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 1 of 5: Verify linear maps work, and extract an equal-distance rule

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|x-y|=|z-y| \implies |f(x)-f(y)|=|f(z)-f(y)|

Detailed analysis

Every affine function $f(x)=ax+b$ with $a\ne0$ clearly satisfies the condition. Conversely, suppose $f$ satisfies the condition. Taking contrapositives in both directions shows that strict distance comparisons are preserved; the remaining equality case gives $|x-y|=|z-y|\implies|f(x)-f(y)|=|f(z)-f(y)|$ .