Brezis' problem 34C: an estimate $\|T u\|_{L^p(0,1)} \leq \frac{p}{p-1} \|u\|_{L^p(0,1)}$

Question

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 $u \in L^p(0,1)$ define $T u$ by $$ T u(x)=\frac{1}{x} \int_0^x u(t) d t \quad \text {for } x \in(0,1] . $$

1. Check that $T u \in C((0,1])$ and that $T u \in L^q(0,1)$ for every $q<p$.
2. Prove that if $u \in C_c((0,1))$ then $$ \|T u\|_{L^p(0,1)} \leq \frac{p}{p-1} \|u\|_{L^p(0,1)} . $$

There are possibly subtle mistakes that I could not recognize in my below attempt of $(2)$. Could you please have a check on it?

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1. The map $(0,1] \to \mathbb R, x \mapsto \frac{1}{x}$ is clearly continuous. By dominated convergence theorem, the map $(0,1] \to \mathbb R, x \mapsto \int_0^x u(t) d t$ is continuous. Then $T u \in C((0,1])$. By Hölder's inequality, $$ \begin{align*} \|T u\|_{L^q}^q &= \int_0^1 \left | \frac{1}{x} \int_0^x u(t) d t \right |^q dx \\ &=\int_0^1 \frac{\|1_{(0, x]} u\|_{L^1}^q}{x^q} dx \\ &\le\int_0^1 \frac{\|1_{(0, x]}\|_{L^{p^*}}^q \|u\|_{L^p}^q}{x^q} dx \\ &=\|u\|_{L^p}^q \int_0^1 x^{-\frac{q}{p}} dx, \end{align*} $$ where $p^*$ is the Hölder conjugate of $p$. If $q<p$ then $\int_0^1 x^{-\frac{q}{p}} dx < \infty$.

2. Let $\varphi (x) = \int_0^x u(t) d t$ for $x \in [0, 1]$. By FTC, $\varphi' (x) = u(x)$. Because $p \in (1, \infty)$, we get $(| \cdot|^p)' (x) =p | x|^{p-1} \operatorname{sgn} (x)$ for all $x \in \mathbb R$. By chain rule, $(| \varphi|^p)' (x) =p | \varphi (x)|^{p-1} \operatorname{sgn} (\varphi (x)) u(x)$ for all $x \in [0, 1]$. Then $| \varphi|^p \in C^1 ([0, 1])$. By change of variable $x \mapsto -x^{1-p}$, we get $$ \begin{align*} \int_0^1|T u(x)|^p d x &=\int_0^1|\varphi(x)|^p \frac{d x}{x^p} \\ &=\frac{1}{p-1} \int_0^1 |\varphi(x)|^p d\left(-x^{1-p}\right). \end{align*} $$

By integration by parts, we get $$ \begin{align*} (p-1) \int_0^1|T u(x)|^p d x &= - \frac{|\varphi(x)|^p}{x^{p-1}} \bigg |_0^1 + \int_0^1 \frac{d(|\varphi(x)|^p)}{x^{p-1}} \\ &=- |\varphi(1)|^p + p \int_0^1 \frac{p|\varphi(x)|^{p-1} \operatorname{sgn} (\varphi (x)) u(x)}{x^{p-1}} dx \\ &= - |\varphi(1)|^p + p \int_0^1 |Tu (x)|^{p-1} u(x) \operatorname{sgn} (\varphi (x)) dx. \end{align*} $$

By Hölder's inequality, $$ \begin{align*} \int_0^1 |Tu (x)|^{p-1} u(x) \operatorname{sgn} (\varphi (x)) dx &\le \left ( \int_0^1 |Tu (x)|^{(p-1) p^*}\right )^{1 / p^*} \|u\|_{L^p} \\ &=\|Tu\|_{L^p}^{p-1} \|u\|_{L^p}. \end{align*} $$

Then $$ (p-1) \|Tu\|_{L^p}^p \le - |\varphi(1)|^p + p \|Tu\|_{L^p}^{p-1} \|u\|_{L^p}. $$

Then $$ 0 \le |\varphi(1)|^p \le \|Tu\|_{L^p}^{p-1} (p \|u\|_{L^p} - (p-1)\|Tu\|_{L^p}). $$

Then $(p-1)\|Tu\|_{L^p} \le p \|u\|_{L^p}$. The claim then follows.