# Supremum of Absolute Value of Difference equals Difference between Supremum and Infimum

## Theorem

Let $f$ be a real function.

Let $S$ be a subset of the domain of $f$.

Let $\displaystyle \sup_{x \mathop \in S} \left\{{f \left({x}\right)}\right\}$ and $\displaystyle \inf_{x \mathop \in S} \left\{{f \left({x}\right)}\right\}$ exist.

Then $\displaystyle \sup_{x, y \mathop \in S} \left\{{\left\vert{f \left({x}\right) - f \left({y}\right)}\right\vert}\right\}$ exists and:

$\displaystyle \sup_{x, y \mathop \in S} \left\{{\left\vert{f \left({x}\right) - f \left({y}\right)}\right\vert}\right\} = \sup_{x \mathop \in S} \left\{{f \left({x}\right)}\right\} - \inf_{x \mathop \in S} \left\{{f \left({x}\right)}\right\}$

## Proof

 $\displaystyle \sup_{x \mathop \in S} \left\{ {f \left({x}\right)}\right\} - \inf_{x \mathop \in S} \left\{ {f \left({x}\right)}\right\}$ $=$ $\displaystyle \sup_{x \mathop \in S} \left\{ {f \left({x}\right)}\right\} + \sup_{x \mathop \in S} \left\{ {- f \left({x}\right)}\right\}$ Negative of Infimum is Supremum of Negatives $\displaystyle$ $=$ $\displaystyle \sup_{x, y \mathop \in S} \left\{ {f \left({x}\right) + \left({- f \left({y}\right)}\right)}\right\}$ Supremum of Sum equals Sum of Suprema $\displaystyle$ $=$ $\displaystyle \sup_{x, y \mathop \in S} \left\{ {f \left({x}\right) - f \left({y}\right)}\right\}$ $\displaystyle$ $=$ $\displaystyle \sup_{x, y \mathop \in S} \left\{ {\left\vert{f \left({x}\right) - f \left({y}\right)}\right\vert}\right\}$ Supremum of Absolute Value of Difference equals Supremum of Difference

$\blacksquare$