The Off-Diagonal Marcinkiewicz Interpolation Theorem

1.4 Lorentz Spaces

1.4.4 The Off-Diagonal Marcinkiewicz Interpolation Theorem

(1.4.16) lies in the spaceL^q(0,∞)with respect to the measuret^q/p−1dt. This fol- lows from the inequality

f

L^p,q = Z _∞

0

t

q p

Z _∞ t/2

s

q0 p0−1

g^∗(s)^q0−1ds s

q

dt t

¹_q

≤C₁(p,q) _{Z ∞}

0

(t

1 p0

g^∗(t))^q0dt t

¹_q

=C₁(p,q) g

q⁰/q L^p0,q0 <∞,

which is a consequence of Hardy’s second inequality in Exercise 1.2.8 withb=q/p.

Using (1.4.15), we conclude that Z ∞

0

f^∗(t)g^∗(t)dt≤ T

f

_Lp,q ≤C₁(p,q) T

g

q⁰/q

L^p0,q0. (1.4.17) On the other hand, we have

Z ∞

0

f^∗(t)g^∗(t)dt ≥ Z ∞

0 Z t

t/2

s

q0 p0−1

g^∗(s)^q0−1ds s g^∗(t)dt

≥ Z ∞

0

g^∗(t)^q0 Z t

t/2

s

q0 p0−1ds

s dt

=C₂(p,q) g

q⁰ L^p0,q0.

(1.4.18)

Combining (1.4.17) and (1.4.18), we obtain g

_Lp⁰,q0≤C(p,q) T

. This estimate is valid only when we have a priori knowledge that

g

_Lp⁰,q0 <∞. Suitably modifying the preceding proof and using that

g _Lp⁰,q0

(K_N)<∞, we obtain that g

_Lp⁰,q0

(K_N)≤ C(p,q)

T

for allN=1,2, . . .. LettingN→∞, we obtain the required conclusion.

(vii) For a complete characterization of this space, we refer to the article of Cwikel [62].

(viii) The dual ofL^∞can be identified with the set of all bounded finitely additive

set functions. See Dunford and Schwartz [77].

Remark 1.4.18.Some parts of Theorem 1.4.17 are false ifXis atomic. For instance, the dual of`^p(Z)containsl^∞when 0<p<1 and thus it is not{0}.

for someK>0,λ∈C, and all functions f,gin the domain ofT. To avoid triviali- ties, we assume thatK≥1.

Theorem 1.4.19.Let0<r≤∞,0<p₀6=p₁≤∞, and0<q₀6=q₁≤∞and let (X,µ)and (Y,ν)be two measure spaces. Let T be either a quasilinear operator defined on L^p0(X) +L^p1(X)and taking values in the set of measurable functions on Y or a linear operator defined on the set of simple functions on X and taking values as before. Assume that for some M₀,M₁<∞the following (restricted) weak type estimates hold:

T(χ_A)

_Lq0,∞ ≤M₀µ(A)^1/p0, (1.4.19) T(χ_A)

_Lq1,∞ ≤M₁µ(A)^1/p1, (1.4.20) for all measurable subsets A of X withµ(A)<∞. Fix0<θ<1and let

1

p=1−θ p0

+ θ p1

and 1

q=1−θ q0

+θ q1

. (1.4.21)

Then there exists a constant M, which depends on K, p0, p1, q0, q1, M0, M1, r, and θ, such that for all functions f in the domain of T and in L^p,r(X)we have

T(f)

L^q,r ≤M f

L^p,r. (1.4.22)

We note thatL^p,∞⊆L^p0+L^p1 (Exercise 1.1.10(c)), and thusT is well defined on L^p,rfor allr≤∞. Ifr<∞andTis linear and defined on the set of simple functions onX, thenT has a unique extension that satisfies (1.4.22) for all f inL^p,r(X), since simple functions are dense in this space.

Before we give the proof of Theorem 1.4.19, we state and prove a lemma that is interesting on its own.

Lemma 1.4.20.Let0<p<∞and0<q≤∞. Let T be either a quasilinear oper- ator defined on L^p(X,µ)and taking values in the set of measurable functions of a measure space(Y,ν), or a linear operator initially defined on the space of simple functions on X and taking values as before. Suppose that there exists a constant L>0such that for all A⊆X of finite measure we have

T(χ_A)

L^q,∞≤Lµ(A)^1/p. (1.4.23)

Fixα0<q with0<α0≤ _{log 2K}^{log 2}. Then for all0<α≤α0 there exists a constant C(p,q,K,α)>0 (depending only on the parameters indicated) such that for all functions f in L^p,α(X)that lie in the domain of T , we have the estimate

T(f)

L^q,∞≤C(p,q,K,α)L f

L^p,α. (1.4.24)

Lemma 1.4.20 is saying that if a quasilinear operator satisfies aL^p,1→L^q,∞es- timate uniformly on all characteristic functions, then it must map a Lorentz space L^p,αtoL^q,∞for someα<1.

Proof. It suffices to prove Lemma 1.4.20 for f ≥0, since we can express a general function f as

f = (f₁−f₂) +i(f₃−f₄), wheref_j≥0, and use quasilinearity.

It follows from the Aoki–Rolewicz theorem (Exercise 1.4.6) that for allf₁, . . . ,f_m we have the pointwise inequality

|T(f₁+· · ·+f_m)| ≤4 m

∑

j=1

|T(f_j)|^α1 _α¹

1

≤4 _m

∑

j=1

|T(f_j)|^α _α¹

,

(1.4.25)

where 0<α≤α₁andα₁satisfies the equation (2K)^α1 =2.

The second inequality in (1.4.25) is a simple consequence of the fact thatα ≤α₁. Fixα0>0 with

α₀≤α₁= log 2

log 2K and α₀<q.

