Schur’s Test

Theorem 1.

(Schur’s Test) Let $\mathrm{(}\mathrm{X}\mathrm{,}\mathrm{\mu }\mathrm{)}$ be a measure space ($\mathrm{\mu }$ a positive measure). Let $\mathrm{K}$ be a positive, measurable function on $\mathrm{X}\mathrm{\times }\mathrm{X}$. Define the operator

$Tf(x)$$:={\displaystyle {\int }_X}K(x,y)f(y)𝑑\mu (y),x\in X$

If for some there exists a measurable, strictly positive function $h$ and a constant $M\mathrm{>}\mathrm{0}$ such that

${\displaystyle {\int }_X}K(x,y)h{(y)}^q𝑑\mu (y)\le Mh{(x)}^q$

${\displaystyle {\int }_X}K(x,y)h{(x)}^p𝑑\mu (x)\le Mh{(y)}^p$

with $p^{\mathrm{-}\mathrm{1}}\mathrm{+}{q}^{\mathrm{-}\mathrm{1}}\mathrm{=}\mathrm{1}$, then $\mathrm{||}T\mathrm{||}\mathrm{\le }M$ in $L^p\mathit{}\mathrm{(}X\mathrm{,}d\mathit{}\mu \mathrm{)}$.

Proof.

Let $f\in L^p(X,d\mu )$. We have

$|Tf(x)|$$\le {\displaystyle {\int }_X}h(y)h{(y)}^{-1}|f(y)|K(x,y)𝑑\mu (y)$

hence by Hoelder’s inequality

$|Tf(x)|$$\le {\left[{\displaystyle {\int }_X}K(x,y)h{(y)}^q𝑑\mu (y)\right]}^{\frac{1}{q}}{\left[{\displaystyle {\int }_X}K(x,y)h{(y)}^{-p}{|f(y)|}^p𝑑\mu (y)\right]}^{\frac{1}{p}}$

By the first inequality in the assumption we have

$|Tf(x)|$$\le M^{\frac{1}{q}}h(x){\left[{\displaystyle {\int }_X}K(x,y)h{(y)}^{-p}{|f(y)|}^p𝑑\mu (y)\right]}^{\frac{1}{p}}$

Evaluating ${||Tf||}_p^p$ by Fubini and the second inequality in the assumption we obtain

${\displaystyle {\int }_X}{|Tf(x)|}^p𝑑\mu (x)$$\le M^p{\displaystyle {\int }_X}{|f(y)|}^p𝑑\mu (y)$

This completes the proof.∎

A noted special case is Young’s Inequality

Corollary 1.

(Young)

Let $K\mathrm{:}{\mathrm{R}}^n\mathrm{\times }{\mathrm{R}}^n\mathrm{\to }\mathrm{C}$ be Borel-measurable such that there is a constant $C\mathrm{>}\mathrm{0}$ with

$\underset{x\in {ℝ}^n}{sup}{\displaystyle {\int }_{{ℝ}^n}}|K(x,y)|𝑑{\lambda }^n(y)\le C$

$\underset{y\in {ℝ}^n}{sup}{\displaystyle {\int }_{{ℝ}^n}}|K(x,y)|𝑑{\lambda }^n(x)\le C$

For $f\mathrm{\in }{L}^p\mathit{}\mathrm{(}{\mathrm{R}}^n\mathrm{)}$ ($\mathrm{1}\mathrm{\le }p\mathrm{\le }\mathrm{+}\mathrm{\infty }$) define

$T(f)(x)$$:={\displaystyle {\int }_{{ℝ}^n}}K(x,y)f(y)𝑑{\lambda }^n(y)$

Then ${\mathrm{||}T\mathit{}f\mathrm{||}}_p\mathrm{\le }C\mathit{}{\mathrm{||}f\mathrm{||}}_p$.