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?



Extension of a smooth function from a convex set
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](http://www.univie.ac.at/EMIS/elibm/journals/index.html)


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.''


