introduction - guillermo rey4 guillermo rey suppose now that q

LORENTZ SPACES AND REAL INTERPOLATION

THE KEEL-TAO APPROACH

GUILLERMO REY

1. Introduction

If an operator T is bounded on two Lebesgue spaces, the theory of complex interpolationallows us to deduce the boundedness of T “between” the two spaces. Real interpolation (ofLp spaces) provides, in a way, a refinement of complex interpolation, this will be shown as weprogress. We will follow Tao’s notes [6]: in particular, we claim no originality over any argumenthere.

Consider the (centered) Hardy-Littlewood maximal operator:

(1) Mf(x) = supr>0

∫Br(x)

|f(y)| dy,

defined for f ∈ L1loc. We would like to obtain Lp boundedness of this operator for 1 ≤ p ≤ ∞,

but alas, this is not the case: although it is trivially bounded from L∞ onto itself, it is notbounded on L1:

Proposition 1.1. If f ∈ L1(Rd) and f is not identically 0 then

Mf(x) ≥ Cd|x|d

for some constant Cd dependent on f and the dimension, in particular f /∈ L1(Rd).

Proof. Since f is not identically 0, there exists an R > 0 such that∫BR(0)

|f | ≥ δ > 0.

Now, if |x| > R, BR(0) ⊆ B(x, 2|x|), so

Mf(x) ≥ 1

|B(x, 2|x|)|

∫BR(0)

|f | ≥ Cdδ

|x|d.

�

Thus M is indeed unbounded as an operator from L1 onto itself. However, We do have thefollowing weaker form of L1 boundedness; proved by Hardy-Littlewood in [1] for d = 1 and byWiener in [7] for d > 1:

Theorem 1.2. There exists a constant C > 0 such that for all f ∈ L1

m({x ∈ Rd : Mf(x) > λ}) ≤ C ‖f‖L1

λ,

where m is the Lebesgue measure on Rd.1

2 GUILLERMO REY

Observe that, by Chebychev’s inequality, we would have had this bound if M had beenbounded on L1, so it is indeed a weaker statement.

This leads us to study the space of functions satisfying Chebychev’s inequality:

(2) µ({x ∈ X : |f(x)| > λ}) ≤‖f‖pLp(X,µ)

λp.

More precisely, define the weak -Lp space Lp,∞(X,µ) to be the set of all measurable functions ffor which its Lp,∞ quasi-norm

(3) ‖f‖Lp,∞ = supλ>0

λµ({x : |f(x)| > λ})1/p

is finite. To abbreviate we will use the notation

df (λ) = µ({x ∈ X : |f(x)| > λ})

when there is no confusion as to what measure we are using. Also, we define by convention L∞,∞

to be L∞.The following lemma is a simple application of the monotone convergence lemma:

Lemma 1.3. Let {fn}n∈N be a sequence of measurable functions such that |fn(x)| ≤ |fn+1(x)|µ-almost everywhere and for all n ∈ N. If we have |fn| → f µ-almost everywhere for somefunction f then

dfn(λ)↗ df (λ)

when n→∞ for all λ > 0.

As we have seen, Chebychev’s inequality gives us the inclusion Lp ↪→ Lp,∞ for p > 0. Thisinclusion is proper as the following example shows:

Example 1.4. If f(x) = |x|−dp then∫

Rd|f(x)|p dx =

∫Rd|x|−d dx =∞,

however

supλ>0

λdf (λ)1/p = ν1/pd <∞,

where νd is the measure of the unit ball in Rd.

As we have noted, ‖ · ‖Lp,∞ is not a priori a norm, but only a quasi-norm:

Proposition 1.5. We can easily see that

df+g(λ) ≤ df (λ/2) + dg(λ/2),

which is just the fact that when two terms sum more than λ, at least one of them has to be greaterthan λ/2. Using this fact we have:

‖f + g‖Lp,∞ = supλ>0

λdf+g(λ)1/p

≤ supλ>0

λ (df (λ/2) + dg(λ/2))1/p

≤ 2Cp (‖f‖Lp,∞ + ‖g‖Lp,∞) .

We will generalize the weak-Lp spaces a bit further, bit first we need the following well-knownidentity:

LORENTZ SPACES AND REAL INTERPOLATION THE KEEL-TAO APPROACH 3

Proposition 1.6. Let 0 < p <∞ and f be a measurable function, then

‖f‖Lp = p1/p

(∫R+

λpdf (λ)dλ

λ

)1/p

.

Proof.

‖f‖pLp(X,µ) =

∫X

|f(x)|p dµ(x)

=

∫X

∫ |f(x)|

0

pλp dλλ dµ(x)

= p

∫X

∫R+

λp1{s: s<|f(x)|}dλλ dµ(x)

= p

∫R+

λp∫X

1{y:|f(y)|>λ} dµ(x) dλλ (Fubini)

= p

∫R+

λpdf (λ) dλλ .

�

So we have:‖f‖Lp,∞ = ‖λdf (λ)1/p‖L∞(R+, dλλ )

and‖f‖Lp = p1/p‖λdf (λ)1/p‖Lp(R+, dλλ ).

From this, we could tentatively define the

Definition 1.7 (Lorentz quasi-norm). Let (X,µ) be a measure space and let 0 < p < ∞ and0 < q ≤ ∞, we define the Lorentz space quasi-norm ‖ · ‖Lp,q(X,µ) as

‖f‖Lp,q(X,µ) = p1/q‖λdf (λ)1/p‖Lq(R+, dλλ ).

Observe that, with this definition, Lp,p (isometrically) coincides with Lp.We have the following useful lemma, which is a kind of Monotone Convergence Theorem.

Lemma 1.8. Let {fn}n∈N be a sequence of measurable functions such that |fn| ↗ |f | µ-almosteverywhere, then

‖f‖Lp,q = limn→∞

‖fn‖Lp,q

Proof. We can obviously assume f to be positive so, by definition, we only have to verify that

limn≥1‖λdfn(λ)1/p‖Lq(R+, dλλ ) = ‖λdf (λ)1/p‖Lq(R+, dλλ ).

Note that, since fn is non-decreasing, dfn(·) is also increasing (in n), so in particular byLemma 1.3

df (λ) = supn≥1

dfn(λ)

for all fixed λ > 0.If q =∞ then the problem is reduced to the easier one of showing that

supn≥1‖λdfn(λ)1/p‖L∞λ (R+) = ‖ sup

n≥1λdfn(λ)1/p‖L∞λ (R+),

which is easily seen to hold true in the more general case

supn≥1‖gn‖L∞ = ‖ sup

n≥1gn‖L∞ .

