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

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

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

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

for almost all $\,x$ and

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

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

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

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

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

Common usage and Young's inequality

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

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

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

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

if

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

where $r\,$ satisfies ${\frac {1}{r}}=1-{\Big (}{\frac {1}{p}}-{\frac {1}{q}}{\Big )}$, for some $1\leq p\leq q\leq \infty$, 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 $\Vert T\Vert _{L^{p}\to L^{q}}\leq C.$

Proof

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

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

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

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

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

See also

• Hardy–Littlewood–Sobolev inequality