Coercive function
# Coercive function

In mathematics, a coercive function is a function that "grows rapidly" at the extremes of the space on which it is defined. Depending on the context different exact definitions of this idea are in use.

## Coercive vector fields

A vector field f : $R^n$ → $R^n$ is called coercive if

${\frac {f(x)\cdot x}{\|x\|}}\to +\infty {\text{ as }}\|x\|\to +\infty ,$

where "$\cdot$" denotes the usual dot product and $\|x\|$ denotes the usual Euclidean norm of the vector x.

A coercive vector field is in particular norm-coercive since $\|f(x)\|\geq (f(x)\cdot x)/\|x\|$ for $x\in \mathbb {R} ^{n}\setminus \{0\}$, by Cauchy–Schwarz inequality. However a norm-coercive mapping f : $R^n$ → $R^n$ is not necessarily a coercive vector field. For instance the rotation f : $R^2$ → $R^2,$ f(x) = $(−x_2,$ $x_1)$ by 90° is a norm-coercive mapping which fails to be a coercive vector field since $f(x)\cdot x=0$ for every $x\in \mathbb {R} ^{2}$.

## Coercive operators and forms

A self-adjoint operator $A:H\to H,$ where $H$ is a real Hilbert space, is called coercive if there exists a constant $c>0$ such that

$\langle Ax,x\rangle \geq c\|x\|^{2}$

for all $x$ in $H.$

A bilinear form $a:H\times H\to \mathbb {R}$ is called coercive if there exists a constant $c>0$ such that

$a(x,x)\geq c\|x\|^{2}$

for all $x$ in $H.$

It follows from the Riesz representation theorem that any symmetric (defined as $a(x,y)=a(y,x)$ for all $x,y$ in $H$), continuous ($|a(x,y)|\leq k\|x\|\,\|y\|$ for all $x,y$ in $H$ and some constant $k>0$) and coercive bilinear form $a$ has the representation

$a(x,y)=\langle Ax,y\rangle$

for some self-adjoint operator $A:H\to H,$ which then turns out to be a coercive operator. Also, given a coercive self-adjoint operator $A,$ the bilinear form $a$ defined as above is coercive.

If $A:H\to H$ is a coercive operator then it is a coercive mapping (in the sense of coercivity of a vector field, where one has to replace the dot product with the more general inner product). Indeed, $\langle Ax,x\rangle \geq C\|x\|$ for big $\|x\|$ (if $\|x\|$ is bounded, then it readily follows); then replacing $x$ by $x\|x\|^{-2}$ we get that $A$ is a coercive operator. One can also show that the converse holds true if $A$ is self-adjoint. The definitions of coercivity for vector fields, operators, and bilinear forms are closely related and compatible.

## Norm-coercive mappings

A mapping $f:X\to X'$ between two normed vector spaces $(X,\|\cdot \|)$ and $(X',\|\cdot \|')$ is called norm-coercive if and only if

$\|f(x)\|'\to +\infty {\mbox{ as }}\|x\|\to +\infty .$

More generally, a function $f:X\to X'$ between two topological spaces $X$ and $X'$ is called coercive if for every compact subset $K'$ of $X'$ there exists a compact subset $K$ of $X$ such that

$f(X\setminus K)\subseteq X'\setminus K'.$

The composition of a bijective proper map followed by a coercive map is coercive.

## (Extended valued) coercive functions

An (extended valued) function $f:\mathbb {R} ^{n}\to \mathbb {R} \cup \{-\infty ,+\infty \}$ is called coercive if

$f(x)\to +\infty {\mbox{ as }}\|x\|\to +\infty .$

A real valued coercive function $f:\mathbb {R} ^{n}\to \mathbb {R}$ is, in particular, norm-coercive. However, a norm-coercive function $f:\mathbb {R} ^{n}\to \mathbb {R}$ is not necessarily coercive. For instance, the identity function on $\mathbb {R}$ is norm-coercive but not coercive.

