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Let $E$ be a Hausdorff topological vector space and $f:\mathbb{R} \to E$ a function. Let us call $f$ differentiable at $x$ if $$ \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} $$ exists (as a limit in $E$). If this limit exists, then $$ \lim_{h \to 0} \frac{e'(f(x+h))-e'(f(x))}{h} $$ exists for any $e' \in E'$, i.e. $e' \circ f$ is differentiable at $x$ in the usual sense.

Is the reverse true? Under which conditions?

My guess would be that if $E'$ separates points and $E$ is complete (or maybe only sequentially complete), then the existence of all these "scalar derivatives" would imply the existence of the derivative as $E$-valued function, but I'm not able to show this or find a reference.

mixotrov
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    I guess Theorem 1.1 of https://www-users.cse.umn.edu/~garrett/m/fun/notes_2016-17/holomorphic_vector-valued.pdf might be useful for you. – Severin Schraven Nov 19 '24 at 20:43
  • That does indeed seem like what I want! On first inspection, though, the proof seems a bit odd/incomplete and seemingly needs the weak regularity to be 2 degrees higher. I'll look at it a bit more carefully, since his notes are usually very clear. – mixotrov Nov 19 '24 at 21:27
  • It is a bit annoying that this notion is called weakly differentiable as one mostly gets results about Sobolev spaces on google. I do not know whether weak differentiability implies strong differentiability. So far I have only used that weakly holomorphic implies holomorphic in Hilbert spaces. I have never worked with topological vectors spaces. – Severin Schraven Nov 19 '24 at 22:09
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    This https://math.stackexchange.com/questions/4255039/for-a-banach-space-e-is-it-sufficient-to-check-differentiability-of-f-u-to seems to answer your question in the negative (even for Hilbert spaces). – Severin Schraven Nov 19 '24 at 22:12
  • This seems to settle it: weak $C^k$ does not imply strong $C^k$, but does imply strong $C^{k-1}$, if $E$ is quasi-complete. – mixotrov Nov 20 '24 at 07:36

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To summarize:

No: being weakly $C^k$ does not imply being strongly $C^k$

Yes: being weakly $C^{k+1}$ does imply being strongly $C^k$, if $E$ is Hausdorff and quasi-complete. A full proof can be found in Schwartz, L. Espaces de fonctions différentiables a valeurs vectorielles. J. Anal. Math. 4, 88–148 (1954), particularly Lemme II on page 146.

mixotrov
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