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.)

Every norm-continuous linear map between normed spaces is weakly continuous (e.g. by universal properties of the weak topologies). Norm continuity follows from weak-sequential continuity because the norm and the weak topology have the same bounded sets.
For locally convex space one has to be more careful.
