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

:$\frac \to + \infty \mbox \, x \, \to + \infty,$

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^n \setminus \$, 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) = (-x₂, x₁)''
by 90° is a norm-coercive mapping which fails to be a coercive vector field since
$f(x) \cdot x = 0$ for every $x \in \mathbb^2$.

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

: $\langle Ax, x\rangle \ge c\, x\, ^2$

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

:$a(x, x)\ge c\, x\, ^2$

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 ($, a(x, y), \le k\, x\, \,\, y\,$ 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, $\langle Ax, x\rangle \ge C\, x\,$ for big $\, x\,$ (if $\, x\,$ is bounded, then it readily follows); then replacing $x$ by $x\, x\, ^$ 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 iff

:$\, f(x)\, ' \to + \infty \mbox \, x\, \to +\infty$.

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
$f:\mathbb^n \to \mathbb \cup \$
is called coercive if
:$f(x) \to + \infty \mbox \, x \, \to + \infty.$
A real valued coercive function $f:\mathbb^n \to \mathbb$
is, in particular, norm-coercive. However, a norm-coercive function
$f:\mathbb^n \to \mathbb$ is not necessarily coercive.
For instance, the identity function on $\mathbb$ is norm-coercive
but not coercive.

See also: radially unbounded functions

References

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{{PlanetMath attribution, id=7154, title=Coercive Function

Functional analysis
General topology
Types of functions