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

 Solution

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