This ensures that the quasinormed spaceL^q/α,∞is normable whenα≤α0. In fact, Exercise 1.1.12 gives that the spaceL^s,∞is normable as long ass>1 and for some equivalent norm









f









L^s,∞ we have

f L^s,∞ ≤









f









L^s,∞ ≤ s s−1

f

L^s,∞. Next we claim that for any f ≥0 we have

T(fχA)

L^q,∞≤C(q,α)Lµ(A)^1/p f

L^∞. (1.4.26)

To prove (1.4.26) first observe that multiplying by a suitable constant, we may as- sume that f≤1. Write

f(x) =

∞

∑

j=1

d_j(x)2^−j in binary expansion, wheredj(x) =0 or 1. Let

B_j={x∈A: d_j(x) =1}. ThenB_j⊆Aand the function fχAcan be written as the sum

∞

∑

j=1

2^−jχBj.

We use (1.4.25) once and (1.4.3) twice in the following argument. We have T(fχA)

_Lq,∞ ≤4

^∞

∑

j=1

2^−jα|T(χ_Bj)|^α_α¹ L^q,∞

=4

∞

∑

j=1

2^−jα|T(χ_Bj)|^α

1 α

L^q/α,∞

≤4



















∞

∑

j=1

2^−jα|T(χ_Bj)|^α

















1 α

L^q/α,∞

≤4 ∞

∑

j=1

2^−jα

















|T(χ_Bj)|^α















L^q/α,∞

_α¹

≤4 q q−α

_α¹ ∞

∑

j=1

2^−jα

|T(χ_Bj)|^α L^q/α,∞

_α¹

=4 q q−α

_α¹ ∞

∑

j=1

2^−jα

T(χ_Bj)

α L^q,∞

_α¹

≤4 q q−α

_α¹ L

∞ j=1

∑

2^−jαµ(B_j)^α/p _α¹

≤2 q q−α

_α¹

(1−2^−α)^−α1Lµ(A)^1/p, sinceB_j⊆A. This establishes (1.4.26) with

C(q,α) =2 q q−α

_α¹

(1−2^−α)^−α1. Now write the function f as

f=

∞ n=−∞

∑

fχAn, whereAnare measurable sets defined by

A_n={x∈X: f^∗(2ⁿ⁺¹)<|f(x)| ≤ f^∗(2ⁿ)}. (1.4.27) Observe that

µ(A_n) =

{t∈R: f^∗(2ⁿ⁺¹)<f^∗(t)≤f^∗(2ⁿ)}

=

[2ⁿ,2ⁿ⁺¹]

=2ⁿ,

since f andf^∗are equidistributed. Next we have

T(f)

L^q,∞ ≤4

^∞

n=−∞

∑

|T(fχAn)|^α_α¹ _Lq,∞

=4

∞ n=−∞

∑

|T(fχ_An)|^α

1 α

L^q/α,∞

≤4



















∞ n=−∞

∑

|T(fχ_An)|^α

















1 α

L^q/α,∞

≤4 ∞

n=−∞

∑

















|T(fχAn)|^α















_Lq/α,∞

_α¹

≤4 q q−α

_α¹ ∞ n=−∞

∑

|T(fχ_An)|^α L^q/α,∞

_α¹

≤4 q q−α

_α¹ ∞ n=−∞

∑

T(fχ_An)

α L^q,∞

_α¹

≤8 q q−α

_α²

(1−2^−α)^−α1L ∞

n=−∞

∑

f^∗(2ⁿ)^α2^nα/p _α¹

≤8 q q−α

_α²

(1−2^−α)^−α1(log 2)^α1L f

L^p,α.

Taking into account the splittingf =f₁−f₂+i f₃−i f₄, wheref_j≥0, we conclude the proof of the lemma with constant

C(p,q,K,α) =C_pK² q q−α

2/α

(log 2)^α1(1−2^−α)^−α1. (1.4.28) Recall that we have been assuming thatα<min _{log 2K}^{log 2} ,q

throughout.

We now continue with the proof of Theorem 1.4.19.

Proof. We assume that p₀<p₁, since if p₀>p₁we may simply reverse the roles ofp₀andp₁. We first consider the casep₁<∞. Lemma 1.4.20 implies that

T(f)

L^q0,∞ ≤A0

f

L^p0,m, T(f)

_Lq1,∞ ≤A₁ f

_Lp1,m, (1.4.29)

wherem=¹

2min q₀,q₁, ^{log 2}

log 2K

,A₀=C(p₀,q₀,K,m)M₀,A₁=C(p₁,q₁,K,m)M₁, andC(p,q,K,α)is as in (1.4.28).

Fix a function f inL^p,r. Split f =f^t+f_tas follows:

f^t(x) =

(f(x) if|f(x)|> f^∗(t^γ), 0 if|f(x)| ≤ f^∗(t^γ),

f_t(x) =

(0 if|f(x)|>f^∗(t^γ), f(x) if|f(x)| ≤ f^∗(t^γ), whereγis the following nonzero real number:

γ=

1 q0−¹_q

1 p₀−¹_p =

1 q−_q¹

1

1 p−_p¹

1

.

Next, observe that the following inequalities are valid:

(f^t)^∗(s) ≤

(f^∗(s) if 0<s<t^γ, 0 ifs≥t^γ, (f_t)^∗(s) ≤

(f^∗(t^γ) if 0<s<t^γ, f^∗(s) ifs≥t^γ.

It follows from these inequalities that f^tlies inL^p0,mandf_tlies inL^p1,mfor allt>0.

