Supremum of Absolute Value of Difference equals Supremum of Difference
Theorem
Let $S$ be a non-empty real set.
Let $\displaystyle \sup_{x, y \mathop \in S} \left({x - y}\right)$ exist.
Then $\displaystyle \sup_{x, y \mathop \in S} \left\lvert{x - y}\right\rvert$ exists and:
- $\displaystyle \sup_{x, y \mathop \in S} \left\lvert{x - y}\right\rvert = \sup_{x, y \mathop \in S} \left({x - y}\right)$.
Proof
Consider the set $\left\{{x - y: x, y \in S, x - y \le 0}\right\}$.
There is a number $x'$ in $S$ as $S$ is non-empty.
Therefore, $0 \in \left\{{x - y: x, y \in S, x - y \le 0}\right\}$ 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 $\left\{{x - y: x, y \in S, x - y \le 0}\right\}$ by definition.
Accordingly:
- $\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \le 0} \left({x - y}\right) = 0$
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:
- $\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \left({x - y}\right) \ge 0$
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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| \(\displaystyle \sup_{x, y \mathop \in S} \left({x - y}\right)\) | \(=\) | \(\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \ge 0 \text { or } x − y \mathop \le 0} \left({x - y}\right)\) | $\quad$ as ($x - y \ge 0$ or $x - y \le 0$) is true | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \max \left({\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \left({x - y}\right), \sup_{x, y \mathop \in S, x − y \mathop \le 0} \left({x - y}\right) }\right)\) | $\quad$ by Maximum of Supremums of Subsets Equals Supremum of Set | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \max \left({\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \left({x - y}\right), 0}\right)\) | $\quad$ as $\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \le 0} \left({x - y}\right) = 0$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \left({x - y}\right)\) | $\quad$ as $\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \left({x - y}\right) \ge 0$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \left\lvert{x - y}\right\rvert\) | $\quad$ as $\left\lvert{x − y}\right\rvert = x − y$ since $x − y \ge 0$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \max \left({\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \left\lvert{x - y}\right\rvert, \sup_{x, y \mathop \in S, x − y \mathop \ge 0} \left\lvert{x - y}\right\rvert }\right)\) | $\quad$ as the two arguments of max are equal | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \max \left({\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \left\lvert{x - y}\right\rvert, \sup_{y, x \mathop \in S, y − x \mathop \ge 0} \left\lvert{y - x}\right\rvert }\right)\) | $\quad$ by renaming variables $x \leftrightarrow y$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \max \left({\sup_{x, y \mathop \in S, x − y \mathop \ge 0} \left\lvert{x - y}\right\rvert, \sup_{x, y \mathop \in S, x − y \mathop \le 0} \left\lvert{x - y}\right\rvert }\right)\) | $\quad$ | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sup_{x, y \mathop \in S, x − y \mathop \ge 0 \text { or } x − y \mathop \le 0} \left\lvert{x - y}\right\rvert\) | $\quad$ by Supremum of Set Equals Maximum of Supremums of Subsets | $\quad$ | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sup_{x, y \mathop \in S} \left\lvert{x - y}\right\rvert\) | $\quad$ as ($x - y \ge 0$ or $x - y \le 0$) is true | $\quad$ |
$\blacksquare$
