Completeness of continuous functions space under $L^{1}$ norm Prove that the space of continuous functions $C([0,1])$ is not complete under the norm
$\|\cdot\|_{L^{1}([0,1])}$.
I have tried the following sequence of functions
$f_{n}(x)=\left\{\begin{array}{cl}1 & \text { if } & 0 \leq x \leqslant \frac{1}{2} \\ 1-n\left(x-\frac{1}{2}\right) & \text { if } \frac{1}{2} \leqslant x \leqslant \frac{1}{2}+\frac{1}{n} \\ 0, & \text { if } \frac{1}{2}+\frac{1}{n} \leqslant x \leqslant 1\end{array}\right.$
I got
$\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}\left|f_{n}(x)-f_{m}(x)\right| d x .$
$=\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}\left|\left(1-n\left(x-\frac{1}{2}\right)\right)-\left(1-m\left(x-\frac{1}{2}\right)\right)\right| d x=\left|\frac{m-n}{2 n^{2}}\right|$
Is the last term bounded by $\frac{1}{n}$? why? and is it ok that my integration limits deals with $n$ and not also $m$?
As last step I thought of taking limit of n to $\infty$ and since 2 limits are obtained - $0,1$ this space is not complete.

Assuming $m<n$, we first show $(f_n)$ is cauchy's sequence.
$\|f_n-f_m\|= \int_0^1 |f_n-f_m|dx$
RHS is equal to
$\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}|m(x-\frac{1}{2})-n(x-\frac{1}{2})|dx+$
$\int_{\frac{1}{2}+\frac{1}{n}}^{\frac{1}{2}+\frac{1}{m}}|1-m(x-\frac{1}{2})|$
Try to simplify, if needed you can use $\frac{n-m}{2n^2} \leq \frac{n+m}{2n^2} <\frac{n+n}{2n^2}=\frac{1}{n}$.
Once you prove it is cauchy's, you can easily show it does not coverge in given space.
