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Consider exercise 34 from chapter 4 ("Hilbert Spaces: An Introduction") of [1] (p. 201):

Let K be a Hilbert-Schmidt kernel which is real and symmetric. Then, as we saw, the operator $T$ whose kernel is $K$ is compact and symmetric. Let $\{\varphi_k(x)\}$ be the eigenvectors (with eigenvalues $\lambda_k$) that diagonalize $T$. Then:

(a) $\sum_k |\lambda_k|^2 < \infty$

(b) $K(x,y) \sim \sum\lambda_k\varphi_k(x)\varphi_k(y)$ is the expansion of $K$ in the basis $\{\varphi_k(x)\varphi_k(y)\}$.

(c) Suppose $T$ is a compact operator which is symmetric. Then $T$ is of Hilbert-Schmidt type if and only if $\sum_n |\lambda_n|^2 < \infty$, where $\{\lambda_n\}$ are the eigenvalues of $T$ counted according to their multiplicities.

I'd appreciate help in solving parts (b) and (c) (I have figured out part (a) on my own). For completeness, here's how [1] defines a Hilbert-Schmidt kernel (pp. 186-187). Note that $\mathcal{H}$ stands for a Hilbert space and that only separable Hilbert spaces are considered throughout [1].

Let $\mathcal{H} = L^2(\mathbb{R}^d)$. If we can define an operator $T:\mathcal{H}\rightarrow\mathcal{H}$ by the formula $$ T(f)(x) = \int_{\mathbb{R}^d} K(x,y)f(y)\ dy,\hspace{1cm} \mbox{whenever } f \in L^2(\mathbb{R}^d), $$ we say that the operator $T$ is an integral operator and $K$ is its associated kernel. [...] Hilbert-Schmidt operators [... are integral operators ...] with a kernel $K$ that belongs to $L^2(\mathbb{R}^d\times \mathbb{R}^d)$.


References

[1] Stein, Elias M. and Shakarchi, Rami. Real Analysis: Measure Theory, Integration, and Hilbert Spaces. Princeton University Press (2005)

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    $\begingroup$ I posted a solution that had additional information about smoothness when $K$ is smooth, but all mention of smoothness can be skipped and the solution to what you want is there: math.stackexchange.com/questions/918902/… $\endgroup$ – DisintegratingByParts Sep 16 '14 at 19:01
  • $\begingroup$ I was not aware that an almost identical post exists...T_T $\endgroup$ – Bombyx mori Sep 16 '14 at 19:10
  • $\begingroup$ @T.A.E.: Thank you. It's a very good answer for part (b). Could you please help me with part (c) too? $\endgroup$ – Evan Aad Sep 16 '14 at 19:22
  • $\begingroup$ @Evan Aad : Square the expansion in (b) and integrate. Use the first equality string for $\|g\|$ in the post I referenced. $\endgroup$ – DisintegratingByParts Sep 16 '14 at 20:29
  • $\begingroup$ @Bombyxmori : Out of the 300,000+ answers on the site, you didn't find that? :) Don't worry, Evan Aad didn't notice that problem either. They're not identical problems, and I knew about the other one only because I spent too much time trying to put together a clean solution to the other problem. $\endgroup$ – DisintegratingByParts Sep 16 '14 at 20:42
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Let $H$ be a separable Hilbert space as desired. For simplicity assuming we are working over the real field. Then being an integral operator, we have $$ K: H\rightarrow H, (Kf)(x)=\int K(x,y)f(y)dy $$ Since $K$ is a symmetric compact operator, it is normal and can be diagonalized. Let us write its eigenvectors as $\phi_{j}$ with eigenvalue $\lambda_{j}$. Then we know $\phi_{j}$s span $H$. Thus $K$ can be defined alternatively by $$ \forall f=\sum a_{j}\phi_{j}, (K(\sum a_{j}\phi_{j}))(x)=\sum a_j \lambda_{j}\phi_{j}(x)=\int_{H} \sum\lambda_{j} \phi_{j}(x)\phi_{j}(y)(\sum a_{i}\phi_{i}(y))dy $$ where we normalized $\langle \phi_{i}, \phi_{j}\rangle=\delta_{ij}$. The equality holds because for each term we have $$ a_{i}\lambda_{j}\phi_{j}(x)\int_{H}\phi_{j}(y)\phi_{i}(y)dy=a_{i}\lambda_{j}\phi_{j}(x)\delta_{ij} $$ and this concludes the proof.

