How to argue this consequence?

Question

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.

Answer 1

This is not a full answer, but maybe can help someone to give the full answer. By using Op's idea, im supposing that $u\leq 1$.

Case $n=1$

As the OP pointed out, if the function $u$ attains its maximum, then its must be constant, so we can suppose that $u\neq 1$. But $u\neq 1$ implies that $u''(x)>0$, or equivalently, $u$ is strictly convex.

Because $u(0)=0$ and $u>0$ we can conclude that $u$ is unlimited, which is a absurd. This concludes the case $n=1$.

Case $n>1$

We have some issues that can not happen. For example:

1 - If $u$ attains a local maximum and a local minimum, this would implies that there is some point $x$ such that $\Delta u(x)=0$.

2 - $u(x)$ can not converges to $0$ as $x\rightarrow \infty$.

Maybe there is a straightforwaard argument, but i think that with 1 and 2 is possible to conclude that $u(x)=1$ for some point.

Edit: Case $n>1$ (complete)

By using some results of Berestycki, Caffarelli and Nirenberg (see referece above and the references therein) we can conclude that $u$ is symmetric i.e. $u=u(x_n)$. This implies in our case that $\displaystyle\frac{\partial^2 u}{\partial x_n^2}=\Delta u>0$. Now, with the help of the case $n=1$ we can conlude.

References:

H. Berestycki - L. Caffarelli - L. Nirenberg, Further qualitative properties for elliptic equations in unbounded domains, Annali della Scuola Normale Superiore di Pisa - Classe di Scienze (1997), Volume: 25, Issue: 1-2, Publisher: Scuola Normale Superiore, page 69-94

Answer 2

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 $u \in H_0^1(\Omega)$