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$M$ 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?

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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? –  Nate Eldredge Jul 13 '13 at 14:35
    
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. –  Nate Eldredge Jul 13 '13 at 14:40
    
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. –  Kofi Jul 13 '13 at 21:14
    
@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.) –  Nate Eldredge Jul 14 '13 at 3:08
    
@Nate Eldredge:It's volume measure.And consider the case when M is bounded. –  jiangsaiyin Jul 15 '13 at 4:59
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1 Answer

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 $c(M \setminus K_n) \le \|g\|_{W^{1,2}} = \|1 - f_n\|_{W^{1,2}} \to 0$.

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