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

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

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    $\begingroup$ link is broken. $\endgroup$ – Charlie Parker Feb 6 '17 at 5:55
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    $\begingroup$ That answer was deleted when its question was. I have copied over the relevant part. $\endgroup$ – robjohn Feb 13 '17 at 15:17
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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 $$\mathrm{sup}(f+h)\ge\mathrm{sup}\left(f+\mathrm{inf}(h)\right)=\mathrm{sup}(f)+\mathrm{inf}(h).$$

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