arg max#Arg min

{{Short description|Inputs at which function values are highest}}

File:Si_sinc.svg functions above have $\operatorname{argmax}$ of {0} because both attain their global maximum value of 1 at x = 0.

The unnormalised sinc function (red) has arg min of {−4.49, 4.49}, approximately, because it has 2 global minimum values of approximately −0.217 at x = ±4.49. However, the normalised sinc function (blue) has arg min of {−1.43, 1.43}, approximately, because their global minima occur at x = ±1.43, even though the minimum value is the same."[http://physics.usyd.edu.au/teach_res/mp/doc/math_sinc_function.pdf The Unnormalized Sinc Function] {{Webarchive|url=https://web.archive.org/web/20170215045226/http://www.physics.usyd.edu.au/teach_res/mp/doc/math_sinc_function.pdf |date=2017-02-15 }}", University of Sydney]]

In mathematics, the arguments of the maxima (abbreviated arg max or argmax) and arguments of the minima (abbreviated arg min or argmin) are the input points at which a function output value is maximized and minimized, respectively.For clarity, we refer to the input (x) as points and the output (y) as values; compare critical point and critical value. While the arguments are defined over the domain of a function, the output is part of its codomain.

Definition

Given an arbitrary set {{nowrap| $X$ ,}} a totally ordered set {{nowrap| $Y$ ,}} and a function, {{nowrap| $f\colon X \to Y$ ,}} the $\operatorname{argmax}$ over some subset $S$ of $X$ is defined by

: $\operatorname{argmax}_S f := \underset{x \in S}{\operatorname{arg\,max}}\, f(x) := \{x \in S ~:~ f(s) \leq f(x) \text{ for all } s \in S \}.$

If $S = X$ or $S$ is clear from the context, then $S$ is often left out, as in $\underset{x}{\operatorname{arg\,max}}\, f(x) := \{ x ~:~ f(s) \leq f(x) \text{ for all } s \in X \}.$ In other words, $\operatorname{argmax}$ is the set of points $x$ for which $f(x)$ attains the function's largest value (if it exists). $\operatorname{Argmax}$ may be the empty set, a singleton, or contain multiple elements.

In the fields of convex analysis and variational analysis, a slightly different definition is used in the special case where $Y = [-\infty,\infty] = \mathbb{R} \cup \{ \pm\infty \}$ are the extended real numbers.{{sfn|Rockafellar|Wets|2009|pp=1-37|ignore-err=yes}} In this case, if $f$ is identically equal to $\infty$ on $S$ then $\operatorname{argmax}_S f := \varnothing$ (that is, $\operatorname{argmax}_S \infty := \varnothing$ ) and otherwise $\operatorname{argmax}_S f$ is defined as above, where in this case $\operatorname{argmax}_S f$ can also be written as:

: $\operatorname{argmax}_S f := \left\{ x \in S ~:~ f(x) = \sup {}_S f \right\}$

where it is emphasized that this equality involving $\sup {}_S f$ holds {{em|only}} when $f$ is not identically $\infty$ on {{nowrap| $S$ .}}{{sfn|Rockafellar|Wets|2009|pp=1-37|ignore-err=yes}}

= Arg min =

The notion of $\operatorname{argmin}$ (or $\operatorname{arg\,min}$ ), which stands for argument of the minimum, is defined analogously. For instance,

: $\underset{x \in S}{\operatorname{arg\,min}} \, f(x) := \{ x \in S ~:~ f(s) \geq f(x) \text{ for all } s \in S \}$

are points $x$ for which $f(x)$ attains its smallest value. It is the complementary operator of {{nowrap| $\operatorname{arg\,max}$ .}}

In the special case where $Y = [-\infty,\infty] = \R \cup \{ \pm\infty \}$ are the extended real numbers, if $f$ is identically equal to $-\infty$ on $S$ then $\operatorname{argmin}_S f := \varnothing$ (that is, $\operatorname{argmin}_S -\infty := \varnothing$ ) and otherwise $\operatorname{argmin}_S f$ is defined as above and moreover, in this case (of $f$ not identically equal to $-\infty$ ) it also satisfies:

