Completeness of $\langle \mathscr{C} [0, 1], \| \cdot \|_1 \rangle$ That's really embarrassing, however I need to ask it. I could not prove that the normed space    $\langle \mathscr{C} [0, 1], \| \cdot \|_1 \rangle $ is complete (as a metric space), where $\| f\|_1 = \int_0^1 |f(x)|dx$. I don't really know how to explicit a continuous functions as a limit of a Cauchy sequence. I tried to prove that the pointwise limit (of the Cauchy sequence) have just a finite number of discontinuities (I don't know if it's true) and then approximate it by a continuous function.
Thanks in advance.

This space is not complete. Consider functions
$$
f_n(t)=\begin{cases}0\qquad\qquad\qquad\qquad t\in\left(0,\frac{1}{2} -\frac{1}{2n}\right)\\\frac{1}{2}+n\left(t-\frac{1}{2}\right)\qquad t\in\left[\frac{1}{2}-\frac{1}{2n},\frac{1}{2}+\frac{1}{2n}\right]\\1\qquad\qquad\qquad\qquad t\in\left(\frac{1}{2}+\frac{1}{2n},1\right)\end{cases}
$$
It is easy to check that $\{f_n\}_{n=1}^\infty\subset \mathscr{C}([0,1])$. Moreover this is a Cauchy sequence. Indeed
$$
\lim\limits_{N\to\infty}\sup\limits_{m,n>N}\Vert f_n-f_m\Vert_1
=\lim\limits_{N\to\infty}\sup\limits_{m,n>N}2^{-2}|n^{-1}-m^{-1}|
\leq\lim\limits_{N\to\infty}2^{-1}N^{-1}=0
$$
Assume that there exist $f\in \mathscr{C}([0,1])$ such that $\lim\limits_{n\to\infty}\Vert f_n-f\Vert_1=0$.
Assume that there exist $t_0\in\left(\frac{1}{2},1\right)$ such that $f(t_0)\neq 1$. Since $f$ is continuous then there exists a neighborhood $U_\delta(t_0)$ such that $|f(t)-f(t_0)|\leq\frac{|1-f(t_0)|}{2}$ for all $t\in U_\delta(t_0)$. In particular, by reverse triangle inequality we have
$$
\begin{align}
|1-f(t)|
&=|(1-f(t_0))-(f(t)-f(t_0))|\\
&\geq\left||1-f(t_0)|-|f(t)-f(t_0)|\right|\\
&\geq\frac{|1-f(t_0)|}{2}.
\end{align}
$$
Now take $\delta'=\min\left(\delta,1-t_0,t_0-\frac{1}{2}\right)$, so we can assume $U_{\delta'}(t_0)\subset \left(\frac{1}{2},1\right)\cap U_\delta(t_0)$. Pick any natural number $N>\frac{1}{2(t_0-\delta')-1}$, then for all $n>N$ we have $U_{\delta'}(t_0)\subset \left(\frac{1}{2}+\frac{1}{2n},1\right)$. Note that for all $n>N$ and $t\in U_{\delta'}(t_0)$ we have $f_n(t)=1$, so
$$
\Vert f_n-f\Vert_1=
\int\limits_{(0,1)}|f_n(t)-f(t)|dt\geq
\int\limits_{(t_0-\delta',t_0+\delta')}|f_n(t)-f(t)|dt=
$$
$$
\int\limits_{(t_0-\delta',t_0+\delta')}|1-f(t)|dt\geq
\int\limits_{(t_0-\delta',t_0+\delta')}\frac{|1-f(t_0)|}{2}dt
=\delta'|1-f(t_0)|>0
$$
and as the consequence
$$
0=\lim\limits_{n\to\infty}\Vert f_n-f\Vert_1\geq\delta'|1-f(t_0)|>0
$$
Contradiction, therefore $f(t_0)=1$ for all $t_0\in\left(\frac{1}{2},1\right)$. Similarly one can show that for all $t_0\in\left(0,\frac{1}{2}\right)$ we also have $f(t_0)=0$. Now note that
$$
\lim\limits_{t\to 1/2-0}f(t)=0\qquad\qquad\lim\limits_{t\to 1/2+0}f(t)=1
$$
so $f$ is not continuous. Contradiction, hence $\mathscr{C}([0,1])$ is not complete.
