Problem 5

Let

\mathbb{Q}_{>0}

be the set of all positive rational numbers. Let

f:\mathbb{Q}_{>0}\to\mathbb{R}

be a function satisfying the following three conditions: (i) for all

x,y\in\mathbb{Q}_{>0}

f(x)f(y)\ge f(xy)

; (ii) for all

x,y\in\mathbb{Q}_{>0}

f(x+y)\ge f(x)+f(y)

; (iii) there exists a rational number

a>1

such that

f(a)=a

. Prove that

f(x)=x

for all

x\in\mathbb{Q}_{>0}

Step 2 of 6: f is increasing and bounded below by x − 1

Previous step Next step

f(x) \ge x - 1 \text{ for } x \ge 1, \qquad f \text{ strictly increasing}

Detailed analysis

From (i) applied with $x=m/n,y=n$ we get $f(m/n)f(n)\ge f(m)$ so $f(q)>0$ for every $q\in\mathbb{Q}_{>0}$ ; then (ii) shows $f(x+y)\ge f(x)+f(y)>f(x)$ , so $f$ is strictly increasing. Combined with $f(n)\ge n$ from the previous step, for $x\ge1$ this gives $f(x)\ge f(\lfloor x\rfloor)\ge\lfloor x\rfloor>x-1$ .