Hardy averaging inequality
= Hardy averaging inequality
{c}
{title2=$\|Au\|_p\leq\frac p{p-1}\|u\|_p$}
For $1<p<\infty$, the <Hardy operator> is bounded on $L^p(0,1)$ with operator norm at most $p/(p-1)$.
= Hardy averaging inequality
{c}
{title2=$\|Au\|_p\leq\frac p{p-1}\|u\|_p$}
For $1<p<\infty$, the <Hardy operator> is bounded on $L^p(0,1)$ with operator norm at most $p/(p-1)$.