I'm trying to find all functions $f : \mathbb{R} \to \mathbb{R}$ such that, for all $n > 1$ and all $x_1, x_2, \cdots, x_n \in \mathbb{R}$:
$$\frac{1}{n} \sum_{t = 1}^n f(x_t) = f \left ( \frac{1}{n} \sum_{t = 1}^n x_t \right )$$
My intuition is that this is only true if $f$ is a linear function. I started from the relation:
$$\sum_{t = 1}^n f(x_t) = f \left ( \sum_{t = 1}^n x_t \right )$$
That is, $f$ is additive. Then, by multiplying by $\frac{1}{n}$ each side, we obtain:
$$\frac{1}{n} \sum_{t = 1}^n f(x_t) = \frac{1}{n} f \left ( \sum_{t = 1}^n x_t \right )$$
And hence, any $f$ which has the property that $f(ax) = a f(x)$ (a linear function) will work. And since all linear functions are trivially additive, any linear function is a solution to my relation.
But all I have done is prove that linear functions are solutions, how should I go about showing that only linear equations are solutions? Is it valid to just "go backwards" in my argument, proving that if $f$ is a solution, then it must be linear? I feel it is not sufficient, since I only have implication and not equivalence. How do I proceed?
I think the $\frac{1}{n}$ term was added to confuse me, since without it, this would be straightforward.
$\endgroup$2 Answers
$\begingroup$First, boil this down to something more familiar. Set $h(x)=f(x)-f(0):$
$$h\left(\frac{1}{n}\sum_1^nx_t\right)=\frac{1}{n}\left(\sum_1^n h(x_t)\right)\quad(\star)$$
Since $h(0)=0$ let $x_2=x_3=\cdots =x_n=0$
$$h\left(\frac{x_1}{n}\right)=\frac{h(x_1)}{n}$$
Let $x_1\to x_1+x_2+\cdots +x_n$ and combine with $(\star):$
$$\sum_1^n h(x_t)=h\left(\sum_1^nx_t\right)$$
Setting $x_3=x_4=\cdots =x_n=0$
$$h(x_1)+h(x_2)=h(x_1+x_2)$$
This is Cauchy's equation, whose only continuous solution is $h(x)=ax,\;\;a\in\mathbb{R}$. Set $f(0)=b:$
$$f(x)=ax+b$$
If we do not impose the restriction of continuity on $f$, there are infinitely many other solutions to your equation. See here for more.
$\endgroup$ 1 $\begingroup$What about function f that:
$f(x+y)=f(x)+f(y)$ $\ \ \ \ \ \ x,y\in \mathbb{R}$
$f(a x)=af(x)$ $ \ \ \ \ \ a\in \mathbb{Q}, x\in \mathbb{R}$
This kind of function does not have to be continuous. (It can be so wild that it does not have to be even measurable. And I guess(not sure at all) if it is measurable than it is linear function(in normal sense) almost everywhere.
Acording to wikipedia
All locally integrable continuous function that satisfy mean value property(in 1D it means $\frac{1}{2}(f(x)+f(y)) = f(\frac{x+y}{2})$) are infinitely differentiable and harmonics.
So I guess this answers your question.
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