Brezis' exercise 8.9.3: check that $u_n \in W^{2, p}(I)$ and $u_n \rightarrow u$ in $W^{2, p}(I)$

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 $n \rightarrow \infty$.

There are possibly subtle mistakes that I could not recognize in my below attempt of $(3)$. Could you please have a check on it?

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We need an auxiliary result (in the same book), i.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 $u(0)=0$.
3. Let $u(x)=(1+|\log x|)^{-1}$. Check that $u \in W^{1,1}(0,1), u(0)=0$, but $\frac{u(x)}{x} \notin$ $L^1(0,1)$.
4. Let $I=(0,1)$ and $u \in W^{1, p} (I)$ with $p \in [1, \infty)$ and $u(0)=0$. Fix a function $\xi \in C^\infty (\mathbb R)$ such that $\xi (x)=0$ for $x \le 1$ and $\xi (x)=1$ for $x \ge 2$. Let $\xi_n (x) := \xi (nx)$ and $u_n (x) = \xi_n (x) u(x)$. Check that $u_n \in W^{1, p} (I)$ and that $u_n \to u$ in $W^{1, p} (I)$.

We have $$ \begin{align*} u_n &= \xi_n u, \\ u'_n &= \xi'_n u + \xi_n u',, \\ u''_n &= \xi''_n u + 2 \xi'_n u' + \xi_n u''. \\ \end{align*} $$

It follows from $\xi''_n, \xi'_n, \xi_n \in L^\infty (I)$ and $u, u', u'' \in L^p (I)$ that $u_n, u'_n, u''_n \in L^p (I)$. Then $u_n \in W^{2, p}(I)$.

We have $u, u' \in W^{1, p}(I)$. By Exercise 8.9.4, $u_n \rightarrow u$ in $W^{1, p}(I)$. It remains to prove $u''_n \rightarrow u''$ in $L^{p}(I)$. We have $\xi_n \to 1$ everywhere on $(0, 1)$. Then $\xi_n u'' \to u''$ in $L^{p}(I)$. I already proved that $u(0)$ implies $\xi_n' u \to 0$ in $L^{p}(I)$. Notice that $u'(0)=0$. With similar reasoning, we can prove that $\xi''_n u \to 0$ and $\xi'_n u' \to 0$ in $L^{p}(I)$. It follows that $u''_n \rightarrow u''$ in $L^{p}(I)$.