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


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

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


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

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

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

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_{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 for integral operators:

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 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 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 $$f\in L^2(Y)$$.