Discontinuous functionals on infinite dimensional topological vector space

Question

Using the Axiom of choice, one can show that (see here) every infinite dimensional normed vector space has discontinuous functionals. My question is: Is this also true for a general topological vector spaces?

Answer 1

No, it is not true for general TVS. Given any vector space $X$, it is possible to construct a Hausdorff locally convex topology $T$ on $X$ that has the property that any convergent sequence in $(X,T)$ is contained in some finite-dimensional subspace of $X$. This guarantees that every linear functional on $(X,T)$ is continuous. A base of neighbourhoods of zero for such a topology can be taken as the family of all absolutely convex absorbing subsets of $X$. (We can choose $X$ to be initially infinite-dimensional)

Such a topology is not metrizable. As already mentioned in the comments, every infinite-dimensional metrizable TVS admits a discontinuous linear functional.

Answer 2

Every linear functional on the vector space $X$ is continuous relative to the topology $\sigma(X,X^*)$, where $X^*$ is the algebraic dual of $X$.