Supremum of Absolute Value of Difference equals Supremum of Difference
Theorem
Let $S$ be a non-empty real set.
Let $\ds \sup_{x, y \mathop \in S} \paren {x - y}$ exist.
Then $\ds \sup_{x, y \mathop \in S} \size {x - y}$ exists and:
- $\ds \sup_{x, y \mathop \in S} \size {x - y} = \sup_{x, y \mathop \in S} \paren {x - y}$
Proof
Consider the set $\set {x - y: x, y \in S, x - y \le 0}$.
There is a number $x'$ in $S$ as $S$ is non-empty.
Therefore, $0 \in \set {x - y: x, y \in S, x - y \le 0}$ as $x = y = x'$ implies that $x - y = 0$, $x, y \in S$, and $x - y \le 0$.
Also, $0$ is an upper bound for $\set {x - y: x, y \in S, x - y \le 0}$ by definition.
Accordingly:
- $\ds \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} = 0$
Consider the set $\left\{{x - y: x, y \in S, x - y \ge 0}\right\}$.
There is a number $x'$ in $S$ as $S$ is non-empty.
Therefore, $0 \in \left\{{x - y: x, y \in S, x - y \ge 0}\right\}$ as $x = y = x'$ implies that $x - y = 0$, $x, y \in S$, $x - y \ge 0$.
Accordingly:
- $\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y} \ge 0$
I can't immediately think of how it would be done, but it would be good if we could devise a neater and more compact notation that what is used here. All the complicated mathematics is being done in the underscript, which makes it not easy to follow. (Improved Dec. 2016.)
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| \(\ds \sup_{x, y \mathop \in S} \paren {x - y}\) | \(=\) | \(\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0 \text { or } x − y \mathop \le 0} \paren {x - y}\) | as ($x - y \ge 0$ or $x - y \le 0$) is true | |||||||||||
| \(\ds \) | \(=\) | \(\ds \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y}, \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} }\) | by Supremum of Set Equals Maximum of Suprema of Subsets | |||||||||||
| \(\ds \) | \(=\) | \(\ds \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y}, 0}\) | as $\ds \sup_{x, y \mathop \in S, x − y \mathop \le 0} \paren {x - y} = 0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y}\) | as $\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \paren {x - y} \ge 0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}\) | as $\size {x − y} = x − y$ since $x − y \ge 0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}, \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y} }\) | as the two arguments of max are equal | |||||||||||
| \(\ds \) | \(=\) | \(\ds \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}, \sup_{y, x \mathop \in S, y − x \mathop \ge 0} \size {y - x} }\) | by renaming variables $x \leftrightarrow y$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \max \set {\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \size {x - y}, \sup_{x, y \mathop \in S, x − y \mathop \le 0} \size {x - y} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{x, y \mathop \in S, x − y \mathop \ge 0 \text { or } x − y \mathop \le 0} \size {x - y}\) | by Supremum of Set Equals Maximum of Suprema of Subsets | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sup_{x, y \mathop \in S} \size {x - y}\) | as ($x - y \ge 0$ or $x - y \le 0$) is true |
$\blacksquare$