The sublinearity of the operatorT and (1.4.9) imply T(f)

L^q,r =

t^1qT(f)^∗(t) L^r(^dt_t)

≤K

t^1q(|T(ft)|+|T(f^t)|)^∗(t) L^r(^dt_t)

≤K

t^1qT(f_t)^∗(₂^t) +t^1qT(f^t)^∗(₂^t) _Lr(^dt_t )

≤Ka_r

t^1qT(f_t)^∗(^t₂)

L^r(^dt_t)+

t^1qT(f^t)^∗(^t₂) L^r(^dt_t)

≤Ka_r t

1 q−_q¹

0t

1

q0T(f_t)^∗(₂^t) L^r(^dt_t)

+ t

1 q−¹

q1t

1

q1T(f^t)^∗(^t₂) _Lr(dt

t)

,

(1.4.30)

where

a_r=

(1 whenr≥1, 2^(1−r)/r whenr≤1.

It follows from (1.4.29) that t

1

q0T(f^t)^∗(^t₂) ≤2^q10sup

s>0

s

1

q0T(f^t)^∗(s)≤2^q10A0

f^t

L^p0,m, (1.4.31) t^q11T(f_t)^∗(^t₂) ≤2^q11sup

s>0

s^q11T(f_t)^∗(s)≤2^q11A₁ f_t

L^p1,m, (1.4.32) for allt>0. Now use (1.4.31) and (1.4.32) to estimate (1.4.30) by

Ka_r2

1 q0A₀

t

1 q−_q¹

0

f^t L^p0,m

L^r(^dt_t)

+Ka_r2^q11A₁

t^1q−q11 f_t

L^p1,m

L^r(^dt_t)

,

which is the same as Ka_r2^q10A₀

t^{−γ(p10−1p)} f^t

L^p0,m

L^r(^dt_t )

+Ka_r2

1 q1A₁

t^γ(

1 p−_p¹

1) f_t

L^p1,m

L^r(^dt_t )

.

(1.4.33)

Next, we change variablesu=t^γin the first term of (1.4.33) to obtain Ka_r2

1 q0A₀

t^−γ(

1 p0−¹_p)

f^t _Lp0,m

_Lr(dt

t)

≤Ka_r2^q10A0

|γ|^1/r

u⁻⁽

1

p0−¹_p)Z _u 0

f^∗(s)^ms

m p0 ds

s ¹_m

L^r(^du_u)

≤Ka_r 2

1 q0A0

m|γ|^1/r r r(_p¹

0−¹_p) _{Z ∞}

0

(s^p10 f^∗(s))^rs^−r(

1 p0−¹_p)ds

s ¹_r

=Ka_r 2

1 q0A0

m|γ|^1/r 1

1 p₀−¹_p

f

L^p,r,

where the last inequality is a consequence of Hardy’s first inequality in Exercise 1.2.8 withp=r/m≥1 andb= (1/p0−1/p)r.

Similarly, change variablesu=t^γin the second term of (1.4.33) to obtain Ka_r2

1 q1A₁

t^γ(

1 p−_p¹

1) f_t

L^p1,m

L^r(^dt_t)

≤Ka_r2

1 q1A₁

|γ|^1/r u

1 p−_p¹

1

Z u 0

f^∗(u)^ms^pm1ds s +

Z ∞

u

f^∗(s)^ms^pm1 ds s

_m¹ L^r(^du_u)

≤Ka²_r2^1−mm 2

1 q1A₁

|γ|^1/r

(p₁ m u

1 p−_p¹

1 f^∗(u)u

1 p1

L^r(^du_u)

+ u

1 p−_p¹

1

Z _∞ u

f^∗(s)^ms

m p1 ds

s _m¹

_Lr(du

u)

)

≤Ka²_r2^1−mm 2

1 q1A₁

|γ|^1/r

(p₁ m f

_Lp,r+ r mr(¹_p−_p¹

1)

u^r(

1 p−¹

p1)

f^∗(u)^ru

r p1

_Lr(du

u)

)

=Ka²_r2^1−mm 2

1 q1A₁

|γ|^1/r

(p₁

m + 1

m(¹_p−_p¹

1) )

f

_Lp,r,

where the last inequality above is Hardy’s second inequality in Exercise 1.2.8 with p=r/m≥1 andb= (1/p−1/p₁)r.

We have now shown that

T(f)

_Lq,r ≤M f

_Lp,r

with constant M=Kar

2

1 q0 +2^q11 m|γ|^1/r

A₀

1

p0−¹_p +a_r2^1−mm A1

p1+ 1

1 p−_p¹

1

!

. (1.4.34)

We have been tacitly assuming thatr<∞. The remaining case is a simple con- sequence of the result just proved by letting r→∞, in which casea_r →1 and

|γ|^1/r→1.

We now turn to the casep₁=∞. Hypotheses (1.4.19) and (1.4.20) together with Exercise 1.1.16 imply that

T(χ_A)

_Lq,∞≤M₀^1−θM₁^θµ(A)^1/p

for all 0≤θ ≤1. We selectλ ∈(0,1)such that the indices p=p_λ andq=q_λ defined by (1.4.21) whenθ=λ satisfyp0<p<p_λ <∞andq_λ is strictly between q0andq1. Then apply the casep1<∞just proved withp0,q0as before andp1=p_λ andq1=q_λ. The result follows withMas in (1.4.34) except thatp1is replaced by

p_λandq₁byq_λ.

Corollary 1.4.21.Let T be as in the statement of Theorem 1.4.19 and let0<p₀6=

p₁≤∞and0<q₀6=q₁≤∞. If T maps L^p0to L^q0,∞and L^p1to L^q1,∞, and for some 0<θ<1we have

1

p=1−θ p₀ + θ

p₁, 1

q=1−θ q₀ +θ

q₁, and p≤q, then T satisfies the strong type estimate

T(f) L^q ≤C

f

L^p for all functions f in the domain of T . Moreover, if T is linear, then it has a bounded extension from L^p(X,µ)to L^q(Y,ν).

Proof. Taker=qin the previous theorem.

