To show that $\sup(f+g)\leq \sup(f) +\sup(g)$, suppose that $\sup(f)=A$ and $\sup(g)=B$, where $A$ and $B$ are the highest point of function $f$ and $g$ respectively.

2019年2月8日 · Denote by $ \sup f $ the supremum of the set of images of a function. I proved that if the values are positive, then $$ \sup \left( f \right)\sup \left( g \right) > \sup \left( {fg} \right). $$ ...

$\begingroup$ Mostly it boils down to writing down the definition of $\sup$ and pointing out that it's satisfied. $\endgroup$ – G Tony Jacobs Commented Mar 9, 2014 at 16:07

2018年2月11日 · "sup" stands for "supremum" or "least upper bound" of the values in a set. When the set is finite, these mean the same thing, and which one you use is a matter of state. When the set is infinite, it may not have a maximum.

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2018年10月22日 · $\begingroup$ Yes, if $\forall x, f(x) \leq a$, then $\sup f(x) \leq a$, by definition of a lowest bound (recall that $\sup$ is the greatest lower bound) $\endgroup$ – Gareth Ma Commented Dec 1, 2023 at 1:04

2023年8月30日 · In the comments you ask about the inequality $$|\sup(f+g)|+|\inf(f+g)|\leq |\sup f|+|\sup g| + |\inf f| + |\inf g|.$$ This inequality also need not hold; it's a bit easier to find counterexamples, because all we need is a pair of nonnegative functions with $\sup(f+g)=\sup(f)+\sup(g)$ and $\inf(f+g)\gt \inf(f)+\inf(g)$, and those are plentiful

2017年1月9日 · If one or both of them equal $+\infty$ then the right hand side of $$\sup(f+g)\le \sup f + \sup g$$ equals $\infty$ and so the inequality is trivially true. $\endgroup$ – Giuseppe Negro Commented Jan 9, 2017 at 11:43

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