Functions satisfying the functional equation $ \big( 1 - f ( x ) f ( y ) \big) f ( x + y ) = f ( x ) + f ( y ) $

Question

How can I prove that there is no real function defined on $ \mathbb R $, continuous at $ 0 $ and not always vanishing satisfying the functional equation $$ \big( 1 - f ( x ) f ( y ) \big) f ( x + y ) = f ( x ) + f ( y ) \text ? \tag E $$

Answer 1

You have no doubt established $f(0)=0$.

Suppose such an $f$ is not bounded above. Then for all $x_0$ with $f(x_0)>0$, there exists a $y_0$ with $f(y_0)=\frac{1}{f(x_0)}$. And then the relation gives $0=f(x_0)+f(y_0)$, a contradiction.

So such an $f$ must be bounded above. (Similarly, it must be bounded below.)

If $f$ ever reaches an output of $1$, then the relation gives $0=2$, so the supreumum of $f$'s outputs, $S$, is at most $1$.

Assume $f$ takes some positive values. Let $X$ be an $x$-value such that $f(X)$ is $\epsilon S$ where $\epsilon$ is some number slightly smaller than $1$. We know that $0<\epsilon S<1$. Then using $x=y=X$, the relation gives that $$f(2X)=\frac{2f(X)}{1-f(X)^2}=\frac{2(\epsilon S)}{1-(\epsilon S)^2}>2\epsilon S>S$$ a contradiction.

So $f$ takes no positive values. Similarly, it takes no negative values. $f$ would have to be the zero function.

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Note that if $f$ is allowed to spike to $\pm\infty$, then this argument fails where we claim that $\left[1-f(x_0)f(y_0)\right]f(x_0+y_0)=f(x_0)+f(y_0)$ yields a contradiction. In that scenario, $f(x_0+y_0)=\pm\infty$ would offer a way out.

Answer 2

Letting $x=y=0$, we get that $f(0)=0$.

Letting $y=-x$, we get that $f(-x)=-f(x)$.

If for some $x$ and $y$, we have $f(x)f(y)=1$ then $f(x)+f(y)=0$ which leads to a contradiction. So for all $x$ and $y$, $f(x)f(y)\ne1$. By $f(-x)=-f(x)$, we also have $f(x)f(y)\ne-1$.

Substituting $x+h$ for $x$ and $-x$ for $y$ in the original equation, we have $f(x+h)-f(x)=\big(1+f(x)f(x+h)\big)f(h)$. Now, because $f$ is continuous at $0$, so for positive $\epsilon$, there exists a positive $\delta$ such that if $|h|<\delta$ then $|f(h)|<\frac\epsilon{|1+f(x)f(x+h)|}$, which shows that $f$ is continuous at every point.

Now, because $f$ is continuous at every $x$, $1-f(0)^2=1>0$ and $1-f(x)^2\ne0$, so we have $1-f(x)^2>0$, or equivalently $|f(x)|<1$, for every $x$. But by letting $y=x$ in the original equation, we get $|f(2x)|=\Big|\frac{2f(x)}{1-f(x)^2}\Big|\ge2|f(x)|$. By induction, we have $|f(2^nx)|\ge2^n|f(x)|$, for every positive integer $n$. This shows that if there is a point $x$ such that $f(x)\ne0$, then there is a positive integer $n$ such that $|f(2^nx)|>1$, which leads to a contradiction. So the only solution to the functional equation is the constant zero function, which you've ruled out.

Answer 3

Perhaps a solution.

I use what @Mathcounterexample.net has found, i.e that the function verify $f(0)=0$, $f(x)\not =0$ if $x\not =0$ and that it is continuous on $\mathbb{R}$. Replace $x$ by $y$ in the functional equation, we get $\displaystyle (1-f(x)^2)f(2x)=2f(x)$. Now if there exist $x$ such that $f(x)^2=1$, we have $x\not = 0$, and $f(x)=0$, a contradiction. Hence $f(x)^2\not =1$ for all $x$. As $f$ is continuous, we must have $(f(x))^2>1$ for all $x$ or $(f(x))^2<1$ for all $x$. But $f(0)=0$; hence we have $(f(x))^2<1$ for all $x$. Now on $[0,+\infty[$, we must have $-1<f(x)\leq 0$ or $0\leq f(x)<1$. Replacing $f$ by $-f$ if necessary, one can suppose that $0\leq f(x)<1$ for all $x\geq 0$. Let $M={\rm Sup}\{f(x), x\geq 0\}$. We have $\displaystyle 2f(x)=(1-f(x)^2)f(2x)\leq M$, hence $f(x)\leq M/2$. This imply $M\leq M/2$, and $M=0$, hence $f=0$ on $[0,+\infty[$. As $f$ is odd, we have $f=0$, a contradiction.