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 5 of 5: Verify the candidate

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In plain words

The threshold $x+y=2$ separates the two verification cases.

f(x)=\begin{cases}\dfrac{2}{2-x},&0\le x<2,\\0,&x\ge2\end{cases}

Detailed analysis

Use the piecewise function above. If $x+y\ge2$ , both sides are zero because the relevant argument lies in $[2,\infty)$ . If $x+y<2$ , direct substitution gives $f(xf(y))f(y)=\frac{2}{2-x-y}=f(x+y)$ . It satisfies all conditions and is unique.