I have been recently reading about clustering validation and came upon the silhouette coefficient, represented by the following formula.
Everywhere I read about this coefficient, it says that it is always between $[-1,1]$, with a value close to $1$ meaning that $x_i$ is close to points of its own cluster, a value close to $0$ meaning that $x_i$ is close to the boundary and a value close to $-1$ meaning that $x_i$ is close to another cluster.
However, what they don't explain is why the coefficient is always between this intervasl. Why does this happen? How can it be mathematically demonstrated?
This reduces to the question "Given nonnegative numbers $u$ and $v$, at least one of them nonzero, why is $$ s = \frac{u - v}{\max(u, v)} $$ always between $-1$ and $1$?
There are a couple of ways to see this. One is to just plot the function using a tool like Desmos. That shows it's true, but not WHY it's true. It may be useful for you to do it anyhow.
Let's look at three possible cases: $u = v$, $u > v$, $u < v$.
In the first case, the numerator is zero, and the denominator is nonzero, so the quotient has the form $\frac{0}{A}$, with $A \ne 0$, and the value is zero.
Case 2: let's assume that $u > v$, or more precisely, that $u > v \ge 0$. The numerator $u- v$ is positive. The denominator is the larger of $u$ and $v$, which is $u$, so we have $$ s = \frac{u-v}{u} = \frac{u}{u} - \frac{v}{u} \le \frac{u}{u} = 1. $$ So our result is never greater than $1$. (I've used the fact that $u$ and $v$ are nonnegative, and $u > 0$, to conclude that $\frac{v}{u}$ is nonnegative, so subtracting it from $\frac{u}{u}$ results in a number smaller than $\frac{u}{u}$).
What about showing that $s \ge -1$? It's similar. We have $$ s = \frac{u}{u} - \frac{v}{u} = 1 - \frac{v}{u}. $$ How small can this number be? The larger the second term, the smaller the number is. Let's look at that second term. We know (because we're in case 2) that $u > 0$ and $v < u$. Dividing through by $u$, we get \begin{align} \frac{v}{u} &< 1\\ -\frac{v}{u} &> -1 & \text{negating an inequality reverses it!}\\ \frac{u}{u}-\frac{v}{u} &>\frac{u}{u} -1 & \text{Add the same thing to both sides}\\ \frac{u}{u}-\frac{v}{u} &>1 -1 & \text{Simplify one of the two fractions.}\\ \frac{u}{u}-\frac{v}{u} &>0. \end{align} Hence in the case where $u > v \ge 0$, we have that $0 < s \le 1$, with $s = 1$ exactly in the case where $v = 0$.
The third case, $v > u \ge 0$ is similar, except that the denominator throughout is $v$ instead of $u$; I'll let you work through that one to discover that in that case, $s$ is between $-1$ and $0$, and is $-1$ only when $u = 0$.