# Functional equation $f(y/x)=xf(y)-yf(x)$

Are there any solutions to $f:\mathbb{R} \backslash \{0\} \rightarrow \mathbb{R}$ with $f(\frac{y}{x})=x \cdot f(y)-y \cdot f(x)$ other than $f(x)=0$ for all $x \in \mathbb{R}$?

I already have found that $f(1)=f(-1)=0$, and $f(-x)=-f(x)=f(\frac{1}{x})$, but I do not know how to find any functions satifying this equation besides plugging in values for $x$ and $y$ and finding constraints for $f$.

I couln't find anything I could use to find a function or a reason why one doesn't exist.

• This possibly has nothing to do with it, but the functional equation reminds me of the quotient rule for differentiation, i.e., if $y$ and $x$ are functions of $t$, then $D(y/x) = \left[x D(y) - y D(x)\right]/x^2$. – TMM Apr 30 '13 at 17:59
• @TMM I already thought about this. Another idea was $ln$, because $ln(y/x)=ln(y)-ln(x)$. – pascalhein Apr 30 '13 at 18:07
• I found one! $f(x)=x-1/x$ – MichalisN Apr 30 '13 at 18:09

We use the property $f(\frac{1}{x})=-f(x)$ that you mentioned. The functional equation with $x'=1/y$ and $y'=1/x$ implies $f(y/x)=f(y'/x')=\frac{1}{y}f(\frac{1}{x})-\frac{1}{x}f(\frac{1}{y})=\frac{1}{x}f(y)-\frac{1}{y}f(x)$. Combining this with the conventional functional equation we get $$\frac{1}{x}f(y)-\frac{1}{y}f(x)=xf(y)-yf(x)$$ so $$(x-\frac{1}{x})f(y)=(y-\frac{1}{y})f(x).$$ Inserting $y=2$ we finally have $$f(x)=\frac{2f(2)(x-\frac{1}{x})}{3}.$$ Conversely any choice of $f(2)$ in $\mathbb{R}$ gives a solution so the set of solutions is $\{f(x)=\lambda(x-\frac{1}{x})|\lambda\in\mathbb{R}\}$.