# When is $\sup(f) \sup (g) = \sup(fg)$?

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## Solution 1

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).

## Solution 2

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.

When the supremum is not attained we need to look at a proper definition of "points where the maximum is attained". These are accumulation points of sequences witnessing the supremum. In other words, these are points where there are arbitrarily close points which are arbitrarily close to the supremum - the correct quantification should be apparent if you try to prove this.

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### Daniel

Updated on February 09, 2020

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?
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 $f=g$.