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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.

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  • $\begingroup$ 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 $D(y/x) = \left[x D(y) - y D(x)\right]/x^2$. $\endgroup$ – TMM Apr 30 '13 at 17:59
  • $\begingroup$ @TMM I already thought about this. Another idea was $ln$, because $ln(y/x)=ln(y)-ln(x)$. $\endgroup$ – pascalhein Apr 30 '13 at 18:07
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    $\begingroup$ I found one! $f(x)=x-1/x$ $\endgroup$ – MichalisN Apr 30 '13 at 18:09
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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 $\{f(x)=\lambda(x-\frac{1}{x})|\lambda\in\mathbb{R}\}$.

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