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.

  • 2
    $\begingroup$ If you know that Sobolev functions are characterized by their ACL property and $p$-summability of partials, the result is immediate. $\endgroup$
    – user357151
    Oct 20 '17 at 18:40
  • $\begingroup$ @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? $\endgroup$
    – 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:


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 $u\in W^{1,p}(\Omega)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.