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 $\,K(x,y)$ , $x\in X$ , $y\in Y$ :

Tf(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\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 $\,p(x)$ , $\,q(y)$ 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 $\,p(x)=q(y)=1.$ 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 for integral operators:[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}\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 $f\in L^{2}(Y)$ .

References

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.
I. Schur, Bemerkungen zur Theorie der Beschränkten Bilinearformen mit unendlich vielen Veränderlichen, J. reine angew. Math. 140 (1911), 1–28.
Theorem 0.3.1 in: C. D. Sogge, Fourier integral operators in classical analysis, Cambridge University Press, 1993. ISBN 0-521-43464-5

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[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.

[2] I. Schur, Bemerkungen zur Theorie der Beschränkten Bilinearformen mit unendlich vielen Veränderlichen, J. reine angew. Math. 140 (1911), 1–28.

[3] Theorem 0.3.1 in: C. D. Sogge, Fourier integral operators in classical analysis, Cambridge University Press, 1993. ISBN 0-521-43464-5

Schur test

Common usage and Young's inequality

Proof

See also

References