Hardy inequality on an interval
= Hardy inequality on an interval
{c}
{title2=$\|u/x\|_p\leq\frac p{p-1}\|u'\|_p$}
{wiki}
If $u\in W^{1,p}(0,1)$ has zero <Sobolev trace theorem>[trace] at zero, then
$$
\left\|\frac ux\right\|_{L^p(0,1)}
\leq\frac p{p-1}\|u'\|_{L^p(0,1)}.
$$
Indeed, the <one-dimensional Sobolev representative> satisfies $u(x)=\int_0^xu'(t)\,dt$, so the claim is the <Hardy averaging inequality> applied to $u'$.