The question is: Suppose $f(x)$ is continuous in $[a,b]$. $x_1,x_2,...,x_n$ are points in $[a,b]$. Show there is a point c in $[a,b]$ and $f(c) = \frac {f(x_1)+f(x_2)+...+f(x_n)}{n}$.
Suppose n tends to infinity (since the case when n is an integer is trivial), then I consider 3 cases:
1) $x_n\to\ c$ ;
2) $x_n$ does not converge but is made of a finite number of convergent subsequences;
3) $x_n$ does not converge and is made of infinitely many convergent subsequences.
The first case I can easily prove with f being continuous and the Cesaro Mean.
As for the second one, I argue that those finite number (say k) of convergent subsequences converge to {$c_1, c_2, ..., c_k$} and say that the Cesaro Mean is equal to $\frac{f(c_1)+f(c_2)+...+f(c_k)}{k}$ (is it okay?). Then I select the maximizer and minimizer in {$c_1, c_2, ..., c_k$} and by Intermediate Theorem, there exists c in between such that f(c) = $\frac{f(c_1)+f(c_2)+...+f(c_k)}{k}$.
In the third case, I argue that there exists a supremum $sup x$ and infinmum $inf x$ for the set of converging points. And the Cesaro mean is $\frac{f(supx)+f(infx)}{2}$ (I am not sure with this). If $f(supx) = f(infx)$ , the proof is complete. Otherwise, because $supx > infx$, by Intermediate Theorem, there exists c in $(infx, supx)$ s.t. $f(c) = \frac{f(supx)+f(infx)}{2}$ .
I think one of the problems with my attempt is that I am not certain about the Cesaro Mean in case 2 and 3. Also, I am not sure whether I can actually do like this. Anyway, can I prove like this? If not, what is the way instead?
Thank you!
You can not "suppose" that $n$ tends to $+\infty$, since $n$ is a given positive integer. A proof goes as follows: let $x_0,~y_0\in [a,b]$ be points where the continuous function $f$ attains it's minimum and maximum values, say, $m$ and $M$, respectively. Since $$m=f(x_0)\leq \frac{1}{n}\Big(f(x_1)+\ldots+f(x_n)\Big)\leq M=f(y_0),$$ by the intermediate value theorem for $f$ on the interval $I$ defined by $x_0,~y_0$, we get $c\in I\subset [a,b]$, such that $\displaystyle f(c)=\frac{1}{n}\Big(f(x_1)+\ldots+f(x_n)\Big)$, as desired.