{c}
{title2=$\|Au\|_p\leq\frac p{p-1}\|u\|_p$}

For $1<p<\infty$, the <Hardy operator> is bounded on $L^p(0,1)$ with operator norm at most $p/(p-1)$.

Hardy averaging inequality

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

{c}
{title2=$Au(x)=x^{-1}\int_0^xu(t)\,dt$}

The Hardy operator is the averaging operator
$$
Au(x)=\frac1x\int_0^xu(t)\,dt
$$
on functions over the positive half-line or an interval beginning at zero.


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