4 GUILLERMO REY

Suppose now that q <∞, then

supn≥1‖λdfn(λ)1/p‖Lq(R+, dλλ ) =

(supn≥1

∫ ∞0

λqdfn(λ)q/p dλλ

)1/q

,

and since, as we noted before, we have λqdfn(λ)q/p ↗ λqdf (λ)q/p for all fixed λ > 0, we can

apply the monotone convergence theorem with the measure dλλ and the desired identity. �

The following Fatou-like lemma shows that the Lorentz space quasi-norm is lower semi-continuous.

Lemma 1.9. Let {fn}n∈N be any sequence of measurable functions, then

‖ lim infn

fn‖Lp,q ≤ lim infn‖fn‖Lp,q

Proof. Let f = lim infn fn, then, by definition, we have to check that

‖λdf (λ)1/p‖Lq(R+,

dλλ )≤ lim inf

n‖λdfn(λ)1/p‖

Lq(R+,dλλ ),

which would follow from the lower-semicontinuity of the usual Lq spaces provided that we have

df (λ) ≤ lim infn

dfn(λ).

This is equivalent to showing

µ({x : lim infn|fn(x)| > λ}) ≤ lim inf

nµ({x : |fn(x)| > λ}),

which follows from Fatou’s lemma. �

Finally, we can prove that the Lp,q spaces are quasi-Banach spaces; which follows the steps ofthe classical proof in the Lp case.

Theorem 1.10. Let (X,µ) be a measure space and let 0 < p <∞, 0 < q ≤ ∞, then Lp,q(X,µ)is a quasi-Banach space, that is, it is complete and it satisfies the quasi-triangle inequality.

Proof. We have already proven that Lp,∞ satisfies the quasi-triangle inequality in Proposition1.5, the proof for Lp,q when q 6=∞ is almost identical so we will omit it.

We are left to prove that it is complete. To this end first note that the Lp,q norm alwayscontrols the Lp,∞ norm:

‖λdf (λ)1/p‖Lq(R+, dλλ ) =

(∫ ∞0

λqdf (λ)q/p dλλ

)1/q

≥(∫ s

0

λqdf (λ)q/p dλλ

)1/q

≥ df (s)1/p

(∫ s

0

λq dλλ

)1/q

= sdf (s)1/pq−1/q,

for any s > 0. Thus, taking the supremum on s, we obtain:

‖f‖Lp,∞ .p,q ‖f‖Lp,q .In particular we have the following generalization of Chebychev’s inequality:

(4) df (λ) .p,q‖f‖pLp,qλp

.

Now let {fn} be a Cauchy sequence in the Lp,q quasi-norm. Estimate (4) allows us to deducethat {fn} is Cauchy in measure, and hence that there exists a µ-almost everywhere convergent


subsequence {fϕ(k)}k. Call f the µ-almost everywhere defined limit of fϕ(k) when k → ∞, wewill show that fn → f in the Lp,q quasi-norm.

Let C be the constant in the quasi-triangle inequality for the ‖ · ‖Lp,q quasi-norm and refinethe subsequence {fϕ(k)} so that we have

‖fϕ(k) − fϕ(k−1)‖Lp,q ≤ (2C)−k

using that fn is an Lp,q-Cauchy sequence and defining fϕ(0) = 0. Observe that we have thefollowing telescoping sum identity:

fϕ(n) =

n∑k=1

fϕ(k) − fϕ(k−1).(5)

Recall that we want to estimate ‖f − fn‖Lp,q and that, using the quasi-triangle inequality, wehave

‖f − fn‖Lp,q ≤ C(‖f − fϕ(N)‖Lp,q + ‖fϕ(N) − fn‖Lp,q

).

The second term can be made arbitrarily small using the fact that fn is an Lp,q-Cauchy sequence,so we choose n and N so large that

‖fϕ(N) − fn‖Lp,q <ε

2C.

For the first term we can use (5) to obtain:

‖f − fϕ(N)‖Lp,q =

∥∥∥∥ ∞∑k=1

(fϕ(k) − fϕ(k−1))−N∑k=1

(fϕ(k) − fϕ(k−1))

∥∥∥∥Lp,q

=

∥∥∥∥ limM→∞

M∑k=N+1

(fϕ(k) − fϕ(k−1)

)∥∥∥∥Lp,q

≤∥∥∥∥ limM→∞

M∑k=N+1

|fϕ(k) − fϕ(k−1)|∥∥∥∥Lp,q

≤ lim infM→∞

∥∥∥∥ M∑k=N+1

|fϕ(k) − fϕ(k−1)|∥∥∥∥Lp,q

,(6)

where in the last line we have used Proposition 1.9. By induction, we can iteratively use thequasi-triangle inequality to exchange

∑and ‖ · ‖Lp,q and bound the last term in (6) by

lim infM→∞

M∑k=N+1

Ck−N‖fϕ(k) − fϕ(k−1)‖Lp,q ≤ lim infM→∞

M∑k=N+1

Ck−N (2C)−k

= C−N∞∑

k=N+1

2−k

≤ (2C)−N .

Choosing N so large that 1(2C)N

< ε2C completes the proof. �

2. The Dyadic Decomposition

It turns out that we will be able to decompose functions in Lp,q into simpler functions:

• Sub-step functions: A sub-step function of height H and width W is any measurablefunction f supported on a set of measure W and which satisfies |f | ≤ H.

6 GUILLERMO REY

• Quasi-step function: A quasi-step function of height H and width W is any measurablefunction f supported on a set of measure W and which satisfies |f | ∼ H in its support.

Example 2.1. Let f be a quasi-step function of height H and width W , then

‖f‖Lp,q ∼ HW 1/p

(p

q

)1/q

.

We will give two different decompositions: a “vertically-dyadic” decomposition; where wedecompose the range of our function into dyadic pieces, and a “horizontally-dyadic” one; wherewe dyadically decompose its support.

