The Dunford-Pettis theorem

1 Weak topology and weak-* topology

If $(E,\tau )$ is a topological vector space, we denote by $E^{*}$ the set of continuous linear maps $E\to ℂ$, the dual space of $E$. The weak topology on $E$, denoted $\sigma (E,E^{*})$, is the coarsest topology on $E$ with which each function $x\mapsto \lambda x$, $\lambda \in E^{*}$, is continuous $E\to ℂ$. Thus, $\sigma (E,E^{*})\subset \tau$. If $(E,\tau )$ is a locally convex space, it follows by the Hahn-Banach separation theorem that $E^{*}$ separates $X$, and hence $|\lambda |,\lambda \in E^{*}$, is a separating family of seminorms on $E$ that induce the topology $\sigma (E,E^{*})$. Therefore, if $(E,\tau )$ is a locally convex space, then $(E,\sigma (E,E^{*}))$ is a locally convex space.

If $(E,\tau )$ is a topological vector space, the weak-* topology on $E^{\mathrm{*}}$, denoted $\sigma (E^{*},E)$, is the coarsest topology on $E^{*}$ with which each function $\lambda \mapsto \lambda x$, $x\in E$, is continuous $E^{*}\to ℂ$. It is a fact that $E^{*}$ with the topology $\sigma (E^{*},E)$ is a locally convex space.

If $E$ is a normed space, then ${\parallel \lambda \parallel }_{op}={sup}_{\parallel x\parallel \le 1}|\lambda x|$ is a norm on the dual space $E^{*}$, and that $E^{*}$ with this norm is a Banach space. The Banach-Alaoglu theorem states that $\{\lambda \in E^{*}:{\parallel \lambda \parallel }_{op}\le 1\}$ is a compact subset of $(E^{*},\sigma (E^{*},E))$.

If $(X,\mathrm{\Sigma },\mu )$ is a $\sigma$-finite measure space, for $g\in L^{\mathrm{\infty }}(\mu )$ define ${\varphi }_g\in {(L^1(\mu ))}^{*}$ by ${\varphi }_g(f)={\int }_{X}fg𝑑\mu$. The map $g\mapsto {\varphi }_g$ is an isometric isomorphism $L^{\mathrm{\infty }}(\mu )\to {(L^1(\mu ))}^{*}$.

Let $(X,\mathrm{\Sigma },\mu )$ be a probability space. If $\mathrm{\Psi }\in {(L^{\mathrm{\infty }}(\mu ))}^{*}$ and $A\mapsto \mathrm{\Psi }({\chi }_A)$ is countably additive on $\mathrm{\Sigma }$, then there is some $f\in L^1(\mu )$ such that

$$\mathrm{\Psi }(g)={\int }_{X}gf𝑑\mu ,g\in L^{\mathrm{\infty }}(\mu ),$$

and ${\parallel \mathrm{\Psi }\parallel }_{op}={\parallel f\parallel }_1$. Also, an additive function $F$ on an algebra of sets $𝒜$ is countably additive if and only if whenever $A_n$ is a decreasing sequence of elements of $𝒜$ with ${\bigcap }_{n=1}^{\mathrm{\infty }}{A}_n=\mathrm{\varnothing }$, we have ${lim}_{n\to \mathrm{\infty }}F(A_n)=0$. Using that $\mu$ is countably additive we get the following.

Theorem 1.

Suppose that $(X,\mathrm{\Sigma },\mu )$ be a probability space and that $\mathrm{\Psi }\in {(L^{\mathrm{\infty }}(\mu ))}^{*}$, and suppose that for each $ϵ>0$ there is some $\delta >0$ such that $E\in \mathrm{\Sigma }$ and $\mu (E)\le \delta$ imply that $|\mathrm{\Psi }({\chi }_A)|\le ϵ$. Then there is some $f\in L^1(\mu )$ such that

$$\mathrm{\Psi }(g)={\int }_{X}gf𝑑\mu ,g\in L^{\mathrm{\infty }}(\mu ).$$

2 Normed spaces

If $E$ is a normed space, its dual space $E^{*}$ with the operator norm is a Banach space, and $E^{**}={(X^{*})}^{*}$ with the operator norm is a Banach space. Define $i:E\to E^{**}$ by

$$i(x)(\lambda )=\lambda (x),x\in E,\lambda \in E^{*}.$$

It follows from the Hahn-Banach extension theorem that $i:E\to E^{**}$ is an isometric linear map.

