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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?

Arctic Char
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    The argument used for normed linear spaces works for mertizable tvs's. – Kavi Rama Murthy Sep 16 '20 at 23:25
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    It seems that there are (non-metrisable, as Kavi Rama Murthy points out) infinite-dimensional locally convex spaces without this property: https://en.wikipedia.org/wiki/Discontinuous_linear_map#Beyond_normed_spaces – Paweł Czyż Sep 16 '20 at 23:27
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    I do not think this question should be closed. In my view it touches upon an interesting topic, it is decently written, it's been upvoted and it has two answers, one of them accepted. Please consider reopening it! – Ruy Sep 17 '20 at 13:33

2 Answers2

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

Ruy
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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.

uniquesolution
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