Schur test

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

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

[math]\displaystyle{ T f(x)=\int_Y K(x,y)f(y)\,dy. }[/math]

If there exist real functions [math]\displaystyle{ \,p(x)\gt 0 }[/math] and [math]\displaystyle{ \,q(y)\gt 0 }[/math] and numbers [math]\displaystyle{ \,\alpha,\beta\gt 0 }[/math] such that

[math]\displaystyle{ (1)\qquad \int_Y K(x,y)q(y)\,dy\le\alpha p(x) }[/math]

for almost all [math]\displaystyle{ \,x }[/math] and

[math]\displaystyle{ (2)\qquad \int_X p(x)K(x,y)\,dx\le\beta q(y) }[/math]

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

[math]\displaystyle{ \Vert T\Vert_{L^2\to L^2} \le\sqrt{\alpha\beta}. }[/math]

Such functions [math]\displaystyle{ \,p(x) }[/math], [math]\displaystyle{ \,q(y) }[/math] are called the Schur test functions.

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

Common usage and Young's inequality

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

[math]\displaystyle{ \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. }[/math]

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

A similar statement about [math]\displaystyle{ L^p\to L^q }[/math] operator norms is known as Young's inequality for integral operators:

if

[math]\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}\le C, }[/math]

where [math]\displaystyle{ r }[/math] satisfies [math]\displaystyle{ \frac 1 r=1-\Big(\frac 1 p-\frac 1 q\Big) }[/math], for some [math]\displaystyle{ 1\le p\le q\le\infty }[/math], then the operator [math]\displaystyle{ Tf(x)=\int_Y K(x,y)f(y)\,dy }[/math] extends to a continuous operator [math]\displaystyle{ T:L^p(Y)\to L^q(X) }[/math], with [math]\displaystyle{ \Vert T\Vert_{L^p\to L^q}\le C. }[/math]

Proof

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

[math]\displaystyle{ \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} }[/math]

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

[math]\displaystyle{ \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. }[/math]

It follows that [math]\displaystyle{ \Vert T f\Vert_{L^2}\le\sqrt{\alpha\beta}\Vert f\Vert_{L^2} }[/math] for any [math]\displaystyle{ f\in L^2(Y) }[/math].

See also

• Hardy–Littlewood–Sobolev inequality