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Let $E$ be a $\mathbb R$-Banach space, $\Theta\subseteq C^{0,\:1}(E,E)$ be a $\mathbb R$-Banach space and $\iota$ be a continuous embedding of $\Theta$ into $C^1(E,E)$.

I would like to show that, given a compact $K\subseteq E$, there is a $c\ge0$ with $$\sup_{x\in K}\left\|{\rm D}(\iota f)(x)\right\|_{\mathfrak L(E)}\le c\left\|f\right\|_{\Theta}\;\;\;\text{for all }f\in\Theta.\tag1$$

If I'm not missing something, $$C^1(K,E):=\left\{\left.g\right|_K: g\in C^1(U,E)\text{ for some open neighborhood }U\text{ of }K\right\}$$ equipped with $$\left\|g\right\|_{C^1(K,\:E)}:=\max\left(\sup_{x\in K}\left\|g(x)\right\|_E,\sup_{x\in K}\left\|{\rm D}g(x)\right\|_{\mathfrak L(E)}\right)\;\;\;\text{for }g\in C^1(K,E)$$ should be a $\mathbb R$-Banach space. If that's true, we may be able to show $$\Theta\ni f\mapsto\left.(\iota f)\right|_K\tag2$$ is a continuous embedding of $\Theta$ into $C^1(K,E)$, from which the desired claim would follow.

Can we show this?

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    $\begingroup$ Re. $C^1(K,E)$ always being complete with the norm you suggest - I might be wrong, but I'm not sure that's even true for all compact subsets of $\mathbb{R}^d$, never mind when $E$ is something more exotic (it is true for all 'nice' compact subsets of $\mathbb{R}^d$ of course). $\endgroup$
    – DCM
    Oct 9, 2020 at 18:20
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    $\begingroup$ Also... what topology do you give to $C^1(E,E)$? Are you mainly interested in the finite dimensional case or do you need to allow $E$ infinite dimensional? $\endgroup$
    – DCM
    Oct 9, 2020 at 18:40
  • $\begingroup$ @DCM I'm interested in the case $E=\mathbb R^d$ as well. But if I'm not missing anything, even in the infinite-dimensional case, $C^1(E,E)$ endowed with the topology induced by compact convergence of the functions and their Fréchet derivatives should always be a Locally convex topological vector space. $\endgroup$
    – 0xbadf00d
    Oct 9, 2020 at 19:41
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    $\begingroup$ Isn't already continuous $C^1(E,E)\ni g\mapsto g_{|K}\in C^1(K,E)$, wrto this topology on $C^1(E,E)$? $\endgroup$ Oct 9, 2020 at 23:45

2 Answers 2

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I agree with Pietro Majer's comment that (1) follows from the continuity of the restrictions $C^1(E,E)\to C^1(K,E)$ -- whatever you mean by this space!

Concerning (2) (in the finite dimensional case $E=\mathbb R^d$): If you define $C^1(K,E)$ as the space of restrictions of $C^1$-functions on open supersets of $K$ then, in general, it isn't complete with respect to the norm $$ \|g\|_K=\sup\{|g(x)|: x\in K\} +\sup\{|Dg(x)|: x\in K\}. $$ This was suspected in the comment of DCM and it is indeed well-known since the work of Whitney.

There is a recent paper of Leonhard Frerick, Laurent Loosveldt and myself (Continuously differentiable functions on compact sets, arXiv:2003.09681) on various definitions of $C^1(K)$. Theorem 5.1 implies that the space of restrictions is complete with respect to the norm above if and only if $K$ has finitely many components which are Whitney regular, i.e., the geodesic distance is equivalent to the euclidean distance. For general $K$ (which is equal to the closure of its interior) one should endow the space of restrictions with the (ugly) quotient norm $$ \|g\|=\inf\{\|f\|_{\mathbb R^d}: f\in C^1(\mathbb R^d), f|_K=g\}. $$ Whitney gave a simpler description for this. For general $K$ one should rather consider the space of Whitney jets $(f|_K, Df|K)$ than just the functions because "the" derivative might not be uniquely defined by a function defined just on $K$.

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  • $\begingroup$ What I mean by $C^1(K,E)$ is the space of functions $K\to E$ which admit continuously differentiable extensions to an open neighborhood of $K$ and I endow this space with $\left\|f\right\|_{C^1(K,\:E)}:=\max\left(\sup_{x\in K}\left\|f(x)\right\|_E,\sup_{x\in K}\left\|{\rm D}f(x)\right\|_{\mathfrak L(E)}\right)$. $\endgroup$
    – 0xbadf00d
    Oct 10, 2020 at 16:54
  • $\begingroup$ Yes, I understood that you mean the spaces of restrictions where it is no difference whether you can extend to some open set or to $\mathbb R^d$ -- just multiply with a cut-off function. But this space isn't Banach in general with the norm you describe. $\endgroup$ Oct 10, 2020 at 18:45
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    $\begingroup$ Upvoted because I like the paper :) $\endgroup$
    – DCM
    Oct 14, 2020 at 18:32
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As an addendum to Jochen Wengenroth's answer: If you are willing to restrict your choice of compact sets somewhat (to those whose interior is dense), then you might find the answers to your questions (for $E$ finite dimensional) in the recent preprint

  • Helge Glockner, Smoothing operators for vector-valued functions and extension operators, arXiv:2006.00254.

Note that the restriction to finite-dimensional spaces here is necessary as on one hand there are no compact sets with nonempty interior in infinite-dimensional spaces and in addition, the differentiability discussed in the paper contains with the Fréchet differentiability you asked for on finite-dimensional spaces.

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