Brezis' exercise 8.22.2: prove that $\rho \in K \iff \mu \in L^2(I)$

Question

Let $I$ be the open interval $(0, 1)$. I'm trying to solve a problem in Brezis' Functional Analysis, i.e.,

Exercise 8.22 Let $K = \{\rho \in H^1(I) : \rho \geq 0 \text{ on } I \text{ and } \sqrt{\rho} \in H^1(I) \}$.

1. Construct an example of a function $\rho \in H^1(I)$ with $\rho \geq 0$ on $I$ such that $\rho \notin K$.
2. Given $\rho \in H^1(I)$ with $\rho \geq 0$ on $I$, set $$ \mu= \begin{cases} \frac{1}{2} \frac{\rho'}{\sqrt{\rho}} & \text { on } [\rho>0], \\ 0 & \text { on } [\rho=0] . \end{cases} $$ Prove that $\rho \in K \iff \mu \in L^2(I)$, and then $(\sqrt{\rho})'=\mu$.

There are possibly subtle mistakes that I could not recognize in my below attempt of (2.). Could you please have a check on it?

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We have $\|\mu\|_{L^2} = \frac{1}{4} \int_{[\rho>0]} \frac{|\rho'|^2}{\rho}$. For $\varepsilon >0$, let $\rho_\varepsilon := \rho + \varepsilon$. Clearly, $\sqrt{\rho_\varepsilon} \to \sqrt{\rho}$ everywhere on $I$ as $\varepsilon \downarrow 0$. Also, $\rho_\varepsilon \in K$ and $(\sqrt{\rho_\varepsilon})'= \frac{\rho'}{2\sqrt{\rho + \varepsilon}}$. By dominated convergence theorem, $\| \sqrt{\rho_\varepsilon} - \sqrt{\rho} \|_{L^2} \to 0$ as $\varepsilon \downarrow 0$. By exercise 8.10 (in the same book), $\rho' =0$ a.e. on the set $[\rho=0]$, so $(\sqrt{\rho_\varepsilon})' \uparrow \mu$ a.e. on $I$ as $\varepsilon \downarrow 0$.

• $\implies$ Let $\rho \in K$. We have $$ \int_I \sqrt{\rho_\varepsilon} \varphi' = - \int_I (\sqrt{\rho_\varepsilon})' \varphi \quad \text{and} \quad \int_I \sqrt{\rho} \varphi' = - \int_I (\sqrt{\rho})' \varphi \quad \forall \varphi \in C^\infty_c (I), $$ and $$ \lim_{\varepsilon \downarrow 0} \int_I \sqrt{\rho_\varepsilon} \varphi' = \int_I \sqrt{\rho} \varphi', \quad \forall \varphi \in C^\infty_c (I), $$ which implies $$ \lim_{\varepsilon \downarrow 0} \int_I (\sqrt{\rho_\varepsilon})' \varphi = \int_I (\sqrt{\rho})' \varphi, \quad \forall \varphi \in C^\infty_c (I), $$ which (by monotone convergence theorem) implies $$ \int_I \mu \varphi = \int_I (\sqrt{\rho})' \varphi, \quad \forall \varphi \in C^\infty_c (I), $$ which implies $\mu = (\sqrt{\rho})'$.

• $\impliedby$ Let $\mu \in L^2(I)$. Then $\| (\sqrt{\rho_\varepsilon})' - \mu \|_{L^2} \to 0$ as $\varepsilon \downarrow 0$. Then $\sqrt\rho \in H^1(I)$ with $(\sqrt\rho)' = \mu$ by the following remark (in the same book)

Remark 4. It is convenient to keep in mind the following fact, which we have used in the proof of Proposition 8.1: let $\left(u_n\right)$ be a sequence in $W^{1, p}$ such that $u_n \rightarrow u$ in $L^p$ and $\left(u_n^{\prime}\right)$ converges to some limit in $L^p$; then $u \in W^{1, p}$ and $\left\|u_n-u\right\|_{W^{1, p}} \rightarrow 0$. In fact, when $1<p \leq \infty$ it suffices to know that $u_n \rightarrow u$ in $L^p$ and $\left\|u_n^{\prime}\right\|_{L^p}$ stays bounded to conclude that $u \in W^{1, p}$ (see Exercise 8.2).