(more) how to prove $C[-1,1]$ with $L^1$-norm is not complete I am studying "real analysis and applications :theory in practice" I have some question in this books'problem. so if you know following ploblem, please teach me ! the ploblem is as follows
excercises for section 7.2 I with N
consider the piecewise linear fuctions in $C[-1,1]$ given by $f_n(x)=0$ for $-1\leq x\leq 0$, $f_n(x)=nx$ for $0\leq x\leq \frac1n$, and $f_n(x)=1$ for $\frac1n\leq x\leq 1$ with $L^1[-1,1]$ norm
(a) show that $f_n$ is Cauchy in the $L^1$-norm (already clear for your help)
(b) show that $f_n$ converges $\chi_\left(0,1\right]$, the characteric function of (0,1], in the $L^1-norm$
(c) show that $\Vert \chi_\left(0,1\right]-h\Vert_1 \gt 0$ for every h in C[-1,1]
I solved above 3 question but problem is as follows
(d) conclude that C[-1,1] is not complete in the $L^1-norm$
this is what i was trying to solve.
(1), what we need to prove this problem suffice to show there is cauchy sequence whose limit is not in C[-1,1].
(2)then suppose (a),(b) is true, (a),(b) said if we choose piecewise function like above fnfn. it converges to a vector $\chi_\left(0,1\right]$ but $\chi_\left(0,1\right]$ is not in C[-1,1] because $\chi_\left(0,1\right]$ is not continuous function on [-1,1].
(3)thus C[-1,1] is not complete.
is there any faults without using problem (c)?
 A: Let $m > n \in \mathbb{N}$. You can calculate $\|f_m-f_n\|_1$ as the area of the shaded triangle with height $1$ and side length $\frac{1}{n} - \frac{1}{m}$.
$$\|f_m-f_n\|_1 = \frac{1}{2n} - \frac{1}{2m} \xrightarrow{m, n \to \infty} 0$$

Your proof of the $(b)$ part is essentially correct. Notice that you don't have to look at $[-1,0]$ and $(0, 1]$ separately, since you can do the separation directly in the integral, exactly like you did in your case $(2)$. And you have a typo: $\chi_{(-1,1]}$ in the integral should be $\chi_{(0,1]}$.
Again, we can use geometry: $\|f_n - \chi_{(0, 1]}\|_1$ is the area of the shaded triangle with side length $\frac{1}{n}$ and height $1$ so:
$$\|f_n - \chi_{(0, 1]}\|_1 = \frac{1}{2n} \xrightarrow{n\to\infty}0$$

A: If $0<m\le n$ we have that
$$\|f_n-f_m\|_1=\int_{-1}^1 |f_n(x)-f_m(x)|dx=\int_{0}^{1/n} (nx-mx)dx+\int_{1/n}^{1/m} (1-mx)dx.$$ That is
$$\|f_n-f_m\|_1=\dfrac{n-m}{2mn}=\dfrac{1}{2m}-\dfrac{1}{2n}<\dfrac{1}{2m}.$$
Now, for any $\epsilon>0$ there exists $N\in\mathbb{N}$ ($N>\dfrac{1}{2\epsilon}$) such that $n,m\ge N\implies \|f_n-f_m\|_1<\epsilon.$ This shows that $(f_n)$ is a Cauchy sequence.
