I'm currently solving the following exercise (in chapter 4 of Stein's Real Analysis / Measure Theory)
Exercise 34
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.
Below are some questions about this exercise
(a)
By spectral thm, $\{\varphi_k\}$ is an orthonormal basis for $\mathcal H = L^2(\mathbb R^d \times \mathbb R^d)$.
Let $f = \sum_k \varphi_k \in \mathcal H$. Then, $T(f) = \sum_k T(\varphi_k) = \sum_k \lambda_k \varphi_k \in \mathcal H$. So,
$$\Vert Tf \Vert ^2 = \Vert \sum_k \lambda_k \varphi_k \Vert ^2 = \sum_k |\lambda_k|^2 \Vert \varphi_k \Vert ^2 = \sum_k |\lambda_k|^2$$
Since $Tf \in \mathcal H$, $\sum_k |\lambda_k|^2 < \infty$
I'd like to check whether this is correct.
I guess this method is totally wrong. What method should I use for $\sum_k |\lambda_k|^2 < \infty$?
(b)
I tried to solve this in the following way, which isn't sure.
Let $Tf(x) = \int_{\mathbb R ^d}K(x,y)f(y)dy$.
$T(\varphi_k(x)) = \int_{\mathbb R ^d}K(x,y)\varphi_k(y)dy = \lambda_k \varphi_k (x)$
$(T(\varphi_k(x)), \varphi_k(x)) = (\lambda_k \varphi_k (x), \varphi_k (x)) = \lambda_k\Vert \varphi_k(x)\Vert ^2 = \lambda_k$
Also, $(T(\varphi_k(x)), \varphi_k(x)) = \int_{\mathbb R ^d}\int_{\mathbb R ^d}K(x,y)\varphi_k(y)dy \overline{\varphi_k(x)} dx = \int_{\mathbb R ^{2d}}K(x,y)\overline{\varphi_k(x)} \varphi_k(y)d(x,y)$
So, $K(x,y) \sim \sum_k \lambda_k \overline{\varphi_k(x)} \varphi_k(y)$
...
I'd like to know whether there is a wrong point. The answer in the textbook has no conjugate form.
Also, the eigenvectors are estimated to be REAL, which I'm not sure about. Some explanation about it would be grateful.
(c)
I referred to this page to understand this problem, and it says that $\Vert T - T_N \Vert \to 0$ implies that $T$ is a Hilbert-Schmidt operator. But I can't figure out what this means. I thought I have to find some $K$ where $Tf(x) = \int K(x,y)f(y)dy$ and $K \in L^2(\mathbb R^d \times \mathbb R^d)$, which isn't clear in the page. Without showing the existence of such $K$, how could I finish this problem?
Any help about this would be appreciated. Thank you.
$(a)$ Why would $f$ be in $\mathcal{H}$ ? in fact, you can be sure that $f$ never exists in $\mathcal{H}$ because $$ \|f\| = \left\|\sum_{n \geqslant 0} \varphi_n\right\| = \sum_{n \geqslant 0} \|\varphi_n\| = \sum_{n \geqslant 0} 1 = +\infty. $$ In fact, $T$ is assumed to be Hilbert-Schmidt which means by definition that $\sum_{n \geqslant 0} \|T\varphi_n\|^2 = \sum_{n \geqslant 0} \lambda_n^2 < +\infty$.
$(b)$ There is no conjugate form because all your functions are real. To prove that $K(x,y) = \sum_{n \geqslant 0} \lambda_n\varphi_n(x)\varphi_n(y)$ in $L^2(\mathbb{R}^{2d})$, begin by proving that the sum $\tilde{K}(x,y) = \sum_{n \geqslant 0} \lambda_n\varphi_n(x)\varphi_n(y)$ converges in $L^2(\mathbb{R}^{2d})$. For this, use point $(a)$, the fact that for all $n$, $(x,y) \mapsto \varphi_n(x)\varphi_n(y) \in L^2(\mathbb{R}^{2d})$ with $L^2$-norm equals $1$ and the Fubini-Tonelli theorem that allows you to exchange the sum and the integral.
Then, verify that for all $n$, $\tilde{T}\varphi_n = \lambda_n\varphi_n$ where $\tilde{T}$ is the Hilbert-Schmidt operator defined using $\tilde{K}$. Use it to deduce that $\tilde{T} = T$ hence $\tilde{K} = K$ a.e..
$(c)$ This is the same idea than for question $(b)$. If $\sum_{n \geqslant 0} \lambda_n^2 < +\infty$, then $K(x,y) = \sum_{n \geqslant 0} \lambda_n\varphi_n(x)\varphi_n(y)$ is in $L^2(\mathbb{R}^{2d})$. Then, define $\tilde{T}$ to be the operator with kernel $K$, prove that for all $n$, $\tilde{T}\varphi_n = \lambda_n\varphi_n$ and deduce that $T = \tilde{T}$ is a Hilbert_Schmidt operator.