Let $X$ be a separable Banach space. Then every pointwise bounded $(f_n) \subset X^*$ has a subsequence that converges uniformly on compact sets

banach-spacesfunctional-analysisnormed-spacesweak-convergenceweak-topology

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In Brezis's Functional Analysis, there is a corollary

Corollary 3.30. Let $E$ be a separable Banach space and let $\left(f_{n}\right)$ be a bounded sequence in $E^*$. Then there exists a subsequence $\left(f_{n_{k}}\right)$ that converges in the weak topology $\sigma\left(E^*, E\right)$.

It seems the corollary still holds for a general (not necessarily complete) normed space, i.e.,

Let $E$ be a separable normed space and let $\left(f_{n}\right)$ be a bounded sequence in $E^*$. Then there exists a subsequence $\left(f_{n_{k}}\right)$ that converges in the weak topology $\sigma(E^*, E)$.

Recently, I have come across a theorem which requires less from $(f_n)$ and gives stronger result if $E$ is a Banach space.

  • Theorem 1: Let $X$ be a Banach space, and $(f_n) \subset X^{*}$ a sequence that converges pointwise on $X$ to a (linear) functional $\ell: X \to \mathbb{R}$. Then $(f_n)$ is bounded, $\ell \in X^{*}$, and the convergence is uniform on compact sets.

  • Theorem 2: Let $E$ be a separable Banach space and let $\left(f_{n}\right)$ be a pointwise bounded sequence in $E^*$, i.e., the set $\{f_n(x) \mid n \in \mathbb N\}$ is bounded for each $x\in X$. Then there exists a subsequence $\left(f_{n_{k}}\right)$ that converges in the weak topology $\sigma(E^*, E)$ and uniformly on compact setss.

Best Answer

I'd like to share a few related facts that you may find useful. You can easily find the proofs in the functional analysis books.

Proposition: Let $K$ be compact and $S=\{g_n:n\in\mathbb{N}\}\subseteq C(K)$ be an equicontinuous set of functions. If $g_n\to g_0$ pointwise, then $g_n\to g$ uniformly on $K$.

Proof: By the Ascoli-Arzela theorem, $S$ is relatively compact in $(C(K),\|\cdot\|_\infty)$. If $g_n\not\to g_0$ uniformly on $K$, then there existed a subsequence $(g_{n_j})$ and a $h\in C(K)\backslash\{g_0\}$ such that $g_{n_j}\to h$ uniformly. But this contradicts the fact that $g_n\to g_0$ pointwise.

For Theorem 1, suppose $f_n\to f_0$ in $\sigma(X^*,X)$ topology and $K\subseteq X$ be compact. As you observed, $(f_n)$ is bounded, say by $c>0$. Let $g_n = f_n|_K$ be the restrictions to $K$. For every $n\in\mathbb{N}$ and $x,y\in K$ $$|g_n(x)-g_n(y)| = |f_n(x-y)|\leq c\|x-y\|$$ Thus, $S=\{g_n:n\in\mathbb{N}\}\in C(K)$ is an equicontinuous set of functions. At this point, we can use the Proposition above.

Theorem 2 can be paraphrased as: If $X$ is separable, then any weak$^{*}$ closed and bounded subset $D\subseteq X^{*}$ is weak$^{*}$ sequentially compact. Compactness is by the Banach-Alaoglu theorem, and sequential compactness is due to the fact that $\sigma(X^*,X)$ topology on $D$ is metrizable if $X$ is separable. More generally, if $X$ is weakly compactly generated, then the dual unit ball of $X^*$ is weak$^*$ sequentially compact (e.g., see https://math.stackexchange.com/a/1246535/922380).

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