MathDB

9

Part of 1981 Austrian-Polish Competition

Problems(1)

|f^n(x)-f^n(y)| < |x-y| for x, y\in [0,1], exists unique x_0 : f (x_0) = x_0

Source: Austrian Polish 1981 APMC

4/29/2020
For a function f:[0,1]→[0,1]f : [0,1] \to [0,1] we define f1=ff^1 = f and fn+1(x)=f(fn(x))f^{n+1} (x) = f (f^n(x)) for 0≤x≤10 \le x \le 1 and n∈Nn \in N. Given that there is a nn such that ∣fn(x)−fn(y)∣<∣x−y∣|f^n(x) - f^n(y)| < |x - y| for all distinct x,y∈[0,1]x, y \in [0,1], prove that there is a unique x0∈[0,1]x_0 \in [0,1] such that f(x0)=x0f (x_0) = x_0.
functioninequalitiesalgebracomposition