Is there an everywhere-defined function that satisfies $f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$

Question

Is there a function $f:\mathbb{R}\to\mathbb{R}$ which is differentiable and satisfies the following:

(1) $f(x+y)=\frac{f(x)+f(y)}{1-f(x)f(y)}$

(2) $f'(0)=1$

(1) is the functional equation for $\tan$ function, but $\tan x$ is not defined at $\frac{\pi}{2}+n\pi$, so $f$ cannot be the $\tan$ function. Then does such $f$ exist?

Answer 1

Here is the outline of a proof.

1. Setting $x = 0$ and $y = 0$ in (1), we find that $f(0) = 0$.

2. By the second condition, we see that $f(a) > 0$ for some small $a > 0$.

3. Letting $x = y$ in (1), we see that $1 - f(x)^2 \ne 0$, so $f(x) \ne \pm 1$. Since also $f(0) = 0$, by the intermediate value theorem $f(x) \in (-1,1)$ for all $x$.

4. Letting $x = y$ in (1), we note first that $0 < 1 - f(x)^2 \leq 1$ by point 3. Thus f(x) > 0 implies $f(2x) > 2f(x)$.

5. By induction from point 4, we find $f(2^n a) > 2^n f(a)$.

6. Point 5 is obviously incompatible with point 3.

So there is no such function $f$.

Answer 2

It is not possible. It is straightforward to show that $f$ satisfies $f'(x) = 1+ f(x)^2$, with $f(0) = 0$ (because $f'(0) = 1$). The unique solution on $(-{\pi \over 2}, {\pi \over 2})$ is $f(x) = \tan x$.

Answer 3

Note that by setting $y=0$ we get $$ f(x) =\frac{f(x)+f(0)}{1-f(0)f(x)} \Rightarrow (1+f(x)^2) f(0) = 0 ~\forall x$$

Hence $f(0)=0$. Differentiating both sides of 1) with respect to $x$ you ge $$ f'(x+y) = \frac{1+f(y)^2}{(1-f(x) f(y))^2} f'(x) $$ similarly, differentiating with respect to $y$ $$ f'(x+y) = \frac{1+f(x)^2}{(1-f(x) f(y))} f'(y) $$

Hence $$ (1+f(x)^2) f'(y) = (1+f(y)^2) f'(x)$$ Substitute $y=0$ to get $$ f'(x) = 1+f(x)^2 $$ We also have $f(0)=0$. Hence $f(x)=\tan(x)$

Added in response to comment if $f$ exists, it has to be $\tan(x)$ and since you want a function other than $\tan(x)$, no other function exists.

Answer 4

No, there is not; the proof is based on three observations:

First, suppose that $f$ passes through $\pm 1$. Near such an $x$, the numerator of the functional equation is near $2$ and the denominator is near $0$, so the function is unbounded (hence discontinuous) at $2x$.

Second, note that $f(0) = 0$. This is immediate from setting $x = y = 0$ in the functional equation.

Third, suppose that $f(x) > 0$ for some small positive $x$. Then $$f(2x) = \frac{2 f(x)}{1 - f(x)^2} > 2f(x)$$

Inducting on this, we find that

$$f(2^n x) > 2^n f(x)$$ for all $n = 1, 2, ...$. Choose an $n$ large enough, and this exceeds $1$, contradicting the first observation. A similar argument shows that if $f(x) < 0$ we arrive at a contradiction; thus $f \equiv 0$ is the only differentiable solution.

Answer 5

Differentiability is not needed to show that all solutions of the functional equation are $\tan Ct$ for constant $C$.

The complex function $g(x) = \frac{1+if(x)}{1-if(x)}$ has $|g|=1$ (for $f$ real-valued) and, from the functional equation on $f$, satisfies the addition formula

$g(x)g(y)=g(x+y)$ for all $x,y \in \mathbb{R}$

The only measurable solutions of that equation with real $f$ (that is, $|g|=1$) are $g(t)=e^{i\alpha t}$ with real $\alpha$, and no other solutions can be proved to exist without some form of the Axiom of Choice for uncountable sets. In that generality, the solution is to take $g(t)=e^{i\alpha t}$, solve for $f(t)$ from $g(t)$ (each is a linear fractional function of the other), and find that $f(t) = \tan \frac{\alpha}{2} t$.

Specializing to differentiable solutions, the only way to have $f'(0)=1$ is $f(t)=\tan t$.