Understanding Hilbert-Schmidt Operators: Eigenvectors and Symmetry

[phreak]

Homework Statement

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. K(x,y) \sim \sum_k \lambda_k \varphi_k(x) \varphi_k(y) is the expansion of K in the basis \{ \varphi_{k,j} \}, and

b. Suppose T is a compact operator which is symmetric. Then T is of Hilbert-Schmidt type if and only if \sum_k |\lambda_k|^2 < +\infty, where \{ \lambda_k \} are the eigenvalues of T counted according to their multiplicities.

Homework Equations

A Hilbert-Schmidt operator is an operator of the form T(f)(x) = \int_{\mathbb{R}^d} K(x,y) f(y) dy, where K(x,y) is in L^2(\mathbb{R}^d).

An earlier problem that I have already done was prove that given \{\varphi_k\}_{k=1}^{\infty} as an orthonormal basis for L^2(\mathbb{R}^d), the set \{ \varphi_{k,j} \}, where \varphi_{k,j}(x,y):=\varphi_k(x) \varphi_j(y), is an orthonormal basis for L^2(\mathbb{R}^d \times \mathbb{R}^d).

Diagonalize in this context means the eigenvectors of the compact operator T which serves as a basis for L^2(\mathbb{R}^d), whose existence is guaranteed by the spectral theorem.

Symmetric here means self-adjoint.

The Attempt at a Solution

There was a previous part of the problem which I've already solved which was that \sum_k |\lambda_k|^2 < +\infty. Therefore, I've got half of the solution to part c. However, I'm not sure how to prove either of the statements, nor can I even find a way to proceed.

What should be the biggest step for part a. would be to write \int K(x,y) \varphi_k(y) dy = \lambda_k \varphi_k(x)\, \forall x\in \mathbb{R}^d, and to somehow extract K(x,y) out of this. However, I don't see how one could do this... it's rather attached to the integral, and I don't know any more expressions in which to start out with K.

As I said, I'm really not sure how to proceed on either of these. I would really appreciate a clue to get me started though. Thanks.

 

[phreak]

Ok, so I think I proved c. It is:

\| K \|_{L^2}^2 = \int |K(x,y)|^2 dxdy = \int | \sum_k \lambda_k \varphi_k(x) \varphi_k(y)|^2 dxdx \le \int \sum_k |\lambda_k|^2 |\varphi_k(x)|^2 |\varphi_k(y)|^2 dx dy = \sum_k |\lambda_k|^2 \int |\varphi_k(x)|^2 |\varphi_k(y)|^2 dxdx =

\sum_k |\lambda_k|^2 \cdot \left( \int |\varphi_k(x)|^2 dx \right) \cdot \left( \int |\varphi_k(y)|^2 dy \right) = \sum_k |\lambda_k|^2. Therefore, as long as \sum_k |\lambda_k|^2 < +\infty, we must have \| K \| < +\infty, which implies that K\in L^2(\mathbb{R}^d \times \mathbb{R}^d).

I'm still stumped on a., though.