood.

Theorem 3.1. Let f, g : Rd → [0,∞) be vanishing at infinity. Then∫f(x)g(x) dx ≤

∫f∗(x)g∗(x) dx.

Moreover, if f is strictly symmetric decreasing, then equality holds if and only if g = g∗ almost everywhere.

Proof. First, we demonstrate the inequality. We compute, using a layer-cake decomposition:∫f(x)g(x) dx =

∫Rd

∫ ∞0

χ{f>λ}(x) dλ

∫ ∞0

χ{g>µ} dµ dx

=

∫ ∞0

∫ ∞0

|{f > λ} ∩ {g > µ}| dλ dµ.

Suppose without loss of generality that |{f > λ}| ≤ |{g > µ}|. Then {f > λ}∗ ⊆ {g > µ}∗. So

|{f > λ} ∩ {g > µ}| ≤ |{f > λ}| = |{f > λ}∗| = |{f > λ}∗ ∩ {g > µ}∗|

=

∫χ{f>λ}∗(x)χ{g>µ}∗(x) dx

25

and hence ∫f(x)g(x) dx ≤

∫Rd

∫ ∞0

χ{f>λ}∗(x) dλ

∫ ∞0

χ{g>µ}∗(x) dµ dx =

∫f∗(x)g∗(x) dx.

Next, we consider the case of equality. Suppose that f is strictly symmetric decreasing and that∫f(x)g(x) dx =∫

f∗(x)g∗(x) dx. Then∫f(x)g(x) dx =

∫f(x)g∗(x) dx, hence∫ ∞

0

∫Rdf(x)χ{g>µ}(x) dx dµ =

∫ ∞0

∫Rdf(x)χ{g>µ}∗(x) dx dµ.

Because ∫Rdf(x)χ{g>µ}(x) dx ≤

∫Rdf(x)χ{g>µ}∗(x) dx

by the demonstrated inequality, it follows that∫Rdf(x)χ{g>µ}(x) dx =

∫Rdf(x)χ{g>µ}∗(x) dx

for almost every µ.We claim that |{g > µ}∆{g > µ}∗| = 0. To see this, first observe that all open balls centered at 0 occur

as super-level sets of f . That is, {f > λ} = B(0, r(λ)) where λ 7→ r(λ) is non-increasing and r(λ) → ∞as λ → 0, since f is strictly symmetric decreasing. Note that r(λ) is continuous. If it were not, then bymonotonicity there would be a jump discontinuity at some point λ0, so that

limε→0|{f > λ0 + ε}| − |{f > λ0 − ε}| < 0

and thuslimε→0|{λ0 − ε ≤ f < λ0 + ε}| > 0.

This implies that |{f = λ0}| > 0, which is a contradiction, as f is strictly symmetric decreasing. Now, bythe above equality, we have∫ ∞

0

∫χ{f>λ}(x)χ{g>µ}(x) dx dλ =

∫ ∞0

∫χ{f>λ}(x)χ{g∗>µ}(x) dx dλ.

If A ⊆ Rd is a Borel measurable function, then the function

FA(λ) :=

∫χ{f>λ}(x)χA(x) dx = |A ∩B(0, r(λ))|

is continuous. By our demonstrated inequality,

F{g>µ}(λ) ≤∫χ{f>λ}(x)χ{g∗>µ}(x) dx = F{g∗>µ}(λ).

The assumed equality implies ∫ ∞0

F{g>µ}(λ) dλ =

∫ ∞0

F{g∗>µ}(λ) dλ

and thus F{g>µ}(λ) = F{g∗>µ}(λ) for almost every λ. Continuity of these functions gives equality for allλ > 0. That is,

|B(0, r(λ)) ∩ {g > µ}| = |B(0, r(λ)) ∩ {g∗ > µ}|

for all λ > 0. Because r(λ) is continuous and r(λ)→∞ as λ→ 0,

|B(0, r) ∩ {g > µ}| = |B(0, r) ∩ {g∗ > µ}|

for all r > 0. From this, it follows that |{g > µ}∆{g > µ}∗| = 0 for almost every µ as desired. From this, wehave g = g∗ almost everywhere.

26

A quick corollary of this inequality is that the symmetric decreasing rearrangement compresses the L2

distance between functions.

Corollary 3.1.1. Let f, g ∈ L2(Rd → [0,∞)). Then ‖f∗ − g∗‖L2 ≤ ‖f − g‖L2 .

Proof. We have

‖f∗ − g∗‖2L2 =

∫(f∗(x)− g∗(x))2 dx = ‖f∗‖2L2 + ‖g∗‖2L2 − 2

∫f∗(x)g∗(x) dx

≤ ‖f‖2L2 + ‖g‖2L2 − 2

∫f(x)g(x) dx

= ‖f − g‖2L2 .

This estimate holds for more general Lp spaces. In fact, we have the following more general theorem.

Theorem 3.2. Let φ : R→ [0,∞) be convex with φ(0) = 0, and suppose that f, g : Rd → [0,∞) are vanishingat infinity. Then ∫

φ(f∗ − g∗)(x) dx ≤∫φ(f − g)(x) dx.

Moreover, if φ is strictly convex and f is strictly symmetric decreasing, then equality holds if and only ifg = g∗ almost everywhere.

Proof. Write φ = φ+ + φ−, where

φ+(t) =

{φ(t) t ≥ 0

0 t < 0; φ−(t) =

{0 t ≥ 0

φ(t) t < 0.

Both φ+ and φ− are convex, and it suffices to prove the inequality for only φ+.Recall that convex functions are Lipschitz on compact domains, and that if c < a < x < y < b < d, we

haveφ+(a)− φ+(c)

a− c≤ φ+(y)− φ+(x)

y − x≤ φ+(d)− φ+(b)

d− b.

Thus, convex functions are absolutely continuous, hence differentiable almost everywhere, and so φ+(t) =∫ t0φ′+(s) ds. Moreover, φ′+ is a nondecreasing function, and strictly increasing if φ is strictly convex. Thus,

φ′+ has countably many (jump) discontinuities, and so we may modify φ′+ on a countable set to ensure thatφ′+ is left-continuous.

So,

φ+(f(x)− g(x)) =

∫ f(x)−g(x)

0

φ′+(t) dt =

∫ f(x)

g(x)

φ′+(f(x)− µ) dµ

=

∫ ∞0

φ′+(f(x)− µ)χ{g≤µ}(x) dµ

where the last equality follows from the fact that φ′+(t) = 0 for t < 0. Thus,∫φ+(f(x)− g(x)) dx =

∫ ∞0

∫φ′+(f(x)− µ)χ{g≤µ}(x) dx dµ.

We now make two claims, which are left as exercises to the reader:

27

1. We have the inequality ∫f(x)χ{g≤µ}(x) dx ≥

∫f∗(x)χ{g∗≤µ}(x) dx.

2. If F is nondecreasing, left continuous, and satisfies F (0) = 0, then (F ◦ f)∗ = F (f∗).

Using these two facts, we have∫φ+(f(x)− g(x)) dx ≥

∫ ∞0

∫[φ′+(f(x)− µ)]∗χ{g∗≤µ}(x) dx dµ

=

∫ ∞0

∫φ′+(f∗(x)− µ)χ{g∗≤µ}(x) dx dµ

where we have adopted the obvious convention that (f − µ)∗ := f∗ − µ to account for the fact that f − µdoes not vanish at∞. Unraveling the above integral, it follows that∫

φ+(f(x)− g(x)) dx ≥∫φ+(f∗(x)− g∗(x)) dx,

which is the desired inequality.Next, we consider the equality case. Assume that φ is strictly convex, f is strictly symmetric decreasing,

and that∫φ+(f(x)− g(x)) dx =

∫φ+(f∗(x)− g∗(x)) dx. Then for almost every µ,∫

φ′+(f(x)− µ)χ{g≤µ}(x) dx =

∫φ′+(f(x)− µ)χ{g∗≤µ}(x) dx

because f = f∗. Let h(x) := φ′+(f(x) − µ). Because f is strictly symmetric decreasing and φ′+ is strictlyincreasing, h is strictly symmetric decreasing on the set {f(x) > µ}. Rewriting the above equality with alayer-cake decomposition then gives∫ ∞

0

∫χ{h>λ}(x)χ{g≤µ}(x) dx dλ =

∫ ∞0

∫χ{h>λ}(x)χ{g∗≤µ}(x) dx dλ.

Let F{g≤µ}(λ) :=∫χ{h>λ}(x)χ{g≤µ}(x) dx and similarly F{g∗≤µ}(λ) :=

∫χ{h>λ}(x)χ{g∗≤µ}(x) dx. By

claim 1, F{g≤µ}(λ) ≥ F{g∗≤µ}(λ). Since∫∞

0F{g≤µ}(λ) dλ =

∫∞0F{g∗≤µ}(λ) dλ, we then have F{g≤µ}(λ) =

F{g∗≤µ}(λ) for almost all λ > 0. Since these functions are continuous in λ, equality in fact holds for allλ > 0. So for all λ > 0,

|{h > λ} ∩ {g ≤ µ}| = |{h > λ} ∩ {g∗ ≤ µ}|

and so

|{f > r} ∩ {g ≤ µ}| = |{f > r} ∩ {g∗ ≤ µ}| ⇒ |{f > r} ∩ {g > µ}| = |{f > r} ∩ {g∗ > µ}|

for all r > µ, since h is strictly symmetric decreasing. Working through the exact same argument with φ−in place of φ+ gives the above equality for all r < µ. As before, it follows that

|{g > µ}∆{g∗ > µ}| = 0

and consequently g = g∗ almost everywhere.

28

3.2 The one-dimensional Riesz rearrangement inequality

Next, we consider a classical rearrangement inequality due to Riesz. The motivation for studying inequali-ties of the kind to follow originates with Poincare and his work on the shapes of fluid bodies in equilibrium.The resulting variational problem deals with integrals of the form∫∫

f(x)h(y)|x− y|−1 dx dy.

More generally, variational problems involving symmetric decreasing functions h(x − y) such as the heat

kernel (4πt)11 e−

|x−y|2t are of interest.

The following inequality holds more generally in Euclidean space of any dimension, but in this we beginwith the one-dimensional case.

Theorem 3.3 (Riesz rearrangement inequality, d = 1). Let f, g, h : R→ [0,∞) be vanishing at infinity. Then

I(f, g, h) :=

∫∫f(x)g(x− y)h(y) dx dy ≤

∫∫f∗(x)g∗(x− y)h∗(y) dx dy = I(f∗, g∗, h∗).

Moreover, if g is strictly symmetric decreasing, then equality holds if and only if there exists x0 ∈ R suchthat f(x) = f∗(x− x0) and h(x) = h∗(x− x0) for almost all x ∈ R.

Proof. Using a layer-cake decomposition on f, g and h, we have

I(f, g, h) =

∫ ∞0

∫ ∞0

∫ ∞0

∫∫χ{f>λ}(x)χ{g>µ}(x− y)χ{h>ν}(y) dx dy dλ dµ dν.

Thus, it suffices to prove the inequality for f = χA, g = χB , and h = χC for finite-measure sets A,B,C. Ap-proximating these sets from above by open setsAn, Bn, andCn, we have I(χAn , χBn , χCn)→ I(χA, χB , χC)

and I(χA∗n , χB∗n , χC∗n) → I(χA∗ , χB∗ , χC∗) by the dominated convergence theorem, hence we may reducefurther and assume without loss of generality that A,B, and C are open finite-measure sets.

Because A is an open subset of R, we can write A =⋃j≥1 Ij where Ij are open, disjoint intervals such

that |Ij | ≥ |Ij+1|. Let Am :=⋃mj=1 Ij . Let Bm and Cm denote the analogous decompositions for B and

C. By the monotone convergence theorem, I(χAm , χBm , χCm) → I(χA, χB , χC) and I(χA∗m , χB∗m , χC∗m) →I(χA∗ , χB∗ , χC∗). Consequently, we may assume without loss of generality that A,B,C are each finiteunions of disjoint open intervals.

Write

χA(x) =

J∑j=1

fj(x− aj)

χB(x) =

K∑k=1

gk(x− bk)

χC(x) =

L∑l=1

hl(x− cl)

where fj , gk, hl are characteristic functions of open intervals which are centered at 0. Then we have

I(χA, χB , χC) =

J∑j=1

K∑k=1

L∑l=1

∫∫fj(x− aj)gk(x− y − bk)hl(y − cl) dx dy.

29

Observe that (fj(x − aj))∗ = f∗j . To demonstrate the desired inequality, our goal is to take each of the

characteristic functions in the integral and shift them towards 0. For t ∈ [0, 1] we define

Ijkl(t) :=

∫∫fj(x− taj)gk(x− y − tbk)hl(y − tcl) dx dy.

We claim that this function is non-increasing with respect to t. To see this, change variables via u = x− tajand u− v = x− y − tbk, so that y = v + t(aj − bk). Then

Ijkl(t) =

∫∫fj(u)gk(u− v)hl(v + t(aj − bk − cl)) du dv.

Let wjk(v) =∫fj(u)gk(u − v) du. Because fj and gk are both characteristic functions of intervals centered

at 0, this is a symmetric decreasing function in v. Thus

Ijkl(t) =

∫wjk(v)hl(v + t(aj − bk − cl)) dv

is non-increasing in t by the same reasoning.Next, define

It(χA, χB , χC) :=

J∑j=1

K∑k=1

L∑l=1

∫∫fj(x− taj)gk(x− y − tbk)hl(y − tcl) dx dy.

As t → 0, the intervals corresponding to the functions fj(x − taj), gk(x − y − tbk), and hl(y − tcl) shifttowards the origin. As t descends from 1 to 0, when two intervals corresponding to the same set A, B, or C,touch, redefine the fj ’s, gk’s, and hl’s accordingly. Carrying this out for finitely many times (because thereare finitely many intervals comprising A,B, and C), we end up with A∗,B∗, and C∗ intervals centered at 0.Moreover, since the functions Ijkl(t) are non-increasing,

I(χA, χB , χC) = I1(χA, χB , χC) ≤ I0(χA, χB , χC).

By construction, I0(χA, χB , χC) = I(χA∗ , χB∗ , χC∗). Thus, we have demonstrated the inequality

I(χA, χB , χC) ≤ I(χA∗ , χB∗ , χC∗)

as desired.Next, we consider the case of equality. Assume that g is strictly symmetric decreasing and that I(f, g, h) =

I(f∗, g∗, h∗). Applying a layer-cake decomposition to f and h yields the equality∫ ∞0

∫ ∞0

∫∫χ{f>λ}(x)g(x−y)χ{h>µ}(y)dxdy dλ dµ =

∫ ∞0

∫ ∞0

∫∫χ{f∗>λ}(x)g(x−y)χ{h∗>µ}(y)dxdy dλ dµ

since g = g∗. Consider the inner dx dy integrals. By the demonstrated inequality,∫∫χ{f>λ}(x)g(x− y)χ{h>µ}(y) dx dy ≤

∫∫χ{f∗>λ}(x)g(x− y)χ{h∗>µ}(y) dx dy.

Hence, equality of the above integrals implies that∫∫χ{f>λ}(x)g(x− y)χ{h>µ}(y) dx dy =

∫∫χ{f∗>λ}(x)g(x− y)χ{h∗>µ}(y) dx dy

for almost all λ, µ > 0. Fix λ and µ for which this holds.Set A = {f > λ} and B = {h > µ}. We wish to show that there exists x0 ∈ R such that f(x) = f∗(x−x0)

and h(x) = h∗(x − x0) for almost all x ∈ R. That is, we wish to show that A and B are intervals (up to a

30

zero-measure set) centered at the same point, and that this central point does not depend on λ or µ. Fixingλ and letting µ vary (and vice versa) shows that the central point will not depend on λ or µ.

Performing a layer-cake decomposition on g in the above equality yields∫ ∞0

∫∫χ{f>λ}(x)χ{g>σ}(x−y)χ{h>µ}(y) dx dy dσ =

∫ ∞0

∫∫χ{f∗>λ}(x)χ{g>σ}(x−y)χ{h∗>µ}(y) dx dy dσ.

Thus, applying our demonstrated inequality to the dx dy integrals as before, it follows that∫∫χ{f>λ}(x)χ{g>σ}(x− y)χ{h>µ}(y) dx dy =

∫∫χ{f∗>λ}(x)χ{g>σ}(x− y)χ{h∗>µ}(y) dx dy

for almost all σ > 0. But

σ 7→∫∫

χ{f>λ}(x)χ{g>σ}(x−y)χ{h>µ}(y) dx dy and σ 7→∫∫

χ{f∗>λ}(x)χ{g>σ}(x−y)χ{h∗>µ}(y) dx dy

are continuous functions in σ, since g is strictly symmetric decreasing. Therefore, equality holds for all σ.The fact that g is strictly symmetric decreasing then gives the equality∫∫

χA(x)χ(− r2 ,r2 )(x− y)χB(y) dx dy =

∫∫χA∗(x)χ(− r2 ,

r2 )(x− y)χB∗(y) dx dy

for all r > 0, with A and B as defined above.We claim that A and B must be intervals. To see this, pick r > |A|+ |B|. With this choice of r,∫∫

χA∗(x)χ(− r2 ,r2 )(x− y)χB∗(y) dx dy = |A| · |B|.

Then for all r > |A|+ |B|, we must have

|A| · |B| =∫∫

χA(x)χ(− r2 ,r2 )(x− y)χB(y) dx dy

=

∫χ(− r2 ,

r2 )(x)

∫χA(x+ y)χB(y) dy dx.

By this equality, the continuous function w(x) :=∫χA(x + y)χB(y) dy must have support in the interval(

− r2 ,r2

). Let IA be the smallest interval such that |IA ∩ A| = |A|, let IB be the smallest interval such that

|IB ∩B| = |B|, and let IAB be the smallest interval containing the support of w.We leave it as an exercise to the reader to show that |IAB | = |IA| + |IB |. With this claim, we then have

|IA| + |IB | < r for all r > |A| + |B|. Taking the infimum over such r gives |IA| + |IB | ≤ |A| + |B|. BecauseIA and IB were chosen to be intervals that essentially contain A and B respectively, it follows that IA = A

and IB = B, up to sets of measure zero, so that A and B are intervals.Thus ∫∫

χA(x)χ(− r2 ,r2 )(x− y)χB(y) dx dy =

∫∫χA∗(x)χ(− r2 ,

r2 )(x− y)χB∗(y) dx dy

for all r > 0, where A and B are intervals. Letting r → 0 yields∫χA(x)χB(x) dx =

∫χA∗(x)χB∗(x) dx.

From here we consider two cases. If |A| = |B|, then∫χA∗(x)χB∗(x) dx = |A|. But if A and B are not

centered at the same point, then∫χA(x)χB(x) dx < |A|, which is a contradiction. Thus, A and B are

intervals centered at the same point, as desired. If |A| > |B|, let r = |A| − |B|. Then∫∫χA∗(x)χ(− r2 ,

r2 )(x− y)χB∗(y) dx dy =

∫B∗

∫A∗∩(y− r2 ,y+ r

2 )dx dy = r|B|.

31

On the other hand, if A and B are not centered at the same point, then∫∫χA(x)χ(− r2 ,

r2 )(x− y)χB(y) dx dy =

∫B

∫A∩(y− r2 ,y+ r

2 )dx dy < r|B|.

This is a contradiction, thus A and B are centered at the same point. This completes the proof.

3.3 Steiner symmetrization; the multi-dimensional Riesz rearrangement inequality

Next, we prove a multi-dimensional analogue of the Riesz rearrangement theorem. In order to do this, weintroduce another notion of symmetrizing sets and functions.

Definition 3.2 (Steiner symmetrization).

