Checking that $(C[0,1], \|\cdot\|_1)$ is not Banach. I want to check that $(C[0, 1], \|\cdot\|_1)$ is not a Banach space, where $$\|f\|_1 = \int_0^1 |f(x)|\,{\rm d}x.$$
I took $(f_n)_{n \geq 1}$ a sequence in $C[0, 1]$ given by: $$f_n(x) = \begin{cases} nx, &\text{ if } x < \frac{1}{n} \\ 1, &\text{ if } x \geq \frac{1}{n} \end{cases} $$
If $m > n$, I computed: $$\|f_m - f_n\|_1 = 1 - \frac{n+1}{m} + \frac{1}{n} + \frac{n}{2}\left( \frac{1}{m^2}-\frac{1}{n^2} \right)$$
It is intuitive that the sequence is Cauchy, but I don't know how to exactly formalize this using the definition of a Cauchy sequence. Sending $m,n$ to $+\infty $ at the same time intuitively gives zero, but it doesn't seem rigorous to me.

How can I formalize this $m,n \to \infty$?

To proving that the sequence does not converge also seems intuitive, we would get a sort of vertical line - not a function. But I'm also not sure of how to approach this - the only thing that comes to mind is some kind of contradiction - suppose that $f_n \to f$, and find $\epsilon > 0 $ such that $\|f_n - f\| > \epsilon $ for infinite values of $n$.

Is the above idea in the right track?

Thanks.
 A: I think you are making your estimate too convoluted, probably because you try to calculate $\|f_n-f_m\|_1$ exactly. What I would do: assume $m>n$; then $f_m=f_n=1$ for $x>1/n$, and so 
$$
\|f_m-f_n\|_1=\int_0^{1/n}|f_m-f_n|\leq\frac2n
$$
since $|f_n|\leq1$, $|f_m|\leq1$. This shows that the sequence is Cauchy, because given $\varepsilon>0$, if $m>n>2/\varepsilon$, then $\|f_m-f_n\|_1<\varepsilon$.
Edit: The problem with your example is that the limit is in C$[0,1]$: it is the constant function $1$. Indeed,
$$
\|1-f_n\|_1=\int_0^{1/n}(1-nx)\,dx=\frac1n-\left.\frac{nx^2}2\right|_0^{1/n}=\frac1{2n}\to0.
$$
Edit2: To achieve what you want, an $\|\cdot\|_1$-Cauchy sequence that does not converge in $C[0,1]$, you can try something like
$$
f_n(t)=\begin{cases}0,&\ 0\leq t\leq 1/2-1/2n,\\ nt\,&\ 1/2-1/2n\leq t\leq 1/2+1/2n,\\ 1,&\ 1/2+1/2n\leq t\leq 1
\end{cases}
$$
A: Hint: consider $m>n, m = n + h$, and prove that the expression converges to $0$ at a speed independent of $h$. In other words: prove that
$$
\sup_{h>0} \left\|
f_{n+h} -  f_n
\right\| \to 0
$$when $n\to\infty$.
