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 5: Rule out a strict lower-bound inequality

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In plain words

If one value were too large, choose a nearby argument so that the product inside $f$ becomes exactly $2$ , forcing a forbidden zero below $2$ .

f(a)>\frac{2}{2-a}\Longrightarrow f(2-\varepsilon)=0

Detailed analysis

Suppose $f(a)>\frac{2}{2-a}$ for some $a<2$ . Choose $\varepsilon>0$ with $f(a)=\frac{2}{2-a-\varepsilon}$ . In the equation with $y=a$ and $x=2-a-\varepsilon$ , the left side is $f(2)f(a)=0$ , while the right side is $f(2-\varepsilon)\ne0$ because $2-\varepsilon<2$ . Contradiction.