Supremum of Function is less than Supremum of Greater Function
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Theorem
Let $f$ and $g$ be real functions.
Let $S$ be a subset of $\Dom f \cap \Dom g$.
Let $\map f x \le \map g x$ for every $x \in S$.
Let $\ds \sup_{x \mathop \in S} \map g x$ exist.
Then $\ds \sup_{x \mathop \in S} \map f x$ exists and:
- $\ds \sup_{x \mathop \in S} \map f x \le \sup_{x \mathop \in S} \map g x$.
Proof
We have:
\(\ds \sup g\) | \(=\) | \(\ds \map \sup {f + \paren {g - f} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sup f + \sup \left({g - f}\right)\) | Supremum of Sum equals Sum of Suprema |
Supremum of Sum equals Sum of Suprema also gives that $\sup f$ and $\sup \paren {g - f}$ exist.
We have:
\(\ds \forall x \in S: \, \) | \(\ds \map g x\) | \(\ge\) | \(\ds \map f x\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall x \in S: \, \) | \(\ds \map g x - \map f x\) | \(\ge\) | \(\ds 0\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall x \in S: \, \) | \(\ds \map \sup {g - f}\) | \(\ge\) | \(\ds \map g x - \map f x \ge 0\) | as $\map \sup {g - f}$ is an upper bound for $\set {\map g x - \map f x: x \in S}$ | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \sup {g - f}\) | \(\ge\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sup f + \map \sup {g - f}\) | \(\ge\) | \(\ds \sup f\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sup g\) | \(\ge\) | \(\ds \sup f\) | as $\sup g = \sup f + \map \sup {g - f}$ |
$\blacksquare$