If $U$ is a bounded domain in $\mathbb R^n$ whose boundary is smooth, and $f$ is a smooth function on $U$ whose partial derivatives of all orders have a continuous extensions to $\bar U$. For an arbitrary domain $V \supseteq \bar U$, is there a smooth function $\tilde f$ on $V$ extending $f$?
[Math] The extension of smooth function
partial differential equationsreal-analysis
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After the question has been edited, the answer is yes, if the boundary of $\Omega$ is sufficiently smooth and there is an open set $G$ with $\overline{\Omega}\subset G\subset\Omega´$. Note at first, that $f$ can be continuously extended to $\overline{\Omega}$ by a classical topological result for uniformly continuous functions. And as the same is true for all partial derivatives, this extension is infinitely continuously differentiable.
The smooth extension to all of $\mathbb{R}^n$ is basically Whitney's extension theorem, see e.g. this question. And multiplication with a suitable function gives the compact support.
I am only three years late, so I am probably just wasting time.... but it is an interesting question, so.... I will only prove that $f$ can be extended to $(-\infty,0)$. The case $(1,\infty)$ is similar. Construct a $C_{c}^{\infty}(\mathbb{R})$ function $\psi$ such that $\psi=1$ in $[0,\frac{1}{2}]$ and $\psi(x)=0$ for $x\geq\frac{3}{4}$ and define the function $g(x):=f(x)\psi(x)$. By extending $g$ to be zero for $x\geq1$, we have that $g$ is $C^{\infty}$ in $[0,\infty)$ and $g=f$ in $[0,\frac{1}{2}]$. Let $\phi:[1,\infty )\rightarrow\mathbb{R}$ be a continuous function be such that $$ \int_{1}^{\infty}\phi(t)\,dt=1,\quad\int_{1}^{\infty}t^{n}\phi(t)\,dt=0,\quad \lim_{t\rightarrow\infty}t^{n}\phi(t)=0 $$ for all $n\in\mathbb{N}$. For $x<0$ define $$ h(x)=\int_{1}^{\infty}\phi(t)g(x(1-t))\,dt. $$ Note that since $t\geq1$, $x-tx>0$ and so $h$ is well-defined. Since $g$ is bounded, by the Lebesgue dominated convergence theorem, $$ \lim_{x\rightarrow0^{\_}}h(x)=\int_{1}^{\infty}\phi(t)\lim_{x\rightarrow 0^{\_}}g(x(1-t))\,dt=g(0)\int_{1}^{\infty}\phi(t)\,dt=1g(0)=f(0). $$ Since all the derivatives of $g$ are bounded, by the Lebesgue dominated convergence theorem, we can differentiate under the integral sign to get that for $x<0$, $$ h^{(n)}(x)=\int_{1}^{\infty}\phi(t)(1-t)^{n}g^{(n)}(x(1-t))\,dt. $$ Again by the Lebesgue dominated convergence theorem, \begin{align*} \lim_{x\rightarrow0^{\_}}h^{(n)}(x) & =\int_{1}^{\infty}\phi(t)(1-t)^{n}% \lim_{x\rightarrow0^{\_}}g^{(n)}(x(1-t))\,dt=g^{(n)}(0)\int_{1}^{\infty }(1-t)^{n}\phi(t)\,dt\\ & =1g^{(n)}(0)=f^{(n)}(0), \end{align*} where by the binomial theorem $$ \int_{1}^{\infty}(1-t)^{n}\phi(t)\,dt=\int_{1}^{\infty}\phi(t)\,dt+\sum _{k}c_{k}\int_{1}^{\infty}t^{k}\phi(t)\,dt=1+0. $$ This shows that $h$ is a $C^{\infty}$ extension of $f$ to $(-\infty,0)$.
Best Answer
We can take $V=\mathbb R^n$ without losing anything. The answer is yes, but this is nontrivial and I'm not going to prove it here. Here are some sources:
1) Short paper by Seeley (1964) covers the case of half-space. If you are interested in local matters, then straighten out a piece of $\partial U$ and apply this reflection-based extension. http://www.ams.org/journals/proc/1964-015-04/S0002-9939-1964-0165392-8/home.html
2) Whitney's classic paper of 1934 treats general closed sets (!) but is not an easy reading. http://www.ams.org/journals/tran/1934-036-01/S0002-9947-1934-1501735-3/home.html
3) In between there was a paper by M.R. Hestenes, "Extension of the range of a differentiable function" (Duke Math. J. 8, (1941) 183--192). Unfortunately, Duke is less generous with old articles than the AMS; the article is behinds a paywall.
4) A very recent 2-volume book by A. Brudnyi and Yu. Brudnyi "Methods of Geometric Analysis in Extension and Trace Problems" is an encyclopedia of extension theorems. See Chapter 2 of volume 1.