$u_n\to u$ weakly in $X$ implies $u_n\to u$ weakly in $Y$, $X\subset Y$ compactly embedded

Question

Let $X$ and $Y$ be (reflexive) Banach spaces. Assume that $X$ is compactly embedded in $Y$. Consider a sequance $(u_n)\subset X$, $u_n\to u$ weakly in $X$. Why compact embedding implies that $u_n\to u$ weakly in $Y$? Where the compactness is used?

Answer 1

Here is how can use the compactness: $u_n \to u$ strongly in Y up to extraction of subsequnce

$ u_n\to u$ weakly in X implies $(u_n) $ is bounded in X.

But X is compactly embedded in Y, therefore there exists a subsequence $u_{n_j}$ which converges strongly in Y to some $y\in Y$

But also $ u_n\to u$ weakly in Y (simply by continuity) Hence, $u=y$

that is $u_{n_j} \to u$ strongly in Y.