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