$\newcommand{\bf}[1]{\mathbb #1}\newcommand{\sc}[1]{\mathscr #1}$
A *duality* between two vector spaces $E$ and $F$ over $\bf K$ ($= {\bf R}$ of ${\bf C}$)
is, by definition, a bilinear form
$$
 \langle \cdot, \cdot\rangle :E\times F\to \bf K,
 $$
such that, if $\langle x, y\rangle =0$ for every $x$ in $E$, then $y=0$. And vice-versa.


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
$$
 \phi(y) = \langle x, y\rangle ,\quad\forall y\in E.
 $$


One could therefore ask:


**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 $\sc F(H)$.



Is every sequentially $\sigma(E',E)$-continuous linear functional on a dual Banach space $E'$ necessarily a point evaluation?
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 $\varphi(\mu\_n) \to 0$.


