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

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 105 / 2 / c / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-105/2/c/ii/solution