# Is the identity map $id: H^2(-\pi,\pi) \to L^2(-\pi,\pi)$ Hilbert-Schmidt?

Let $H_1, H_2$ be Hilbert spaces.

A linear operator $A: H_1 \to H_2$ is Hilbert-Schmidt iff for some orthonormal basis $\lbrace e_n : ~ n \in \mathbb{N} \rbrace$ of $H_1$ the sum $\sum_{n \in \mathbb{N}} \Vert A e_n \Vert^2_{H_2}$ is finite.

It is easy to see that if $H_1 = H_2$ the identity operator $id: H_1 \to H_1$ is Hilbert-Schmidt if and only if $H$ is finite-dimensional, since otherwise $\sum_{n \in \mathbb{N}} \Vert e_n \Vert^2_{H_2}= \sum_{n \in \mathbb{N}} \Vert e_n \Vert^2_{H_1}= \sum_{n \in \mathbb{N}} 1$ clearly diverges.

But what if $H_1$ is a real subset of $H_2$? Then the situation changes somehow, because $\Vert \cdot \Vert_{H_1} = \Vert \cdot \Vert_{H_2}$ needs not to hold anymore.

More specific:

Is the identity map $id: H^2(-\pi,\pi) \to L^2(-\pi,\pi)$ Hilbert-Schmidt?

and if not:

Is there any chance that $id: H^p(-\pi,\pi) \to L^2(-\pi,\pi)$ is Hilbert-Schmidt for any $p$?

EDIT: We equip $L^2$ and $H^2$ with the standard norms $\Vert f \Vert_{L^2}^2 = \int \vert f(x) \vert^2 dx$ and $\Vert f \Vert_{H^2}^2 = \int \vert f(x) \vert^2 dx + \int \vert D f(x) \vert^2 dx + \int \vert D^2f(x) \vert^2 dx$.

-
Do we know an orthonormal basis of $H^2(-\pi,\pi)$? – Berci Feb 15 '13 at 11:14

The answer to both questions is affirmative. Letting $\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}$, we have on $H^1(\mathbb{T})$ the following orthonormal basis: $$e_n= \frac{e^{i n x}}{\sqrt{2\pi(1+n^2)}},\quad n\in \mathbb{Z}.$$ Specializing the sum $\sum_n \lVert I e_n\rVert_{L^2}^2$ to this basis we get $$\sum_{n\in \mathbb{Z}}\lVert I e_n\rVert_{L^2}^2 = \frac{1}{2\pi}\sum_{n \in \mathbb{Z}}\frac{1}{1+n^2}<\infty.$$ Since this particular sum converges, we can prove via standard arguments that the sum $\sum \lVert I e_n\rVert_{L^2}^2$ will converge for any choice of an orthonormal basis $\{e_n\}$ of $H^1(\mathbb{T})$. We have thus proven that the embedding of $H^1(\mathbb{T})$ into $L^2(\mathbb{T})$ is Hilbert-Schmidt.
The argument for $H^k(\mathbb{T})$ is similar. In this case the convergence of the series will be even faster. As a side note, we can consider the spaces $$H^s(\mathbb{T})=\left\{ f\in L^2(\mathbb{T})\ :\ \sum_{m\in \mathbb{Z}} \lvert\hat{f}(m)\rvert^2\left(1+m^2\right)^s < \infty\right\},\quad s\in \mathbb{R},$$ with inner product $$(f, g)_{H^s}=\sum_{n \in \mathbb{Z}}\hat{f}(n)\overline{\hat{g}(n)}\left(1+n^2\right)^{s},$$ which generalize the spaces $H^k(\mathbb{T})$ with integer $k$. Here we have the orthonormal basis $$e_n=\frac{e^{inx}}{\sqrt{(2\pi)\left(1+n^2\right)^s}},\qquad n \in \mathbb{Z}.$$ So $$\sum_n \lVert I e_n\rVert_{L^2}^2=\frac{1}{2\pi}\sum_{n \in \mathbb{Z}}\frac{1}{\left(1+n^2\right)^s},$$ and the last series converges if and only if $s>\frac{1}{2}$. So the identity operator $$I\colon H^s(\mathbb{T})\to L^2(\mathbb{T})$$ is Hilbert-Schmidt if and only if $s>\frac{1}{2}$. I find it interesting to note that $I$ is compact if and only if $s>0$. So for $0<s\le\frac{1}{2}$ we have an example of a compact operator which is not Hilbert-Schmidt.
Sorry, what is the inner product in $H^p(\Bbb T)$? – Berci Feb 15 '13 at 11:26
@Berci: I edited the text adding the inner product of $H^s(\mathbb{T})$. – Giuseppe Negro Feb 15 '13 at 14:14
One more question: In the first part of your answer you gave a basis for $H^1(\mathbb{T})$. I checked, that it is orthonormal, but why is it a basis? Any hint how to check that? Thanks a lot! – mjb Feb 15 '13 at 15:22
It is essentially a consequence of the definition. The space $H^s(\mathbb{T})$ is defined in such a way that its Fourier transform (that is, the map which passes from a function on $\mathbb{T}$ to its Fourier series) is an isomorphism onto the space $\ell_w^2$ where the weight function is $$w(n)=(1+n^2)^s.$$ The family $e_n$ given in the text is just the pull-back of the standard orthonormal basis of this weighted $\ell^2_w$-space. – Giuseppe Negro Feb 16 '13 at 1:57