1. Let A ⊆ Rd be a Borel measurable set of finite measure. Let e ∈ Sd−1 ⊆ Rd. The Steiner symmetriza-tion of A along e is the set A∗e which is obtained from A by symmetrizing along lines parallel toe.

2. Let f : Rd → [0,∞) be Borel measurable and vanishing at infinity. For e ∈ Sd−1 ⊆ Rd, let ρ : Rd → Rd

be the rotation taking e to e1. Write ρf(x) := f(ρ−1(x)), and let (ρf)∗1 be the one-dimensional sym-metric decreasing rearrangement of ρf along x1, keeping x2, . . . , xd fixed. Then the Steiner sym-metrization of f along e is f∗e := ρ−1(ρf)∗1.

Observe that these quantities are well-defined by the usual considerations of product measures andmeasurability of slices. To clarify the definition of the Steiner symmetrization of a function, note that f∗1

can be written as follows:f∗1(x1, x2, . . . , xd) = (x 7→ f(x1, x2, . . . , xd))

∗.

Theorem 3.4 (Riesz rearrangement inequality, d-dimensions). Let f, g, h : Rd → [0,∞) be vanishing atinfinity. Then

I(f, g, h) :=

∫∫f(x)g(x− y)h(y) dx dy ≤

∫∫f∗(x)g∗(x− y)h∗(y) dx dy = I(f∗, g∗, h∗).

Moreover, if g is strictly symmetric decreasing, then equality holds if and only if there exists x0 ∈ Rd suchthat f(x) = f∗(x− x0) and g(x) = g∗(x− x0) for almost all x ∈ R.

Proof. The proof proceeds as follows. We consider the inequality when d = 2 separately, then prove thegeneral inequality by induction. Finally, we consider the case of equality.

The case d = 2.Suppose that d = 2. By the usual layer-cake decomposition argument that we have given multiple times,

it suffices to prove the inequality when f = χA, g = χB , and h = χC for finite measure setsA,B,C ⊆ R2. Tosimplify notation, we write I(A,B,C) := I(χA, χB , χC). First, observe that I(A,B,C) ≤ I(A∗e, B∗e, C∗e)

for any e ∈ S1. This follows from Fubini’s theorem and the one-dimensional Riesz rearrangement inequal-ity, since in this case Steiner symmetrization is essentially symmetric rearrangement one direction.

Next, we define setsAk, Bk, andCk for k ≥ 1 by more or less iterating Steiner symmetrization in variousdirections, with the goal of showing Ak → A∗, Bk → B∗, and Ck → C∗. So let α be an irrational multiple of

32

2π and let Rα denote the rotation around the origin by α. Let X and Y denote the Steiner symmetrizationsalong the x and y axes, respectively. Note then that Xf = f∗1 and Y f = Rπ/2

(R−π/2f

)∗1. For k ≥ 1, define

Ak = (Y XRα)k(A)

Bk = (Y XRα)k(B)

Ck = (Y XRα)k(C).

In words, we obtain Ak by rotating A, symmetrizing in the X direction and then symmetrizing in the Ydirection, and then iterating this process k times. Because these operations are invariant under measure,|Ak| = A. Moreover,Ak has double-reflection symmetry over both the x and y axes, and hence we can write

Ak ={

(x, y) ∈ R2 : |y| < ωAk (|x|)}

where ωAk : [0,∞)→ [0,∞) is non-increasing, symmetric, and lower-semicontinuous. The analogous state-ments obviously hold for B and C.

Our goal is to show that χAkL2

−−→ χA∗ , and similarly for B and C. To see why this is sufficient, firstobserve that

|I(A∗, B∗, C∗)− I(Ak, Bk, Ck)| ≤∣∣∣∣∫∫ (χA∗ − χAk)(x)χB∗(x− y)χC∗(y) dx dy

∣∣∣∣+

∣∣∣∣∫∫ χAk(x)(χB∗ − χBk)(x− y)χC∗(y) dx dy

∣∣∣∣+

∣∣∣∣∫∫ χAk(x)χBk(x− y)(χC∗ − χCk)(y) dx dy

∣∣∣∣≤ ‖χA∗ − χAk‖L2 ‖χB∗ ∗ χC∗‖L2 + ‖χB∗ − χBk‖L2 ‖χAk ∗ χC∗‖L2

+ ‖χC∗ − χCk‖ ‖χAk ∗ χBk‖L2

≤ ‖χA∗ − χAk‖L2 ‖χB∗‖L1 ‖χC∗‖L2 + ‖χB∗ − χBk‖L2 ‖χAk‖L2 ‖χC∗‖L1

+ ‖χC∗ − χCk‖ ‖χAk‖L1 ‖χBk‖L1 .

Because A, B, and C are finite measure sets, if we can show that χAk → χA∗ and likewise for B and C, thenthis shows that I(Ak, Bk, Ck)→ I(A∗, B∗, C∗). Next, observe that

I(Ak+1, Bk+1, Ck+1) = I(Y XRαAk, Y XRαBk, Y XRαCk)

≥ I(XRαAk, XRαBk, XRαCk)

≥ I(RαAk, RαBk, RαCk)

= I(Ak, Bk, Ck).

The first and second inequalities follow from the one-dimensional Riesz rearrangement inequality, andthe last equality is because the quantity I(f, g, h) is invariant under rotation, via the obvious change ofvariables. Hence, I(Ak, Bk, Ck) is an increasing sequence which converges to I(A∗, B∗, C∗). In particular,

I(A,B,C) = I(A0, B0, C0) ≤ I(A∗, B∗, C∗)

as desired.With this goal in mind, we make the following observations. Suppose that F and G are finite measure

sets. Then ‖RαχF −RαχG‖L2 = ‖χF − χG‖L2 , and by the inequality ‖f∗ − g∗‖L2 ≤ ‖f − g‖L2 we also have

‖XχF −XχG‖L2 ≤ ‖χF − χG‖L2

‖Y χF − Y χG‖L2 ≤ ‖χF − χG‖L2 .

33

Consequently, we may assume without loss of generality that A,B, and C are bounded sets. Indeed, forevery ε > 0 there exists an r > 0 and a set A′ ⊆ Br(0) such that ‖χA − χA′‖L2 < ε. Let A′k = (Y XRα)k(A′).Then by the above inequalities we have∥∥∥χAk − χA′k∥∥∥L2

≤ ‖χA − χA′‖L2 < ε.

Thus, it is enough to consider convergence of the finite measure sets A′k, and similarly for B and C.From now on we suppose that A,B, and C are bounded. This implies that the functions ωAk , ω

Bk , and

ωCk are uniformly bounded. For now we focus on A, as the arguments for B and C are exactly the same.Using a diagonalization argument and passing to a subsequence l(k), we see that {ωAl(k)} converges at everysingle rational number. By monotonicity, {ωAl(k)} converges everywhere except at countably many points, atmost. This implies that χAl(k)

→ χA almost everywhere for some set A. Necessarily, A has double reflectionsymmetry across the x and y axes. To conclude that A = A∗, which then concludes the proof of the d = 2

inequality, it suffices to show that |A| is a ball, since |A| = limk→∞ |Al(k)| = |A|.We will accomplish this by essentially showing that RαA = A, and using that fact that α is an irrational

multiple of 2π and thus A has rotational symmetry on a dense set of angles. To do this, we take advantageof the equality case in the ‖f∗ − g∗‖L2 ≤ ‖f − g‖L2 , which presumes f to be strictly symmetric decreasing.Let

γ(x, y) = e−|x|2−|y|2

be a Gaussian on R2, which is indeed strictly symmetric decreasing. Let ak = ‖γ − χAk‖L2 . Note thatRαγ = γ, Xγ = γ, and Y γ = γ. So

ak+1 =∥∥γ − χAk+1

∥∥L2 = ‖Y XRαγ − Y XRαχAk‖L2 ≤ ‖γ − χAk‖L2

= ak.

Thus, {ak} is a decreasing sequence. Because ak ≥ 0, a := limk→∞ ak = limk→∞ al(k) exists. Moreover, bythe dominated convergence theorem, a = ‖γ − χA‖L2 . Next, note that∣∣al(k)+1 − ‖γ − Y XRαχA‖L2

∣∣ =∣∣∣∥∥γ − χAl(k)+1

∥∥L2− ‖γ − Y XRαχA‖L2

∣∣∣≤∥∥χAl(k)+1

− Y XRαχA∥∥L2

=∥∥Y XRαχAl(k)

− Y XRαχA∥∥L2

≤∥∥χAl(k)

− χA∥∥L2

which→ 0 as k →∞. On the other hand,∣∣al(k)+1 − ‖γ − Y XRαχA‖L2

∣∣→ ∣∣a− ‖γ − Y XRαχA‖L2

∣∣as k →∞. Thus, a = ‖γ − Y XRαχA‖L2 . But we have

a = ‖γ − Y XRαχA‖L2 = ‖Y γ − Y XRαχA‖L2 ≤ ‖γ −XRαχA‖L2

= ‖Xγ −XRαχA‖L2 ≤ ‖γ −RαχA‖L2

= ‖γ − χA‖L2

= a.

Therefore, we must have ‖γ −RαχA‖L2 = ‖γ −XRαχA‖L2 , i.e.,∫∫|γ −RαχA|

2 dx dy =

∫∫|γ −XRαχA|

2 dx dy.

34

Because∫|γ − RαχA|2 dx ≤

∫|γ − XRαχA|2 dx, it follows that

∫|γ − RαχA|2 dx =

∫|γ − XRαχA|2 dx

for almost every y. Thus, for almost every y, XRαχA = RαχA for almost every x. By Fubini’s theorem,XRαχA = RαχA almost everywhere in R2. The same argument shows that Y XRαχA = XRαχA almost ev-erywhere in R2, and hence Y XRαχA = RαχA almost everywhere in R2. Thus, RαχA has double reflectionsymmetry about both axes. If P denotes reflection across the y-axis, then

RαχA = PRαχA = R−αPχA = R−αχA

which implies that χA = R2αχA. Because α, and hence 2α, is an irrational multiple of 2π, the set {n(2α)

mod 2π} is dense in [0, 2π). Thus, the function

µ(θ) = ‖χA −Rθ‖

has dense set of zeroes. To conclude that A is a circle, and thus conclude the proof of the Riesz rearrange-ment inequality in the d = 2 case, it suffices to prove that µ is continuous in θ. By writing and expandingµ2 as an inner product, it suffices to show continuity of the map

θ 7→∫∫

χA ·RθχA dx dy.

To see this, let fn ∈ C∞ such that fn → χA. Then∣∣∣∣∫∫ χA ·RθχA dx dy −∫∫

fn ·RθχA dx dy∣∣∣∣ ≤ ‖χA − fn‖L2 · |A|

12 → 0

uniformly in θ. Because θ 7→∫∫

fn · RθχA dx dy is continuous, it then follows that θ 7→∫∫

χA · RθχA dx dyis a uniform limit of continuous functions, hence continuous. Thus, µ is continuous, and so A is a ball andtherefore A = A∗.

The case d > 2.Let e ∈ Sd−1 ⊆ Rd. Recall that the Steiner symmetrization of f : Rd → [0,∞) along e is f∗e(x) =

(f ◦ ρ−1)∗1(ρx), where ρ is a rotation defined by ρe = e1. Define the Schwartz symmetrization along directionsperpendicular to e by

f∗e⊥

(x) := (f ◦ ρ−1)∗e⊥1 (ρx),

where ∗e⊥1 indicates symmetrization in the variables x2, . . . , xd, keeping x1 fixed.As before, to prove the d > 2 case it suffices to consider f = χA, g = χB , and h = χC for finite

and bounded measure sets A,B,C. Let R be a rotation such that Red = ed−1. Let Y denote the Steinersymmetrization along ed, and let X denote the Schwartz symmetrizations perpendicular to ed. Let

Ak = (Y XR)kA

Bk = (Y XR)kB

Ck = (Y XR)kC.

Write x = (x′, xd) where x′ = (x1, . . . , xd−1). By induction, we have∫∫f(x′, xd)g(x′ − y′, xd − yd)h(y′, yd) dx

′ dy′ ≤∫∫

(Xf)(x′, xd)(Xg)(x′ − y′, xd − yd)(Xh)(y′, yd) dx′ dy′.

In particular, I(Ak+1, Bk+1, Ck+1) ≥ I(Ak, Bk, Ck), so by the same argument as before it suffices to prove

χAkL2

−−→ χA∗ , and likewise for B and C.

35

By construction, we can write

Ak ={x ∈ Rd : |xd| < ωAk (|x′|)

}where ωAk is a symmetric and non-increasing function. The ωAk ’s are uniformly bounded by boundednessof A. Exactly as before, we may pass to a subsequence to get χAl(k)

→ χA almost everywhere. As eachAl(k) is rotationally symmetric around the xd axis, the limiting set A is also rotationally symmetric aroundthe xd axis. Also as before, define the Gaussian γ and set ak = ‖γ − χAk‖L2 . Running through the samecomputation gives Y XRA = RA almost everywhere. Hence, RA is rotationally symmetric along the xdaxis, which implies that A is rotationally symmetric around the xd axis and the xd−1 axis. We will show thatthis implies that A is a ball. Then because |A| = |A| = |A∗| it will follow that A = A∗ almost everywhere, asdesired.

Let ϕε be a radial smooth approximation to the identity and set χε(x) = (ϕε ∗ χA)(x). Then χε and χAhave the same symmetry properties. As such, we can write

χε(x1, . . . , xd) = u(√x2

1 + · · ·+ x2d−1, xd) = v(

√x2

1 + · · ·+ x2d−2 + x2

d, xd−1)

for smooth functions u and v of two variables. Writing ρ2 = x21 + · · ·+ x2

d−1, we have u(√ρ2 + x2

d−1, xd) =

v(√ρ2 + x2

d, xd−1) for all xd−1, xd ∈ R and for all ρ > 0. In particular, setting xd = 0 we have

u(√ρ2 + x2

d−1, 0) = v(√ρ2, xd−1)

for all xd−1 ∈ R and for all ρ > 0. Choosing ρ2 = x21 + · · · + x2

d−2 + x2d gives χε(x) = u(|u|, 0), hence χε is

spherically symmetric and so A is a ball.This completes the proof of the inequality for d > 2.

The case of equality.Finally, we consider the equality statement. Suppose that g is strictly symmetric decreasing and that

equality holds. As before, we assume that f = χA and h = χB for finite measure sets A and B. Using the(x′, xd) notation from above, we have∫

R

∫R

∫Rd−1

∫Rd−1

f(x′, xd)g(x′ − y′, xd − yd)h(y′, yd) dx′ dy′ dxd dyd

=

∫R

∫R

∫Rd−1

∫Rd−1

f∗(x′, xd)g∗(x′ − y′, xd − yd)h∗(y′, yd) dx′ dy′ dxd dyd.

By the inequality in the statement of the theorem,∫Rd−1

∫Rd−1

f(x′, xd)g(x′ − y′, xd − yd)h(y′, yd) dx′ dy′

≤∫Rd−1

∫Rd−1

f∗(x′, xd)g∗(x′ − y′, xd − yd)h∗(y′, yd) dx′ dy′.

Hence, the assumed equality gives∫Rd−1

∫Rd−1

f(x′, xd)g(x′ − y′, xd − yd)h(y′, yd) dx′ dy′

=

∫Rd−1

∫Rd−1

f∗(x′, xd)g∗(x′ − y′, xd − yd)h∗(y′, yd) dx′ dy′

for almost all (xd, yd).

36

By induction we get sets Axd and Byd corresponding to the functions f(·, xd) and h(·, yd) which areballs in Rd−1 centered at the same point, by the equality case of the one-dimensional Riesz rearrangementinequality. Also, by the same argument as before, this central point is independent of xd and yd. Thus, Aand B are rotationally symmetric with respect to a line L1 which is parallel to the xd axis. Similarly, A andB must also be rotationally symmetric with respect to another line L2 parallel to the ed−1 axis.

We claim that L1 ∩ L2 6= ∅. In d = 2, this is obvious, but the existence of skew-symmetric lines in higherdimensions makes the d ≥ 3 case more complicated. This can be seen via a ping pong argument:2 if L1 andL2 do not intersect, then by rotating aroundL2 180 degrees it follows that the sets are rotationally symmetricabout L1 and and an affine shift of L1. By rotating around this shift of L1, the sets are then also rotationallysymmetric around an affine shift of L2. Continuing this process, we contradict the finite measure of A andB.

Thus, A and B are rotationally symmetric about two axes. As we argued previously, it follows that Aand B are balls.

3.4 The Polya-Szego inequality

Our first application of the Riesz rearrangement inequality is a special case of the Polya-Szego inequality.We begin with a lemma.

Lemma 3.5. Let f : Rd → C be such that f ∈ L1loc and ∇f ∈ L1

loc. Then

∇|f |(x) =

{(u∇u+v∇v)(x)

|f(x)| if f = u+ iv 6= 0

0 if f = u+ iv = 0

in the sense of distributions. In particular, |∇|f || ≤ |∇f | almost everywhere.

Proof. First we demonstrate the “in particular” statement, assuming the formula for∇|f |. For f(x) 6= 0 (upto a choice of representative), we have by Cauchy-Schwarz

|∇|f(x)||2 =

[u2|∇u|2 + v2|∇v|2 + 2uv∇u∇v

](x)

|f(x)|2

≤[u2|∇u|2 + v2|∇v|2 + u2|∇v|2 + v2|∇u|2

](x)

|f(x)|2

=(u2 + v2)(|∇u|2 + |∇v|2)

|f |2(x)

= |∇f(x)|2

as desired.To prove the formula in the statement, let ψ ∈ C∞c . We wish to show∫

∇ψ |f | dx = −∫{f 6=0}

ψu∇u+ v∇v|f |

dx.

The first issue we need to deal with is moving the derivative onto |f |. To do this, for δ > 0 consider√δ2 + |f |2. Note that for small δ,

√δ2 + |f |2 ≤ 1 + |f |, so by the dominated convergence theorem∫∇ψ

√δ2 + |f |2 dx→

∫∇ψ |f | dx

2Drawing a picture here is helpful.

37

as δ → 0. Next, let ϕε be an approximation to the identity, uε = u ∗ ϕε, and vε = v ∗ ϕε. Then uε → u andvε → v almost everywhere and in L1

loc. Also,∇(uε + ivε)→ ∇f almost everywhere and in L1loc. Note that∫

∇ψ(√

δ2 + u2 + v2 −√δ2 + u2

ε + v2ε

)dx =

∫∇ψ (u− uε)(u+ uε) + (v − vε)(v + vε)√

δ2 + u2 + v2 +√δ2 + u2

ε + v2ε

dx

so that∣∣∣∣∫ ∇ψ (√δ2 + u2 + v2 −√δ2 + u2

ε + v2ε

)dx

∣∣∣∣ ≤ ∫ |∇ψ| |u− uε|(|u|+ |uε|) + |v − vε|(|v|+ |vε|)√δ2 + u2 + v2 +

√δ2 + u2

ε + v2ε

dx

≤ 2

∫|∇ψ| (|u− uε|+ |v − vε|) dx.

Because uε → u and vε → v in L1loc, since ψ has compact support, it follows that∫∇ψ√δ2 + u2

ε + v2ε dx→

∫∇ψ√δ2 + u2 + v2 dx

as ε→ 0. Integrating by parts, we then have∫∇ψ√δ2 + u2 + v2 dx = lim

ε→0

∫∇ψ√δ2 + u2

ε + v2ε dx

= − limε→0

∫ψuε∇uε + vε∇vε√δ2 + u2

ε + v2ε

dx

= − limε→0

∫ψuε∇u+ vε∇v√δ2 + u2

ε + v2ε

dx− limε→0

∫ψuε(∇uε −∇u) + vε(∇vε −∇v)√

δ2 + u2ε + v2

ε

dx.

