Brezis' exercise 8.9.4: prove that $\frac{u(x)}{x^m} \in L^p(I)$ and $\frac{u(x)}{x^{m-1}} \in X_1$

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 $v(0)=0$.
3. Let $u$ be as in question $(1)$. Set $u_n=\zeta_n u$, where $\zeta_n$ is defined in question 4 of Exercise 8.8. Check that $u_n \in W^{2, p}(I)$ and $u_n \rightarrow u$ in $W^{2, p}(I)$ as $n \rightarrow \infty$.
4. More generally, let $m \geq 1$ be an integer, and let $1<p<\infty$. Assume that $u \in X_m$, where $$ X_m=\left\{u \in W^{m, p}(I) : u(0)=D u(0)=\cdots=D^{m-1} u(0)=0\right\} . $$ Show that $\frac{u(x)}{x^m} \in L^p(I)$ and that $\frac{u(x)}{x^{m-1}} \in X_1$.
5. Assume that $u \in X_m$ and prove that $$ v=\frac{D^j u(x)}{x^{m-j-k}} \in X_k \quad \forall j, k \text { integers, } j \geq 0, k \geq 1, j+k \leq m-1. $$
6. Let $u \in X_m$ and $\zeta \in C^\infty (\mathbb R)$ such that $\zeta (x)=0$ for $x \le 1$ and $\zeta (x)=1$ for $x \ge 2$. Let $\zeta_n (x) := \xi (nx)$ and $u_n (x) = \zeta_n (x) u(x)$. Prove that $\zeta_n u \in W^{m, p}(I)$ and $\zeta_n u \rightarrow u$ in $W^{m, p}(I)$, as $n \rightarrow \infty$.

There are possibly subtle mistakes that I could not recognize in my below attempt of $(4)$. 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 $u(0)=0$.

• First, we prove that $u \in X_m$ implies $\frac{u(x)}{x^m} \in L^p(I)$. We have $$ \begin{align*} \left | \frac{u(t_1)}{t_1^m} \right | &= \left | \frac{1}{t_1^m} \int_0^{t_1} u'(t_2) dt_2 \right | \\ &= \left | \frac{1}{t_1^m} \int_0^{t_1} \int_0^{t_2} u''(t_3) dt_3 dt_2 \right | \\ &= \cdots \\ &= \left | \frac{1}{t_1^m} \int_0^{t_1} \int_0^{t_2} \cdots \int_0^{t_{m}} D^m u(t_{m+1}) dt_{m+1} \cdots dt_3 dt_2 \right | \\ &\le \left | \frac{1}{t_1} \int_0^{t_1} \frac{1}{t_2} \int_0^{t_2} \cdots \frac{1}{t_m} \int_0^{t_{m}} D^m u(t_{m+1}) dt_{m+1} \cdots dt_3dt_2 \right | \\ &= \left | \frac{1}{t_1} \int_0^{t_1} \frac{1}{t_2} \int_0^{t_2} \cdots \frac{1}{t_{m-1}} \int_0^{t_{m-1}} T(D^m u)(t_m) dt_{m} \cdots dt_3dt_2 \right | \\ &= \left | \frac{1}{t_1} \int_0^{t_1} \frac{1}{t_2} \int_0^{t_2} \cdots \frac{1}{t_{m-2}} \int_0^{t_{m-2}} T(T(D^m u))(t_{m-1}) dt_{m-1} \cdots dt_3 dt_2 \right | \\ &= \cdots \\ &= \left | \frac{1}{t_1} \int_0^{t_1} T^{m-1} (D^m u) (t_2) dt_2 \right | \\ &= | T^{m} (D^m u) (t_1) |. \end{align*} $$ Let $c_p := \frac{p}{p-1}$. By Problem 34C.3, $$ \| T^{m} (D^m u) \|_{L^p(I)} \leq c_p \| T^{m-1} (D^m u) \|_{L^p(I)} \leq \cdots \le c_p^m \| D^m u \|_{L^p(I)}. $$ Then $$ \left\|\frac{u(x)}{x^m}\right\|_{L^p(I)} \le c_p^m \| D^m u \|_{L^p(I)} < \infty. $$

• Second, we prove that $v=\frac{u(x)}{x^{m-1}} \in X_1$. We proceed by induction in $m \ge 2$. The base case has been proved in $(1, 2)$. Assume that the statement holds for $m-1$. We are going to prove that it also holds for $m$. Fix $u \in X_m$. Then $u' \in X_{m-1}$. We have $v = \frac{x u(x)}{x^{m}}$ and $$ v ' (x) = \frac{u'(x)}{x^{m-1}} - \frac{(m-1)u(x)}{x^m}. $$ By inductive hypothesis, $\frac{u'(x)}{x^{m-1}} \in L^p (I)$. By the first part of this question, $\frac{u(x)}{x^m} \in L^p (I)$. Then $v, v' \in L^p (I)$ and thus $v \in W^{1, p} (I)$. It remains to prove $v(0)=0$. This actually follows from exercise 8.8.2 and the fact that $$ \frac{v(x)}{x} = \frac{u(x)}{x^m} \in L^p(I). $$