I am reading a paper and have trouble understanding the following step on page 581:
Assume we know that $w: \mathbb{R}^n \to \mathbb{R}$ is smooth and bounded. Furthermore $\Vert w \Vert_{L^\infty}>1$ and $w(x) \to 1$ whenever $\vert x \vert \to \infty.$ Finally the following inequality holds \begin{equation} \Delta w + 2 w(1+\frac{c^2}{4}-w) \geq 0. \end{equation}The author then applys the weak maximum principle to obtain \begin{equation} w \leq 1+\frac{c^2}{4}. \end{equation}
While I found several versions of a "weak maximum principle" (e.g. here), I can't figure out how to apply them to this situation. They all either assume that the domain is bounded and/or there are only derivatives of $w$ (but not $w$ itself) in the PDE.
So my question is:
Can someone give me a hint which version of the weak maximum principle is applicable in this situation and how to apply it?
All help is much appreciated!
Suppose $w(x)>1+c^2/4$ for some $x$. Since the set $\{x: w(x)\ge 1+c^2/4 \}$ is bounded and closed, it is compact. Let $x_0$ be a point of this set where $w$ attains its maximum. Since it is a global maximum of $w$, we have $w(x_0)>1+c^2/4$.
On the other hand, $$\Delta w(x_0)\ge 2w(w-1-c^2/4)>0 \tag1$$ Therefore there is $j\in \{1,\dots,n\}$ such that $$\frac{\partial^2}{\partial x_j^2}w(x_0)>0 \tag2$$ Inequality (2) is incompatible with $w$ (considered as a function of $j$th variable only) having a maximum at $x_0$.