The L p -Spaces

Chapter 6 Basic Inequalities and Lebesgue Spaces

6.2 The Lebesgue Spaces

6.2.1. The L p -Spaces

(ii) Assume thatg :C −→[0,∞) is a bounded, continuous, concave func- tion, and set ˆC ={(x, t)∈R^N ×R : x∈C andt≤g(x)}. Show that ˆC is a closed, convex subset of R^N+1. Next, define ˆF:E −→ Cˆ by ˆF = _F

g◦F

, note that ˆFisµ-integrable, and apply the first part to see that itsµ-integral is an element of ˆC. Finally, notice that

Z Fˆdµ∈Cˆ =⇒ Z

g◦Fdµ≤g Z

Fdµ

. Exercise 6.1.10. Suppose that u∈C² [0,1];R

satisfies u(0) = 0 =u(1).

The goal of this exercise is to show that

(∗) u(t) =−

Z

[0,1]

(s∧t−st)u⁰⁰(s)ds fort∈[0,1].

In particular, ifu⁰⁰≤0, thenu≥0.

(i) Use integration by parts to show that u(t) =tu⁰(0) +

Z

[0,t]

(t−s)u⁰⁰(s)ds fort∈[0,1].

(ii) Using (i), show thatu⁰(0) =−R

[0,1](1−s)u⁰⁰(s)dsand therefore that (∗) holds.

6 Basic Inequalities and Lebesgue Spaces

the collection of equivalence classes [f]^∼µ ofR-valued, measurable functions f satisfyingkfk_Lp(µ;R)<∞. Once again, I will consistently abuse notation by using f to denote its own equivalence class [f]^∼µ.

Obviously kαfkL^p(µ;R) = |α|kfkL^p(µ;R) for all p ∈ [1,∞), α∈ R, and B- measurable f’s. Also, by (3.1.10) and Minkowski’s inequality, we have the triangle inequality

kf1+f2k_Lp(µ;R)≤ kf1k_Lp(µ;R)+kf2k_Lp(µ;R)

for all p∈ [1,∞) and f₁, f₂ ∈L^p(µ;R). Moreover, it is a simple matter to check that these relations hold equally well when p=∞. Thus, each of the spacesL^p(µ;R) is a vector space. In addition, because of our convention and Markov’s inequality (Theorem 3.1.6),kfk_Lp(µ;R)= 0 if and only iff = 0 as an element ofL^p(µ;R). Hence,kf₂−f₁k_Lp(µ;R)determines a metric onL^p(µ;R), and I will write f_n −→ f in L^p(µ;R) when {f_n : n≥ 1} ∪ {f} ⊆L^p(µ;R) andkf_n−fk_Lp(µ;R)−→0.

The following theorem simply summarizes obvious applications of the re- sults in §§3.1 and 3.2 to the present context. The reader should verify that each of the assertions here follows from the relevant result there.

Theorem 6.2.1. Let(E,B, µ)be a measure space. Then, for anyp∈[1,∞]

andf, g∈L^p(µ;R),

kgk_Lp(µ;R)− kfk_Lp(µ;R)

≤ kg−fk_Lp(µ;R).

Next suppose that {fn : n≥1} ⊆L^p(µ;R)for some p∈[1,∞]and thatf is anR-valued measurable function on(E,B).

(i)Ifp∈[1,∞) andfn −→f in L^p(µ;R), thenfn −→f in µ-measure. If fn−→f in L^∞(µ;R), then fn−→f uniformly off of a set ofµ-measure 0.

(ii)Ifp∈[1,∞]andf_n−→f in µ-measure or(a.e.,µ), then kfkL^p(µ;R)≤ lim_n→∞kfnkL^p(µ;R). Moreover, if p ∈ [1,∞) and, in addition, there is a g ∈ L^p(µ;R) such that|f_n| ≤ g (a.e.,µ)for each n∈Z⁺, then f_n −→f in L^p(µ;R).

(iii)If p∈[1,∞]andlimm→∞sup_n≥mkfn−fmkL^p(µ;R)= 0, then there is an f ∈L^p(µ;R) such that fn −→ f in L^p(µ;R). In other words, the space L^p(µ;R)is complete with respect to the metric determined byk · k_Lp(µ;R).

