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Dunford--Pettis for real $L^{1}$ on a finite measure space

Statement

Assume the Axiom of Choice. Let $(S, A, μ)$ be a finite measure space, and let $K$ be a subset of the real Banach space $L^{1} (μ)$ of almost-everywhere equivalence classes. Then $K$ is relatively weakly compact if and only if it is uniformly integrable. Here uniform integrability is equivalently the conjunction of $L^{1}$ boundedness and uniform absolute continuity:

$\sup_{f \in K} ∥ f ∥_{1} < \infty, \forall ε > 0 \exists δ > 0 \forall f \in K \forall E \in A μ (E) < δ ⟹ \int_{E} ∣ f ∣ d μ < ε .$

Facts & Assumptions

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[L1]

AC supplies Dependent Choice and Countable Choice, the ultrafilter lemma, and the real Hahn--Banach principle (AC supplies the countable and dependent choices used in Banach integration, The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter, The real dominated-extension principle as an additional hypothesis over ZF, Hahn-Banach dominated extension theorem for real vector spaces).

[L2]

Real $L^{1} (μ)$ is the almost-everywhere quotient with its integral norm and is complete under Countable Choice, hence is Banach (The space $L^{p} (μ)$ as the quotient by null functions, The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ , Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ , Banach space).

[L3]

On a finite measure space, uniform integrability is exactly $L^{1}$ boundedness plus the displayed uniform absolute continuity (On a finite measure space, uniform integrability is equivalent to L^1-boundedness plus uniform absolute continuity).

[L4]

Under the principles supplied by [L1], Eberlein--Smulian identifies relative weak compactness with relative weak sequential compactness, and every weakly convergent sequence is norm bounded (Relative weak compactness and three sequential notions, Eberlein–Šmulian theorem, Weakly convergent sequences are norm bounded, A strictly increasing index map satisfies $n_{k} \geq k$ ).

[L5]

An individual integrable function has absolutely continuous integral, and under DC a nonempty complete metric space is a Baire space (Absolute continuity of the integral, Under Dependent Choice, a nonempty complete metric space is not a countable union of closed sets with empty interior).

[L6]

Under Countable Choice, real $L^{2} (μ)$ is reflexive. Under the ultrafilter lemma and HB, its closed ball is weakly compact (Reflexivity of Lp for one less p less infinity, Reflexive iff unit ball weakly compact).

[L7]

Holder's inequality gives the bounded inclusion $L^{2} ↪ L^{1}$ on a finite measure space. Since a finite measure is sigma-finite, every member of $(L^{1})^{*}$ is integration against a member of $L^{\infty}$ (Holder's inequality for integrals, including the endpoint cases, On a sigma-finite measure space, every bounded linear functional on $L^{p}$ is integration against a unique $L^{q}$ function).

[L8]

Under HB the canonical map $J_{E} : E \to E^{* *}$ is linear and isometric. The weak and weak-star topologies are their evaluation initial topologies, and weak-star addition and scalar multiplication are continuous; weak-star space is Hausdorff (Relative Hahn–Banach makes the canonical bidual map an isometry, Weak topology on a normed space, The weak-star topology from finite evaluations, Basic weak star neighborhoods).

[L9]

Under the ultrafilter lemma, bidual balls are weak-star compact. Closed subsets and continuous images of compact spaces are compact, finite products of compact spaces are compact, and compact subsets of Hausdorff spaces are closed (Banach–Alaoglu, A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact, A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, A product of finitely many compact spaces is compact in the product topology, In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones).

Proof

technique · direct

Given: AC, a finite measure space, $E = L^{1} (μ; R)$ , and $K \subseteq E$ .

