Solve the functional equation $f(y)f(xf(y))=f(x+y)$ 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.
 A: 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 $y$.
