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

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

No idea how to think of this.

-
It's been very long, so I'm not quite up to snuff with the terminology anymore, does "compactly embedded" mean that the embedding is a compact operator? In that case iff the space is finite dimensional. – Daniel Fischer Jun 26 '13 at 12:59
It might help to know that $H \cong H^*$. They are isometrically isomorphic thorough the Fréchet-Riesz representation theorem. Also check the answer to (this question)[math.stackexchange.com/questions/350022/…. – scjorge May 3 at 17:47

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.

-
It's a bounded sequence that doesn't converge in $H$. So the answer is no. – aere Jun 26 '13 at 13:02
It does not change the argument, but why would the enbedding have to be the identity? – 1015 Jun 26 '13 at 13:12
You are right @julien, it could be any surjective embedding. Do you think I have to change it? – Tomás Jun 26 '13 at 13:13
Maybe, yes. Just assume $T:H\longrightarrow H$ is injective and compact. Then the range is finite dimensional and isomorphic to $H$. – 1015 Jun 26 '13 at 13:16
Ok, your idea is better than mine @julien, let me change it. – Tomás Jun 26 '13 at 13:19

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.

-