Understanding Hilbert-Schmidt Operators: Eigenvectors and Symmetry

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In summary, we have seen that a Hilbert-Schmidt kernel, K, which is real and symmetric, gives rise to a compact and symmetric operator, T. The eigenvectors, \varphi_k(x), of T with eigenvalues, \lambda_k, form an orthonormal basis for L^2(\mathbb{R}^d). The expansion of K in this basis is given by K(x,y) \sim \sum_k \lambda_k \varphi_k(x) \varphi_k(y). Additionally, for T to be of Hilbert-Schmidt type, it is necessary and sufficient for \sum_k |\lambda_k|^2 < +\infty. This can be shown through the spectral
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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 [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.

Homework 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.

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.
 
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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.
 

FAQ: Understanding Hilbert-Schmidt Operators: Eigenvectors and Symmetry

What are Hilbert-Schmidt Operators?

Hilbert-Schmidt Operators are linear transformations between Hilbert spaces that satisfy a specific condition called the Hilbert-Schmidt norm. They are commonly used in functional analysis and have applications in quantum mechanics, signal processing, and differential equations.

How are Hilbert-Schmidt Operators defined?

A Hilbert-Schmidt Operator is defined as a bounded linear operator on a Hilbert space that has a finite Hilbert-Schmidt norm. The norm is calculated by taking the square root of the trace of the operator's adjoint multiplied by the operator itself.

What is the significance of the Hilbert-Schmidt norm?

The Hilbert-Schmidt norm is significant because it allows us to measure the size of a Hilbert-Schmidt Operator and determine whether it is finite. This is important in applications where we need to ensure that the operator is well-behaved and its effects on the Hilbert space are finite.

How are Hilbert-Schmidt Operators used in quantum mechanics?

In quantum mechanics, Hilbert-Schmidt Operators are used to represent physical observables such as position, momentum, and energy. They also play a role in the mathematical formulation of quantum mechanical systems and their dynamics.

Can Hilbert-Schmidt Operators be diagonalized?

Yes, Hilbert-Schmidt Operators can be diagonalized if they are compact, meaning they map bounded sets to relatively compact sets. This is known as the spectral theorem for compact operators and is an important result in the theory of Hilbert spaces.

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