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Arg max

## Summary

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.[note 1] While the arguments are defined over the domain of a function, the output is part of its codomain.

## Definition

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

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

If ${\displaystyle S=X}$  or ${\displaystyle S}$  is clear from the context, then ${\displaystyle S}$  is often left out, as in ${\displaystyle {\underset {x}{\operatorname {arg\,max} }}\,f(x):=\{x~:~f(s)\leq f(x){\text{ for all }}s\in X\}.}$  In other words, ${\displaystyle \operatorname {argmax} }$  is the set of points ${\displaystyle x}$  for which ${\displaystyle f(x)}$  attains the function's largest value (if it exists). ${\displaystyle \operatorname {Argmax} }$  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 ${\displaystyle Y=[-\infty ,\infty ]=\mathbb {R} \cup \{\pm \infty \}}$  are the extended real numbers.[2] In this case, if ${\displaystyle f}$  is identically equal to ${\displaystyle \infty }$  on ${\displaystyle S}$  then ${\displaystyle \operatorname {argmax} _{S}f:=\varnothing }$  (that is, ${\displaystyle \operatorname {argmax} _{S}\infty :=\varnothing }$ ) and otherwise ${\displaystyle \operatorname {argmax} _{S}f}$  is defined as above, where in this case ${\displaystyle \operatorname {argmax} _{S}f}$  can also be written as:

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

where it is emphasized that this equality involving ${\displaystyle \sup {}_{S}f}$  holds only when ${\displaystyle f}$  is not identically ${\displaystyle \infty }$  on ${\displaystyle S}$ .[2]

### Arg min

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

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

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

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

${\displaystyle \operatorname {argmin} _{S}f:=\left\{x\in S~:~f(x)=\inf {}_{S}f\right\}.}$ [2]

## Examples and properties

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

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

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

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

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

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

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

We can rearrange to give the simple identity[note 3]

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

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

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

(rather than the singleton set ${\displaystyle \{5\}}$ ), since the maximum value of ${\displaystyle x(10-x)}$  is ${\displaystyle 25,}$  which occurs for ${\displaystyle x=5.}$ [note 4] However, in case the maximum is reached at many points, ${\displaystyle \operatorname {argmax} }$  needs to be considered a set of points.

For example

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

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

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

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

2. ^ Due to the anti-symmetry of ${\displaystyle \,\leq ,}$  a function can have at most one maximal value.
3. ^ This is an identity between sets, more particularly, between subsets of ${\displaystyle Y.}$
4. ^ Note that ${\displaystyle x(10-x)=25-(x-5)^{2}\leq 25}$  with equality if and only if ${\displaystyle x-5=0.}$