determining every function $f$ defined for positive numbers satisfying $f\big(xf(y)\big)f(y)=f(x+y)$, $f(2)=0$ and $f(x)\ne 0$ for $0\le x<2$

Question

determine every function $f$ defined for positive numbers , having positive values, such that:

• $f\big(xf(y)\big)f(y)=f(x+y)$;
• $f(2)=0$;
• $f(x)\ne 0$ for every $0\le x<2$.

I proved that $f(x)=0$ for every $x$ greater than $2$

Answer 1

Step 1. Substitution $y = 2$ gives $f(x+2) = 0$, so $f(x) = 0$ for all $x \ge 2$. Since $f(x) \neq 0$ for $0\le x < 2$ we conclude $$ f(x) = 0 \iff x \ge 2. $$

Step 2. Substitution $x \to 2 - x$ and $y \to x$ gives $$ f((2-x)f(x))f(x) = 0, $$ so for $0\le x < 2$ (since $f(x) \neq 0$) we have $f((2-x)f(x)) = 0$ or (by step 1) $(2-x)f(x) \ge 2$, so $$ f(x) \ge \frac{2}{2-x} \iff 0\le x < 2. $$

Step 3. Substitution $x \to \frac{2}{f(x)}$ and $y \to x$ gives for $x < 2$ $$ 0 = f\left(\frac{2}{f(x)}f\left(x\right)\right)f\left(x\right) = f\left(\frac{2}{f(x)} + x\right), $$ so by step 1 $$ x + \frac{2}{f(x)} \ge 2 $$ which implies $$ f(x) \le \frac{2}{2-x}. $$ Comparing this with result of step 2 we get, finally $$ f(x) = \begin{cases} \dfrac{2}{2-x}, & 0\le x < 2, \\ 0, & 2 \le x. \end{cases} $$