Solution
= Solution
The <one-dimensional Sobolev representative> of $u$ is absolutely continuous, and its zero <Sobolev trace theorem>[trace] gives
$$
u(x)=\int_0^xu'(t)\,dt.
$$
Consequently $u(x)/x=A(u')(x)$ for the <Hardy operator>. Applying the <Hardy averaging inequality> to $u'$ gives the <Hardy inequality on an interval>:
$$
\boxed{\left\|\frac ux\right\|_{L^p(0,1)}
\leq\frac p{p-1}\|u'\|_{L^p(0,1)}}.
$$