# Is every sequentially $\sigma(E',E)$-continuous linear functional on a dual Banach space $E'$ necessarily a point evaluation?


Given a duality as above, one defines the weak topology on $$F$$, usually denoted $$\sigma (F,E)$$, to be the coarsest topology according to which the linear functionals $$y\in F\mapsto \langle x, y\rangle \in \bf K$$ are continuous for every $$x$$ in $$E$$.

It is a classical fact that every $$\sigma (F,E)$$-continuous linear functional $$\varphi :F\to \bf K$$, may be represented by a vector in $$E$$ in the sense that there exists a (necessarily unique) $$x$$ in $$E$$ such that $$\phi(y) = \langle x, y\rangle ,\quad\forall y\in E.$$

Question. Does the above still hold if continuity is replaced by sequential continuity. In other words, must every sequentially $$\sigma (F, E)$$-continuous linear functional on $$F$$ be represented by a vector in $$E$$.

Before the reader jumps to the task of proving or disproving it, let me say that unfortunately the answer is negative, a counter example being presented below.

So let me specialize this a bit by restricting to the situation in which $$E$$ is a Banach space and $$F$$ is its topological dual, with the canonical duality $$\langle x, \varphi \rangle = \varphi (x), \quad \forall x\in E, \quad \forall \varphi \in E'.$$

To be precise:

Question. Let $$E$$ be a Banach space and let $$\varphi$$ be a linear functional on $$E'$$ which is sequentially $$\sigma (E',E)$$-continuous. Is $$\varphi$$ necessarily represented by a vector in $$E$$?

This is obviously true if $$E$$ is reflexive and I think I can also prove it for $$E=c_0$$, as well as for $$E=\ell ^1$$.

A COUNTER EXAMPLE

Let $$E=\sc F(H)$$ be the set of all finite-rank operators on Hilbert's space, and $$F=\sc B(H)$$, with duality defined by means of the trace, namely $$\langle S, T\rangle = \text{tr}(ST), \quad\forall S\in \sc F(H), \quad\forall T\in \sc B(H).$$

In this case $$\sigma \big (\sc B(H),\sc F(H)\big )$$ turns out to be the weak operator topology (WOT), which coincides with the sigma weak operator topology ($$\sigma$$-WOT) on bounded subsets of $$\sc B(H)$$.

Since WOT-convergent sequences are bounded by Banach-Steinhauss, we have that the WOT-convergent sequences are the same as the $$\sigma$$-WOT convergent ones. It follows that every $$\sigma$$-WOT-continuous linear functional on $$\sc B(H)$$ is also WOT-continuous. Making a long story short, for every trace class operator $$S$$ on $$H$$ of infinite rank, the linear functional $$T\in \sc B(H) \mapsto \text{tr}(ST)\in {\bf C}$$ is sequentially WOT-continuous, but it is not represented by an operator in $$\sc F(H)$$.

• This is at least true if $E$ is separable. Indeed, in that case, by the Krein-Smulian theorem, a convex subset of $E'$ is $\sigma(E',E)$-closed if and only if it is $\sigma(E',E)$ sequentially closed. See Corollary V.12.7 in Conway's course in functional analysis. Apply this to the kernel of $\varphi$. – Mikael de la Salle Jan 10 at 19:42
• @Mikael, thanks for pointing it out. In fact the following Corollary in Conway's book, namely V.12.8, is exactly what I am asking! – Ruy Jan 10 at 19:58
• Indeed, I had missed this Corollary. Note however that separability of $E$ is assumed, and used essentially in the proof. – Mikael de la Salle Jan 10 at 20:03
• Related: mathoverflow.net/q/284276/61785 – Robert Furber Jan 11 at 7:57
• Side note: These questions can bump into set-theoretic undecidability. Consider $\ell^\infty(\mathbb{R})$ (the bounded, not required to be measurable, functions $\mathbb{R} \rightarrow \mathbb{R}$) as the dual space of $\ell^1(\mathbb{R})$. There is a sequentially weak-* continuous functional that is not weak-* continuous (i.e. a point evaluation) if and only if there is an atomless real-valued measurable cardinal (i.e. $(\mathbb{R}, \mathcal{P}(\mathbb{R}))$ admits a probability measure vanishing on singletons). – Robert Furber Jan 11 at 8:36

Mikael de la Salle points out this is true when $$E$$ is separable, as shown in Corollary V.12.8 of Conway, A Course in Functional Analysis, 2e.
For a non-separable counterexample, consider the uncountable ordinal space $$[0, \omega_1]$$, which is compact Hausdorff, and $$E = C([0, \omega_1])$$. By the Riesz representation theorem, $$E'$$ is the space of signed Radon measures $$\mu$$ on $$[0, \omega_1]$$ with its total variation norm. Let $$\varphi(\mu) = \mu(\{\omega_1\})$$. This is clearly not represented by any vector in $$E$$ since the function $$1_{\{\omega_1\}}$$ is not continuous, but I claim $$\varphi$$ is sequentially $$\sigma(E', E)$$ continuous.
Let $$\mu_n$$ be a sequence converging to 0 in $$\sigma(E', E)$$ and fix $$\epsilon > 0$$. Since each $$\mu_n$$ is Radon, so is its total variation measure $$|\mu_n|$$, and thus we can approximate $$\{\omega_1\}$$ in $$|\mu_n|$$-measure from outside by open sets. So there exists $$\alpha_n < \omega_1$$ such that $$|\mu_n|((\alpha_n, \omega_1)) < \epsilon$$. Let $$\alpha = \sup_n \alpha_n < \omega_1$$; then $$|\mu_n((\alpha, \omega_1))| \le |\mu_n|((\alpha, \omega_1)) < \epsilon$$ for every $$n$$.
Define $$f : [0, \omega_1] \to \mathbb{R}$$ by $$f(x) = \begin{cases} 0, & x \le \alpha \\ 1, & x > \alpha \end{cases}$$ and note that $$f$$ is continuous. Now $$\varphi(\mu_n) = \mu_n(\{\omega_1\}) = \mu_n((\alpha, \omega_1]) - \mu_n((\alpha, \omega_1)) = \int f\,d\mu_n - \mu_n((\alpha, \omega_1)).$$
But by assumption $$\int f\,d\mu_n \to 0$$, and $$|\mu_n((\alpha, \omega_1))| < \epsilon$$, so we conclude $$\varphi(\mu_n) \to 0$$.