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Hilbert-Schmidt Operators

  1. Mar 8, 2009 #1
    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 [tex]\varphi_k(x)[/tex] be the eigenvectors (with eigenvalues [tex]\lambda_k[/tex]) that diagonalize T . Then:

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

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

    2. Relevant equations

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

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

    Diagonalize in this context means the eigenvectors of the compact operator T which serves as a basis for [tex]L^2(\mathbb{R}^d)[/tex], 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 [tex]\sum_k |\lambda_k|^2 < +\infty[/tex]. 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 [tex]\int K(x,y) \varphi_k(y) dy = \lambda_k \varphi_k(x)\, \forall x\in \mathbb{R}^d[/tex], 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. jcsd
  3. Mar 8, 2009 #2
    Ok, so I think I proved c. It is:

    [tex] \| 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 =[/tex]

    [tex]\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[/tex]. Therefore, as long as [tex]\sum_k |\lambda_k|^2 < +\infty[/tex], we must have [tex]\| K \| < +\infty[/tex], which implies that [tex]K\in L^2(\mathbb{R}^d \times \mathbb{R}^d)[/tex].

    I'm still stumped on a., though.
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