Why are coercive functions called coercive, and why is it useful?

Question

A function $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is called coercive if $$ \frac{f(x)\cdot x}{\|x\|} \rightarrow \infty \;\; \text{as} \;\;\|x\|\rightarrow \infty.$$

I came across this requirement in calculus of variations, where a coercivity condition is needed to show that a sequence of functions gamma-converges to some limiting function. It seems related to showing that the compactness condition that usually accompanies Gamma-convergence - does coerciveness imply compactness? I don't see how they are related, but they seem to be.

Edit: The "compactness condition" I refer to is that any sequence of arguments $\{u_n\}_{n\in \mathbb{N}}$ has a convergent subsequence. I think that this compactness condition is implies equi-coerciveness of the sequence $\{u_n\}$. I don't know why coerciveness is used in proofs of gamma-convergence instead of compactness directly. Anyways, I still want to know why the functions are called "coercive".

Answer 1

$\bf{Definition}$: A function $f : \mathbb{R}^n \to \mathbb{R^n}$ is coercive if and only if: $$\lim_{\| x \| \to + \infty} \frac{f(x) \cdot x}{\| x \|} \to + \infty,$$

In terminology, this is named a coercive function because of its literal latin meaning according to Wikitionary:

From Latin coercere (“to surround, encompass, restrain, control, curb”), from co- (“together”) + arcere (“to inclose, confine, keep off”);

The term "coercive" comes from the idea that the function "coerce" or forces its values to increase as the input gets larger and it relates to the behavior of a function as it approaches critical points. Here, it is forced (coerced) to grow sufficiently fast as the absolute value of its argument grows.

Here, coercivity does implies compactness which is a correct claim, where it must also be recognize that compactness does not necessarily implies coercive functions. In the context of Gamma-convergence, the connection between coercivity and compactness is important because coercivity ensures that you have control over the behavior of the functionals and their minimizers as they approach certain limits. This control is essential for proving convergence results and understanding the behavior of functionals in the limit.

If you have a sequence of functions that minimizes a coercive functional, then this sequence is necessarily bounded. This follows from the definition of coercivity, which prevents the energy (functional value) from going to infinity as the functions in the sequence become unbounded. Boundedness is a crucial property for establishing compactness, as a bounded set in a function space is relatively compact.

User @Rdizzl3 has solved the implications and has stated:

Let $f:\mathbb{R}^n\rightarrow \mathbb{R}\color{red}{^n}$ be continuous on all of $\mathbb{R}^n$.

$f$ is coercive $\iff \forall \alpha\in\mathbb{R}.\{x|f(x)\leq\alpha\}$ is compact.

Here is his proof for this claim (in my understandings):

Since $f$ is continuous, it implies the closedness of the sets $\{x|f(x)\leq\alpha\}$. Our task now is to establish that any set of the form $\{x|f(x)\leq\alpha\}$ is bounded.

Let's assume there exists an $\alpha \in \mathbb{R}^n$ such that the set $S = \{x|f(x)\leq\alpha\}$ is unbounded. Then, there must exist a sequence $\{x_n\} \subset S$ with $\|x_n\| \rightarrow \infty$. However, due to the coercivity of $f$, we must also have $f(x_n) \rightarrow \infty$. This contradicts the fact that $f(x_n) \leq \alpha$ for all $n = 1, 2, \ldots$. Therefore, we conclude that $S$ must be bounded.

Now, let's assume that each of the sets $\{x|f(x)\leq\alpha\}$ is bounded, and consider a sequence $\{x_n\} \subset \mathbb{R}^n$ where $\|x_n\| \rightarrow \infty$. Suppose there exists a subsequence of integers $J \subset \mathbb{N}$ such that the set $\{f(x_n)\}$ is bounded from above. This would imply the existence of an $\alpha \in \mathbb{R}^n$ such that $\{x_n\}_J \subset \{x|f(x)\leq\alpha\}$. However, this cannot hold true since each of these sets is bounded, while every subsequence of $\{x_n\}$ is unbounded by definition. Consequently, the set $\{f(x_n)\}_J$ cannot be bounded, and thus, $\{f(x_n)\}$ contains no bounded subsequence, meaning that $f(x_n) \rightarrow \infty$.

This satisfies what we stated about about how coercivity implies compactness. Coercivity also implies injectivity and is useful for showing uniqueness.

Answer 2

I don't know if this can be helpful, but I'll provide an insight about the definition.

As already pointed out, a function $f:\mathbb{R^n} \rightarrow \mathbb{R}$ is usually said to be coercive "if it is $+\infty$ at infinity", namely if $$\lim_{\|x\|\rightarrow+\infty} f(x) = +\infty $$ The word comes indeed from the Latin verb; this condition is used for example in convex analysis to ensure that a convex function attains a minimum (for example $e^{t}$ is convex but has no minimzer on $\mathbb{R}$, and indeed it is not coercive, whereas $e^{|t|}$ is convex and coercive, and thus attains a minimum at $0$).

However, there is no natural generalization of this definition to vector-valued functions $F:\mathbb{R^n} \rightarrow \mathbb{R}^n$, basically because in $\mathbb{R}^n$ there is no notion of "$+$" or "$-$" infinity.

Instead, we could weaken this notion of coercivity: for $f:\mathbb{R^n} \rightarrow \mathbb{R}$, $f$ is norm-coercive if $|f|$ is coercive, namely if $$\lim_{\|x\|\rightarrow+\infty} |f(x)| = +\infty $$

This means that at infinity, the function takes infinite (regardless of the sign) values (for example now $t\mapsto t$ is norm-coercive), and this can be generalised to the vector-valued case $F:\mathbb{R^n} \rightarrow \mathbb{R}^n$ as $$ \lim_{\|x\|\rightarrow+\infty} \|F(x)\| = +\infty $$

However, for vector-valued functions the most usual definition of coercivity is yours: $$ \lim_{\|x\|\rightarrow+\infty} F(x)\cdot \frac{x}{\|x\|} = +\infty $$

I don't know what properties this carries actually, but the intuition is that we are requesting the function to "grow at infinity, only in the direction that we are considering for the limit". For example, if $n=2$ and $f(x,y) = R(x,y) = (-y, x)$, i.e. $f$ is a $90°$ rotation (represented by an orthogonal matrix $R$), then clearly $$\|R(x,y)\| = \|(x,y)\| \rightarrow +\infty $$ but $$ R(x,y)\cdot \frac{(x,y)}{\|(x,y)\|} = \frac{-yx+xy}{\|(x,y)\|} = 0 $$ (if we consider the limit on the $x-$axis $(x,0)$, then $R(x,0) = (0,x)$ and thus we are not growing at infinity in the direction of our limit).

This definition of coercivity is stronger than the norm-coercivity, because $\|F(x)\| = F(x) \cdot \frac{F(x)}{\|F(x)\|}\ge F(x) \cdot \frac{x}{\|x\|}$.