IMC 2011 · Day 1 · P1

(a) We prove by contradiction. Suppose that exists a point $c \in (a, b)$ such that $f(c) > f(b)$.

By Weierstrass' theorem, $f$ has a maximal value $m$ on $[c, b]$; this value is attained at some point $d \in [c, b]$. Since $f(d) = \max\limits_{[c,b]} f \ge f(c) > f(b)$, we have $d \ne b$, so $d \in [c, b) \subset (a, b)$. The point $d$, lying in $(a, b)$, is a shadow point, therefore $f(y) > f(d)$ for some $y > d$. From combining our inequalities we get $f(y) > f(d) > f(b)$.

Case 1: $y > b$. Then $f(y) > f(b)$ contradicts the assumption that $b$ is not a shadow point.

Case 2: $y \le b$. Then $y \in (d, b] \subset [c, b]$, therefore $f(y) > f(d) = m = \max\limits_{[c,b]} f \ge f(y)$, contradiction again.

(b) Since $a < b$ and $a$ is not a shadow point, we have $f(a) \ge f(b)$.

By part (a), we already have $f(x) \le f(b)$ for all $x \in (a, b)$. By the continuity at $a$ we have $$f(a) = \lim_{x \to a+0} f(x) \le \lim_{x \to a+0} f(b) = f(b)$$ Hence we have both $f(a) \ge f(b)$ and $f(a) \le f(b)$, so $f(a) = f(b)$.