1.1

Expose every choice principle used below. By [L1], AC supplies Countable Choice for [L2] and [L6], DC for Baire and Eberlein--Smulian, the ultrafilter lemma for Alaoglu and the compactness forms in [L4], [L6], and [L9], and HB for [L4], [L6], and [L8]. No additional choice principle will be left implicit.

givenA1L1

1.2

Fix the Banach and weak-topology conventions. By [L2], $E$ is the real Banach space of classes, not the raw class of integrable representatives. Its weak topology is $σ (E, E^{*})$ . By [L8], $J_{E}$ is an isometry and a homeomorphism from weak $E$ to its image in $E^{* *}$ with the relative weak-star topology: the identity $(J_{E} f) (Λ) = Λ (f)$ makes the two evaluation families identical.

givenL2L8

1.3

Dispose of the empty and null cases. If $K = \emptyset$ , it is relatively weakly compact and uniformly integrable vacuously. If $μ (S) = 0$ , then $E = {0}$ and every subset of $E$ is finite, weakly compact, and uniformly integrable. Hence below we may assume $K \neq \emptyset$ and $μ (S) > 0$ .

givenL2L3L4

2.1

A relatively weakly compact family is norm bounded. Suppose $K$ is relatively weakly compact but not norm bounded. Using AC choose $f_{n} \in K$ with $∥ f_{n} ∥_{1} > n$ . By Eberlein--Smulian in [L4], a subsequence $f_{n_{j}}$ converges weakly in $E$ . The weak-sequence boundedness theorem in [L4] makes its norms bounded, whereas strict increase of the indices gives $n_{j} \geq j$ and hence $∥ f_{n_{j}} ∥_{1} > n_{j} \geq j$ , a contradiction.

A1L4step 1.1step 1.2choose

2.2

Set up the Baire argument for a weakly null sequence. Let $g_{n} ⇀ 0$ in $E$ , and let $X = {[1_{A}] : A \in A} \subseteq E$ with the $L^{1}$ metric. This set is closed: if a sequence of indicator classes converges in $L^{1}$ , [L2] supplies a subsequence of representatives converging almost everywhere to a representative $u$ of the limit; outside the countable union of the null sets on which those representatives differ from their indicators, $u$ is a pointwise limit of zeros and ones and therefore equals an indicator. Thus $X$ is complete and nonempty.

L2L5L7step 1.1step 1.2

For $η > 0$ define

$X_{m} = {[1_{A}] \in X : ∣ \int_{A} g_{n} d μ ∣ \leq η for every n \geq m} .$

Each $X_{m}$ is closed. Indeed, $L^{1}$ convergence of indicators means $μ (A_{j} △ A) \to 0$ , and [L5] applied to the fixed $g_{n}$ gives $\int_{A_{j} △ A} ∣ g_{n} ∣ \to 0$ for every $n$ . Also $⋃_{m} X_{m} = X$ , because $h \mapsto \int_{A} h$ is a bounded functional by the endpoint Holder inequality in [L7], and hence weak nullity gives $\int_{A} g_{n} \to 0$ for each fixed $A$ .

2.3

Uniform integrability gives weakly compact truncation approximants. For the reverse implication assume $K$ is uniformly integrable and fix $ε > 0$ . By [L3] choose $M > 0$ so that $\int_{{∣ f ∣ > M}} ∣ f ∣ < ε$ for all $f \in K$ . The truncation $T_{M} f = \max (- M, \min (f, M))$ is well defined on classes, measurable, and

L3L6L7step 1.1step 1.2step 1.3construct

$∥ f - T_{M} f ∥_{1} \leq \int_{{∣ f ∣ > M}} ∣ f ∣ d μ < ε .$

Every $T_{M} f$ belongs to $L^{2}$ and has $L^{2}$ norm at most $R = M \sqrt{μ (S)}$ . By [L6], $R B_{L^{2}}$ is weakly compact. The inclusion $I : L^{2} \to E$ satisfies $∥ I h ∥_{1} \leq \sqrt{μ (S)} ∥ h ∥_{2}$ by [L7] and is weak-to-weak continuous: if $Λ \in E^{*}$ , [L7] writes $Λ (h) = \int h g$ for some $g \in L^{\infty}$ ; finiteness of $μ$ puts $g \in L^{2}$ , so this is an $L^{2}$ -continuous functional. Therefore $C_{ε} = I (R B_{L^{2}})$ is weakly compact and $K \subseteq C_{ε} + ε B_{E}$ .

