MathLabs

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 2 of 5: ff is injective
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f(x)=f(y), y≠z  ⟹  ∣f(x)−f(y)∣=0<∣f(y)−f(z)∣, taking x=y gives a contradictionf(x)=f(y),\ y\ne z \implies |f(x)-f(y)|=0<|f(y)-f(z)|,\ \text{taking }x=y\text{ gives a contradiction}
Detailed analysis

Suppose f(y)=f(z)f(y)=f(z) for some y≠zy\ne z. Taking x=yx=y in the given condition gives 0<∣y−z∣0<|y-z|, so the condition requires ∣f(x)−f(y)∣<∣f(y)−f(z)∣|f(x)-f(y)|<|f(y)-f(z)|. But both sides are 00 since f(y)=f(z)f(y)=f(z), a contradiction. Thus the function is injective.