$C^\infty(R^n)$ is a Banach Space when equipped with topology of uniform convergence Prove $C^\infty(\Bbb R^n)$ is a Banach Space when equipped with topology of uniform convergence.
$C^\infty(\Bbb R^n)$ is space of all continuous functions that converge to $0$ at $\infty$.
And, the topology of uniform convergence is defined by the norm
$f \mapsto \sup_{x\in\Bbb  R^n}|f(x)|$. 
 A: The statement is false. Norms have to be finite, but if we let $n=1$ and take $$f(x)=\begin{cases} 1/x & x \neq 0 \\ 0 & x=0\end{cases}$$ then $f$ is in your space but has infinite norm.
Norms also have to be positive, but if we take $f(x)=-1$ then the norm is $-1$.
A: It's not difficult to show $\|f\|=\sup_{x\in \mathbb{R}^n}|f(x)|$ is a norm. We need to show the space is complete.
Suppose $\{f_n\}_{n=1}^\infty$ is a Cauchy sequence in the space $i.e.\ \|f_n-f_m\|\to0$ as $m,n\to\infty$. For any fixed $x\in \mathbb{R}^n,\ \{f_n(x)\}$ is Cauchy in $\mathbb{R}$(I hope I didn't misunderstand the map $f: \mathbb{R}^n\to\mathbb{R}$).
Define $f(x):=\lim_{n\to\infty}f(x)$, this is a uniform convergence $i.e. \|f_n-f\|\to 0$ as $n\to\infty$. Hence, $f$ is continuous and moreover,
$$ |f(x)|\le |f_n(x)-f(x)|+|f_n(x)|\le \|f_n-f\|+|f_n(x)|
$$
$\forall \epsilon>0,\ \exists\ M>0\ s.t.\ \|f_n-f\|\le\epsilon$ when $n>M$. For a fixed $n>M,\ \exists\ N>0,\ s.t.|f_n(x)|<\epsilon$ if $|x|>N$. Therefore, there is an $N$ such that for any $|x|>N$
$$ |f(x)|\le \|f_n-f\|+|f_n(x)|<2\epsilon\  \Rightarrow\ f\ converges\ to\ 0\ at\ \infty
$$
So $f$ is an element in your space.