Definition 1.4.22.Let 0<p,q≤∞. We call an operatorT ofrestricted weak type (p,q)if it satisfies

T(χ_A)

_Lq,∞≤Cµ(A)^1/p

for all subsetsAof a measure space(X,µ)with finite measure. Using this terminol- ogy, Corollary 1.4.21 says that if a quasilinear operatorT is of restricted weak types (p₀,q₀)and(p₁,q1)for some p06=p1andq06=q1, then it is bounded fromL^pto L^qwhen p≤q.

We now give examples to indicate why the assumptions p₀6=p₁andq₀6=q₁ cannot be dropped in Theorem 1.4.19.

Example 1.4.23.LetX=Y =Rand T(f)(x) =|x|^−1/2

Z ₁ 0

f(t)dt.

Thenα|{x: |T(χ_A)(x)|>α}|^1/2=2^1/2|A∩[0,1]|and thusT is of restricted weak types(1,2)and(3,2). But observe that T does not mapL²=L^2,2 toL^q,2. Thus Theorem 1.4.19 fails if the assumptionq₀6=q₁is dropped. The dual operator

S(f)(x) =χ_[0,1](x) Z +∞

−∞ f(t)|t|^−1/2dt

satisfiesα|{x: |S(χ_A)(x)|>α}|^1/q≤c|A|^1/2whenq=1 or 3, and thus it furnishes an example of an operator of restricted weak types(2,1)and(2,3)that is notL² bounded. Thus Theorem 1.4.19 fails if the assumptionp₀6=p₁is dropped.

As an application of Theorem 1.4.19, we give the following strengthening of Theorem 1.2.13.

Theorem 1.4.24.(Young’s inequality for weak type spaces) Let G be a locally com- pact group with left Haar measureλ that satisfies (1.2.12) for all measurable sub- sets A of G. Let1<p,q,r<∞satisfy

1

q+1=1 p+1

r. (1.4.35)

Then there exists a constant B_pqr>0such that for all f in L^p(G)and g in L^r,∞(G) we have

f∗g

L^q(G)≤B_pqr g

L^r,∞(G)

f

L^p(G). (1.4.36) Proof. We fix 1<p,q<∞. Sincepandqrange in an open interval, we can find p₀<p<p₁,q₀<q<q₁, and 0<θ<1 such that (1.4.21) and (1.4.35) hold. Let T(f) = f∗g, defined for all functions f onG. By Theorem 1.2.13,T extends to a bounded operator fromL^p0 toL^p1,∞ and fromL^q0 toL^q1,∞. It follows from the Marcinkiewicz interpolation theorem thatT extends to a bounded operator from

L^p(G)toL^q(G).

Exercises

1.4.1.(a) Letgbe a nonnegative simple function on(X,µ)and letAbe a measur- able subset ofX. Prove that

Z

A

g dµ≤ Z µ(A)

0

g^∗(t)dt.

(b) (G. H. Hardy and J. E. Littlewood) Forf andgmeasurable on(X,µ), prove that

Z

X

|f(x)g(x)|dµ(x)≤ Z ∞

0

f^∗(t)g^∗(t)dt.

Compare this result to the classical Hardy–Littlewood result asserting that ifa_j,b_j>

0, the sum∑ja_jb_jis greatest when botha_jandb_jare rearranged in decreasing order (for this see Hardy, Littlewood, and P´olya [122, p. 261]).

1.4.2.Prove that if f ∈L^q0,∞∩L^q1,∞for some 0<q₀<q₁≤∞, then f ∈L^q,s for all 0<s≤∞andq₀<q<q₁.

1.4.3.(Hunt [134]) Given 0<p,q<∞, fix anr=r(p,q)>0 such thatr≤1,r≤q andr<p. Fort≤µ(X)define

f^∗∗(t) = sup

µ(E)≥t

1 µ(E)

Z

E

|f|^rdµ 1/r

,

while fort>µ(X)(ifµ(X)<∞) let f^∗∗(t) =

1 t Z

X

|f|^rdµ 1/r

.

Also define











f









_Lp,q= _{Z ∞}

0

t^1p f^∗∗(t)qdt t

¹_q . (The function f^∗∗and the functionalf →









f









L^p,q depend onr.) (a) Prove that the inequality

(((f+g)^∗∗)(t))^r≤(f^∗∗(t))^r+ (g^∗∗(t))^r is valid for allt≥0. Sincer≤q, conclude that the functional

f →









f











r L^p,q

is subadditive and hence it is a norm whenr=1 (this is possible only ifp>1).

(b) Show that for all f we have

f _Lp,q≤









f









_Lp,q ≤ p

p−r 1/r

f

_Lp,q.

(c) In conjunction with Exercise 1.1.12, conclude thatL^p,qis metrizable whenever 0<p,q≤∞and also normable when 1<p<∞and 1≤q≤∞.

1.4.4.(a) Show that on aσ-finite measure space(X,µ)the set of countable linear combinations of simple functions is dense inL^p,∞(X).

(b) Prove that simple functions are not dense inL^p,∞(R)for any 0<p≤∞.

Hint: Part (b): Show that the function h(x) =x^−1/pχx>0 cannot be approxi- mated by a sequence of simple functions L^p,∞. To see this, partition the inter-

val(0,∞)into small subintervals of lengthε>0 and let f_ε be the step function

∑^[1/ε]_−[1/ε]f(kε)χ[kε,(k+1)ε](x). Show that for somec>0 we havekfε−fk_L^p,∞≥c.

1.4.5.Let(X,µ)be a nonatomic measure space. Prove the following facts:

(a) IfA0⊆A1⊆X, 0<µ(A₁)<∞, andµ(A₀)≤t≤µ(A₁), then there exists an E_t⊆A₁withµ(E_t) =t.

(b) Givenϕ(t)continuous and decreasing on[0,∞), there exists a measurable func- tion f onXwith f^∗(t) =ϕ(t)for allt>0.

