Assume $x_1$,...,$x_n$ are positive real numbers - Is there a measurable function such that $f(x_1\cdot ... \cdot x_n)=max(f(x_1),...,f(x_n))$?

Question

Assume $x_1$,...,$x_n$ are positive real numbers - Is there a measurable function $f:\mathbb{R_+^n}\to \mathbb{R}_+$ such that $f(x_1\cdot ... \cdot x_n)=max(f(x_1),...,f(x_n))$?

Here with $x_1 \cdot x_j$ I mean just the product and max the maximum over all numbers

Answer 1

Let $x>0$. Then $$ f(x)=f(x\cdot 1\cdot\ldots \cdot 1)=\max\{f(x),f(1),\ldots,f(1)\}\ge f(1)$$ and $$ f(1)=f(x\cdot\tfrac1x\cdot 1\cdot\ldots\cdot 1)=\max\{f(x),f(\tfrac1x),f(1),\ldots f(1)\}\ge f(x).$$ Hence $f(x)=f(1)$ for all $x$.

Answer 2

Let $\sup_{\{x>0\}} f(x) = f(x*)$.

$f(x) = f(x*.x/x*)= \max(f(x*),f(x/x*)) = f(x*)$.

Hence $f$ is a constant function (assuming $x*$ exists).

if $x*$ does not exist then let $f(x_n) \rightarrow \sup_{\{x>0\}} f(x)$ with $x_n >0$ then

$f(x) = f(x_n.x/x_n)= \max(f(x_n),f(x/x_n))$.

Hence $f(x) = \lim_{n \rightarrow \infty} \max(f(x_n),f(x/x_n)) = \sup_{\{x>0\}} f(x)$.

Hence $f$ is a constant function.