Function Spaces: Cc, C0, Cb

This scene is about function spaces. See how the three families of continuous functions

C_c

C_0

, and

C_b

nest and split.

C_b

is all bounded continuous functions.

C_0

holds the ones that fade to

0

far away, and

C_c

the ones that are exactly

0

outside a finite interval. Both inclusions

C_c \subset C_0 \subset C_b

are strict.

For example, a ⛺ tent stands on a finite patch of ground. Fog thins out with distance but lies everywhere. An endless plain never fades away at all.

Examples & Exercises

Prove completely that $C_0(\mathbb{R})$ is a closed subspace of $C_b(\mathbb{R})$ in the sup norm.

That is, show that if $f_n \in C_0(\mathbb{R})$ , $f \in C_b(\mathbb{R})$ , and $\|f_n - f\|_\infty \to 0$ , then $f \in C_0(\mathbb{R})$ .

(You may use the fact that a uniform limit of continuous functions is continuous.)

|f(x)| \le |f_n(x)| + \|f - f_n\|_\infty < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \quad (|x| > M)

The three representatives: the tent (

C_c

e^{-|x|}

(

C_0

), and the constant

1

(

C_b

) — closedness says the middle layer is sealed under uniform limits.

Show proof

Fix $\varepsilon > 0$ . Since $\|f_n - f\|_\infty \to 0$ , there is an $n$ with $\|f - f_n\|_\infty < \frac{\varepsilon}{2}$ . Fix this $n$ .
Since $f_n \in C_0(\mathbb{R})$ , it vanishes at infinity: there is $M$ such that $|x| > M$ implies $|f_n(x)| < \frac{\varepsilon}{2}$ .
For $|x| > M$ , the triangle inequality gives
$|f(x)| \le |f_n(x)| + |f(x) - f_n(x)| \le |f_n(x)| + \|f - f_n\|_\infty < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$ .
So $|x| > M$ implies $|f(x)| < \varepsilon$ : $f$ vanishes at infinity, hence $f \in C_0(\mathbb{R})$ .
Hence $C_0(\mathbb{R})$ contains all its sequential limits inside $C_b(\mathbb{R})$ : it is closed.
Since $C_b(\mathbb{R})$ is complete in the sup norm (uniformly Cauchy sequences converge uniformly to bounded continuous limits), the closed subspace $C_0(\mathbb{R})$ is complete as well.

Using the cutoff functions $\varphi_k(x) = 1$ for $|x| \le k$ , $\varphi_k(x) = k + 1 - |x|$ for $k \le |x| \le k+1$ , and $\varphi_k(x) = 0$ for $|x| \ge k+1$ , prove completely that $C_c(\mathbb{R})$ is dense in $C_0(\mathbb{R})$ in the sup norm.

That is, show that for every $f \in C_0(\mathbb{R})$ and every $\varepsilon > 0$ there is a $k$ with $\|f - \varphi_k f\|_\infty < \varepsilon$ .

|f(x) - \varphi_k(x) f(x)| = (1 - \varphi_k(x))\,|f(x)| \le \sup_{|t| \ge k} |f(t)|

k = 1

f_1 = \varphi_1 f

is a

C_c

function vanishing outside

[-(k+1), k+1] = [-2, 2]

, with sup error

e^{-2} \approx 0.1353

Show proof

Fix $f \in C_0(\mathbb{R})$ and $\varepsilon > 0$ . First, $f_k = \varphi_k f$ lies in $C_c(\mathbb{R})$ : $\varphi_k$ is piecewise linear, hence continuous, and $f$ is continuous, so the product $f_k$ is continuous. For $|x| \ge k+1$ , $\varphi_k(x) = 0$ , so $f_k(x) = 0$ . Hence $\operatorname{supp} f_k \subseteq [-(k+1), k+1]$ , a compact set, and $f_k \in C_c(\mathbb{R})$ .
Estimate the error. If $|x| \le k$ then $\varphi_k(x) = 1$ , so $f_k(x) = f(x)$ and the error is $0$ .
If $|x| > k$ then $0 \le 1 - \varphi_k(x) \le 1$ , so
$|f(x) - f_k(x)| = (1 - \varphi_k(x))\,|f(x)| \le |f(x)| \le \sup_{|t| \ge k} |f(t)|$ .
Hence $|f(x) - f_k(x)| \le \sup_{|t| \ge k} |f(t)|$ for every $x$ , and $\|f - f_k\|_\infty \le \sup_{|t| \ge k} |f(t)|$ .
Since $f \in C_0(\mathbb{R})$ , it vanishes at infinity. Applying this with $\frac{\varepsilon}{2}$ gives a $k$ with $|f(t)| < \frac{\varepsilon}{2}$ for all $|t| \ge k$ .
For this $k$ , $\sup_{|t| \ge k} |f(t)| \le \frac{\varepsilon}{2}$ , and the previous step gives $\|f - f_k\|_\infty \le \frac{\varepsilon}{2} < \varepsilon$ .
So every $\varepsilon$ -neighborhood of $f$ contains the element $f_k$ of $C_c(\mathbb{R})$ .
Therefore $C_0(\mathbb{R}) \subseteq \overline{C_c(\mathbb{R})}$ .
Conversely, $C_c(\mathbb{R}) \subseteq C_0(\mathbb{R})$ , and $C_0(\mathbb{R})$ is closed by Problem 1, so $\overline{C_c(\mathbb{R})} \subseteq C_0(\mathbb{R})$ .
Hence $\overline{C_c(\mathbb{R})} = C_0(\mathbb{R})$ : $C_c(\mathbb{R})$ is dense in $C_0(\mathbb{R})$ .

