# How to find a $c \in \mathbb [a, b]$ such that $f(c) = [f(x_1) + f(x_2) + … + f(x_n)] / n$?

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.

• $[f(x_1)+f(x_2)]/2$ 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 $b$). – Nick Oct 3 '19 at 4:44
• Are the square brackets to be interpreted simply as brackets? Seems to be so from your inductive step, just want to make sure. – Certainly not a dog Oct 3 '19 at 4:55

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!

• 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? – Tim Oct 3 '19 at 5:49
• 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. – Luyw Oct 3 '19 at 8:03

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 $$c$$.