Consider the second integral. We have∣∣∣∣∣∫ψuε(∇uε −∇u) + vε(∇vε −∇v)√

δ2 + u2ε + v2

ε

dx

∣∣∣∣∣ ≤∫|ψ| (|∇uε −∇u|+ |∇vε −∇v|) dx

Again by L1loc convergence and compact support of ψ, this integral converges to 0 as ε → 0. In the first

integral, observe that ∣∣∣∣∣∫ψuε∇u+ vε∇v√δ2 + u2

ε + v2ε

dx

∣∣∣∣∣ ≤∫|ψ| (|∇u|+ |∇v|) dx <∞.

Thus, by the dominated convergence theorem,∫∇ψ√δ2 + u2 + v2 dx = − lim

ε→0

∫ψuε∇u+ vε∇v√δ2 + u2

ε + v2ε

dx

= −∫ψ

u∇u+ v∇v√δ2 + u2 + v2

dx.

Letting δ → 0 gives ∫∇ψ|f | dx = −

∫ψu∇u+ v∇v|f |

dx

as desired.

Theorem 3.6 (Polya-Szego). Let f : Rd → C be vanishing at infinity such that f ∈ L1loc and ∇f ∈ L2. Then

‖∇f∗‖L2 ≤ ‖∇f‖L2 .

38

Proof. First, note that ∇f ∈ L2 implies ∇f ∈ L1loc, hence by the previous lemma we have |∇|f || ≤ |∇f |.

Consequently, it suffices to consider functions f satisfying f ≥ 0.Next, we make a further reduction. For ε > 0, define

fε := min

{1

ε, [f − ε]+

}=

0 f ≤ ε

f − ε ε ≤ f ≤ ε+ ε−1

1ε f ≥ ε+ ε−1

= ϕε ◦ f

where

ϕε(x) =

0 x ≤ ε

x− ε ε ≤ x ≤ ε+ ε−1

1ε x ≥ ε+ ε−1

.

Note that ∫|fε| dx =

1

ε

∣∣{f ≥ ε+ ε−1}∣∣+

∫{ε≤f≤ε+ε−1}

|f − ε| dx

≤ 1

ε

∣∣{f ≥ ε+ ε−1}∣∣+

1

ε|{f ≥ ε}| .

Because f vanishes at infinity, the above super-level sets have finite measure, hence∫|fε| dx < ∞ and so

fε ∈ L1. Similarly, it can be shown that fε ∈ L2. Also note that

∇fε = ∇fχ{ε≤f≤ε+ε−1}L2

−−→ ∇f

as ε→ 0 by the monotone convergence theorem.Next, recall that because ϕε is symmetric and non-decreasing, (fε)

∗ = (ϕε ◦ f)∗ = ϕε ◦ f∗ = (f∗)ε. Thus,

we have∇(fε)∗ = ∇(f∗)ε

L2

−−→ ∇f∗. This implies that it suffices to consider f ∈ L1 ∩ L2 and∇f ∈ L2.By Plancharel and the monotone convergence theorem, we have∫

|∇f |2 dx =

∫|ξ|2|f(ξ)|2 dξ = lim

t→0

∫1− e−t|ξ|2

t|f(ξ)|2 dξ.

Because f ∈ L1 and f ≥ 0, we can distribute and use Fubini to get∫|∇f |2 dx = lim

t→0

1

t

[‖f‖2L2 − (2π)−d

∫e−t|ξ|

2

∫e−ix·ξf(x) dx

∫eiy·ξf(y) dy dξ

]= limt→0

1

t

[‖f‖2L2 − (2π)−d

∫∫f(x)f(y)

∫e−t|ξ|

2−i(x−y)·ξ dξ dx dy

]= limt→0

1

t

[‖f‖2L2 − (2π)−d

∫∫f(x)f(y)

∫e−t(ξ+

i(x−y)2t )

2

e−|x−y|2

4t dξ dx dy

]where we are using the convention that for z ∈ Cd, z2 =

∑dj=1 z

2j . Changing variables and integrating the

resulting ξ-Gaussian then gives∫|∇f |2 dx = lim

t→0

1

t

[‖f‖2L2 − (2π)−dt−

d2 π

d2

∫∫f(x) e−

|x−y|24t f(y) dx dy

].

By the Riesz rearrangement inequality, and by equi-measurability of f∗ with f ,∫|∇f |2 dx ≥ lim

t→0

1

t

[‖f∗‖2L2 − (2π)−dt−

d2 π

d2

∫∫f∗(x) e−

|x−y|24t f∗(y) dx dy

].

Beginning with the expression on the right and undoing all of the previous calculations, we recover theinequality ‖∇f‖2L2 ≥ ‖∇f∗‖2L2 .

39

We make passing observation that the heat kernel (4πt)−d2 e−

|x−y|24t appears in the above proof. This

use of the heat kernel is possible because the stated estimate is in L2. A generalization of the Polya-Szegoinequality holds for other values of p, but the above method does not give this result.

3.5 The hydrogen atom energy functional

As an application of the Polya-Szego inequality, we consider the energy functional

E(f) :=

∫R3

|∇f(x)|2 − |f(x)|2

|x|dx

for f ∈ H1(R3). Such a functional can be taken to represent the energy of a hydrogen atom. We willdemonstrate that

inf‖f‖L2=1

E(f)

is achieved by a symmetric decreasing function. Via a scaling argument, we show that the infimum isnegative. Let ϕ ∈ C∞c (R3) with ‖ϕ‖L2 = 1. For λ > 0, let ϕλ(x) = λ−

32ϕ(xλ

). Then ‖ϕλ‖L2 = ‖ϕ‖2 = 1, and

E(ϕλ) = λ−2λ−3

∫ ∣∣∣(∇ϕ)(xλ

)∣∣∣2 dx− λ−3

∫ ∣∣ϕ (xλ)∣∣2|x|

dx

= λ−2 ‖∇ϕ‖2L2 − λ−1

∫|ϕ(x)|2

|x|dx

Because ‖∇ϕ‖2L2 and |ϕ(x)|2|x| dx are fixed numbers, for large λ the above quantity is negative. Hence,Emin :=

inf‖f‖2=1E(f) < 0. To show that the problem is well-posed, we also must prove that the infimum is notidentically −∞. To do this, we first recall Hardy’s inequality:

Proposition 3.4 (Hardy’s Inequality). Let f ∈ S(Rd) and 0 ≤ s < d. Then∥∥∥∥f(x)

|x|s

∥∥∥∥Lp

. ‖|∇|sf‖Lp

for all 1 < p < ds .

By this inequality, we have∥∥∥∥f(x)

|x| 12

∥∥∥∥L2

.∥∥∥|∇| 12 f(x)

∥∥∥L2

=∥∥∥|ξ| 12 f(ξ)

∥∥∥L2

=∥∥∥|ξ| 12 |f(ξ)| 12 |f(ξ)| 12

∥∥∥L2≤∥∥∥|ξ| 12 |f(ξ)| 12

∥∥∥L4

∥∥∥|f(ξ)| 12∥∥∥L4

= ‖∇f‖12

L2 ‖f‖12

L2 .

So for ‖f‖L2 = 1, E(f) ≥ ‖∇f‖2L2 − C ‖∇f‖L2 where the constant C comes from Hardy’s inequality. SoEmin ≥ infx>0 x

2 − Cx > −∞ as desired.Next, we show that the infimum is achieved. Let fn ∈ H1(R3) be a decreasing minimizing sequence,

that is, ‖fn‖L2 = 1 for all n and E(fn) → Emin. First, we claim that we can find a minimizing sequence ofradially symmetric decreasing functions. Indeed, consider f∗n. We have ‖f∗n‖L2 = ‖fn‖L2 = 1, and by thePolya-Szego inequality, ‖∇f∗n‖L2 ≤ ‖∇fn‖L2 . By our very first rearrangement estimate,∫

|fn(x)|2

|x|dx =

⟨1

|x|, |fn(x)|2

⟩≤⟨

1

|x|, (|fn(x)|2)∗

⟩=

⟨1

|x|, |f∗n(x)|2

⟩=

∫|f∗n(x)|2

|x|dx.

40

Hence,E(f∗n) ≤ E(fn), which implies that f∗n is also a minimizing sequence. Consequently, we may assumefrom now on that each fn is radially symmetric and decreasing.

Next, we claim that the sequence {fn} is bounded inH1(R3). If it were not, then because ‖fn‖L2 = 1, thequantities ‖∇f‖L2 are unbounded. But since E(f) ≥ ‖∇f‖2L2 −C ‖∇f‖L2 , this implies that lim supE(fn) =

∞. This contradicts the assumption that E(fn) decreases to Emin < 0. Hence, {fn} is indeed bounded inH1(R3). Because H1(R3) can be reformulated on the Fourier side as an L2-space with a weighted measure,and bounded sets in L2 spaces are weakly compact, it follows that (passing to a subsequence) fn ⇀ f

weakly in H1(R3).We need to demonstrate that the weak limit f is an optimizer. First, we will show that

E(f) ≤ lim inf E(fn) = Emin

as desired. By weak lower semi-continuity of the L2 norm, ‖∇f‖L2 ≤ lim inf ‖∇fn‖L2 . This is great, but itdoesn’t help in demonstrating the energy inequality because it tells us nothing about the potential energyterm −

∫ |f(x)|2|x| dx. To remedy this, we invoke the following fact.

Lemma 3.7 (Strauss). The space H1(R3) embeds compactly into Lp(R3) for 2 < p < 6.

The proof of this lemma will come in the future. Here we only remark that the upper bound 6 for p comesfrom the scaling condition of the Sobolev embedding theorem.

In particular, this lemma implies that because fn ⇀ f weakly inH1(R3), then fn → f strongly in Lp(R3)

for 2 < p < 6. With this in mind, we estimate as follows:∣∣∣∣∫ |fn(x)|2

|x|dx−

∫|f(x)|2

|x|dx

∣∣∣∣ ≤ ∣∣∣∣∫ (|fn| − |f |)(|fn|+ |f |)|x|

dx

∣∣∣∣ ≤ ∫ |fn − f |(|fn|+ |f |)|x|dx.

In order to use the fact that fn → f strongly in certain Lp spaces, we want to apply Holder’s inequality. Butsince 1

|x| is only in weak Lp spaces, we split up the integral:∣∣∣∣∫ |fn(x)|2

|x|dx−

∫|f(x)|2

|x|dx

∣∣∣∣ ≤ ∫|x|≤1

|fn − f |(|fn|+ |f |)|x|

dx+

∫|x|≥1

|fn − f |(|fn|+ |f |)|x|

dx.

In the first integral, 1|x| ∈ L

p(B(0, 1) ⊆ R3) for p < 3. Applying Holder’s inequality twice gives∫|x|≤1

|fn − f |(|fn|+ |f |)|x|

dx ≤∥∥∥∥ 1

|x|

∥∥∥∥L2(B(0,1))

‖|fn − f |(|fn|+ |f |)‖L2(B(0,1))

≤∥∥∥∥ 1

|x|

∥∥∥∥L2(B(0,1))

‖fn − f‖L4 (‖fn‖L4 + ‖f‖L4) .

The first norm is finite by the above comment. We claim that the quantities ‖fn‖L4 + ‖f‖L4 are bounded inn. To see this, recall the Gagliardo-Nirenberg ineqaulity:

Proposition 3.5 (Gagliardo-Nirenberg inequality). Fix d ≥ 1 and 0 < p <∞ for d = 1, 2, or 0 < p < 4d−2 for

d ≥ 3. Then for all f ∈ S(Rd),

‖f‖p+2Lp+2 . ‖f‖p+2− pd2

L2 ‖∇f‖pd2

L2 .

When p = 2 and d = 3, the inequality above takes the form ‖f‖4L4 . ‖f‖L2 ‖∇f‖3L2 . Because {fn} is a

41

bounded sequence in H1(R3), it follows that ‖fn‖L4 + ‖f‖L4 is bounded. Then because ‖fn − f‖L4 → 0,∫|x|≤1

|fn − f |(|fn|+ |f |)|x|

dx→ 0.

Similarly,∫|x|≥1

|fn − f |(|fn|+ |f |)|x|

dx ≤∥∥∥∥ 1

|x|

∥∥∥∥L4(|x|≥1)

‖fn − f‖L4 (‖fn‖L2 + ‖f‖L2) . ‖fn − f‖L4 .

Thus, ∫|fn − f |(|fn|+ |f |)

|x|dx→ 0.

From this and the fact that ‖∇f‖L2 ≤ lim inf ‖∇fn‖L2 it follows that

E(f) ≤ lim inf E(fn) = Emin

as desired. Therefore, f does achieve the energy value Emin. It only remains to verify that f is admissible,i.e., that ‖f‖L2 = 1. First observe that f is not identically 0, because E(f) = Emin < 0. Next, by weaklower semicontinuity of the L2 norm, ‖f‖L2 ≤ lim inf ‖fn‖L2 = 1. Assume for the sake of contradiction that‖f‖L2 < 1. Let g = f

‖f‖L2. Then E(g) = 1

‖f‖2L2E(f). Because E(f) < 0 and ‖f‖L2 < 1, this implies that g

is a super-optimizer of the problem in question, which is a contradiction. Thus, ‖f‖L2 = 1, and so we havedemonstrated that Emin is achieved, and it is achieved by a radially symmetric decreasing function.

In fact, a stronger statement is true. All of the optimizers of the energy functional E(f) above areradially symmetric; in particular, any optimizer satisfies f = eiθf∗. We sketch the argument here. If f is anoptimizer, then as seen above it follows that f∗ is also an optimizer. From E(f) = E(f∗) and Polya-Szego

we deduce the equalities ‖∇f‖L2 = ‖∇f∗‖L2 and∥∥∥∥ f(x)

|x|12

∥∥∥∥L2

=

∥∥∥∥ f∗(x)

|x|12

∥∥∥∥L2

. By the equality case of our first

rearrangement inequality, |f | = f∗ almost everywhere. Unfortunately this does not rule out the possibilitythat f itself is not radial. However, we then have

‖∇f‖L2 = ‖∇f∗‖L2 = ‖∇|f |‖L2 .

Having previously shown |∇|f || ≤ |∇f | almost everywhere, it follows that |∇|f || = |∇f | almost every-where. Then if f = u + iv, we have u∇v = v∇u almost everywhere. Writing f = reiθ where r and θ arefunctions of x, we then have r∇θ = 0 almost everywhere. Because f is not identically 0, θ is constant asdesired.

42

4 5/3 — 5/8: Compactness in Lp Spaces

4.1 The Riesz compactness theorem

In variational problems like the one encountered with the hydrogen atom above, and more generally in thestudy of function spaces, the question often arises whether a sequence or family of functions is precompact.For continuous functions, the answer is given by the well-known Arzela-Ascoli theorem. Here, we presentan analogous result for Lp spaces and discuss some of its consequences.

Theorem 4.1 (Riesz compactness). Fix 1 ≤ p < ∞. A family F ⊆ Lp(Rd) is precompact in Lp(Rd) if andonly if it satisfies the following three properties:

1. (Boundedness) There exists A > 0 such that ‖f‖Lp ≤ A for all f ∈ F .

2. (Equicontinuity) For every ε > 0 there exists a δ > 0 such that∫|f(x+y)−f(x)|p dx < ε for all |y| < δ

and for all f ∈ F .

3. (Tightness) For every ε > 0, there exists an R > 0 such that∫|x|≥R |f(x)|p dx < ε for all f ∈ F .

Proof. The forward direction is fairly straightforward. Suppose that F is precompact in Lp(Rd). Then forevery ε > 0, there exists f1, . . . , fn ∈ F such that F ⊆

⋃nj=1B(fj ,

ε10 ). Because any function f ∈ F lies in

one of these balls, ‖f‖Lp ≤ε10 +maxj ‖fj‖Lp . Thus, F is bounded. Next, for |y| < δ = δ(ε) << 1 sufficiently

small, we have

‖f(·+ y)− f‖Lp ≤ 2ε

10+ max

j‖fj(·+ y)− fj‖Lp < ε,

so that F is equicontinuous. Finally,

‖f‖Lp(|x|≥R) ≤ε

10+ max

j‖fj‖Lp(|x|≥R) < ε

for R sufficiently large. Hence, F is tight.The other direction is more involved. Suppose that F is bounded, equicontinuous, and tight. Fix ε > 0.

We wish to show that there exist f1, . . . , fn ∈ F so that F ⊆⋃nj=1B(fj , ε).

We will approximate F by a continuous family of functions and then apply Arzela-Ascoli. Let ϕ ∈ C∞cbe a radial bump function satisfying

∫ϕdx = 1 and

ϕ(x) =

{cd |x| ≤ 1

2

0 |x| > 1.

for some dimensional constant cd. For R > 0 and f ∈ F , define

fR(x) = ϕ( xR

)∫f(x− x

R

)ϕ(y) dy = ϕ

( xR

)∫f(y)Rdϕ(R(x− y)) dy.

The passage from f to fR “fuzzes” f at a scale 1/R. Define FR = { fR : f ∈ F }. Because fR is definedas convolution with a smooth function multiplied by a compactly supported function, FR is a continuousfamily of functions supported on B(0, R). Also,

‖f − fR‖Lp =∥∥∥(1− ϕ

( xR

))f(x)

∥∥∥Lp

+

∥∥∥∥ϕ( xR)(f(x)−

∫f(x− x

R

)ϕ(y) dy

)∥∥∥∥Lp.

43

Because F is tight, we can estimate the first norm by∥∥∥(1− ϕ( xR

))f(x)

∥∥∥Lp

= ‖f‖Lp(|x|>R) <ε

10

for R = R(ε) >> 1 sufficiently large. For the second norm, we use the fact that ϕ has unit mass to estimate∥∥∥∥ϕ( xR)(f(x)−

∫f(x− x

R

)ϕ(y) dy

)∥∥∥∥Lp≤∥∥∥∥∫ (f(x)− f

( xR

))ϕ(y) dx

∥∥∥∥Lpx

≤∫|ϕ(y)|

∥∥∥f(x)− f( xR

)∥∥∥Lpx

dy

≤ ε

10

for R sufficiently large, by equicontinuity of F . Thus, for R sufficiently large we have ‖f − fR‖Lp <ε5 .

Next, we claim that FR is uniformly bounded and equicontinuous. To see boundedness, by Young’sinequality and a change of variables we have

‖fR‖L∞ .∥∥∥f (x− y

R

)∥∥∥Lpy‖ϕ‖Lp′ . R

dp ‖f‖Lp . R

dpA.

To see equicontinuity, write

|fR(x+ y)− fR(x)| ≤∣∣∣∣[ϕ(x+ y

R

)− ϕ

( xR

)]∫f(x+ y − z

R

)ϕ(z) dz

∣∣∣∣+

∣∣∣∣ϕ( xR)∫ [

f(x+ y − z

R

)− f

(x− z

R

)]ϕ(z) dz

∣∣∣∣.

∣∣∣∣ϕ(x+ y

R

)− ϕ

( xR

)∣∣∣∣R dp ‖f‖Lp ‖ϕ‖Lp′

+∥∥∥f (x+ y − z

R

)− f

(x− z

R

)∥∥∥Lpz‖ϕ‖Lp′

.

∣∣∣∣ϕ(x+ y

R

)− ϕ

( xR

)∣∣∣∣R dpA+R

dp ‖f(·+ y)− f(·)‖Lp .

Because ϕ is smooth and because F is tight, for |y| < δ = δ(ε,R) with δ sufficiently small, |fR(x + y) −fR(x)| < ε uniformly in f .

