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question -

Find all continuous functions $f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy the equation $$ f(x+y)=\frac{f(x)+f(y)+2 f(x) f(y)}{1-f(x) f(y)} $$ for all $x, y$

my try - i proved that f(0)=0 ..then hint says substitute $g(x)=f(x) /(1+f(x))$.. i am not getting how to use this hint...i substitute in original equation and simplified lots of thing but nothing seems useful..

any hints ??? ...and how one can thought that we have to substitute this...

Ishan
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  • Use the first principle to find f'(x) (i.e. f'(x) = f(x+h) - f(x)/h) as h-> 0 . You can substitute the value of f(x+h) using the given functional equation then apply limits. You should get a neat differential equation. – Scilife Apr 11 '20 at 05:39
  • You have already proved that f(0) = 0. – Scilife Apr 11 '20 at 05:41
  • without differentiation ??? – Ishan Apr 11 '20 at 05:41
  • that might get a bit hairier. – Scilife Apr 11 '20 at 05:42
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    If you tried to use the hint, you would get $g(x+y)=g(x)+g(y)$. By continuity, $g(x)=kx$. However, you should first prove that $f(x)\ne -1$ for all $x$ to be able to use the hint without a hiccup (without division by $0$, that is). – Batominovski Apr 11 '20 at 05:42

2 Answers2

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Using the given hint, we get, $$g(x)=\dfrac{f(x)}{1+f(x)}\\ \implies f(x)=\dfrac{g(x)}{1-g(x)}\\ \implies \dfrac{g(x+y)}{1-g(x+y)}=\dfrac{\dfrac{g(x)}{1-g(x)}+\dfrac{g(y)}{1-g(y)}+\dfrac{2g(x)g(y)}{(1-g(x))(1-g(y))}}{1-\dfrac{g(x)g(y)}{(1-g(x))(1-g(y))}}\\ \implies \dfrac{g(x+y)}{1-g(x+y)}=\dfrac{g(x)(1-g(y))+g(y)(1-g(x))+2g(x)g(y)}{1-g(x)-g(y)}\\ \implies\dfrac{g(x+y)}{1-g(x+y)}=\dfrac{g(x)+g(y)}{1-g(x)-g(y)}\\ \implies \dfrac{g(x+y)}{1-g(x+y)}+1=\dfrac{g(x)+g(y)}{1-g(x)-g(y)}+1\\ \implies \dfrac{1}{1-g(x+y)}=\dfrac{1}{1-g(x)-g(y)}\\ \implies \boxed{g(x+y)=g(x)+g(y)}$$

Now use this.

Martund
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As suggested, let $g(x)=\frac{f(x)}{1+f(x)}$. Then $f(x)=\frac{g(x)}{1-g(x)}$. Thus $$\frac{g(x+y)}{1-g(x+y)}=\frac{\frac{g(x)}{1-g(x)}+\frac{g(y)}{1-g(y)}+2\frac{g(x)}{1-g(x)} \cdot \frac{g(y)}{1-g(y)}}{1-\frac{g(x)}{1-g(x)} \cdot\frac{g(y)}{1-g(y)}}=\frac{g(x)+g(y)}{1-(g(x)+g(y))}.$$ So $$\frac{g(x+y)}{1-g(x+y)}=\frac{g(x)+g(y)}{1-(g(x)+g(y))}.$$ Since the function $\frac{t}{1-t}$ is monotone, we conclude from the above equality that $$g(x+y)=g(x)+g(y).$$ Thus $g(x)$ is a linear function with $g(0)=0$, and therefore $g(x)=mx$ for some $m \in \mathbb{R}$. Therefore if $$f(x)=\frac{mx}{1-mx}, m \in \mathbb{R},$$ then the identity in the statement of the problem will hold for points in the domain of the functions in the left and right hand sides, i.e. $$x,y \neq \frac{1}{m}, \ \ x+y \neq \frac{1}{m}.$$ If the problem really wants the identity to be satisfied for all $x,y \in \mathbb{R}$, then the domain of $f$ has to be all of $\mathbb{R}$, and hence the only function satisfying the assumptions is $f=0$ (m=0).

Mike Williams
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