TITLE: Supremum of the Difference of Two Functions
QUESTION [16 upvotes]: Given two real-valued functions $f$ and $g$, is it true that $\sup(f-g) \geq \sup(f) - \sup(g)$

REPLY [3 votes]: I'm just going to add another answer which may provide some different intuition. We want to show
$$\mathrm{sup}(f-g)\ge\mathrm{sup}(f)-\mathrm{sup}(g).$$
To make everything in terms of addition we bring the minus sign inside the last supremum:
$$-\mathrm{sup}(g)=+\mathrm{inf}(-g).$$
If we now define $h(x)=-g(x)$, the problem them becomes to show
$$\mathrm{sup}(f+h)\ge\mathrm{sup}(f)+\mathrm{inf}(h).$$
This however is clear, since for all $x$ we have $h(x)\ge\mathrm{inf}(h)$, so
$$\mathrm{sup}(f+h)\ge\mathrm{sup}\left(f+\mathrm{inf}(h)\right)=\mathrm{sup}(f)+\mathrm{inf}(h).$$