# Sobolev spaces on $B_1^2(0)$ don't change if we remove a point from the domain

FIRST: Let $W^{1,p}(B_1)$ be a Sobolev space on the open unit ball $B_1\subset \mathbb{R}^2$ with $p<+\infty$. Let $\omega=B_1\setminus\{0\}$ and $W^{1,p}(\omega)$ the Sobolev space with the same $p<+\infty$ as before. The question is to show that $W^{1,p}(B_1)=W^{1,p}(\omega)$. I think that the question is to be intended as follows: show that for all $\varphi\in C^\infty_c(B_1)$ if $u\in W^{1,p}(\omega)$ and $\partial_i u$ is a weak derivative, then $$\int_\omega u \,\partial_i\varphi=-\int_\omega\partial_iu\,\varphi,$$ since the converse (that is taking $\varphi\in C^\infty_c(\omega)$ and $u\in W^{1,p}(B_1)$ and verify the definition of weak derivative) is trivial because $C^\infty_c(\omega)\subset C^\infty_c(B_1)$.

I tried this: take $\varphi\in C^\infty_c(B_1)$ and $u\in W^{1,p}(\omega)$. Take the function $\eta\in C^\infty_c(B_1)$ given by $$\eta(x)=e^{\frac{|x|^2}{|x|^2-1}}$$ and consider the functions $\eta_r(x)=\eta(x/r)$ for $r\in(0,1)$ extended to zero outside the ball $B_r$. If my calculation are right we have: $$\|\eta_r\|_p=\|\eta\|_p r^{\frac{2}{p}}$$ $$\|\nabla\eta_r\|_p=\|\nabla\eta\|_pr^{\frac{2}{p}-1}$$ Take $\psi_r\in C^\infty_c(B_1)$ a function such that

$$\|\psi_r-\eta_r\|_{W^{1,p}}<r \qquad supp(\psi_r)=B_r \qquad \psi_r|_{B_{\epsilon(r)}}=1$$

for some $\epsilon(r)>0$ sufficiently small. So the function $\tilde{\varphi}_r=\varphi-\psi_r\varphi\in C^\infty_c(\omega)$ is such that $$\tilde{\varphi}_r\to\varphi \mbox{ in }L^p\qquad \mbox{as }r\to0$$ $$\nabla\tilde{\varphi}_r\to\nabla\varphi \mbox{ in }L^p\qquad \mbox{as }r\to0\qquad \mbox{if } p<2$$ in fact for $p<2$ we have that $\eta_r\to0$ in $W^{1,p}(\omega)$. Now for all $r$ we have $$\int_\omega u \,\partial_i\tilde{\varphi}_r=-\int_\omega\partial_iu\,\tilde{\varphi}_r,$$ then taking $\lim_{r\to0}$ we get the claim for $p<2$.

My questions: Is my procedure right? What is a better proof in order to achieve the claim for all $p<+\infty$?

SECOND: This should be more advanced. Let $\Omega\subset\mathbb{R}^n$ open and let $K\subset\Omega$ closed such that $\mathcal{H}^{n-1}(K)=0$ where $\mathcal{H}^{n-1}$ is the $(n-1)$-dimensional Hausdorff measure. Show that $W^{1,p}(\Omega)=W^{1,p}(\Omega\setminus K)$ for $p<+\infty$ (in the same sense of the first question I think). The only hint I had is that here it is needed some (I hope basic) result of geometric measure theory.

• If you know that Sobolev functions are characterized by their ACL property and $p$-summability of partials, the result is immediate. – user357151 Oct 20 '17 at 18:40
• @Desire thank you! I didn't know this property, but I found it online. Do you have any good reference book for it? Perhaps with this characterization we can also solve the second part, what do you think? – Pozz Oct 20 '17 at 20:21

With the hint of @Desire in the comments above, I give an answer to the SECOND part, which will imply the FIRST part.

Let us give the definition of ACL property: we say that a function $u:\Omega\to\mathbb{R}$ is absolutely continuous on lines in $\Omega$ (and we write $u\in ACL(\Omega)$) if, for every subspace $V_i :=\{x = (x_1,...,x_n) : x_i = 0\}\subset\mathbb{R}^n\}$, $i = 1,...,n$, there is a set of Lebesgue $(n −1)$-measure zero, called $N_i \subset V_i$, with the following property: for every $y \in V_i \setminus N_i$ the function $t \mapsto u(y + te_i)$ is absolutely continuous on every compact interval $[a,b]$ such that $y +te_i \in \Omega$ for $a \le t \le b$, where $e_i$ is the $i^{th}$ element of the standard basis of $\mathbb{R}^n$. If $u\in ACL(\Omega)$ it is well defined its gradient calculated almost everywhere, that we denote $\nabla^{ae}u$. Also, we write $u\in ACL^p(\Omega)$ if $u\in ACL(\Omega)$ and $\nabla^{ae}u\in L^p(\Omega)$. It turns out that:

$$u\in W^{1,p}(\Omega) \Longleftrightarrow u\in ACL^p(\Omega)\cap L^p(\Omega),$$ and weak gradients coincide with gradients almost everywhere.

I used as reference the book Heinonen, Koskela, Shanmugalingam, Tyson - Sobolev Spaces on Metric Measure Spaces.

Now take $u\in W^{1,p}(\Omega\setminus K)=ACL^p(\Omega\setminus K)\cap L^p(\Omega\setminus K)$ as in the hypotheses of the SECOND part of the question. We want to show that $u\in ACL^p(\Omega)\cap L^p(\Omega)$. Let $N_i$ be the sets corresponding to the fact that $u\in ACL(\Omega\setminus K)$ as in the definition above mentioned. For $i=1,...,n$ let:

$$\pi_i:\mathbb{R}^n\to V_i$$

be the orthogonal projection and let:

$$\tilde{N}_i:=\pi_i(K).$$

Since $\pi_i$ is a Lipschitz function with Lipschitz constant equal to one, it is known that:

$$0\le \mathcal(H)^{n-1}(\pi_i(K))\le (1)^{n-1}\mathcal(H)^{n-1}(K)=0,$$

hence the Lebesgue $(n-1)$-measure of $\tilde{N}_i$ is zero. Now define:

$$M_i:=N_i\cup\tilde{N}_i \qquad i=1,...,n.$$

By construction we have that for all $y\in V_i\setminus M_i$ the function $t\mapsto u(y+t e_i)$ is absolutely continuous on every compact interval $[a,b]$ such that $y+te_i\in \Omega$ (in fact for such $y$ we have that: $y+te_i\in\Omega$ if and only if $y+te_i\in\Omega\setminus K$). Hence the sets $M_i$ can take the place of sets $N_i$ in the definition of $ACL(\Omega)$, then we get $u\in ACL(\Omega)$. Moreover $u,\nabla u\in L^p(\Omega)$, then we conclude that $u\in W^{1,p}(\Omega)$.