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I'm considering a situation where I have the linear restriction map of Fréchet spaces $$ C^\infty(C_1) \to C^\infty(C_2) $$ where $C_2 \hookrightarrow C_1$ are a pair of compact, connected subsets of $\mathbb{R}^n$ homeomorphic to closed balls, and interiors diffeomorphic to open balls. I believe I can assume that $C_2$ is a manifold with at most codimension 3 corners and $C_2$ with at most codimension 2 corners.

What I'm interested in is whether this has a (Edit: continuous!) linear section.

The case of $n=1$, restriction along $[a,b] \hookrightarrow [c,d]$, I believe I have the requisite understanding to extract as a corollary from a theorem of Seeley (use $n=0$ in the result that the restriction $C^\infty(\mathbb{R}^{n+1}) \to C^\infty(\mathbb{R}^n\times\mathbb{R}_{\geq 0})$ has a linear section, using the usual Fréchet topologies -- thanks to Andrew Stacey for pointing this out), but I don't know how one would go about the more general case.

In looking around I find a lot of work by Fefferman on the case of $C^k$ maps, and also a lot of work by people considering general extension problems for inclusions $A \hookrightarrow \mathbb{R}^n$ and arbitrary functions $A \to \mathbb{R}$ for all different sorts of subsets $A$, including very diverse examples.

Nothing I've found though seems to be the sort of thing I'd need, but that may be my unfamiliarity with this sort of analysis. Ideally, the necessary result is right under my nose, and it just needs someone to say "oh, that clearly follows from so-and-so's theorem".

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  • $\begingroup$ For instance, this paper of Fefferman and Ricci ("Some examples of $C^\infty$ extension by linear operators") considers much more wild subsets than I'm interested in. $\endgroup$
    – David Roberts
    Feb 29, 2016 at 1:15
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    $\begingroup$ Jochen Wengenroth seems to be our resident expert on this sort of thing, see his answer to this question. If he doesn't see this question you could try asking him directly. $\endgroup$
    – Nik Weaver
    Feb 29, 2016 at 4:13
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    $\begingroup$ I'm sure I missed something. It suffices to have a continuous linear extension operator $\mathcal{E}: C^\infty(C_2) \rightarrow C^\infty(\mathbb{R}^n)$, because you can then just restrict to $C_1$. Does the extension operator constructed in Stein's book "Singular Integrals and Differentiability Properties of Functions" not suffice? $\endgroup$
    – Deane Yang
    Feb 29, 2016 at 4:19
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    $\begingroup$ It's Theorem 5 in section VI.3 (page 181). It works on Lipschitz domains, which I believe includes your case. $\endgroup$
    – Deane Yang
    Feb 29, 2016 at 5:06
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    $\begingroup$ @DeaneYang ok, now I have educated myself about the literature surrounding the problem, and I agree I was hasty about brushing off your comment. Thanks for your patience! $\endgroup$
    – David Roberts
    May 5, 2016 at 9:49

1 Answer 1

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In general, there are several candidates for the definition of $C^\infty(K)$: One is the space $\lbrace f|_K: f\in C^\infty(\mathbb R^n)\rbrace$ of all restrictions (endowed with the quotient topology), another is the intersection $\bigcap\limits_{k\in\mathbb N_0} \lbrace f|_K: f\in C^k(\mathbb R^n)\rbrace$ (which is equal to the former for $n=1$ due to Merrien but different in general — an elementary example is in Wieslaw Pawlucki, On the algebra of functions $\scr C^k$-extendable for each $k$ finite, Proc. Amer. Math. Soc. 133 (2005), no. 2, 481–484), and finally the probably best understood definition is that of Whitney jets, i.e. families $(f^{(\alpha)})_{\alpha \in \mathbb N_0^d}$ of continuous functions which satisfy the correct Taylor approximations on $K$ as if $f^{(\alpha)}=\partial^{\alpha} f$ for some $f\in C^\infty(\mathbb R^n)$.

If $K$ is the closure of its interior the definitions coincide and you should consult the literature about extension of Whitney jets. The article Leonhard Frerick, Extension operators for spaces of infinite differentiable Whitney jets, J. Reine Angew. Math. 602 (2007), 123–154, contains a lot of information. As mentioned by Deane Yang. Lipschitz boundary is enough for having a continuous linear extension operator (this is due to E.M. Stein). However, a sharp cusp like $K=\lbrace (x,y)\in [0,1]^2: y\le \exp(-1/x)\rbrace$ does not have such an extension. For general $K$ and the space of all restrictions, the question is wide open, besides the examples of Fefferman and Ricci mentioned by David Roberts there are some results of Dietmar Vogt, Restriction spaces of $A^\infty$, Rev. Mat. Iberoam. 30 (2014), no. 1, 65–78.

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  • $\begingroup$ I've always wondered whether Stein was the first to do this for a domain with smooth boundary or whether his Lipschitz boundary theorem was the first to do this case. $\endgroup$
    – Deane Yang
    Feb 29, 2016 at 15:15
  • $\begingroup$ I am hoping my case is the one where the smaller set is the closure of its interior, and these closed sets arise from rather flexible geometric considerations, so they can be tweaked a little to avoid cusps. $\endgroup$
    – David Roberts
    Feb 29, 2016 at 20:49
  • $\begingroup$ @DeaneYang Extensions from intervals were done by Mityagin in 1961. In 1964 Seeley constructed extension operators for half spaces. Although he did not write it explicitly, this covers more or less the case of smooth boundaries. Stein's result for Lipschitz boundaries is from 1970. $\endgroup$ Mar 1, 2016 at 7:19

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