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Consider locally convex vector spaces $X,Y$ and let $A: X\to Y$ be a linear operator with closed graph. Closed Graph Theorem states that if $Y$ is Fréchet then $A$ is continuous.

What is a counterexample to this in the case when $Y$ is not Fréchet?

EDIT: Thank you very much for your comments and shots!;) My question was of course about a counterexample where $X$ is "good" (let it be barrelled).

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    $\begingroup$ The Closed Graph Theorem requires some conditions on $X$ as well, e.g. that it is a barrelled space. Otherwise, for a counterexample take any closed unbounded (partially defined) operator $A$ in a Banach space $Y$, and let $X = {\mathscr D}(A)$ with the topology of $Y$. $\endgroup$
    – Robert Israel
    Commented Nov 10, 2014 at 18:03
  • $\begingroup$ Robert, you shot first! ;-) $\endgroup$
    – Alain Valette
    Commented Nov 10, 2014 at 18:21
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    $\begingroup$ In Proposition 4.13 and Example 4.14 of that paper: arxiv.org/pdf/math/0612398.pdf you have a very explicit counter-example to the open mapping theorem, where $X$ is a (non-separable) Hilbert space, $Y$ is locally convex, and $A:X\rightarrow Y$ is linear continuous, bijective, with non-continuous inverse. $\endgroup$
    – Alain Valette
    Commented Nov 10, 2014 at 18:24
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    $\begingroup$ For an infinite-dimensional Hilbert space $X$ consider $Y=X$ endowed with the finest locally convex topology and the identical map. On the other hand, if the domain $X$ is ultrabornologial the closed graph theorem holds for so-called webbed spaces introduced by de Wilde. This is a very large class of locally convex spaces which contains all Banach spaces and is stable w.r.t. closed subspaces, separated quotients, and countable products and direct sums. See, e.g., Introduction to Functional Analysis of Meise and Vogt. $\endgroup$
    – Jochen Wengenroth
    Commented Nov 11, 2014 at 7:50

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