Problem 3

Let

\mathbb{R}_+

denote the set of positive real numbers. Determine all functions

f:\mathbb{R}_+\to\mathbb{R}

such that for all

x,y,z\in\mathbb{R}_+

|x-y|<|y-z|\ \text{if and only if}\ |f(x)-f(y)|<|f(y)-f(z)|.

Step 3 of 5: Derive the midpoint property

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y=\tfrac{x+z}{2} \implies |y-x|=|y-z| \implies |f(x)-f(y)|=|f(y)-f(z)| \implies f(y)=\tfrac{f(x)+f(z)}{2}

Detailed analysis

For $x\ne z$ , take $y=\tfrac{x+z}{2}$ , so $|y-x|=|y-z|$ . By Step 1, $|f(x)-f(y)|=|f(y)-f(z)|$ . Since $f(x)\ne f(z)$ , exactly one of $f(x),f(y),f(z)$ lies between the other two, and combined with injectivity this forces $f(y)=\tfrac{f(x)+f(z)}{2}$ .