Consider the sequence of functions $f_n$ on the unit interval defined as follows. For $x\in[0,\frac{1}{2}-\frac{1}{3n}]$, $f_n(x)=-1$; for $x\in [\frac{1}{2}-\frac{1}{3n},\frac{1}{2}+\frac{1}{3n}]$, $f_n(x)=3n(x-\frac{1}{2})$; for $x\in [\frac{1}{2}+\frac{1}{3n},1]$, $f_n(x)=1$. I'm trying to show that this sequence is Cauchy w.r.t. the norm $||f||=\int_0^1|f(t)|dt$.
Let $n>m$. Then $$0<A:=\frac{1}{2}-\frac{1}{3m}<B:=\frac{1}{2}-\frac{1}{3n}<\frac{1}{2}<C:=\frac{1}{2}+\frac{1}{3n}<D:=\frac{1}{2}+\frac{1}{3m}<1.$$ Further, $\int_0^1|f_n(x)-f_m(x)|dx=\int_{A}^{D}|f_n(x)-f_m(x)|dx$ since $f_n-f_m=0$ outside the interval $[A,D]$. Now
\begin{multline}\int_A^D|f_n(x)-f_m(x)|dx=\int_A^B\left|1+3m\left(x-\frac{1}{2}\right)\right|dx+\\+\int_B^C\left|3n\left(x-\frac{1}{2}\right)-3m\left(x-\frac{1}{2}\right)\right|dx+\int_C^D\left|1-3m\left(x-\frac{1}{2}\right)\right|dx\end{multline} In fact, as I write it, I realize that the previous computation might me unnecessary (but I'll leave it just in case). If we know that $|f_n-f_m|$ is bounded on $[A,D]$ by $M$, we can conclude that $||f_n-f_m||\le\frac{2M}{3m}\rightarrow 0$. Is it true that $|f_n-f_m|$ is bounded? If so, how to see this? If not, how to finish the proof?
The graph of $f_n$ is something like:
Therefore, if $m>n$, the graph of $f_n-f_m$ looks like:
So, $\|f_n-f_m\|$ is twice the area of a triangle whith base $\frac1{3n}-\frac1{3m}$ and whose height is $\frac12$; that is $\|f_m-f_n\|=\frac16\left(\frac1n-\frac1m\right)$. This quantity becames arbitrarily small when $m>n\gg1$.