0
$\begingroup$

This question relates a theory of Mosco convergence.

Let $X$ be a compact metric space, and $\mu$ a Borel measure on $X$.

A symmetric bilinear form $(\mathcal{E},\text{Dom}(\mathcal{E}))$ on $L^2(X,\mu)$ is called a Dirichlet form if the following conditions are satisfied:

  • $\text{Dom}(\mathcal{E})$ is a dense subspace of $L^2(X,\mu)$.
  • $\text{Dom}(\mathcal{E})$ is a Hilbert space under $\mathcal{E}(f,g)+\int_{X}fg\,d\mu$, $f,g \in \text{Dom}(\mathcal{E})$. For any $f \in \text{Dom}(\mathcal{E})$, $\mathcal{E}(f,f)\ge 0$.
  • For any $f \in \text{Dom}(\mathcal{E})$, we have $\hat{f}:= \max(0,\min(f,1))\in \text{Dom}(\mathcal{E})$ and $\mathcal{E}(\hat{f},\hat{f}) \le \mathcal{E}(f,f)$.

Let $\{(\mathcal{E}_n,\text{Dom}(\mathcal{E}_n))\}_{n=1}^ {\infty}$ be a sequence of Dirichlet forms on $L^2(X,\mu)$. We assume that each embedding $\text{Dom}(\mathcal{E}_n) \subset L^2(X,\mu)$ is compact.

My question

Let $\{u_n\}_{n=1}^\infty$ be a bounded sequence in $L^2(X,\mu)$ such that \begin{align*} \liminf_{n \to \infty}\mathcal{E}_n(u_n,u_n)<\infty. \end{align*} We assume that there exits a subsequence of $\{u_n\}_{n=1}^\infty$ which converges to $u \in \text{Dom}(\mathcal{E})$ in $L^2(X,\mu)$. Here, $\text{Dom}(\mathcal{E})$ is the domain of a Dirichlet form on $L^2(X,\mu)$. We assume moreover that the injection $\text{Dom}(\mathcal{E}) \subset L^2(X,\mu)$ is compact,

Then, can we show that there exists a subsequence of $\{u_{n}\}_{n=1}^\infty$ which strongly converges in $L^2(X,\mu)$?

In fact, I don't feel this claim is correct (although I have no counter examples). If there are sufficient conditions for this claim to hold, please let me know.

$\endgroup$
2
  • 2
    $\begingroup$ a) Your definition of Dirichlet form is not complete, you additionally need $\mathcal{E}(f\wedge 1,f\wedge 1)\leq \mathcal{E}(f,f)$ (in fact, this makes you third bullet point superfluous). b) Your conditions are too weak. You could simply take $\mathcal{E}_n=\frac 1 n \mathcal{E}$, in which case your condition reduces to $\liminf_n \frac 1 n\mathcal{E}(u_n,u_n)<\infty$. Of course this is not strong enough to guarantee compactness in $L^2$ in general. $\endgroup$
    – MaoWao
    Commented Feb 13, 2021 at 6:07
  • $\begingroup$ @MaoWao Thank you for your comment. I modified the definition of the Dirichlet form. Do you know a nice condition to show the $L^2$ convergence? $\endgroup$
    – sharpe
    Commented Feb 13, 2021 at 8:16

1 Answer 1

0
$\begingroup$

Disclaimer: The question has been edited several times so that this may not apply to actual version.

This seems obvious (after the edit, you consider only one Dirichlet form, right?): Take a subsequence (still denoted by $u_n$) such that $\mathcal E(u_n,u_n)$ is bounded. Then, this subsequence is bounded in $\mathcal E$ (with the norm defined by $\|u\|_{\mathcal E}^2 =\mathcal E(u,u) + \|u\|^2_{L_2(X,\mu)})$. Since the inclusion into $L_2(X,\mu)$ is compact, there is a further subsequence which converges in $L_2(X,\mu)$.

$\endgroup$
3
  • $\begingroup$ Sorry. My edits weren't appropriate. A subsequence of $\{u_n\}_{n=1}^\infty$ which converges to $u \in \text{Dom}(\mathcal{E})$ in $L^2(X,\mu)$. Hence, this does not imply that $\sup_{n \in \mathbb{N}}\mathcal{E}(u_n,u_n)<\infty.$ $\endgroup$
    – sharpe
    Commented Feb 14, 2021 at 15:49
  • $\begingroup$ What is the relation between $\mathcal E_n$ and $\mathcal E$? $\endgroup$ Commented Feb 15, 2021 at 8:17
  • $\begingroup$ There are no other conditions. Rather, I am asking what are sufficient conditions for the strong convergence to hold. For example, it is known that it is sufficient for the semigroups of $\{(\mathcal{E}_n,\text{Dom}(\mathcal{E}_n))\}_{n=1}^\infty$ to have a "uniform" ultracontractivity in a sense. However, this condition is not interesting. $\endgroup$
    – sharpe
    Commented Feb 15, 2021 at 9:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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