**Definition (Median)**

Let $D=\{x_1,\dots, x_n\}$ be a set of values with $x_1\leq \cdots\leq x_n$. The median of the set is defined to be the following number:

- If $n$ is odd and $n=2k+1$ then

- If $n$ is even and $n=2k$ then

**Theorem.**

The minimum of the sum
$$g(c)=\sum\limits_{i=1}^n|x_i-c|$$
is attained at the median.

**Proof.**

Let us take the case of only two points $D=\{x_1, x_2\}$. Using the triangle inequality

$$|x_1-c|+|x_2-c|\geq |x_1-x_2|=x_2-x_1.$$Moreover, for any $c\in [x_1,x_2]$,

$$|x_1-c|+|x_2-c|=(c-x_1)+(x_2-c)=x_2-x_1.$$So any point in $[x_1,x_2]$ minimizes $|x_1-c|+|x_2-c|$.

**When n is even**

Then rewrite

$$\sum\limits_{i=1}^{2k}|x_i-c|= \color{blue}{(|x_1-c|+|x_{2k}-c|)}+\color{blue}{(|x_2-c|+|x_{2k-1}-c|)}+\cdots + \color{blue}{(|x_{k}-c|+|x_{k+1}-c|)}$$

According to the discussion above the first term is minimal when $c\in [x_1, x_{2k}]$, the second term is minimal when $c\in [x_2, x_{2k-1}]$ and so on. Thus, when $c\in [x_k,x_{k+1}]$ all the terms have their smallest values. In particular, the median is in that interval: $\frac{x_{k}+x_{k+1}}{2}\in [x_k,x_{k+1}]$ so we have that the median the sum takes its minimal value.

**When n is odd**

Then rewrite

$$\sum\limits_{i=1}^{2k+1}|x_i-c|= \color{blue}{(|x_1-c|+|x_{2k+1}-c|)}+\color{blue}{(|x_2-c|+|x_{2k}-c|)}+\cdots + \color{blue}{(|x_{k}-c|+|x_{k+2}-c|)}+\color{green}{|x_{k+1}-c|}.$$

Again, according to the discussion above the first term is minimal when $c\in [x_1, x_{2k+1}]$, the second term is minimal when $c\in [x_2, x_{2k}]$ and so on. Fiinally, the last term $|x_{k+1}-c|$ is minimal when $c=x_{k+1}$. Thus, when $c=x_{k+1}$ all the terms have their smallest values since it belongs to all the above intervals. And we concude thay the minimal value of their sum is at the median. $\blacksquare$