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Find all functions f defined over the positive reals.

$f(y)f(xf(y))=f(x+y)$

I proved that $0<f(x)<1$;

If a positive number $t$ such that $f(t)>1$ exists,

By setting $x$ as $\frac{t}{f(t)-1}$, and $y$ as $t$,

We get $f(t)=1$, which is a contradiction.

So for all $x$, $f(x) \le 1$.

$f(x) \not = 1$ can easily be proved, so

for all $x$,

$0 < f(x) < 1$.

Also got;

By setting $y$ to $x$ and vice versa, $f(x)f(yf(x))=f(x+y)=f(y)f(xf(y))$.

Please help.

HJS
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1 Answers1

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I assume that the function $f$ maps positive real numbers to positive real numbers.

Step 1: We show that $f(y) \leq 1$ for any $y$.

(This has been worked out in the question. I give proof for completeness.)

Suppose there is $y$ such that $f(y) > 1$. Then putting $x = \frac y {f(y) - 1}$ in the original equation gives $f(y) = 1$, contradiction.

As a corollary, the function $f$ is non-increasing: for any $y < z$, putting $x = z - y$ in the original equation gives $f(y) \geq f(z)$.

Step 2: If there exists $a$ such that $f(a) = 1$, then $f$ is constantly equal to $1$.

Putting $y = a$ in the original equation gives $f(x + a) = f(x)$. Thus by induction, we get $f(ka) = 1$ for any $k \in \Bbb Z_{> 0}$.

For any $x$, we may choose sufficiently large $k$ such that $x < ka$. Since $f$ is non-increasing, this leads to $f(x) \geq 1$ and hence $f(x) = 1$.

Step 3: The remaining case.

From now on, we suppose that $f(x) < 1$ for all $x$. The same argument as above shows that $f$ is strictly decreasing, and in particular is injective.

We replace $x, y$ in the original equation with $x + y - xf(y), xf(y)$ (note that these are positive real numbers). This gives: $$f(xf(y))f((x + y - xf(y))f(xf(y))) = f(x + y).$$ Together with the original equation, we get $f((x + y - xf(y))f(xf(y))) = f(y)$. By injectivity of $f$, this implies $$(x + y - xf(y))f(xf(y)) = y.$$ We now set $x = \frac z{f(y)}$ in the above equation, which becomes $$\left(\frac z{f(y)} + y - z\right)f(z) = y.$$ After algebraic transformation, we get $$\frac 1 y \left(\frac 1{f(y)} - 1\right) = \frac 1 z \left(\frac 1{f(z)} - 1\right).$$ This being valid for all $y, z$, we see that $\frac 1 y \left(\frac 1{f(y)} - 1\right)$ is a constant, independent of the value of $y$.

If we denote this constant by $c$, then we get $f(y) = \frac 1{cy + 1}$. It is easy to verify that this is a possible solution to the original equation.

Thus either $f(y) = 1$ for all $y$, or there exists a constant $c$ such that $f(y) = \frac 1{cy + 1}$ for all $y$.

WhatsUp
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  • Sorry, but $f$ is not yet proved to be a injective function. Please check the comments of the problem. Thanks for writing down this long solution. – HJS Sep 29 '21 at 16:47
  • @HJS Please read the whole answer, from very beginning. In Step 3 I treat the remaining case where $f(x)$ is never equal to $1$ and have shown that $f$ must be injective in that case. – WhatsUp Sep 29 '21 at 17:17
  • Yes, now I understand. Thanks for the long answer and the comment. – HJS Sep 29 '21 at 17:22
  • I'm sorry to question a correct solution, but is there some reason(motivation) why you set $x,y$ to $x+y-f(x), xf(y)$ at STEP 3? – HJS Sep 29 '21 at 17:29
  • This more or less comes from experience. The aim is to try to get something that is "closely related" to the original equation. Setting these parameters would allow us to get two identical terms in the equations. – WhatsUp Sep 29 '21 at 17:43
  • Oh, so it's basically intuition with a bit of logic. – HJS Sep 29 '21 at 17:45