Roots of equation

Let $n \geq 1$ and $a_1, a_2 , \dots, a_n \in [0, 1]$ . Prove that

$\left | x-a_1 \right | + \left | x - a_2 \right | + \cdots + \left | x - a_n \right | = \frac{n}{2}$

has solution in $[0, 1]$ .

Solution

Consider the function $f(x) = \sum \limits_{k=1}^{n} \left | x - a_k \right |$ which is clearly continuous. It holds that

(1) $\begin{equation*} n = \sum_{k=1}^{n} |1-a_k+a_k| \leq \sum_{k=1}^{n} |1-a_k| + \sum_{k=1}^{n} |a_k|= f(1)+f(0) \end{equation*}$

If $f(1)=f(0)= \frac {n}{2}$ then $x=0$ and $x=1$ are roots. If one of $f(1)$ , $f(0)$ is strictly less than $\frac {n}{2}$ then it follows from $(1)$ that the other is strictly greater than $\frac{n}{2}$ and vice versa. It follows from Bolzano , that somewhere in between there exists an $x_0$ such that $f (x_0) =\frac {n}{2}$ . The result follows.