weak convergence of $L^2$ implies weak convergence of $W_0^{1,2}$ (up to a subsequence)?

Question

In the paper that I am reading, it says that if $\{u_n\}$ are bounded in $W_0^{1,2} (\Omega)$ (bounded $\Omega\subset \mathbb{R}^N$) and $u_n \rightharpoonup u$ weakly in $L^2 (\Omega)$, then there exist a further subsequence such that $u_{n_k} \rightharpoonup u$ in $W_0^{1,2}(\Omega)$. How do we know it is still the same limit?

I know that $I: W_0^{1,p}(\Omega) \rightarrow L^2(\Omega)$ is continuous and dense for bounded $\Omega$ and $p\geq 2$. And here is my second question, given $X,Y$ are Banach spaces and $T:X\rightarrow Y$ is continuous and dense, can we say if $x_n \rightharpoonup x$ weakly in $X$, then $T(x_n) \rightharpoonup T(x)$ weakly in $Y$? If this is true, then the above is clear, since the subsequence also converges to $u$ weakly in $L^2$. If not what other conditions we need? I know compact is too strong since we actually get if $x_n \rightharpoonup x$ weakly in $X$, then $T(x_n) \rightarrow T(x)$ strongly in $Y$.

Thank you very much!

Answer 1

Let's prove your assertion. Let $f$ be a bounded linear functional on $Y$. Then $f \circ T$ is a bounded linear functional on $X$ and as $x_n$ converges to $x$ weakly, we have

$$ \lim _{n \to \infty} f\circ T(x_n) = f\circ T(x), $$

that is,

$$\lim_{n\to \infty} f(T(x_n)) = f(T(x))$$

for all bounded linear functional $f$. Thus $T(x_n)$ converges weakly to $T(x)$.

(Only the continuity of $T$ is needed).