Mixed Lebesgue Spaces

Chapter 6 Basic Inequalities and Lebesgue Spaces

6.2 The Lebesgue Spaces

6.2.2. Mixed Lebesgue Spaces

6 Basic Inequalities and Lebesgue Spaces

Finally, to check (iv), first note that the right side dominates the left. To get the opposite inequality, define gM = _µ(|f|≥M^1[M,∞]◦f₎ for M ∈[0,∞) satisfying µ(|f| ≥ M)∈ (0,∞). Obviously, kg_Mk_L1(µ;R) = 1 and kf g_Mk_L1(µ;R) ≥M. If R = kfk_L∞(µ;R), take M = R to get kf g_Rk_L1(µ;R) ≥ kfk_L∞(µ;R). If R <

kfkL^∞(µ;R), get the same conclusion by considering M ∈

R,kfkL^∞(µ;R)

. Thus, the left side dominates the right.

§6.2.2. Mixed Lebesgue Spaces: For reasons that will become clearer

E2, fn(·, x2) −→ f(·, x2) (a.e.,µ1), |fn(·, x2)| ≤ g(·, x2) (a.e.,µ1), and g(·, x2) ∈ L^p1(µ1;R). Thus, by part (ii) of Theorem 6.2.1, for µ2-almost every x2∈E2,kfn(·, x2)−f(·, x2)kL^p1(µ₁;R)−→0. In addition,

kf_n(·, x₂)−f(·, x₂)k_Lp1(µ₁;R)≤2kg(·, x₂)k_Lp1(µ₁;R)

forµ2-almost everyx2∈E2 and, by (∗) withg replacingf,

kg(·1,·2)kL^p1(µ₁;R)

_Lp2(µ2;R)<∞.

Hence the required result follows after a second application of (ii) in Theorem 6.2.1.

We turn now to the final part of the lemma, in which both the measures µ₁ andµ₂ are assumed to be finite. In fact, without loss of generality, we will assume that they are probability measures. In addition, by the preceding, it is clear that, for eachf ∈L^(p1,p2) (µ₁, µ₂);R

,

kf −fnk_L(p1,p2)((µ1,µ2);R)−→0 where fn≡f1_[−n,n]◦f.

Thus, we need only consider f’s that are bounded. Finally, becauseµ1×µ2

is also a probability measure, Jensen’s inequality and (∗) imply that kf−ψk_L(p1,p2 )(µ₁,µ₂)≤ kf −ψkL^q(µ₁×µ2) where q=p1∨p2. Hence, all that remains is to show that, for every bounded,B1×B2-measurable f :E₁×E2−→Rand >0, there is aψ∈ Gfor whichkf−ψkL^q(µ₁×µ2)< . But, by part (iv) of Theorem 6.2.1, the class of simple functions having the form

ψ=

n

X

m=1

am1Γ_1,m×Γ_2,m

with Γi,m ∈ Bi is dense in L^q(µ1×µ2;R). Thus, we will be done once we check that such a ψis an element of G. To this end, set I = {0,1}ⁿ

and, forη∈ I, define Γ_1,η =Tn

m=1Γ_1,m^(ηm) where Γ⁽⁰⁾≡Γ^{ and Γ⁽¹⁾≡Γ. Then ψ(x1, x2) =

n

X

m=1

am



 X

η∈I

ηm1Γ_1,η(x1)



1Γ_2,m(x2) =X

η∈I

1Γ_1,η(x1)ϕη(x2),

where

ϕη =

n

X

m=1

ηmam1Γ_2,m.

Since the Γ_1,η’s are mutually disjoint, this completes the proof.

We can now prove the following continuous version of Minkowski’s inequality.

6 Basic Inequalities and Lebesgue Spaces

Theorem 6.2.7. Let (Ei,Bi, µi), i ∈ {1,2}, be σ-finite measure spaces.

Then, for any measurable functionf on(E1×E2,B1× B2),

kfk_L(p1,p2 )((µ₁,µ₂);R)≤ kfk_L(p2,p1 )((µ₂,µ₁);R) if1≤p₁≤p₂<∞.

