# Schur test

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

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

${\displaystyle Tf(x)=\int _{Y}K(x,y)f(y)\,dy.}$

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

${\displaystyle (1)\qquad \int _{Y}K(x,y)q(y)\,dy\leq \alpha p(x)}$

for almost all ${\displaystyle \,x}$ and

${\displaystyle (2)\qquad \int _{X}p(x)K(x,y)\,dx\leq \beta q(y)}$

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

${\displaystyle \Vert T\Vert _{L^{2}\to L^{2}}\leq {\sqrt {\alpha \beta }}.}$

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

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

## Common usage and Young's inequality

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

${\displaystyle \Vert T\Vert _{L^{2}\to L^{2}}^{2}\leq \sup _{x\in X}\int _{Y}|K(x,y)|\,dy\cdot \sup _{y\in Y}\int _{X}|K(x,y)|\,dx.}$

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

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

if

${\displaystyle \sup _{x}{\Big (}\int _{Y}|K(x,y)|^{r}\,dy{\Big )}^{1/r}+\sup _{y}{\Big (}\int _{X}|K(x,y)|^{r}\,dx{\Big )}^{1/r}\leq C,}$

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

## Proof

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

{\displaystyle {\begin{aligned}|Tf(x)|^{2}=\left|\int _{Y}K(x,y)f(y)\,dy\right|^{2}&\leq \left(\int _{Y}K(x,y)q(y)\,dy\right)\left(\int _{Y}{\frac {K(x,y)f(y)^{2}}{q(y)}}dy\right)\\&\leq \alpha p(x)\int _{Y}{\frac {K(x,y)f(y)^{2}}{q(y)}}\,dy.\end{aligned}}}

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

${\displaystyle \Vert Tf\Vert _{L^{2}}^{2}\leq \alpha \int _{Y}\left(\int _{X}p(x)K(x,y)\,dx\right){\frac {f(y)^{2}}{q(y)}}\,dy\leq \alpha \beta \int _{Y}f(y)^{2}dy=\alpha \beta \Vert f\Vert _{L^{2}}^{2}.}$

It follows that ${\displaystyle \Vert Tf\Vert _{L^{2}}\leq {\sqrt {\alpha \beta }}\Vert f\Vert _{L^{2}}}$ for any ${\displaystyle f\in L^{2}(Y)}$.

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