# 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.

• You should specify what you mean by continuity of $T$. Should $T$ be continuous with respect to the weak topologies? – Sergei Akbarov Oct 24 '17 at 10:39
• I have already said that $X^*$ and $Y^*$ are equipped with weak* topology. So everything above is with respect to these topology, be it the continuity of $T$ or the metrizability of closed balls of $X^*$. – Manish Kumar Oct 24 '17 at 11:29
• This is Corollary 4.14 in my book. – Nik Weaver Oct 24 '17 at 18:04

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 $X^*$.
• @DJA I use the separability to go from sequentially closed to weak-* closed. What I'm using is the fact that the weak-* topology on bounded subsets of the dual of a separable Banach space is metrizable, so the intersection of the kernel of $f(y)$ with each closed bounded set is closed. Krein-Šmulian/Banach-Dieudonné can then be applied to deduce that the kernel of $f(y)$ is weak-* closed. – Robert Furber Nov 13 '19 at 17:46
• @DJA It is not true in general that sequential continuity implies continuity. There is a nice counterexample in Solovay's model where all sets are Lebesgue measurable: the linear map $\int : \ell^\infty([0,1]) \rightarrow \mathbb{R}$ defined by Lebesgue integration is sequentially continuous with $\ell^\infty([0,1])$ given the weak-* topology as the dual of $\ell^1([0,1])$, essentially by the dominated convergence theorem, but not continuous. – Robert Furber Nov 13 '19 at 17:56
• @DJA An example that works in ordinary set theory is to take $X = \aleph_1$ with the order topology. Then by the Riesz representation theorem and the fact that $X$ is a scattered space, $C_0(X)^* \cong \ell^1(X)$, which defines a weak-* topology on $\ell^1(X)$. Then the mapping defined by summation $\ell^1(X) \rightarrow \mathbb{R}$ is sequentially continuous but not continuous. – Robert Furber Nov 13 '19 at 18:01