Boundedness Tests and Further Examples

Here we prove two important sufficient conditions for boundedness of integral operators: The Schur test and the Hilbert-Schmidt test. Using these conditions, we discuss boundedness of some concrete classes of operators.

Integral Operators

Definition 3.1: Let $(\Omega, \mu)$ be a measure space, and $k : \Omega \times \Omega \rightarrow \mathbb{C}$ be a measureable function. Then, the formula:

[Kf](x) \coloneqq \int_{\Omega} k(x,y)f(y)d \mu(y)

formally defines a linear operator. Operators of this type are called integral operators and $k$ is called the integral kernel of $K$ .

Discussion:

The reason why we state that the formula above defines formally a linear operator is that we haven't defined its domain or codomain, which can depend on the integral kernel $k$ . To say that the operator is linear in such a case should be understood in the sense that "if $f$ and $g$ are such that $Kf$ and $Kg$ are well-defined, then for all $\alpha \in \mathbb{C}$ , $K(\alpha f + g)$ is also well-defined and is equal to $\alpha Kf + Kg$ ".

The aim of this section is to give conditions on the integral kernel $k$ so that we can identify the domain and codomain on which $K$ is well-defined as a bounded operator. In examples the measure space $(\Omega,\mu)$ will often be either $\mathbb{N}$ or $\mathbb{Z}$ with the counting measure; or $\mathbb{R}^d$ with the Lebesgue measure. In applications, it can be a good idea to translate the definition of some operator into one where it looks like an integral operator for an appropriate measure space $(\Omega,\mu)$ .

Schur's Test

Theorem 3.2: Let $(\Omega, \mu)$ be a measure space, $k: \Omega \times \Omega \rightarrow \mathbb{C}$ be a measurable function such that:

m_1 \coloneqq \text{ess } \sup_{x \in \Omega} \int_{\Omega} |k(x,y)|d \mu(y) < \infty, \\ m_2 \coloneqq \text{ess } \sup_{y \in \Omega} \int_{\Omega} |k(x,y)|d \mu(x) < \infty.

Then $K \in \mathcal{B}(L^2(\Omega))$ , where $K$ is the integral operator with integral kernel $k$ . Furthermore, it satisfies Schur's bound:

\|K\| \leq m_1^{\frac{1}{2}}m_2^{\frac{1}{2}}.

Remark: As usual, the use of the essential supremum in 3.2 is due to the measure-theoretic nature of the definition. One often computes these quantities as a supremum, when $k$ is a nice enough function.

Proof:

Applying the Cauchy-Schwarz inequality, we have that:

|[Kf](x)| \leq \int_{\Omega} |k(x,y)||f(y)|d \mu(y) \\ \leq \left(\int_{\Omega} |k(x,y)|d \mu(y) \right)^{\frac{1}{2}} \left(\int_{\Omega}|k(x,y)||f(y)|^2 d\mu(y)\right)^{\frac{1}{2}} \\ \leq m_1^{\frac{1}{2}} \left(\int_{\Omega} |k(x,y)| |f(y)|^2 d \mu(y) \right)^{\frac{1}{2}} .

Squaring this inequality and integrating with respect to the $x$ variable we obtain that:

\|Kf\|_2^2 \leq m_1 \int_{\Omega} \int_{\Omega} |k(x,y)| |f(y)|^2 d\mu(y) d\mu(x) \\ = \int_{\Omega} |f(y)|^2 \int_{\Omega} |k(x,y)| d\mu(x) d\mu(y) \\ \leq m_1 m_2 \|f\|_2^2;

where changing the integral order in the equality is justified from 3.2 and using Hölder's inequality with exponents $(1,\infty)$ in the last line. $\blacksquare$

Let us now see a few examples where Schur's test allows us to prove that an operator is bounded.

Example 3.3: Let $\{k_{ij} : i,j \in \mathbb{N}\}$ be an "infinite matrix" of complex numbers. Then if we take $\Omega = \mathbb{N}$ and $\mu$ be the counting measure, we can define an operator with integral kernel $k$ as:

[Kx]_i = \sum_j k_{ij}x_j.

If:

m_1 = \sup_i \sum_j |k_{ij}| < \infty \ \text{and} \ m_2 = \sup_j \sum_i |k_{ij}| < \infty,

then Schur's test tells us that $K: \ell^2 \rightarrow \ell^2$ and:

\|K\| \leq m_1^{\frac{1}{2}}m_2^{\frac{1}{2}}.

Note: $\ell^2 = L^2(\mathbb{N})$ with the counting measure.

Example 3.4: Let $t \in \ell^1$ . Consider the operator on $\ell^2$ defined by the "matrix" $\{t_{i+j}\}_{i,j=0}^{\infty}$ (A matrix of this type is called a Hankel matrix). By the Schur test, the condition $t \in \ell^1$ ensures the boundedness of this operator.

Example 3.5: Let $\Omega \subset \mathbb{R}^d$ be a bounded domain, let $\alpha > 0$ , and let $k_0 \in L^{\infty}(\Omega \times \Omega)$ . Define $K_{\alpha}: L^2(\Omega) \rightarrow L^2(\Omega)$ as the integral operator given by:

[Kf](x) = \int_{\Omega} \frac{k_0(x,y)}{|x-y|^{\alpha}}f(y) dy.

Then Schur's test tells us that $K_{\alpha}$ is bounded whenever $\alpha \in (0,d)$ . Operators of this form appear naturally in many problems in analysis. They are called integral operators with weak singularity. For $\alpha = d$ operators of this type are, in general, no longer bounded. However one has: Example 3.6: Let $H \in L^2(\mathbb{R})$ be defined by:

(Hf)(x) = \frac{1}{\pi i} \ \text{p.v.} \int_{\mathbb{R}} \frac{f(y)}{x-y} dy.

where:

\text{p.v. } \int_{\mathbb{R}}\frac{f(y)}{x-y} dy = \lim_{\varepsilon \rightarrow 0}\int{\mathbb{R} - [-\varepsilon,\varepsilon]} \frac{f(y)}{x-y} dy

Then $H$ is bounded; this is a very non-trivial statement which we will not prove in this course. It is proven using Fourier analysis. $H$ is the Hilbert transform. In fact, it is not only bounded but unitary (i.e. $\|Hf\| = \|f\|$ ).

Example 3.7: Let $r \in L^1(\mathbb{R}^d)$ . Consider the operator $T \in L^2(\mathbb{R}^d)$ given by:

[f \star r](x) \coloneqq [Rf](x) = \int_{\mathbb{R}^d} r(x-y)f(y)dy.

$Tf$ is called the convolution of $r$ and $f$ . Operators of this type are called the convolution operators. In this case, Schur's test applies and $R$ is bounded with:

\|R\| \leq \|r\|_1.

Remark 3.8: There is a little subtlety in defining some of the above operators. It can be illustrated for the convolution operator. Let $f \in L^2(\mathbb{R}^d)$ ; then a priori it is not clear why the integral above would exist for any $x$ at all. One should define $T$ on some dense set $D$ for example $D=L^2(\mathbb{R}^d) \cap L^{\infty}(\mathbb{R}^d)$ , and then use the Schur bound to prove that there exists an extension of $T$ to the whole of $L^2(\mathbb{R}^2)$ .