Home > Analysis, Olympiad > IMC 2011 Day 1 Problem 1

## IMC 2011 Day 1 Problem 1

Let $f: \Bbb{R} \to \Bbb{R}$ be a continuous function. A point $x$ is called a shadow point if there is a point $y \in \Bbb{R}$ with $y>x$ such that $f(y)>f(x)$. Let $a be real numbers and suppose that

• all points in $(a,b)$ are shadow points;
• $a,b$ are not shadow points.

Prove that

a) $f(x) \leq f(b),\ \forall a;

b) $f(a)=f(b)$.

Solution: Pick $c \in (a,b)$. Then $f: [c,b] \to \Bbb{R}$ is continuous on a compact set and therefore it is bounded and it reaches its bounds in $f(x_0)$. If $x_0 \neq b$ then we contradict that $b$ is not a shadow point. This proves a). Since $a$ is not a shadow point, b) follows now easily.

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