Let {γm}m∈Z be a sequence in R+, we say that it is of lower exponential growth if we have

C1γm ≤ γm+1

for some constant C1 > 1. We will say that it is of exponential growth if we additionally have

γm+1 ≤ C2γm

for some C2 > 1.Note that the condition of being of lower exponential growth can easily be tested by the

equivalent condition of

infm∈Z

γm+1

γm> 1,

while that of being of exponential growth is in turn equivalent to being of lower exponentialgrowth and

supm∈Z

γm+1

γm<∞.

We can now state the main theorem of this section:

Theorem 2.2. Let f ≥ 0 be a measurable function, {γm}m∈Z be a sequence of exponential growthand 0 < p <∞, 1 ≤ q ≤ ∞. Then the following statements are equivalent:

(a) There exists a decomposition f =∑m∈Z fm, where each fm is a quasi-step function of

height γm and width Wm, the fm’s have pairwise disjoint supports and we have

(7)∥∥γmW 1/p

m

∥∥`q.p,q 1.

(b) We have the estimate

‖f‖Lp,q .p,q 1.

(c) There exists a decomposition f =∑m∈Z fm, where each fm is a sub-step function of

width γm and height Hm, the fm’s have pairwise disjoint supports, Hm is non-increasing,Hm+1 ≤ |fm| ≤ Hm in the support of fm and we have

(8)∥∥Hmγ

1/pm ‖`q .p,q 1.

Proof. Observe that, since {γm}Z is of lower exponential growth, we have

(9) γm−k ≤ C−k1 γm and γm+k ≥ Ck1 γmfor all m ∈ Z and k ≥ 0. This implies in particular that γm → 0 when m → −∞ and γm → ∞when m→∞.

Let us first prove that (a) =⇒ (b): By monotonicity we can assume that

f =∑m∈Z

γm1Em


where Em is the support of fm. Since the sequence γm is (strictly) increasing, we have

df (γm) =

∞∑k=1

µ(Em+k).

Let us first suppose that q =∞, we have to prove that

supλ>0

λdf (λ)1/p .p 1.

To this end observe that

supλ>0

λdf (λ)1/p = supm∈Z

supλ∈(γm,γm+1]

λdf (λ)1/p

≤ supm∈Z

γm+1df (γm)1/p

= supmγm+1

( ∞∑k=1

Wm+k

)1/p

≤

( ∞∑k=1

supm∈Z

γpm+1Wm+k

)1/p

=

( ∞∑k=1

supm∈Z

γpm+1−kWm

)1/p

≤

(supm∈Z

γpmWm

∞∑k=1

C−p(k−1)1

)1/p

(9)

.p supm∈Z

γmW1/pm

. 1.

Now suppose q <∞, then we can do something similar:

‖f‖qLp,q ∼p,q ‖λdf (λ)1/p‖qLq(R+, dλλ )

=∑m∈Z

∫ γm+1

γm

λqdf (λ)q/p dλλ

≤∑m∈Z

df (γm)q/p∫ γm+1

γm

λq dλλ

.q∑m∈Z

df (γm)q/pγqm+1

=∥∥df (γm)γpm+1

∥∥q/p`q/p

=

∥∥∥∥ ∞∑k=1

γpm+1Wm+k

∥∥∥∥q/p`q/p

.

We would like to exchange ‖ · ‖`p/q with∑

, but this is not possible in general unless q/p ≥ 1.We could fix this using the q/p-convexity of `q/p for 0 < q/p < 1, but we will treat it in a unifiedway using Lemma 5.1:

8 GUILLERMO REY

From (9) and (7), we have:

‖γpm+1Wm+k‖`q/pm= ‖γpm−(k−1)Wm‖`q/pm

≤ C−p(k−1)1 ‖γpmWm‖`q/pm

.C1C−pk1 ‖γpmWm‖`q/pm

= C−pk1 ‖γmW 1/pm ‖p

`qm

.p C−pk1 .

Hence, using Lemma 5.1, we obtain:∥∥∥∥ ∞∑k=1

γpm+1Wm+k

∥∥∥∥q/p`q/p.C1,p,q 1.

Let us now see that (b) =⇒ (c):Define

Hm = inf{λ > 0 : df (λ) ≤ γm},

Clearly Hm is non-increasing and, by definition, we have

λ < Hm =⇒ df (λ) > γm.

Observe also that, from the monotone convergence theorem, this infumum is achieved and sodf (Hm) = γm. Since ‖f‖Lp,q is finite, df (λ) has to be finite for λ > 0 (otherwise df (s) would beinfinite for 0 < s < λ). Thus, if Hm 6= 0, then the sets

Em = {x : f(x) > Hm}

are of finite measure. Using the monotone convergence theorem again we conclude that the set

E = {x : f(x) ≥ supmHm}

has measure 0. By the definition of Hm

df (λ) ≤ γm =⇒ Hm ≤ λ,

thus if df (ε) ≤ γm (which is true for m sufficiently large since γm → ∞ when m → ∞), thenHm ≤ ε, that is Hm → 0 when m→∞.

These remarks show that

f =∑m∈Z

fm,

where fm = f1Hm+1<f≤Hm .If q <∞ then

‖λdf (λ)1/p‖qLq((Hm+k+1,Hm+k], dλλ )

=

∫ Hm+k

Hm+k+1

λqdf (λ)q/p dλλ

≥ γq/pm+k

∫ Hm+k

Hm+k+1

λq dλλ

∼q γq/pm+k

(Hqm+k −H

qm+k+1

)&C1,p,q C

kq/p1 γq/pm

(Hqm+k −H

qm+k+1

).


Since Hm → 0 when m→∞, we can sum telescopically as follows:

Hmγ1/pm =

(Hqmγ

q/pm

)1/q

=

( ∞∑k=0

(Hqm+k −H

qm+k+1)γq/pm

)1/q

.C1,p,q

( ∞∑k=0

C−kq/p1 ‖λdf (λ)1/p‖q

Lq((Hm+k+1,Hm+k], dλλ )

)1/q

,

so taking `q norms

‖Hmγ1/pm ‖

q`q .C1,p,q

∑m∈Z

∞∑k=0



=

∞∑k=0

C−kq/p1

∑m∈Z‖λdf (λ)1/p‖q


=

∞∑k=0


Lq(R+, dλλ )

.C1,p,q 1.

