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



Sequentially weak-continuous and continuous functions under the weak-topology of a Banach space
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](https://math.stackexchange.com/a/2030014/822) 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 $f(0)=0$.


