# Sufficient conditions for an asymptotic compactness

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.

• 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. Feb 13, 2021 at 6:07
• @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? Feb 13, 2021 at 8:16

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 $$L_2(X,\mu)$$.
• 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 $\sup_{n \in \mathbb{N}}\mathcal{E}(u_n,u_n)<\infty.$ Feb 14, 2021 at 15:49
• What is the relation between $\mathcal E_n$ and $\mathcal E$? Feb 15, 2021 at 8:17
• 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. Feb 15, 2021 at 9:53