How do I show this function sequence is Cauchy in $C([0,1])$ with the norm $||f||_{L^1} = \int_0^1|f(x)|$ dx?

Question

The function is the following:

\begin{equation} f_n(x) = \begin{cases} 0 & \quad\text{if} \ \ 0 \leq x \leq \frac{1}{2} - \frac{1}{n} \\ nx + 1 - \frac{n}{2} & \quad\text{if} \ \ \frac{1}{2} - \frac{1}{n} \leq x \leq \frac{1}{2} \\ 1 & \quad\text{if} \ \ \frac{1}{2} \leq x \leq 1 \end{cases} \end{equation}

I know that I have to prove that, given $\varepsilon > 0$, $\exists N \in \mathbb{N}$ such that $||f_m - f_n|| < \varepsilon, \ \forall m, n \geq N$.

I have tried to do develop the norm

\begin{equation} ||f_m - f_n|| = \int_0^1 |f_m(x) - f_n(x)| \ dx \end{equation}

but I think this will just make things more complicated...

What would be the right approach?

Answer 1

Fix $N$. If $n,m \ge N$ then $f_n$ and $f_m$ both vanish on the interval $[0,\frac 12 - \frac 1N]$ and both equal $1$ on $[\frac 12,1]$. Thus $$ n,m \ge N \implies \int_0^1 |f_n(x) - f_m(x)| = \int_{1/2-1/N}^{1/2} |f_n(x) - f_m(x)| \, dx \le \frac 2N$$ because $|f_n(x)| \le 1$ and $|f_m(x)| \le 1$ for all $x$.