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<b 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<x<b;

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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