: $\operatorname{argmin}_S f := \left\{ x \in S ~:~ f(x) = \inf {}_S f \right\}.$ {{sfn|Rockafellar|Wets|2009|pp=1-37|ignore-err=yes}}

Examples and properties

For example, if $f(x)$ is $1 - |x|,$ then $f$ attains its maximum value of $1$ only at the point $x = 0.$ Thus

: $\underset{x}{\operatorname{arg\,max}}\, (1 - |x|) = \{ 0 \}.$

The $\operatorname{argmax}$ operator is different from the $\max$ operator. The $\max$ operator, when given the same function, returns the {{em|maximum value}} of the function instead of the {{em|point or points}} that cause that function to reach that value; in other words

: $\max_x f(x)$ is the element in $\{ f(x) ~:~ f(s) \leq f(x) \text{ for all } s \in S \}.$

Like $\operatorname{argmax},$ max may be the empty set (in which case the maximum is undefined) or a singleton, but unlike $\operatorname{argmax},$ $\operatorname{max}$ may not contain multiple elements:Due to the anti-symmetry of $\,\leq,$ a function can have at most one maximal value. for example, if $f(x)$ is $4 x^2 - x^4,$ then $\underset{x}{\operatorname{arg\,max}}\, \left( 4 x^2 - x^4 \right) = \left\{-\sqrt{2}, \sqrt{2}\right\},$ but $\underset{x}{\operatorname{max}}\, \left( 4 x^2 - x^4 \right) = \{ 4 \}$ because the function attains the same value at every element of $\operatorname{argmax}.$

Equivalently, if $M$ is the maximum of $f,$ then the $\operatorname{argmax}$ is the level set of the maximum:

: $\underset{x}{\operatorname{arg\,max}} \, f(x) = \{ x ~:~ f(x) = M \} =: f^{-1}(M).$

We can rearrange to give the simple identityThis is an identity between sets, more particularly, between subsets of $Y.$

: $f\left(\underset{x}{\operatorname{arg\,max}} \, f(x) \right) = \max_x f(x).$

If the maximum is reached at a single point then this point is often referred to as {{em|the}} $\operatorname{argmax},$ and $\operatorname{argmax}$ is considered a point, not a set of points. So, for example,

: $\underset{x\in\mathbb{R}}{\operatorname{arg\,max}}\, (x(10 - x)) = 5$

(rather than the singleton set $\{ 5 \}$ ), since the maximum value of $x (10 - x)$ is $25,$ which occurs for $x = 5.$ Note that $x (10 - x) = 25 - (x-5)^2 \leq 25$ with equality if and only if $x - 5 = 0.$ However, in case the maximum is reached at many points, $\operatorname{argmax}$ needs to be considered a {{em|set}} of points.

For example

: $\underset{x \in [0, 4 \pi]}{\operatorname{arg\,max}}\, \cos(x) = \{ 0, 2 \pi, 4 \pi \}$

because the maximum value of $\cos x$ is $1,$ which occurs on this interval for $x = 0, 2 \pi$ or $4 \pi.$ On the whole real line

: $\underset{x \in \mathbb{R}}{\operatorname{arg\,max}}\, \cos(x) = \left\{ 2 k \pi ~:~ k \in \mathbb{Z} \right\},$ so an infinite set.

Functions need not in general attain a maximum value, and hence the $\operatorname{argmax}$ is sometimes the empty set; for example, $\underset{x\in\mathbb{R}}{\operatorname{arg\,max}}\, x^3 = \varnothing,$ since $x^3$ is unbounded on the real line. As another example, $\underset{x \in \mathbb{R}}{\operatorname{arg\,max}}\, \arctan(x) = \varnothing,$ although $\arctan$ is bounded by $\pm\pi/2.$ However, by the extreme value theorem, a continuous real-valued function on a closed interval has a maximum, and thus a nonempty $\operatorname{argmax}.$

Notes

References

{{Rockafellar Wets Variational Analysis 2009 Springer}}

External links

{{PlanetMath|urlname=argminandargmax|title=arg min and arg max}}

Category:Elementary mathematics

Category:Inverse functions