Brezis' exercise 8.20.4: prove that $\frac{u(x)}{x^2} \in L^2(I), \frac{u(x)}{x} \in H^1(I)$, and $\frac{u'(x)}{x} \in L^2(I)$

Question

I'm trying to solve a problem in Brezis' Functional Analysis, i.e.,

Exercise 8.20 Let $I$ be the open interval $(0, 1)$. Let $V = \{v \in H^1 (I) : v(0) = 0\}$.

1. Given $f \in L^2(I)$ such that $\frac{1}{x} f(x) \in L^2(I)$, prove that there exists a unique $u \in V$ satisfying $$ (1) \quad \int_0^1 u^{\prime}(x) v^{\prime}(x) d x+\int_0^1 \frac{u(x) v(x)}{x^2} d x=\int_0^1 \frac{f(x) v(x)}{x^2} d x \quad \forall v \in V. $$
2. What is the minimization problem associated with $(1)$?

In what follows we assume that $\frac{1}{x^2} f(x) \in L^2(I)$.

3. In $(1)$ choose $v(x)=\frac{u(x)}{(x+\varepsilon)^2}, \varepsilon>0$, and deduce that $$ (2) \quad \int_0^1\left|\frac{d}{d x} \left(\frac{u(x)}{x+\varepsilon}\right) \right|^2 d x \leq \int_0^1 \frac{f(x)}{x^2} \frac{u(x)}{(x+\varepsilon)^2} d x . $$
4. Prove that $\frac{u(x)}{x^2} \in L^2(I), \frac{u(x)}{x} \in H^1(I)$, and $\frac{u'(x)}{x} \in L^2(I)$.

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 an auxiliary result (in the same book), i.e.,

Exercise 8.8 Let $I$ be the open interval $(0, 1)$.

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

Clearly, $v(x) := \frac{u(x)}{x+\varepsilon} \in H^1 (I)$ with $v(0)=0$. By Exercise 8.8.1, $$ \left \| \frac{u(x)}{x(x+\varepsilon)} \right \|_{L^2}^2 \leq 2 \left\|v'\right\|_{L^2}^2. $$

Let $C := \left \| \frac{f(x)}{x^2} \right \|_{L^2} < \infty$. By $(2)$ and C-S inequality, $$ \left\|v'\right\|_{L^2}^2 \leq C \left \| \frac{u(x)}{(x+\varepsilon)^2} \right \|_{L^2} \le C \left \| \frac{u(x)}{x(x+\varepsilon)} \right \|_{L^2}. \quad (*) $$

Combining above inequalities, we get $\left \| \frac{u(x)}{x(x+\varepsilon)} \right \|_{L^2} \le 2 C$ for all $\varepsilon >0$. By Fatou's lemma, $\left \| \frac{u(x)}{x^2} \right \|_{L^2} \le 2 C$, which implies $\frac{u(x)}{x^2} \in L^2(I)$.

We have $v'(x) = \frac{u'(x)}{x+\varepsilon} - \frac{u(x)}{(x+\varepsilon)^2}$. It follows from $(*)$ that $$ \left \| \frac{u'(x)}{x+\varepsilon} - \frac{u(x)}{(x+\varepsilon)^2} \right \|_{L^2}^2 \le C \left \| \frac{u(x)}{x^2} \right \|_{L^2}. $$

By Fatou's lemma, $\left \| \left (\frac{u(x)}{x} \right)' \right \|_{L^2}^2 = \left \| \frac{u'(x)}{x} - \frac{u(x)}{x^2} \right \|_{L^2}^2 \le C \left \| \frac{u(x)}{x^2} \right \|_{L^2}$, which implies $\frac{u'(x)}{x} \in L^2(I)$ and thus $\left (\frac{u(x)}{x} \right)' \in L^2(I)$. The claim then follows.