Brezis' exercise 8.9.5: $u \in X_m$ implies $\frac{D^j u(x)}{x^{m-j-k}} \in X_k$

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 $$ v=\frac{D^j u(x)}{x^{m-j-k}} \in X_k \quad \forall j, k \text { integers, } j \geq 0, k \geq 1, j+k \leq m-1. $$

There are possibly subtle mistakes that I could not recognize in my below attempt of $(5)$. Could you please have a check on it?

---

We proceed by induction on $m \ge 2$. The base case $m=2$ (and thus $j=0$ and $k=1$) clearly follows from $(2)$. Assume the statment holds for $m$, i.e., $$ \frac{D^j u(x)}{x^{m-j-k}} \in X_k \quad \forall j, k \text { integers, } j \geq 0, k \geq 1, j+k \leq m-1. $$

We will prove that it holds for $m+1$. Let $j, k$ be integers such that $j \geq 0, k \geq 1, j+k \leq m$. We want to prove that $$ v = \frac{D^j u(x)}{x^{(m+1)-j-k}} \in X_k. $$

• Assume $j \ge 1$. Then $(j-1) + k \le m-1$. Notice that $u \in X_{m+1}$ implies $Du \in X_m$. By inductive hypothesis, $$ v= \frac{D^{j-1} (Du)(x)}{x^{m-(j-1)-k}} \in X_k. $$

• Assume $j=0$. Then $v= \frac{u(x)}{x^{m+1-k}}$. By (4), $u \in X_{m+1}$ implies $\frac{u(x)}{x^{m}} \in X_1$. Then $v =x^{k-1} \frac{u(x)}{x^{m}} \in X_{1 + (k-1)} = X_k$.

This completes the proof.