# Sequential continuity implies continuity in the weak topology on a normed space

Let $X$ be a normed vector space (over $\mathbb{R}$ or $\mathbb{C}$) and let $f$ be a linear functional on $X$ that is not necessarily continuous. If for any sequence $(x_n)$ that converges to $x$ weakly, we have $\lim f(x_n)=f(x)$, does it follow that $f$ is continuous in the weak topology?

In other words, does sequential continuity imply continuity in the weak topology?

(We know that the weak topology need not be first countable, so a priori we cannot characterise continuity of linear functionals in terms of sequences.)

• I deleted my earlier comment because I feel I didn't understand the question at first. I think you mean that $f$ is a linear functional that is not necessarily continuous, but it is weakly sequentially continuous. Then your question is whether such a function is automatically weakly continuous. Is this the right interpretation? – Josse van Dobben de Bruyn Mar 10 '16 at 7:33
• What is $f$? What is its domain and codomain? Is it linear? – Eric Wofsey Mar 10 '16 at 7:43
• @JossevanDobbendeBruyn yes, that is what I mean! – yoyostein Mar 10 '16 at 7:46
• Great! :-) I hope you will consider making the question a little clearer (and more precise), even if you feel it has been sufficiently answered. I recognise the questions raised by @EricWofsey: at first I thought that $f$ was an element of $X^*$, and Jochen still doesn't know whether the question pertains to normed spaces or locally convex (or even arbitrary) topological vector spaces. – Josse van Dobben de Bruyn Mar 10 '16 at 8:14
• I feel that this was a perfectly legitimate question to ask on this site, and I think that it should be reopened now that the details are there. But I haven't earned the privilege "vote to reopen" yet, and in any case we would need 5 such votes. I think we need at least one reopen vote to get it in the reopen queue. Who would be so kind? :-) – Josse van Dobben de Bruyn Apr 13 '16 at 16:01