# the supremum of a set

Denote by $\sup f$ the supremum of the set of images of a function. I proved that if the values are positive then $$\sup \left( f \right)\sup \left( g \right) > \sup \left( {fg} \right)$$ When does equality hold?

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Could you fix the problem? Equality very seldom holds, since you need, intuitively, for $f$ and $g$ to approach their suprema "in the same place". That said, you cannot possibly have proven that the strict inequality always holds, since equality holds if $f=g$. –  Arturo Magidin Aug 15 '11 at 20:02
@Yuval: I don't know if that is an accurate rendering of whathever it was that Daniel was in the middle of writing (and which you simply chopped off). Perhaps you might have waited a bit to see what the OP was trying to say in the sentence that got truncated? –  Arturo Magidin Aug 15 '11 at 20:16
In that case, my apologies, I just wanted to correct the first supremum. –  Yuval Filmus Aug 15 '11 at 20:28
I retagged as there is not much set theory here. Since the functions were not specified, I went with real-analysis. –  Asaf Karagila Aug 15 '11 at 20:57
and how can i prove the equality when f=g, i did it but im not sure –  Daniel Aug 17 '11 at 6:16

Indeed $\sup(fg)\ge\sup(f)\sup(g)$ and in general the inequality is strict. A situation where $\sup(fg)=\sup(f)\sup(g)$ is when $f$ and $g$ are both monotone functions of a common variable. That is, when there exists a function $h$ and some nondecreasing functions $\varphi$ and $\psi$ such that $f=\varphi\circ h$ and $g=\psi\circ h$ (the proof is direct since $f$ and $g$ reach their respective supremum where $h$ does).
Suppose the supremum was actually attained for all of $f,g,fg$ (for example, maybe the functions are defined on a closed interval and continuous there). In that case, it is intuitively clear that equality holds when the set of points where $f$ attains the maximum and the set of points where $g$ attains the maximum intersect.