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?



2 Answers 2


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 $f(0)=0$.


Added in proof: After writing this answer and before posting I noticed that Nate Elderge already answered the OP. I decided to post my answer anyway, as it contains a "particulr easy example", as requested in the OP.

So, here it is:

Your argument is right. It is closely related to the Eberlin-Smulian theorem. Note that by Banach-Steinhaus every weakly converges sequence is actually bounded, so a function which is weakly sequentially continuous is exactly (a function which is weakly sequentially continuous on any ball, hence) a function which is weakly continuous on any ball.

Thus, your question translates to "can one find a function which is weekly continuous on every ball but not weekly continuous". Here is an example: $$ \ell^2\to \mathbb{R},\quad (a_n)_{n=1}^\infty \mapsto -\sum_{n=1}^\infty \min\{n-\sum_{k=1}^n |a_k|^2,0\}. $$ Note that on every ball this sum is finite, hence the restricted function is continuous. However this function is not globally continuous, as one can observes using the unbounded converging net constructed in another MSE answer of Nate Elderge.


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