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 4 of 6: Equality is forced exactly at powers of a

Previous step Next step

a^n = f(a)^n \ge f(a^n) \ge a^n \implies f(a^n) = a^n

Detailed analysis

Since $a>1$ , the previous step gives $f(a^n)\ge a^n$ ; on the other hand (i) gives $f(a)^n\ge f(a^n)$ , i.e. $a^n\ge f(a^n)$ because $f(a)=a$ . The two inequalities force $f(a^n)=a^n$ for every positive integer $n$ .