The Hilbert-Schmidt Test

Theorem 3.9: Let $(\Omega, \mu)$ be any measure space and let $k \in L^2(\Omega \times \Omega)$. Then the operator $K$ with integral $k$ is bounded from $L^2(\Omega) \rightarrow L^2(\Omega)$, and:

Proof:

Using the Cauchy-Schwarz inequality:

Squaring over and integrating over $x$ yields:

$\blacksquare$

Remark: This is called the Hilbert-Schmidt bound, and $\|k\|_2$ is called the _Hilbert-Schmidt norm of $K$:

Operators such that $\|K\|_{HS} < \infty$ are called Hilbert-Schmidt operators, and the norm notation is justified by the fact that this is the $L^2$-norm of the integral kernel, and therefore the axioms of the norm are satisfied on the class of Hilbert-Schmidt operators. Later, we will see that this norm has a deeper significance and that Hilbert-Schmidt operators are not only bounded, but also compact.

Example 3.10: For an "infinite matrix" $\{k_{ij} : i,j \in \mathbb{N}\}$ and the corresponding operator $K \in \mathcal{B}(\ell^2)$ we have:

Example 3.11: Consider the Hankel matrix $T = \{t_{ij}\}_{i,j=0}^{\infty}$ on $\ell^2$. It is straightforward to see that $T$ is of Hilbert-Schmidt class if and only if:

Further Examples

It is not the case that every bounded integral operator satisfies the Schur test or the Hilbert-Schmidt test.

Example 3.12: The operator $\mathcal{F}$ on $L^2(\mathbb{R}^d)$, defined by:

is the operator of the Fourier transform. It is an integral operator which satisfies neither the Schur nor the Hilbert-Schmidt test. Yet, the Fourier transform is bounded and even unitary, this will be proven in the exercises.

Usually operator theory studies not only individual operators, but classes of operators, labelled by some functional parameter. This parameter is usually a function or a sequence. The simplest examples are the multiplication operators and the convolution operators. Here are more examples:

Example 3.13: Let $a,b \in \ell^{\infty}$ be sequences such that $a_n > 0$ and $b_n \in \mathbb{R}$ for all $n$. Consider the operator $J$ on $\ell^2$ given by:

Then $J$ is called the Jacobi matrix. The sequences $a,b$ are the Jacobi parameters of $J$. Using Schur's test, one can see that the assumptions $a,b \in \ell^{\infty}$ imply the boundedness of $J$ on $\ell^2$, and one can show that it is also a necessary condition.

Example 3.14: In $L^2(0,2\pi)$, one has the orthonormal basis:

Let $H^2 \subset L^2$ be the span of $\{e_n\}_{n=0}^{\infty}$; this is the Hardy space; it can be identified with the space of functions analytic in the unit disc in $\mathbb{C}$ having $L^2$ boundary values on the boundary circle of the disc. Let $P_+$ be the orthogonal projection onto $H^2$ and $P_-$ be the orthogonal projection onto $(H^2)^{\perp}$. For a bounded function $\varphi$ on $[0,2\pi]$, the Toeplitz operator with the symbol $\varphi$ can be defined as $P_+ \varphi P_+$. The Hankel operator with the symbol $\varphi$ can be defined as $P_- \varphi P_+$.

Example 3.15: Let $a : \mathbb{R}^d \times \mathbb{R}^d \rightarrow \mathbb{C}$ be a bounded function such that all of its partial derivatives are bounded on $\mathbb{R}^d \times \mathbb{R}^d$. The pseudodifferential operator with the Weyl symbol $a$ is defined as:

The integral above usually has to be understood in some regularised sense. It is a very non-trivial fact (known as the Calderon-Vaillancourt theorem) that the boundedness of $a$ and its derivatives implies that the operator $A$ is bounded on $L^2(\mathbb{R}^d)$.