Locally convex topological subspace on dual space

Question

This is an exercise in Lax. Let $S=\{l_\alpha\}$ be a set of functionals that separate points, show that the weakest topology that all functionals in S are continuous is locally convex. I know this looks like weak star topology, but S may not be the whole dual space.

Answer 1

By definition a basic neighborhood of $0$ is of the type $\{x: |l_{\alpha_1 }(x)<\epsilon_1,|l_{\alpha_2 }(x)|<\epsilon_2,...,|l_{\alpha_n }(x)| <\epsilon_n)\}$ and this set is convex. Hence any neighborhood of $ 0$ contains a convex neighborhood of $0$.