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 $\sup g = \sup f + \map \sup {g - f}$

$\blacksquare$