(c) GivenA⊆Xwith 0<µ(A)<∞andgan integrable function onX, there exists a subsetAeofX withµ(A) =e µ(A)such that

Z

Ae

g dµ= Z µ(A)

0

g^∗(s)ds.

(d) Given f andgmeasurable functions onX, we have sup

h:d_h=d_f

Z

X

h g dµ

= Z _∞

0

f^∗(s)g^∗(s)ds, where the supremum is taken over allhequidistributed with f.

Hint: Part (a): Reduce matters to the situation in which A₀= /0. Consider first the case that for all A⊆X there exists a subset B of X satisfying ₁₀¹ µ(A)≤ µ(B)≤ ₁₀⁹ µ(A). Then we can find subsets of A₁ of measure in any arbitrar- ily small interval, and by continuity the required conclusion follows. Next con- sider the case in which there is a subset A₁ of X such that every B⊆A₁ satis- fiesµ(B)<₁₀¹ µ(A₁)orµ(B)>₁₀⁹ µ(A₁). Without loss of generality, normalizeµ so that µ(A₁) =1. Let µ1=sup{µ(C): C⊆A₁,µ(C)<₁₀¹} and pickB₁⊆A₁ such that ¹₂µ1≤µ(B₁)≤µ1. SetA₂=A₁\B₁and define µ2=sup{µ(C): C⊆ A₂,µ(C)<₁₀¹}. Continue in this way and define setsA₁⊇A₂⊇A₃⊇ · · ·and num- bers₁₀¹ ≥µ1≥µ2≥µ3≥ · · ·. IfC⊆A_n+1withµ(C∪A_n+1)<₁₀¹, thenC∪B_n⊆A_n withµ(C∪B_n)<¹₅<₁₀⁹, and hence by assumption we must haveµ(C∪B_n)<₁₀¹. Conclude that µn+1≤ ¹₂µn and that µ(A_n)≥ ⁴₅ for alln=1,2, . . .. Then the set T_∞

n=1A_n must be an atom. Part (b): First show that whend is a simple right con- tinuous decreasing function on[0,∞)there exists a measurable f on X such that f^∗=d. For general continuous functions, use approximation. Part (c): Lett=µ(A) and defineA₁={x: |g(x)|>g^∗(t)}andA₂={x: |g(x)| ≥g^∗(t)}. ThenA₁⊆A₂ andµ(A₁)≤t≤µ(A₂). Pick Aesuch thatA₁⊆Ae⊆A₂ andµ(A) =e t=µ(A)by part (a). Then ^R

Aeg dµ=^R_Xgχ

Aedµ=^R₀^∞(gχ

Ae)^∗ds=^R₀^µ(eA)g^∗(s)ds. Part (d): Let f =∑^N_j=1a_jχ_Aj wherea₁>a₂>· · ·>a_N >0 and theA_j are pairwise disjoint.

Write f as∑^N_j=1b_jχ_Bj, whereb_j= (a_j−aj+1)andB_j=A1∪ · · · ∪A_j. PickBe_jas in part (c). ThenBf₁⊆ · · · ⊆Bf_N and the function f₁=∑^N_j=1b_jχ

Bf_j has the same distri- bution function as f. It follows from part (c) that^R_X f₁g dµ=^R₀^∞f^∗(s)g^∗(s)ds.The case of a general function f follows from that in which f is simple using Exercise 1.4.1 and approximation.

1.4.6.(Aoki [5]/ Rolewicz [224]) Let ·

be a nonnegative functional on a vector spaceXthat satisfies

x+y ≤K

x +

y

for allxandyinX. (To avoid trivialities, assume thatK≥1.) Then forαdefined by the equation

(2K)^α=2 (α≤1), we have

x1+· · ·+xn

α≤4( x1

α+· · ·+ xn

α) for alln=1,2, . . .and allx1,x2,. . .,xninX.

Hint:Quasilinearity implies that kx₁+· · ·+x_nk ≤max1≤j≤n[(2K)^jkx_jk]for all

x₁, . . . ,x_ninX(use thatK≥1). DefineH: X→Rby settingH(0) =0 andH(x) =

2^j/αif 2^j−1<kxk^α≤2^j. Thenkxk ≤H(x)≤2^1/αkxkfor allx∈X. Prove by induc- tion thatkx₁+· · ·+x_nk^α≤2(H(x₁)^α+· · ·+H(x_n)^α).Suppose that this statement is true whenn=m. To show its validity forn=m+1, without loss of generality assume thatkx₁k ≥ kx₂k ≥ · · · ≥ kx_m+1k. ThenH(x₁)≥H(x₂)≥ · · · ≥H(x_m+1).

Assume that all theH(x_j)’s are distinct. Then sinceH(x_j)^α are distinct powers of 2, they must satisfyH(x_j)^α≤2^−j+1H(x₁)^α. Then

kx₁+· · ·+x_m+1k^α ≤

1≤j≤m+1max (2K)^jkx_jkα

≤

1≤j≤m+1max (2K)^jH(x_j)α

≤

1≤j≤m+1max (2K)^j2^1/α2^−j/αH(x₁)α

=2H(x₁)^α

≤2(H(x₁)^α+· · ·+H(x_m+1)^α).

We now consider the case thatH(x_j) =H(x_j+1)for some 1≤j≤m. Then for some integerrwe must have 2^r−1<kx_j+1k^α≤ kx_jk^α≤2^r andH(x_j) =2^r/α. Next note that

kx_j+x_j+1k^α≤K^α(kx_jk+kx_j+1k)^α≤K^α(2 2^r)^α≤2^r+1. This implies

H(x_j+xj+1)^α≤2^r+1=2^r+2^r=H(x_j)^α+H(xj+1)^α.

