$C[0,1]$ with $L^1$ norm is not Banach space.

Question

I want to check that $(C[0,1],∥⋅∥_1)$ is not a Banach space, where $\|f\|_1 = \int_0^1 |f(x)|\,{\rm d}x$.I took $(f_n)_{n \geq 1}$ a sequence in $C[0,1]$ given by:$f_n: [0,1] \rightarrow \mathbb{R}, \ x\rightarrow \begin{cases} 0 & \text{, $\ 0 ≤ x ≤ \frac{1}{2}$}\\[2ex] (nx-n/2) &\text{, $\frac{1}{2} < x ≤ \frac{1}{2} + \frac{1}{n+1}$} \\[2ex] 1 & \text{, $\frac{1}{2} + \frac{1}{n+1} < x ≤ 1$} \end{cases}$
I want to show that $f_n$ is Cauchy but also that the limit function $f$ is not continuous and hence $∉C([0,1])$. The second part is obvious, since $f$ is not continous at $1/2$. What I tried is that: Let $n≥m:$ $\begin{align} \|f_n - f_m\| &= \int_0^1 | f_n(x) - f_m(x) | \ dx \\ &= \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{n+1}} | f_n(x) - f_m(x) | \ dx + \int_{\frac{1}{2}+\frac{1}{n+1}}^{\frac{1}{2}+\frac{1}{m+1}} | f_n(x) - f_m(x) | \ dx\\ &= \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{n+1}} | (m-n) |\left(x-\frac{1}{2}\right) \ dx + \int_{\frac{1}{2}+\frac{1}{n+1}}^{\frac{1}{2}+\frac{1}{m+1}} (1-mx+m/2) )\ dx \end{align}$
I'd appreciate any help.

Answer 1

It's easier than you think, remember that in order to prove that a sequence is Caushy, you don't need the exact distance between two terms, you just need a bound.

You're integrating $|f_n(x) - f_m(x)|$ from $1/2$ to $1/2 + \max\left(\frac{1}{n+1}, \frac{1}{m+1}\right)$. We have that the integrand is $$ |f_n(x) - f_m(x)| \leq |f_n(x)|+|f_m(x)| \leq 1+1 \leq 2 $$ and if you let $m, n \geq N$ for some natural $N$, then the interval you're integrating over is shorter than $1/N$, which means that the result is smaller than $2/N$. Now, given any $\epsilon > 0$, pick an $N\in \Bbb N$ such that $2/N < \epsilon$. The rest should do itself.