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 $\operatorname{Dom} \left({f}\right) \cap \operatorname{Dom} \left({g}\right)$.

Let $f \left({x}\right) \le g \left({x}\right)$ for every $x \in S$.

Let $\displaystyle \sup_{x \mathop \in S} g \left({x}\right)$ exist.


Then $\displaystyle \sup_{x \mathop \in S} f \left({x}\right)$ exists and:

$\displaystyle \sup_{x \mathop \in S} f \left({x}\right) \le \sup_{x \mathop \in S} g \left({x}\right)$.


Proof

We have:

\(\displaystyle \sup g\) \(=\) \(\displaystyle \sup \left({f + \left({g - f}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \sup f + \sup \left({g - f}\right)\) Supremum of Sum equals Sum of Suprema

Supremum of Sum equals Sum of Suprema also gives that $\sup f$ and $\sup \left({g - f}\right)$ exist.

We have:

\(\, \displaystyle \forall x \in S: \, \) \(\displaystyle g \left({x}\right)\) \(\ge\) \(\displaystyle f \left({x}\right)\)
\(\displaystyle \iff \ \ \) \(\, \displaystyle \forall x \in S: \, \) \(\displaystyle g \left({x}\right) - f \left({x}\right)\) \(\ge\) \(\displaystyle 0\)
\(\displaystyle \implies \ \ \) \(\, \displaystyle \forall x \in S: \, \) \(\displaystyle \sup \left({g - f}\right)\) \(\ge\) \(\displaystyle g \left({x}\right) - f \left({x}\right) \ge 0\) as $\sup \left({g - f}\right)$ is an upper bound for $\left\{ {g \left({x}\right) - f \left({x}\right): x \in S}\right\}$
\(\displaystyle \implies \ \ \) \(\displaystyle \sup \left({g - f}\right)\) \(\ge\) \(\displaystyle 0\)
\(\displaystyle \iff \ \ \) \(\displaystyle \sup f + \sup \left({g - f}\right)\) \(\ge\) \(\displaystyle \sup f\)
\(\displaystyle \iff \ \ \) \(\displaystyle \sup g\) \(\ge\) \(\displaystyle \sup f\) as $\sup g = \sup f + \sup \left({g - f}\right)$

$\blacksquare$