# Features pleasing an offered problem

Allow $m,n \geq 2$ be integers. Locate all features $f:[0,\infty) \to \mathbb{R}$ continual at at the very least one factor in $[0,\infty)$ and also such that $$f\biggl(\frac{1}{n} \sum\limits_{i=1}^{n} x_{i}^{m} \biggr)=\frac{1}{n} \sum\limits_{i=1}^{n}(f(x_i))^{m} \quad \text{for} \ x_{i} \geq 0, i=1,2,...,n$$

I am not obtaining any kind of suggestion to continue.

Let $x_1=x_2=\ldots=x_n=x$, after that $f(x^m)=f(x)^m$, and also $f(0)=f(0)^m$, so $f(0)$ is either 0,1 or - 1.

Currently allow $y_i=x_i^m$, after that we have $f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$ for any kind of $y_i\in \mathbb{R}$.

If $f(0)=0$, after that $y_1=y$, $y_2=\ldots=y_n=0$ indicates that $f(\frac{1}{n} y)=\frac{1}{n}f(y)$. This indicates $\frac{1}{n}f(\sum_{i=1}^n y_i)=f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$.

Allowing $y_3=\ldots=y_n=0$, we get that $f(y_1+y_2)=f(y_1)+f(y_2)$, which is Cauchy's Functional Equation. Given that $f$ is continual at one factor, it's straight, so $f(x)=ax$.

We get $a=a^m$, so the remedies in this instance are $f(x)=0$, $f(x)=x$ and also $f(x)=-x$ (if $m$ is weird).

If $f(0)\neq 0$, WLOG allowed $f(0)=1$ (otherwise change $f$ by $-f$ if $m$ is weird).

Analogously $f(\frac{1}{n}y)=\frac{f(y)+n-1}{n}$, and also $\frac{f(\sum_{i=1}^n y_i)+n-1}{n}=f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$, to make sure that $f(\sum_{i=1}^n y_i)+n-1=\sum_{i=1}^n f(y_i)\Rightarrow g(\sum_{i=1}^n y_i)=\sum_{i=1}^n g(y_i)$ where $g(y)=f(y)-1$.

From this we have that $g$ pleases Cauchy's useful formula, to make sure that $f(y)=ay+1$. Yet $ay^m+1=f(y^m)=f(y)^m=(ay+1)^m$, which indicates $a=0$ by contrasting coefficients. So in this instance the remedies are $f(x)=1$ and also $f(x)=-1$ (if $m$ is weird).

We end that the remedies are $f(x)=0$, $f(x)=x$, $f(x)=1$ and also if $m$ is weird $f(x)=-x$ and also $f(x)=-1$.

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