Do there exist continuous linear maps $T\colon V \to V$ such that $T^{-1}\colon V\to V$ exists but is not continuous? Clearly if $V$ has finite dimension the answer is no.
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By the open mapping theorem, that can't happen when $V$ is a complete metrizable TVS (over $\mathbb{R}$ or $\mathbb{C}$), but it is easy to construct examples for (some) other situations, e.g. let $V = \mathbb{C}^\infty$ the space of all complex sequences with only finitely many nonzero terms. Endow $V$ with any $\ell^p$ norm you particularly like, and let
$$T(x) = \left(\frac{x_k}{2^k}\right)_{k\in \mathbb{N}}.$$
Since every sequence in $V$ has only finitely many nonzero terms, $T$ is bijective, but $T^{-1}$ is not bounded.
Daniel Fischer
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(+1) Let me just point out to the OP that this sort of thing is no idle curiosity either. Somewhat more generally, if $A$ is a strictly positive unbounded operator with compact resolvent on a separable Hilbert space $H$, then $A^{-1}$ will be a strictly positive compact operator on $H$; this turns out to be an extremely important fact when dealing with say, (generalised) Laplacians on compact domains/manifolds. – Branimir Ćaćić Jul 22 '13 at 14:57