# Solve the functional equation $\frac{f(x)}{f(y)}=f\left( \frac{x-y}{f(y)} \right)$

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.

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

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.

• Moreover, $ax+1$ is a solution for any $a$ and there is no other linear solution. Jun 6, 2015 at 11:34