When $f$ is a function on a set $A$, the notation: $\arg\max_{x\in A} f(x)$
denotes the set of elements of $A$ for which $f$ attains its maximum value. This set may be empty, for example, if $f(x)=x$ and $A=(0,1)$, then $f$ has no maximum on $A$, so:
$$\arg\max_{x\in (0,1)} f(x) = \emptyset$$
However, $f$ always has a supremum (that can be $\infty$ if $f$ is unbounded), so apparently we can define an "argument-supremum". In this case, this would be:
$$\arg\sup_{x\in (0,1)} f(x) = \{1\}$$
Can this "arg sup" operator be defined formally?
I thought to define it in the following way:
$$\arg\sup_{x\in A} f(x) = \arg \max_{x\in \text{closure}(A)}f(x)$$
Is this definition meaningful? Is it used anywhere in mathematics?
Best Answer
I think the main problem is not non-closedness but non-compactness. Consider $f(x) =-e^x$. Its supremum is zero, but the function is nowhere zero. However, as every supremum, you can approximate its supremum by function values, i. e. there is a sequence $x_n$ such that $f(x_n) $ converges to 0, but the sequence itself does not converge. This problem would not occur if the base set would be compact. Taking some compactification of the base set may do the trick, but the problem is that you need to ensure some continuity of the extension of the function to the compactification.