Schur test explained

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 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 ) \int _{Y} K(x,y)q(y)dy\le\alpha p(x) x and( 2 ) \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 .^[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 ^{2} _{L ^{2\to} L ^{2}} \le \sup _{x\in} \int _{Y|K(x,y)|} dy ⋅ \sup _{y\in} \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(\int} ^{rdy )} ^{Y|K(x,y)|} ^{1/r} + \sup _{y(\int} ^{rdx )} ^{X|K(x,y)|} ^{1/r} \le C,

where

r satisfies (1)/(r =1- ( (1)/(p - (1)/(q )))) , for some 1 \le p\le q\le inf ty , 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} (K(x,y)f(y) ^{2})/(q(y)) dy\right)\\ &\le\alpha p(x)\int _{Y} (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 ^{2} _{L ^{2}} \le \alpha \int _{Y} \left(\int _{X} p(x)K(x,y)dx\right) (f(y) ^{2})/(q(y)) dy \le\alpha\beta \int _{Y} f(y) ^{2} dy = \alpha\beta\Vert f\Vert ^{2 .} _{L ^{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)} .

See also

• Hardy–Littlewood–Sobolev inequality

References

1. [Paul Richard Halmos]
2. [I. Schur]
3. Theorem 0.3.1 in: C. D. Sogge, Fourier integral operators in classical analysis, Cambridge University Press, 1993.