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 4 of 4: Rule out mixed fixed points and conclude

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f(x)=x+D,\ f(y)=y,\ D>0\ \Longrightarrow\ |x-y|>D\ \Longrightarrow\ f(t)=t+D\ \text{for all }t>0

Detailed analysis

If $f$ is the identity, $c=0$ works; otherwise $D>0$ and some $x>0$ has $f(x)=x+D$ . Suppose some $y>0$ has $f(y)=y$ . If $y<x\le y+D$ , the left-hand inequality at $(x,y)$ gives $x>\sqrt{\tfrac{x^2+y^2}{2}}=\sqrt{\tfrac{x^2+f(y)^2}{2}}\ge\tfrac{f(x)+y}{2}=\tfrac{x+(y+D)}{2}\ge x$ , a contradiction; if $x<y\le x+D$ , the right-hand inequality at $(y,x)$ gives $y>\tfrac{y+x}{2}=\tfrac{f(y)+x}{2}\ge\sqrt{yf(x)}=\sqrt{y(x+D)}\ge y$ , again a contradiction. Thus $f(x)=x+D$ forces $f(t)=t+D$ for every $t\in[x-D,x+D]\cap\mathbb{R}_{>0}$ , and propagating this interval step by step covers all of $\mathbb{R}_{>0}$ . Therefore the solutions are precisely $f(x)=x+c$ for any constant $c\ge0$ .