Finally, we have the following variants of Theorem 3.2.14 and Corollary 3.2.15.

(iv)Assume thatµ(E)<∞ and thatp, q∈[1,∞). Referring to Theorem 3.2.14, define S as in that theorem. Then, for eachf ∈L^p(µ;R)∩L^q(µ;R), there is a sequence{ϕn: n≥1} ⊆ S such thatϕn−→f both inL^p(µ;R)and in L^q(µ;R). In particular, if µ isσ-finite andB is generated by a countable collection C, then each of the spacesL^p(µ;R),p∈[1,∞), is separable.

(v) Let (E, ρ) be a metric space, and suppose that µ is a measure on (E,BE)for which there exists a non-decreasing sequence of open setsEn %E satisfying µ(En)<∞ for each n≥1. Then, for each pair p, q ∈[1,∞)and f ∈ L^p(µ;R)∩L^q(µ;R), there is a sequence {ϕn : n ≥ 1} of bounded, ρ- uniformly continuous functions such thatϕ_n≡0off ofE_nandϕ_n−→f both in L^p(µ;R)and inL^q(µ;R).

The version of Lieb’s variation on Fatou’s Lemma forL^p-spaces withp6= 1 is not so easy as the assertions in Theorem 6.2.1. To prove it we will need the following lemma.

Lemma 6.2.2. Letp∈(1,∞), and suppose that {fn : n≥1} ⊆L^p(µ;R) satisfies sup_n≥1kfnk_Lp(µ;R) <∞ and that fn −→ 0 either in µ-measure or (a.e.,µ). Then, for everyg∈L^p(µ;R),

n→∞lim Z

|f_n|^p−1|g|dµ= 0 = lim

n→∞

Z

|f_n| |g|^p−1dµ.

Proof: Without loss of generality, we will assume that all of thef_n’s as well as gare non-negative. Givenδ >0, we have that

Z

f_n^p−1g dµ= Z

{fn≤δg}

f_n^p−1g dµ+ Z

{fn>δg}

f_n^p−1g dµ

≤δ^p−1kgk^p_Lp(µ;R)+ Z

{fn≥δ²}

f_n^p−1g dµ+ Z

{g≤δ}

f_n^p−1g dµ.

Applying H¨older’s inequality to each of the last two terms, we obtain Z

f_n^p−1g dµ≤δ^p−1kgk^p_Lp(µ;R)

+kfnk^p−1_Lp(µ;R)



 Z

{fn≥δ²}

g^pdµ

!¹_p +

Z

{g≤δ}

g^pdµ

!¹_p

. Since, by Lebesgue’s Dominated Convergence Theorem, the first term in the final brackets tends to 0 asn→0, we conclude that

n→∞lim Z

f_n^p−1g dµ≤δ^p−1kgk^p_Lp(µ;

R)+ sup

n≥1

kf_nk^p−1_Lp(µ;

R)k1_{g≤δ}gk_Lp(µ;R)

for every δ > 0. Thus, after another application of Lebesgue’s Dominated Convergence Theorem, we get the first equality upon lettingδ&0.

To derive the other equality, apply the preceding with f_n^p−1 and g^p−1 re- placing, respectively,f_n andg and withp⁰= _p−1^p in place ofp.

6 Basic Inequalities and Lebesgue Spaces

Theorem 6.2.3 (Lieb). Let(E,B, µ)be a measure space,p∈[1,∞), and {fn : n≥1} ∪ {f} ⊆L^p(µ;R). If sup_n≥1kfnkL^p(µ;R)<∞ andfn −→f in µ-measure or(a.e.,µ), then

n→∞lim Z

|f_n|^p− |f|^p− |f_n−f|^p dµ= 0;

and thereforekfn−fkL^p(µ;R)−→0if kfnkL^p(µ;R)−→ kfkL^p(µ;R).

Proof: The case p= 1 is covered by Theorems 3.2.3 and 3.2.5, and so we will assume thatp∈(1,∞). Given such ap, we will first check that there is a K_p<∞such that

(∗)

|b|^p− |a|^p− |b−a|^p

≤Kp |b−a|^p−1|a|+|a|^p−1|b−a|

, a, b∈R. Since it is clear that (∗) holds for all a, b∈ R if it does for all a ∈ R\ {0}

and b∈R, we can assume thata6= 0 and divide both sides by |a|^p, thereby showing that (∗) is equivalent to

|c|^p−1− |c−1|^p

≤K_p |c−1|^p−1+|c−1|

, c∈R.

