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 5 of 5: Normalize and pin down ff by density
f(1)=1,f(2)=2  ⟹  f ⁣(n2k)=n2k ∀n∈Z>0,k≥0  ⟹  f(x)=x ∀x∈R+f(1)=1,f(2)=2 \implies f\!\left(\tfrac{n}{2^k}\right)=\tfrac{n}{2^k}\ \forall n\in\mathbb{Z}_{>0},k\ge0 \implies f(x)=x\ \forall x\in\mathbb{R}_+
Detailed analysis

Because ff is injective, f(2)≠f(1)f(2)\ne f(1). Define g(x)=1+(f(x)−f(1))/(f(2)−f(1))g(x)=1+(f(x)-f(1))/(f(2)-f(1)). Affine changes of the range preserve the condition, so gg also satisfies it and has g(1)=1,g(2)=2g(1)=1,g(2)=2. Applying the midpoint property repeatedly gives g(n/2k)=n/2kg(n/2^k)=n/2^k for every positive integer nn and k≥0k\ge0. These values are dense in R+\mathbb R_+, so monotonicity forces g(x)=xg(x)=x; undoing the normalization gives the original form f(x)=ax+bf(x)=ax+b.