# Prove there are no other solutions of functional equation $f(x+y) = \frac{f(x)+f(y)}{1-f(x)f(y)}$

I have the following functional equation. Find all continuous functions $f:(-1,1) \to \mathbb R$ such that $$f(x+y)=\frac{f(x) + f(y)}{1 - f(x)f(y)}$$ The first obvious solution is $f(x) \equiv 0$. Another one I guessed, it is $f(x) = \pm \tan x$. I suspect, that there are no more solutions. The problem is that I don't know how to prove that.

Since we have a rational equation, I have no idea how to make any substitutions in order to get expression for $\tan x$ (as it is not a rational expression).

P.S. I can also show that it is true that (a) $f(0) = 0$ (take $x=y=0$), and (b) $f(-x) = -f(x)$ (take $y=-x$).

Update: there is a solution of this problem also here: https://artofproblemsolving.com/community/c6h386060

• Calculate $$\frac{f(x+h)-f(x)}h$$ using the functional equation for $f(x+h)$. Let $h\to0$. Assuming that $f'(0)$ exists you get that $$f'(x)=f'(0)(1+f(x)^2).$$ I'm sure you can take it from here. – Jyrki Lahtonen Oct 30 '17 at 14:58
• @JyrkiLahtonen Very nice edit: maybe make it an answer? – Zubin Mukerjee Oct 30 '17 at 15:01
• How about $f(x) = \tan (cx)$ where $c$ is constant? – Michael Hardy Oct 30 '17 at 15:40
• Indeed, if we assume existence of $f'(0)$ we obtain solution $f(x) = \tan(c x + \pi n)$ for any $c \in \mathbb R, n \in \mathbb Z$. But can we say something without this assumption (as original problem does not say anything about derivatives)?.. – Yauhen Yakimenka Oct 30 '17 at 16:06
• I have a feeling that without this assumption you'll get a whole gang of pathological solutions. – Ivan Neretin Oct 30 '17 at 17:58

Define $g(x)=\arctan(f(x))$, so that $f(x)=\tan(g(x))$. Then $g:(-1,1)\to(-\pi/2,\pi/2)$ is continuous. The functional equation becomes $$\tan(g(x+y))=\frac{\tan(g(x))+\tan(g(y))}{1-\tan(g(x))\tan(g(y))}=\tan(g(x)+g(y)).$$ The latter equality used the tangent angle addition formula. It follows that $g(x+y)-g(x)-g(y)\in \mathbb{Z}\cdot\pi$. The function $w(x,y):=g(x+y)-g(x)-g(y)$ is continuous and discretely-valued on the connected set $\{(x,y)\in(-1,1)^2:x+y\in(-1,1)\}$, so $w$ must be constant. Since $w(0,0)=0$, we have $w\equiv 0$, and we conclude that $$g(x+y)=g(x)+g(y),$$ for all $x$, $y\in(-1,1)$ such that $x+y\in(-1,1)$. It is well-known that every continuous real-valued functions on $\mathbb{R}$ that preserves addition is of the form multiplication by a constant, and essentially the same proof works for functions on $(-1,1)$. I won't write out the details.

Since $g$ takes $(-1,1)$ into $(-\pi/2,\pi/2)$, we must have $g(x)=Cx$ for some $C\in [-\pi/2,\pi/2]$. We conclude that the only solutions to the original equation are $$f(x)=\tan(Cx)$$ for $C\in[-\pi/2,\pi/2]$. The cases $C=0,\pm1$ are the solutions you found.

• But what if $g(x) + g(y) \in (\pi/2, \pi)$ - haven't you lost this case when applied arctan? – Yauhen Yakimenka Nov 1 '17 at 10:46
• @YauhenYakimenka Good point. In fact this cannot happen, I will edit the answer to explain why – Julian Rosen Nov 1 '17 at 13:59
• Okay, but I think $f(x) = \tan(Cx+\pi k)$ should be also solution for any $k \in \mathbb Z$, formally speaking. – Yauhen Yakimenka Nov 1 '17 at 16:19
• Tangent has period $\pi$, so we don't get anything new. – Julian Rosen Nov 1 '17 at 18:49