Simple question about extension of Sobolev functions in one dimension

Question

Given $v\in {\rm H}^{2}(0,1)$, is it true that there exists an extension $\overline{v}\in {\rm H}^2(\mathbf{R})$ such that $\overline{v}=v$ on $(0,1)$? If so, what is the usual choice of such an extension? The point here is that we require that $\overline{v}$ belongs to the class ${\rm H}^2$ on the whole space $\mathbf{R}$. I think that the classical extension theorem requires that extension domain is bounded and open (smoothness of boundary of such extension domain is not a problem in one dimension). If it is so, then we should be able to obtain the extension $\overline{v}\in{\rm H}^2_{loc}(\mathbf{R})$. But can we obtain the extension $\overline{v}$ which belongs to ${\rm H}^2(\mathbf{R})$? I think this is an elementary question, the answer can probably found in the literature, but I would appreciate an elementary answer. Thanks in advance.

Answer 1

The answer can be found in the book G. Leoni: First Course in Sobolev Spaces, Exercise 12.7 and, in particular, Remark 12.8. This remark provides one example of the extension operator $P:{\rm W}^{m,p}(\Omega)\rightarrow {\rm W}^{m,p}(\mathbf{R}^n)$, where $m\in\mathbf{N}$, provided that $\Omega\subseteq\mathbf{R}^n$ is an extension domain with Lipschitz boundary of the special type (see Theorem 12.3 therein). This result is due to E. M. Stein. In the case $n=1$, the domain $(0,1)$ is such an extension domain, so everthing works fine. Also, a further reference is the book E.M. Stein: Singular Intergals and Differentiability Properties of Functions. Final remark: idea of antiperiodic extension outside of $(0,1)$ does not work (with the exception of the case $m=1$), the simple counterexample is: $v(s):=s$, $s\in (0,1)$. Then the antiperiodic extension is not smooth enough at points $0$ and $1$ to belong to ${\rm H}^2_{loc}(\mathbf{R})$ (although it does belong to ${\rm H}^1_{loc}(\mathbf{R})$).

Answer 2

If you already have $\bar{v}\in H^2_{loc}(\mathbb{R})$ you can multiply it with the following function: $$\varphi(t)=1 \mbox{ if } t\in (0,1),\ \varphi(t)=0 \mbox{ if } t\in \mathbb{R}\setminus(-1,3).$$ and $\varphi$ should be smooth. Then $\varphi\cdot \bar{v}$ has compact support and is therefore in $H^2$. You can construct such $\varphi$ by examining $$t\mapsto\left\{\begin{array}{c}\exp(-\frac{1}{t^2}), t>0\\ 0, t\leq 0\end{array}\right.$$ See also mollifiers: https://en.wikipedia.org/wiki/Mollifier

Answer 3

Based on previous useful comments, I will try to answer my question and make some points more clear.

@andrija: What we actually need is Theorem 12.15 from Leoni's book, which is states that it is possible to extend $g\in {\rm W}^{1,p}(\Omega)$ to $\overline{g}\in {\rm W}^{1,p}(\mathbf{R})$, provided that $\Omega\subseteq\mathbf{R}$ is open set with uniformly Lipschitz boundary. But this is not particularly elementary argument, as it requires the partition of unity construction. Now the question is whether or not we can extend such result to the case $g\in {\rm W}^{m,p}(\Omega)$, where $m\in\mathbf{N}$, under suitable assumptions on $\Omega$. In Leoni's book there is comment to that effect, and also in book of Brezis from 1983, there is a short comment that the theorem holds provided $\Omega\subseteq\mathbf{R}$ is of class ${\rm C}^m$, although I find interesting noting that there is no such comment in more recent version of book of Brezis. But the question still stands: is there an elementary formula for extension (say, in the case $\Omega=[0,1],$ $p=2$ and $m=2$)? In the case $m=1$ the formula is very simple (extension by antiperiodicity), and I would like to have, if possible, the analogous of such formula in the case $m=2$.

@humanStampedist: Your construction solves the problem provided that we already have an extension to $\overline{g}\in {\rm H}^2_{loc}(\mathbf{R})$. But as andrija pointed out, $\overline{g}$ can not be constructed by reflection. You proposed a different construction which should work. But I have some reservations about this as well: the construction in question definitely gives $\overline{g}\in {\rm C}^1(-\infty,1)$, but I am not sure that we can achieve $g''\in {\rm L}^2(B_r(0))$, where $B_r(0)$ is open ball with small radius $r$ centered at $0$. Maybe this is obviously true, but I do not see it. It is known that a partition od unity can be used to extend ${\rm C}^r$-functions from some chosen open set in $\mathbf{R}$ to an ${\rm C}^r(\mathbf{R})$-function (see, for instance, If $f: S\to \mathbb{R}$ and $f$ is differentiable of class $C^r$ at each point $x_0$ of $S$, then $f$ may be extended to a $C^r$ function) but this sounds to me like a variant of the Urysohn's Lemma, not the extension theorem for Sobolev functions. So I do not believe that such a result can be used to obtain $\overline{g}\in {\rm H}^2_{loc}(\mathbf{R})$, which seems to be your idea.

In conclusion, I would appreciate if anyone could provide a reference for more explicit construction of such extension (even to ${\rm H}^2_{loc}(\mathbf{R})$, if not to ${\rm H}^2(\mathbf{R})$, since humanStampedist showed how to truncate the function to achieve global integrability). Is this construction really so intricate that it requires the partition of unity in the case $\Omega=[0,1]$? Here is my idea: since $g\in {\rm H}^2(0,1)$, we have $g'\in {\rm H}^1(0,1)$, and so we can extend $g'$ to $h\in {\rm H}^1_{loc}(\mathbf{R})$ by antiperiodicity. What remains to be done is to construct the function $\overline{g}\in {\rm H}^2_{loc}(\mathbf{R})$ such that $\overline{g}'=h$. Can we do it? I am open to any remarks.