Given the functional equation $$f(x+(1+x)f(y))=y+(1+y)f(x)$$

Such that $f:(-1,\infty) \to (-1,\infty)$ and the function $g(x):=\frac{f(x)}{x}$ is strictly increasing in $I=(-1,0)\cup(0,+\infty),$ how can one show that for every $t \in I$ we have $f(t)\neq t$?

Plugging $x=y=0,$ that makes $f(f(0))=f(0).$ Does that mean that $f(0)=0$?

Making $x=y,$ $f(x+(1+x)f(x))=x+(1+x)f(x)$ so we make $x+(1+x)f(x)=t$ that makes $f(t)=t$ but we can't make $t=0$ because $t$ is not in $I$? So do I have to use the fact that $\frac{f(x)}{x}$ is increasing?

Then after all this work we have to deduce that $f(0)=0$ and $f(x)=\frac{-x}{x+1}$ for $x \in (-1,\infty).$

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    $\begingroup$ Do $F$ and $f$ denote the same function here? $\endgroup$ – JimmyK4542 Dec 15 '15 at 23:20
  • $\begingroup$ @JimmyK4542 fixed $\endgroup$ – user233658 Dec 15 '15 at 23:21
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    $\begingroup$ To answer your question in line 5, $f(f(0)) = f(0)$ does not imply that $f(0) = 0$ unless you have other facts. For instance, $f(x) = 1$ satisfies $f(f(0)) = f(0)$ but not $f(0) = 0$. $\endgroup$ – JimmyK4542 Dec 15 '15 at 23:28
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    $\begingroup$ it was me, and I'm not familiar with these notations, so I guess not only me.... but then you have empty set, perhaps you want union $\cup$? $\endgroup$ – Michael Medvinsky Dec 15 '15 at 23:32
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    $\begingroup$ Is $g$ increasing on $I$ or strictly increasing? $\endgroup$ – user293794 Dec 15 '15 at 23:58

I think that this will do the trick for showing that f(0) = 0:

Let's start by setting $x = 0$. We get

$$f(f(y)) = f(0) + (1+f(0))y.$$

By setting $y = 0$, we get

$$f(x) = f(x + (1+x)f(0)).$$

Apply now $f$ to expression above and use the first equation:

$$f(0) + (1+ f(0))x = f(0) + (1+f(0))(x + (1+x)f(0)).$$

This simplifies to

$$(1+x)(1+f(0))f(0) = 0$$ and thus $f(0) = 0$. We can also conclude that $f(f(y)) = y $, so f is its' own inverse.

Let suppose that there is numbers $a,b$ such that $a < b$ and $f(a) = b$ and $f(b) = a$ ($f$ is its' own inverse). Since function

$$ g(x) = \frac{f(x)}{x}$$ is increasing by the hypothesis. This means that $g(a) \le g(b)$, so

$$\frac{f(a)}{a} \le \frac{f(b)}{b} \iff \frac{b}{a} \le \frac{a}{b} \iff$$ $$\frac{b^2 - a^2}{ab} \le 0 \iff \frac{(b-a)(b+a)}{ab} \le 0 \iff $$ $$ \frac{(b+a)}{ab} \le 0. $$

so we can say that there can't be such $a,b$ with same sign. So $x$ and $f(x)$ must have different signs.

As OP noticed the equation $$ f(x+(1+x)f(x)) = x + (1+x)f(x) $$ holds. Let denote $x+ (1+x)f(x)$ as $t$. Then the equation says $f(t) = t$ but $f(t)$ and $t$ have different signs, therefore, $t$ must be zero. Thus, we conclude that $x + (1+x)f(x) = 0$ and the function $f$ is:

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

  • $\begingroup$ Thank you for your answer but I edited my question ! We should show that f(t)#t $\endgroup$ – user233658 Dec 17 '15 at 20:23
  • $\begingroup$ By substituting $f(t) = t$ to the functional equation, we will see that functional equation holds. furthermore, g is increasing with $f(t) = t $ (but not strictly). So $f(t) = t$ fits the given conditions. $\endgroup$ – Wekku Dec 17 '15 at 21:15
  • $\begingroup$ good point, I will fix that error $\endgroup$ – Wekku Dec 17 '15 at 22:27

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