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When introducing differentiable functions between locally convex spaces, many authors (e.g. Bastiani, Keller, Kriegl-Michor) notice that the evaluation map $$ E\times E^*\to\mathbb R,\qquad (x,L)\mapsto L(x) $$ is not continuous with respect to any locally convex topology on $E^*$, where $E$ is a non normable locally convex space and $E^*$ is the set of all continuous linear functionals $E\to\mathbb R$. Then they argue that, for this reason, it is not good to define $f\colon E\to\mathbb R$ to be continuously differentiable by requiring $$ x\mapsto Df(x)\in E^*$$ to be continuous with respect to some locally convex topology on $E^*$ (say, for example, the finest locally convex topology, in order to make the strongest assumption).

My question is: what's the problem with this definition? More precisely, what is an example of missing properties of $f\in C^1$ defined as above? I was thinking about continuity of $f$ being not implied by $f\in C^1$, or maybe the failure of the chain rule, but I didn't find an explicit issue. For example, the definition is strong enough to have a mean value theorem $$ f(x+h)-f(x)=\int_0^1 Df(x+th)h\,dt$$

Moreover, which of these classical properties actually require the evaluation map to be continuous?

I am particularly interested in the case of $E$ Fréchet space.

[cross-posted from MSE]

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  • $\begingroup$ Without remembering any details, one would like to have continuity of the directional derivatives $(x,v)\mapsto Df(x)(v)$ which would not follow from the continuity of $x\mapsto Df(x)$. $\endgroup$ Commented May 16, 2022 at 14:20
  • $\begingroup$ Yes, that is actually the definition used by Bastiani, Hamilton etc; I'm interested in what we lose by asking this "weak" continuity. For example, if the dual topology is finer than the weak* one (that is, always), we have continuity separately in the two arguments, which suffices for a number of proofs (e.g. the mean value theorem) $\endgroup$ Commented May 17, 2022 at 8:23
  • $\begingroup$ I have to comment that the theorem (from Baire, etc.) that separate continuity for bilinear maps on Frechet spaces implies joint continuity, which is indeed often handy, does also allow people (me, for example) to become lazy or sloppy in keeping track of issues such as you mention. $\endgroup$ Commented Jun 2, 2022 at 21:51
  • $\begingroup$ For spaces of distributions the bilinear pairing is not continuous but it is hypocontinuous. Why isn't this enough to do differential calculus? $\endgroup$ Commented Jun 2, 2022 at 23:45
  • $\begingroup$ @paulgarrett in this case the dual of a Fréchet space is NEVER a Fréchet space, hence that theorem cannot be of use $\endgroup$ Commented Jun 6, 2022 at 19:32

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