TITLE: Comparison principle for functions $u,v$ such that $\Delta u-f(u)\ge \Delta v-f(v)$
QUESTION [1 upvotes]: Let $\Omega \in {{R}^{n}}$ and $u,v\in {{C}^{2}}\left( \Omega  \right)\cap C\left( {\bar{\Omega }} \right),\text{ }f\in {{C}^{1}}\left( R \right)$ such that ${f}'\left( t \right)\ge 0$, for all $t\in R$.  Assume  
$$\left\{ \begin{align}
  & \Delta u-f\left( u \right)\ge \Delta v-f\left( v \right)\text{   on   }\Omega  \\ 
 & u\le v\text{   on   }\partial \Omega \text{ }, \\ 
\end{align} \right.$$
Show that $u\le v\text{   on   }\Omega \text{ }.$  
My attempt:
Assume $u\gt v\text{   on   }\Omega \text{ }.$  Let $w=u-v$.  Then $w>0\text{  on  }\Omega ,\text{ and }w\le0\text{ on }\partial \Omega $
Since ${f}'\left( t \right)\ge 0$, we have $f\left( u \right)-f\left( v \right)\ge 0$.  
Because  $\Delta w=\Delta u-\Delta v\ge f\left( u \right)-f\left( v \right)\ge 0$, $w$ is subhamonic.  
Then $w\le \frac{1}{\left| \partial B \right|}\int_{\partial B}{wdS}=\frac{1}{\left| \partial B \right|}\int_{\partial B}{\left( u-v \right)dS}\le 0$, a contraditciton.
Hence $u\gt v\text{   on   }\Omega \text{ }.$
I don't feel confident with my approach.  Please help.

REPLY [1 votes]: The claim $\frac{1}{\left| \partial B \right|}\int_{\partial B}{\left( u-v \right)dS}\le 0$ was not justified, and need not be true. 

Instead, I would proceed as follows: assuming $w=u-v>0$ somewhere, let $G$ be a connected component of the set $\{x:w(x)>0\}$. Note that $w=0$ on $\partial G$: this is where the assumption $w\le 0$ on $\partial \Omega$ is used. Within $G$, we have
$$
\Delta w = f(u)-f(v)\ge 0
$$ 
meaning $w$ is subharmonic. By the maximum principle for subharmonic functions, $w\le 0$ in $G$, a contradiction.
{"set_name": "stack_exchange", "score": 1, "question_id": 1562451}