Supremum of Absolute Value of Difference equals Difference between Supremum and Infimum
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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 |
$\blacksquare$