# Is the topological dual of a Banach space weakly* closed in its algebraic dual?

The question is completely contained in the title :) I can only add, that it is not difficult to give a counterexample for normed spaces, and also Banach-Steinhaus theorem implies the sequential closeness. However, I don't think that this argument can be extended beyond sequences. Thank you.

• More precisely, the topological dual is weak$^*$ dense in the algebraic dual. The axiom of choice thus implies that it is not closed. – Jochen Wengenroth Mar 10 '16 at 7:19