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Let M be an Riemannian manifold with the Dirichlet form $$\varepsilon (u,v) =-\int_M \langle \nabla u,\nabla v \rangle$$ for $u,v \in W^{1,2}_0(M)$. Let $\Delta^M:D(\Delta^M) \to L^2(M)$ denote the generator induced from $(\varepsilon, W^{1,2}_0(M)$. It's hard for me to describe $D(\Delta^M)$.

Let $f \in L^2(M)$ and $u \in W_0^{1,2}(M)$. If $$L_u(\phi)=\varepsilon (u,\phi)= \int_\Omega {f\phi dvol} $$ for all nonnegative $\phi \in W^{1,2}_0(M)$. In this case, we know the functional $L_u$ is a signed radon measure without sigular part w.r.t vol. And the absolute part is $fvol$. Then can we get $u \in D(\Delta^M)$ and $\Delta^M u=f$?

If this is right, can we extend this result to general metric measure space?

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The answers to your questions are yes and yes. I'll explain in terms of a general Dirichlet space $(\varepsilon, D(\varepsilon))$ on $(M,\mu)$ a locally compact Hausdorff space.

In the general Dirichlet form setting, it is often taken that $D(\Delta)$ is defined as $u\in D(\varepsilon)$ such that there is $f\in L^2(\mu)$ with $\varepsilon(u,\phi) = -\langle f, \phi \rangle_\mu$, and $f = \Delta u$ . This should be implied by your condition with $\mu = vol$ and $D(\varepsilon) = {W}^{1,2}$.

Assume $\varepsilon$ is regular, i.e. $C_0(M)\cap D(\varepsilon)$ is dense in $C_0(M)$ and in $D(\varepsilon)$ ($C_0(M)$: continuous functions vanishing at $\infty$). One could define $\Delta$ by your condition (replacing $vol$ with $\mu$ and $W^{1,2}$ with $D(\varepsilon)$). The problem is that we would not know a priori that $D(\Delta)\neq \emptyset$, or that it is densely defined, etc. But assuming the standard proof of the existence and nice properties of $\Delta$, the definitions should coincide.

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  • $\begingroup$ :$D(\Delta^M)=\{f:lim \frac{P_tf-f}{f} exists\}$. When $$L_u(\phi)=\varepsilon (u,\phi)= \int_\Omega {f\phi dvol} $$, how can you prove $u \in D(\Delta^M)$? $\endgroup$ – wang mu Oct 21 '13 at 5:25
  • $\begingroup$ You need to show that $\int \lambda^2 d(E_\lambda u,u) = ||f||_{L^2}$ where $E_\lambda$ is the spectral resolution of $\Delta$. I was hoping to give and explanation without too much spectral calculus. $\endgroup$ – D. Kelleher Oct 23 '13 at 3:22

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