# Supremum of Function is less than Supremum of Greater Function

## 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 $\sup g = \sup f + \sup \left({g - f}\right)$

$\blacksquare$