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determine every function $f$ defined for positive numbers , having positive values, such that : $f(xf(y))f(y)=f(x+y)$

$f(2)=0$

$f(x)\neq 0$ for every $0\le x<2$

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

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    $\begingroup$ Please do not add tags without reading the tag description carefully. This question has absolutely nothing to do with functional analysis. The "functions" tag is also used specifically for questions about the basic set theory of functions. $\endgroup$ – Eric Wofsey Sep 11 '18 at 21:10
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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 $$ f(x) = \begin{cases} \dfrac{2}{2-x}, & 0\le x < 2, \\ 0, & 2 \le x. \end{cases} $$

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