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.

  • 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


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 $L_2(X,\mu)$.

  • $\begingroup$ 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

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