TITLE: How to argue this consequence?
QUESTION [8 upvotes]: Suppose that $\Omega=\mathbf{R}^n_+$ and consider a function $0<u<\sup\limits_\Omega u=M<\infty$ such that:
$$\Delta u+u-1=0 \ \ \text{in} \ \ \Omega,$$
$$u=0 \ \ \text{on} \ \ \partial\Omega.$$
If $u$ exists, then $M>1$.
I don't know to argue this. My idea is to try by contradiction. Suppose that $M\leq1$, so
$$\Delta u=1-u\geq0,$$
that is, $u$ is a subharmonic function. If $u$ attains a maximum in the interior of $\Omega$, for the maximum principle, $u$ should be a constant function and this would be the contradiction. But I don't know how to prove that the maximum is attain in interior.

REPLY [2 votes]: Suppose that $u \in H_0^1(\Omega)$. Then we have
$$ -\int_\Omega \nabla u \nabla \phi +\int_\Omega u\phi =\int_\Omega \phi, \forall \phi \in C_c^\infty(\Omega). $$
We can find a sequence of smooth functions $\phi_n$ which converges to $u$ in $H_0^1(\Omega)$. Then we have
$$ -\int_\Omega |\nabla u|^2 +\int_\Omega u^2 =\int_\Omega u. $$
Suppose that $M \leq 1$. Then $u^2 \leq u$ everywhere, and therefore 
$$ \int_\Omega |\nabla u|^2 =0$$
This implies that $u$ is constant, and therefore zero. Contradiction.
Maybe this can be adapted to be used without the assumption that $u \in H_0^1(\Omega)$