Problem 5

Find all functions

f

mapping the non-negative reals onto the non-negative reals such that

f(xf(y))f(y)=f(x+y)

for all non-negative reals

x,y

, with

f(2)=0

and

f(x)\ne0

for every

0\le x<2

Step 3 of 4: Get the upper bound

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In plain words

Choose $x$ so that $xf(y)=2$ ; the left side then contains the known zero $f(2)$ .

x=\frac{2}{f(y)}\Longrightarrow f\left(y+\frac{2}{f(y)}\right)=0\Longrightarrow f(y)\le\frac{2}{2-y}

Detailed analysis

For $0\le y<2$ , $f(y)>0$ , so set $x=2/f(y)$ . Then $f(y+2/f(y))=f(2)f(y)=0$ , which forces $y+2/f(y)\ge2$ . Rearranging gives $f(y)\le\frac{2}{2-y}$ .