Suppose that the function $f:[a,b]\rightarrow \mathbb{R}$ is continuous. For a natural number $k$, let $x_1,\cdots,x_k$ be points in $[a,b]$. Prove there is a point $z\in[a,b]$ at which $f(z)=\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}$.
This can be proved by showing
$\hspace{130pt}f(a)<\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}<f(b)$
and using the intermediate value theorem to say there exists a point $z\in [a,b]$ such that the statement is true.
I'm not sure how to show this though. Any suggestions?
Best Answer
Pick $x_u, x_v$ among $x_1, \ldots, x_n$ such that
$$\begin{align} f(x_u) = & \min( f(x_1), f(x_2), \ldots, f(x_n) )\\ f(x_v) = & \max( f(x_1), f(x_2), \ldots, f(x_n) ) \end{align}$$ and let $x_l = \min(x_u, x_v)$ and $x_r = \max(x_u, x_v)$. We have
$$\begin{align} & f(x_u) \le f(x_1), f(x_2), \ldots, f(x_n) \le f(x_v)\\ \implies & f(x_u) \le \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n} \le f(x_v) \end{align}$$ Since $f(x)$ is continuous over $[a,b]$ and $x_u, x_v \in [a,b]$, $f(x)$ is continuous over the closed interval $[x_l, x_r]$. By Intermediate value theorem, there is a number $z \in [ x_l, x_r] \subseteq [a,b]$ such that $$f(z) = \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}$$