prove $C_c({\mathbb{R}})$ is not a Banach space I want to prove that

The collection of all continuous complex functions on ${\mathbb{R}}$ whose support is compact is denoted by $C_c({\mathbb{R}})$.  Then the space $(C_c(\mathbb{R}), \lVert\,\cdot\,\rVert_u)$ is not a Banach space. 

Please help me. Thanks
 A: Hint: 
$\mathbb{C}$ is a Banach space so 
$C_b(\mathbb{R})$ is a Banach space. This means that $C_c(\mathbb{R})$ is a Banach space if and only if $C_c(\mathbb{R})$ is closed in $C_b(\mathbb{R})$ but it is not closed because there exists a sequence of continuos functions of compact support that converge to a function who support is not compact.
For example, you can consider the sequence of continuos functions on compact support $\{f_n\}_n$ defined in the following way:
$f_n(x):=\frac{1}{1+x^2}sin(x)\mathbb{1}_{[-2n\pi, 2n\pi]}$ 
This sequence converges to the bounded function $f(x)=\frac{1}{1+x^2}sin(x)$ that has not compact support:
$sup_{x\in \mathbb{R}}|f(x)-f_n(x)|=$
$=sup_{x\in (-\infty, -2n\pi)\cup (2n\pi,\infty)}\frac{1}{1+x^2}|sin(x)|=$
$=\frac{1}{1+(\frac{\pi}{2}+2n\pi)^2}\to_{n\to \infty}0$
A: As pointed out by Federico Falluca, the way to do this is to show that $\newcommand{\R}{\mathbb{R}}C_c(\R)$ is not closed in $C_b(\R)$ (or alternatively in $C_0(\R)$). Here is a simple way to build an explicit example of a function in the closure of $C_c(\R)$ but not in $C_c(\R)$.
For each $a \in \R$, we can define a "spike function" $f_a : \R \rightarrow \R$ by
$$
f_a(x) = 1 - 2| a - x |
$$
for $x \in [a - \frac{1}{2}, a + \frac{1}{2}]$, and
$$
f_a(x) = 0
$$
otherwise. If you draw the graph of this, you'll see that it's continuous with support $[a - \frac{1}{2},a + \frac{1}{2}]$. I'll leave that to you. So $f_a \in C_c(\R)$ for all $a \in \R$.
Define 
$$
f = \sum_{i=1}^\infty 2^{-i}f_{i + \frac{1}{2}}
$$
This sum converges in $\|\cdot\|_u$ (in fact we have "absolute convergence"), and defines a function in $C_0(\R)$ (again - draw the graph of this function, it looks like a sequence of spikes decaying exponentially as they go to the right). Since it is defined by a sum of elements of $C_c(\R)$ it is in the closure of $C_c(\R)$. But $f$ does not have compact support - the support of $f$ contains the points $i + \frac{1}{2}$ for all nonzero natural numbers $i$, so is unbounded.
A: Consider $f(x) = \frac1{1+x^2}$ and
$$
f_n(x)=\begin{cases}
\frac1{1+x^2} &\text{for }|x|<n\\
0 &\text{for }|x|>n+1 \\
linear &\text{otherwise}
\end{cases}
$$
You have $f_n \rightarrow f$, but $f$ does not have compact support.
