Understanding a geometric argument in the proof of the strong maximum principle for elliptic operators in Evans's PDE Here is the strong maximum principle in Evans's Partial Differential Equations:




Here $U\subset\mathbb{R}^n$ is open and bounded. Also, 

The proof is very short once one has Hopf's lemma. 



Here is my question: 

Would anyone elaborate the underlined sentence that why such $y$ exists? (I don't have any intuition at all why this should be true.)


We are assuming $u$ attains its maximum inside $U$, so $C$ is nonempty, and we are assuming it's non-constant, so $V$ is nonempty.  And $V$ is open by the continuity of $u$.  Since $U$ is connected and $U = C \cup V$, $C$ must not be open.  So there is a point $z \in C$ which is not an interior point of $C$.  Since $z \in U$ and $U$ is open, we can find $r>0$ such that $B(z,r) \subset U$.  In particular, $d(z,\partial U) \ge r$.  And since $z$ is not an interior point of $C$, we can find $y$ with $|z-y| < r/2$ and $y \notin C$.  Since $y \in U$ and $U=C \cup V$, we must have $y \in V$.  And we now have $d(y,C) \le |y-z| < r/2 < d(y, \partial U)$.
