Solve the functional equation $$ \frac{f(x)}{f(y)}=f\left( \frac{x-y}{f(y)} \right), $$ here $f: \mathbb{R} \to \mathbb{R}$ and $f$ is differentiable at $x=0.$

By set $x=y$ we get $f(0)=1$.

Differentiate $$ \frac{f'(x)}{f(y)}=f'\left( \frac{x-y}{f(y)} \right) \cdot \frac{1}{f(y)} $$ and set $x=0$ get $$ f'(0)=f'\left( \frac{-y}{f(y)} \right). $$ So we reduce the problem to the problem to describe all function $g$ $$ g\left( -\frac{y}{g(y)} \right)=const. $$

I have no more ideas.

  • $\begingroup$ Leox if $f$ is only differentiable at $x=0$ I don't think you can differentiate it like that. $\endgroup$
    – pointer
    Jun 6, 2015 at 11:42
  • $\begingroup$ What is the right differentiation? $\endgroup$
    – Leox
    Jun 6, 2015 at 11:44
  • $\begingroup$ Define a variable $h$ , where $y=x$ and $x=x+h$ and look for derivative as $h\to 0$ at $x=0$ $\endgroup$
    – Someone
    Jun 6, 2015 at 11:46
  • $\begingroup$ @Mann Do you mean I have to find this limit at $h \to 0$ $$ \frac{1}{h}\left(f\left(\frac{x+h-y}{f(y)}\right) - f\left(\frac{x-y}{f(y)}\right) \right)? $$ $\endgroup$
    – Leox
    Jun 6, 2015 at 12:21

2 Answers 2


Let us define our variables as $x=x+h$ and $y=x$, for convenience.

Then our functional equation becomes,

$$\frac{f(x+h)}{f(x)}=f\left( \frac{h}{f(x)} \right)$$

With $h=0$ and $x=x_0$, we can get that $f(0)=1$. (If $f$ exists and is non-zero as well for a single $x_{0}$, otherwise not). $x_0=0$ is then one such point.

Using $f(x+h)=f(x)\times f\left( \frac{h}{f(x)} \right) $ and the definition of derivative,

$$f'(x)=\lim_\limits{h \to 0}\frac{f(x+h)-f(x)}{h}=\lim_\limits{h \to 0}\;f(x) \times \frac{f\left( \frac{h}{f(x)} \right)-1}{h}=\lim_\limits{h \to 0}\frac{f\left( \frac{h}{f(x)} \right)-1}{\frac{h}{f(x)}}$$

Then, $\lim_\limits{x \to x_{0}} f'(x)$ exists for arbitrary $x_{0}$ and is equal to $f'(0)$. Hence, $f'(x)$ is continuous 'almost everywhere' or everywhere and is equal to $f'(0)$. We can integrate $f'(x)$ to obtain a linear straight line, because Null sets have measure $0$. The null set only consists the point where $f(x) $ doesn't exist.

Under the assumption that $f(x)$ is finite at the point of inspection $x$ and also the fact that $f(0)=1$. We see that this limit is simply $f'(0)$ which exists. We get $$f'(x)=f'(0)$$

Which on integrating gives general solution, $$f(x)=f'(0)x+c=ax+c$$

also $f(0)=1\implies c=1$ which gives $$f(x)=ax+1$$ except possibly on a Null set which contains the points $x$ where $f(x)$ doesn't exist or $f(x)=0$


Let $f(x)=x+1$. It is obviously a solution.

  • $\begingroup$ Moreover, $ax+1$ is a solution for any $a$ and there is no other linear solution. $\endgroup$
    – pointer
    Jun 6, 2015 at 11:34
  • $\begingroup$ what about nonlinear solutions? $\endgroup$
    – Leox
    Jun 6, 2015 at 11:37

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