If $E$ and $F$ are normed spaces and $T:E\to F$ is a bounded linear map, we define the transpose $T^{*}:F^{*}\to E^{*}$ by $T^{*}\lambda =\lambda \circ T$ for $\lambda \in F^{*}$. If $T$ is an isometric isomorphism, then $T^{*}:F^{*}\to E^{*}$ is an isometric isomorphism, where $E^{*}$ and $F^{*}$ are each Banach spaces with the operator norm. In particular, we have said that when $(X,\mathrm{\Sigma },\mu )$ is a $\sigma$-finite measure space, then the map $\varphi :L^{\mathrm{\infty }}(\mu )\to {(L^1(\mu ))}^{*}$ defined for $g\in L^{\mathrm{\infty }}(\mu )$ by

$${\varphi }_g(f)={\int }_{X}fg𝑑\mu ,f\in L^1(\mu ),$$

is an isometric isomorphism, and hence ${\varphi }^{*}:{(L^1(\mu ))}^{**}\to {(L^{\mathrm{\infty }}(\mu ))}^{*}$ is an isometric isomorphism. Therefore, for $E=L^1(\mu )$ we have that

$${\varphi }^{*}\circ i:L^1(\mu )\to {(L^{\mathrm{\infty }}(\mu ))}^{*}$$ (1)

is an isometric linear map. For $f\in L^1(\mu )$ and $g\in L^{\mathrm{\infty }}(\mu )$,

$({\varphi }^{*}\circ i)(f)(g)$ $=({\varphi }^{*}(i(f)))(g)$

$=(i(f)\circ \varphi )(g)$

$=i(f)({\varphi }_g)$

$={\varphi }_g(f).$

The Eberlein-Smulian theorem states that if $E$ is a normed space and $A$ is a subset of $E$, then $A$ is weakly compact if and only if $A$ is weakly sequentially compact.

3 Equi-integrability

Let $(X,\mathrm{\Sigma },\mu )$ be a probability space and let $ℱ$ be a subset of $L^1(\mu )$. We say that $ℱ$ is equi-integrable if for every $ϵ>0$ there is some $\delta >0$ such that for any $A\in \mathrm{\Sigma }$ with $\mu (A)\le \delta$ and for all $f\in ℱ$,

$${\int }_A|f|𝑑\mu \le ϵ.$$

If $ℱ$ is a bounded subset of $L^1(\mu )$, it is a fact that $ℱ$ being equi-integrable is equivalent to

$$\underset{C\to \mathrm{\infty }}{lim}\underset{f\in ℱ}{sup}{\int }_{\{|f|>C\}}|f|𝑑\mu =0.$$ (2)

The following theorem gives a condition under which a sequence of integrable functions is bounded and equi-integrable.

Theorem 2.

Let $(X,\mathrm{\Sigma },\mu )$ be a probability space and let $f_n$ be a sequence in $L^1(\mu )$. If for each $A\in \mathrm{\Sigma }$ the sequence ${\int }_A{f}_n𝑑\mu$ has a finite limit, then $\{f_n\}$ is bounded in $L^1(\mu )$ and is equi-integrable.

4 The Dunford-Pettis theorem

A subset $A$ of a topological space $X$ is said to be relatively compact if $A$ is contained in some compact subset of $X$. When $X$ is a Hausdorff space, this is equivalent to the closure of $A$ being a compact subset of $X$.

The following is the Dunford-Pettis theorem.

Theorem 3 (Dunford-Pettis theorem).

Suppose that $(X,\mathrm{\Sigma },\mu )$ is a probability space and that $ℱ$ is a bounded subset of $L^1(\mu )$. $ℱ$ is equi-integrable if and only if $ℱ$ is a relatively compact subset of $L^1(\mu )$ with the weak topology.

Proof.

Suppose that $ℱ$ is equi-integrable, and let $T={\varphi }^{*}\circ i:L^1(\mu )\to {(L^{\mathrm{\infty }}(\mu ))}^{*}$ be the isometric linear map in (1), for which

$$T(f)(g)={\int }_{X}fg𝑑\mu ,f\in L^1(\mu ),g\in L^{\mathrm{\infty }}(\mu ).$$

Then $T(ℱ)$ is a bounded subset of ${(L^{\mathrm{\infty }}(\mu ))}^{*}$, so is contained in some closed ball $B$ in ${(L^{\mathrm{\infty }}(\mu ))}^{*}$. By the Banach-Alaoglu theorem, $B$ is weak-* compact, and therefore the weak-* closure $\mathscr{H}$ of $T(ℱ)$ is weak-* compact. Let $F\in \mathscr{H}$. There is a net $F_{\alpha }=T(f_{\alpha })$ in $T(ℱ)$, $\alpha \in I$, such that for each $g\in L^{\mathrm{\infty }}(\mu )$, $F_{\alpha }(g)\to F(g)$, i.e.,

$${\int }_X{f}_{\alpha }g𝑑\mu \to F(g),g\in L^{\mathrm{\infty }}(\mu ).$$ (3)

Let $ϵ>0$. Because $ℱ$ is equi-integrable, there is some $\delta >0$ such that when $A\in \mathrm{\Sigma }$ and $\mu (A)\le \delta$,

$$\underset{\alpha \in I}{sup}{\int }_A|f_{\alpha }|𝑑\mu \le ϵ,$$

which gives

$$|F({\chi }_A)|=\underset{\alpha }{lim}\left|{\int }_X{f}_{\alpha }{\chi }_A𝑑\mu \right|=\underset{\alpha }{lim}\left|{\int }_A{f}_{\alpha }𝑑\mu \right|\le \underset{\alpha \in I}{sup}{\int }_A|f_{\alpha }|𝑑\mu \le ϵ.$$

