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Is $(C_{0}(X), ||.||_{\infty})$ a Banach Algebra?

Given $$C_{0}(\mathbb{R}^{n})=\{f\in C(\mathbb{R^n} \ | \ \ \exists R \ge 0 \ \text{such that } f(x)=0 \ \text{for} \ ||x||\ge R \}$$ and $$||f(x)||_{\infty} = \max_{x\in R^n}|f(x)| $$

What I have tried:

$C_{0}(\mathbb{R}^n)$ is the space of all continous functions whose values can be restricted by a constant R by restricting the largest possible coordinate in a vector from the domain set.

Let $(f_n)_{n\in \mathbb{N}}$ be a cauchy sequence in $C_{0}(\mathbb{R}^n)$: $$|| f_n-f_m||_{\infty} < \frac{\epsilon}{2} \\ \text{for n,m greater than an index N}$$

because of the completeness of $\mathbb{R}^n$ there exists a $N_x \ge N$ for each $x\in \mathbb{R^n}$ so that: $$|f_{N_x}-f(x)| < \frac{\epsilon}{2}$$

this leads to : $$|f_n-f| \le ||f_n-f_{N_x}||_\infty+|f_{N_x}-f | \le \epsilon $$

and if one takes the max norm of it: $$|| f_n- f ||_\infty \le \epsilon$$ which is uniform convergence of $f_n$ to $f$ .

$(C_{0}(X),||.||_\infty)$ is a Banach space. To be a Banach Algebra, closure under multiplication for all elements of $C_{0}(X)$ is also needed (otherwise cant check the Algebra multiplication conditions), if one function has $R_1=2$ and the other $R_2=4$ , then multiplying f(x)f(y) it would not work with the closure under multiplication, so it can not be a Banach Algebra ...

Does $C_{0}(X)$ have a common name ?

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up vote 2 down vote accepted

Your $C_0(\mathbb{R}^n)$ is often called the space of compactly supported continuous functions. In many books this is denoted $C_c(\mathbb{R}^n)$. $\newcommand{\supp}{\operatorname{supp}}$

As you've already noted, convergence in your $C_0$ is the same as uniform convergence on $\mathbb{R}^n$. If $f$ and $g$ are in $C_0$, then $fg \in C_0$, since if $f$ and $g$ vanish for $\|x\| \ge R_f$ and $\|x\| \ge R_g$ , respectively, then $fg$ vanishes for $\| x \| \ge \min\{R_f, R_g\}$. (The ''$R$'' in the definition of $C_0$ is not fixed, but can be different for different functions.)

However, $C_0$ is not a Banach space. Let

$$ f_n(x) = \begin{cases} \dfrac{1}{1+x^2}- \dfrac{1}{1+n^2}, & -n \le x \le n \\[1ex] 0, & \text{otherwise}. \end{cases}$$

Then $f_n \in C_0(\mathbb{R})$ and $f_n$ converges uniformly to $f(x) = \dfrac{1}{1+x^2}$, but $f \notin C_0(\mathbb{R})$, since the support of $f$ is the whole line $\mathbb{R}$. You can adapt the example to higher dimensions yourself.

The problem here is that each function has compact support, but their support grows larger and larger with $n$, so that the support of the limit function is no longer compact.

Note: In the literature, $C_0(X)$ is often used for the set of continuous functions tending to $0$ at infinity, and this is a Banach algebra. In fact this version of $C_0$ is the closure of $C_c$.

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Your functions $f_n$ are not continuous at $x = \pm n$. If you replace $\frac{1}{1+x^2}$ by $$\frac{1}{1+|x|^2} - \frac{1}{1+n^2} \quad \text{for } |x| \leq n$$ in the definition then you get a sequence of continuous functions that works (even on $\mathbb{R}^k$). Note also that OP knows from previous questions that the space of bounded functions is complete, so it might be worthwhile to point out that the space of functions vanishing at infinity is the closure of $C_c$. – Martin Jan 22 '13 at 12:34
@Martin Thanks, I forgot about the continuity when writing down the example. – mrf Jan 22 '13 at 12:40
Thank you !!!!! – bakabakabaka Jan 22 '13 at 12:51

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