Solution
= Solution
For $u\in H$, the trace $u(0)$ vanishes. Apply the <Hardy inequality on an interval> with $p=2$ to obtain
$$
\boxed{\left\|\frac ux\right\|_{L^2(0,1)}\leq2\|u'\|_{L^2(0,1)}}.
$$
Thus $u/x\in L^2(0,1)$.
= Solution
For $u\in H$, the trace $u(0)$ vanishes. Apply the <Hardy inequality on an interval> with $p=2$ to obtain
$$
\boxed{\left\|\frac ux\right\|_{L^2(0,1)}\leq2\|u'\|_{L^2(0,1)}}.
$$
Thus $u/x\in L^2(0,1)$.