Proof: Since it is easy to reduce the general case to the one in which both µ1 andµ2 are finite, we will take them to be probability measures from the outset.

Let G be the class described in the last part of Lemma 6.2.5. Given ψ = Pn

11Γ_1,m(·1)ϕm(·2) from G, note that, since the Γ1,m’s are mutually dis- joint, |Pn

1am1Γ_1,m|^r=Pn

1|am|^r1Γ_1,m for anyr∈[0,∞) anda1, . . . , an∈R. Hence, by Minkowski’s inequality withp= ^p_p²

1, kψk_L(p1,p2)((µ1,µ2);R)=

"

Z

E2

ⁿ X

m=1

µ1(Γ1,m)|ϕm(x2)|^p1

p2 p1

µ2(dx2)

#p¹2

=

n

X

m=1

µ1(Γ1,m)|ϕm(·2)|^p1

1 p1

L

p2 p1(µ₂;R)

≤

" _n X

m=1

µ1(Γ1,m)

|ϕm|^p1

L

p2 p1(µ₂;R)

#_p¹₁

=

" _n X

1

µ1(Γ1,m)kϕmk^p_L¹p2(µ2;R)

#_p¹₁

=



 Z

E1

n

X

m=1

1Γ_1,m(x1)kϕmk^p_L¹p2(µ₂;R)µ1(dx1)





1 p1

=



 Z

E₁ n

X

m=1

1Γ1,m(x1)kϕmk^p_L²p2(µ2;R)

!

p1 p2

µ1(dx1)





1 p1

=



 Z

E1

Z

E2

n

X

1

1_Γ1,m(x₁)ϕ_m(x₂)

p₂

µ₂(dx₂) ^p_p¹₂

µ₁(dx₁)





1 p1

=kψk_L(p2,p1 )((µ₂,µ₁);R).

Therefore we are done when the functionf is an element ofG.

To complete the proof, letf be a measurable function on (E1×E2,B1×B2).

Clearly we may assume thatkfk_L(p2,p1 )((µ2,µ1);R)<∞. Using the last part of Lemma 6.2.5, choose{ψn: n≥1} ⊆ Gsuch thatkψn−fk_L(p2,p1 )((µ2,µ1);R)−→

0. Then, by Jensen’s inequality, one has thatkψ_n−fk_L1((µ₁×µ₂);R)−→0, and therefore thatψn−→f inµ1×µ2-measure. Hence, without loss of generality, assume that ψn −→f (a.e.,µ1×µ2). In particular, by Fatou’s Lemma and Exercise 4.1.9, this means that

Z

E₁

|f(x₁, x₂)|^p1µ₁(dx₁)≤ lim

n→∞

Z

E₁

|ψ_n(x₁, x₂)|^p1µ₁(dx₁)

for µ2-almost every x2 ∈ E2; and so, by the result for elements of G and another application of Fatou’s Lemma, the required result follows forf.

Exercises for §6.2

Exercise 6.2.8. Let (E,B, µ) be a measure space and f : E −→ R a B- measurable function.

(i) Ifµis finite, show thatkfkL^p(µ;R)≤µ(E)^p1−1qkfkL^q(µ;R)for 1≤p < q≤

∞. In particular, whenµis a probability measure, this means thatkfkL^p(µ;R)

is non-decreasing as a function ofp.

(ii) WhenE is countable,B=P(E), andµis thecounting measure on E (i.e., µ({x}) = 1 for eachx∈E), show thatkfk_Lp(µ;R) is a non-increasing function ofp∈[1,∞].

Hint: First reduce to the case whenE={1, . . . , n}for somen∈Z⁺. Second, show that it suffices to prove that Pn

m=1a^p_m ≤ 1 for every p ∈ [1,∞) and {a_m: 1≤m≤n} ⊆[0,1] withP

m=1a_m= 1, and apply elementary calculus to check this.

(iii) Ifµis finite orf isµ-integrable, show that, asp→ ∞,kfkL^p(µ;R)−→

kfkL^∞(µ;R) for anyB-measurablef :E −→R.

