Let $Y$ be an NLS, $Y'$ its dual space, $K \subset Y$ a compact subset. Show that the restriction $I_K: Y' \to C(K)$ is a compact operator

Question

Let $X$ and $Y$ be normed vector spaces. Let $K$ be a non-empty subset of $Y$

If $\psi$ is any function on $Y$ define $I_K \psi$ to be its restriction in $K$. Prove that, if $K$ is compact, then $I_K: Y' \to C(K)$ is compact.

($Y'$ is the dual space of $Y$. Many authors denote the dual space as $Y^*$)

Since $I_K$ is defined on the dual space of $Y$, then I presume $\psi$ is a functional $\psi: Y \to \mathbb{F}$ rather than merely any function on $Y$.

I presume we must show that $I_K$ is a compact in the sense of a compact operator, that it maps any bounded set to a totally bounded set. Or equivalently, it maps the unit ball to a set with a compact closure. $I_K$ is clearly linear, but the functions in the image of $I_K$ are not. $K$ is not necessarily a linear subspace, so functions defined only on $K$ are not generally linear functions.

I suspect we Schauder's theorem (maybe not). Schauder's theorem says if we have a compact operator $T$ than its adjoint $T^\top$ is also compact.

Any tips are greatly appreciated.

Answer 1

Let $B$ be the unit ball of $Y'.$ Then, $I_K(B)$ is:

• equicontinuous since it is even $1$-Lipschitz-continuous,
• pointwise bounded since it is even globally bounded (because $K$ is bounded).

By the Arzelà–Ascoli theorem, $I_K(B)$ is then relatively compact in $C(K),$ q.e.d.

Answer 2

This follows from Arzela-Ascoli Theorem. Let $(f_n)$ be a norm bounded sequence in $Y'$. The inequality $|f_n(x)-f_n(y)| \leq M \|x-y\|$ where $M=\sup_n \|f_n\|$ shows that $(f_n)$ is equi-continuous. It is also uniformly bounded (because $K$ is norm bounded ), so it has a subsequence which converges uniformly on $K$.