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Problem 3

Let R+\mathbb{R}_+ denote the set of positive real numbers. Determine all functions f:R+→Rf:\mathbb{R}_+\to\mathbb{R} such that for all x,y,z∈R+x,y,z\in\mathbb{R}_+, ∣x−y∣<∣y−z∣ if and only if ∣f(x)−f(y)∣<∣f(y)−f(z)∣.|x-y|<|y-z|\ \text{if and only if}\ |f(x)-f(y)|<|f(y)-f(z)|.
Step 4 of 5: ff is monotone
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∃ y<x<z: sign(f(x)−f(y))≠sign(f(z)−f(x))  ⟹  ∣f(y)−f(z)∣<max⁡(∣f(x)−f(y)∣,∣f(z)−f(x)∣)\exists\, y<x<z:\ \text{sign}(f(x)-f(y))\ne \text{sign}(f(z)-f(x)) \implies |f(y)-f(z)|<\max(|f(x)-f(y)|,|f(z)-f(x)|)
Detailed analysis

Suppose ff is not monotone: there exist y<x<zy<x<z such that f(x)−f(y)f(x)-f(y) and f(z)−f(x)f(z)-f(x) have opposite signs. Then ∣f(y)−f(z)∣<max⁡(∣f(x)−f(y)∣,∣f(z)−f(x)∣)|f(y)-f(z)|<\max(|f(x)-f(y)|,|f(z)-f(x)|), whereas the hypothesis with (y,x,z)(y,x,z) in the roles of (x,y,z)(x,y,z) gives ∣f(y)−f(z)∣>max⁡(∣f(x)−f(y)∣,∣f(z)−f(x)∣)|f(y)-f(z)|>\max(|f(x)-f(y)|,|f(z)-f(x)|), a contradiction. Hence ff is monotone.