Problem
Consider the space $C[-1,1]$, together with the norm defined by $\|f\|_1 = \int_{-1}^1|f|d\lambda$ (where $\lambda$ is the Lebesgue measure). For each $n$ define a function $f_n:[-1,1]\to\mathbb{R}$ by\begin{align*} f_n(x)= \begin{cases} 0\quad &\text{if $-1\leq x\leq 0$},\\ nx\quad &\text{if $0<x\leq\frac{1}{n}$},\\ 1\quad &\text{if $\frac{1}{n}<x\leq1$}. \end{cases}\end{align*}Prove that $\{f_n\}$ is a Cauchy sequence in $C[-1,1]$. Then show there is now continuous function $f$ such that $\lim_{n\to\infty}\|f_n-f\|_1=0$.
My Attempt So Far
- Here is my attempt so far to prove the sequence $\{f_n\}$ is Cauchy:
Let $\epsilon$ be any positive number. We want to show that there is a positive integer $N$ such that $d(f_m,f_n)<\epsilon$ whenever $m\geq N$ and $n\geq N$. Write\begin{align}d(f_m,f_n) &= \|f_m-f_n\|_1\\&= \int_{-1}^1|f_m-f_n|d\lambda\\&= \int|f_m-f_n|\chi_{[a,b]}d\lambda\\&= \sup\left\{\int hd\lambda:h\in\mathscr{S}_+\ \text{and}\ h\leq|f_m-f_n|\chi_{[a,b]}\right\}\\&\dots\end{align}
I got stuck here, don't know how to proceed next.
- For the nonexistence of a continuous function, I drew a diagram. I can see that the function would eventually goes to\begin{align*}f(x) = \begin{cases}0\quad &\text{if $-1\leq x\leq0$},\\1\quad &\text{if $0<x\leq1$}.\end{cases}\end{align*}which is clearly not continuous. But I don't know how to prove it rigorously.
Could someone please help me out? Thanks a lot in advance!