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
DottedCaculator
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 $g(x)=mx-\frac nx+C$.
VicKmath7
23.06.2024 11:35
Basically the same idea as in ISL 2018 A5.
The solutions are $f(x)=a-\frac{b} {x}$ and $g(x)=ax+\frac{b} {x}+c$ for some constants $a, b, c$. Let $P(x, y)$ denote the given assertion.
Observe that we can find reals $a, b, c$, such that $g(1)=a+b+c$, $g(2)=2a+\frac{b} {2}+c$, $g(4)=4a+\frac{b} {4}+c$. Then we can consider $f_1(x)=f(x)-a+\frac{b} {x}$ and $g_1(x)=g(x)-ax-\frac{b} {x}-c$, which are easily checked to work and satisfy $g_1(1)=g_1(2)=g_1(4)=0$. Hence, we can assume WLOG that $g(1)=g(2)=g(4)=0$ and we will show that $f, g \equiv 0$, which will imply the desired solution set.
Observe that $P(x, 1)$, $P(x, 2)$, $P(x, 4)$ give us $$g(x)=(x-1)f(x)=(x-2)f(2x)=(x-4)f(4x). $$On the other hand, $P(2x, 1)$ and $P(2x, 2)$ give us $$g(2x)=(2x-1)f(2x)=(2x-2)f(4x).$$Fix some $x \neq \frac{1}{2}, 1, 2, 4$ and assume that $f(x) \neq 0$. This means that $f(x), f(2x), f(4x) \neq 0$, so $$\frac{f(2x)}{f(4x)}=\frac{2x-2}{2x-1}=\frac{x-4}{x-2},$$i.e. $x=0$, contradiction. Thus, $f(x)=g(x)=0$ for all $x \neq \frac{1}{2}, 1, 2, 4$. This easily implies that $f, g \equiv 0$: for $k \in \{\frac{1}{2}, 1, 2, 4\}$, $P(3, \frac{k} {3})$ implies $f(k)=0$ and $P(1, \frac{1}{2})$ implies $g(\frac{1}{2})=0$, so we are done.
idkk
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