(iv) Let (E1,B1) and (E2,B2) be a pair of measurable spaces, and letµ2be aσ-finite measure on (E₂,B2). Iff :E₁×E2−→RisB1×B2-measurable, use (iii) to show thatx₁∈E₁ 7−→ kf(x₁,·)kL^∞(µ₂;R)∈[0,∞] isB1-measurable.

Hence, we could have defined L^(p1,p2) (µ1, µ2);R

for allp1, p2∈[1,∞].

Exercise 6.2.9. Let a measure space (E,B, µ) and 1 ≤ q0 ≤ q1 ≤ ∞ be given. Iff ∈L^q0(µ;R)∩L^q1(µ;R), show that for everyt∈(0,1)

kfkLqt(µ;R)≤ kfk^t_Lq0(µ;R)kfk^1−t_Lq1(µ;R) where 1 qt

= t q0

+1−t q1

. Note that this can be summarized by saying that p −logkfkL^p(µ;R) is a concave function of ¹_p.

Exercise 6.2.10. If (E₁,B₁, µ₁) and (E₂,B₂, µ₂) are a pair of σ-finite mea- sure spaces andp∈[1,∞), show that the set of functions that can be written in the form Pn

m=1f_1,m(x₁)f_2,m(x₂), where n ≥ 1, {f_1,m : 1 ≤ m ≤ n} ⊆ L^p(µ1;R), and{f2,m: 1≤m≤n} ⊆L^p(µ2;R), is dense inL^p(µ1×µ2;R).

Hint: First reduce to the case in whichµ₁ andµ₂ are finite, and then apply part (iv) of Theorem 6.2.1 to handle this case.

Exercise 6.2.11. Let (E,B, µ) be a measure space,ga non-negative element ofL^p(µ;R) for somep∈(1,∞), andf a non-negative,B-measurable function for which there exists aC∈(0,∞) such that

(∗) µ(f ≥t)≤ C

t Z

{f≥t}

g dµ, t∈(0,∞).

6 Basic Inequalities and Lebesgue Spaces

The purpose of this exercise is to show that (∗) allows one to estimate theL^p- norm off in terms of that ofg whenp >1. The result here is a very special case of a general result known as Marcinkiewicz’s Interpolation Theorem.

(i) Setν(Γ) =R

Γg dµfor Γ∈ B, note that (∗) is equivalent to µ(f > t)≤ C

tν(f > t), t∈(0,∞), and use (5.1.7) or Exercise 5.1.11 to justify

kfk^p_Lp(µ;R)=p Z

(0,∞)

t^p−1µ(f > t)dt

≤Cp Z

(0,∞)

t^p−2ν(f > t)dt= Cp p−1

Z

f^p−1dν.

Finally, note that R

f^p−1dν =R

f^p−1g dµ, and apply H¨older’s inequality to conclude that

(∗∗) kfk^p_Lp(µ;R)≤ Cp

p−1kfk^p−1_Lp(µ;R)kgkL^p(µ;R).

(ii) Under the condition thatkfk_Lp(µ;R)<∞, it is clear that (∗∗) implies (6.2.12) kfk_Lp(µ;R)≤ Cp

p−1kgk_Lp(µ;R).

Now suppose that µ(E) <∞. After checking that (∗) forf implies (∗) for fR≡f ∧R, conclude that (6.2.12) holds first withfR replacingf and then, afterR% ∞, forf itself. In other words, whenµis finite, (∗) always implies (6.2.12).

(iii) Even ifµis not finite, show that (∗) implies (6.2.12) if, for every >0, µ(f > )<∞.

Hint: Given >0, considerµ=µB

{f > }

, note that (∗) withµimplies itself with µ, and use (ii) to conclude that (6.2.12) holds withµ in place of µ. Finally, let&0.

Exercise 6.2.13. In §3.3.2 I derived the Hardy–Littlewood inequality (3.3.7) and pointed out the one cannot pass from such an estimate to an estimate on the L¹-norm. Nonetheless, use (3.3.7) together with Exercise 6.2.11 to show that

kM fkL^p(λ_R;R)≤ 2p

p−1kfkL^p(λ_R;R) for allp∈(1,∞).

§6.3 Some Elementary Transformations on Lebesgue Spaces