Effective number of bits

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

$Tf(x)=\underset{Y}{\int }K(x,y)f(y)dy.$

If there exist functions $p(x)>0$ and $q(x)>0$ and numbers $\alpha ,\beta >0$ such that

$(1)\underset{Y}{\int }K(x,y)q(y)dy\le \alpha p(x)$

for almost all $x$ and

$(2)\underset{X}{\int }p(x)K(x,y)dx\le \beta q(y)$

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 $\alpha =\beta =1$.

Common usage and Young's inequality

A common usage of the Schur test is to take $p(x)=q(x)=1.$ Then we get:

$\Vert T{\Vert }_{L^2\to L^2}^2\le {\sup }_{x\in X}\underset{Y}{\int }|K(x,y)|dy\cdot {\sup }_{y\in Y}\underset{X}{\int }|K(x,y)|dx.$

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:

if

${\sup }_x\left(\underset{Y}{\int }\right|K(x,y){|}^{r}dy{)}^{1/r}+{\sup }_y\left(\underset{X}{\int }\right|K(x,y){|}^{r}dx{)}^{1/r}\le C,$

where $r$ satisfies $\frac{1}{r}=1-(\frac{1}{p}-\frac{1}{q})$, for some $1\le p\le q\le \mathrm{\infty }$, then the operator $Tf(x)=\underset{Y}{\int }K(x,y)f(y)dy$ 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 the inequality (1), we get:

$\begin{array}{rl}|Tf(x){|}^2={|\underset{Y}{\int }K(x,y)f(y)dy|}^2& \le (\underset{Y}{\int }K(x,y)q(y)dy)\left(\underset{Y}{\int }\frac{K(x,y)f(y{)}^2}{q(y)}dy\right)\\ & \le \alpha p(x)\underset{Y}{\int }\frac{K(x,y)f(y{)}^2}{q(y)}dy.\end{array}$

Integrating the above relation in $x$, using Fubini's Theorem, and applying the inequality (2), we get:

$\Vert Tf{\Vert }_{L^2}^2\le \alpha \underset{Y}{\int }(\underset{X}{\int }p(x)K(x,y)dx)\frac{f(y{)}^2}{q(y)}dy\le \alpha \beta \underset{Y}{\int }f(y{)}^{2}dy=\alpha \beta \Vert f{\Vert }_{L^2}^2.$

It follows that $\Vert Tf{\Vert }_{L^2}\le \sqrt{\alpha \beta }\Vert f{\Vert }_{L^2}$ for any $f\in L^2(Y)$.

See also

• Hardy–Littlewood–Sobolev inequality