Finally, the existence of a Kp < ∞ for which this inequality holds can be easily verified with elementary consideration of what happens when cis near 1 and when|c|is near infinity.

Applying (∗) witha=fn(x) andb=f(x), we see that

|fn|^p− |f|^p− |fn−f|^p

≤Kp |fn−f|^p−1|f|+|fn−f||f|^p−1 pointwise. Thus, by Lemma 6.2.2 withf_nandgthere replaced by, respectively, f_n−f andf here, our result follows.

Before applying H¨older’s inequality to the L^p-spaces, it makes sense to complete the definition of the H¨older conjugatep⁰that was given in Theorem 6.1.5 only for p∈(1,∞). Namely, I will take the H¨older conjugate of 1 to be

∞and that of ∞to be 1. Notice that this is completely consistent with the equation ¹_p+_p¹0 = 1 used before.

Theorem 6.2.4. Let(E,B, µ)be a measure space.

(i)Iff andgare measurable functions on(E,B), then for everyp∈[1,∞], kf gk_L1(µ;R)≤ kfk_Lp(µ;R)kgk_Lp0(µ;R).

In particular, iff ∈L^p(µ;R)andg∈L^p0(µ;R), thenf g∈L¹(µ;R).

(ii)Ifp∈[1,∞)andf ∈L^p(µ;R), then kfkL^p(µ;R)= sup

kf gkL¹(µ;R):g∈L^p0(µ;R) andkgk_Lp0

(µ;R)≤1 .

In fact, if kfk_Lp(µ;R)>0, then the supremum on the right is achieved by the function

g= |f|^p−1 kfk^p−1_Lp(µ;R)

.

(iii)More generally, for anyf that is measurable on (E,B), kfk_Lp(µ;R)≥sup

kf gk_L1(µ;R):g∈L^p0(µ;R) andkgk_Lp0(µ;R)≤1 , and equality holds ifp= 1orp∈(1,∞)and eitherµ(|f| ≥δ)<∞for every δ >0orµisσ-finite.

(iv)Iff :E −→Ris measurable andµ(|f| ≥R)∈(0,∞)for someR≥0, then

kfkL^∞(µ;R)= sup

kf gkL¹(µ;R):g∈L¹(µ;R) andkgkL¹(µ;R)≤1 . Proof: Part (i) is an immediate consequence of H¨older’s inequality when p ∈ (1,∞). At the same time, when p ∈ {1,∞}, the conclusion is clear without any further comment. Given (i), (ii) as well as the inequality in (iii) are obvious.

Whenp= 1, equality in (iii) is trivial, since one can takeg=1. Moreover, in view of (ii), the proof of equality in (iii) forp∈(1,∞) reduces to showing that, under either one of the stated conditions, kfk_Lp(µ;R)=∞implies that the supremum on right-hand side is infinite. To this end, first suppose that µ(|f| ≥δ)<∞for every δ >0. Then, for eachn≥1, the function given by

ψn≡ |f|^p−1

1[n¹,n]◦ |f|

+n1_{∞}◦f

is an element ofL^p0(µ;R). Moreover, ifkfk_Lp(µ;R)=∞, then, by the Mono- tone Convergence Theorem,kψnk_Lp0(µ;R)−→ ∞. Thus, sincekf ψnk_L1(µ;R)≥ kψnk^p0

L^p0(µ;R),we see that

kf g_nk_L1(µ;R)−→ ∞ ifkfk_Lp(µ;R)=∞andg_n≡ ψ_n 1 +kψnk_Lp0

(µ;R)

. To handle the case µ isσ-finite andµ(|f| ≥δ) =∞for some δ >0, choose {En: n≥1} ⊆ Bsuch that En%E andµ(En)<∞for every n≥1. Then it is easy to see that lim_n→∞kf gnk_L1(µ;R)=∞when

g_n≡ 1_Γn

1 +µ(Γn)1−¹_p with Γ_n=E_n∩ {|f| ≥δ}.

Sincekgnk_Lp0

(µ;R)≤1, this completes the proof of (iii).

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