I am studying the above proposition $(i)$ which is as follows $:$
If $u$ is subharmonic in a domain $\Omega$, it satisfies strong maximum principle in $\Omega$; and if $v$ is superharmonic in a bounded domain $\Omega$ with $v \ge u$ on $\partial \Omega$, then either $v>u$ throughout $\Omega$ or $v \equiv u$.
To prove the fact the author assumed in contrary that $\exists$ some $x_0 \in \Omega$ such that $(u-v)(x_0) = \underset {\Omega} {\mathrm {sup}}(u-v)= M \ge 0$. In addition the author asserted that we may assume there is ball $B=B(x_0)$ such that $u-v \not \equiv M$ on $\partial B$.
I have failed to catch this reasoning. Why should it be so?
Would anybody please help me understanding this? My exam on this paper is on tomorrow so please try to answer it (if possible) as early as anybody can.
Thank you very much.
First note that if the result holds for $\Omega$ for the condition $v \ge u$ in $\partial \Omega$ the result holds for any sufficiently small subdomain of $\Omega$ for the same condition. Also if the result is valid for $u$ and $v$ (where $u$ and $v$ are subharmonic and superharmonic on $\Omega$ respectively) the result is valid for $u$ and $v+M$ since $v+M$ is also superharmonic. So assume in contrary $u \not \equiv v+M$ on some sufficiently small subdomain of $\Omega$ with $u \geq v+M$ on that small subdomain.Then you will find $x_0,y_0$ in that subdomain with $(u-v)(x_0) \ge M$ and $u(y_0)\neq v(y_0)+M$. SInce $M = \underset {\Omega} {\mathrm {sup}}\ (u-v)$ so it follows that $u(x_0)-v(x_0)=M$ and $u(y_0) < v(y_0)+M$. SInce we are in a sufficiently small subdomain so we can take a ball $B$ centered at $x_0$ and radius $r=\| y_0-x_0\|$ such that $B \Subset \Omega$. Thus we can find a ball $B$ around $x_0$ such that $u-v \not \equiv M$ on $\partial B$.