Operator into the dual space is compact

Question

I want to solve the following

Let $X,Y$ be Banach spaces, with compact embedding $X\hookrightarrow Y$. Define the bilinear form $b:X\times Y\to\mathbb{R}$ that satisfy $$b(u,v)\leq C\|u\|_X \|v\|_Y, \forall u \in X, v\in Y$$ For $w\in X$ define the operator $A_\omega$ by $$A_\omega : X\to \mathbb{R}, v\to b(w,v)$$ Show that the operator $A:X\to X', \omega \to A_\omega$ is well defined and compact.

My thoughts so far: So we know that $$\frac{b(w,v)}{\|u\|_X}\leq C\|v\|_Y$$ So maybe we can conclude $$\frac{\|b(w,v)\|}{\|u\|_X}\leq C\|v\|_Y$$ meaning $A_{\omega}$ is a bounded operator. Now I want to show that for a bounded sequence $(a_n)\in X$ there exist $a_{n_j}$ such that $Aa_{n_j}$ converges in $X'$.

Answer 1

You have a continuous linear map $T:X\to Y'$, $w\mapsto A(w,\cdot)$ (the difference to $A$ is the range $Y'$ instead of $X'$). A theorem of Schauder says that the transposed $J^t: Y'\to X'$ of the compact inclusion $J:X\hookrightarrow Y$ is again compact ($J^t$ is the restriction). Hence $A=J^t\circ T$ is compact.