Average Sum of Distances

 Given $n$ numbers $0 \le x_i \le 1$, define the function $f:[0,1] \to \mathbb{R}$ as

$$f(x) = \frac{1}{n} \sum_{i=1}^{n} |x - x_i|$$

A) Show that there is some $a \in [0,1]$ such that $$f(a) = \frac{1}{2}$$ 

B) Show that there is some $b \in [0,1]$ such that 

$$f(b) = \frac{1}{2} - \frac{1}{n} \sum_{i=1}^{n} (x_i- x_i^2)$$

Scroll down for a solution.

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Solution:

A) Notice that $f(0) + f(1) = 1$.  If $f(0) \leq \frac{1}{2}$ then $f(1) \geq \frac{1}{2}$ and by the intermediate value theorem, there is some $a \in [0,1]$ such that $f(a) = \frac{1}{2}$. Other cases are similar.

B) By the mean value theorem, there is some $b \in [0,1]$ such that

$$f(b) = \int_{0}^{1} f(x) dx$$

Turns out that

$$\int_{0}^{1} f(x) dx = \frac{1}{2} - \frac{1}{n} \sum_{i=1}^{n} (x_i- x_i^2)$$

It is the sum of right angled triangle areas of legs $x_i, x_i$ and $(1-x_i), (1-x_i)$. 

$\textbf{Additional Remark:}$ Result in part B can easily be used to show that if $f(x) \ge \frac{1}{2} \forall x \in [0,1]$, then $n$ must be even, with half the $x_i = 0$ and the rest $ = 1$, and $f$ a constant $= \frac{1}{2}$.