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 6 of 6: Extend from x > 1 to all positive rationals

Previous step Next step

n f(x) = f(n) f(x) \ge f(nx) \ge n f(x) \implies f(nx) = n f(x) \implies f(x) = x \text{ on } \mathbb{Q}_{>0}

Detailed analysis

For every $x\in\mathbb{Q}_{>0}$ and $n\in\mathbb{Z}_{>0}$ , (i) and the first step give $nf(x)=f(n)f(x)\ge f(nx)\ge nf(x)$ , so $f(nx)=nf(x)$ . Choosing $n$ so that $nx>1$ , the previous step gives $f(nx)=nx$ , hence $nf(x)=nx$ and $f(x)=x$ . Thus $f(x)=x$ for every $x\in\mathbb{Q}_{>0}$ .