Let us consider the following elliptic problem $$ \begin{cases} -\Delta u = \frac{u^p}{|x|^2} \mbox{ in } \Omega \\ u >0 \mbox{ in } \Omega \\ u = 0 \mbox{ on } \partial \Omega. \end{cases} $$ with $N \geq 3$, $\Omega$ a bounded domain in $\mathbb{R}^N$ with smooth boundary,$1 < p < \frac{N+2}{N-2}$ and $0 \in \partial \Omega$.

We say that $u$ is an energy solution of the preceding problem if $u \in H^1_0(\Omega)$, $u >0$ such that $$ \int_\Omega \nabla u \nabla \varphi \, dx = \int_\Omega \frac{u^p}{|x|^2} \varphi \, dx \quad \forall \varphi \in \mathcal{C}^\infty_0(\Omega). $$

The authors proved the following Lemma.

If $u$ is an energy solution of the preceding problem then $u \in L^\infty(\Omega)$.

The proof, they said, is just a consequence of the a priori estimates of Gidas and Spruck, "A priori bounds for positive solutions of nonlinear elliptic equations", Comm. Partial Differential Equations 6 (1981), after suitable scaling. Moreover they gave this hint. Suppose $x_0 \in \Omega$, $x\neq 0$. Let $r:= \frac{|x_0|}{2}$ and define the function $$ v(y) := u(x_0 + ry) \quad \mbox{in } \frac{\Omega - x_0}{r} $$ Then $v$ satisfies the equation $$ - \Delta v = \frac{r^2}{|x_0 + ry|^2}v^p \quad \mbox{in } \frac{\Omega - x_0}{r} $$ Now restrict the equation to $\frac{\Omega - x_0}{r} \cap B_1(0)$. In this region the weight $\frac{r^2}{|x_0 + ry|^2}$ is bounded and smooth. Now using the a priori estimates of Gidas and Spruck we deduce that $v(0) \leq C$ for some universal $C$. Finally since $u(x_0) = v(0)$ we have the thesis, since $C$ is not depending on $x_0$.

Howewer I don't understand how the result of Gidas and Spruck could be useful in this setting. Indeed although in the region $\frac{\Omega - x_0}{r} \cap B_1(0)$ the weight is bounded and smooth, the boundary conditions of the problem for the function $v$ are "free" ( at least in some part of the boundary of that region). So, applying the Gidas and Spruck results we have that there exists a constant, but I don't get how can this constant be universal, since it seems, to me, depending on the boundary conditions and hence on the point $x_0$.