Sufficient conditions for an asymptotic compactness

Question

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.

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 $L_2(X,\mu)$.