Pointwise limit and riemann integral 
I'm fine with defining the norm, but the second part and third part has stumped me. I've tried inputting fn(x) and rearranging the integral but it doesn't lead anywhere.
Any help would be appreciated.
 A: You are able to show it defines a norm. We are now pondering whether or not the space of continuous functions on $[0,2]$ is complete under this norm. We will show it is not by giving a sequence of continuous functions that converge to a non-continuous function under this norm.
We start with $f_n(x) = \begin{cases} x^n & 0 \leq x \leq 1 \\ 1 & 1 \leq x \leq 2 \end{cases},$ and we want to find its pointwise limit. Clearly $f_n(x) \to 1$ for $1 \leq x \leq 2$, as that's what it always is. For $0 \leq x < 1$, since we know that $x^n \to 0$, we see that $f_n(x) \to 0$ there. So $f_n(x) \to f(x)$, where $f(x)$ is the indicator function on $[1,2]$ (i.e. it's $0$ on $[0,1)$ and $1$ on $[1,2]$.)
What is $f_n(x) - f(x)$? It's $0$ on $[1,2]$, so when we're integrating, we can ignore that part. So 
$$||f_n - f|| = \int_0^1x(2-x)|x^n - 0|\mathrm{d}x = \int_0^12x^{n+1} - x^{n+2} \mathrm{d}x \to 0$$
as $n \to \infty$. Thus $f_n \to f$ in this norm. But $f$ is not continuous. Thus the space is not complete under this norm.