Therefore, asFR is a continuous family of uniformly bounded and equicontinuous compactly supportedfunctions, it follows by the Arzela-Ascoli theorem that FR is precompact in the L∞ topology. So for everyε > 0 there exists f1, . . . , fn ∈ F such that FR ⊆

⋃nj=1BL∞ ((fj)R, ε).

Fix f ∈ F . There exists a j such that fR ∈ BL∞((fj)R, ε). By the triangle inequality, we have

‖f − fj‖Lp ≤ ‖f − fR‖Lp + ‖fj − (fj)R‖Lp + ‖fR − (fj)R‖Lp

<ε

5+ε

5+ cd ‖fj − (fj)R‖L∞ R

dp .

Because ‖fj − (fj)R‖L∞ < ε, by choosing ε sufficiently small it follows that ‖f − fj‖Lp < ε, and henceF ⊆

⋃nj=1B(fj , ε) as desired.

Our first corollary of the Riesz compactness theorem is an alternative characterization of precompactfamilies when p = 2. In particular, in this case we can replace the equicontinuity condition by tightness inthe Fourier domain.

Corollary 4.1.1. A family F ⊆ L2(Rd) is precompact if and only if it satisfies the following two properties:

44

1. There exists A > 0 such that ‖f‖L2 ≤ A for all f ∈ F .

2. For every ε > 0, there exists R > 0 such that∫|x|≥R

|f(x)|2 dx+

∫|ξ|≥R

|f(ξ)|2 dξ < ε

for all f ∈ F .

Proof. The forward direction is almost exactly the same as in the proof of the Riesz compactness theorem;tightness in the Fourier domain (i.e. the estimate

∫|ξ|≥R |f(ξ)|2 dξ < ε) comes from the fact that the Fourier

transform is an isometry on L2(Rd).Conversely, suppose the two conditions hold. By the Reisz compactness theorem, it suffices to verify

that F is equicontinuous.Fix ε > 0. Using the fact that the Fourier transform is an isometry on L2(Rd), for anyR > 0 we can write

‖f(x+ y)− f(x)‖L2x

=∥∥∥eiy·ξ f(ξ)− f(ξ)

∥∥∥L2ξ

≤∥∥∥eiy·ξ f(ξ)− f(ξ)

∥∥∥L2(|ξ|≥R)

+∥∥∥f(ξ)

[eiy·ξ − 1

]∥∥∥L2(|ξ|≤R)

.

When |ξ| is large, there is a lot of oscillation coming from the eiy·ξ phase. Thus, we simply use the trianglequality and tightness of f to estimate the first term:∥∥∥eiy·ξ f(ξ)− f(ξ)

∥∥∥L2(|ξ|≥R)

≤ 2∥∥∥f∥∥∥

L2(|ξ|≥R)<ε

2

for R sufficiently large. For the second term, we have∥∥∥f(ξ)[eiy·ξ − 1

]∥∥∥L2(|ξ|≤R)

≤∥∥∥f · |y| · |ξ|∥∥∥

L2(|ξ|≤R)≤ |y|RA.

Then∥∥∥f(ξ)

[eiy·ξ − 1

]∥∥∥L2(|ξ|≤R)

< ε/2 for |y| < δ where δ(R, ε) is sufficiently small. For such a choice of δ,

‖f(x+ y)− f(x)‖L2x< ε uniformly in f and hence F is equicontinuous.

Before continuing with applications of the Riesz compactness theorem, we (again) recall the Gagliardo-Nirenberg inequality.

Proposition 4.1 (Gagliardo-Nirenberg). Fix d ≥ 1. Let p satisfy 2 ≤ p <∞ if d = 1 or d = 2, and 2 ≤ p < 2dd−2

if d ≥ 3. Then for f ∈ H1(Rd),

‖f‖Lp . ‖f‖2d−p(d−2)

2p

L2 ‖∇f‖d(p−2)

2p

L2 .

Proof. This inequality can be proven in a number of ways. Here, we present a short proof using theLittlewood-Paley projections and Bernstein’s inequalities.

It suffices to consider Schwartz functions f . Because the Littlewood-Paley projections of f converge tof in Lp for the prescribed values of p, we have

‖f‖Lp ≤∑N∈2Z

‖fN‖Lp .

By Bernstein’s first inequality,

‖fN‖Lp . Nd2−

dp ‖fN‖L2 . N

d2−

dp ‖f‖L2 .

45

On the other hand, by Bernstein’s second inequality we have

‖fN‖Lp . Nd2−

dpN−1 ‖∇fN‖L2 . N

d2−

dp−1 ‖∇f‖L2 .

Together, this implies that

‖f‖Lp .∑N∈2Z

min{N

d2−

dp ‖f‖L2 , N

d2−

dp−1 ‖∇f‖L2

}.

Because the Gagliardo-Nirenberg inequality is trivial when p = 2, we may assume that p > 2. Consequently,d2 −

dp > 0. Also, d2 −

dp − 1 = d−2

2 −dp < 0. This latter statement is clear when d = 1 or d = 2, and when

d ≥ 3 this follows from the fact that p < 2dd−2 . Using the positivity and negativity of the above exponents,

we split the sum into two summable series to get the desired inequality. Explicitly,

‖f‖Lp .∑

N.‖∇f‖

L2‖f‖

L2

Nd2−

dp ‖f‖L2 +

∑N&

‖∇f‖L2

‖f‖L2

Nd2−

dp−1 ‖∇f‖L2

. ‖∇f‖d2−

dp

L2 ‖f‖1−d2 + d

p

L2 + ‖∇f‖d2−

dp

L2 ‖f‖1+ dp−

d2

L2

. ‖f‖2d−p(d−2)

2p

L2 ‖∇f‖d(p−2)

2p

L2 .

Theorem 4.2 (Rellich-Kondrachov). Let B1 denote the unit ball in H1(Rd). Let χR be a cutoff function onBR(0) ⊆ Rd, smooth or not. Then the family F := {χRf : f ∈ B1 } is compact in Lp for2 ≤ p <∞ d = 1, 2

2 ≤ p < 2dd−2 d ≥ 3

.

Proof. First, observe that F is closed. Indeed, suppose that fn ∈ B1 is a sequence of functions such thatχRfn converges in Lp. Because fn ∈ B1, the sequence fn is bounded in L2. Passing to a subsequence, wehave fn ⇀ f weakly in L2. This implies that χRfn ⇀ χRf weakly in L2. As weak limits are unique, itfollows that χRfn → χrf in Lp and hence F is closed. Thus, it is sufficient to prove that F is precompact.

By Gagliardo-Nirenberg, ‖χRf‖Lp ≤ ‖f‖Lp . ‖f‖H1 , hence the family F is bounded. Tightness of F isimmediate, because each function in F has compact support.

It remains to show equicontinuity. Fix ε > 0. We have

‖χRf(x+ y)− χRf(x)‖Lpx ≤ ‖[χR(x+ y)− χR(x)]f(x+ y)‖Lpx + ‖χR(x)[f(x+ y)− f(x)]‖Lpx .

To estimate the first term, we want to use Holder, the Lp continuity property of translations on χR toget smallness, and then Gagliardo-Nirenberg on the f term to get uniformity in f . However, becausetranslation of χR is not continuous in the L∞ norm, we have to be careful. We have

‖[χR(x+ y)− χR(x)]f(x+ y)‖Lpx ≤ ‖χR(x+ y)− χR(x)‖Lrx ‖f(x+ y)‖Lqx

for r and q satisfying

p < q <

∞ d = 1, 2

2dd−2 d ≥ 3

;1

p=

1

r+

1

q.

46

By Gagliardo-Nirenberg, ‖f(x+ y)‖Lqx is uniformly bounded in f . For |y| < δ(ε) << 1 where δ is suffi-ciently small, we have ‖χR(x+ y)− χR(x)‖Lrx < ε.

To estimate the second term, we simply ignore the χR and use Gagliardo-Nirenberg:

‖χR(x)[f(x+ y)− f(x)]‖Lpx ≤ ‖f(x+ y)− f(x)‖Lpx. ‖f(x+ y)− f(x)‖θL2

x‖∇x[f(x+ y)− f(x)]‖1−θL2

x

where θ is the correct exponent dictated by Gagliardo-Nirenberg. By the triangle inequality,

‖∇x[f(x+ y)− f(x)]‖1−θL2x

. ‖∇f‖1−θL2

and hence is uniformly bounded in f . By the fundamental theorem of calculus, ‖f(x+ y)− f(x)‖θL2x.

(|y| ‖∇f‖L2)θ. Thus, for |y| sufficiently small, ‖χR(x)[f(x+ y)− f(x)]‖Lpx < ε.These two estimates give equicontinuity of F . By the Riesz compactness theorem, the proof is complete.

4.2 The Strauss lemma

In this section, we provide a proof of the Strauss lemma that was used in our analysis of the energy of thehydrogen atom.

First, we recall the homogeneous Sobolev space H1(Rd), which is the completion of the Schwartz classunder ‖∇ϕ‖L2 . That is, in the homogeneous Sobolev space H1(Rd), we only assume that the gradient,rather than both the function and its gradient, is in L2.

Proposition 4.2 (Radial Sobolev Embedding). Fix d ≥ 2 and 1 ≤ p <∞. For a radial function f ∈ H1 ∩ Lp,we have

r2(d−1)p+2 |f(r)| . ‖f‖

p2+p

Lp ‖∇f‖2

2+p

L2

for almost all r > 0.

Proof. It suffices to prove the inequality for Schwartz functions. Indeed, S(Rd) is dense in H1(Rd)∩Lp(Rd),and by passing to a subsequence we can ensure almost everywhere convergence. As |∇|f || ≤ |∇f | almosteverywhere, we can further assume that f ≥ 0.

By the fundamental theorem of calculus we have

rd−1f(r)1+ p2 =

∣∣∣∣rd−1

∫ ∞r

f ′(σ)f(σ)p2 dσ

∣∣∣∣ . ∫ ∞r

σd−1|f ′(σ)|f(σ)p2 dσ

.

(∫ ∞r

σd−1|f ′(σ)|2 dσ) 1

2(∫ ∞

r

σd−1|f(σ)|p dσ) 1

2

. ‖∇f‖L2 ‖f‖p2

Lp .

The final inequality here comes from the fact that f is radial and noting that σd−1 dσ describes a polarcoordinate transformation. Raising both sides of this inequality to the power 2

p+2 completes the proof.

In words, the radial Sobolev embedding theorem describes the decay of radial H1 functions at infinity.Observe that as p increases in the above inequality, the decay condition weakens. Thus, given a radial H1,the best decay behavior we can expect is precisely this case.

47

Corollary 4.2.1. If f ∈ H1rad(Rd) for d ≥ 2, then

rd−1

2 |f(r)| . ‖f‖12

L2 ‖∇f‖12

L2

for almost all r > 0.

Using the radial Sobolev embedding theorem, we can finally give a proof of the Strauss lemma that wehave previously invoked. We recall the (more general) statement of the lemma here.

Lemma 4.3 (Strauss). Fix d ≥ 2 and 2 < p < ∞ if d = 2 or 2 < p < 2dd−2 if d ≥ 3. Then H1

rad(Rd) embedscompactly into Lp(Rd).

Proof. By renormalizing appropriately, it suffices to prove that the unit ball B1 ⊆ H1rad is precompact in Lp.

Naturally, we will apply the Riesz compactness theorem.The arguments for boundedness and equicontinuity are similar to those used in the proof of the Rellich-

Kondrachov theorem. For f ∈ B1, Gagliardo-Nirenberg gives

‖f‖Lp . ‖f‖θL2 ‖∇f‖1−θL2 . 1

for some the appropriate exponent θ. Thus, B1 is Lp-bounded. Equicontinuity follows from

‖f(x+ y)− f(x)‖Lpy . ‖f(x+ y)− f(x)‖θL2 ‖∇x(f(x+ y)− f(x))‖1−θL2 . (|y| ‖∇f‖L2)θ ‖∇f‖1−θL2

. |y|θ

where we have again used Gagliardo-Nirenberg, followed by the fundamental theorem of calculus.It remains to show tightness. We wish to use the radial Sobolev embedding theorem; or rather, its

immediate corollary. If we try to use the theorem directly, we get∫|x|≥R

|f(x)|p dx .∫|x|≥R

r−d−1

2 p dx =

∫ ∞R

r−d−1

2 prd−1 dr

. Rd−d−1

2 p

provided that d− d−12 p < 0, hence p > 2d

d−1 . This does not yield the full range of 2 < p <∞. To fix this, weuse the fact that f ∈ L2 and write∫

|x|≥R|f(x)|p dx =

∫|x|≥R

|f(x)|2 |f(x)|p−2 dx.

Using the radial Sobolev embedding theorem on |f(x)|p−2 rather than |f(x)|p then gives∫|x|≥R

|f(x)|p dx . ‖f‖2L2 R− d−1

2 (p−2) ‖f‖2L2 ,

which can be made uniformly small for 2 < p <∞ by choosing R sufficiently large.Thus, by the Riesz compactness theorem, we are done.

The endpoint cases in the Strauss lemma, namely p = 2 and p = 2dd−2 when d ≥ 3, do not hold. That

is, H1rad(Rd) does not embed compactly into L2(Rd) or L

2dd−2 (Rd). To see that H1

rad(Rd) does not embed

48

compactly into L2(Rd), we choose a radial bump function and rescale on a large scale. Explicitly, let ϕ ∈ C∞cbe radial, and define ϕλ(x) := λ−

d2ϕ(xλ ). Then ‖ϕλ‖L2 = ‖ϕ‖L2 . Also,

‖∇ϕλ‖L2 = λ−d2 λ−1λ

d2 ‖∇ϕ‖L2 = λ−1 ‖∇ϕ‖L2

which is . 1 for λ >> 1, hence ϕλ ∈ H1(Rd). Note that

| 〈ϕλ, ψ〉 | . λ−d2 ‖ϕ‖L∞ ‖ψ‖L1

and so ϕ ⇀ 0as λ→∞. However, as the L2-norm of ϕλ is constant in λ, the ϕλ’s cannot converge stronglyto 0 in L2(Rd). The argument for showing that H1

rad(Rd) does not embed compactly into L2dd−2 (Rd) is

similar; here, we rescale to small scales. Let ϕλ(x) = λ−d−2

2 ϕ(xλ ). Then ‖ϕλ‖L

2dd−2

= ‖ϕ‖L

2dd−2

, and in fact‖ϕλ‖H1 = ‖ϕ‖H1 . Also,

‖ϕλ‖L2 = λ−d−2

2 λd2 ‖ϕ‖L2 = λ ‖ϕ‖L2

which is . 1 for λ << 1. We have ϕλ ⇀ 0 as λ→ 0, but ϕλ cannot converge to 0 in L2dd−2 (Rd).

49

5 5/8 — 5/19: Optimal Constants I: Gagliardo-Nirenberg

Next, we revisit the Gagliardo-Nirenberg inequality, with the goal of determining its optimal implicit con-stant.

5.1 Motivation: the focusing cubic NLS

First, we include a discussion as to why we might care about the optimal constant in the Gagliardo-Niremberg inequality. Our motivation comes from the focusing cubic nonlinear Schrodinger equationin d = 2, given by {

i∂tu+ ∆u = −|u|2uu(0) = u0

. (4)

Here, t ∈ R and x ∈ R2, and for now we take the initial data u0 to be a complex-valued Schwartz function.This is one of the most commonly studied nonlinear Schrodinger equations.

Solutions of (4) conserve a number of quantities over time, but in particular they conserve the mass andenergy of the initial data:

M(u) :=

∫R2

|u(t, x)|2 dx;

E(u) :=

∫R2

1

2|∇u(t, x)|2 − 1

4|u(t, x)|4 dx.

To see why solutions conserve mass, we will show that the time derivative of M(u) is 0.

∂tM(u) = ∂t

∫R2

|u(t, x)|2 dx = ∂t

∫R2

uu dx

= 2 Re

∫R2

utu dx.

Because u is a solution of (4), we have

iut = −∆u− |u|2u ⇒ ut = i∆u+ i|u|2u.

Thus,

∂tM(u) = 2 Re

∫R2

u(i∆u+ i|u|2u) dx = 2 Re i

∫R2

u∆u+ |u|2|u| dx = −2 Im

∫R2

u∆u+ |u|3 dx

= −2 Im

∫R2

u∆u dx.

Integrating by parts gives

∂tM(u) = 2 Im

∫R2

∇u · ∇u dx = 2 Im

∫R2

|∇u|2 dx = 0.

As ∂tM(u) = 0, mass is conserved over time. The computation for energy is similar, and is left as an exerciseto the reader.

Another remark that we make is that large-mass initial data can have negative energy. To see this, letuλ0 (x) := λu0(x). Then for sufficiently large λ (and hence sufficiently large mass),

E(uλ0 ) =λ2

2‖∇u0‖2L4 −

λ4

4‖u0‖4L4 < 0.

50

What about initial data with small mass? First, by the d = 2 Gagliardo-Nirenberg inequality, we have

‖u0‖L4 ≤ CGN ‖u‖12

L2 ‖∇u0‖12

L2

where CGN is the optimal constant in the Gagliardo-Nirenberg inequality. Consequently,

E(u0) ≥ 1

2‖∇u0‖2L2 −

1

4C4GN ‖u0‖2L2 ‖∇u0‖2L2

= ‖∇u0‖2L2

(1

2− 1

4C4GNM(u0)

).

Thus, if M(u0) < 2C4GN

, then E(u0) > 0. So for sufficiently small mass, the energy of the initial data canbe positive. As such, the question of finding the largest mass for which the energy is positive translates tofinding the optimal constant CGN in the Gagliardo-Nirenberg inequality.

Positivity and negativity of the initial data energy is important because if E(u0) < 0, the correspondingsolution to (4) blows up in finite time in both directions; this calculation is done explicitly below. Onthe other hand, it can be shown that a priori bounds on the H1-norm of a solution lead to a time-globalsolution. In other words, if no blow-up occurs, a solution can be constructed forever. Conservation ofmass provides an a priori bound on the L2-norm of the solution; energy conservation provides an a prioribound on the H1-norm (and hence the H1-norm) of the solution if M(u0) < 2

C4GN

. So in summary, thesharp Gagliardo-Nirenberg inequality is important for understanding the finite-time blow-up behavior ofthe nonlinear Schrodinger equation given in (4).

We finish this section by proving that if E(u0) < 0, the solution to (4) blows up in finite time in bothdirections. Let u be a solution and consider the second spatial-moment

V (u) :=

∫R2

|x|2|u(x, t)|2 dx.

Clearly, V (u) measures the concentration of mass in some weighted sense. We will show that V (u) becomesnegative for some finite time t, which represents the “blow-up” of the solution. Towards this goal, wecompute ∂ttV (u). The following computation is similar to the mass-conservation computation from above.

First, we have

∂tV (u) = 2 Re

∫R2

|x|2utu dx = 2 Re

∫R2

|x|2u(i∆u+ i|u|2u) dx = −2 Im

∫R2

|x|2u(∆u+ |u|2u) dx

= −2 Im

∫R2

|x|2u∆u dx

= 2 Im

∫R2

2xu∇u+ |x|2|∇u|2 dx

= 4 Im

∫R2

xu∇u dx

where the last few equalities come from eliminating real functions and integrating by parts. Then

∂ttV (u) = 4 Im

∫R2

xut∇u+ xu∇ut dx

= 4 Im

∫R2

x∇u(−i∆u− i|u|2u) + xu∇(i∆u+ i|u|2u) dx

= −4 Re

∫R2

x∇u(∆u+ |u|2u)− xu∇(∆u+ i|u|2u) dx.

