Operator compact on $H^1 (0,\pi)$

Question

Consider the operator $K\colon H^1(0,\pi)\to H^1(0,\pi)$ defined by duality (Riesz. Theorem) as $$ \langle K\phi,\psi\rangle = \int_{0}^{\pi}{\phi(x)\psi(x)\,dx} $$ for all $\psi \in H^1(0,\pi)$, where obviously $\langle\cdot,\cdot\rangle$ is the product in $H^1(0,\pi)$. Prove that $K$ is compact from $H^1(0,\pi)$ to itself: the suggestion is to use the compact embedding of $H^1(0,\pi)\hookrightarrow L^2(0,\pi)$. I try to solve this excercise but i can only prove that exists a subsequence weakly convergent .

Answer 1

Notice that if $f$ is an element of $H^1$, then $$\lVert K(f)\rVert_{H^1}^2=\int_0^{\pi}f(x)(Kf)(x)\mathrm dx\leqslant \lVert f\rVert_{L^2}\cdot \lVert K(f)\rVert_{L^2}.$$ Since $\lVert K(f)\rVert_{L^2}\leqslant \lVert K(f)\rVert_{H^1}$, we obtain that $$\tag{1}\lVert K(f)\rVert_{H^1}\leqslant \lVert f\rVert_{L^2}.$$ If $(f_n)_{n\geqslant 1}$ is a bounded sequence of elements of $H^1(0,\pi)$, then we extract a subsequence $(f_{n_k})_{k\geqslant 1}$ such that $\lVert f_{n_k}-f\rVert_2\to 0$ for some $f\in\mathbb L^2$ (in fact $f\in H^1$) and we conclude by (1) that $\lVert K(f_{n_k})-K(f)\rVert_{H^1}\to 0$ as $k$ goes to infinity.