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 4 of 4: Squeeze and verify

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In plain words

The lower and upper bounds coincide, leaving exactly one candidate; substitution verifies it.

f(y)=\frac{2}{2-y}\ (0\le y<2),\qquad f(y)=0\ (y\ge2)

Detailed analysis

The two bounds force $f(y)=\frac{2}{2-y}$ on $[0,2)$ . Together with the zero tail, the unique function is $f(y)=\frac{2}{2-y}$ for $y<2$ and $f(y)=0$ for $y\ge2$ . Direct substitution verifies the equation in the cases $x+y<2$ and $x+y\ge2$ .