3.1

Apply Baire to obtain one uniform tail neighborhood. The Baire theorem applied to the complete nonempty space $X$ and its closed cover $(X_{m})$ gives $N$ , an indicator $[1_{A_{0}}]$ , and $ρ > 0$ such that every indicator whose $L^{1}$ distance from $[1_{A_{0}}]$ is below $ρ$ belongs to $X_{N}$ .

L5step 1.1step 2.2

3.2

Put the uniformly integrable family into a compact bidual closure. Uniform integrability gives a bound $C$ for $∥ f ∥_{1}$ , $f \in K$ . Let $G = {\overline{J_{E} (K)}}^{w^{*}}$ in $E^{* *}$ . Every $z \in G$ satisfies $∥ z ∥ \leq C$ : for $Λ \in E^{*}$ , every weak-star neighborhood of $z$ meets $J_{E} (K)$ , so $∣ z (Λ) ∣ \leq C ∥ Λ ∥$ by letting the neighborhood radius tend to zero. Hence $G \subseteq C B_{E^{* *}}$ . Banach--Alaoglu and [L9] make that ball weak-star compact; $G$ , being closed in it, is weak-star compact.

L3L8L9step 1.1step 1.2step 2.3

4.1

Derive uniform absolute continuity for every weakly null sequence. Fix a desired $ε > 0$ and run steps 2.2--3.1 with $η = ε / 8$ . If $μ (A) < ρ$ , put $B_{1} = A_{0} \cup A$ and $B_{2} = B_{1} ∖ A$ . Both indicators are within $μ (A) < ρ$ of $1_{A_{0}}$ , so for $n \geq N$ ,

L5step 3.1

$∣ \int_{A} g_{n} ∣ = ∣ \int_{B_{1}} g_{n} - \int_{B_{2}} g_{n} ∣ \leq 2 η .$

Apply this to $A \cap {g_{n} \geq 0}$ and $A \cap {g_{n} < 0}$ . Their measures are below $ρ$ , and the two signed integrals have absolute value at most $2 η$ , whence $\int_{A} ∣ g_{n} ∣ \leq 4 η = ε / 2$ for $n \geq N$ . For the finitely many $n < N$ , [L5] supplies a common positive $δ \leq ρ$ for which every corresponding integral is below $ε$ . Thus $μ (A) < δ$ implies $\int_{A} ∣ g_{n} ∣ < ε$ for every $n$ : every weakly null sequence has uniformly absolutely continuous integrals.

4.2

Trap the bidual closure in compact neighborhoods of the canonical image. For each $ε > 0$ , the set $J_{E} (C_{ε})$ is weak-star compact, because $C_{ε}$ is weakly compact and $J_{E}$ is weak-to-weak-star continuous. The product $J_{E} (C_{ε}) \times ε B_{E^{* *}}$ is compact by [L9], and weak-star addition is continuous by [L8]. Hence

L8L9step 1.2step 2.3step 3.2

$S_{ε} : = J_{E} (C_{ε}) + ε B_{E^{* *}}$

is weak-star compact and therefore weak-star closed in the Hausdorff weak-star space. Step 2.3 gives $J_{E} (K) \subseteq S_{ε}$ , so its weak-star closure satisfies $G \subseteq S_{ε}$ for every $ε > 0$ .

