Sequentially continuous but not continuous linear map $(X^*, w^*)$ to $(Y^*,w^*)$

Question

Let $X, Y$ be Banach spaces and let $T : (X^*, w^*) \rightarrow (Y^*,w^*)$ be a linear map. Suppose that $T$ is sequentially continuous. Must $T$ be weak*-to-weak*-continuous ?

Answer 1

The answer is sometimes "no", even when $Y$ is one-dimensional.

Let's say the scalars are $\mathbb R$; the same argument will work for $\mathbb C$.

Let $\omega_1$ be the least uncountable ordinal. Let $K = [0,\omega_1]$ with the order topology, a compact Hausdorff space. This is chosen so that the subset $[0,\omega_1)$ is sequentially closed but not closed. The characteristic function of the singleton $\{\omega_1\}$ is a sequentially continuous but not continuous function from $K$ to $\mathbb R$.

Now Banach spaces $X = C(K)$ and $Y = \mathbb R$ will be used in our example.

Note that every $f \in C(K)$ is constant in some neighborhood of the endpoint $\omega_1$. Compute the dual $X^* = M(K) = l^1(K)$; that is: countably-additive signed measures $\mu$ on $K$ with countable support. Define the map $T : M(K) \to \mathbb R$ mapping the signed measure $\mu$ to the value $\mu(\{\omega_1\})$. So $T$ takes the $\omega_1$-component of the $l_1(K)$ function.

Claim: $T$ is w* sequentially continuous. Let $\mu_n, \mu \in M(K)$ with $\mu_n \to \mu$ for the w* topology. We must show $T(\mu_n) \to T(\mu)$. There exists a countable ordinal $\alpha$ so that $\mu$ and all $\mu_n$ have support in $[0,\alpha] \cup \{\omega_1\}$. [The supremum of countably many countable ordinals is countable.] There is a continuous function $f \in C(K)$ that vanishes on $[0,\alpha]$ and has $f(\omega_1)=1$. By w* convergence we have $\int f\;d\mu_n \to \int f\;d\mu$. Thus we get $T(\mu_n) \to T(\mu)$.

Finally we claim that $T$ is not w* continuous. For each $\alpha \in [0,\omega_1]$, let $e_\alpha \in X^*$ be the point-mass at $\alpha$. That is, $e_\alpha(f) = f(\alpha)$ for all $f \in C(K)$. Then the net $(e_\alpha)_{\alpha < \omega_1}$ converges w* to $e_{\omega_1}$. [Because the elements of $C(K)$ are continuous at the point $\omega_1$.] But $T(e_\alpha)=0$ for all $\alpha < \omega_1$ and $T(e_{\omega_1}) = 1$. So $T$ is not w* continuous.