Lastly, let q =∞. Then we immediately have

df (λ) . λ−p,

so if λ > γ−1/pm , then df (λ) . γm, that is: Hm . γ

−1/pm . This implies

‖Hmγ1/pm ‖`∞ . 1,

which is what we wanted.We now prove the implications in the reverse order. Let’s start with (c) =⇒ (b):By hypothesis and the estimate (9) we have

df (Hm) ≤∞∑k=1

γm−k ≤ γm∞∑k=1

C−k1 .C1γm.

Observe also that it follows from (7) that Hm → 0 when m→∞. Additionally:

λ < supm∈Z

Hm =⇒ λ ∈⋃m∈Z

[Hm+1, Hm).

So, if q <∞:

‖λdf (λ)1/p‖qLq(R+, dλλ )

=∑m∈Z

∫ Hm

Hm+1

λqdf (λ)q/p dλλ

.C1,q

∑m∈Z

γq/pm+1

(Hqm −H

qm+1

).C2,p

∑m∈Z

γq/pm Hm

= ‖γ1/pm Hm‖q`q

. 1.

10 GUILLERMO REY

If q =∞ then we can similarly estimate:

supλ>0

λdf (λ)1/p = supm∈Z

supλ∈[Hm+1,Hm)

λdf (λ)1/p

≤ supm∈Z

Hmdf (Hm+1)1/p

.C1,p supm∈Z

Hmγ1/pm+1

.C2,p supm∈Z

Hmγ1/pm

. 1.

The final stage of the proof is showing that

(b) =⇒ (a).

Define

fm = f1γm<|f |≤γm+1,

then clearly Wm = df (γm)− df (γm+1) ≤ df (γm).If q <∞ we proceed as follows:

1 & ‖λdf (λ)1/p‖qLq(R+, dλλ )

=∑m∈Z

∫ γm+1

γm

λqdf (λ)q/p dλλ

≥∑m∈Z

df (γm)q/p(γqm+1 − γqm

)&C2

∑m∈Z

df (γm)q/pγqm

≥∑m∈Z

W q/pm γq

=∥∥γmW 1/p

m

∥∥q`q.

If q =∞, then we have to show that

supmγmW

1/pm . 1,

but by hypothesis df (λ)1/p . λ−1, so

W 1/pm ≤ df (γm)1/p . γ−1

m

and the proof follows. �

Remark 2.3. Observe that the proof of (a) =⇒ (b) works verbatim when the sequence {γm}m∈Zis only of lower exponential growth.

3. Quantitative Properties of Lp,q spaces

As a first application of the dyadic decomposition theorems of the last section we can provethe following generalization of Holder’s inequality, originally proven by O’Neil in [4].

Theorem 3.1. Let 0 < p, p1, p2 <∞ and 0 < q, q1, q2 ≤ ∞ exponents satisfying

1

p=

1

p1+

1

p2and

1

q=

1

q1+

1

q2,


then for every measurable function we have

‖fg‖Lp,q .p1,q1,p2,q2 ‖f‖Lp1,q1‖g‖Lp2,q2 .

Proof. First observe that we can assume ‖f‖Lp1,q1 = ‖g‖Lp2,q2 = 1 and that f, g ≥ 0, so we onlyhave to prove

‖fg‖Lp,q . 1.

We will omit the dependence on the exponents for the rest of the proof.Decompose f and g using the horizontally dyadic decomposition with γm = 2m, i.e.:

f =∑m

fm, g =∑m

gm

where fm and gm are sub-step functions of height Hm and H ′m respectively and have supportsof measure ≤ 2m.

To estimate the fg we proceed as follows:

fg ≤∑m

∑n

fmgn

=∑k

∑m

fmgm+k

=∑k<0

∑m

fmgm+k +∑k≥0

∑m

fmgm+k,

since the Lp,q quasi-norm satisfies the quasi-triangle inequality it suffices to estimate the k < 0and k ≥ 0 sums separately. We will only show

‖∑k≥0

∑m

fmgm+k‖Lp,q . 1

since the other sum is estimated in the same way.Observe that fmgm+k is a sub-step function of height HmH

′m+k and width ≤ min(2m, 2m+k),

so ∑m

fmgm+k ≤∑m

HmH′m+k1Em ,

where Em is a set of measure 2m. We can now employ the implication (c) =⇒ (b) in Theorem2.2 to obtain:

‖∑m

fmgm+k‖Lp,q .∥∥HmH

′m+k2m/p

∥∥`q

=∥∥2m/p1Hm2m/p2H ′m+k

∥∥`q

≤∥∥2m/p1Hm

∥∥`q1

∥∥2m/p2H ′m+k

∥∥`q2

. 2−k/p2 ,

where we have used Holder’s inequality in the last line.To get ∥∥∥∥∑

k≥0

∑m

fmgm+k

∥∥∥∥Lp,q. 1

we just use Lemma 5.1. �

The following dual characterization of the Lp,∞ space will be useful.

Theorem 3.2. Let (X,µ) be a σ-finite measure space, 1 < p <∞ and f ∈ Lp,∞. Then

12 GUILLERMO REY

(i) If f ∈ Lp,∞ then

(10) ‖f‖Lp,∞ ∼p sup0<µ(E)<∞

1

µ(E)1/p′

∣∣∣∣∫X

f1E dµ

∣∣∣∣.(ii) If f1E is integrable for all sets E of finite measure and

Mp = sup0<µ(E)<∞

1

µ(E)1/p′

∣∣∣∣∫X

f1E dµ

∣∣∣∣ <∞,then f ∈ Lp,∞ and ‖f‖Lp,∞ .p Mp.

Proof. Let’s prove (i) first. The &p part is a simple consequence of Holder’s inequality for Lorentzspaces, so we will concentrate on showing the reverse inequality.

Let us first assume f to be non-negative. Let Eλ = {x ∈ X : f(x) > λ}, the assumptionf ∈ Lp,∞ guarantees that µ(Eλ) <∞. Then, if µ(Eλ) 6= 0:

λµ(Eλ)1/p =1

µ(Eλ)1/p′λµ(Eλ) ≤ 1

µ(Eλ)1/p′

∫X

f1Eλ dµ ≤ sup0<µ(E)<∞

1

µ(E)1/p′

∣∣∣∣∫X

f1E dµ

∣∣∣∣,taking the supremum in λ > 0 yields (10).

