Showing that $f_n$ is Cauchy in $C[-1,1]$, but it doesn't converge 
Equip $C[-1,1]$ with $L^1$ norm. Define $$f_n(x)= \begin{cases}0, -1\le x \le0 \\nx, 0\le x \le \frac1n \\ 1, \frac1n \le x \le 1 \end{cases}$$ Show that $f_n$ is Cauchy, but that it doesn't converge.

To show that $(f_n)$ is Cauchy I've computed $$||f_n-f_m\|_1 = \int_{0}^{\frac1n}|f_n(x)-f_m(x)| \ dx + \int_{\frac1n}^{1}|f_n(x)-f_m(x)| \ dx = \frac{|n-m|}{2|n|}$$ so in order to show that $\|f_n-f_m\|_1 < \varepsilon$, when $n,m \ge K \in \mathbb{N}$ I would need to bound $\frac{|n-m|}{2|n|}$? I have that $\frac{|n-m|}{2|n|} = \frac{n+m}{2|n|} \le \frac{n+m}{2} = \frac{n}{2}+\frac{m}{2}$. So picking $K$ such that $\frac{n}{2} < \frac{\varepsilon}{2}$ and $\frac{m}{2} < \frac{\varepsilon}{2}$ I get that $$\|f_n-f_m\|_1 = \frac{|n-m|}{2|n|} = \frac{n+m}{2|n|}< \frac{n+m}{2} = \frac{n}{2} + \frac{m}{2} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$ so $(f_n)$ is Cauchy? How can I show that it doesn't converge e.g $C[-1,1]$ isn't complete?
 A: If $m\ge n$, then $f_m$ and $f_n$ agree in $[-1,0]\cup [1/n,1]$, while
$$
0\le f_n(x)\le f_m(x)\le 1, \quad x\in [0,1/n].
$$
Thus
$$
\int_{-1}^1 |f_m(x)-f_n(x)|\,dx=\int_{0}^{1/n} \big(f_m(x)-f_n(x)\big)\,dx\le \frac{1}{n}.
$$
Hence $\{f_n\}$ is a Cauchy sequence, with respect to the $\|\cdot\|_{L^1}-$norm.
However, $\{f_n\}$ does not converge to an $f\in C[-1,1]$.
Assume it did. Say $f_n\to f\in C[-1,1]$. We shall show that $f(x)=0$, if $x<0$ and $f(x)=1$, if $x>0$, and hence, such continuous $f$ does not exist.
Define
$$
\varphi_n(x)=\left\{
\begin{array}{ccc}
0 & \text{if}& x<-1/n \\
n^2x+n &\text{if}& -1/n\le x\le  0 \\
n-n^2x &\text{if}& 0\le x\le 1/n \\
0 & \text{if}& x >1/n.
\end{array}
\right.
$$
Then $\|\varphi_n\|_{L^1}=1$ and
$$
\lim_{n\to\infty}\int_{-1}^1 f(x)\varphi_n(x-t)\,dx=f(t),
$$
for all $t\in (-1,1)$.
In particular, if $\,x_0>0,\,$ then $f_m(x_0)=1$, for $m>1/x_0$, and hence
$$
\lim_{n\to\infty}\int_{-1}^1 \varphi_{n}(x-x_0)f_m(x)\,dx=f_m(x_0)=1.
$$
But
$$
f(x_0)=\lim_{n\to\infty}\int_{-1}^1 \varphi_{n}(x-x_0)f(x)\,dx \\
=\lim_{n\to\infty}\int_{-1}^1 \varphi_{n}(x-x_0)\big(f(x)-f_m(x)\big)\,dx+
\lim_{n\to\infty}\int_{-1}^1 \varphi_{n}(x-x_0)f_m(x)\,dx \\
\lim_{n\to\infty}\int_{-1}^1 \varphi_{n}(x-x_0)\big(f(x)-f_m(x)\big)\,dx+1,
$$
and
$$
\int_{-1}^1 \varphi_{n}(x-x_0)\big(f(x)-f_m(x)\big)\,dx=\int_{x_0-1/n}^{x_0+1/n}\varphi_{n}(x-x_0)\big(f(x)-f_m(x)\big)\,dx.
$$
But $f_m(x)=f(x)$, for $x\in (x_0-1/n,x_0+1/n)$, if $x_0-1/n>1/m$ or equivalently $n> (x_0-1/m)^{-1}$, and thus
$$
\lim_{n\to\infty}\int_{x_0-1/n}^{x_0+1/n}\varphi_{n}(x-x_0)\big(f(x)-f_m(x)\big)\,dx=0,
$$
which means that $f(x_0)=1$.
We can similarly prove that $f(x_0)=0$, for all $x_0<0$.
Thus $f$ can not be continuous at $0$.
