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 3 of 6: Taking n-th roots forces f(x) ≥ x

Previous step Next step

f(x)^n \ge f(x^n) > x^n - 1 \implies f(x) \ge x \text{ for } x > 1

Detailed analysis

Induction on (i) gives $f(x)^n\ge f(x^n)$ for every positive integer $n$ ; combined with the previous step, $f(x)^n>x^n-1$ , so $f(x)>\sqrt[n]{x^n-1}$ for all $n$ . Letting $n\to\infty$ (using that $x>1$ fixed and comparing growth rates) yields $f(x)\ge x$ for every rational $x>1$ .