# 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.

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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

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.

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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? –  julien 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 $H$. –  julien 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
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.