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Let $\mathcal{F}$ be a Fréchet space (locally convex, Hausdorff, metrizable, with a family of seminorms ${\|~\|_n}$).

I've read that the dual $\mathcal{F}^*$ is never a Fréchet space, unless $\mathcal{F}$ is actually a Banach space. I'd like to know in which ways the dual can fail to be Fréchet in general; for example, is it always incomplete as a metric space? Or is it always non-metrizable? What are the real issues here?

If there's a relatively easy proof of this fact (that $\mathcal{F}^*$ is not Fréchet unless $\mathcal{F}$ is Banach), I would appreciate it as well.


[EDIT: The reference cited by Dirk contains a Theorem whose proof is inaccessible to me, so I'd upvote/accept an answer that at least sketches such a proof, or provides another way of settling the question.

I'd also be interesting in any further explanations regarding the statement that "(...) a LCTVS cannot be a (non-trivial) projective limit and an inductive limit of countably infinite families of Banach spaces at the same time.", made by Andrew Stacey in that link, which is related to my question.]

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There is a related question on mathoverflow:… – Dirk Feb 21 '12 at 16:06
@Dirk thanks for pointing that out! – student Feb 21 '12 at 16:08
up vote 6 down vote accepted

The answer depends a bit on what you mean by "is a Frechet space": If you consider $F^*$ just as a vector space there might be complete norms on it which, however, have nothing to do with the duality. If $F$ itself is not normed there is no complete norm $p$ on $F^*$ whose topology is finer than the weak-star-topology: Since every continuous linear functional on $F$ is continuous with respect to some $\|\cdot\|_n$ one has $F^*=\bigcup_n A_n$ where $A_n=\{f\in F^*: |F| \le n\|\cdot\|_n \}$ is weak-star-compact by Alaoglu's theorem and hence $p$-closed. Now, Baire's theorem implies that some $A_N$ has $p$-interior points and the convexity of $A_N$ implies that $0$ is an interior point. This implies $F^* = \bigcup_m m A_N$, that is, every $f\in F^*$ is continuous with respect to the same seminorm $\| \cdot\|_N$. This only holds for normed spaces.

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What I mean by "is a Fréchet space" is, of course, the usual definition: a locally convex space which is Hausdorff, metrizable and complete. I'm confused about what you're trying to prove here. You say $\mathcal{F}^*$ has no complete norm whose topology is finer than $weak^*$; what does this imply? – student Feb 24 '12 at 15:38
@Leandro: Dear Leandro, Jochen's comments are related to the question of which topology you endow $\mathcal F^*$ with. The point is that $\mathcal F^*$ can be endowed with several natural topologies, the coarsest of which is the weak-star topology (the topology of pointwise convergence). Jochen is showing that none of these natural topologies can be Frechet. Regards, – Matt E Feb 24 '12 at 16:02

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