Let $I=(0,1)$. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Exercise 8.9.8 Assume now that $u \in W^{2,1}(I)$ with $u(0)=u^{\prime}(0)=0$. Set $$ v(x)= \begin{cases}\frac{u(x)}{x} & \text { if } x \in(0,1], \\ 0 & \text { if } x=0 .\end{cases} $$ Check that $v \in C([0,1])$. Prove that $v \in W^{1,1}(I)$.
There are possibly subtle mistakes that I could not recognize in my below attempt. Could you please have a check on it?
First, we prove $v \in C([0,1])$. Because $u'(0)=0$, we get $$ v(x) = \frac{1}{x} \int_0^xu'(t) dt. $$
By dominated convergence theorem, $v \in C((0, 1])$. It remains to prove that $\lim_{x \downarrow 0} v(x) =0$. Because $u \in W^{2,1}(I)$, we get $u' \in C([0, 1])$ and thus $$ \lim_{x \downarrow 0} \left ( \frac{1}{x} \int_0^xu'(t) dt \right ) = u'(0) =0. $$
Second, we prove $v \in W^{1,1}(I)$. We have $$ \begin{align*} v'(x) &= \frac{u'(x)}{x} - \frac{u(x)}{x^2} \\ &= \frac{x u'(x)}{x^2} - \frac{u(x)}{x^2} \\ &= \frac{1}{x^2} \int_0^x (tu'(t))' dt - \frac{1}{x^2} \int_0^x u'(t) dt \\ &= \frac{1}{x^2} \int_0^x tu''(t) dt. \end{align*} $$
Then $$ \begin{align*} \int_0^1 |v'(x)| dx &\le \int_0^1 \frac{1}{x^2} \int_0^x t |u''(t)| dt dx \\ &= \int_0^1 \int_0^1 \frac{1}{x^2} t 1_{[0, x]} (t) |u''(t)| dt dx \\ &= \int_0^1 \left ( \int_t^1 \frac{1}{x^2} dx \right ) t |u''(t)|dt \\ &= \int_0^1 (1-t) |u''(t)|dt \\ &\le \| u'' \|_{L^1} < \infty. \end{align*} $$