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 4: Get the lower bound

Previous step Next step

In plain words

The zero-region characterization turns a product equal to zero into an inequality on its argument.

f(xf(2-x))f(2-x)=f(2)=0\Longrightarrow f(x)\ge\frac{2}{2-x}

Detailed analysis

For $0\le x<2$ , substitute $y=2-x$ . Since $f(2-x)\ne0$ , the equation gives $f(xf(2-x))=0$ , so $xf(2-x)\ge2$ . Renaming $2-x$ as $y$ yields $f(y)\ge\frac{2}{2-y}$ .