51

Consider the terms in the above integral with three derivatives. Let xj denote the jth component of x,and let uj be the partial derivative of u in the xj direction. Using implicit summations over correspondingindices, we then have

Re

∫R2

x∇u∆u− xu∇∆u dx = Re

∫R2

xjuj ukk − xjuujkk dx

= Re

∫R2

xjuj ukk + 2uukk + xjuj ukk dx

where the last equality comes from the integrating the latter term in the integral by parts with respect to xj .Integrating the uukk by parts and combining the other two terms yields

Re

∫R2

x∇u∆u− xu∇∆u dx = 2 Re

∫R2

xjuj ukk dx− 2

∫R2

|∇u|2 dx.

From here, we integrate by parts in the first integral to get

Re

∫R2

x∇u∆u− xu∇∆u dx = −2 Re

∫R2

δjkuj uk + xj∂j|uk|2

2dx− 2

∫R2

|∇u|2 dx

= −2

∫R2

|∇u|2 dx+ 2

∫R2

|∇u|2 dx− 2

∫R2

|∇u|2 dx

= −2

∫R2

|∇u|2 dx.

This takes care of the terms in ∂ttV (u) with three derivatives. The other terms are handled similarly, andwe leave to the reader to compute. At the end of the day, finishing this computation gives

∂ttV (u) = 4

∫R2

2|∇u|2 + |u|4 dx = 16E(u).

Thus, if E(u0) < 0, then ∂ttV (u) < 0, hence the graph of V (u) as a function of t is concave down. Conse-quently, V (u) becomes negative in finite time in both directions, hence the solution blows up.

5.2 The sharp Gagliardo-Nirenberg inequality

Having motivated the need for the optimal constant in the Gagliardo-Nirenberg inequality, we now stateand prove the result.

Theorem 5.1 (Sharp Gagliardo-Nirenberg). Fix d ≥ 1, and 2 < p < ∞ if d = 1, 2, or 2 < p < 2dd−2 if d ≥ 3.

Then

‖f‖pLp ≤2p

2d− (d− 2)p

[2d− (d− 2)p

p− 2

] (p−2)d4

‖Q‖−(p−2)L2 ‖f‖

2d−p(d−2)2

L2 ‖∇f‖d(p−2)

2

L2 (5)

where Q is the unique non-negative non-zero radial H1-solution to the equation

∆Q+ |Q|p−2Q = Q. (6)

Moreover, equality in (5) holds if and only if f(x) = αQ(λ(x− x0)) for some α ∈ C, λ > 0, and x0 ∈ Rd.

Before proving the theorem, we make a few preliminary remarks. First, let θ = 2d−(d−2)p2p . Then 1− θ =

d(p−2)2p , and (5) becomes:

‖f‖Lp ≤(

1

θ

) 1p(

θ

1− θ

) 1−θ2

‖Q‖−p−2p

L2 ‖f‖θL2 ‖∇f‖1−θL2 . (7)

52

Even with this reformulation of the inequality, it is not clear why the function Q (and any translation orrescaling of Q) is an optimizer for the sharp Gagliardo-Nirenberg inequality. We claim that inequality (7)can be written as:

‖f‖Lp ≤‖Q‖Lp

‖Q‖θL2 ‖∇Q‖1−θL2

‖f‖θL2 ‖∇f‖1−θL2 . (8)

Under this formulation, it is immediate that Q is an optimizer. In order to connect (8) to (7), we need tocompute ‖Q‖Lp and ‖∇Q‖L2 . We will use the following Pohozaev identities for the equation (6), which relatethe mass, kinetic energy, and potential energy of solutions Q.

Lemma 5.2 (Pohozaev identities). Suppose thatQ is a non-negative non-zero radialH1-solution to (6). Then

−∫|∇Q|2 dx+

∫Qp dx =

∫Q2 dx

andd− 2

2

∫|∇Q|2 dx− d

p

∫Qp dx = −d

2

∫Q2 dx.

Proof. We obtain the first identity by multiplying (6) by Q, integrating, and then integrating by parts:∫∆QQdx+

∫Qp dx =

∫Q2 dx ⇒ −

∫|∇Q|2 dx+

∫Qp dx =

∫Q2 dx.

The second identity comes from multiplying (6) by x ·∇Q and integrating. Using implicit summations overindices, this yields ∫

QkkxjQj dx+

∫Qp−1xjQj dx =

∫QxjQj dx.

Consider the first integral. Integrating by parts yields∫QkkxjQj dx = −

∫δjkQkQj dx−

∫xjQjkQk dx = −

∫δjkQkQj dx−

∫xj

1

2∂j(Q

2k) dx

= −∫δjkQkQj dx+

1

2

∫Q2k dx

= −∫|∇Q|2 dx+

d

2

∫|∇Q|2 dx

=d− 2

2

∫|∇Q|2 dx.

Similarly, ∫Qp−1xjQj dx =

∫xj

1

p∂j(Q

p) dx = −dp

∫Qp dx

and ∫QxjQj dx =

∫xj

1

2∂j(Q

2) dx = −d2

∫Q2 dx.

Together, these computations give

d− 2

2

∫|∇Q|2 dx− d

p

∫Qp dx = −d

2

∫Q2 dx

as desired.

53

Solving for the kinetic energy∫|∇Q|2 dx in the first Pohozaev identity and inserting this into the second,

we have (d− 2

2− d

p

)∫Qp dx =

(d− 2

2− d

2

)∫Q2 dx.

Hence ∫Qp dx =

2p

2d− p(d− 2)

∫Q2 dx =

1

θ

∫Q2 dx.

Similarly, solving for the mass∫Qp dx in the first identity, inserting this into the second, and solving the

corresponding equation yields∫|∇Q|2 dx =

d(p− 2)

2d− p(d− 2)

∫Q2 dx =

1− θθ

∫Q2 dx.

Plugging the computations∫Qp dx = 1

θ

∫Q2 dx and

∫|∇Q|2 dx = 1−θ

θ

∫Q2 dx into (8) yields (7), as desired.

Now, we prove Theorem 5.1.

Proof. For nonzero f ∈ H1(Rd) and for θ as above, let

J(f) :=‖f‖pLp

‖f‖pθL2 ‖∇f‖p(1−θ)L2

.

We seek Jmax := sup{J(f) : f ∈ H1(Rd) \ {0}

}. By the regular Gagliardo-Nirenberg inequality, Let fn

be an optimizing sequence so that J(fn) → Jmax. Consider the symmetric decreasing rearrangements f∗nof fn. By equimeasurability of rearrangements, ‖f∗n‖L2 = ‖fn‖L2 and ‖f∗n‖Lp = ‖fn‖Lp . By Polya-Szego,‖∇f∗n‖L2 ≤ ‖fn‖L2 . Hence, J(f∗n) ≥ J(fn), and so f∗n is also an optimizing sequence. Consequently, wemay assume from now on that each fn is symmetric decreasing.

As usual in optimization problems like this, we want to pass to a subsequence to extract a weak H1-limit. Unfortunately, the optimizing sequence fn may be unbounded in H1. This is remedied by defining anew sequence gn of appropriately rescaled functions. Let

gn(x) := αnfn(λnx)

where αn and λn are chosen so that ‖gn‖L2 = ‖∇gn‖L2 = 1. These two conditions force

anλ− d2n ‖fn‖L2 = anλnλ

− d2n ‖∇fn‖L2 = 1 ⇒

λn =‖fn‖L2

‖∇fn‖L2

αn = ‖fn‖L2

(‖∇fn‖L2

‖fn‖L2

) d2

.

Because the J functional is scale-invariant, J(gn) = J(fn). On the other hand, because ‖gn‖L2 = ‖∇gn‖L2 =

1, J(gn) = ‖gn‖pLp . From this it follows that gn is an optimizing sequence and that ‖gn‖pLp → Jmax.As the gn sequence is bounded in H1(Rd), we may pass to a subsequence and assume that gn ⇀ g

weakly in H1. We need to show that g is an optimizer. By weak lower semicontinuity of the H1-norm,‖g‖L2 ≤ 1 and ‖∇g‖L2 ≤ 1. It remains to consider the numerator of J(g).

For d ≥ 2, the Strauss lemma gives compact embedding ofH1rad(Rd) into Lp(Rd), hence ‖gn‖Lp → ‖g‖Lp

(after possibly passing to another subsequence) and so

J(g) ≥ limn→∞

‖gn‖pLp = Jmax

as desired. This also implies ‖g‖L2 = ‖∇g‖L2 = 1, for otherwise g would be a super optimizer.For d = 1, we use the fact that F := { f∗ : ‖f‖L1 ≤ 1, ‖∇f‖L2 ≤ 1 } is precompact in Lp for 2 < p <

∞. This follows from the Riesz compactness theorem as usual. Boundedness is clear, and equicontinuity

54

follows from Gagliardo-Nirenberg and the fundamental theorem of calculus. We leave tightness as anexercise to the reader; this is a straightforward consequence of the fact that each element of F is symmetricdecreasing.

Having found an optimizer g for the functional J , we now proceed to actually computing the value ofJmax, which is precisely the optimal constant in the Gagliardo-Nirenberg inequality. This is accomplishedby deriving the Euler-Lagrange equation corresponding to J . Fix a positive Schwartz function ϕ. For εsmall, consider J(g + εϕ) as a function of ε. Because g is an optimizer for J , ε 7→ J(g + εϕ) has a criticalpoint at ε = 0. Hence,

d

dε

∣∣∣∣∣ε=0

J(g + εϕ) = 0.

To compute the left hand side, we first differentiating each term in the expression

J(g + εϕ) = ‖g + εϕ‖pLp ‖g + εϕ‖−pθL2 ‖∇g + ε∇ϕ‖−p(1−θ)L2 .

First,

d

dε

∣∣∣∣∣ε=0

‖g + εϕ‖pLp =d

dε

∣∣∣∣∣ε=0

∫|g + εϕ|p dx = p

∫gp−1ϕdx.

Next,

d

dε

∣∣∣∣∣ε=0

‖g + εϕ‖−pθL2 =d

dε

∣∣∣∣∣ε=0

(‖g + εϕ‖2L2

)− pθ2= −pθ

2‖g‖−pθ−2

L2 · 2∫gϕ dx

= −pθ∫gϕ dx

and

d

dε

∣∣∣∣∣ε=0

‖∇g + ε∇ϕ‖−p(1−θ)L2 =d

dε

∣∣∣∣∣ε=0

(‖∇g + ε∇ϕ‖2L2

)− p(1−θ)2

= −p(1− θ)2

‖∇g‖−p(1−θ)−2L2 · d

dε

∣∣∣∣∣ε=0

∫|∇g + ε∇ϕ|2 dx

= −p(1− θ)2

2

∫∇g · ∇ϕdx

= −p(1− θ)∫

(−∆g)ϕdx.

Using the product rule and the fact that ‖g‖L2 = ‖∇g‖L2 = 1, we then have

0 =d

dε

∣∣∣∣∣ε=0

J(g + εϕ)

= p

∫gp−1ϕdx− ‖g‖pLp pθ

∫gϕ dx− ‖g‖pLp p(1− θ)

∫(−∆g)ϕdx.

Dividing by p and using the fact that ‖g‖pLp = Jmax, this equality becomes∫ (gp−1 − θJmaxg + (1− θ)Jmax∆g

)ϕdx = 0.

This means that g is a distributional solution to the equation

(1− θ)Jmax∆g + gp−1 = θJmaxg.

55

Up to some constants, this is (6). It remains to rescale g in an appropriate way to recover (6) exactly. Writeg(x) = αQ(λx). Then

(1− θ)Jmaxλ2α∆Q+ αp−1Qp−1 = θJmaxαQ.

Towards the goal of making all constants equal 1, divide by αp−1 to get

(1− θ)Jmaxλ2

αp−2∆Q+Qp−1 =

θJmaxαp−2

Q.

We require (1 − θ)Jmaxλ2

αp−2 = 1 and θJmaxαp−2 = 1. From these equations we get (1 − θ)λ2 = θ, hence

λ =(

θ1−θ

) 12

. Also, α = (θJmax)1p−2 .

To find Jmax, recall that ‖g‖L2 = 1. Hence αλ−d2 ‖Q‖L2 = 1. Using the previous λ and α expressions

yields

(θJmax)1p−2

(1− θθ

) d4

‖Q‖L2 = 1

and thus

Jmax =1

θ

(θ

1− θ

) d(p−2)4

‖Q‖−(p−2)L2 .

Comparing with (7), this is indeed the constant given in the statement of the theorem.Having demonstrated the sharp inequality, we now consider the case of equality. Assume that f ∈

H1(Rd) is an optimizer. We want to see that f(x) = αQ(λ(x − x))); the converse is true by the discussionfollowing the statement of Theorem 5.1. If f is an optimizer, then as we have seen, f∗ is an optimizer. Also,‖∇f‖L2 = ‖∇f∗‖L2 , for otherwise f∗ would be a super-optimizer. As f∗ solve the Euler-Lagrange equationcorresponding to J , a rescaling of f∗ solves the equation ∆u+|u|p−2u = u. By a uniqueness result of Kwong(1989), it follows that this rescaling is exactly Q. Thus, we want to see that f(x) = eiθf∗(x − x0) for someθ ∈ [0, 2π) and x0 ∈ Rd.

Ideally this would follow from the fact that ‖∇f‖L2 = ‖∇f∗‖L2 , but this is not the case in general.Consider the following example. Let ϕ ∈ C∞c (Rd) be a radial bump function satisfying ϕ(x) = 1 for|x| ≤ 1 and ϕ(x) = 0 for |x| ≥ 2. Choose some x0 satisfying |x0| = 1

2 . Then ϕ(4(x − x0)) is a tighterbump function centered at x0. Let f(x) = ϕ(x) + ϕ(4(x − x0)). By drawing a picture, it is easy to see thatf∗(x) = ϕ(x) + ϕ(4x). Then ‖∇f‖L2 = ‖∇f∗‖L2 , but f∗ is not a translation of f . What is true is a result dueto Brothers and Zimmer, which say that if in addition ∇f∗ does not vanish on any set of positive measure,then indeed f(x) = eiθf∗(x− x0).

Applying this result to our optimizer f∗, it suffices to show that ∇f∗ does not vanish on any set ofpositive measure. Here we use the additional data that f∗ satisfies the Euler-Lagrange equation. Let udenote the rescaling of f∗ which solves

∆u+ |u|p−2u = u.

As f∗ is nonnegative and radial, so is u. Hence, we can write the above equation in polar coordinates as

urr +d− 1

rur + up−1 − u = 0.

Suppose for the sake of contradiction that ur vanishes on a set of positive measure. Let r0 be an accumu-lation point of this set. As such, urr(r0) = 0 and hence up−1(r0) − u(r0) = 0. As u is nonnegative, it iseither the case that u(r0) = 0 or u(r0) = 1. In the first case, uniqueness of ODE’s gives u ≡ 0, which is acontradiction because u is an optimizer. In the second case, then uniqueness of ODE’s give u ≡ 1, which isa contradiction because u ∈ L2(Rd).

Thus,∇f∗ does not vanish on any set of positive measure, which completes the proof.

56

5.3 Concentration compactness for Gagliardo-Nirenberg

As we have seen, the issue of compactness in function spaces is delicate. In the analysis of variational andminimization problems, finding a solution often involves working in a topology which balances continuityof the functional with compactness in some careful way. As such, our next topic is a principle known asconcentration compactness. For now we specifically discuss concentration compactness in the context of theGagliardo-Nirenberg inequality.

Historically, the concentration compactness method was introduced by Lions in 1984. Motivation forthe principle comes from the observation that, given a weak limit of probability measures, there are threescenarios:

i. Vanishing: the weak limit is 0.

ii. Compactness: the weak limit is a probability measure.

iii. Dichotomy: something else happens, i.e., the weak limit is a measure which is not a probability mea-sure.

5.3.1 Motivation

Here, we motivate concentration compactness by sketching an alternative proof of the sharp Gagliardo-Niremberg inequality. For the sake of simplicity, we consider the case d = 1.

Proof sketch. Fix 2 < p < ∞. Let fn be an optimizing sequence for the J functional, normalized so that‖fn‖L2 = 1 and ‖∇fn‖L2 = 1 as before.

The first key observation is that fn cannot converge to 0 in the L∞-norm. Indeed, if this were the case,then Holder’s inequality gives

limn→∞

‖fn‖Lp ≤ limn→∞

‖fn‖2p

L2 ‖fn‖1− 2

p

L∞ = 0.

As fn is an optimizing sequence, this is a contradiction.Thus, passing to a subsequence, there exists ε > 0 such that ‖fn‖L∞ > ε. For each n, let xn be a Lebesgue

point of fn with |fn(xn)| > ε. To remedy the possibility that the xn’s march off to ∞, define gn(x) =

fn(x + xn). Then ‖gn‖L2 = 1 and ‖∇gn‖L2 = 1, so by passing to another subsequence Alaoglu’s theoremguarantees a weak H1-limit ϕ. By Rellich-Kondrachov, and by uniqueness of weak limits, gnχBR(0) →ϕχBR(0) strongly in Lp for 2 < p <∞.

Next, we claim that gn being an optimizing sequence prevents the following two scenarios:

i. vanishing, i.e. ϕ ≡ 0, and

ii. dichotomy, i.e. ‖ϕ‖L2 < 1 or ‖∇ϕ‖L2 < 1.

This leaves only the compactness scenario, which implies that ϕ is an optimizer. With an optimizer, theremainder of the argument is as before (using Euler-Lagrange, etc.).

We begin by ruling out the vanishing scenario. The key observation is one of regularity, namely, that thefn’s are uniformly (in n) 1

2 -Holder continuous, that is,

‖fn‖L∞ + supx 6=y

|fn(x)− fn(y)||x− y| 12

. 1

57

uniformly in n. To see boundedness of the L∞ norm, decompose fn to high and low frequencies, and usethe fact that low frequencies contribute to smoothness:

‖fn‖L∞ ≤ ‖P≤Nfn‖L∞ +∑M>N

‖PMfn‖L∞

. N12 ‖fn‖L2 +

∑M>N

M12−1 ‖∇fn‖L2

. N12 ‖fn‖L2 +N−

12 ‖∇fn‖L2 .

Optimizing in N yields ‖fn‖L∞ . ‖fn‖12

L2 ‖∇fn‖12

L2 . 1. Next, by the fundamental theorem of calculus andCauchy-Schwarz,

|fn(x)− fn(y)| ≤ |x− y|∫ 1

0

|f ′n(x+ θ(y − x))| dθ ≤ |y − x|(∫ 1

0

|f ′n(x+ θ(y − x))|2 dθ) 1

2

.

Changing variables gives |fn(x)−fn(y)| ≤ |x−y| 12 ‖f ′n‖L2 . 1 as desired. With this, we can show that eachfn being large in a neighborhood of xn yields ϕ being large in some neighborhood. Explicitly,

|fn(x+ xn)| ≤ |fn(xn)| − |fn(x+ xn)− fn(xn)| ≥ ε−O(|x| 12 ) ≥ ε

2

for |x| . ε2. So, by Rellich-Kondrachov,∫|x|.ε2

|ϕ(x)|p dx = limn→∞

∫|x|.ε2

|gn(x)|p dx & εpε2 > 0.

This implies that ϕ is not identically 0.Next, we rule out the dichotomy scenario. Define rn(x) = fn(x)−ϕ(x−xn). First, we make the following

observation.

