# Supremum of the Difference of Two Functions

Given two real-valued functions $f$ and $g$, is it true that $\sup(f-g) \geq \sup(f) - \sup(g)$

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 $$\sup(f-g)\ge\sup(f)-\sup(g)$$

• link is broken. – Charlie Parker Feb 6 '17 at 5:55
• That answer was deleted when its question was. I have copied over the relevant part. – robjohn Feb 13 '17 at 15:17

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 $$\mathrm{sup}(f+h)\ge\mathrm{sup}\left(f+\mathrm{inf}(h)\right)=\mathrm{sup}(f)+\mathrm{inf}(h).$$