Now apply the inductive hypothesis tox₁, . . . ,xj−1,x_j+xj+1,x_j+1, . . . ,x_m and use the previous inequality to obtain the required conclusion.

1.4.7.(Stein and Weiss [264]) Let(X,µ)and(Y,ν)be measure spaces. LetZbe a Banach space of complex-valued measurable functions onY. Assume thatZ is closed under absolute values and satisfies

f _Z=

|f|

_Z. Suppose thatTis a linear operator defined on the space of measurable functions on(X,µ)and taking values in Z. Suppose that for some constantA>0 we have the restricted weak type estimate

T(χ_E)

Z≤Aµ(E)^1/p

for allE measurable subsets ofX and some 0<p<∞. Then there is a constant C(p)>0 such that

T(f)

_Z≤C(p)A f

_Lp,1

for all f in the domain ofT.

Hint:Let f =∑^N_j=1a_jχ_Ej ≥0, wherea1>a2>· · ·>a_N >0, µ(E_j)<∞pair- wise disjoint. Let F_j=E1∪ · · · ∪E_j, B0=0, and B_j =µ(F_j) for j≥1. Write

f =∑^N_j=1(a_j−aj+1)χ_Fj, whereaN+1=0. Then T(f)

Z =

|T(f)|

Z

≤

N

∑

j=1

(a_j−aj+1) T(χ_Fj)

Z

≤A

N

∑

j=1

(a_j−a_j+1)(µ(F_j))^1/p

=A

N−1

∑

j=0

a_j+1(B^1/p_j+1−B^1/p_j )

=p⁻¹A f

L^p,1,

where the penultimate equality follows summing by parts; see Appendix F.

1.4.8.Let 0<p<∞and 0<q₁<q₂≤∞. Letα,β,q>0.

(a) Show that the functionf(t) =t^−α(logt⁻¹)^−βχ_(0,1)(t)lies inL^p,q(R)if and only if eitherp>1/α or bothp=1/αandq>1/β.

(b) Show that the function t^−1p(logt⁻¹)⁻

1

q1χ_(0,1)(t) lies in L^p,q2(R) but not in L^p,q1(R).

(c) OnRⁿconstruct examples to show thatL^p,q1 $L^p,q2.

(d) On a general nonatomic measure space(X,µ)prove that theredoes notexist a constantC(p,q₁,q₂)>0 such that for allf inL^p,q2(X)the following is valid:

f

_Lp,q1 ≤C(p,q₁,q₂) f

_Lp,q2.

1.4.9.(Stein and Weiss [263]) LetL^p(ω)denote the weightedL^pspace with mea- sureω(x)dx. LetT be a sublinear operator that maps

T: L^p0(ω₀)→L^q0,∞(w), T: L^p1(ω₁)→L^q1,∞(w),

for somep₀6=p₁, where 0<p₀,p₁,q₀,q₁≤∞andω₀,ω₁,ωare positive functions.

Suppose that

1

p_θ =1−θ p₀ + θ

p₁, 1

q_θ =1−θ q₀ + θ

q₁.

ThenT maps

L^pθ ω

1−θ p0 p_θ

0 ω

pθ1p_θ 1

→L^qθ,pθ(ω). Hint:Define

L(f) = (ω₁/ω₀)^{p1−p1 0} f and observe that for eachθ∈[0,1],Lmaps

L^pθ ω

1−θ p0 p_θ

0 ω

pθ1p_θ 1

→L^pθ

(ω₀^p1ω₁^−p0)^{p1−p1 0}

isometrically. Then apply the classical Marcinkiewicz interpolation theorem to the sublinear operatorT◦L⁻¹,and the required conclusion easily follows.

1.4.10.(Kalton [147]/Stein and Weiss [266]) Letλnbe a sequence of positive num- bers with∑nλn≤1 and∑nλnlog(¹

λn) =K<∞.

(a) Letf_nbe a sequence of complex-valued functions inL^1,∞(X)such that f_n

L^1,∞≤ 1 uniformly inn. Prove that∑_nλ_nf_nlies in L^1,∞(X)with norm at most 2(K+2).

(This property is referred to as thelogconvexityofL^1,∞.)

(b) Let Tn be a sequence of sublinear operators that mapL¹(X)to L^1,∞(Y) with norms

Tn

L¹→L^1,∞ ≤Buniformly in n. Use part (a) to prove that ∑nλnTn maps L¹(X)toL^1,∞(Y)with norm at most 2B(K+2).

(c) Givenδ>0 pick 0<ε<δ and use the simple estimate µ {

∞ n=1

∑

2^−δnf_n>α}

≤

∞ n=1

∑

µ {2^−δnf_n>(2^ε−1)2^−εnα}

to obtain a simple proof of the statements in part (a) and (b) whenλn=2^−δn,n= 1,2, . . ., and zero otherwise.

Hint:Part (a): For fixedα >0, write f_n=u_n+v_n+w_n, whereu_n= f_nχ_|fn|≤^α

2, v_n=f_nχ_|fn|> ^α

2λn, andw_n= f_nχ^α

2<|f_n|≤_2λn^α . Letu=∑nλnu_n,v=∑nλnv_n, andw=

∑_nλ_nw_n. Clearly|u| ≤α/2. Also{v6=0} ⊆^S_n{|f_n|> ^α

2λn}; henceµ({v6=0})≤_α². Finally,

Z

X

|w|dµ ≤

∑

n

λn Z

X

|f_n|χα 2<|f_n|≤ ^α

2λndµ

≤

∑

n

λ_n

_{Z α/(2λn)}

α/2

d_fn(β)dβ+ Z α/2

0

d_fn(α/2)dβ

≤K+1.

Usingµ({|u+v+w|>α})≤µ({|u|>α/2}) +µ({|v| 6=0}) +µ({|w|>α/2}), deduce the conclusion.

