Is a Hilbert space $H$ compactly embedded in its dual?

Question

Is a Hilbert space $H$ compactly embedded in its dual? Is it compactly embedded in itself?

No idea how to think of this.

Answer 1

Hint: Let $e_k$ be a orthonormal basis (if it exists!). What can you say about it?

As suggested by @Julien, another approach is: Suppose that there exist $T:H\to H$ injective and compact, then $T(H)$ is finite dimensional (why?) and isomorphic to $H$, which is an absurd.

Remark: I am considering spaces with infinite dimension.

Answer 2

In addition to other interesting remarks and answers, it may be important to emphasize that there is substantial ambiguity in the question. We don't have to go as far as looking at the map from a Hilbert space (with Hilbert-space norm topology) to the weak topology on it (and apply Banach-Alaoglu), although that is worth keeping in mind.

For example, Levi-Sobolev spaces on the circle, $H^k$ the completion of finite Fourier series with respect to the norm $|f|_k^2=\sum_n |\hat{f}(n)|^2\cdot (1+n^2)^k$. The $0$th one is $L^2$, which we identify with its own strong dual. Then $H^{-k}$ and $H^{+k}$ are in natural duality by extending Plancherel. And/but $H^{k}\rightarrow H^{k-1}$ is Hilbert-Schmidt (so compact).

Indeed, this nice property is what makes the projective limit of the $H^k$ on the circle (=all smooth functions on the circle, by Sobolev imbedding) "nuclear Frechet", which basically means that we can prove a Schwartz kernel theorem: every continuous linear map from smooth functions on the circle to distributions on the circle (the colimit of the $H^k$'s) is given by a distribution on the product of two circles.