# Arg max

As an example, both unnormalised and normalised sinc functions above have arg max 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 of the domain of some function at which the function values are maximized.[note 1] In contrast to global maxima, referring 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 a function, ${\displaystyle f\colon X\rightarrow Y}$, the arg max over some subset, S, of X is defined by

${\displaystyle {\underset {x\in S\subseteq X}{\operatorname {arg\,max} }}\,f(x):=\{x\mid x\in S\wedge \forall y\in S:f(y)\leq f(x)\}.}$

If S = X or S is clear from the context, then S is often left out, as in ${\displaystyle {\underset {x}{\operatorname {arg\,max} }}\,f(x):=\{x\mid \forall y:f(y)\leq f(x)\}.}$ In other words, arg max is the set of points, x, for which f(x) attains the function's largest value. Arg max may be the empty set, a singleton, or contain multiple elements. For example, if f(x) is 1−|x|, then f attains its maximum value of 1 if and only if x = 0, implying

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

The arg max operator is the natural complement of the max operator which, given the same function, returns the maximum value instead of the point or points that reach that value; in other words

${\displaystyle \max _{x}f(x)}$ is the element in ${\displaystyle \{f(x)\mid \forall y:f(y)\leq f(x)\}.}$

Like arg max, max may be the empty set (in which case the maximum is undefined) or a singleton, but unlike arg max, max may not contain multiple elements: for example, if f(x) is 4x2 - x4, then ${\displaystyle {\underset {x}{\operatorname {arg\,max} }}\,(4x^{2}-x^{4}):=\{-{\sqrt {2}},{\sqrt {2}}\}}$, but ${\displaystyle {\underset {x}{\operatorname {max} }}\,(4x^{2}-x^{4}):=\{4\}}$ because the function attains the same value at every element of arg max.

Equivalently, if M is the maximum of f, then the arg max is the level set of the maximum:

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

If the maximum is reached at a single point then this point is often referred to as the arg max, meaning we define the arg max as 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 {5}), since the maximum value of x(10 − x) is 25, which occurs for x = 5.[note 2] However, in case the maximum is reached at many points, arg max is a set of points.

Then, we have for example

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

since the maximum value of cos(x) is 1, which occurs on this interval for x = 0, 2π or 4π. On the whole real line, the arg max is ${\displaystyle \{0,2\pi ,-2\pi ,4\pi ,\dots \}.}$

Note also that functions do not in general attain a maximum value, and hence the arg max is sometimes the empty set; for example, ${\displaystyle {\underset {x\in \mathbb {R} }{\operatorname {arg\,max} }}\,x^{3}=\emptyset }$, since ${\displaystyle x^{3}}$ is unbounded on the real line. However, by the extreme value theorem (or the classical compactness argument), a continuous function on a compact interval has a maximum, and thus a nonempty arg max.

## Arg min

arg min (or argmin) stands for argument of the minimum, and is defined analogously. For instance,

${\displaystyle {\underset {x}{\operatorname {arg\,min} }}\,f(x)}$

are points x for which f(x) attains its smallest value. The complementary operator is, of course, min.

2. ^ Note that ${\displaystyle x(10-x)=25-(x-5)^{2}\leq 25}$ with equality if and only if ${\displaystyle x-5=0}$.