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I'm reading through some functional analysis lecture notes and there the closed graph theorem was stated in the following form:

Let $X$ be a Baire locally convex space and $Y$ a Frechet space. If the graph of a linear map between $X$ and $Y$ is closed in $X\times Y$, then this map is continuous.

The text goes than on to say that if just one of these spaces in Banach, the theorem doesn't hold anymore. But how can this be true ? Banach spaces are in particular locally convex and Frechet, so the theorem would have to hold.
If someone indeed somehow comes up with what I'm not seeing here could he/she please provide the construction of this counterexample ?

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  • $\begingroup$ Here is a related question. $\endgroup$
    – David Mitra
    Commented May 21, 2014 at 14:08
  • $\begingroup$ The point is: "if just one of these spaces is Banach". It's not formulated immaculately clearly. The meaning is that if one of the two spaces is a Banach space, and the other arbitrary, then there can be linear maps with closed graphs that are nevertheless discontinuous. If $Y$ is Banach and $X$ is Baire and locally convex, the conclusion holds. If $X$ is Banach, it is in particular Baire and locally convex, so then if $Y$ is Fréchet, the conclusion holds. If $Y$ is Banach and $X$ arbitrary, or $X$ Banach and $Y$ arbitrary, it need not hold. $\endgroup$
    – Daniel Fischer
    Commented May 21, 2014 at 14:13
  • $\begingroup$ @DavidMitra Thanks. If someone could provide a counterexample for this theorem I would be most grateful... $\endgroup$
    – user38178
    Commented May 21, 2014 at 17:23
  • $\begingroup$ Here's one. $\endgroup$
    – David Mitra
    Commented May 21, 2014 at 17:35

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