Determine all the continuous functions $f:(-1,1)\rightarrow \mathbb{R}$ such that $\forall x,y \in (-1,1)$, $f(x)+f(y)=f(\frac{x+y}{1+xy})$

Question

Determine all the continuous functions $f:(-1,1)\rightarrow \mathbb{R}$ such that $\forall x,y \in (-1,1)$, $f(x)+f(y)=f(\frac{x+y}{1+xy})$.

My attempt: For $x=y=0$, $2f(0)=f(0) \iff f(0)=0$ or $f(0)=1$. However, $f(0)=1 \Rightarrow f(x)=f(\frac{x+0}{1+x\cdot0})=f(x)+f(0)=f(x)+1$, and this gives $0=1$, false.

So, $f(0)=0$. Also, for $y=-x$, $f(x)+f(-x)=0 \Rightarrow f(x)=-f(-x)$, so $f$ is an odd function. Also I observed that $f(x)+f(-y)=f(x)-f(y)$, so $f(x)-f(y)=f(\frac{x-y}{1-xy})$.

I observe that there is a resemblance with the tangent of the sum: $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$, but that minus in $1-xy$ just does not fit with the plus in $1+xy$. So I think $\arctan$ is not a response.

Any help, please?

Answer 1

Actually, consider $g(x) = f(\tanh x):$ $$g(x) + g(y) = f\left(\dfrac{\tanh x + \tanh y}{1+\tanh x\tanh y} \right) = f(\tanh(x+y)) = g(x+y),$$ so $g$ solves the Cauchy equation. It's well-known that when you add continuity requirement on Cauchy equation, the only solution is $g(x) = ax$, which have been solved thoroughly on this site. In fact, only the continuity at $x = 0$ would be enough.