Does strong convergence in $H$ (or $L^2$) imply convergence in $V$ (or $W_0^{1,2}$)?

Question

Suppose:

• $\mathbf{V}$ and $\mathbf{H}$ are Hilbert spaces.
• $\mathbf{V} \hookrightarrow \mathbf{H}$ is compact embedding.
• $\mathbf{V}$ is dense in $\mathbf{H}$.

For example $\mathbf{V} = W_0^{1,2}(\Omega)$ and $\mathbf{H} = L^2(\Omega)$ for bounded $\Omega \in \mathbb{R}^n$.

Let $\mathbf{u}_n,\mathbf{u} \in \mathbf{V}$. It's implied from the Rellich-Kondrachov embeddidng theorem that bounded $\mathbf{u}_n$ in $\mathbf{V}$ has convergent subsequence in $\mathbf{H}$.

I am asking the opposite question:

Consider a bounded sequence $\mathbf{u}_n \in \mathbf{V}$. If $\mathbf{u}_n \to \mathbf{u}$ strongly in $\mathbf{H}$, does it weakly convergent in $\mathbf{V}$? Does it converge to the same $\mathbf{u}$?

A more general question:

What requirements the sequence $(\mathbf{u}_n)_{n \in \mathbb{N}}$ should have in $\mathbf{H}$, so that $\mathbf{u}_n$ is

• weakly convergenent in $\mathbf{V}$?
• Strongly convergent in $\mathbf{V}$?

Here is a similar question, but I do not know how this is shown.

Answer 1

From ${u_n},u \in H$ we can't imply ${u_n},u \in V$.

Answer 2

If $(u_n)$ is bounded in $V$ and $u_n\to u$ in $H$, then $u_n\rightharpoonup u$ in $V$:

As $(u_n)$ is bounded in $V$, it has a weakly converging subsequence $u_{n_k}\rightharpoonup v$ in $V$. The continuity of the embedding into $H$ implies $v=u$ and $u\in V$.

Using this argument, one can prove that every subsequence of $(u_n)$ has a subsequence that converges weakly to $u$ in $V$. Hence $u_n\rightharpoonup u$ in $V$.