‖fn‖2L2 − ‖rn‖2L2 − ‖ϕ‖2L2 = ‖fn‖2L2 − 〈fn(x)− ϕ(x− xn), fn(x)− ϕ(x− xn)〉 − ‖ϕ‖2L2

= ‖fn‖2L2 − 〈fn(x+ xn)− ϕ(x), fn(x+ xn)− ϕ(x)〉 − ‖ϕ‖2L2

= ‖fn‖2L2 −(‖f‖L2 − 〈ϕ(x), fn(x+ xn)〉 − 〈fn(x+ xn), ϕ(x)〉+ ‖ϕ‖2L2

)− ‖ϕ‖2L2

= 2 Re 〈fn(x+ xn), ϕ(x)〉 − 2 ‖ϕ‖2L2 .

As fn(x+ xn) ⇀ ϕ in H1, we have

‖fn‖2L2 − ‖rn‖2L2 − ‖ϕ‖2L2 → 0 (9)

as n → ∞. The limiting statement (9) is know as asymptotic L2 decoupling, as we have shown that the thesquared L2-norm of the sum rn +ϕ “decouples” in the limit to the sum of the squared L2-norms. The sameargument shows ‖∇fn‖2L2 − ‖∇rn‖2L2 − ‖∇ϕ‖2L2 → 0.

Asymptotic decoupling in Lp is due to the following refinement of Fatou’s lemma, whose proof will begiven after finishing the sketch of the sharp Gagliardo-Nirenberg proof.

Lemma 5.3 (Brezis-Lieb). Fix 1 ≤ p < ∞, and let fn is an Lp-bounded sequence of functions. Suppose thatfn → f almost everywhere. Then ∫

| |fn|p − |f − fn|p − |f |p | dx = 0.

58

This lemma implies that‖fn‖pLp = ‖f‖pLp + ‖f − fn‖pLp + o(1)

as n→∞. In particular, if fn → f almost everywhere and ‖fn‖Lp → ‖f‖Lp , then fn → f in Lp.Returning to the sketch of the proof, Rellich-Kondrachov givesFINISH

FINISH...

5.3.2 The concentration compactness principle

Inspired by the above argument, we now state and prove the main results regarding the concentrationcompactness principle.

A stepping stone towards the concentration compactness principle for the Gagliardo-Nirenberg inequal-ity is given by the following lemma. Broadly, the lemma describes how to find a bubble profile ϕ and a se-quence of locations xn ∈ Rd for the bubble profile for which a given sequence of H1 functions concentrates.Alternatively, the lemma suggests that Gagliardo-Nirenberg (which bounds the Lp-norm by the H1-norm)is a good estimate in that the loss between the Lp-norm and H1-norm is small. As such, we refer to it as theinverse Gagliardo-Nirenberg inequality.

Lemma 5.4 (Inverse Gagliardo-Nirenberg). Fix d ≥ 1, and let 2 < p < ∞ if d = 1, 2, and 2 < p < 2dd−2 if

d ≥ 3. Suppose that fn ∈ H1(Rd) satisfies lim inf ‖fn‖Lp > ε and lim sup ‖fn‖H1 ≤ A. Then there existsϕ 6= 0 ∈ H1(Rd) and xn ∈ Rd such that, passing to a subsequence,

1. fn(x+ xn) ⇀ ϕ(x) in H1,

2. ‖fn‖2L2 − ‖fn − ϕ(x− xn)‖2L2 − ‖ϕ‖2L2 → 0 as n→∞,

3. ‖∇fn‖2L2 − ‖∇fn −∇ϕ(x− xn)‖2L2 − ‖∇ϕ‖2L2 → 0 as n→∞, and

4. ‖fn‖pLp − ‖fn − ϕ(x− xn)‖pLp − ‖ϕ‖pLp → 0 as n→∞.

Before proving Lemma 5.4, we briefly discuss its content. As previously remarked, the lemma describeshow to find a bubble profile ϕ, possibly travelling around Rd to various locations xn over time, for which anH1-bounded and Lp-large sequence concentrates to. Statement 1 says that this concentration occurs weaklyin H1. However, this fact alone does not have much content. As fn is bounded in H1, we can always finda weak limit! The interesting content is the nature of the bubble profile as a weak limit, and is describedby the asymptotic decoupling statements 2-4. Indeed, note that fn(x + xn) ⇀ ϕ in H1 does not implystrong convergence in H1, or even the convergence of H1 norms. What is guaranteed by the lemma is theasymptotic relationship

‖fn‖2H1 ∼ ‖ϕ‖2H1 + ‖rn‖2H1

where rn = fn − ϕ(· − xn) is the remainder of the function fn after extracting the bubble profile. TheH1-norms of the remainder terms need not vanish, but they do decouple (i.e., are pushed away) from thebubble profile over time. Statement 4 says that this asymptotic decoupling also happens in the Lp sense.

59

Proof of Lemma 5.4. First, by definition of lim inf and lim sup we may pass to a subsequence and assume that‖fn‖Lp >

ε2 and ‖fn‖H1 ≤ 2A.

As ‖fn‖Lp >ε2 , each fn must large somewhere. For low frequencies of fn, Berstein’s inequalities give

‖P≤Nfn‖Lp . Nd2−

dp ‖fn‖L2 . N

d2−

dpA <

ε

10

when N is sufficiently small. At high frequencies,

‖P>Nfn‖Lp .∑M>N

‖PMfn‖Lp .∑M>N

Md2−

dp−1 ‖∇fn‖Lp . N

d2−

dp−1A .

ε

10

when N is sufficiently large. Thus, largeness in the Lp-norm of fn comes from the middle frequencies.In particular, there exist N1, N2, dependent on ε and A, such that ‖PN1≤·≤N2

fn‖Lp ≥ε4 . Passing to a

subsequence, there is an N0 satisfying N1(ε,A) < N0 < N2(ε,A) such that ‖PN0fn‖Lp ≥ c(ε,A) > 0, wherec(ε,A) is a constant dependent on ε and A.

As the Lp-norm of PN0fn is large and the L2-norm is bounded, this implies that the L∞-norm is large.

Explicitly, by Holder,

‖PN0fn‖Lp ≤ ‖PN0

fn‖1− 2

p

L∞ ‖PN0fn‖

2p

L2 . ‖PN0fn‖

1− 2p

L∞ A2p .

As ‖PN0fn‖Lp ≥ c(ε,A), it follows that ‖PN0

fn‖L∞ ≥ c(ε,A).3 Let xn ∈ Rd be a point such that PN0fn(xn) ≥

c(ε,A). This xn will be the position of the bubble corresponding to fn. Next, we need a bubble profile. As fnis bounded in H1, the translated functions x 7→ fn(x+ xn) are bounded in H1, so passing to a subsequenceyields a weak H1-limit fn(x+ xn) ⇀ ϕ.

To verify statement 1 in the lemma it remains to check that ϕ is not identically 0. Let ψN0 denote theusual Littlewood-Paley multiplier. Then by Holder’s inequality,

|⟨ϕ, ψN0

⟩| ≤ ‖ϕ‖Lp

∥∥ψN0

∥∥Lp′

= ‖ϕ‖Lp∥∥Nd

0 ψ(N0·)∥∥Lp′

= ‖ϕ‖Lp Nd− dp0

∥∥ψ∥∥Lp′

. c(ε,A) ‖ϕ‖Lp .

Thus, to show that ϕ 6= 0, we will show that |⟨ϕ, ψN0

⟩| is bounded below. By definition of ϕ as a weak limit

and the fact that the Fourier transform preserves L2 inner products,

|⟨ϕ, ψN0

⟩| = lim

n→∞

∣∣⟨fn(x+ xn), ψN0

⟩∣∣= limn→∞

∣∣∣∣∫ eixnξ fn(ξ)ψN0(ξ) dξ

∣∣∣∣ .Note that fn(ξ)ψN0(ξ) = PN0fn(ξ). Then∫

eixnξ fn(ξ)ψN0(ξ) dξ =

(PN0

fn

)(xn).

Therefore,|⟨ϕ, ψN0

⟩| = lim

n→∞|PN0

fn(xn)| > c(ε,A)

by the choice of xn. Hence, ‖ϕ‖Lp ≥ c(ε,A), and so ϕ 6= 0.The proofs of statements 2 and 3 are direct computations using the Hilbert space structure ofH1; in fact,

the computation is exactly the same as the computation of (9) in the sketch of the sharp Gagliardo-Nirenbergproof.

3Here we are implicitly changing the constant c(ε,A).

60

Statement 4 follows from the a change of variables in the refined Fatou’s lemma. To invoke the refinedFatou’s lemma, we need almost everywhere convergence of fn(x + xn) to ϕ(x), and Lp-boundedness ofthe fn(x + xn) sequence. Almost everywhere convergence follows from Rellich-Kondrachov and diago-nalization as usual, and Lp-boundedness is immediate from Gagliardo-Nirenberg and the fact that fn isH1-bounded.

Finally, we state and prove our main result.

Theorem 5.5 (Concentration compactness principle for Gagliardo-Nirenberg). Fix d ≥ 1, and let 2 < p <∞if d = 1, 2, and 2 < p < 2d

d−2 if d ≥ 3. Let fn ∈ H1(Rd) be an H1-bounded sequence. There existsJ∗ ∈ {0, 1, 2, . . . } ∪ {∞}, {ϕj}J∗j=1 ∈ H1(Rd) \ {0}, and xjn ∈ Rd such that, after passing to a subsequence,we have the decomposition

fn(x) =

J∑j=1

ϕj(x+ xjn) + rJn(x) (10)

for all finite 0 ≤ J ≤ J∗, satisfying the following properties:

1. The Lp-norm of the remainder terms tend towards 0, i.e.,

limJ→J∗

lim supn→∞

∥∥rJn∥∥Lp = 0.

2. There is asymptotic decoupling in H1 uniformly in J , i.e.,

supJ

lim supn→∞

‖fn‖2H1 −J∑j=1

∥∥ϕj∥∥2

H1 −∥∥rJn∥∥2

H1

= 0.

3. There is asymptotic decoupling in Lp uniformly in J , i.e.,

supJ

lim supn→∞

‖fn‖pLp − J∑j=1

∥∥ϕj∥∥pLp−∥∥rJn∥∥pLp

= 0.

4. For all j ≤ J , rJn(·+ xjn) ⇀ 0 in H1 as n→∞.

5. For all j 6= k, |xjn − xkn| → ∞ as n→∞.

As the statement of Theorem 5.5 is long and technical, we precede the proof with a few comments. Theconcentration compactness principle says that given a sequence of bounded H1 functions, we can pass toa subsequence and decompose the limiting behavior of the sequence as a sum of travelling bubble profiles.The parameter J∗ counts the number of bubble profiles ϕj , and the xjn’s are locations for the bubble profiles.Statements 4 and 5 describe how the remainder terms extract everything from a bubble profile, and that thelocations of two different profiles disperse over time.

Proof. We argue inductively, extracting bubbles one at a time via the inverse Gagliardo-Nirenberg inequal-ity.

We start with the sequence fn, having extracted no bubbles. Set r0n = fn. For the inductive step, suppose

that we have a decomposition of the form (10) satisfying properties 2 and 3, and 4 for j = J . We will prove4 for j < J and 5 at the end.

61

Passing to a subsequence, let AJ = limn→∞∥∥rJn∥∥H1 and εJ = limn→∞

∥∥rJn∥∥Lp . If εJ = 0, then stopand set J∗ = J . In this case, property 1 is immediately satisfied. If εJ > 0, then we may apply theinverse Gagliardo-Nirenberg inequality to the sequence rJn to get a bubble profile ϕJ+1 6= 0 ∈ H1 andcorresponding locations xJ+1

n ∈ Rd such that, after passing to a subsequence,

rJn(·+ xn) ⇀ ϕJ+1

weakly in H1. Set rJ+1n = rJn − ϕJ+1(· − xJ+1

n ). Then by construction,

rJ+1n (·+ xJ+1

n ) = rJn(·+ xJ+1n )− ϕJ+1 ⇀ 0

weakly in H1, hence 4 is satisfied for j = J + 1.Next, we consider the asymptotic decoupling. Also by the inverse Gagliardo-Nirenberg inequality,∥∥rJn∥∥2

H1 −∥∥rJ+1n

∥∥2

H1 −∥∥ϕJ+1

∥∥2

H1 → 0 as n→∞

and ∥∥rJn∥∥pLp − ∥∥rJ+1n

∥∥pLp−∥∥ϕJ+1

∥∥pLp→ 0 as n→∞.

Using the inductive hypothesis, specifically the fact that statements 2 and 3 hold up to J , this gives

supJ

lim supn→∞

‖fn‖2H1 −J∑j=1

∥∥ϕj∥∥2

H1 −∥∥rJ+1n

∥∥2

H1 −∥∥ϕJ+1

∥∥2

H1

= 0,

which gives property 2 up to J + 1. We obtain property 3 up to J + 1 in the same way.The process continues: after passing to a subsequence, set AJ+1 = limn→∞

∥∥rJ+1n

∥∥H1 and εJ+1 =

limn→∞∥∥rJ+1n

∥∥Lp

. If εJ+1 = 0, stop and set J∗ = J + 1. Then as before, property 1 is immediate. IfεJ+1 > 0, repeat the above procedure using the inverse Gagliardo-Nirenberg inequality as before. If thisprocess does not terminate in finitely many steps, set J∗ =∞. In this case, to see that 1 holds, note that

AJ+1 ≤ AJ(

1− c(εJAJ

)c)and likewise for εJ , where the constants are dimensional. From this, property 1 follows. In words, this isbecause at each step we are extracting a nontrivial bubble profile.

It remains to verify 4 for j < J , and 5.First, consider 5. Suppose by way of contradiction that the conclusion fails for some pair (j, k) with

j < k, and moreover that this is the first instance of failure in the sense that 5 holds for all (j, l) with l < k.Thus, as |xkn − xjn| is bounded in n, passing to a subsequence we have xkn − xjn → x0 for some x0 ∈ Rd. But|xjn − xln| → ∞ for all j + 1 ≤ l ≤ k − 1. By definition of the remainder terms,

rk−1n (x) = rjn(x)−

k−1∑l=j+1

ϕl(x− xln).

Thus,

rk−1n (x+ xkn) = rjn(x+ xkn)−

k−1∑l=j+1

ϕl(x+ xkn − xln)

= rjn(x+ xjn + xkn − xjn)−k−1∑l=j+1

ϕl(x+ xjn − xln + xkn − xjn).

62

As xkn − xjn → x0 and rjn(x+ xjn) tends weakly to 0, it follows that

rjn(x+ xjn + xkn − xjn) ⇀ 0

as translation by x0 preserves this weak limit. Similarly, because xkn − xjn → x0 and |xjn − xln| → ∞,

ϕl(x+ xjn − xln + xkn − xjn) ⇀ 0

asϕl is a fixedH1 bubble profile. All of this implies that rk−1n (x+xkn) ⇀ 0 weakly inH1. But by construction,

rk−1n (x+ xkn) ⇀ ϕk(x). Uniqueness of weak limits implies that ϕk = 0, which is a contradiction.

The proof that 4 holds for j < J is more or less the same computation, and the details are left as anexercise to the reader.

As an application of the concentration compactness principle, we provide an alternate proof for thesharp Gagliardo-Nirenberg inequality.

Proof of Theorem 5. As before, let fn ∈ H1(Rd) be an optimizing sequence of the J functional normalized sothat ‖fn‖L2 = 1 and ‖∇fn‖L2 = 1. We will use the concentration compactness principle to show existenceof an optimizer. The rest of the argument is the same as in the original proof.

Passing to a subsequence and invoking Theorem 5.5, there is a decomposition fn(x) =∑Jj=1 ϕ

j(x+xjn)+

rJn for every finite 0 ≤ J ≤ J∗ satisfying properties 1-5. In particular, this means that the decompositionsatisfies limJ→J∗ limn→∞

∥∥rJn∥∥Lp = 0, and asymptotic decoupling in both H1 and Lp.Since the sequence fn is H1-normalized, Jmax = limn→∞ ‖fn‖pLp . By asymptotic decoupling in Lp,

Jmax = limn→∞

‖fn‖pLp = limn→∞

J∑j

∥∥ϕj∥∥pLp

+∥∥rJn∥∥pLp .

Letting n → ∞ and J → J∗, we then have Jmax =∑J∗

j

∥∥ϕj∥∥pLp

. Applying Gagliardo-Nirenberg followedby the standard inequality ab ≤ 1

qaq + 1

q bq′ ,

Jmax ≤J∗∑j

Jmax∥∥ϕj∥∥pθ

L2

∥∥∇ϕj∥∥p(1−θ)L2

≤ JmaxJ∗∑j

[θ∥∥ϕj∥∥p

L2 + (1− θ)∥∥∇ϕj∥∥p

L2

].

As ‖fn‖L2 = 1, asymptotic decoupling in L2 shows that∥∥ϕj∥∥

L2 < 1. Then p > 2 implies∥∥ϕj∥∥p

L2 ≤∥∥ϕj∥∥2

L2 .

Similarly,∥∥∇ϕj∥∥p

L2 ≤∥∥∇ϕj∥∥2

L2 . Therefore,

Jmax ≤ Jmax

θ J∗∑j

∥∥ϕj∥∥2

L2 + (1− θ)J∗∑j

∥∥∇ϕj∥∥2

L2

.Invoking asymptotic decoupling in H1 and the normalization of fn yet again tells us that

∑J∗

j

∥∥ϕj∥∥2

L2 ≤ 1,

and likewise∑J∗

j

∥∥∇ϕj∥∥2

L2 ≤ 1. Therefore, the quantity in the square brackets above is ≤ 1. This givesJmax ≤ Jmax. But Jmax = Jmax, so at every step in the above computation we have equality. Thus,

∑J∗

j=1

∥∥ϕj∥∥pL2 =

∑J∗

j=1

∥∥ϕj∥∥2

L2 = 1∑J∗

j=1

∥∥∇ϕj∥∥pL2 =

∑J∗

j=1

∥∥∇ϕj∥∥2

L2 = 1.

63

This is only possible if all but one of the ϕj are identically 0. In other words, there must be exactly onebubble profile.

Hence, we have the decomposition fn(x) = ϕ(x+xn)+ rn(x) and ‖rn‖Lp → 0. This implies that fn → ϕ

in Lp and therefore ϕ is the desired optimizer.

64

6 5/19 — 5/26: Optimal Constants II: Sobolev Embedding

In the previous section, we discussed the optimal constant in the Gagliardo-Nirenberg inequality in thecontext of a concentration compactness principle. There, the concentration compactness principle measuredthe defect of compactness due to translation.

In this section, we will have a similar discussion about the optimal constant in the Sobolev embeddingtheorem. However, this time around there is an added layer of complexity. As before, a defect in com-pactness exists due to translation. But now a second defect comes from scaling. This will soon be madeprecise.

For now, we state and discuss the main theorem.

Theorem 6.1 (Sharp Sobolev embedding). Fix d ≥ 3, and let f ∈ H1(Rd). Then

‖f‖L

2dd−2≤ CSob ‖∇f‖L2 , CSob =

‖W‖L

2dd−2

‖∇W‖L2

(11)

where

W (x) =

(1 +

1

d(d− 2)|x|2)− d−2

2

(12)

is the unique (up to scaling) nonzero nonnegative radial H1-solution to the elliptic equation

∆W +Wd+2d−2 = 0. (13)

Moreover, equality in (11) holds if and only if f(x) = αW (λ(x− x0)) for some α ∈ C, λ > 0, and x0 ∈ Rd.

It is tempting to try to adapt our original proof of the sharp Gagliardo-Nirenberg inequality to proveTheorem 6.1. However, this argument does not work. In the Gagliardo-Nirenberg case, we showed exis-tence of an optimizer by taking an optimizing sequence of radial H1 functions and used the fact that H1

rad

embeds compactly in Lp to extract a strong Lp limit. This argument fails for homogeneous Sobolev spaces,because H1

rad does not embed compactly in L2dd−2 .

