Let $X$ be an infinite dimensional, reflexive and separable real Banach space. Consider a function $f: X \to \mathbb{R}$, and assume $f$ is sequentially continuous with respect to the weak topology, that is, if $a_n \rightharpoonup a$ then $f(a_n) \to f(a)$.
What conditions are needed to show that $f$ is also continuous with respect to the weak topology?
Moreover, let $B_R = \{a \in X \mid \|a\|_X \leq R \}$ endowed with the weak topology, this is a Polish space, and after a metric is defined is a compact metric space. Convergence in the metric is weak convergence.
If $f$ is sequentially weak continuous on $X$, the restriction of $f$ to $B_1$ should be sequentially weak continuous as well, which is sequentially continuous with respect to the metric. In turn, this means that $f$ is continuous on the topology induced by the metric, which is the inherited weak topology on $B_1$. Hence, if $f$ is sequentially weak continuous, the restriction of $f$ to $B_1$ is weak-continuous.
Is the argument right? This is enough for my application, but I am afraid I may be overlooking something.
I guess there are functions that are weak continuous when restricted to closed balls of any radius, but only sequentially continuous in the whole space $X$. Can you think of some particular easy example of this?
Thanks
Best Answer
Your argument is right. Indeed, a function is weakly sequentially continuous iff its restriction to every ball is weakly continuous. You have proved one direction; the converse is given by the fact that every weakly convergent sequence is bounded (uniform boundedness principle).
As to your last question, this Math.SE answer constructs a countable set of points $S = \{x_{k,j} : 1 \le j \le n_k, k \ge 1\}$, with $\|x_{k,j}\| = k$, such that $0$ is in the weak closure of $S$. Let $B_k$ denote the closed ball of radius $k$, which is also weakly closed. By induction and the Tietze extension theorem, we can find a sequence of weakly continuous functions $f_k : B_k \to \mathbb{R}$ such that $f_k(0) = 0$, $f_k |_{B_{k-1}} = f_{k-1}$, and $f_k(x_{k,j}) = 47$ for $1 \le j \le n_k$. If we let $f : X \to \mathbb{R}$ be given on $B_k$ by $f_k$ (which is well defined), then by construction $f$ is weakly continuous on each $B_k$, hence weakly sequentially continuous as noted above. But $f$ is not weakly continuous since $f = 47$ on $S$ but $f(0)=0$.