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  • $\begingroup$ Thank you. Which part did you solve? (b) or (c)? $\endgroup$ – Evan Aad Sep 16 '14 at 18:57
  • $\begingroup$ This is enough to prove (b) and (c). $\endgroup$ – Bombyx mori Sep 16 '14 at 18:59
  • $\begingroup$ I see why your answer proves part (b). However regarding part (c), I still don't see why $\sum |\lambda_k|^2 < \infty$ implies that $T$ is a Hilbert-Schmidt operator. $\endgroup$ – Evan Aad Sep 16 '14 at 19:21
  • $\begingroup$ @EvanAad: In the case of (c), the only real thing left to do is to show $K$ is a Hilbert-Schimidt operator. The condition given guarantees its norm is finite. $\endgroup$ – Bombyx mori Sep 16 '14 at 19:23
  • $\begingroup$ So the condition $\sum|\lambda_k|^2 < \infty$ guarantees that $\int_{\mathbb{R}^d\times \mathbb{R}^d}|K(x,y)|^2 d(x,y) < \infty$? How? $\endgroup$ – Evan Aad Sep 16 '14 at 19:30
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The following is a solution to part (c). One direction was proved in part (a). Therefore, I will only prove the other direction: assuming $\sum_{n = 1}^\infty |\lambda_n|^2 < \infty$, I will show that $T$ is a Hilbert-Schmidt operator (where $\lambda_n$ are the eigenvalues corresponding to some orthonormal basis of eigenvectors, such that $T(f) = \sum_{n = 1}^\infty \lambda_n \left<f, \varphi_n\right> \varphi_n$ for all $f \in L^2(\mathbb{R}^d)$).

There is a natural isometry $\Psi$ from $L^2(\mathbb{R}^d\times \mathbb{R}^d)$ to $L^2(\mathbb{R}^d)\rightarrow L^2(\mathbb{R}^d)$, namely $\Psi(k) := T_k$, where $$ T_k(f) := \int_{\mathbb{R}^d}k(x,y)f(y)\,dy $$ Note that $\Psi(L^2(\mathbb{R}^d\times\mathbb{R}^d))$ is precisely the set of Hilbert-Schmidt operators.

Now, let $T \in L^2(\mathbb{R}^d) \rightarrow L^2(\mathbb{R}^d)$ be compact and symmetric. Since $L^2(\mathbb{R}^d\times \mathbb{R}^d)$ is complete and $\Psi$ is norm-preserving, $\Psi(L^2(\mathbb{R}^d\times\mathbb{R}^d))$ is complete. It suffices, therefore, to show that $T \in \overline{\Psi(L^2(\mathbb{R}^d\times\mathbb{R}^d))}$.

$T$ being compact and symmetric, by the spectral theorem there is an orthonormal basis of eigenvectors of $L^2(\mathbb{R}^d)$, $\varphi_n \in L^2(\mathbb{R}^d)$, with $\lambda_n$ being the eigenvalue corresponding to $\varphi_n$, such that for every $f \in L^2(\mathbb{R}^d)$, $$ T(f) = \sum_{n = 1}^\infty \lambda_n \left<f,\varphi_n\right>\varphi_n $$ We suppose that $\sum_{n = 1}|\lambda_n|^2 < \infty$.

For every $N \in \mathbb{N}_1$, define $$ T_N(f) := \sum_{n = 1}^N \lambda_n \left<f,\varphi_n\right>\varphi_n $$ We have $T_N = \Psi(k_N)$ with $$ k_N(x,y) := \sum_{n = 1}^N\lambda_n\phi_n(x)\overline{\phi_n(y)} $$ (recall that $\{\phi_k(x)\overline{\phi_m(y)}\}_{k,m \in \mathbb{N}_1}$ form a basis for $L^2(\mathbb{R}^d\times\mathbb{R}^d)$).

It remains to show: $\|T - T_N\| \underset{N \rightarrow \infty}{\longrightarrow} 0$. Let $f \in L^2(\mathbb{R}^d)$ with $\|f\| = 1$. Then $$ \begin{aligned} \|(T - T_n)(f)\|^2 & = \left\|\sum_{n = N+1}^\infty \lambda_n \left<f,\varphi_n\right>\varphi_n\right\|^2 \\ & = \sum_{n = N + 1}^\infty\left|\lambda_n \right|^2\left|\left<f,\varphi_n\right>\right|^2 \\ & \leq \sum_{n = N + 1}^\infty \left|\lambda_n\right|^2 \left\|f\right\|^2\left|\varphi_n\right\|^2 \\ & = \sum_{n = N + 1} \left|\lambda_n\right|^2 \end{aligned} $$ Therefore, since by assumption $\sum_{n = 1}^\infty \left|\lambda_n\right|^2 < \infty$, $\|T - T_N\| \underset{N \rightarrow \infty}{\longrightarrow} 0$, Q.E.D.

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