# Functions satisfying a given condition

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.

• 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). Aug 8 '10 at 10:41
• Is it true for all m, n? Or f depends on m, n? Aug 8 '10 at 11:26
• This problem came from a problem book with solutions (a book several other of the OP's questions have come from): books.google.com/… Nov 2 '10 at 1:13

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 $$f(x)=-1$$.

• I had a few mistakes in my solution, but now I think it's correct :P Aug 8 '10 at 14:05
• -1: Detailed answer to homeworklike question. Aug 9 '10 at 8:39
• 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. Aug 12 '10 at 15:57