Q: $C([0, 1])$ is not complete with respect to the norm $\lVert f\rVert _1 = \int_0^1 \lvert f (x) \rvert \,dx$ Consider $C([0, 1])$, the linear space of continuous complex-valued functions
on the interval $[0, 1]$, with the norm
$$\displaystyle\lVert f\rVert_1 = \int_0^1 \lvert f(x)\rvert \,dx.$$
I have to show that $C([0, 1])$ is not complete with respect to this norm. I have found the following example from a book.
Let $f_n \in C[0,1]$ be given by
$$f_n(x) := \begin{cases}
 0   & \text{if $0 \le x \le \frac1{2}$}\\
 n(x-\frac{1}{2}) & \text{if $\frac {1}{2} < x \le \frac {1}{2} + \frac {1}{n}$}\\
 1              & \text{if $ \frac {1}{2} + \frac {1}{n} <x \leq 1 $}
\end{cases}$$
How to prove that $f_n$  is a Cauchy sequence with respect to $\lVert \cdot\rVert_1$? 
If I use basic definition then I have to prove that $\lVert f_n - f_m\rVert_1 < \epsilon$ $\forall n, m > N$. But I am finding it difficult to prove this.
Please help me to understand how to prove that $f_n$ is Cauchy sequence in $C([0, 1])$.
Thanks

A: Let $m\leq n$ both natural numbers, then
$$\|f_n-f_m\|_1 = \int_0^1 |f_n(x)-f_m(x)|\,\mathrm{d}x $$
$$ = \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{n}}(n-m)\left(x-\frac{1}{2}\right)\,\mathrm{d}x + \int_{\frac{1}{2}+\frac{1}{n}}^{\frac{1}{2}+\frac{1}{m}}\left(1-m\left(x-\frac{1}{2}\right)\right)\,\mathrm{d}x.$$
Now try to bound these integrals for $n,m\geq N$.

A: Its correct but some part is still left. You have to show that it is not complete.
For this we need to show that limit point is not continuous. Suppose function f(x) is the limit. Observe that it is not right continuous at x=1/2. hence it is not complete
