Problem 5

Let

\mathbb{R}_{>0}

be the set of positive real numbers. Determine all functions

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

such that

\sqrt{\tfrac{x^2+f(y)^2}{2}}\ge\tfrac{f(x)+y}{2}\ge\sqrt{xf(y)}

for every

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

Step 2 of 4: Every orbit under f is an arithmetic progression

Previous step Next step

x=f(t),\ y=t\ \Longrightarrow\ f(t)\ge\frac{f(f(t))+t}{2}\ge f(t)\ \Longrightarrow\ f^n(t)=t+n\,d(t),\quad d(t):=f(t)-t\ge0

Detailed analysis

Setting $x=f(t)$ and $y=t$ makes $x=f(y)$ , so both outer expressions equal $f(t)$ and squeeze $\tfrac{f(f(t))+t}{2}=f(t)$ , i.e. $f(f(t))-f(t)=f(t)-t$ . Applying this to $f^{n-1}(t)$ for all $n\ge1$ shows that the orbit $(t,f(t),f^2(t),\ldots)$ is an arithmetic progression with common difference $d(t):=f(t)-t$ , so $f^n(t)=t+n\,d(t)$ . Since $f^n(t)>0$ for every $n\ge1$ , we must have $d(t)\ge0$ for all $t>0$ .