Now suppose f is not necessarily non-negative. We can decompose f = u+0 −u

−0 + i(u+

1 −u−1 )

where u±i ≥ 0 for i ∈ {0, 1}, so by the quasi-triangle inequality:

‖f‖Lp,∞ .p ‖u+0 ‖Lp,∞ + ‖u−0 ‖Lp,∞ + ‖u+

1 ‖Lp,∞ + ‖u−1 ‖Lp,∞ .It thus suffices to show that

‖u±i ‖Lp,∞ ≤ sup0<µ(E)<∞

1

µ(E)1/p′

∣∣∣∣∫X

f1E dµ

∣∣∣∣,but since u±i is non-negative, we can use what we already know to reduce the problem to showingthat

sup0<µ(E)<∞

1

µ(E)1/p′

∣∣∣∣∫X

u±i 1E dµ

∣∣∣∣ ≤ sup0<µ(E)<∞

1

µ(E)1/p′

∣∣∣∣∫X

f1E dµ

∣∣∣∣.Finally, observe that u±i = eiθf1F for some F ⊆ X, so for any given set E with 0 < µ(E) <∞we have that

1

µ(E)1/p′

∣∣∣∣∫X

u±i 1E dµ

∣∣∣∣ = 0 ≤ sup0<µ(E′)<∞

1

µ(E′)1/p′

∣∣∣∣∫X

f1E′ dµ

∣∣∣∣if µ(E ∩ F ) = 0 and

1

µ(E)1/p′

∣∣∣∣∫X

u±i 1E dµ

∣∣∣∣ =1

µ(E)1/p′

∣∣∣∣∫X

eiθf1E∩F dµ

∣∣∣∣≤ 1

µ(E ∩ F )1/p′

∣∣∣∣∫X

f1E∩F dµ

∣∣∣∣≤ sup

0<µ(E′)<∞

1

µ(E′)1/p′

∣∣∣∣∫X

f1E′ dµ

∣∣∣∣if µ(E ∩ F ) > 0. Taking the supremum over these sets yields the proof.

Let us now prove (ii). The idea is to approximate f by functions fn which are a priori inLp,∞, use (i) on these functions and then try to pass to the limit.

Indeed, since (X,µ) is σ-finite we can find an increasing sequence of sets {Fn}n∈N of finitemeasure and such that

X =⋃n

Fn.


We now define fn = f1En with En = Fn ∩ {x : |f(x)| ≤ n}. These functions are qualitatively inLp,∞:

‖fn‖Lp,∞ . nµ(En)1/p,

and also |fn| ↗ |f |, so we can use Proposition 1.9 to get:

(11) ‖f‖Lp,∞ ≤ lim infn→∞

‖fn‖Lp,∞ .

We would have finished if the right hand side of (11) were finite, because then we would havef ∈ Lp,∞ a priori. But observe that we can use (i) to bound each of these functions:

‖fn‖Lp,∞ .p sup0<µ(E)<∞

1

µ(E)1/p′

∣∣∣∣∫X

fn1E dµ

∣∣∣∣.Now we use a procedure similar to that in the proof of (i):

sup0<µ(E)<∞

1

µ(E)1/p′

∣∣∣∣∫X

f1E∩En dµ

∣∣∣∣ ≤ sup0<µ(E)<∞

1

µ(E ∩ En)1/p′

∣∣∣∣∫X

f1E∩En dµ

∣∣∣∣= supµ(E′)>0, E′⊆En

1

µ(E′)1/p′

∣∣∣∣∫X

f1E′ dµ

∣∣∣∣≤ sup

0<µ(E′)<∞

1

µ(E′)1/p′

∣∣∣∣∫X

f1E′ dµ

∣∣∣∣= Mp

�

We can also prove a duality theorem for Lp,q spaces:

Theorem 3.3. Let 1 < p <∞ and 1 ≤ q ≤ ∞, then for all f ∈ Σ+:

(12) ‖f‖Lp,q ∼p,q sup

{∫X

fg dµ : g ∈ Σ+, ‖g‖Lp′,q′ ≤ 1

}=: Mp,q(f),

where Σ+ is the set of all non-negative simple functions with support of finite measure.

Proof. As before, the &p,q inequality follows from Holder’s inequality for Lorentz spaces. Also,if q = ∞, then the result follows from Theorem 3.2. We will thus prove the inequality .p,qrestricted to q <∞.

Homogeneity allows us to normalize ‖f‖Lp,q = 1, we thus have to show that

Mp,q(f) &p,q 1.

Using Theorem 2.2 we can decompose f as

f =∑m∈Z

fm,

where each fm is a quasi-step function of height 2m and width Wm, and such that∥∥2mW 1/pm

∥∥`q∼ 1.

Observe that it follows from the proof of Theorem 2.2 that each fm must also be a simple functionand, also, that fm is 0 for |m| sufficiently large.

Define the function g by

g =∑m∈Z

gm

14 GUILLERMO REY

where gm = armfp−1m , am = 2mW

1/pm and r = max(0, q − p). Clearly g ∈ Σ+, and furthermore,

since the fm′s have disjoint supports Em:∫

X

fg dµ =

∫X

∑m

armfpm dµ

=∑m

arm

∫X

fpm dµ.

The functions fm are quasi-step functions of height 2m and width Wm, thus∫X

fpm dµ ∼p 2mpWm = apm,

that is ∫X

fg dµ ∼p∑m∈Z

ar+pm

(∗)≤ ‖am‖r+p`q ∼p,q ‖f‖Lp,q = 1,

where in (∗) we have used that ‖ · ‖`p1 ≤ ‖ · ‖`p2 when p1 ≥ p2 and that r + p ≥ q. So we onlyhave to show that

‖g‖Lp′,q′ ≤ 1.

Observe that

gm .p,q arm2m(p−1)

1Em

with µ(Em) = Wm, so a direct application of Theorem 2.2 would yield the result. The sequencearm2m(p−1), however, can fail to be of lower exponential growth (at least a priori). We can fixthis by defining

γm = 2mp−12 sup

k≤mark2k

p−12 = 2m(p−1) sup

k≥0arm−k2−k

p−12 .

Obviously γm ≥ arm2m(p−1), and furthermore:

γm+1 = 2p−12 2m

p−12 sup

k≤m+1ark2k

p−12

≥ 2p−12 2m

p−12 sup

k≤mark2k

p−12

= 2p−12 γm,

that is, {γm}m∈Z is of sub-exponential growth with constant C1 = 2p−12 > 1. Now we can use

Theorem 2.2 and the remark that follows it to reduce the problem to showing∥∥γmW 1/p′

m

∥∥`q′.p,q 1.

