# Schur test

In mathematical analysis, the Schur test, named after German mathematician Issai Schur, is a bound on the $L^2\to L^2$ operator norm of an integral operator in terms of its Schwartz kernel (see Schwartz kernel theorem).

Here is one version.[1] Let $X,\,Y$ be two measurable spaces (such as $\mathbb{R}^n$). Let $\,T$ be an integral operator with the non-negative Schwartz kernel $\,K(x,y)$, $x\in X$, $y\in Y$:

$T f(x)=\int_Y K(x,y)f(y)\,dy.$

If there exist functions $\,p(x)>0$ and $\,q(x)>0$ and numbers $\,\alpha,\beta>0$ such that

$(1)\qquad \int_Y K(x,y)q(y)\,dy\le\alpha p(x)$

for almost all $\,x$ and

$(2)\qquad \int_X p(x)K(x,y)\,dx\le\beta q(y)$

for almost all $\,y$, then $\,T$ extends to a continuous operator $T:L^2\to L^2$ with the operator norm

$\Vert T\Vert_{L^2\to L^2} \le\sqrt{\alpha\beta}.$

Such functions $\,p(x)$, $\,q(x)$ are called the Schur test functions.

In the original version, $\,T$ is a matrix and $\,\alpha=\beta=1$.[2]

## Common usage and Young's inequality

A common usage of the Schur test is to take $\,p(x)=q(x)=1.$ Then we get:

$\Vert T\Vert^2_{L^2\to L^2}\le \sup_{x\in X}\int_Y|K(x,y)| \, dy \cdot \sup_{y\in Y}\int_X|K(x,y)| \, dx.$

This inequality is valid no matter whether the Schwartz kernel $\,K(x,y)$ is non-negative or not.

A similar statement about $L^p\to L^q$ operator norms is known as Young's inequality:[3]

if

$\sup_x\Big(\int_Y|K(x,y)|^r\,dy\Big)^{1/r} + \sup_y\Big(\int_X|K(x,y)|^r\,dx\Big)^{1/r}\le C,$

where $r\,$ satisfies $\frac 1 r=1-\Big(\frac 1 p-\frac 1 q\Big)$, for some $1\le p\le q\le\infty$, then the operator $Tf(x)=\int_Y K(x,y)f(y)\,dy$ extends to a continuous operator $T:L^p(Y)\to L^q(X)$, with $\Vert T\Vert_{L^p\to L^q}\le C.$

## Proof

Using the Cauchy–Schwarz inequality and the inequality (1), we get:

\begin{align} |Tf(x)|^2=\left|\int_Y K(x,y)f(y)\,dy\right|^2 &\le \left(\int_Y K(x,y)q(y)\,dy\right) \left(\int_Y \frac{K(x,y)f(y)^2}{q(y)} dy\right)\\ &\le\alpha p(x)\int_Y \frac{K(x,y)f(y)^2}{q(y)} \, dy. \end{align}

Integrating the above relation in $x$, using Fubini's Theorem, and applying the inequality (2), we get:

$\Vert T f\Vert_{L^2}^2 \le \alpha \int_Y \left(\int_X p(x)K(x,y)\,dx\right) \frac{f(y)^2}{q(y)} \, dy \le\alpha\beta \int_Y f(y)^2 dy =\alpha\beta\Vert f\Vert_{L^2}^2.$

It follows that $\Vert T f\Vert_{L^2}\le\sqrt{\alpha\beta}\Vert f\Vert_{L^2}$ for any $f\in L^2(Y)$.

1. ^ Paul Richard Halmos and Viakalathur Shankar Sunder, Bounded integral operators on $L^{2}$ spaces, Ergebnisse der Mathematik und ihrer Grenzgebiete (Results in Mathematics and Related Areas), vol. 96., Springer-Verlag, Berlin, 1978. Theorem 5.2.