$f:\mathbb R^2\to\mathbb R$ is a continuous function. $x,y\in\mathbb R^2$.
Define $\overline{xy}$ as the segment connecting the points $x,y$, excluding the end points.
Given: $f(x)>f(z)$ for all $z\in\overline{xy}$.
Can we claim that, there exists a neighborhood open set $S$ of $x$, such that for any $x'$ in the set $S$, we have $f(x')>f(z')$, for any $z'\in\overline{x'y}$?
By the definition of continuity, it is obvious that for any $a\in\overline{xy}$, there exists an open epsilon ball $B_\epsilon(x)$ and $B_\epsilon(a)$, such that for any $x'\in B_\epsilon(x)$ and $a'\in B_\epsilon(a)$, we have $f(x')>f(a')$
However, the difficulty is, when $a$ is moving from $y$ to $x$, the $\epsilon\to 0$
I don't know how to get over this. Would be very grateful if you could help. Are there any named theorems on this?
Hint: Consider $g:\mathbb R\to \mathbb R:t\mapsto f((1-t)x+ty).$