Weak and Weak* convergences implying reflexivity

Question

Let $X$ be a Banach space. Suppose that for any sequence of functionals $(\phi_n) \subseteq X^*$ we have that $\phi_n$ converges weakly to some $\phi \in X^*$ if and only if $\phi_n$ converges weakly* to $\phi.$

We know that the weak and weak* topologies on $X^*$ coincide if and only if $X$ is reflexive. Since two topological spaces that have the same convergent sequences do not need to be equal, our $X$ above does not need the weak and weak* topologies to coincide.

So what can we say about such an $X$ above? Is it still reflexive?

Answer 1

Here is an argument for the following case:

Suppose $X$ is separable and it admits a predual $X_\#$. Then $X$ is reflexive iff every weak* convergent sequence in $X^*$ is weak convergent.

Begin with the image of the unit ball $X_\#$ in $X^*$, this image is weak closed (since its norm closed + convex) but weak* dense in the unit ball of $X^*$ (by the Goldstine theorem). Weak closed clearly implies weak sequentially closed.

Since $X$ is separable the weak* topology on the unit ball of $X^*$ is metrisable, so every element in the unit ball can be weak* approximated by a sequence in $X_\#$. So unless $X_\#=X^*$ (ie unless $X$ is reflexive) there are sequences that are weak* convergent but not weakly convergent.

Answer 2

Here is an expansion of the comment by GEdgar:

Let $X$ be separable and suppose that weak and weak-* convergence of sequences coincides on $X^*$. Let $B \subset X^*$ be the (closed) unit ball.

Since $X$ is separable, the Sequential Banach–Alaoglu theorem implies that $B$ is sequentially weak-* compact. Since weak and weak-* convergent sequences coincide in $X^*$, it follows that $B$ is sequentially weakly compact. Now, the Eberlein–Šmulian theorem implies that $B$ is weakly compact. Finally, Kakutani's theorem yields that $X^*$ (and, consequently, $X$) is reflexive.

Moreover, it is possible to substitute the application of the Eberlein–Šmulian theorem and Kakutani's theorem by James' theorem: Let $f \in X^{**}$ be a linear and continuous functional on $X^*$. Since $B$ is sequentially weakly compact, it is easy to see that the supremum of $f$ is attained on $B$. Thus, James' theorem implies the reflexivity of $X^*$ (and, consequently of $X$).

Answer 3

This, too, is a partial answer. I originally wrote it to argue that GEdgar's suggestion of $C(\beta\mathbb{N})$ and similar are not counterexamples (i.e., your hypothetical equivalence holds for them), but flubbed a key point. Nevertheless, my argument shows (in rather elementary fashion) that a counterexample should have to be rather exotic.

Specifically, let $K$ be any infinite compact Hausdorff space — say, $\beta\mathbb{N}$, $[0,1]^{\mathbb{R}}$, or just $\omega_1+1$ with the ordinal topology. I will show that if $K$ admits a nontrivial convergent sequence in the sense of this paper — that is, a convergent sequence that is not eventually constant — then $K$ cannot be a counterexample. Not all infinite compact Hausdorff spaces admit such a sequence because they need not be sequentially compact; in particular, such is the case for $\beta\mathbb{N}$.

Anyways, on to the proof!

Pick a maximal set $S$ of mutually singular measures on the Borel $\sigma$-algebra associated with the topology on $K$; $S$ always exists by Zorn's Lemma. Endow $S$ with the discrete topology and place the Borel $\sigma$-algebra on $S\times K$; the latter has a natural measure, given by $$\lambda(A)=\sum_{s\in S}{s(\{x:(s,x)\in A\})}$$

Now, $C(K)^*$ is the space of Borel-regular measures on $K$. By Lebesgue decomposition and Radon-Nikodym, $T:C(K)^*\to L^1(S\times K)$; $$T(\mu)(s,x)=\frac{d\mu}{ds}(x)$$ (where the Radon-Nikodym derivative vanishes if $\mu\perp s$) is an isometric isomorphism. Correspondingly $C(K)^{**}\cong L^{\infty}(S\times K)$, and I now unabashedly identify across both isomorphisms.

In particular, pick any sequence $\{\mu_n\}_{n=0}^{\infty}\in L^1(S\times K)^{\omega_0}$. $\{\mu_n\}_n\overset{*}{\rightharpoonup}0$ iff, for all $f\in C(K)$, we have $$\lim_{n\to\infty}{\iint_{S\times K}{f(x)\mu_n(s,x)\,d^2(s,x)}}=0$$ and $\{\mu_n\}_n\rightharpoonup0$ iff, for all $f\in L^{\infty}(S\times K)$, we have $$\lim_{n\to\infty}{\iint_{S\times K}{f(s,x)\mu_n(s,x)\,d^2(s,x)}}=0$$

With those preliminaries in hand, an easy construction now shows that the topologies of weak and weak-$*$ convergence do not coincide on sequences in $C(K)^*$. Choose any sequence $\{q_n\}_{n=0}^{\infty}\in K^{\omega_0}$ of distinct elements, tending to a limit $q_{\infty}$. W/oLoG, for each $n$, the Dirac mass $\delta_{q_n}\in S$. So let $$\mu_n(s,x)=\begin{cases} 1 & s=\delta_{q_{2n}} \\ -1 & s=\delta_{q_{2n+1}} \\ 0 & \text{otherwise} \end{cases}$$ Then, since any $f\in C(K)$ is continuous at $q_{\infty}$, we can compute $$\iint_{S\times K}{f(x)\mu_n(s,x)\,d^2(s,x)}=f(q_{2n})-f(q_{2n+1})\to0$$ as $n\to\infty$. But just as clearly the pointwise sum $\sum_n{\mu_n}\in L^{\infty}(S\times K)$, so that $$\iint_{S\times K}{\left(\sum_n{\mu_n}\right)(s,x)\mu_n(s,x)\,d^2(s,x)}=2\not\to0$$ as $n\to\infty$.