Find all functions $f$ mapping the non-negative reals onto the non-negative reals such that:
- $f\big(xf(y)\big)f(y)=f(x+y)$ for all non-negative $x,y$,
- $f(2)=0$,
- and $f(x)\ne 0$ for all $0\le x <2$.
I do not think any such functions exist. My reasoning is as follows:
Proof: Assume there is a function $f$ satisfying the above. A few simple calculations show that $f(0)=1$ and $f(1)=2$.
Assume $x>2$, then there exists $y$ such that $y+2=x$. It follows then that $$f\big(yf(2)\big)f(2)=f(x) \implies 0=f(x)\text.$$
Thus $f(x)=0$, for all $x\ge2$.
Next, consider some $y\in (0,1)$. There exists $x\in(1,2)$, such that $x+y=2$, and so $$f\big(xf(y)\big)f(y)=0 \implies f\big(xf(y)\big)=0 \implies f(y)>1\text.$$
Thus, for all $y\in (0,1)$, we have $f(y)\in (1,2)$.
Since $f$ is onto, for any $a\in(0,1)$, there must exist a non-negative $x$ such that $f(x)=a$. Further, by above we must have $x\in (1,2)$. Arguing as before, there exists $y\in(0,1)$ such that $x+y=2$. Hence $$f\big(yf(x)\big)f(x)=0 \implies f(ya)=0\text.$$
However, by our choice of $y$ we have $ 0<ya<1$, and so $f(ya)$ cannot equal zero. Thus, there are no functions $f$ which satisfy these criteria. $\quad \square$