But Wm = 2−mpapm, so∥∥γmW 1/p′

m ‖`q′ =∥∥γmap−1

m 2−m(p−1)∥∥`q′

=∥∥ap−1

m 2−m(p−1)2m(p−1) supk≥0

arm−k2−kp−12

∥∥`q′

=∥∥supk≥0

ap−1m arm−k2−k

p−12

∥∥`q′

≤∥∥∥∥∑k≥0

ap−1m arm−k2−k

p−12

∥∥∥∥`q′

≤∞∑k=0

2−kp−12

∥∥ap−1m arm−k

∥∥`q′.


If q ≤ p =⇒ r = 0, but then q′ ≥ p′ =⇒ ‖ · ‖`q′ ≤ ‖ · ‖`p′ , so∥∥ap−1m

∥∥`q′≤∥∥ap−1

m

∥∥`p′

= ‖am‖p−1`p ≤ ‖am‖p−1

`q =∥∥2mW 1/p

m

∥∥p−1

`q.p,q 1,

which completes the proof if q ≤ p.If q > p, then r = q − p and

p− 1

q+q − pq

=1

q′,

so using Holder’s inequality:∥∥ap−1m aq−pm−k‖`q′ ≤

∥∥ap−1m

∥∥`qp−1

∥∥aq−pm−k∥∥`qq−p

= ‖am‖p−1`q ‖am−k‖

q−p`q

.p,q 1.

This finishes the proof. �

We can now strengthen this theorem to eliminate the qualitative assumptions of being simple:

Corollary 3.4. Let 1 < p <∞ and 1 ≤ q ≤ ∞, then

(13) ‖f‖Lp,q ∼p,q sup

{∣∣∣∣∫X

fg dµ

∣∣∣∣ : g ∈ Σ, ‖g‖Lp′,q′ ≤ 1

}for all f ∈ Lp,q and where Σ is the set of all finite combinations of characteristic functions ofsets of finite measure. Furthermore, if the measure space is σ-finite and if f is not a priori inLp,q but the right hand side is well defined and finite, t hen the left hand side is also finite andwe have (13).

Proof. We will only show this in the case f ∈ Lp,q and f ≥ 0, the complete proof follows thesame pattern as in the proof of Theorem 3.2.

From Holder’s inequality we can reduce to showing only the .p,q inequality. Approximate(a.e.) f by an increasing sequence {fn}n∈N of positive simple functions of finite support (thiscan be achieved if the measure space is σ-finite or if f is a priori in Lp,q, we omit the details).

Using the Monotone Convergence Theorem for Lorentz spaces 1.8:

‖f‖Lp,q = supn‖fn‖Lp,q

∼p,q supn

sup

{∫X

fng dµ : g ∈ Σ+, ‖g‖Lp′,q′ ≤ 1

}= sup

{supn

∫X

fng dµ : g ∈ Σ+, ‖g‖Lp′,q′ ≤ 1

}= sup

{∫X

fg dµ : g ∈ Σ+, ‖g‖Lp′,q′ ≤ 1

},

where we have used the (usual) monotone convergence theorem in the last line. �

This shows that there is a norm (the one defined using the dual characterization) which isequivalent to the Lp,q quasi-norm defined at the beginning when 1 < p <∞.

4. Marcinkiewicz Interpolation Theorem

In this section we will present a proof of the Marcinkiewicz interpolation theorem using thedecomposition introduced in section 2.

We will use duality from the outset, indeed, we will need the following extension of Theorem3.2:

16 GUILLERMO REY

Theorem 4.1. Let (X,µ) be a measure space and let 0 < p < ∞, the following statements areequivalent for f ∈ Lp,∞:

(1) ‖f‖Lp,∞ .p 1.(2) For every set E of finite measure there exists a subset E′ ⊆ E such that µ(E) ≥ 1

2µ(E)such that ∣∣∣∣∫

X

f1E′ dµ

∣∣∣∣ .p µ(E′)1/p′ .

Furthermore, the theorem holds without the a priori assumption of f ∈ Lp,∞ provided (X,µ) isσ-finite.

Proof. We will assume 0 < p ≤ 1, we will also assume f ≥ 0 and ‖f‖Lp,∞ = 1 without loss ofgenerality. The proof of (2) =⇒ (1) is almost identical to the one in Theorem 3.2, let us showthe other one.

Let E ⊂ X be a set of finite (and positive) measure and let α =(

2µ(E)

)1/p. Define E′ =

E \ {x : f(x) > α}, then, since we have the estimate

µ({x : f(x) > α}) ≤ µ(E)

2,

we can bound µ(E) ≥ µ(E′) ≥ 12µ(E). Now we can estimate the integral∫X

f1E′ dµ ≤(

2

µ(E)

)1/p

µ(E′)

≤ 21/pµ(E′)1/p′ .

�

The Marcinkiewicz interpolation theorem, as in complex interpolation, uses the boundednessof an operator in two “extremal spaces” to provide boundedness in the “intermediate spaces”.However, while complex interpolation of Lp spaces requires the extremal spaces to also be Lp

spaces, the Marcinkiewicz Interpolation Theorem only needs them to be weak-Lp spaces. Itactually requires a bit less, the following definitions will make the exposition easier:

Definition 4.2. Let T be an operator defined on a subset of the set of measurable functionsof a measure space (X,µ) and that takes values in the measurable functions of another measurespace (Y, ν).

• We will say that T is of strong type (p, q) if T extends to a bounded operator fromLp(X,µ) to Lq(Y, ν).

• We will say that T is of weak type (p, q) if T extends to a bounded operator from Lp(X,µ)to Lq,∞(Y, ν).

In what follows we will always assume T to be an operator defined on a set D closed underaddition, multiplication by scalars and containing Σc(X,µ): the set of finite linear combinationsof characteristic functions of sets of finite measure of (X,µ).

We will call T sublinear if∣∣T (f + g)∣∣ ≤ |T (f)|+ |T (g)|, |T (λf)| = |λ||T (f)|

for all f, g in the domain of T and λ ∈ C.We say that a sublinear operator T is of restricted weak type (p, q) if

(14) ‖Tf‖Lq,∞ . HW 1/p

for all sub-step functions f ∈ D of height H and width W .


