in a book (Brezis' Functional Analysis book) I saw a proof for showing (strong-to-weak*) continuity of an operator

$$T : (E, \|\cdot\|) \to (E^*, \sigma(E^*,E))$$

where $E$ was a generic Banach space (not necessarily reflexive or separable), and $\sigma(E^*,E)$ is the weak-* topology.

The argument started with "We can argue with sequences (why?)" and then he proceeded to show that for any $x_n\to x$ we obtain $Tx_n \stackrel{*}{\rightharpoonup} T x $.

Why is that possible? In this generic case the weak-* topology does not even allow for metrization of the unit sphere, let alone the whole topology. The only idea I had is the following:

In order to prove that $T$ is continous, it suffices to look at the evaluation functionals on $(E^*, \sigma(E^*, E)$ of the form $\phi_x: E^*\to \mathbb R$ with $\phi_x(f) = \langle f, x\rangle$. Then $T$ is continous iff all $\phi_x \circ T$, for any $x\in E$ are continous as a mapping $(E,\|\cdot\|) \to (\mathbb R, \mathcal B(\mathbb R))$. And this is

$$\phi_x\circ T (y) = \langle Ty, x\rangle,$$

and continuity of this mapping is proven by showing that

$$ y_n \to y \Rightarrow \langle T y_n, x\rangle \to \langle T y, x\rangle ~ \forall x\in E \Leftrightarrow T y_n \stackrel{*}{\rightharpoonup} T y$$

Is this correct? If so, how does this fit together with the platitude "never try to prove continuity with sequences in a non-metrizable topology"?


1 Answer 1


If $X$ is a sequential space then a function $f: X \to Y$ is continuous iff $f$ is sequentially continuous, also for non-sequential $Y$.

In your case the domain has the strong, metrisable topology, and so is sequential.

A proof for the first statement: let $f$ be sequentially continuous, and let $F$ be closed in $Y$; to show $f^{-1}[F]$ is closed in $X$ take a sequence $x_n \to x$ where all $x_n \in f^{-1}[F]$. By definition, all $f(x_n) \in F$, and by sequential continuity of $f$, $f(x_n) \to f(x)$, so $f(x) \in F$ (in all spaces closed implies sequentially closed) and so $x \in f^{-1}[F]$. So $f^{-1}[F]$ is sequentially closed and thus closed (as $X$ is assumed to be sequential and "sequentially closed implies closed for all subsets" is the definition of a sequential space). This shows that for continuity on a sequential space it suffices to check sequential continuity.


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