TITLE: What is an example of a bounded, discontinuous linear operator between topological vector spaces?
QUESTION [6 upvotes]: I am thinking there might be an example between the space of compactly supported smooth functions on the real line (chosen because it is non-metrizable under the standard topology for this space of test functions) and $L^{1/2}[0,1]$ (chosen because it is not locally convex).

REPLY [12 votes]: Let $E$ be an infinite dimensional Banach space or Fréchet space whose dual has uncountable Hamel basis. Let $F$ be the same space endowed with its weak topology. The identity $\operatorname{id} \colon F \to E$ is bounded (every weakly bounded set is strongly bounded by Mackey's theorem - Banach and Fréchet spaces carry their Mackey topology), but not continuous (the strong topology is strictly finer than the weak topology; for Banach spaces it follows directly because every weak neighbourhood of $0$ contains an infinite dimensional subspace, for Fréchet spaces with big enough dual, you can for every countable family $\mathcal{U} = (U_n)$ of weak $0$-neighbourhoods find a continuous linear form $\lambda$ that is not in the span of the forms used to determine the $U_n$, and $\bigcap \mathcal{U}$ then contains a nontrivial subspace not contained in $\ker \lambda$, whence $\{x\colon \lvert \lambda(x)\rvert < 1\}$ does not contain any $U_n$ [that reasoning applies of course also to infinite dimensional Banach spaces, their dual is an infinite dimensional Banach space, hence has uncountable Hamel basis]).