The next result shows that, usually, an operator which is of restricted weak type on charac-teristic functions is also of restricted weak type on Σc. More precisely:

Proposition 4.3. Let 0 < p ≤ ∞, 0 < q ≤ ∞, A > 0 and T be a sublinear operator defined onΣc. Then the following statements are equivalent:

(1) T satisfies

‖Tf‖Lq,∞ . AHW 1/p

for all sub-step functions f of height H and width W in Σc.(2) For every set F ⊆ Y of finite measure there exists a subset F ′ ⊆ F with ν(F ′) ≥ 1

2ν(F )and such that for all E ⊆ X of finite measure∫

Y

|T1E |1F ′ dν . Aµ(E)1/pν(F ′)1/q′ .

Proof. That (1) =⇒ (2) is an easy consequence of Theorem 4.1 and Holder’s inequality forLorentz spaces, so we will prove the reverse implication.

First assume that f takes the form

fN =

N∑j=1

2−j1Ej

where µ(Ej) ≤ W . In this case ‖TfN‖Lq,∞ . AW 1/p (the constant being independent of N).To show this first observe that using sublinearity and Lemma 5.1 it suffices to show

(15) ‖T1Ej‖Lq,∞ . AW 1/p.

Now let F ⊆ Y be a set of finite measure. By hypothesis there exists a subset F ′ ⊆ F whichsatisfies ν(F ′) ≥ 1

2ν(F ) and ∫Y

1F ′T (1Ej ) dν . Aµ(Ej)1/pν(F ′)1/q′

≤ ν(F ′)1/q′AW 1/p.

So by Theorem 4.1 we obtain (15).Now let 0 ≤ f ≤ 1 be an arbitrary function in Σc of width W :

f =

M∑j=1

aj1Ej ,

the Ej ’s disjointly supported. If dj(x) denotes the j-th digit in the binary expansion of f(x) andwe define

fN =

N∑j=1

2−jdj =

N∑j=1

2−j1Fj ,

then we easily see that

f − fN =

M∑j=1

bj,N1Ej

18 GUILLERMO REY

with 0 ≤ bj,N ≤ 2−N . So:

‖Tf‖Lq,∞ . ‖TfN‖Lq,∞ + ‖T (f − fN )‖Lq,∞

. AW 1/p +

∥∥∥∥ M∑j=1

bj,N |T1Ej |∥∥∥∥Lq,∞

. AW 1/p + 2−N∥∥∥∥ M∑j=1

|T1Ej |∥∥∥∥Lq,∞

,

since the second term can be made arbitrarily small we conclude that

‖Tf‖Lq,∞ . AW 1/p.

By homogeneity and sublinearity we obtain the same bound (with a possibly larger constant)for arbitrary f ∈ Σc. �

To prove the Marcinkiewicz interpolation theorem we first need a weaker result:

Proposition 4.4. Let 0 < p0, p1, q0, q1 ≤ ∞, A0, A1 > 0 and suppose that T is a sublinearoperator defined on Σc which is of restricted weak type (pi, qi) with constants Ai for i = 0, 1.Then it is of restricted weak type (pθ, qθ), i.e.:

(16) ‖Tf‖Lqθ,∞ .p0,q0,p1,q1 AθHW 1/pθ ∀f ∈ Σc,

where1

pθ=

1− θp0

+θ

p1,

1

qθ=

1− θq0

+θ

q1

and

Aθ = A1−θ0 Aθ1

Proof. Theorem 4.3 reduces matters to showing that for every set F ⊆ Y of finite measure thereexists a subset F ′ ⊆ F with ν(F ′) ≥ 1

2ν(F ) and such that for all sets E ⊆ X of finite measure:∫Y

1F ′T (1E) dν .p0,q0,p1,q1 Aθµ(E)1/pθν(F ′)1/q′θ ,

but, using Theorem 4.3 and the hypotheses, we have that there exists a subset F ′ of F withν(F ′) ≥ 1

2ν(F ) and that for all sets E of finite measure∫Y

1F ′T (1E) dν .p0,q0,p1,q1 Aθµ(E)1/piν(F ′)1/q′i

for i = 0, 1. The result now follows from the trivial scalar interpolation inequality:

X ≤ Y ∧ X ≤ Z =⇒ X ≤ Y 1−θZθ

for X,Y, Z ≥ 0 and 0 ≤ θ ≤ 1, and an application of Theorem 4.3 once more. �

We need one last lemma:

Lemma 4.5. Let

Λλ(x, y) = (1− λ)x+ λy,

then

Λα(Λβ(x, y),Λγ(x, y)) = ΛΛα(β,γ)(x, y).

We are now ready to state and prove the main theorem of this section:


Theorem 4.6. Let T be a sublinear operator defined on Σ+. Suppose that 0 < pi, qi ≤ ∞ andq0 6= q1. If T satisfies

‖T1E‖Lqi,∞ . Ai‖1E‖Lpi,1 i = 0, 1,

for all E of finite measure, then for all 1 ≤ r ≤ ∞ and 0 < θ < 1 such that qθ > 1 we have

‖Tf‖Lqθ,r .p0,p1,q0,q1,r,θ Aθ‖f‖Lpθ,r

for all f in Σc.

Proof. Using Theorem 4.3 we can assume that T is of restricted weak type (pi, qi) with constantsAi.

We will suppose first that qi ≥ 1. Let f ∈ Σ+, which by homogeneity we may assume tobe normalized so that ‖f‖Lpθ,r = 1. By duality (Theorem 3.3), it will be enough to verify theestimate ∫

Y

|Tf |g dν .p0,q0,p1,q1,r,θ Aθ

for all g ∈ Σ+ with ‖g‖Lq′θ,r′ = 1. In what follows all constants will be assumed to depend on

p0, q0, p1, q1, r and θ and will be omitted.By hypothesis and Holder’s inequality we have

(17)

∫Y

|Tu|v dν . AiHH ′W 1/piW ′1/q′i ≤ AθHH ′ min

i=0,1

(W 1/piW ′1/q

′i

)for all sub-step functions u ∈ Σ+(X,µ) of height H and width W and for all sub-step functionsv ∈ Σ+(Y, ν) of height H ′ and width W ′.

By Theorem 2.2 we can decompose f and g as

f =∑m

fm and g =∑n

gn

where each fm is a sub-step function of height Hm and width 2m and where each gn is a sub-stepfunction of height H ′n and width 2n. These decompositions also satisfy∥∥Hm2m/pθ

∥∥`r∼∥∥H ′n2n/q

′θ

∥∥`r′∼ 1.

