Showing a Cauchy sequence does not have a limit 
Okay basically I've managed to work through parts 1 a and b (with some help) now I'm a little stuck on part c). I think I can show fn is a cauchy sequence by virtue of the fact that fn-f tends to 0, but I'm stumped as to how I can then prove it does not have a limit under the norm.
Any help would be appreciated.
 A: *

*To show $\| \cdot \|$ is a norm on $\mathscr{C}\big( [0,2] ;  \mathbb{R}  \big)$ :


Clearly, for any $f\in \mathscr{C}\big( [0,2] ;  \mathbb{R}  \big)$, $\| f \|$ is nonnegative. 
Since for $x\in[0,2]$, $x(2-x)\vert f(x) \vert \geq 0$, then $\| f\| = 0$ implies 
 $$ 0 = \int_0^2 x(2-x) \big\vert f(x) \big\vert\, dx \Longrightarrow f \equiv 0 $$
that is, $\| f\| = 0$ if and only if $f = 0$. 
For any $\lambda\in\mathbb{R}$, 
 $$ \big\| \lambda f \big\| = \int_0^2 x(2-x) \big\vert \lambda f(x) \big\vert\, dx = \vert \lambda \vert  \int_0^2 x(2-x) \big\vert f(x) \big\vert\, dx = \vert \lambda \vert \cdot \big\| f\big\| .$$
It remains to verify the triangle inequality:  given $f, g\in \mathscr{C}\big( [0,2] ;  \mathbb{R}  \big)$,
 $$\big\| f + g \big\| =  \int_0^2 x(2-x) \big\vert f(x) + g(x) \big\vert\, dx \leq \int_0^2 x(2-x)\Big[  \big\vert f(x) \big\vert + \big\vert g(x) \big\vert \Big]\, dx = \int_0^2 x(2-x) \big\vert f(x)  \big\vert\, dx +  \int_0^2 x(2-x) \big\vert g(x) \big\vert\, dx = \big\| f\big\| + \big\| g \big\|$$
Hence $\| \cdot \|$ is a norm on $ \mathscr{C}\big( [0,2] ;  \mathbb{R}  \big)$.
The second question is to show that $\Big( \mathscr{C}\big( [0,2] ;  \mathbb{R}  \big) , \| \cdot \| \Big)$ is not complete.
Question(a) is simple: the pointwise limit is 
\begin{align}
 f(x) = \begin{cases}
 0\,\, ,\quad\text{if $x\in[0,1)$} ;\\
 1\,\, ,\quad\text{if $x\in[1,2]$} .
\end{cases}
\end{align}
$f$ is Riemann-integrable, since it has only one jump.(For details, you can see Rudin's Principles of mathematical analysis.)
Question(b)
$$\big\| f_n - f\big\| =  \int_0^2 x(2-x) \big\vert f_n(x) - f(x) \big\vert\, dx = \int_0^1 x(2-x) x^n = \frac{2}{n+2} - \frac{1}{n+3}\xrightarrow{n\to+\infty} 0 \,\, . $$
Question (c)
It follows from  question (b) that the sequence $f_n$ in $ \mathscr{C}\big( [0,2] ;  \mathbb{R}  \big) $ is Cauchy, that is, 
 $$\lim_{n,m\to+\infty} \big\| f_n - f_m \big\| = 0  \,\, .$$
If $ \mathscr{C}\big( [0,2] ;  \mathbb{R}  \big) $ is a complete space, then the sequence is convergent to some $\widehat{f}\in  \mathscr{C}\big( [0,2] ;  \mathbb{R}  \big)$, but due to the uniqueness of limit in a metric space, we have $f = \widehat{f}$. But $f$ is not continuous hence not in $ \mathscr{C}\big( [0,2] ;  \mathbb{R}  \big) $, leading to a contradiction. We conclude that this normed space is not complete.  
Q.E.D.
