I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Problem 34C In this part we set $E=L^p(0,1)$ with $1<p<\infty$. Given $u \in E$ define $T u$ by $$ T u(x)=\frac{1}{x} \int_0^x u(t) d t \quad \text {for } x \in(0,1] . $$
- Check that $T u \in C((0,1])$ and that $T u \in L^q(0,1)$ for every $q<p$.
- Prove that if $u \in C_c((0,1))$ then $$ \|T u\|_{L^p(0,1)} \leq \frac{p}{p-1} \|u\|_{L^p(0,1)} . \label{a}\tag{$\ast$} $$
- Prove that \eqref{a} holds for $u \in E$.
There are possibly subtle mistakes that I could not recognize in my below attempt of $(3)$. Could you please have a check on it?
Because $p < \infty$, we get $C_c((0,1))$ is dense in $E$. Let $(u_n) \subset C_c((0,1))$ such that $u_n \to u$ in $L^p$. Then $Tu_n \to Tu$ everywhere. By $(3)$, $$ \|T (u_n-u_m)\|_{L^p(0,1)} \leq \frac{p}{p-1} \|u_n-u_m\|_{L^p(0,1)} . $$
Notice that $T$ is linear. Then $(Tu_n)$ is a Cauchy sequence in $E$. So there is $v \in E$ such that $Tu_n \to v$ in $E$. Clearly, $v=Tu$. The claim then follows.