Recall that, since f and g are simple, only a finite number of fm’s and gn’s are non-zero.Using (17) and sublinearity we have∫

Y

|Tf |g dν . Aθ∑m,n

HmH′n mini=0,1

(2m/pi2n/q

′i).

Call am = Hm2m/pθ and bn = H ′n2n/q′θ , we then have to obtain the estimate

(18)∑m,n

ambn mini=0,1

(2m(1/pi−1/pθ)2n(1/qθ−1/qi)

). 1,

where ‖am‖`r = ‖bn‖`r′ = 1.After reorganizing terms we can write the third factor as

min

(m

(1

p0− 1

pθ

)+ n

(1

qθ− 1

q0

),−m

(1

pθ− 1

p1

)− n

(1

q1− 1

qθ

)),

which can be further simplified to

min

(αθ

(m− β

αn

),−α(1− θ)

(m− β

αn

)),

where α = 1p0− 1

p1and β = 1

q0− 1

q1, which is easily seen to be true. Let

ϕ(m− γn) = min (αθ (m− γn) ,−α(1− θ) (m− γn))

20 GUILLERMO REY

where γ = βα , then we have to show that∑

m,n

ambnϕ(m− γn) . 1.

We can change variables and let m 7→ m+ bγnc and reduce to showing∑m,n

am+bγncbnϕ(m+ bγnc − γn) . 1.

Observe that |bγnc − γn| ≤ 1, so we easily see that

ϕ(m+ bγnc − γn) ∼ ϕ(m),

thus we end up with ∑m

ϕ(m)∑n

am+bγncbn .∥∥∥∑n

am+bγncbn

∥∥∥`∞m

since ϕ is in `1. Applying Holder’s inequality with exponents r and r′ we reduce to showing that

‖am+bγnc‖`rn . 1

uniformly in m, but this is just the observation that the sum of arm+bγnc is bounded by a constant

depending on γ times the sum of arm+n, which is of course bounded by a constant independentof m.

Finally, we show how to drop the assumption of qi > 1. Observe that we may interpolateusing Theorem 4.4 to obtain restricted weak type (pθ, qθ) boundedness of T for 0 ≤ θ ≤ 1. Sincewe are imposing qθ > 1, we may assume that qi > 1 for at least one i ∈ {0, 1} (otherwise thetheorem is void), thus, there must be a τ ∈ (0, 1) such that 1 < qτ < qθ and such that T isof restricted weak type (pτ , qτ ). We can now interpolate using what we have just proved from(pτ , qτ ) to the (pi, qi) which has qi > 1 and use Lemma 4.5 to ensure that (pθ, qθ) is in between(pτ , qτ ) and (pi, qi).

�

Corollary 4.7 (Marcinkiewicz). Let T be a sublinear operator that is of weak type (pi, qi) withconstants Ai > 0 and such that 1 ≤ pi ≤ pi ≤ ∞ for i ∈ {0, 1}, suppose also that q0 6= q1. Then,

for 0 < θ < 1, T is of strong type (pθ, qθ) with constant .p0,q0,p1,q1,r,θ Aθ = A1−θ0 Aθ1 where

1

pθ=

1− θp0

+θ

p1and

1

qθ=

1− θq0

+θ

q1.

Proof. One just needs to recall that the Lp,q are nested and that Lp,p coincides with Lp, we leavethe easy details to the reader. �

5. Appendix

Lemma 5.1. Let X be a vector space endowed with a quasi-norm ‖ · ‖, that is, there exists aconstant C1 > 0 such that

(19) ‖f + g‖ ≤ C1(‖f‖+ ‖g‖).Then, for any sequence {fn}n∈N of elements of X verifying

(20) ‖fj‖ . AC−j2

for some A > 0 and C2 > 1, we have

(21)

∥∥∥∥ N∑j=1

fj

∥∥∥∥ ≤ AC4


where C3 does not depend on N or A.

Proof. Choose κ ∈ N so large that C1C−κ2 < 1 and write N as

N = κd+ r

where d ∈ N and 0 ≤ r < κ. Observe that κ does not depend on N .We can now split the sum as follows

N∑j=1

fj =

κd∑j=1

fj +

r∑j=1

fkd+j .

The norm of the second sum is trivially bounded by A times some constant depending on C1

and C2 (since r < κ,) so we can center our efforts on the first sum. To this end we factorize thesum as

κd∑j=1

fj =

d−1∑m=0

κ∑j=1

fκm+j .

Now observe that, by induction, we have the following estimate:∥∥∥∥ n∑j=1

gj

∥∥∥∥ ≤ n∑j=1

Cj1‖gj‖

for any sequence {gj} of elements of X. Using this we can estimate the first sum as follows:∥∥∥∥ κd∑j=1

fj +

r∑j=1

fkd+j

∥∥∥∥ ≤ d−1∑m=0

Cm+11

∥∥∥∥ κ∑j=1

fκm+j

∥∥∥∥≤ A

d−1∑m=0

Cm+11

κ∑j=1

Cj1C−(κm+j)2

= A

d−1∑m=0

(C1C

−κ2

)m κ∑j=1

Cj+11 C−j2

≤ A∞∑m=0

(C1C

−κ2

)m · κ∑j=1

Cj+11 C−j2

.C1,C2 A

κ∑j=1

Cj+11 C−j2

.C1,C2A.

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References

[1] G. Hardy and J. Littlewood. A maximal theorem with function-theoretic applications. Acta Mathematica,54(1):81–116, 1930.

[2] R. A. Hunt. On L(p, q) spaces. Enseignement Math. (2), 12:249–276, 1966.[3] M. Keel and T. Tao. Endpoint Strichartz estimates. Amer. J. Math., 120(5):955–980, 1998.

[4] R. O’Neil. Convolution operators and L(p, q) spaces. Duke Mathematical Journal, 30:129–142, 1963.[5] E. Stein and G. Weiss. Introduction to Fourier Analysis on Euclidean Spaces. Princeton University Press,

1990.

[6] T. Tao. Lecture notes 1 for 245C: Interpolation of Lp spaces.

[7] N. Wiener. The ergodic theorem. Duke Mathematical Journal, 5(1):1–18, 1939.

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