$M$ is an $n$-dimensional Riemannian manifold. Consider the Dirichlet form $$\varepsilon \left( {u,v} \right) = \int_M {\left\langle {\nabla u,\nabla v} \right\rangle }, \quad u ,v \in {W^{1,2}}\left( M \right).$$ The capacity $c$ of an open set $A \subset M$ is defined by $$c\left( A \right): = \inf \{ {\left\| u \right\|_{{W^{1,2}}}}:u \ge 1 \text{ a.e. in $A$}\}. $$ And it can be extended to arbitrary sets $B \subset M$ by $$c\left( B \right): = \inf \{ c\left( A \right):B \subset A,A \text{ open}\}. $$ We say the Dirichlet form $\varepsilon $ is tight if there exist compact sets ${K_n} \subset M$, such that $c\left( {M\setminus {K_n}} \right) \to 0$. Is the Dirichlet form defined above on the Riemannian manifold tight?

  • $\begingroup$ Is the integral in the first display taken with respect to volume measure, or something else? If volume measure, this is trivially false for any manifold $M$ with infinite volume (e.g. $M = \mathbb{R}^n$), since then for any compact $K$, $c(M \setminus K) \ge \operatorname{Vol}(M \setminus K) = \infty$. Or are there other assumptions missing? $\endgroup$ – Nate Eldredge Jul 13 '13 at 14:35
  • $\begingroup$ Also, I cleaned up the formatting in your question. Punctuation marks in sentences such as . , ? ! : ; should always be followed by a space. And when a display equation comes at the end of a sentence, put the period inside the display ($$a+b.$$ not $$a+b$$.), otherwise it appears on the next line and looks bad. $\endgroup$ – Nate Eldredge Jul 13 '13 at 14:40
  • $\begingroup$ I don't understand. $W^{1,2}$ functions are not continuous unless $\dim(M) =1$, so that $c(A)$ does not make sense. Even if you would restrict it to smooth functions, say, then it would be always zero, as you can just take constant functions. $\endgroup$ – Matthias Ludewig Jul 13 '13 at 21:14
  • $\begingroup$ @Kofi: note the definition of $c(A)$ only requires $u \ge 1$ almost everywhere on $A$, which is a well-defined statement for $u \in W^{1,2}$. Also, the $W^{1,2}$ norm is usually $\|u\|_{W^{1,2}}^2 = \|u\|_{L^2}^2 + \|\nabla u\|_{L^2}^2$, so that for $u=1$ we get the volume of $M$, not zero. (And if $M$ has infinite volume, constants are not in $W^{1,2}$ at all.) $\endgroup$ – Nate Eldredge Jul 14 '13 at 3:08
  • $\begingroup$ @Nate Eldredge:It's volume measure.And consider the case when M is bounded. $\endgroup$ – jiangsaiyin Jul 15 '13 at 4:59

As discussed in comments, $W^{1,2}$ is interpreted here as $W^{1,2}_0$, the completion of $C^\infty_c(M)$ in the $W^{1,2}$ norm $\|f\|_{W^{1,2}}^2 = \int_M (f^2 + |\nabla f|^2)\,dVol$. Also, the usual definition of capacity involves $\|u\|_{W^{1,2}}^2$ (your version is then the square root of this); I'll use the usual convention, though it doesn't affect the answer.

In general the capacity $c$ need not be tight (my comment was mistaken). Consider the one-dimensional manifold $M = (0,1)$. By Sobolev embedding (which has an elementary proof in this case), each function in $W^{1,2}$ is absolutely continuous and vanishes at the boundary $\{0,1\}$. In particular, if $K$ is compact then there is no $f \in W^{1,2}$ with $f\ge 1$ on $(0,1) \setminus K$, so $c((0,1) \setminus K) = \infty$.

Indeed, $c$ is tight if and only if $1 \in W^{1,2}(M)$.

For the forward direction, recall that since $\varepsilon$ is Dirichlet, if $f \in W^{1,2}$ then $f \wedge 1, f \vee 0 \in W^{1,2}$ as well. If $c$ is tight then in particular there is a compact set $K$ with $c(M \setminus K) < \infty$, i.e. there exists $f \in W^{1,2}(M)$ with $f \ge 1$ a.e. on $M \setminus K$. We can also find a function $g \in W^{1,2}$ with $g \ge 1$ on $K$ (indeed, we could take $g \in C^\infty_c(M)$ by a standard cutoff function construction). Then $1 = ((f \vee 0) + (g \vee 0)) \wedge 1 \in W^{1,2}$.

Conversely, if $1 \in W^{1,2}(M)$ there is a sequence $f_n \in C^\infty_c(M)$ with $f_n \to 1$ in $W^{1,2}$-norm. If $K_n$ is the support of $f_n$ then $g_n := 1 - f_n$ is a $W^{1,2}$ function with $g_n = 1$ on $M \setminus K_n$. So $c(M \setminus K_n) \le \|g\|_{W^{1,2}} = \|1 - f_n\|_{W^{1,2}} \to 0$.


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.