5.1

Every weakly convergent sequence is uniformly integrable. If $f_{n} ⇀ f$ , then $g_{n} = f_{n} - f$ is weakly null. Step 4.1 gives uniform absolute continuity of $(g_{n})$ , and [L4] gives norm boundedness. The individual function $f$ has absolutely continuous integral by [L5], so $\int_{A} ∣ f_{n} ∣ \leq \int_{A} ∣ g_{n} ∣ + \int_{A} ∣ f ∣$ makes $(f_{n})$ uniformly absolutely continuous as well. It is norm bounded by the triangle inequality. Thus [L3] makes the entire sequence $(f_{n})$ uniformly integrable, including its finite initial segment.

L3L4L5step 1.1step 4.1

5.2

Show that the compact bidual closure actually lies in $J_{E} (E)$ . First $J_{E} (E)$ is norm closed. Indeed, if $y$ is in its norm closure, AC chooses $x_{n} \in E$ with $∥ y - J_{E} x_{n} ∥ < 1 / (n + 1)$ . Isometry makes $(x_{n})$ Cauchy; completeness of $E$ gives $x_{n} \to x$ , and then $J_{E} x_{n} \to J_{E} x = y$ .

A1L2L8step 1.1step 4.2choose

Now let $z \in G$ . From $G \subseteq S_{1 / (n + 1)}$ , AC chooses $c_{n} \in J_{E} (C_{1 / (n + 1)}) \subseteq J_{E} (E)$ with $∥ z - c_{n} ∥ \leq 1 / (n + 1)$ . Hence $z$ belongs to the norm closure of $J_{E} (E)$ , which is $J_{E} (E)$ . Therefore $G \subseteq J_{E} (E)$ .

6.1

Complete the relatively-weakly-compact-to-UI implication. Assume $K$ is relatively weakly compact. If its integrals were not uniformly absolutely continuous, AC would supply $ε_{0} > 0$ , $f_{n} \in K$ , and $A_{n} \in A$ with $μ (A_{n}) < 1 / (n + 1)$ but $\int_{A_{n}} ∣ f_{n} ∣ \geq ε_{0}$ . By [L4], a subsequence $f_{n_{j}}$ converges weakly. Step 5.1 makes that subsequence uniformly integrable and hence uniformly absolutely continuous by [L3]. But $n_{j} \geq j$ makes $μ (A_{n_{j}}) \to 0$ , contradicting the displayed lower bound. Thus $K$ is uniformly absolutely continuous; step 2.1 supplies norm boundedness, so [L3] makes $K$ uniformly integrable.

A1L3L4step 1.1step 2.1step 5.1choose

6.2

Complete the UI-to-relatively-weakly-compact implication. Assume $K$ is uniformly integrable. Step 3.2 makes $G$ weak-star compact and step 5.2 puts it inside $J_{E} (E)$ . Since $J_{E}$ is the weak-to-relative-weak-star homeomorphism of step 1.2, $J_{E}^{- 1} (G)$ is weakly compact. Because ambient weak-star closure agrees with relative closure once $G \subseteq J_{E} (E)$ , this inverse is exactly ${\overline{K}}^{w}$ . Thus $K$ is relatively weakly compact in the sense of [L4].

L4L8L9step 1.2step 3.2step 5.2

7.1

Combine both directions and account for all boundaries. [A1, L3, step 1.3, step 6.1, step 6.2] Steps 6.1 and 6.2 prove the two implications; step 1.3 covers empty $K$ and null measure spaces. Zero truncation levels are unnecessary because uniform integrability permits positive $M$ , and arbitrary positive $ε$ is retained in the bidual intersection argument. The theorem is specifically for real $L^{1}$ ; no complex-duality conclusion is silently used. AC is spent only as itemized in step 1.1 and for the explicit countable selections in the norm-boundedness argument, step 5.2, and step 6.1.

A1step 1.1step 1.3step 6.1step 6.2∎

Depends on

Used by

Dunford--Pettis: dominated and concentrating families Example

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Sources

Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (standard reference, not scraped)

Dunford--Pettis for real L1L1 on a finite measure space