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Let $m,n \geq 2$ be integers. Find all functions $f:[0,\infty) \to \mathbb{R}$ continuous at at least one point in $[0,\infty)$ and 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 getting any idea to proceed.

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Can you see any functions $f$ which do work? This is a pretty strong conditition on $f$: one might expect only the "obvious" $f$ would work. Also one might "warm up" with the case $m=1$ (even though your conditions exclude it). –  Robin Chapman Aug 8 '10 at 10:41
    
Is it true for all m, n? Or f depends on m, n? –  KennyTM Aug 8 '10 at 11:26
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This problem came from a problem book with solutions (a book several other of the OP's questions have come from): books.google.com/… –  Jonas Meyer Nov 2 '10 at 1:13

1 Answer 1

up vote 3 down vote accepted

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

Now let $y_i=x_i^m$, then we have $f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$ for any $y_i\in \mathbb{R}$.

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

Letting $y_3=\ldots=y_n=0$, we get that $f(y_1+y_2)=f(y_1)+f(y_2)$, which is Cauchy's Functional Equation. Since $f$ is continuous at one point, it's linear, so $f(x)=ax$.

We get $a=a^m$, so the solutions in this case are $f(x)=0$, $f(x)=x$ and $f(x)=-x$ (if $m$ is odd).

If $f(0)\neq 0$, WLOG let $f(0)=1$ (or else replace $f$ by $-f$ if $m$ is odd).

Analogously $f(\frac{1}{n}y)=\frac{f(y)+n-1}{n}$, and $\frac{f(\sum_{i=1}^n y_i)+n-1}{n}=f(\frac{1}{n}\sum_{i=1}^n y_i)=\frac{1}{n}\sum_{i=1}^n f(y_i)$, so 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 $g$ satisfies Cauchy's functional equation, so that $f(y)=ay+1$. But $ay^m+1=f(y^m)=f(y)^m=(ay+1)^m$, which implies $a=0$ by comparing coefficients. So in this case the solutions are $f(x)=1$ and $f(x)=-1$ (if $m$ is odd).

We conclude that the solutions are $f(x)=0$, $f(x)=x$, $f(x)=1$ and if $m$ is odd $f(x)=-x$ and $f(x)=-1$.

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I had a few mistakes in my solution, but now I think it's correct :P –  Jorge Miranda Aug 8 '10 at 14:05
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-1: Detailed answer to homeworklike question. –  Charles Stewart Aug 9 '10 at 8:39
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I'm sorry, I didn't notice the OP had posted a lot of this unmotivated contest-like problems. To me it didn't seem like an homework question, so I posted a solution. –  Jorge Miranda Aug 12 '10 at 15:57

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