If $u_n \rightharpoonup u$ in $W$ and $W \subset C$ then $u_n \rightharpoonup u$ in $C$?

Question

Let $W \subset C$ be Banach spaces with continuous embedding here.

If $u_n \rightharpoonup u$ in $W$ and $W \subset C$ then is it true that $u_n \rightharpoonup u$ in $C$ for the same sequence (not a subsequence)?

I saw this result used where $W=L^2(0,T;H^1) \cap H^1(0,T;H^{-1})$ and $C=C([0,T];L^2)$.

Answer 1

Yes. To be explicit, let $T:W\to C$ be a bounded linear operator (it need not be injective). For every linear functional $f\in C^*$ we have $f\circ T\in W^*$. Since $u_n$ converge weakly, $f(T(u_n))\to f(T(u))$. But this says precisely that $T(u_n)$ converge to $T(u)$ weakly.

Answer 2

Yes. If the embedding $j \colon W \hookrightarrow C$ is continuous when the two spaces are endowed with their respective norm topologies, it is also continuous when both spaces are endowed with their weak topologies, and that means weakly convergent nets are mapped to weakly convergent nets. In particular, if the weakly convergent net is a sequence, its image is a weakly convergent sequence.