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 2 of 5: Obtain a lower bound below 2

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In plain words

For $y<2$ , the nonzero hypothesis removes the factor $f(y)$ , so the other factor must have an argument in the zero region.

f((2-y)f(y))f(y)=f(2)=0\Longrightarrow(2-y)f(y)\ge2

Detailed analysis

For $0\le y<2$ , put $x=2-y$ . Then $f((2-y)f(y))f(y)=f(2)=0$ . Since $f(y)\ne0$ , we have $f((2-y)f(y))=0$ , so $(2-y)f(y)\ge2$ , or $f(y)\ge\frac{2}{2-y}$ .