Let $U$ be a bounded, open set in $\mathbb{R}^n$. For any $u \in \mathcal{C}^2(\overline{U})$ show that $0\leq u \leq 1$ if it satisfies the Poisson problem with Dirichlet B.C. as \begin{equation} \Delta u = f =u^{2k+1} \quad \text{in} \ U \quad \text{and} \quad u = 1 \quad \text{on} \ \partial U \end{equation} and $k \in \mathbb{Z}^+$.
Now, I think I am stuck with this problem that it seemed to me simple but I cannot get my way through it properly (I just started studying PDE). Firstly I thought that I will use a subharmonic function such as the one produced using $\Delta v = \pm v^{2k}$ with same B.C. for which we obtain an upper bound such as $u \leq 1$. But I cannot properly justify that $\Delta u \leq \Delta v$ since $f$ changes sign. Then I proceeded with the more general approach, i.e. that \begin{equation} \max_{\overline{U}} |u| \leq C \max_{\overline{U}}|f| + \max_{\partial U} |g| \end{equation} with C a constant depending only on the domain $U$ and $g$ the value of $u$ at the boundary $\partial U$. For this case this translates to \begin{equation} \max_{\overline{U}} |u| \leq C \max_{\overline{U}}|u^{2k+1}| + 1 \end{equation} But I don't know what to argue about the term $C \max_{\overline{U}}|u^{2k+1}|$, I was trying to think of a way to drive this to zero..
I am not asking for a solution but could you please provide me some hints or some guidelines for what I am missing?
Thanks!
Let us show first that $u \geq 0$ in $U$. Assume, for contradiction that the set $\{x\in U: u(x) < 0\}$ is not empty. Since $u \in C(\overline{U})$, $\overline{U}$ is compact (in view of boundedness of $U$), and $u>0$ in $\partial U$, it follows that $u$ attains a minimum in the interior of $U$, say at $x_0$. By our assumption, we get $u(x_0)<0$. Since $D^2 u(x_0)$ (the Hessian) must be positive definite at the point of minimum, we obtain $$0\leq \mathrm{tr}(D^2 u(x_0)) = \Delta u(x_0) = u^{2k+1}(x_0) < 0, $$ which is a contradiction (notice that we used the fact that $2k+1$ is odd, hence $u(x_0)<0$ implies $u^{2k+1}(x_0)<0$). Hence, $u\geq 0 $ everywhere in $U$.
But then, we have a subharmonic function at hand, i.e. $\Delta u \geq 0$ everywhere in $U$, and the ordinary maximum principle implies that $u\leq \max_{\partial U} u =1 $ on $U$, completing the desired estimate.