# Hilbert-Schmidt Operators

1. Mar 8, 2009

### phreak

1. The problem statement, all variables and given/known data

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.

2. Relevant 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.

3. 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.

Last edited: Mar 8, 2009
2. Mar 8, 2009

### 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.