Example 6.1. With W as in (12), define fn(x) := nd−2

2 W (nx). Then ‖fn‖H1 = ‖W‖H1 and ‖fn‖L

2dd−2

=

‖W‖L

2dd−2

. From the definition of CSob in (11) it follows that fn is a sequence of radial optimizers. However,as fn lives at increasingly small spatial scales, fn ⇀ 0.

The remedy for this will be a concentration compactness principle for the Sobolev embedding theorem.In the Gagliardo-Nirenberg case, the concentration compactness principle relied on being able to locatelargeness in the form of spatial bubbles; here, we will additionally need to locate largeness in the frequencydomain. Naturally, we will employ Littlewood-Paley theory extensively.

6.1 Motivation: the focusing energy-critical NLS

Before developing the concentration compactness principle necessary to prove Theorem 6.1, we motivatethe need for the optimal constant in the Sobolev embedding as before by considering certain nonlinearSchrodinger equations.

65

First, consider the following focusing nonlinear Schrodinger equation:

iut + ∆u = −|u|p−2u for 2 < p <2d

d− 2. (14)

A soliton solution to a partial differential equation such as (14) is, broadly, a solution which retains its spatialprofile over time. Explicitly, a soliton solution takes the form u(t, x) = eitQ(x). Note that a soliton solutionof (14) of this form satisfies

−eitQ+ eit∆Q = −|Q|p−2eitQ.

Dividing through by eit yields the equation ∆Q+ |Q|p−2Q = Q, which is precisely (6), the equation whichdetermines the optimal constant in the sharp Gagliardo-Nirenberg inequality.

It is a fact which we do not prove here (reference?) that if p > 2dd−2 , soliton solutions of (14) are stable,

in the sense that initial data in a neighborhood of a rescaled and translated soliton converges to a solitonsolution as time tends to∞. On the other hand, if p < 2d

d−2 , it is possible to have blow up in finite time (forexample, our earlier analysis of the focusing cubic NLS (4) in d = 2 exemplifies this).

What about the endpoint case p = 2dd−2 ? Then p − 2 = 2d

d−2 − 2 = 4d−2 , and we have the focusing

energy-critical nonlinear Schrodinger equation:{iut + ∆u = −|u|

4d−2u

u(0) = u0

. (15)

The first observation we make regarding (15) deals with the energy of a solution u, defined as

E(u) :=

∫1

2|∇u(t, x)|2 − d− 2

2d|u(t, x)|

2dd−2 dx.

As with the cubic NLS (4), the energy of of the initial data u0 is conserved over time. In our analysis of (4),we computed conservation of mass and left conservation of energy to the reader; here, we will perform theenergy conservation computation. Explicitly, we show that if u solves (15), then ∂tE(u(t)) = 0. Indeed, bythe chain rule and product rule,

∂tE(u(t)) =

∫1

2∂t(∇u · ∇u)− d− 2

2d∂t(uu)

dd−2 dx

=

∫Re∇ut · ∇u− (uu)

dd−2−1 Reutu dx

= Re

∫∇ut · ∇u− |u|

4d−2utu dx.

Integrating by parts,

= −Re

∫ut∆u+ |u|

4d−2utu dx

= −Re

∫ut

(∆u+ |u|

4d−2u

)dx.

As u is a solution to (15), ut = i(

∆u+ |u|4d−2u

). Thus,

∂tE(u(t)) = −Re i

∫ ∣∣∣∆u+ |u|4d−2u

∣∣∣2 dx = 0

as desired.

66

This fact, together with the Sobolev embedding theorem, implies that if u0 ∈ H1 then |E(u(t))| =

|E(u0)| <∞. Also note that the class of solutions is left invariant by the rescaling

u(t, x) 7→ uλ(t, x) := λ−d−2

2 u

(t

λ2,x

λ

).

Indeed, if u solves (15), then

iuλt + ∆uλ = iλ−d−2

2 λ−2ut + λ−d−2

2 λ−2∆u = λ−d−2

2 λ−2|u|4d−2u = λ−

4d−2 ·

d−22 |u|

4d−2λ−

d−22 u

= |uλ|4d−2uλ

so that uλ is also a solution.The most important fact about (15), and the reason why (15) is called energy-critical, is thatE(uλ) = E(u).

In words, scaling leaves the energy unchanged. Indeed,

E(uλ) = E(uλ(0)) =

∫1

2|λ−

d−22 λ−1∇u(0, x/λ)|2 − d− 2

2d|λ−

d−22 u(0, x/λ)|

2dd−2 dx

=

∫1

2λ−d|∇u(0, x/λ)|2 − d− 2

2dλ−d|u(0, x/λ)|

2dd−2 dx

=

∫1

2|∇u(0, x)|2 − d− 2

2d|u(0, x)|

2dd−2 dx = E(u0)

= E(u).

Among other things, scaling invariance of energy is inconvenient for the reason that scaling cannot detectblow-ups in energy.

Now we describe the importance of the optimal constant CSob in Theorem 6.1. Invoking the theoremgives the following estimate on the energy:

E(u) =

∫1

2|∇u|2 − d− 2

2d|u|

2dd−2 dx

≥ 1

2‖∇u‖2L2 −

d− 2

2dC

2dd−2

Sob ‖∇u‖2dd−2

L2 .

Consider the function f(y) = 12y

2 − d−22d C

2dd−2

Sob y2dd−2 . As 2d

d−2 > 2, this polynomial attains its maximum on ahump between its two nonnegative roots. To locate the maximizing critical point, we compute:

0 = f ′(y) = y − C2dd−2

Sob yd+2d−2 ⇒ y

d+2d−2−1 = C

− 2dd−2

Sob ⇒ y4d−2 = C

− 2dd−2

Sob

and so the maximum occurs at y = C− d2Sob . It will be convenient to solve for CSob strictly in terms of the

kinetic energy of W . Towards this goal, multiplying the equation (13) by W and integrating gives∫(∆W )W +W

d+2d−2 +1 dx = 0 ⇒

∫|∇W |2 dx =

∫W

2dd−2 dx.

So ‖W‖L

2dd−2

= ‖∇W‖d−2d

L2 and therefore the critical point is

C− d2Sob =

(‖W‖

L2dd−2

‖∇W‖L2

)− d2=(‖∇W‖

d−2d −1

L2

)− d2= ‖∇W‖L2 .

Furthermore, the maximum value is

f(‖∇W‖L2) =1

2‖∇W‖2L2 −

d− 2

2dC

2dd−2

Sob ‖∇W‖2dd−2

L2 =1

2‖∇W‖2L2 −

d− 2

2d‖W‖

2dd−2

L2dd−2

= E(W ).

67

The importance of W , and consequently the optimal constant CSob, can be seen as follows. Suppose thatE(u0) < E(W ) and ‖∇u0‖L2 < ‖∇W‖L2 . Because energy is conserved and also because the energy isbounded below by the polynomial f(‖∇u(t)‖L2), the solution is forced to live on the dotted line in thefigure below for all times t, i.e., ‖∇u(t)‖L2 < ‖∇W‖L2 for all times of existence.

‖∇u0‖L2 ‖∇W‖L2

E(u(t)) = E(u0)

E(W ) f(y)

In this scenario (namely, the NLS (15) with E(u0) < E(W ) and ‖∇u0‖L2 < ‖∇W‖L2 ), global well-posedness and scattering4 have been established for d ≥ 5 by Killip and Visan and for d = 4 by Dodson. Inthe d = 3 case, Kenig and Merle established global well-posedness and scattering for radial initial data. Thed = 3 case for non-radial initial data is still an open problem.

We end this discussion with a few comments on the elliptic equation (13). Given the energy critical NLS(15), it is natural to seek soliton solutions of the form u(t, x) = eitQ(x). A solution of this form must thensatisfy

−i · ieitQ+ eit∆Q = −|Q|4d−2 eitQ ⇒ ∆Q+Q

d+2d−2 = Q.

However, this equation only has trivial solutions. This can be shown by using the Pohozaev identities fromLemma 5.2 with p = 2d

d−2 to solve for the mass, which is evidently 0. We leave the details to the reader.Hence, no soliton solutions of this form exist. Next, we might seek solutions which are completely timeindependent, i.e., a solution of the form u(t, x) = W (x). In this case, a solution must satisfy the equation

∆W +Wd+2d−2 = 0,

which is precisely (13).In summary, the optimal constant in the Sobolev embedding theorem is important because the solution

W to the energy-critical NLS (15) is as unstable as possible. At energy levels below that ofW , there is globalwell-posedness and scattering. Above, it is possible for solutions to blow up in finite time.

6.2 Motivation: existence of bound states

Before proving the sharp Sobolev embedding theorem, we discuss another application, this time in thecontext of linear partial differential equations.

4Scattering here means that solutions to the nonlinear problem behave asymptotically like solutions to the linear Schrodinger equa-tion. Explicitly, an equation exhibits scattering if there exists u± ∈ H1 such that

∥∥u(t)− eit∆u±∥∥H1 → 0 as t→ ±∞.

68

Consider the operator H := −∆ + V , where V ≤ 0 is a smooth potential which decays at ∞. Aneigenfunction of H corresponding to a negative eigenvalue is called a bound state. Our motivating questionis the following: how large does V have to be so that H admits a bound state?

The following proposition offers a step in the right direction towards the answer to this question.

Proposition 6.1. If ‖V ‖Ld2≤ C−2

Sob, then H = −∆ + V does not admit a bound state.

Proof. Suppose for the sake of contradiction that H does admit a bound state, i.e., there exists a λ < 0 and anonzero ψ ∈ H2(Rd) such that (−∆ + V )ψ = λψ. Integrating this equation against ψ gives∫

|∇ψ|2 dx+

∫V |ψ|2 dx = λ

∫|ψ|2 dx.

Because λ∫|ψ|2 dx < 0, we have∫

|∇ψ|2 dx <∫|V ||ψ|2 dx ≤ ‖V ‖

Ld2

∥∥|ψ|2∥∥L

dd−2≤ C−2

Sob ‖ψ‖2

L2dd−2

.

Applying the sharp Sobolev embedding theorem,∫|∇ψ|2 dx ≤ C−2

SobC2Sob ‖∇ψ‖

2L2 = ‖∇ψ‖2L2 .

But ‖∇ψ‖2L2 < ‖∇ψ‖2L2 is a contradiction.

Next, we will show that there are potentials V with Ld2 -norm larger that C−2

Sob which do admit boundstates. In fact, we have the following.

Proposition 6.2. For any η > 0, there exists a smooth and decaying potential V ≤ 0 satisfying C−2Sob <

‖V ‖Ld2< C−2

Sob + η such that H = −∆ + V admits a bound state.

Proof. Take V = −(1 + ε)W4d−2 , where W is the solution to (13) as in the statement of the sharp Sobolev

embedding theorem. Then

‖V ‖Ld2

= (1 + ε)∥∥∥W 4

d−2

∥∥∥Ld2

= (1 + ε) ‖W‖4d−2

L2dd−2

.

Integrating (13) against W as in the previous section, we get∫|∇W |2 dx =

∫W

2dd−2 dx ⇒ ‖∇W‖L2 = ‖∇W‖

dd−2

L2dd−2

(16)

and consequently

CSob = ‖W‖1−dd−2

L2dd−2

= ‖W‖−2d−2

L2dd−2

.

Thus, ‖V ‖Ld2

= (1 + ε)C−2Sob, hence C−2

Sob < ‖V ‖L d2 < C−2Sob + η for ε sufficiently small.

We have produced a potential V with the desired properties. It remains to show that the operator−∆ − (1 + ε)W

4d−2 admits a bound state, that is, that there exists a λ < 0 and a nonzero ψ ∈ H2(Rd)

such that −∆ψ − (1 + ε)W4d−2ψ = λψ. Integrating the left hand side of this equation against ψ gives the

expression for the energy of ψ:

E(ψ) =

∫|∇ψ|2 − (1 + ε)W

4d−2 |ψ|2 dx.

69

Note thatE(W ) =

∫|∇W |2 − (1 + ε)W

2dd−2 dx = −ε

∫W

2dd−2 dx < 0

by (16). Also observe that the map ψ 7→ E(ψ) is continuous on H1(Rd).We want to use this information to conclude thatH has a negative eigenvalue. We present two methods.

Method 1. The first method invokes some tools from functional analysis. Because ψ 7→ E(ψ) is continuouson H1(Rd) and E(W ) < 0, there exists ϕ ∈ C∞c (Rd) such that E(ϕ) < 0. Because E(ϕ) = 〈Hϕ,ϕ〉, thisimplies that H has negative spectrum.

We want to deduce that H has a negative eigenvalue within this spectrum. Recall (see Reed and Simon?)Weyl’s criterion, which says if H is a relatively compact perturbation of of −∆, then the essential spectrumofH coincides with the essential spectrum of−∆, that is, σess(H) = σess(−∆) = [0,∞]. From this it followsthat the negative spectrum of H isolated contains eigenvalues of finite multiplicities.

Method 2. More directly, we can show existence of an eigenvalue by solving a minimization problem. Con-sider

Emin := infψ∈H1, ‖ψ‖2=1

E(ψ).

We will show that Emin is the negative eigenvalue of H .First, we claim that −∞ < Emin < 0. For ψ ∈ H1(Rd) with ‖ψ‖L2 = 1,

E(ψ) ≥ −(1 + ε)

∫W

4d−2 |ψ|2 dx.

Note (see (12)) that |W | ≤ 1. Thus,

E(ψ) ≥ −(1 + ε) ‖ψ‖2L2 = −(1 + ε) > −∞.

This shows that Emin > −∞. Next we show that Emin < 0. We already know that E(W ) < 0, so we willuse this to our advantage. Note that W . 〈x〉−(d−2), and that 〈x〉−(d−2) ∈ L2(Rd) provided 2(d − 2) > d.This holds for d > 4. Consequently, will consider d ≥ 5 separately from d = 3 and d = 4. For d ≥ 5, simplylet ψ = W

‖W‖L2. Then E(ψ) = ‖W‖−2

L2 E(W ) < 0, and so Emin < 0. If d = 3 or d = 4, ‖W‖2L2 is not finite, sothe above argument has to be modified. Let

ψ =Wϕ(·/R)

‖Wϕ(·/R)‖L2

where ϕ ∈ C∞c (Rd) is a bump function of unit height and R >> 1 is to be determined. Then

E(ψ) =1

‖Wϕ(·/R)‖L2

[∫|∇(W (x)ϕ(x/R))|2 dx− (1 + ε)

∫W

2dd−2 |ϕ(x/R)|2 dx

].

By the monotone convergence theorem,∫W

2dd−2 |ϕ(x/R)|2 dx→

∫W

2dd−2 dx as R →∞. Expanding the first

integral, we have∫|∇(W (x)ϕ(x/R))|2 dx =

∫|∇W (x)|2 |ϕ(x/R)|2 dx+R−2

∫|W (x)|2 |∇ϕ(x/R)|2 dx

+ 2R−1

∫∇W (x)∇ϕ(x/R)W (x)ϕ(x/R) dx.

Since W ∈ H1, the first integral converges to ‖∇W‖2L2 as R → ∞, by the monotone convergence theorem.In the second integral, observe that the support of ∇ϕ(x/R) is an annulus of radius ∼ R. Thus, we canapply Holder’s inequality by only integrating over this annulus:

R−2

∫|W (x)|2 |∇ϕ(x/R)|2 dx ≤ R−2 ‖W‖2

L2dd−2 (|x|∼R)

‖(∇ϕ)(·/R)‖Ld2.

70

By the above comment, ‖(∇ϕ)(·/R)‖Ld2. Rd·

2d = R2. Thus, the second term is bounded by ‖W‖2

L2dd−2 (|x|∼R)

,

which → 0 as R → ∞ by the decay of W . We handle the third term similarly, using the annular supportof ∇ϕ(·/R) and the decay of W to show that this tends towards 0 as R → ∞. Hence, E(ψ) → E(W ) asR→∞, so for R sufficiently large it follows that E(ψ) < 0, and hence Emin < 0. This proves the claim.

We continue. Let ψn ∈ H1(Rd) with ‖ψn‖L2 = 1, be an optimizing sequence, that is, E(ψn)→ Emin. Weclaim that, passing to a subsequence, ψn → ψ in H1(Rd).

Assume this claim for now. Let ϕ ∈ C∞c (Rd) with ‖ϕ‖L2 = 1 satisfying 〈ϕ,ψ〉 = 0. Consider

ψ + tϕ√1 + t2

∈ H1(Rd).

This function is L2-normalized. Thus,

E

(ψ + tϕ√

1 + t2

)≥ Emin = E(ψ).

This implies thatd

dtE

(ψ + tϕ√

1 + t2

) ∣∣∣∣∣t=0

= 0.

Expanding this out yields

0 =d

dt

{1

t2

[∫|∇(ψ + tϕ)|2 dx− (1 + ε)

∫W

4d−2 (ψ + tϕ)2 dx

]} ∣∣∣∣∣t=0

.

Note that ddt

(1

1+t2

) ∣∣∣t=0

= 0. Thus,

0 =d

dt

[∫|∇(ψ + tϕ)|2 dx− (1 + ε)

∫W

4d−2 (ψ + tϕ)2 dx

] ∣∣∣∣∣t=0

= 2

∫∇ψ · ∇ϕdx− 2(1 + ε)

∫W

4d−2ψϕdx

= 2

∫ [−∆ψ − (1 + ε)W

4d−2ψ

]ϕdx.

Thus.−∆ψ − (1 + ε)W

4d−2ψ⊥ Spanϕ

for any such ϕ, and hence−∆ψ − (1 + ε)W

4d−2ψ⊥ (Spanψ)⊥.

This implies that−∆ψ − (1 + ε)W

4d−2ψ = λψ

for some λ ∈ C. To see that λ must be a negative real number, integrate this equation against ψ to getE(ψ) = λ

∫|ψ|2 dx. But as E(ψ) < 0, it follows that λ < 0. In fact, by normalization of ψ, λ = Emin.

We return to the claim that the optimizing sequence converges strongly in H1(Rd). Because W ≤ 1, wehave

‖∇ψn‖2L2 = E(ψn) + (1 + ε)

∫W

4d−2 |ψn|2 dx ≤ E(ψn) + 1 + ε.

Because the ψn sequence is optimizing, E(ψn) is bounded above, and thus the above computation showsthat ψn is H1-bounded. Passing to a subsequence we extract a weak H1-limit ψ. FINISH

71

6.3 The concentration compactness principle for Sobolev embedding

In this section, we develop the concentration compactness principle necessary to prove Theorem 6.1, thesharp Sobolev embedding theorem. The structure of what follows is similar to the development of the con-centration compactness principle for the Gagliardo-Nirenberg inequality: first, we describe how to locatebubble profiles, and then inductively construct a profile decomposition.

As we remarked at the beginning of this section, in H1(Rd) degenerate weak limits can exist due totranslation (as in H1(Rd)) but also due to scaling. Thus, to develop a concentration compactness principlefor the Sobolev embedding theorem, it is necessary to identify not only locations for a bubble profile, butalso spatial scales. The following lemma helps us achieve this.

Lemma 6.2 (Refined Sobolev embedding). Fix d ≥ 3. Then

‖f‖L

2dd−2

. supN∈2Z

‖PNf‖2d

L2dd−2‖∇f‖

d−2d

L2 .

Broadly, this lemma says that if the potential energy of f (its L2dd−2 -norm) is large, then this must be due

to a particular frequency annulus.

Proof. Assume that d ≥ 4. The d = 3 case involves a few more subtleties, and we leave the details to thereader.

