Supremum of the Difference of Two Functions Given two real-valued functions $f$ and $g$, is it true that $\sup(f-g) \geq \sup(f) - \sup(g)$
 A: 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
$$\mathrm{sup}(f+h)\ge\mathrm{sup}\left(f+\mathrm{inf}(h)\right)=\mathrm{sup}(f)+\mathrm{inf}(h).$$
A: Copied from this answer to a deleted question:

I think this question is asking how to show something like
$$
\sup_{x\in A}(f(x)+g(x))\le\sup_{x\in A}f(x)+\sup_{x\in A}g(x)\tag{1}
$$
This is an instance of the fact that the supremum over a set is no smaller than the supremum over a subset. The left hand side of $(1)$ is
$$
\sup_{\substack{x,y\in A\\x=y}}(f(x)+g(y))\tag{2}
$$
whereas the right hand side of $(1)$ is
$$
\sup_{x,y\in A}(f(x)+g(y))\tag{3}
$$
The set of $x$ and $y$ being considered in $(2)$ is a subset of the $x$ and $y$ being considered in $(3)$, so $(1)$ follows.

Using the result above, we have
$$
\sup((f-g)+g) \le \sup(f-g) + \sup(g)
$$
which becomes
$$
\sup(f-g)\ge\sup(f)-\sup(g)
$$
