To find all functions $f: (0,\infty) \rightarrow (0, \infty)$ such that $f(x+y) = f(x)f(yf(x))$

Question

$f: (0,\infty) \rightarrow (0, \infty)$

and for all $x,y >0$ we have:

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

Because it is very similar to $f(x+y) = f(x) f(y)$

it seems good to define $g:(0,\infty) \rightarrow \mathbb R$

$f(x) = e^{g(x)}$

Answer 1

We shall prove that all solutions are: the constant function $f(x)=1$ and $f(x)=\dfrac{1}{1+bx}$ with an arbitrary $b>0$.

Claim 1. $f$ is bounded from above by $1$.

Proof. For the sake of contradiction suppose that $f(t)>1$ for some $t>0$. Note that $\dfrac{t}{f(t)-1}>0$ so we may substitute $x=t$ and $y=\dfrac{t}{f(t)-1}$ in the equation. Observe that $x+y=yf(x)$ so $f(x+y)$ and $f(yf(x))$ cancel out leading to $f(t)=1$. This is a contradiction as $t$ was assumed to satisfy $f(t)>1$. $\square$

Claim 2. $f$ is nonincreasing.

Proof. By Claim 1. we have $f(yf(x))\le 1$ for all $x,y>0$ and therefore $$f(x+y) = f(x)f(yf(x)) \le f(x) \cdot 1 = f(x),$$ which shows that $f$ is nonincreasing. $\square$

Now suppose that there exists $t>0$ such that $f(t)=1$. Substitute $x=t$. We obtain $f(t+y)=f(t)f(yf(t))=f(y)$ for all $y>0$. So $f$ is periodic and also nonincreasing by Claim 2. Hence $f$ must be constant and since $f(t)=1$ it must be $f(x)=1$ for all $x>0$. This function clearly works.

From now on assume that $f(x)<1$ for all $x>0$.

Claim 3. $f$ is strictly decreasing. In particular, $f$ is injective.

Proof. For all $x,y>0$ we have $$f(x+y) = f(x) f(yf(x)) < f(x) \cdot 1 = f(x). \square$$

Substituting $y:=\dfrac{y}{f(x)}$ we get $$f\left(x+\frac{y}{f(x)}\right)=f(x)f(y).$$ Changing the roles of $x$ and $y$ in the above equation we get $$f\left(y+\frac{x}{f(y)}\right)=f(y)f(x)$$ which implies that for all $x,y>0$ we have $$f\left(x+\frac{y}{f(x)}\right)=f\left(y+\frac{x}{f(y)}\right)$$ and, by injectivity, $$x+\frac{y}{f(x)}=y+\frac{x}{f(y)}.$$ Fixing $y$ and calculating $f(x)$ from this we see that $f(x)$ is of the form $\dfrac{1}{a+bx}$ for some constants $a,b$. It remains to substitute $f(x)=\dfrac{1}{a+bx}$ into the original equation and calculate the values of $a,b$.

\begin{align*} \frac{1}{a+b(x+y)} &= \frac{1}{a+bx}\cdot \frac{1}{a+by\frac{1}{a+bx}} \\ a+bx+by &= (a+bx)(a+by\frac{1}{a+bx}) \\ a+bx+by &= a^2+abx+by \\ (a-a^2) + (b-ab) x &= 0 \end{align*}

It follows that $a-a^2=0$ and $b-ab=0$. This means $a=1$ and $b>0$ arbitrary or $a=b=0$. First case leads to $f(x)=\dfrac{1}{1+bx}$ and these functions clearly work. Second case leads to division by $0$ and is therefore impossible.