Schur’s condition for a matrix to be a bounded operator on $l^2$

Theorem 0.1

Let $B$ be a matrix defined on $T\mathrm{\times }T$ for some countable set $T$. If there exists a positive number $C$ such that

then $B$ is a bounded operator on $l^{\mathrm{2}}\mathit{}\mathrm{(}T\mathrm{)}$ with its operator norm $\mathrm{\parallel }B\mathrm{\parallel }$ less than or equal to $C$.

• Let $x$ be a sequence in $l^2(T)$. We have

  ${\parallel Bx\parallel }_2^2$$={\displaystyle \sum _{s\in T}}{\left|{\displaystyle \sum _{t\in T}}b(s,t)x(t)\right|}^2$

  $\le {\displaystyle \sum _{s\in T}}{\left[{\displaystyle \sum _{t\in T}}\sqrt{|b(s,t)|}\left(\sqrt{|b(s,t)|}|x(t)|\right)\right]}^2$

  $\le {\displaystyle \sum _{s\in T}}\left[\left({\displaystyle \sum _{t\in T}}|b(s,t)|\right)\left({\displaystyle \sum _{t\in T}}|b(s,t)|{|x(t)|}^2\right)\right]$

  $\le C{\displaystyle \sum _{s\in T}}{\displaystyle \sum _{t\in T}}|b(s,t)|{|x(t)|}^2$

  $\le C{\displaystyle \sum _{t\in T}}{|x(t)|}^2{\displaystyle \sum _{s\in T}}|b(s,t)|$

  $\le C^2{\displaystyle \sum _{t\in T}}{|x(t)|}^2.$

  Therefore we have ${\parallel Bx\parallel }_2\le C{\parallel x\parallel }_2$ for all $x\in l^2(T)$, hence $\parallel B\parallel \le C$. $\mathrm{\square }$