{c}
{title2=$\|u/x\|_p\leq\frac p{p-1}\|u'\|_p$}
{wiki}

If $u\in W^{1,p}(0,1)$ has zero <Sobolev trace theorem>[trace] at zero, then
$$
\left\|\frac ux\right\|_{L^p(0,1)}
\leq\frac p{p-1}\|u'\|_{L^p(0,1)}.
$$
Indeed, the <one-dimensional Sobolev representative> satisfies $u(x)=\int_0^xu'(t)\,dt$, so the claim is the <Hardy averaging inequality> applied to $u'$.

Hardy inequality on an interval

Set
$$
\phi(x)=\int_0^xu(t)\,dt,
\qquad
Au(x)=\frac{\phi(x)}x.
$$
The <Holder inequality> gives
$$
|\phi(x)|^p x^{1-p}
\leq\int_0^x|u(t)|^p\,dt\longrightarrow0
\quad(x\downarrow0).
$$
Using $x^{-p}\,dx=-(p-1)^{-1}d(x^{1-p})$ and <integration by parts>, while discarding the nonpositive boundary term at $x=1$, yields
$$
\begin{aligned}
\|Au\|_p^p
&=\int_0^1|\phi|^px^{-p}\,dx\\
&\leq\frac p{p-1}\int_0^1|\phi|^{p-1}x^{1-p}|u|\,dx\\
&=\frac p{p-1}\int_0^1|Au|^{p-1}|u|\,dx.
\end{aligned}
$$
Another application of <Holder inequality>[Hölder's inequality] gives
$$
\|Au\|_p^p
\leq\frac p{p-1}\|Au\|_p^{p-1}\|u\|_p.
$$
After cancellation, with the zero case immediate,
$$
\boxed{\|Au\|_{L^p(0,1)}\leq\frac p{p-1}\|u\|_{L^p(0,1)}}.
$$
This is the <Hardy averaging inequality>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 105 / 2 / a / 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>:
$$
\boxed{\left\|\frac ux\right\|_{L^p(0,1)}
\leq\frac p{p-1}\|u'\|_{L^p(0,1)}}.
$$

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

{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 $p/(p-1)$.

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