Using $f(x) = e^{-|x|}$ and $f_j = \varphi_j f$ for $j = 1, 2, 3, \dots$ , where $\varphi_j$ is the cutoff of Problem 2, prove completely that $C_c(\mathbb{R})$ is not complete in the sup norm.

That is, show that $(f_j)$ is a uniformly Cauchy sequence in $C_c(\mathbb{R})$ whose sup-norm limit $f$ does not belong to $C_c(\mathbb{R})$ .

\|f - f_j\|_\infty = e^{-(j+1)}, \qquad \|f_i - f_j\|_\infty \le e^{-(i+1)} + e^{-(j+1)}

k = 2

: widening the cutoff shrinks the sup error

e^{-(k+1)}

e^{-3} \approx 0.0498

, but the limit

e^{-|x|}

itself is zero nowhere.

Show proof

Each $f_j = \varphi_j f$ lies in $C_c(\mathbb{R})$ by the first step of Problem 2: it is continuous, with $\operatorname{supp} f_j \subseteq [-(j+1), j+1]$ .
The key identity is $\|f - f_j\|_\infty = e^{-(j+1)}$ , checked region by region.
For $|x| \le j$ : $\varphi_j(x) = 1$ , so the error is $0$ .
For $j \le |x| \le j+1$ : with $s = |x| - j \in [0, 1]$ , the error is $s\, e^{-|x|} = e^{-(j+1)} \cdot s e^{1-s}$ . Since $e^{s-1} \ge 1 + (s - 1) = s$ (convexity of the exponential), $s e^{1-s} \le 1$ , so the error is at most $e^{-(j+1)}$ .
For $|x| \ge j+1$ : $\varphi_j(x) = 0$ , so the error is $e^{-|x|} \le e^{-(j+1)}$ .
At $|x| = j+1$ the error equals $e^{-(j+1)}$ exactly, so the supremum is that value.
The sequence $(f_j)$ is uniformly Cauchy: for all $i, j$ , the triangle inequality gives
$\|f_i - f_j\|_\infty \le \|f_i - f\|_\infty + \|f - f_j\|_\infty = e^{-(i+1)} + e^{-(j+1)}$ .
The right side tends to $0$ as $i, j \to \infty$ , so for every $\varepsilon > 0$ , $\|f_i - f_j\|_\infty < \varepsilon$ once $i, j$ are large: $(f_j)$ is Cauchy in the sup norm.
Also $\|f_j - f\|_\infty = e^{-(j+1)} \to 0$ , so the sup-norm limit of $(f_j)$ (its limit inside $C_b(\mathbb{R})$ ) is $f$ .
We have $f \in C_0(\mathbb{R})$ : for $\varepsilon > 0$ , $|x| > \ln\frac{1}{\varepsilon}$ implies $|f(x)| = e^{-|x|} < \varepsilon$ .
But $f \notin C_c(\mathbb{R})$ : $f(x) = e^{-|x|} > 0$ for every $x$ , so $\{x : f(x) \ne 0\} = \mathbb{R}$ and $\operatorname{supp} f = \mathbb{R}$ is not compact.
Suppose $C_c(\mathbb{R})$ were complete. Then the Cauchy sequence $(f_j)$ would converge in sup norm to some $g \in C_c(\mathbb{R})$ .
But sup-norm limits are unique inside $C_b(\mathbb{R})$ : if $\|f_j - g\|_\infty \to 0$ and $\|f_j - f\|_\infty \to 0$ , then $\|f - g\|_\infty \le \|f - f_j\|_\infty + \|f_j - g\|_\infty \to 0$ , so $f = g$ .
Since $f \notin C_c(\mathbb{R})$ , this is a contradiction. Hence $C_c(\mathbb{R})$ is not complete, and it is closed neither in $C_0(\mathbb{R})$ nor in $C_b(\mathbb{R})$ (together with the density of Problem 2, $\overline{C_c(\mathbb{R})} = C_0(\mathbb{R})$ ).