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>:
$$
\boxed{\left\|\frac ux\right\|_{L^p(0,1)}
\leq\frac p{p-1}\|u'\|_{L^p(0,1)}}.
$$

Solution

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

 Solution

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