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 $\ds \sup_{x \mathop \in S} \set {\map f x}$ and $\ds \inf_{x \mathop \in S} \set {\map f x}$ exist.

Then $\ds \sup_{x, y \mathop \in S} \set {\size {\map f x - \map f y} }$ exists and:

$\ds \sup_{x, y \mathop \in S} \set {\size {\map f x - \map f y} } = \sup_{x \mathop \in S} \set {\map f x} - \inf_{x \mathop \in S} \set {\map f x}$

Proof

\(\ds \sup_{x \mathop \in S} \set {\map f x} - \inf_{x \mathop \in S} \set {\map f x}\)

\(=\)

\(\ds \sup_{x \mathop \in S} \set {\map f x} + \sup_{x \mathop \in S} \set {-\map f x}\)

Negative of Infimum is Supremum of Negatives

\(\ds \)

\(=\)

\(\ds \sup_{x, y \mathop \in S} \set {\map f x + \paren {-\map f y} }\)

Supremum of Sum equals Sum of Suprema

\(\ds \)

\(=\)

\(\ds \sup_{x, y \mathop \in S} \set {\map f x - \map f y}\)

\(\ds \)

\(=\)

\(\ds \sup_{x, y \mathop \in S} \set {\size {\map f x - \map f y} }\)

Supremum of Absolute Value of Difference equals Supremum of Difference

$\blacksquare$