The space of continuous functions from $\mathbb{R}$ to $\mathbb{R}$ Is the space of continuous function from the set of the number $\mathbb{R}$ to $\mathbb{R}$ (usually denoted by $C(\mathbb{R},\mathbb{R})$ is a Banach space?
With the norm $\mathop {\sup }\limits_{x \in \mathbb{R}} \left| {f\left( x \right)} \right|$.
 A: Of course the original version of the question, "is $C(\Bbb R,\Bbb R)$ a Banach space?" makes no sense unless we specify a norm. And of course the supremum is not a norm.
There is a related question that may be interesting: "Does there exist a norm on $C(\Bbb R, \Bbb R)$ making it into a Banach space?" Except that's not the right version of that question - the answer to that question is yes on Axiom of Choice grounds, but the norm we get has nothing to do with the fact that we're talking about continuous functions on $\Bbb R$.
The correct version of the possibly interesting question is this: "Does there exist a norm on $C(\Bbb R, \Bbb R)$ making it into a Banach space, and such that $||f_n||\to0$ if and only if $f_n\to 0$ uniformly on compact sets?"
The answer is no. This must be clear from some theorem in functional
analysis of which I'm unaware, saying that a topological 
vector space can be normed if and only if something. It's easy to show directly:
For $A>0$ say $X_A$ is the space of continuous functions that vanish off $[-A,A]$. Now, uniform convergence on compact sets, restricted to $X_A$, is the same as uniform convergence. So if $||.||$ were a norm as above then the restriction of $||.||$ to $X_A$ would be equivalent to the supremum norm; in particular $$||f||\ge c_A\sup|f(t)|\quad(f\in X_A).$$But now it's easy to construct a function $f$ so that $||f||=\infty$. (For example take $f=\sum f_n$ where $f_n$ is supported in $[2n,2n+1]$, and note that the partial sums converge to $f$ uniformly on compact sets, hence in "norm". But it's easy to give $f_n$ so that the "norms" of the partial sums are unbounded.)
