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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?

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  • $\begingroup$ Wait, wouldn't $\arg \sup_{x \in (0, \, 1)} f(x) = \mathbb{R}_{\ge 1}$? $\endgroup$ – MathMajor Jul 18 '16 at 6:40
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    $\begingroup$ It's not always true that $f$ has a supremum, just take an unbounded function $f$. No upper bounds, therefore no smallest upper bound. The very last equality on your question has other issues, namely that $f$ is not be defined on the closure of $A$, so $f(x)$ where $x$ is in this closure, is meaningless. $\endgroup$ – Git Gud Jul 18 '16 at 6:46
  • $\begingroup$ @GitGud For an unbounded function it is possible to define the supremum as $\infty$ $\endgroup$ – Erel Segal-Halevi Jul 18 '16 at 13:10
  • $\begingroup$ If the sup is not attained then the "arg sup" would presumably be something like a sequence of points $x_n$ with $f(x_n)$ increasing to the sup. $\endgroup$ – Ian Jul 18 '16 at 13:13
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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.

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