By Theorem 1, this tells us that there is some $f\in L^1(\mu )$ for which

$$F(g)={\int }_{X}gf𝑑\mu ,g\in L^{\mathrm{\infty }}(\mu ),$$

and hence $F=T(f)$. This shows that $\mathscr{H}\subset T(L^1(\mu ))$, and

$${\int }_X{f}_{\alpha }g𝑑\mu \to {\int }_{X}fg𝑑\mu ,g\in L^{\mathrm{\infty }}(\mu )$$

tells us that $f_{\alpha }\to f$ in $\sigma (L^1(\mu ),{(L^1(\mu ))}^{*})$, in other words $T^{-1}(F_{\alpha })$ converges weakly to $T(F)$. Thus $T^{-1}:\mathscr{H}\to L^1(\mu )$ is continuous, where $\mathscr{H}$ has the subspace topology ${\tau }_{\mathscr{H}}$ inherited from ${(L^{\mathrm{\infty }}(\mu ))}^{*}$ with the weak-* topology and $L^1(\mu )$ has the weak topology. $(\mathscr{H},{\tau }_{\mathscr{H}})$ is a compact topological space, so $T^{-1}(\mathscr{H})$ is a weakly compact subset of $L^1(\mu )$. But $ℱ\subset T^{-1}(\mathscr{H})$, which establishes that $ℱ$ is a relatively weakly compact subset of $L^1(\mu )$.

Suppose that $ℱ$ is a relatively compact subset of $L^1(\mu )$ with the weak topology and suppose by contradiction that $ℱ$ is not equi-integrable. Then by (2), there is some $\eta >0$ such that for all $C_0$ there is some $C\ge C_0$ such that

$$\underset{f\in ℱ}{sup}{\int }_{\{|f|>C\}}|f|𝑑\mu >\eta ,$$

whence for each $n$ there is some $f_n\in ℱ$ with

$${\int }_{\{|f_n|>n\}}|f_n|𝑑\mu \ge \eta ,$$ (4)

On the other hand, because $ℱ$ is relatively weakly compact, the Eberlein-Smulian theorem tells us that $ℱ$ is relatively weakly sequentially compact, and so there is a subsequence $f_{a(n)}$ of $f_n$ and some $f\in L^1(\mu )$ such that $f_{a(n)}$ converges weakly to $f$. For $A\in \mathrm{\Sigma }$, as ${\chi }_A\in L^{\mathrm{\infty }}(\mu )$ we have

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_A{f}_{a(n)}𝑑\mu ={\int }_{A}f𝑑\mu ,$$

and thus Theorem 2 tells us that the collection $\{f_{a(n)}\}$ is equi-integrable, contradicting (4). Therefore, $ℱ$ is equi-integrable. ∎

Corollary 4.

Suppose that $(X,\mathrm{\Sigma },\mu )$ is a probability space. If $\{f_n\}\subset L^1(\mu )$ is bounded and equi-integrable, then there is a subsequence $f_{a(n)}$ of $f_n$ and some $f\in L^1(\mu )$ such that

$${\int }_X{f}_{a(n)}g𝑑\mu \to {\int }_{X}fg𝑑\mu ,g\in L^{\mathrm{\infty }}(\mu ).$$

Proof.

The Dunford-Pettis theorem tells us that $\{f_n\}$ is relatively weakly compact, so by the Eberlein-Smulian theorem, $\{f_n\}$ is relatively weakly sequentially compact, which yields the claim. ∎

5 Separable topological spaces

It is a fact that if $E$ is a separable topological vector space and $K$ is a compact subset of $(E^{*},\sigma (E^{*},E))$, then $K$ with the subspace topology inherited from $(E^{*},\sigma (E^{*},E))$ is metrizable. Using this and the Banach-Alaoglu theorem, if $E$ is a separable normed space it follows that $\{\lambda \in E^{*}:{\parallel \lambda \parallel }_{op}\le 1\}$ with the subspace topology inherited from $(E,\sigma (E^{*},E))$ is compact and metrizable, and hence is sequentially compact. In particular, when $E$ is a separable normed space, a bounded sequence in $E^{*}$ has a weak-* convergent subsequence.

If $X$ is a separable metrizable space and $\mu$ is a $\sigma$-finite Borel measure on $X$, then the Banach space $L^p(\mu )$ is separable for each .

Theorem 5.

Suppose that $X$ is a separable metrizable space and $\mu$ is a $\sigma$-finite Borel measure on $X$. If $\{g_n\}$ is a bounded subset of $L^{\mathrm{\infty }}(\mu )$, then there is a subsequence $g_{a(n)}$ of $g_n$ and some $g\in L^{\mathrm{\infty }}(\mu )$ such that

$${\int }_{X}f{g}_{a(n)}𝑑\mu \to {\int }_{X}fg𝑑\mu ,f\in L^1(\mu ).$$