1.4.11.Construct a sequence of functions f_k inL^1,∞(Rⁿ)and a function f ∈L^1,∞

such that f_k−f

L^∞→0 but f_k

L^1,∞→∞ask→∞.

1.4.12.(a) Suppose thatXis a quasi-Banach space and letX^∗be its dual (which is always a Banach space). Prove that for allT∈X^∗we have

T

_X∗= sup

x∈X kxk_X≤1

|T(x)|.

(b) Now suppose thatX is a Banach space. Use the Hahn–Banach theorem to prove that for everyx∈Xwe have

x

X= sup

T∈X^∗ kTk_X∗≤1

|T(x)|.

Observe that this result may fail for quasi-Banach spaces. For example, ifX=L^1,∞, every linear functional onX^∗vanishes on the set of simple functions.

(c) TakeX =L^p,1 andX^∗=L^p0,∞. Then for 1<p<∞both of these spaces are normable. Conclude that

f

_Lp,1 ≈ sup

kgkLp0,∞≤1

Z

X

f g dµ ,

f

L^p,∞ ≈ sup

kgkLp0,1≤1

Z

X

f g dµ .

1.4.13.Let 0<p,q<∞. Prove that any function inL^p,q(X,µ)can be written as f =

+∞

n=−∞

∑

c_nf_n,

wheref_nis a function bounded by 2^−n/p, supported on a set of measure 2ⁿ, and the sequence{c_k}_klies in`^qand satisfies

2^−1p(log 2)^1q {c_k}_k

_`q≤ f

_Lp,q≤ {c_k}_k

_`q2^1p(log 2)^1q. Hint:Letc_n=2^n/pf^∗(2ⁿ)andf_n=c⁻¹_n fχ_An whereA_nis as in (1.4.27).

1.4.14.(T. Tao) Let 0<p<∞, 0<γ<1,A>0, and let fbe a measurable function on a measure space(X,µ).

(a) Suppose that f

_Lp,∞≤A. Then for every measurable setE of finite measure there exists a measurable subsetE⁰ofEwithµ(E⁰)≥γ µ(E)such that

Z

E⁰

f dµ

≤CγAµ(E)^1−1p, whereC_γ= (1−γ)^−1/p.

(b) Conversely, if the last condition holds for someC_γ,A<∞and all measurable

subsetsEof finite measure, then f

L^p,∞≤c_γA, wherec_γ=C_γ4^1/pγ⁻¹

√ 2.

(c) Conclude that f

_Lp,∞≈ sup

E⊆X 0<µ(E)<∞

inf

E⁰⊆E µ(E⁰)≥¹₂µ(E)

µ(E)^−1+1p Z

E⁰

f dµ .

Hint:Part (a): TakeE⁰=E\ {|f|>A(1−γ)^−1pµ(E)^−1p}. Part (b): Givenα>0, note that the set

|f|>α is contained in Re f>^√α

2 ∪

Imf>^√α

2 ∪

Ref<−^√α

2 ∪

Imf<−^√α

2 .

For E any of the preceding four sets, let E⁰ be a subset of it with measure at leastγ µ(E)such as in the hypothesis. Then

R

E⁰f dµ ≥^√α

2γ µ(E), which gives

f

_Lp,∞≤4^1/pγ⁻¹C_γ√ 2A.

1.4.15.Given a linear operator T defined on the set of measurable functions on a measure space(X,µ)and taking values in the set of measurable functions on a measure space(Y,ν), define its “transpose”T^tvia the identity

Z

Y

T(f)g dν= Z

X

T^t(g)f dµ

for all measurable functions f onX andgonY, whenever the integrals converge.

LetT be such a linear operator given in the form T(f)(y) =

Z

X

K(y,x)f(x)dµ(x),

whereKis measurable and bounded by some constantM>0. Suppose thatT maps L¹(X)toL^1,∞(Y)with norm

T

, andT^t mapsL¹(Y)toL^1,∞(X)with norm T^t

. Show that for all 1<p<∞there exists a constantC_pthat depends only onpand is independent ofMsuch thatT mapsL^p(X)toL^p(Y)with norm

T

_Lp(X)→L^p(Y)≤C_p T

1 p

T^t

1−¹_p

.

Hint:ForR>0, letBRbe the set of all(A,B), whereAis a measurable subset of X withµ(A)≤RandBis a measurable subset ofY withν(B)≤R. LetBR,M be the set of all(A,B)inBRsuch that|K(x,y)| ≤Mfor allx∈Aandy∈B. Also let M_p=M_p(R,M)<∞be the smallest constant such that for all(A,B)∈BR,Mwe have

R

BT(χ_A)dν

≤Mpµ(A)^1pν(B)

1

p0. Letδ >0 and(A,B)∈BR,M. Ifµ(A)≤δ ν(B), use Exercise 1.4.14 to find a B⁰ with ν(B⁰)≥¹₂ν(B)such that

R

B⁰T(χ_A)dν ≤ c

T

µ(A)≤cδ

1 p0

T

µ(A)^1pν(B)

1

p0. Thenν(B\B⁰)≤¹₂ν(B)and we have

Z

B\B⁰T(χ_A)dν

≤M_p2⁻

1

p0µ(A)^1pν(B)^p10.

