Hardy averaging inequality ($\Vert Au{\Vert }_p\le \frac{p}{p-1}\Vert u{\Vert }_p$)

For $1<p<\infty$, the Hardy operator is bounded on $L^p(0,1)$ with operator norm at most $p/(p-1)$.

If $u\in W^{1,p}(0,1)$ has zero trace at zero, thenIndeed, the one-dimensional Sobolev representative satisfies $u(x)={\int }_0^x{u}^{\prime }(t)dt$, so the claim is the Hardy averaging inequality applied to $u^{\prime }$.