I am reading a paper of Brezis and Oswald about existence and uniqueness of positive solutions to sublinear elliptic equations: \begin{equation} - \Delta u = f(x, u), \ u \geq 0, \ u \not\equiv 0 \text{ in } \Omega, \qquad u = 0 \text{ on } \partial \Omega. \end{equation} As usual, $\Omega$ is a bounded smooth domain of $\mathbb R^N$ and there are certain hypotheses on $f$, but I think they are not important for my question.
They prove in the paper that any solution of the problem satisfy $$ (*) \qquad u > 0 \text{ in } \Omega, \qquad \frac{\partial u}{\partial \nu} < 0 \text{ on } \partial \Omega, $$ where $\nu$ is the outer unit normal. Then they claim that from this it follows that if $u_1, u_2$ are two solutions of the problem, then $u_1/u_2 \in L^\infty(\Omega)$. Radulescu also presents this result, and explains that it follows by L'Hospital's rule, but I also don't understand this.
Why does $u_1/u_2 \in L^\infty(\Omega)$?
Does the claim still hold if $u$ is a solution of the problem $$ - \Delta u = f(x, u) \text{ in } \Omega, \qquad u = 0 \text{ on } \Gamma, \qquad \frac{\partial u}{\partial \nu} = 0 \text{ on } \Gamma_1, $$ where $\partial \Omega = \Gamma \cup \Gamma_1$, both parts being smooth? It is known that $(*)$ still holds in this case with $\Gamma$ in the place of $\partial \Omega$.
For simplicity let's assume $u_1,u_2 \in C^2(\Omega) \cap C^1(\overline \Omega)$ (if $f$ is sufficiently regular then you get this automatically from standard elliptic regularity theory). Also, let $d(x) = \mathrm {dist} (x,\partial \Omega)$.
First question: Since $u_1,u_2$ are continuous and positive, we immediately obtain from the extreme value theorem that $ u_1/u_2 \in L^\infty(\Omega_\varepsilon)$ for all $\varepsilon>0$ where $\Omega_\varepsilon :=\{ x\in \Omega \text{ s.t. } d(x)>\varepsilon \}$ i.e. $u_1/u_2$ is bounded away from the boundary.
Since $\Omega$ is smooth and bounded there exists some $\varepsilon_0$ small such that $d \in C^\infty(\overline{\Omega \setminus \Omega_{\varepsilon_0}})$. Moreover, since $u=0$ on $\partial \Omega$, $$ \lim_{x\to x_0} \frac{u(x)}{d(x)} = -\frac{\partial u}{\partial \nu}(x_0), \qquad x_0 \in \partial \Omega .$$ Thus, for $i=1,2$, $u_i/d \in C^0(\overline {\Omega \setminus \Omega_{\varepsilon_0/2}})$. Then, since $\partial u_i / \partial \nu<0$ on $\partial \Omega$, it follows that we also have $u_i/d >0 $ in $\overline {\Omega \setminus \Omega_{\varepsilon_0/2}}$. By the extreme value theorem, $$ 0<C^{-1}_i\leqslant \frac{u_i}{d} \leqslant C_i \qquad \text{in } {\Omega \setminus \Omega_{\varepsilon_0/2}}. $$ Finally, since $$ \frac{u_1}{u_2} = \frac{u_1}{d} \cdot \frac{d}{u_2},$$ we obtain $$ 0\leqslant C^{-1}<\frac{u_1}{u_2}\leqslant C\qquad \text{in } {\Omega \setminus \Omega_{\varepsilon_0/2}}.$$ In particular, $u_1/u_2 \in L^\infty (\Omega \setminus \Omega_{\varepsilon_0/2})$ (i.e. $u_1/u_2$ is also bounded close to the boundary) which further implies $u_1/u_2 \in L^\infty (\Omega)$.
Second Question: I'm not sure about this but I'm sceptical. The above argument shows that we will definitely have $u_1/u_2$ is bounded away from $\partial \Omega$ and close to $\Gamma$. However, I don't see why you couldn't have $x_0\in\Gamma_1$ and $u_1,u_2$ such that $u_1\to c$ with $c>0$ but $u_2\to 0$. I'll have a think about a possible counter-example.