Brezis' exercise 8.9.2: deduce that $v(x)=\frac{u(x)}{x} \in W^{1, p}(I)$ with $v(0)=0$

Question

Let $I=(0,1)$. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,

Exercise 8.9

1. Let $u \in W^{2, p}(I)$ with $1<p<\infty$. Assume that $u(0)=u^{\prime}(0)=0$. Show that $\frac{u(x)}{x^2} \in L^p(I)$ and $\frac{u^{\prime}(x)}{x} \in L^p(I)$ with (1) $$ \left\|\frac{u(x)}{x^2}\right\|_{L^p(I)}+\left\|\frac{u^{\prime}(x)}{x}\right\|_{L^p(I)} \leq C_p\left\|u^{\prime \prime}\right\|_{L^p(I)} $$
2. Deduce that $v(x)=\frac{u(x)}{x} \in W^{1, p}(I)$ with $v(0)=0$.

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

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We need auxiliary results (in the same book), 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 $$ T u(x)=\frac{1}{x} \int_0^x u(t) d t \quad \text {for } x \in(0,1] . $$

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 $$ \|T u\|_{L^p(0,1)} \leq \frac{p}{p-1} \|u\|_{L^p(0,1)} . \label{a}\tag{$\ast$} $$
3. Prove that \eqref{a} holds for $u \in E$.

Exercise 8.8:

1. Let $u \in W^{1, p}(0,1)$ with $1<p<\infty$. Show that if $u(0)=0$, then $\frac{u(x)}{x} \in L^p(0,1)$ and $$ \left\|\frac{u(x)}{x}\right\|_{L^p(0,1)} \leq \frac{p}{p-1}\left\|u^{\prime}\right\|_{L^p(0,1)} . $$
2. Conversely, assume that $u \in W^{1, p}(0,1)$ with $1 \leq p<\infty$ and that $\frac{u(x)}{x} \in$ $L^p(0,1)$. Show that $u(0)=0$.

1. It follows from $u(x)=u'(x)=0$ that $$ \begin{align*} \left | \frac{u'(x)}{x} \right | &= \left | \frac{1}{x} \int_0^x u''(t) dt \right | \\ &= T u'' (x), \end{align*} $$ and $$ \begin{align*} \left | \frac{u(x)}{x^2} \right | &= \left | \frac{1}{x^2} \int_0^x u'(t) dt \right | \\ &= \left | \frac{1}{x^2} \int_0^x \int_0^t u'' (s) ds dt \right | \\ &\le \left | \frac{1}{x} \int_0^x \frac{1}{t} \int_0^t u'' (s) ds dt \right | \\ &= \left | \frac{1}{x} \int_0^x (Tu'')(t) dt \right | \\ &= T((Tu''))(x) =: T^2 u'' (x). \end{align*} $$

Let $c_p := \frac{p}{p-1}$. By Problem 34C.3, $$ \|T^2 u''\|_{L^p(I)} \leq c_p \|Tu''\|_{L^p(I)} \leq c_p^2 \|u''\|_{L^p(I)}. $$

Then $$ \left\|\frac{u(x)}{x^2}\right\|_{L^p(I)} + \left\|\frac{u^{\prime}(x)}{x}\right\|_{L^p(I)} \leq \left ( c_p + c_p^2 \right ) \left\|u^{\prime \prime}\right\|_{L^p(I)}. $$

2. By exercise 8.8.1, $v \in L^p (I)$. On the other hand, $v' \in L^p (I)$ by $(1)$ and the fact that $$ v ' (x) = \frac{u^{\prime}(x)}{x} - \frac{u(x)}{x^2}. $$

We have $v(0) = 0$ by $(1)$ and exercise 8.8.2 and the fact that $$ \frac{v(x)}{x} = \frac{u(x)}{x^2}. $$