What is the meaning of the $c$ in $C_c^{\infty}(\mathbb{R})$?

Solution 1:

Parsing the Notation

The notation $C_c^{k}(\mathbb{R})$ denotes the space of compactly supported, $k$-times continuously differentiable functions on $\mathbb{R}$. Tackling this rather long phrase one piece at a time, we have:

• $C_c^k(X)$ is a "space of ... functions on $X$." The space $X$ needn't have a great deal of structure^[1], though from the point of view of pragmatics, let's assume that $X$ is a subset of $\mathbb{R}^n$ for some $n$. The codomain of a function in $C_c^k(X)$ is typically either $\mathbb{R}$ (or $\mathbb{C}$). In other words, a typical element of $C_c^k(X)$ is, first and foremost, a function $$ u : X \to \mathbb{R}. $$

• The $k$ indicates that elements of $C_c^k(X)$ are $k$-times continuously differentiable. Roughly speaking, if $u \in C_c^k(X)$, then for any multiindex $\alpha = (\alpha_1, \dotsc, \alpha_n)$ (where $n$ is the dimension of $X$, as a vector space) with $$|\alpha| = \alpha_1 + \dotsb + \alpha_n \le k,$$ the $\alpha$-th derivative $$ D^{\alpha} u = \frac{\partial^{\alpha_1}}{{\partial x_1}^{\alpha_1}} u + \dotsb + \frac{\partial^{\alpha_n}}{{\partial x_1}^{\alpha_n}} u $$ exists and is continuous. Functions in $C_c^{\infty}(X)$ are smooth, meaning that they have continuous derivatives of all orders.

• Finally, the $c$ indicates that elements of $C_c^{\infty}(X)$ have compact support. In functional analysis, the support of a function $u$ is the set $$ \DeclareMathOperator{\supp}{supp}\supp(u) := \overline{ \{x\in X \mid u(x) \ne 0\} }. $$ That is, it is the closure of of the set on which $u$ is nonzero.^[2] To say that a function $u$ is compactly supported or that it has compact support is to say that $\supp(u)$ is compact. Equivalently, a function $u:X\to \mathbb{R}$ is compactly supported if there exists a compact set $K \subseteq X$ such that $$ \{ x\in X \mid u(x) \ne 0 \} \subseteq K, $$ i.e. the nonvanishing set of $u$ is contained in a compact set.

  From a practical point of view, many of the spaces which are considered in functional analysis satisfy the Heine-Borel property (that is, sets are compact if and only if they are closed and bounded). In such a space, a function is compactly supported if it is zero away from a bounded set. That is, $u$ has compact support if and only if there is some $R > 0$ such that $$ \|x\| > R \implies u(x) = 0, $$ where $\|\cdot\|$ is the norm on $X$.

Examples

• If $K \subseteq X$ is any compact set, then the characteristic function $\chi_K$, defined by $$ \chi_K(x) = \begin{cases} 1 & \text{if $x\in K$, and} \\ 0 & \text{otherwise} \end{cases} $$ is compactly supported. As a concrete example, the the function $$ \chi_{(-1,1)} : \mathbb{R} \to \mathbb{R} : x \mapsto \begin{cases} 1 & \text{if $|x| < 1$, and} \\ 0 & \text{if $|x| \ge 1$.} \end{cases} $$ is compactly supported. Note that the nonvanishing set of $\chi_{(-1,1)}$ is the open interval $(-1,1)$, but the support of this function is the closed interval $[-1,1]$.

  Note that characteristic functions are not (generally speaking) differentiable (there are going to be problems at the boundaries of the set). So, while these functions may have compact support, they will not generally be elements of $C_c^{k}(X)$.

• In $\mathbb{R}^n$, the you will often encounter smooth bump functions. In general, such a function smooth and supported on specified set. A typical example is a function $u : \mathbb{R}^n \to \mathbb{R}$ of the form $$ u(x) := \begin{cases} \exp\left( -\frac{1}{1-\|x\|^2} \right) & \text{if $\|x\| < 1$, and} \\ 0 & \text{otherwise.} \end{cases} $$ This function is supported on the closed unit ball, and it is not too difficult to verify that it is smooth (it is smooth away from $\|x\|=1$ by elementary theory, and some basic calculus will show that derivatives of all orders exist along set $\|x\|=1$). Smooth bump functions show up frequently in the construction of mollifiers (which are useful, for example, in obtaining "weak" solutions to PDEs).

Other Notation

While the notation $C_c$ is pretty common for compactly supported (continuous) functions, there are other notations which occur in the literature, or in particular settings.

For example, some authors will use $C_0(X)$ to denote the space of compactly supported (continuous) functions. My guess is that this is a French / Bourbakist convention, as I have mostly encountered it in lectures from my advisor, who did his Ph.D. work in France. However, the notation $C_0(X)$ is also used by other authors to denote the set of (continuous) functions which vanish at infinity, i.e. $u : X \to \mathbb{R}$ such that $$ \lim_{\|x\|\to\infty} u(x) = 0. $$ As such, it is probably a bad idea to use $C_0(X)$ to denote compactly supported function.

One might also encounter the notation $C_{00}(X)$ for the space of (continuous) compactly supported functions. This notation is consistent with the notation $C_0(X)$ for functions which vanish at infinity, and also parallels common notation for analogous sequence spaces.

[1] At the very least, we expect $X$ to be a topological space, since notions of continuity don't make sense outside of that context. In addition, if we want to take derivatives, then $X$ must also be equipped with some smooth structure. As such, $X$ could be taken to be a smooth manifold. However, such generality likely confuses more than clarifies in the context of the question asked.

[2] In this answer, Eric Towers defines the support of a function $u$ to be $$ \supp(u) := \{ x\in X \mid u(x) \ne 0 \}. $$ This definition is may be correct in some circumstances (say, for a function $u : X \to Y$, where $X$ is a set with no additional structure). Indeed, the Wikipedia article on this topic suggests this definition. However, in the field of functional analysis, the support of a function is always the closure of the nonvanishing set. See, for example, Springers's Encyclopedia of Mathematics (which cites Rudin's classic text) and MathWorld.

Solution 2:

The "$c$" means "compact support" but there is a small caveat involving taking the closure of the support. Let $f:\mathbb{R} \rightarrow \mathbb{R}$ and $S = \{x \in \mathbb{R} : f(x) \neq 0 \}$. Then $f \in C_c^\infty(\mathbb{R})$ if $f$ is infinitely differentiable and the closure of its support, $\overline{S}$, is compact.