Compactness of the identity operator

Question

As far as I know, by Rellich-Kondrachov theorem, we can say $I:H_{0}^{k}\to H_{0}^{m}$, for $m<k$ is a compact operator, where $H_{0}^{k}=\{f\in H_{{}}^{k}|f(0)={f}'(0)=\cdots ={{f}^{(k)}}(0)=0\}$.

I am wondering if it is possible to say that the operator $I:H_{0}^{k}\to H_{0}^{k}$ is a compact operator?

Thanks for your help in advance.

Answer 1

Rellich Kondrachov gives different embedding theorem; be ware of compact and continuous embedding. for some k the embedding is continuous but not compact.

Answer 2

No. The identity operator is not compact in any infinite dimensional Banach space.