Sequential Continuity in dual spaces of separable Banach Spaces Is the following true?
Let $X$ and $Y$ be separable Banach spaces and consider their dual spaces $X^*$ and $Y^*$ equipped with weak* topology. Suppose that a linear map $T:X^*\to Y^*$ is sequentially continuous. Then, is it true that $T$ is continuous?
I suspect this may be true using the following two facts:
(i) Any closed ball in $X^*$ (and in $Y^*$) is metrizable as $X$ and $Y$ are separable.
(ii) A convex set in $X^*$ (and in $Y^*$) is weak* closed iff it is sequentially closed (by Krein-Smulian theorem).
From (i), we have that $T$, when restricted to a closed ball, is continuous. But I am not able to go further from this.
Any help or hint would be appreciated. Thank you.
 A: This is true. To show it, in the following I will use $\langle \mbox{-}, \mbox{-} \rangle$ for the pairing between a space and its dual (with the vectors from the space on the left, the dual on the right).
For each $y \in Y$, if we define $f(y) = \langle y, \mbox{-}\rangle \circ T$, then $f(y)$ is a sequentially continuous linear functional on $X^*$. Its kernel is therefore a sequentially closed convex set, so is weak-* closed in $X^*$ (by, according to taste, Krein-Šmulian or Banach-Dieudonne or Grothendieck's completeness theorem), and therefore $f(y)$ is actually weak-* continuous. Therefore there exists a $g(y) \in X$ such that $\langle g(y), \mbox{-}\rangle = f(y)$. It is then not difficult to show that $g(y)$ is a linear map $Y \rightarrow X$ such that $\langle g(y), \phi \rangle = \langle y, T(\phi) \rangle$ for all $y \in Y$ and $\phi \in X^*$. This implies that $T$ is continuous from the weak-* of $X^*$ to the weak-* of $Y^*$ (not hard to prove directly, but a reference is Schaefer's Topological Vector Spaces IV.2.1).
Note, however, that the analogous statement for incomplete normed spaces is false -- any element $x$ of the completion of a normed space $X$ that is not part of $X$ defines a linear functional that is weak-* continuous on the ball of $X^*$ but not on $X^*$.
