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Let $C\subset \mathbb{R}^{n}$, $C'\subset\mathbb{R}^{m}$ be two convex sets with a non-empty interior. A function $F\: : \: C\to C'$ is said to be differentiable at $x\in C$ if there exists a linear map $dF_{x}\: : \: \mathbb{R}^{n}\to\mathbb{R}^{m}$ such that

$$\quad \frac{|| F(y)-F(x)-dF_{x}(x-y)||}{||y-x||}\to 0$$

as $y\to x$, for $y\in C$. $f$ is smooth $(*)$ if all its higher order derivatives are differentiable. I would like to know under which conditions such a function has a smooth (local) extension on a open set, i.e I want to (dis)prove the followings

1)(local extension) for any $p\in \partial C$ there exists an open set $U\subset \mathbb{R}^{n}$ such that $p\in U$ and a smooth function $H\: : \: U\to \mathbb{R}^{m}$ such that $$ H|_{U\cap C}=F|_{U\cap C}. $$

2)(global extension)there exists an open set $A\subset \mathbb{R}^{n}$ such that $C\subset A$ and a smooth function $G\: : \: A\to \mathbb{R}^{m}$ such that $$ G|_{C}=F. $$ 3)Now assume that $C$, $C'$ are two manifolds with boundary (or with corners) as well, are 1), 2) still true/false?

Here a small remark. I'm definitely not an expert but I know that using a Theorem due to Seeley we can say that: Given a set $\Omega$ in $\mathbb{R}^{n}$ with smooth boundary, then any smooth function (as defined above) $f\: : \: \bar{\Omega}\to \mathbb{R}$ can be extended to a function $g\: : \: \mathbb{R}^{n}\to \mathbb{R}$. I wonder if there is a version of this result where $\Omega$ has a piecewise smooth boundary. If the answer is yes then point 3) is true for $F\: : \: C\to \mathbb{R}$.

(Edit:) The nature of this question comes from the following problem. Let $A\subset \mathbb{R}^{n}$ be any subset. In this case the definition $(*)$ of smoothness sounds bad, but we have two other choices

Definition $(**)$: a map $f\: : \: A\to \mathbb{R}$ is smooth if for any $p\in A$ there exists an open set $U\subset \mathbb{R}^{n}$ such that $p\in U$ and a smooth function $F\: : \: U\to \mathbb{R}$ such that $$ F|_{U\cap A}=f|_{U\cap A}. $$

Definition$(***)$: a map $f\: : \: A\to \mathbb{R}$ is smooth if there exists an open set $V\subset \mathbb{R}^{n}$ such that $A\subset V$ and a smooth function $F\: : \: V\to \mathbb{R}$ such that $$ F|_{ A}=f. $$
Assume that $A$ is convex, then we have $(***)$ implies $(**)$ which implies $(*)$. But when are they equal? For examples this is true when $A$ is closed with non empty interior and smooth boundary. But what happen when the boundary is piecewise smooth?

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    $\begingroup$ Stein's book, "Singular Integrals and Differentiability Properties of Functions" shows how to extend a smooth function on a domain with boundary (including any convex domain) to an open neigbborhood of the domain. This is easily extended to a manifold with boundary using partitions of unity. $\endgroup$
    – Deane Yang
    Jun 11, 2015 at 16:07
  • $\begingroup$ Lipschitz boundary is okay, see Stein's book. A simultaneous smoothing for all smoothing classes at once is done in a paper of there was a paper of Bierstone in Inventeones which I cannot find at the moment. The general subject is called Whitney extension theorem, see en.wikipedia.org/wiki/Whitney_extension_theorem. For finite smoothing classes I think any domain will work according to results of Feferman (see his homepage or link at the above wiki page). $\endgroup$ Jun 12, 2015 at 22:07

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Some answers are in the paper

  • MR1474553 (98i:46040) Kriegl, A.(A-WIEN) Remarks on germs in infinite dimensions. (English summary) Acta Math. Univ. Comenian. (N.S.) 66 (1997), no. 1, 117–134. pdf available via EMIS

From the review:

"Now what happens if one changes to X⊆Rn? For closed convex sets with nonempty interior the conditions corresponding to the one-dimensional situation still agree. In the case of holomorphic and real-analytic maps the germ on such a subset is already defined by the values on the subset. Hence we are actually speaking about germs in this situation.

"In infinite dimensions we consider maps on just those convex subsets. Thus we do not claim greatest achievable generality, but rather restrict ourselves to a situation which is quite manageable. We show that even in infinite dimensions the conditions above often coincide, and that real-analytic and holomorphic maps on such sets are often germs of that class. Furthermore, we have exponential laws for all three classes; more precisely, the maps on a product correspond uniquely to maps from the first factor into the corresponding function space on the second.''

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  • $\begingroup$ Since $\mathbb{R}^{n}$ is a convenient vector space as well, I can apply the results of this paper to my problem. But i don't understand why in the proposition 1.9 they show that the extension exists if the convex set is closed, with non-empty interior, and with smooth boundary. Assume that $C$ is a triangle in $\mathbb{R}^{2}$, and consider $f\: : \: C\to \mathbb{R}$, does the extension exists ? $\endgroup$
    – Cepu
    Jun 12, 2015 at 8:28

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