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So suppose I have a continuous function on, say, $[a,b]$ with $x_1, x_2, ..., x_n \in [a, b]$. I want to prove that there exists a $c \in [a, b]$ such that $f(c) = [f(x_1) + f(x_2) + ... + f(x_n)] / n$. I thought about using mathematical induction at first for n. In this case, $n = 1$ is trivial since $f(x_1) = f(x_1)$. But then I need to begin with $n = 2$, as in finding a $c \in [a, b]$ such that $f(c) = [f(x_1) + f(x_2)]/n$. However, all the theorems I've tried so far don't seem to work. For instance, the Bolzano Intermediate Value Theorem doesn't work since there is a possibility that $f(b) \lt [f(x_1) + f(x_2)]/n$ or $[f(x_1) + f(x_2)]/n \lt f(a)$. I'm very stuck on this and I'm not sure where to begin. Any help would be much appreciated.

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  • $\begingroup$ $[f(x_1)+f(x_2)]/2$ should be between $f(x_1)$ and $f(x_2)$. So if $f$ is continuous, the intermediate value theorem should apply. The intermediate value theorem just says that this particular $c$ will be in the smaller interval between $x_1$ and $x_2$ (instead of between $a$ and $b$). $\endgroup$ – Nick Oct 3 '19 at 4:44
  • $\begingroup$ Are the square brackets to be interpreted simply as brackets? Seems to be so from your inductive step, just want to make sure. $\endgroup$ – Certainly not a dog Oct 3 '19 at 4:55
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The key points are that $f$ is continuous, $[a,b]$ is compact and connected, and continuous functions on compact connected sets attain extreme values, as well as all values between those extreme values on compact connected sets.

Somewhat more formally:

Since $f$ is continuous, and $[a,b]$ is compact, it follows that there exist $x_{m},x_{M} \in [a,b]$ such that for all $x \in [a,b]$, $f(x_{m}) \leq f(x) \leq f(x_{M})$. Now, let $\lbrace x_{1}, \ldots, x_{n} \rbrace \subset [a,b]$, then $nf(x_{m}) \leq \sum_{j=1}^{n}f(x_{j}) \leq nf(x_{M})$. Since $f$ is continuous, and $[a,b]$ is connected, it follows that $\frac{1}{n}\sum_{j=1}^{n}f(x_{j}) \in f([a,b])$ (this is nothing more than the intermediate value theorem!); that is, there exists $c \in [a,b]$ such that $f(c) = \frac{1}{n}\sum_{j=1}^{n}f(x_{j})$.

Hope this helps!

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  • $\begingroup$ I get most of it, but how did you use the intermediate value theorem to deduce what you stated? The intermediate value theorem has to involve $f(x_m)$ and $f(x_M)$ having opposite signs, doesn't it? $\endgroup$ – Tim Oct 3 '19 at 5:49
  • $\begingroup$ no, the IVT says that for two given $x$s, the function takes every value between their images for all $x$ between said $x$s provided it is continuous in that interval. $\endgroup$ – Luyw Oct 3 '19 at 8:03
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Hint:

  • Just consider $g(x) = f(x)-\frac{1}{n}\sum_{k=1}^nf(x_k)$ on $[a,b]$ and evaluate $g$ at the points where $f$ attains its minimum and maximum, resp.
  • Since $g$ is continuous on [a,b], this gives a point $c \in [a,b]$ with $g(c) = 0$. So, you found your $c$.
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