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Find all functions ${\rm f}:{\mathbb R}_{+} \to {\mathbb R}_{+}$ , such that $\forall\ x,y \in \mathbb R_+$ the equation
$$ \big(1 + y{\rm f}\left(x\right)\big)\big(1 - y{\rm f}\left(x + y\right)\big) = 1 $$ is fulfilled.

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2 Answers 2

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Assume $f$ is of the given form. Let $g(x)=\frac{1}{f(x)}$.

Plugging this into the equation gives

$$\left(1+\frac{y}{g(x)}\right)\left(1-\frac{y}{g(x+y)}\right)=1$$

Multiplying by $g(x)g(x+y)$, factoring out and subtracting $g(x)g(x+y)$ we obtain

$$yg(x+y)-y^2-yg(x)=0$$

Dividing by $y>0$ gives

$$g(x+y)=g(x)+y$$

This implies $\frac{g(x+y)-g(x)}{y}=1$ for all $x,y\in\mathbb{R}_+$.

In particular, $\lim_{y\rightarrow 0}\frac{g(x+y)-g(x)}{y}=1$ (*).

Hence $g$ is differentiable on $\mathbb{R}^+$ with derivative $1$.

By the fundamental theorem of calculus, $g(x)=x+c$ for some $c\ge 0$, that is

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

for $c\in\mathbb{R}_{\ge 0}$. These are also really solutions.

Important note (*):

We have the equation $g(x+y)-g(x)=y$ only for positive $y$. Therefore, the limit is a priori only a right-sided limit. This is a problem, because we can then only conclude right-differentiability, which is useless.

However, we can extend the equation to small negative $y$, thus making the limit two-sided.

Claim: Let $x>0$ be fixed. For all $y$ with $-x<y<0$ (that is: $y$ is negative and has small enough absolute value), we have $$\frac{g(x+y)-g(x)}{y}=1$$

Proof: Plugging in $-y>0$ for $y$ and $x+y>0$ for $x$ into the original equation gives

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

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  • $\begingroup$ Just curious, the note in the end; Since $x>0$, and $y \rightarrow 0$: just taking $y \neq 0$ values satisfying $x-y>0$ is enough for the definition of a limit. No? $\endgroup$
    – user76568
    Commented Feb 12, 2014 at 21:21
  • $\begingroup$ I don't follow (sorry..), I still can't see the potential problem.. $\endgroup$
    – user76568
    Commented Feb 12, 2014 at 21:26
  • $\begingroup$ @Dror: Yes, I also got confused, I wrote it down with wrong signs. Now I made it more clear I hope. It is clear that one does need to check that the limit is two-sided right? Since the original equation only holds for positive $y$. $\endgroup$
    – J.R.
    Commented Feb 12, 2014 at 21:32
  • $\begingroup$ @Dror: I just saw, you also forgot to check your limit from both sides. Your $\Delta x$ is a priori only allowed to be positive. $\endgroup$
    – J.R.
    Commented Feb 12, 2014 at 21:34
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Correct me if I'm very wrong. $$(1+y f(x))(1-y f(x+y))=1 \implies \frac{f(x+y)-f(x)}{y}=-f(x)f(x+y) < 0$$ So $f$ is strictly decreasing.
If $f$ was not continous at some point $x_1 \in \mathbb{R}_+$, it would have to be a jump discontinuity downwards and from the restirction it must follow that there's a jump discontinuity at every positive $y$ greater than $x_1$. But $x_1=x+y$ for many $x,y \in \mathbb{R}_+$, and again it follow from the restriction that there's a jump discontinuity at $x$. given $x$ is arbitrary and smaller than $x_1$, it follows that there's a jump discontinuity at every positive $x$ smaller than $x_1$. Finally, we conclude $f$ is jump discontinous in all It's domain, and coupled with the fact $f$ is bounded above (strictly decreasing), and bellow, we have a contradiction (To be elaborated on later).
SO... $f$ is continous! And: $$lim_{\Delta x \rightarrow 0^{+}} \frac{f(x+\Delta x)-f(x)}{\Delta x} = lim_{\Delta x \rightarrow 0^{+}} -f(x)f(x+\Delta x)=-f(x)^2$$ Again, $f$ is continous, so for $x$ taken as a positive constant, $-f(x)f(x+\Delta x)$ is a continous function of $\Delta x$ for all $\Delta x > -x$. Hence: $$lim_{\Delta x \rightarrow 0^{-}} \frac{f(x+\Delta x)-f(x)}{\Delta x} = lim_{\Delta x \rightarrow 0^{-}} -f(x)f(x+\Delta x) = lim_{\Delta x \rightarrow 0^{+}} -f(x)f(x+\Delta x) = lim_{\Delta x \rightarrow 0^{+}} \frac{f(x+\Delta x)-f(x)}{\Delta x}$$
So $f$ is even differentiable, and: $$f'(x)=-f(x)^2$$ And the only solutions are $f(x)=\frac{1}{x+a}$, for any constant $a \geq 0$ .

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  • $\begingroup$ The limit as $\Delta x\rightarrow 0$ needs to consider also negative $\Delta$, otherwise you cannot conclude differentiability (you only showed that the right derivative exists). See our discussion in the comments of my answer below. $\endgroup$
    – J.R.
    Commented Feb 12, 2014 at 21:34
  • $\begingroup$ Typo: negative $\Delta x$. $\endgroup$
    – J.R.
    Commented Feb 12, 2014 at 21:40
  • $\begingroup$ @TooOldForMath I think my modification corrects the issue. Is this correct? $\endgroup$
    – user76568
    Commented Feb 12, 2014 at 22:28
  • $\begingroup$ That's close enough. $\endgroup$
    – J.R.
    Commented Feb 12, 2014 at 22:30
  • $\begingroup$ @TooOldForMath hah. Don't be like that.. Tell me the error! :) $\endgroup$
    – user76568
    Commented Feb 12, 2014 at 22:31

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