# Arg max

(Redirected from Argmax)
As an example, both unnormalised and normalised sinc functions above have ${\displaystyle \operatorname {argmax} }$ 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.[1]

In mathematics, the arguments of the maxima (abbreviated arg max or argmax) are the points, or elements, of the domain of some function at which the function values are maximized.[note 1] In contrast to global maxima, which refers to the largest outputs of a function, arg max refers to the inputs, or arguments, at which the function outputs are as large as possible.

## 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 ${\displaystyle x-5=0.}$