Compactness of unit ball in WOT of B(X)

Question

It is known that the unit ball in $\mathcal{B}(H)$, where $H$ is a separable Hilbert space is compact in the weak operator topology. Is it the same true if instead of $H$ we have any separable Banach space?

Answer 1

It holds if and only if $X$ is reflexive.

First, note by Equivalence of reflexive and weakly compact that $X$ is reflexive iff $\newcommand{\ball}{\operatorname{ball}}\ball X$ is weakly compact.

Suppose $X$ is not reflexive, so that $\ball X$ is not weakly compact. Choose an arbitrary $h \in X$ with $\|h\| = 1$. By the Hahn-Banach theorem there exists $\ell \in X^*$ with $\|\ell\|=1$ and $\ell(h) = 1$. Then define $F : (B(X), \mathrm{WOT}) \to (X, \mathrm{wk})$ by $F(T) = Th$ which, by definition of WOT, is continuous. To see it maps $\ball B(X)$ onto $\ball X$: given $f \in \ball X$, define $T_f$ by $T_f g = \ell(g) f$, so that $\|T_f\| = \|\ell\| \|f\| \le 1$ and $T_f h = f$. So $F$ maps $(\ball B(X), \mathrm{WOT})$ continuously onto a non-compact set.

For the converse, suppose $X$ is reflexive so that $\ball X$ is weakly compact. We mimic the proof of Alaoglu's theorem. Let $K = (\ball X, \mathrm{wk})^{\ball X}$ with its product topology, which is compact by Tychonoff. Then there is an obvious injection of $(\ball B(X), \mathrm{WOT})$ into $K$, which by definition of WOT is a homeomorphism onto its image. If we show $\ball B(X)$ is closed in $K$, we will be done. Let $T_\alpha$ be a net in $\ball B(X)$ and suppose it converges in $K$ (i.e. pointwise) to some function $T : \ball X \to \ball X$. By scaling, we extend $T$ to a function $T : X \to X$. It's easy to check that $T$ is linear. To see that $T$ is bounded, note that if $f \in \ball X$ we have $T_\alpha f \to Tf$ weakly. Since the Norm is weakly lower semicontinuous we get $\|Tf\| \le \liminf \|T_\alpha f\| \le 1$. So indeed $T \in \ball B(X)$, showing that $\ball B(X)$ is closed in $K$, hence WOT-compact.

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For a concrete counterexample, let $X = C([0,1])$ and consider the evaluation map $F : (B(X), \mathrm{WOT}) \to (X, \mathrm{wk})$ defined by $F(T) = T1$ where $1 \in C([0,1])$ is the constant function 1. It's easy to check $F$ is continuous (this is basically the definition of WOT). Moreover, I claim we have $F(\ball B(X)) = \ball X$; given $f \in \ball X$, define $T \in B(X)$ by $Tg(x) = g(0) f(x)$. Clearly $\|T\| = \|f\| \le 1$ and $T1 = f$. Hence if $\ball B(X)$ is WOT compact then $\ball X$ must be weakly compact.

In this case we can see explicitly that $\ball X$ is not weakly compact. Let $f_n(x) = x^n$; I claim this sequence has no weak-* accumulation point. Since $\{f_n\}$ converges pointwise and point evaluation is a continuous linear functional on $C([0,1])$, any weak accumulation point of $f_n$ must equal the pointwise limit. But the pointwise limit is not continuous so no such accumulation point can exist.