How to argue this consequence?

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.

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What is the purpose of the function, $f$? – djws Nov 20 '12 at 6:45
You don't need suposse that. I will delete this. Sorry. – José Carlos Nov 20 '12 at 6:58
Do you have any ideia how to argue this? – José Carlos Nov 20 '12 at 7:00
Please! Stop adding [Solved] to the title! – Asaf Karagila Nov 25 '12 at 22:21

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

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The article you mention proves that this problem has no solution for $n=2,3$. – Beni Bogosel Nov 25 '12 at 23:18
yes, i know. But before he proves it, he shows that if there exist a solution to the problem then $u> 2$. Here we just prove that $u>1$ and this is sufficient for what they want in the article. @BeniBogosel – Tomás Nov 26 '12 at 0:04
I haven't read the paper. Please, give a little help, because I haven't found where they proved $u>2$. What I found is "For this problem it is not difficult to verify that if there were a solution, then $M>2$". – vesszabo Nov 28 '12 at 16:38
Well haha @vesszabo, come to the party. They dont prove it and i dont know how to prove, nevertheless we have proved that $u>1$. Do you have any idea? – Tomás Nov 28 '12 at 16:40
Absolutely not :-( Luckily, as you said, $u>1$ is enough. Could it be a typo? However a direct proof without citing that article would be interesting. – vesszabo Nov 28 '12 at 16:56

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

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Interesting calculation Beni, but i think the idea is equivalently to the item 2 of case $n>1$. If $u\in H_0^1(\Omega)$ then $u=0$, because in this case $u$ must attain a maximum value in $\Omega$. What do you think? – Tomás Nov 25 '12 at 22:47
It's just an idea. I suppose that this assumption is quite restrictive... – Beni Bogosel Nov 25 '12 at 22:51
I have tried to show that $u\in H_0^1(\Omega)$ without result. Did you tried? – Tomás Nov 25 '12 at 22:55
I don't think it is possible... Note that if the second equality can be obtained locally, then the conclusion also follows. The problem is that if we try to prove the equality for $\omega \subset \subset \Omega$ then a boundary term appears, and we have no sign control on it. – Beni Bogosel Nov 25 '12 at 23:03
but $u>0$ and maybe we can use the fact that $H_{loc}^1(\Omega)$ ? – Tomás Nov 26 '12 at 10:27