Schur test

From Wikipedia, the free encyclopedia
Jump to: navigation, search

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:

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

If there exist functions \,p(x)>0 and \,q(x)>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(x) 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[edit]

A common usage of the Schur test is to take \,p(x)=q(x)=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:[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}\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[edit]

Using the Cauchy–Schwarz inequality and the 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 the 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).

See also[edit]

References[edit]

  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