Summing, we obtainM_p≤M_p2⁻

1 p0+cδ

1 p0

T

. Wheneverν(B)≤δ⁻¹µ(A), write

R

BT(χ_A)dν =

R

AT^t(χ_B)dµ

and use Exercise 1.4.14 to find a set A⁰ with µ(A⁰)≥¹₂µ(A)and

R

A⁰T^t(χ_B)dµ ≤c

T^t

ν(B). Argue similarly to obtainM_p≤ M_p2^−1p+cδ⁻

1 p

T^t

. Pick a suitableδ to optimize both expressions. Obtain that Mp is independent of R and M. Considering B+=B∩ {T(χ_A)>0} and B−= B∩ {T(χ_A)<0}, obtain that^R_B

T(χ_A)

dν≤2M_pµ(A)^1pν(B)^p10 for(A,B)∈BR,M. Use Fatou’s lemma to remove the restriction that(A,B)∈BR,M. Finally, use the characterization of

·

L^p,∞ obtained in Exercise 1.1.12 withr=1 to conclude that T(χ_A)

L^p,∞≤C_p T

1 p

T^t

1−¹_p

µ(A)^1p.

1.4.16.(Bourgain [29]) Let 0<p₀<p₁<∞and 0<α,β,A,B<∞. Suppose that a family of sublinear operatorsT_kis of restricted weak type(p₀,p₀)with constant A2^−kαand of restricted weak type(p₁,p₁)with constantB2^−kβ for allk∈Z. Show that there is a constantC=C(α,β,p₀,p₁)such that∑k∈ZT_k is of restricted weak type(p,p)with constantC A^1−θB^θ, whereθ=α/(α+β)and

1

p =1−θ p0

+ θ p1

.

Hint:Estimateµ({|T(χ_E)|>λ})by the sum∑k≥k₀µ({|T_k(χ_E)|>cλ2^α0(k0−k)})+

∑k≤k₀µ({|T_k(χ_E)|>cλ2^β0(k−k0)}), wherecis a suitable constant and 0<α⁰<α, β<β⁰<∞. Apply the restricted weak type(p₀,p₀)hypothesis on each term of the first sum, the restricted weak type(p₁,p₁)hypothesis on each term of the second sum, and choosek0to optimize the resulting expression.

APPENDIX: SOME MULTILINEAR INTERPOLATION

Multilinear maps are defined on products on linear spaces and take values in another linear space. We are interested in the situation that these linear spaces are function spaces. Let(X₁,µ1), . . . ,(X_m,µm)be measure spaces, letDjbe spaces of measurable functions onX_j, and letT be a map defined onD1× · · · ×Dmand taking values in the set of measurable functions on another measure space(Z,σ). ThenT is called multilinear if for all f_j,g_jinDjand all scalarsλ we have

|T(f₁, . . . ,λf_j, . . . ,f_m)| = |λ| |T(f₁, . . . ,f_j, . . . ,f_m)|,

T(f₁, . . . ,f_j+g_j, . . . ,f_m) = T(f₁, . . . ,f_j, . . . ,f_m) +T(f₁, . . . ,g_j, . . . ,f_m).

If Dj are dense subspaces of L^pj(X_j,µj) andT is a multlinear map defined on

∏^mj=1Djand satisfies

T(f1, . . . ,fm)

L^p(Z)≤C f1

L^p1(X1)· · · fm

L^pm(Xm),

for allf_j∈Dj, thenT has a bounded extension fromL^p1× · · · ×L^pm→Z. The norm of a multilinear mapT : L^p1× · · · ×L^pm→Z is the smallest constantCsuch that the preceding inequality holds and is denoted by

T

_Lp1×···×L^pm→Z.

Suppose thatT is defined on∏^m_j=1Dj, where eachDjcontains the simple func- tions. We say thatTis quasimultilinear if there is aK>0 such that for all 1≤j≤m, all fj,gjinDj, and allλ∈Cwe have

|T(f1, . . . ,λfj, . . . ,fm)| = |λ| |T(f1, . . . ,fj, . . . ,fm)|,

|T(f1, . . . ,fj+g_j, . . . ,fm)| ≤ K |T(f1, . . . ,fj, . . . ,fm)|+|T(f1, . . . ,gj, . . . ,fm)|

. In the special case in whichK=1,T is called multisublinear.

1.4.17.LetT be a multilinear map defined on the set of simple functions of the product ofmmeasure spaces(X₁,µ₁)× · · · ×(X_m,µ_m)and taking values in the set of measurable functions on another measure space(Z,σ). Let 1≤p_jk≤∞for 1≤ k≤mandj∈ {0,1}and also let 1≤pj≤∞for j∈ {0,1}. Suppose thatT satisfies

T(f1, . . . ,fm)

L^{p j} ≤Mj

f1

L^{p j1}· · ·

fm

L^{p jm}, j=0,1,

for all simple functions f_konX_k. Let(1/q,1/q₁, . . . ,1/q_m)lie on the open line seg- ment joining(1/p₀,1/p₀₁, . . . ,1/p_0m)and(1/p₁,1/p₁₁, . . . ,1/p_1m)inR^m+1. Then for some 0<θ<1 we have

1

q=1−θ p₀ + θ

p₁, 1

q_k =1−θ p_0k + θ

p_1k, 1≤k≤m.

Prove thatT has a bounded extension fromL^q1× · · · ×L^qm toL^qthat satisfies T(f1, . . . ,fm)

L^q≤M₀^1−θM₁^θ f1

L^q1· · ·

fm

L^qm

for all f_k∈L^qk(X_k).

Hint:Adapt the proof of Theorem 1.3.4.

1.4.18.Let (X₁,µ₁), . . . ,(X_m,µ_m) be measure spaces, let Dj be spaces of mea- surable functions on X_j that contain the simple functions, and let T be a quasi- multilinear map defined onD1× · · · ×Dmthat takes values in the set of measurable functions on another measure space (Z,σ). Let 0<p_jk ≤∞ for 1≤ j≤m+1 and 1≤k≤m, and also let 0<pj≤∞for 1≤ j≤m+1. Suppose that for all 1≤j≤m+1,T satisfies

T(χ_E1, . . . ,χEm)

L^{p j,∞} ≤Mµ1(E₁)

1

p j1· · ·µm(E_m)

1 p jm

for all setsE_kof finiteµ_kmeasure. Assume that the system