Let $\mathbb R^+$ denote the set of positive real numbers. Find all functions $f:\mathbb R^+\to\mathbb R$ and $g:\mathbb R^+\to\mathbb R$ such that for all $x,y\in\mathbb R^+$, $g(x)-g(y)=(x-y)f(xy)$. Linus Tang
Problem
Source: ELMO Shortlist 2024/A6
Tags: Elmo, algebra
23.06.2024 02:10
Rewrite as $\sum_{\text{cyc}}a(b-c)f(a)=0$. If $f(a)=m+\frac na$ and $f(b)=m+\frac nb$, then we get $f(c)=m+\frac nc$, which gives $g(x)=mx-\frac nx+C$.
23.06.2024 11:35
Basically the same idea as in ISL 2018 A5.
23.06.2024 20:28
Trivial . ig. Let $S$ be the set of all $g$ for such $f$ exists. Claim: For any two elements $x,y \in S$ and real nos $a,b$ $ax+by \in S$. Proof: trivial Claim: For an element in $g \in S$ which is not injective , then $g$ is a linear combination of $x$ and $\frac{1}{x}$ . and a constant. Proof: Say $g(a) = g(b)$ , $a \neq b$ observe $g(x)=g(\frac{ab}{x}) \forall x \in \mathbb{R}$ . Observe $f(xy)(x-y)=(g(x)-g(y))$ and $f(\frac{dx}{y})(x-\frac{d}{y})=g(x)-g(y)$ divide the two [ u can check divisible So $\frac{f(xy)}{f(\frac{dx}{y})}=\frac{x-y}{x-\frac{d}{y}} \forall xy \neq d$ from here u can get an explicit formula for $f$ and $g$ Now observe any constant function is in $S$ and of form $cx$ and $\frac{c}{x}$ , say we have another solution $g'$ which is not a linear combination of these . construct a linear combination of ocnstant $x$ and $\frac{1}{x}$ and $g'$ which is not injective and use last claim