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 $\mathbb^n$). 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 real functions $\,p(x)>0$ and $\,q(y)>0$ 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

:$\Vert T\Vert_ \le\sqrt.$

Such functions $\,p(x)$, $\,q(y)$ 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 $\,p(x)=q(y)=1.$ Then we get:

:$\Vert T\Vert^2_\le \sup_\int_Y, K(x,y), \, dy \cdot \sup_\int_X, K(x,y), \, dx.$

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 for integral operators:Theorem 0.3.1 in: C. D. Sogge, ''Fourier integral operators in classical analysis'', Cambridge University Press, 1993.

if

:$\sup_x\Big(\int_Y, K(x,y), ^r\,dy\Big)^ + \sup_y\Big(\int_X, K(x,y), ^r\,dx\Big)^\le C,$

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 $\Vert T\Vert_\le C.$

Proof

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

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

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

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

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

See also

* Hardy–Littlewood–Sobolev inequality

References

Inequalities