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 smoothdifferentiable at $x\in C$ if there exists a linear map $dF_{x}\: : \: \mathbb{R}^{n}\to\mathbb{R}^{m}$ such that
$$(*)\quad \frac{|| F(y)-F(x)-dF_{x}(x-y)||}{||y-x||}\to 0$$$$\quad \frac{|| F(y)-F(x)-dF_{x}(x-y)||}{||y-x||}\to 0$$
as $y\to x$, for $y\in C$. I$f$ 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?
