# The Dunford-Pettis theorem

Jordan Bell
April 19, 2015

## 1 Weak topology and weak-* topology

If $(E,\tau)$ is a topological vector space, we denote by $E^{*}$ the set of continuous linear maps $E\to\mathbb{C}$, 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 $E\to\mathbb{C}$. 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 $E^{*}$, denoted $\sigma(E^{*},E)$, is the coarsest topology on $E^{*}$ with which each function $\lambda\mapsto\lambda x$, $x\in E$, is continuous $E^{*}\to\mathbb{C}$. It is a fact that $E^{*}$ with the topology $\sigma(E^{*},E)$ is a locally convex space.

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

If $(X,\Sigma,\mu)$ is a $\sigma$-finite measure space, for $g\in L^{\infty}(\mu)$ define $\phi_{g}\in(L^{1}(\mu))^{*}$ by $\phi_{g}(f)=\int_{X}fgd\mu$. The map $g\mapsto\phi_{g}$ is an isometric isomorphism $L^{\infty}(\mu)\to(L^{1}(\mu))^{*}$.11 1 Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, second ed., p. 190, Theorem 6.15.

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

 $\Psi(g)=\int_{X}gfd\mu,\qquad g\in L^{\infty}(\mu),$

and $\left\|\Psi\right\|_{op}=\left\|f\right\|_{1}$.22 2 V. I. Bogachev, Measure Theory, volume I, p. 263, Proposition 4.2.2. Also, an additive function $F$ on an algebra of sets $\mathscr{A}$ is countably additive if and only if whenever $A_{n}$ is a decreasing sequence of elements of $\mathscr{A}$ with $\bigcap_{n=1}^{\infty}A_{n}=\emptyset$, we have $\lim_{n\to\infty}F(A_{n})=0$.33 3 V. I. Bogachev, Measure Theory, volume I, p. 9, Proposition 1.3.3. Using that $\mu$ is countably additive we get the following.

###### Theorem 1.

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

 $\Psi(g)=\int_{X}gfd\mu,\qquad g\in L^{\infty}(\mu).$

## 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

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

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 $(X,\Sigma,\mu)$ is a $\sigma$-finite measure space, then the map $\phi:L^{\infty}(\mu)\to(L^{1}(\mu))^{*}$ defined for $g\in L^{\infty}(\mu)$ by

 $\phi_{g}(f)=\int_{X}fgd\mu,\qquad f\in L^{1}(\mu),$

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

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

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

 $\displaystyle(\phi^{*}\circ i)(f)(g)$ $\displaystyle=(\phi^{*}(i(f)))(g)$ $\displaystyle=(i(f)\circ\phi)(g)$ $\displaystyle=i(f)(\phi_{g})$ $\displaystyle=\phi_{g}(f).$

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.44 4 Robert E. Megginson, An Introduction to Banach Space Theory, p. 248, Theorem 2.8.6.

## 3 Equi-integrability

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

 $\int_{A}|f|d\mu\leq\epsilon.$

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

 $\lim_{C\to\infty}\sup_{f\in\mathscr{F}}\int_{\{|f|>C\}}|f|d\mu=0.$ (2)

The following theorem gives a condition under which a sequence of integrable functions is bounded and equi-integrable.55 5 V. I. Bogachev, Measure Theory, volume I, p. 269, Theorem 4.5.6.

###### Theorem 2.

Let $(X,\Sigma,\mu)$ be a probability space and let $f_{n}$ be a sequence in $L^{1}(\mu)$. If for each $A\in\Sigma$ the sequence $\int_{A}f_{n}d\mu$ 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.66 6 V. I. Bogachev, Measure Theory, volume I, p. 285, Theorem 4.7.18; Fernando Albiac and Nigel J. Kalton, Topics in Banach Space Theory, p. 109, Theorem 5.2.9; R. E. Edwards, Functional Analysis: Theory and Applications, p. 274, Theorem 4.21.2; P. Wojtaszczyk, Banach Spaces for Analysts, p. 137, Theorem 12; Joseph Diestel, Sequences and Series in Banach Spaces, p. 93; François Trèves, Topological Vector Spaces, Distributions and Kernels, p. 471, Theorem 46.1.

###### Theorem 3 (Dunford-Pettis theorem).

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

###### Proof.

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

 $T(f)(g)=\int_{X}fgd\mu,\qquad f\in L^{1}(\mu),\quad g\in L^{\infty}(\mu).$

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

 $\int_{X}f_{\alpha}gd\mu\to F(g),\qquad g\in L^{\infty}(\mu).$ (3)

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

 $\sup_{\alpha\in I}\int_{A}|f_{\alpha}|d\mu\leq\epsilon,$

which gives

 $|F(\chi_{A})|=\lim_{\alpha}\left|\int_{X}f_{\alpha}\chi_{A}d\mu\right|=\lim_{% \alpha}\left|\int_{A}f_{\alpha}d\mu\right|\leq\sup_{\alpha\in I}\int_{A}|f_{% \alpha}|d\mu\leq\epsilon.$

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

 $F(g)=\int_{X}gfd\mu,\qquad g\in L^{\infty}(\mu),$

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

 $\int_{X}f_{\alpha}gd\mu\to\int_{X}fgd\mu,\qquad g\in L^{\infty}(\mu)$

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 $(L^{\infty}(\mu))^{*}$ 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 $\mathscr{F}\subset T^{-1}(\mathscr{H})$, which establishes that $\mathscr{F}$ is a relatively weakly compact subset of $L^{1}(\mu)$.

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

 $\sup_{f\in\mathscr{F}}\int_{\{|f|>C\}}|f|d\mu>\eta,$

whence for each $n$ there is some $f_{n}\in\mathscr{F}$ with

 $\int_{\{|f_{n}|>n\}}|f_{n}|d\mu\geq\eta,$ (4)

On the other hand, because $\mathscr{F}$ is relatively weakly compact, the Eberlein-Smulian theorem tells us that $\mathscr{F}$ 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 $A\in\Sigma$, as $\chi_{A}\in L^{\infty}(\mu)$ we have

 $\lim_{n\to\infty}\int_{A}f_{a(n)}d\mu=\int_{A}fd\mu,$

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

###### Corollary 4.

Suppose that $(X,\Sigma,\mu)$ 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

 $\int_{X}f_{a(n)}gd\mu\to\int_{X}fgd\mu,\qquad g\in L^{\infty}(\mu).$
###### 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 $\{\lambda\in E^{*}:\left\|\lambda\right\|_{op}\leq 1\}$ with the subspace topology inherited from $(E,\sigma(E^{*},E))$ is compact and metrizable, and hence is sequentially compact.77 7 A second-countable $T_{1}$ space is compact if and only if it is sequentially compact: Stephen Willard, General Topology, p. 125, 17G. 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 $1\leq p<\infty$.88 8 René L. Schilling, Measures, Integrals and Martingales, p. 270, Corollary 23.20.

###### 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 $L^{\infty}(\mu)$, then there is a subsequence $g_{a(n)}$ of $g_{n}$ and some $g\in L^{\infty}(\mu)$ such that

 $\int_{X}fg_{a(n)}d\mu\to\int_{X}fgd\mu,\qquad f\in L^{1}(\mu).$