The natural way to introduce frequencies into the L2dd−2 -estimate is via the Littlewood-Paley square

function. We have

‖f‖2dd−2

L2dd−2∼

∥∥∥∥∥∥(∑

N

|fN |2) 1

2

∥∥∥∥∥∥2dd−2

L2dd−2

=

∫ (∑N

|fN |2) dd−2

dx

=

∫ (∑N

|fN |2) d

2(d−2)(∑

N

|fN |2) d

2(d−2)

dx.

If p ≤ 1, then ‖·‖`1 ≤ ‖·‖`p . Thus, if d2(d−2) ≤ 1, we have the estimate

(∑N

|fN |2) d

2(d−2)

≤

(∑N

|fN |2·d

2(d−2)

)=∑N

|fN |dd−2 .

The condition d2(d−2) ≤ 1 is equivalent to d ≥ 4, which true by assumption. Thus,

‖f‖2dd−2

L2dd−2

.∫ (∑

N

|fN |dd−2

)(∑N

|fN |dd−2

)dx .

∑M≤N

∫|fM |

dd−2 |fN |

dd−2 dx.

We wish to use Holder’s inequality. By Bernstein’s inequality, we have the heuristic that low frequenciesare small in high Lebesgue spaces, and high frequencies are small in low Lebesgue spaces. By our targetestimate, we want a Holder decomposition with an L

2dd−2 -norm. By the latter heuristic, we want to put a

copy of fN in a low Lebesgue space (say L2), as N represents high frequencies, and a copy of fM in a highLebesgue space. So consider∫

|fM |dd−2 |fN |

dd−2 dx =

∥∥∥|fM | |fM | 2d−2 |fN |

2d−2 |fN |

∥∥∥L1.

72

Note that∥∥∥|fN | 2

d−2

∥∥∥Ld

= ‖fN‖2d−2

L2dd−2

. Towards our goal estimate, we put the second and third terms in

Ld. If we put the fourth term in L2, then the first term necessarily lives in the Lebesgue space dictated by1− 1

d −1d −

12 = 2d−2−2−d

2d = d−42d . Thus,

‖f‖2dd−2

L2dd−2

.∑M≤N

‖fM‖L

2dd−4‖fM‖

2d−2

L2dd−2‖fN‖

2d−2

L2dd−2‖fN‖L2

. supK‖fK‖

4d−2

L2dd−2

∑M≤N

‖fM‖L

2dd−4‖fN‖L2 .

With the goal of summing in both M and N , we use Bernstein’s inequalities as follows:

‖f‖2dd−2

L2dd−2

. supK‖fK‖

4d−2

L2dd−2

∑M≤N

Md( d−22d −

d−42d ) ‖fM‖

L2dd−2

N−1 ‖∇fN‖L2

= supK‖fK‖

4d−2

L2dd−2

∑M≤N

M

N‖fM‖

L2dd−2‖∇fN‖L2 .

By Schur’s test,

∑M≤N

M

N‖fM‖

L2dd−2‖∇fN‖L2 .

(∑M

‖fM‖2L

2dd−2

) 12(∑

N

‖∇fN‖2L2

) 12

.

The latter quantity is comparable to ‖∇f‖L2 . By the Sobolev embedding theorem,(∑M

‖fM‖2L

2dd−2

) 12

.

(∑M

‖∇fM‖2L2

) 12

∼ ‖∇f‖L2 .

Putting all of this together gives

‖f‖2dd−2

L2dd−2

. supK‖fK‖

4d−2

L2dd−2‖∇f‖2L2 .

Raising both sides to the power of d−22d gives the desired estimate.

Next, we prove the analogue of Lemma 5.4. Given a sequence of functions which is bounded in H1(Rd)and has large potential energy, the following lemma describes how to locate a bubble profile, as well aslocations and spatial scales for the profile.

Lemma 6.3 (Inverse Sobolev embedding). Fix d ≥ 3, and suppose fn ∈ H1(Rd) satisfies lim sup ‖∇fn‖L2 =

A <∞ and lim inf ‖fn‖L

2dd−2

= ε > 0. Then by passing to a subsequence, there exists a ϕ ∈ H1(Rd), xn ∈ Rd,and λn ∈ (0,∞), such that

1. The sequence fn concentrates to the bubble profile ϕ in the sense that

λd−2

2n fn(λnx+ xn) ⇀ ϕ(x)

in H1(Rd).

2. There is nontrivial asymptotic decoupling in H1(Rd):

limn→∞

‖fn‖2H1 −∥∥∥fn(x)− λ−

d−22

n ϕ(λ−1n (x− xn))

∥∥∥2

H1= ‖ϕ‖2H1 & A2

( εA

) d2

2

.

73

3. There is nontrivial asymptotic decoupling in L2dd−2 (Rd):

limn→∞

‖fn‖2dd−2

L2dd−2−∥∥∥fn(x)− λ−

d−22

n ϕ(λ−1n (x− xn))

∥∥∥ 2dd−2

L2dd−2

= ‖ϕ‖2dd−2

L2dd−2

& ε2dd−2

( εA

) d(d+2)2

.

Proof. By passing to a subsequence we may assume that ‖fn‖H1 ≤ 2A and ‖fn‖L

2dd−2

> ε2 .

First, we identify the spatial scales λn. By the refined Sobolev embedding theorem, the largeness of thepotential energy of fn is due to a particular frequency scale. By the uncertainty principle, identifying thisfrequency scale will give us a corresponding spatial scale. Explicitly, by the refined Sobolev embeddingtheorem,

ε

2. sup

N‖PNfn‖

2d

L2dd−2‖∇fn‖

d−2d

L2 . supN‖PNfn‖

2d

L2dd−2

Ad−2d .

This implies that

supN‖PNfn‖

L2dd−2

&[εA

2−dd

] d2

= A( εA

) d2

.

By definition of the supremum, there must be a frequency which witnesses this inequality. In other words,there must be a frequency responsible for the nontrivial potential energy. Let Nn be such a frequency, i.e., adyadic number such that

‖PNnfn‖L

2dd−2

& A( εA

) d2

.

With the uncertainty principle in mind, let λn be the spatial scale defined by λn = N−1n .

Having identified spatial scales λn, we now identify spatial locations xn ∈ Rd. As in the inverseGagliardo-Nirenberg inequality, we seek a witness show for the fact that the L∞-norm of PNnfn is large.We can insert the L∞-norm with a clever use5 of Holder’s inequality:

A( εA

) d2

. ‖PNnfn‖L

2dd−2

=∥∥∥|PNnfn| d

d−2

∥∥∥ d−2d

L2=∥∥∥|PNnfn| d

d−2−1 |PNnfn|∥∥∥ d−2

d

L2

≤∥∥∥|PNnfn| d

d−2−1∥∥∥ d−2

d

L∞‖PNnfn‖

d−2d

L2

= ‖PNnfn‖1− d−2

d

L∞ ‖PNnfn‖d−2d

L2 .

We have‖PNnfn‖

d−2d

L2 . ‖fn‖d−2d

L2 . N− d−2

dn ‖∇fn‖

d−2d

L2 . N− d−2

dn A

d−2d .

Thus,

A( εA

) d2

. N− d−2

dn A

d−2d ‖PNnfn‖

2d

L∞ .

This implies that

N− d−2

2n ‖PNnfn‖L∞ & A

( εA

) d2

4

.

Let xn be a witness of this inequality. Then N−d−2

2n |PNnfn(xn)| & A

(εA

) d2

4 .

Having identified spacial scales and spatial locations, define gn(x) = λd−2

2n fn(λnx+ xn). Then

‖gn‖H1 = λd−2

2n λ

− d2n λn ‖∇fn‖L2 = ‖∇fn‖L2 . A.

so that gn is H1-bounded. Passing to a subsequence, we extract a weak H1 limit ϕ. This proves 1.

5Or rather, simply working out the exponents in the interpolation version of Holder’s inequality.

74

Next, we prove 2 and 3. First, we show that the bubble profile ϕ is nontrivial in the desired H1- sense.Let ψ = P1δ0 be the usual Littlewood-Paley multiplier. First,

|⟨ϕ, ψ

⟩| . ‖ϕ‖

L2dd−2

∥∥ψ∥∥L

2dd+2

. ‖ϕ‖L

2dd−2

. ‖∇ϕ‖L2

so that ‖ϕ‖H1 & |⟨ϕ, ψ

⟩|. Next, we produce a lower bound for the inner product. As ϕ is a weak H1-limit

of the gn’s,

|⟨ϕ, ψ

⟩| = lim

n→∞

∣∣∣⟨λ d−22

n fn(λnx+ xn), ψ(x)⟩∣∣∣ = lim

n→∞N− d−2

2n

∣∣∣∣∫ fn(N−1n x+ xn) ψ(x) dx

∣∣∣∣= limn→∞

N− d−2

2n

∣∣∣∣∫ fn(y + xn)Ndnψ(Ny) dy

∣∣∣∣= limn→∞

N− d−2

2n

∣∣[fn ∗Ndnψ(N ·)

](xn)

∣∣ .Thus,

‖ϕ‖H1 & |⟨ϕ, ψ

⟩| = lim

n→∞N− d−2

2n |PNnfn(xn)| & A

( εA

) d2

4

.

This is precisely the lower bound that we need for statement 2. The asymptotic decoupling statement in 2is immediate from the Hilbert space structure of H1(Rd), just as with the H1-asymptotic decoupling in theinverse Gagliardo-Nirenberg inequality. In particular,

‖gn‖2H1 − ‖gn − ϕ‖2H1 − ‖ϕ‖2H1 = 2 Re 〈gn, ϕ〉H1 − 2 ‖ϕ‖2H1 → 0

as n→∞. This establishes the entirety of statement 2.To prove statement 3, we will ultimately use the refined Fatou’s lemma, as in the proof of the inverse

Gagliardo-Nirenberg inequality. To invoke the lemma, we need to ensure that the gn’s are L2dd−2 -bounded

and also that they converge almost everywhere to ϕ. Boundedness in L2dd−2 follows immediately from the

Sobolev embedding theorem. To get almost everywhere convergence, we wish to use Rellich-Kondrachovas we did before. However, Rellich-Kondrachov as stated in Theorem 4.2 holds for H1(Rd), not H1(Rd).However, the same result for H1(Rd). This can be proven by repeating the proof of Theorem 4.2 andreplacing all uses of the Gagliardo-Nirenberg inequality by the Sobolev embedding theorem and a fewextra applications of Holder’s inequality.6 With this, we have (after passing to a subsequence) that gn → ϕ

strongly in L2(K) for any compact set K. Passing to another subsequence, we have almost everywhereconvergence on K. Exhausting Rd by compact sets and using a diagonalization argument, it follows thatgn → ϕ almost everywhere in Rd. Then, by the refined Fatou’s lemma,

‖gn‖2dd−2

L2dd−2− ‖gn − ϕ‖

2dd−2

L2dd−2− ‖ϕ‖

2dd−2

L2dd−2→ 0

as n→∞. Changing variables gives the desired result.

At last, we state and prove the concentration compactness principle for the Sobolev embedding theorem.With the inverse Sobolev embedding theorem, proof is almost the same as the proof in the Gagliardo-Nirenberg case, with the necessary scaling condition being the only new computation.

6I worked through the details on paper; the computations are subtly different. I may later add the H1(Rd) Rellich-Kondrachovtheorem as a separate result in the Compactness in Lp section.

75

Theorem 6.4 (Concentration compactness principle for Sobolev embedding). Fix d ≥ 3 and let fn ∈ H1(Rd)be an H1-bounded sequence. There exists J∗ ∈ {0, 1, 2, . . . } ∪ {∞}, {ϕj}J∗j=1 ∈ H1(Rd) \ {0}, λjn ∈ (0,∞),and xjn ∈ Rd such that, after passing to a subsequence, we have a decomposition

fn(x) =

J∑j=1

(λjn)− d−2

2 ϕj(x− xjnλjn

)+ rJn(x) (17)

for every finite 0 ≤ J ≤ J∗, satisfying the following properties:

1. The potential energy of the remainder term tends to 0, i.e.,

lim supJ→J∗

limn→∞

∥∥rJn∥∥L

2dd−2

= 0.

2. There is asymptotic decoupling in H1(Rd), i.e.,

supJ

limn→∞

‖fn‖2H1 −J∑j=1

∥∥ϕj∥∥2

H1 −∥∥rJn∥∥2

H1

= 0.

3. There is asymptotic decoupling in L2dd−2 (Rd), i.e.,

supJ

limn→∞

‖fn‖ 2dd−2

L2dd−2−

J∑j=1

∥∥ϕj∥∥ 2dd−2

L2dd−2−∥∥rJn∥∥ 2d

d−2

L2dd−2

= 0.

4. For all j ≤ J ,(λjn) d−2

2 rJn(λjnx+ xjn) ⇀ 0 in H1(Rd) as n→∞.

5. For all j 6= k, λjn

λkn+

λknλjn

+|xjn−x

kn|

2

λjnλkn→∞ as n→∞.

Much of the discussion following the statement of the Gagliardo-Nirenberg concentration compactnessprinciple applies in this case. The primary difference is that statement 5 in the former case described howdifferent bubble profiles disperse in location over time. In this case, statement 5 describes how either thelocations or the spatial scales of two different bubble profiles disperse over time.

Proof. As before, we inductively extract bubbles using the inverse Sobolev embedding theorem.Set r0

n = fn, and suppose that there exists a decomposition of the form (17) up to a level J ≥ 0 satisfyingproperties 2 and 3, and 4 for j = J . We will prove properties 1, 5, and 4 for j < J at the end.

Passing to a subsequence, we may set AJ = limn→∞∥∥rJn∥∥H1 and εJ = limn→∞

∥∥rJn∥∥L

2dd−2

. If εJ = 0,then set J∗ = J . In this case, property 1 is immediate. If εJ > 0, then by the inverse Sobolev embeddingtheorem there exists a nonzero ϕJ+1 ∈ H1(Rd), λJ+1

n ∈ (0,∞), and xJ+1n ∈ Rd such that(

λJ+1n

) d−22 rJn(λJ+1

n x+ xJ+1n ) ⇀ ϕJ+1(x)

in H1(Rd) as n→∞. Set

rJ+1n (x) = rJn(x)−

(λJ+1n

)− d−22 ϕJ+1

(x− xJ+1

n

λJ+1n

).

By construction, (λJ+1n

) d−22 rJ+1

n (λJ+1n x+ xJ+1

n ) ⇀ ϕJ+1(x)

76

in H1(Rd) as n→∞. This establishes property 4 for j = J + 1.Next, we verify 2 and 3. The inverse Sobolev embedding theorem gives the following asymptotic de-

coupling in the H1-sense: ∥∥rJn∥∥2

H1 −∥∥ϕJ+1

∥∥2

H1 −∥∥rJ+1n

∥∥2

H1 → 0.

Substituting this into the assumed level-J asymptotic decoupling in statement 2 gives the desired asymp-totic decoupling for J + 1. This verifies 2. The verification of 3 is exactly the same.

Next, passing to a subsequence we set AJ+1 = limn→∞∥∥rJ+1n

∥∥H1 and εJ+1 = limn→∞

∥∥rJ+1n

∥∥L

2dd−2

. IfεJ+1 = 0, stop and set J∗ = J + 1 and continue as above. If this process does not terminate in finitely manysteps, then set J∗ = ∞. By the non-degeneracy estimates of the inverse Sobolev embedding theorem wehave

A2J+1 ≤ A2

j

1− c(εJAJ

) d2

2

ε

2dd−2

J+1 ≤ ε2dd−2

J

1− c(εJAJ

) d(d+22

.

It can be shown using these estimates that εJ → 0, in which case property 1 holds.Next, we verify 5. Assume for the sake of contradiction that 5 fails. Let (j, k) with j < k witness

this failure, and moreover suppose that this is the first instance of failure in the sense that 5 holds for(j, l) with j + 1 ≤ l ≤ k − 1. Then because 5 does not hold for (j, k), by passing to a subsequence we haveλknλjn→ λ0 ∈ (0,∞) and xkn−x

jn√

λjnλkn→ x0 ∈ Rd. As in the Gagliardo-Nirenberg case, we will reach a contradiction

by showing that ϕk = 0. By construction,

rk−1n (x) = rjn(x)−

k−1∑l=j−1

(λln)− d−2

2 ϕl(x− xlnλln

).

Also by construction, (λkn) d−2

2 rk−1n (λknx+ xkn) ⇀ ϕk(x).

Combining these two facts tells us that

(λkn) d−2

2 rjn(λknx+ xkn) −k−1∑l=j−1

(λknλln

) d−22

ϕl(λknλlnx+

xkn − xlnλln

)⇀ ϕk(x).

To reach the desired contradiction, we will show that each of these pieces tends weakly to 0. Consider thefirst piece. By property 4, we know that

rnj(x) :=

(λjn) d−2

2 rjn(λjnx+ xjn) ⇀ 0.

To use this to our advantage, note that

(λkn) d−2

2 rjn(λknx+ xkn) =

(λkn

λjn

) d−22

rnj

(λkn

λjnx+

xkn − xjnλjn

).

By assumption, λkn

λjn→ λ0 and xkn−x

jn

λjn→ x0

√λ0. Hence, asymptotically,

(λkn) d−2

2 rjn(λknx + xkn) is a rescaling

and translation of rnj , and consequently converges weakly to 0. For the rest of the terms, it obviously

77

suffices to show that only one of the summands tends weakly to 0. Furthermore, by density we may replaceϕl by ϕ ∈ C∞c (Rd). So it remains to show that the quantity⟨(

λknλln

) d−22

ϕ

(λknλlnx+

xkn − xlnλln

), ψ(x)

⟩H1

=

⟨∇

(λknλln

) d−22

ϕ

(λknλlnx+

xkn − xlnλln

) ,∇ψ(x)

⟩L2

tends to 0. If we try to blindly use Holder’s inequality, the H1-invariance of ϕ under the prescribed scalingwill prevent this from being useful. Instead, we first integrate by parts and then use Holder. As there aretwo ways to integrate by parts in the above pairing, we then have

⟨(λknλln

) d−22

ϕ

(λknλlnx+

xkn − xlnλln

), ψ(x)

⟩H1

. min

(λknλln

) d−22 +2− d2

‖∆ϕ‖L2 ‖ψ‖L2 ,

(λknλln

) d−22 −

d2

‖ϕ‖L2 ‖∆ψ‖L2

.

The exponent d−22 + 2 − d

2 in the first term comes from differentiating ϕ twice by the Laplacian, followedby changing variables. In the second term, the exponent d−2

2 −d2 arises because we only change variables.

Thus, simplifying these exponents gives⟨(λknλln

) d−22

ϕ

(λknλlnx+

xkn − xlnλln

), ψ(x)

⟩H1

. min

{λknλln,λlnλkn

}.

Because 5 holds for (l, k), we have λknλln

+λlnλkn

+|xln−x

kn|

2

λlnλkn→∞. Hence, it is either the case that λ

kn

λln+

λlnλkn→∞

or |xln−x

kn|

2

λlnλkn→ ∞. In the former case it is clear that min

{λknλln,λlnλkn

}→ 0, in which case we are done. In the

latter case, if this does not hold then by passing to a subsequence we may assume λknλln→ λ1 ∈ (0,∞). Then

because |xln−x

kn|

2

λlnλkn→∞, the function

(λknλln

) d−22

ϕ

(λknλlnx+

xkn − xlnλln

)behaves asymptotically like a rescaling of ϕ whose profile location marches off to ∞, and hence clearlyconverges weakly to 0.

All of this, together with uniqueness of weak limits, implies that ϕk is identically 0. This is is a contra-diction, and hence 5 holds.

The argument for statement 4 for j < J is similar to the argument just given for 5, and the details areleft to the reader.

78