📂Hilbert Space

The Support of Functions and the Class of Continuous Function Spaces

Definitions

Let’s consider a function $f : \mathbb{R} \to \mathbb{C}$ in the function space $\mathbb{C}^{\mathbb{R}}$.

• The support of a function $f$ is defined as the closed set obtained by taking the closure of the set of points where the function value is not $0$. $$\text{supp} f = \overline{\left\{ x \in \mathbb{R} : f(x) \ne 0 \right\}}$$

• If $\text{supp} f$ is bounded, then $f$ is said to have a compact support because the closure is a closed set, and a set that is closed and bounded in the real number space is compact.

• $U\Subset V$ is $\overline{U} \subset V$ and $\overline{U}$ being compact means that $\mathrm{supp}(f) \Subset U$ means that $f$ has a compact support in $U$. It is also written as $\subset \subset$.

• The set of continuous functions forms a vector space and is called a space of continuous functions, denoted as follows:

  $$C(\mathbb{R}) := \left\{f \text{ is continuous} \right\}$$

  If there’s confusion with $C^{1}$, it is sometimes written as $C^{0}$.

• The vector space of continuous functions that have a compact support is denoted as follows:

  $$C_{c} (\mathbb{R}) := \left\{ f \in C(\mathbb{R}) : f \text{ has compact support} \right\}$$

• The vector space of continuous functions whose function value converges to $0$ when $x \to \pm \infty$ is denoted as follows:

  $$C_{0} ( \mathbb{R} ) := \left\{ f \in C(\mathbb{R}) : f(x) \to 0 \text{ as } x \to \pm \infty \right\}$$

• The vector space of continuous functions that are differentiable up to $m$ times, and all of its derivatives are continuous is denoted as follows:

  $$C^{m}(\mathbb{R}) :=\left\{ f \in C(\mathbb{R}) : f^{(n)} \text{ is continuous } \forall n \le m \right\}$$

  Here, $C^{0}(\mathbb{R})$ means $C(\mathbb{R})$. A function that is an element of $C^{m}$ is called a $m$-times continuously differentiable function.

• The vector space of infinitely differentiable functions, all of whose derivatives are continuous, is denoted as follows: $$C^{\infty}(\mathbb{R})=\bigcap _{m=0}^{\infty}C^{m}(\mathbb{R})$$ An element of $C^{\infty}$ is referred to as a smooth function.

※ Depending on the author, $C_{0}$ is sometimes used in the sense of $C_{c}$, so be sure to check the notation defined in the textbook.

Explanation

In Sobolev spaces, theory of distributions, etc., $C_{c}^{\infty}$ is mainly dealt with.

Naturally, $C_{c} (\mathbb{R})$ is a subspace of $C_{0} (\mathbb{R})$. Although both are superior spaces compared to the mere space of continuous functions $C (\mathbb{R})$, one must be cautious that they do not become a Banach space with respect to the operator norm $\left\| \cdot \right\|_{\infty}$. For example, consider the following $\left\{ f_{k} \right\}_{k \in \mathbb{N}} \subset C_{c} (\mathbb{R})$

$$f_{k} (x) := \begin{cases} {{ \sin x } \over { x }} \chi_{[ - k \pi , k \pi ]} (x) & , x \ne 0 \\ 1 & , x = 0 \end{cases}$$

$f_{k}$ converges to the following sinc function $\sinc \in C_{0} (\mathbb{R}) \setminus C_{c} (\mathbb{R})$ while having a compact support $[-k \pi , k \pi]$ for all $k \in \mathbb{N}$.

$$\sinc x = \begin{cases} {{ \sin x } \over { x }} & , x \ne 0 \\ 1 & , x = 0 \end{cases}$$

As a Metric Space¹

Let’s refer to the set of continuous real-valued functions on the interval $[0, 1]$ as $X = C[0, 1]$. And let’s define the [metric] $d$ as follows.

$$d(x, y) := \int\limits_{0}^{1} \left| x(t) - y(t) \right| dt \qquad \forall x, y \in X$$

Then, the [metric space] $(X, d)$ is not a complete space. Let’s consider the function $x_{m}$ as shown in the image (a) below.

If we say $n \gt m$, for any $\varepsilon \gt 0$, whenever $m \gt 1/\varepsilon$, $1 \cdot \frac{1}{m} \lt \varepsilon$ holds; hence, by $d(x_{m}, x_{n}) \lt \varepsilon$, $\left\{ x_{m} \right\}$ is a Cauchy sequence.

However, since $x_{m}(t) = 0$ and $(t \in [0, 1/2])$, and $x_{m}(t) = 1$ and $(t \in [a_{m}, 1])$, the following holds.

$$\begin{align*} d(x_{m}, x) &= \int\limits_{0}^{1} \left| x_{m(t)} - x(t) \right| dt \\ &= \int\limits_{0}^{\frac{1}{2}} \left| 0 - x(t) \right| dt + \int\limits_{\frac{1}{2}}^{a_{m}} \left| x_{m(t)} - x(t) \right| dt + \int\limits_{a_{m}}^{1} \left| 1 - x(t) \right| dt \\ &= \int\limits_{0}^{\frac{1}{2}} \left| x(t) \right| dt + \int\limits_{\frac{1}{2}}^{a_{m}} \left| x_{m(t)} - x(t) \right| dt + \int\limits_{a_{m}}^{1} \left| 1 - x(t) \right| dt \\ \end{align*}$$

Since each of the integrands is greater than or equal to $0$, for $d(x_{m}, x)$ to converge to $0$, each of the integrand functions must be $0$. Hence, $x$ at $t\in[0, \frac{1}{2})$ is $x(t) = 0$, and at $t\in (\frac{1}{2}, 1]$ is $x(t) = 1$. Clearly, it is not continuous, thus $x \notin X$, and $\left\{ x_{m} \right\}$ does not converge to $X$.

As a Normed Space²

The space of continuous functions $C[0, 1]$ becomes a complete space, i.e., a complete normed (Banach) space, when given the maximum value as a norm, rather than an integral. In other words, $\left\| \cdot \right\|$ is a Banach space with the norm $(C[0, 1], \left\| \cdot \right\|)$ defined as follows.

$$\left\| f \right\| := \max\limits_{t \in [0, 1]} \left| f(t) \right|,\qquad f \in C[0, 1]$$