Consider $C([0, 1])$, the linear space of continuous complex-valued functions
on the interval $[0, 1]$, with the norm
$$\displaystyle\lVert f\rVert_1 = \int_0^1 \lvert f(x)\rvert \,dx.$$
I have to show that $C([0, 1])$ is not complete with respect to this norm. I have found the following example from a book.
Let $f_n \in C[0,1]$ be given by
$$f_n(x) := \begin{cases}
0 & \text{if $0 \le x \le \frac1{2}$}\\
n(x-\frac{1}{2}) & \text{if $\frac {1}{2} < x \le \frac {1}{2} + \frac {1}{n}$}\\
1 & \text{if $ \frac {1}{2} + \frac {1}{n} <x \leq 1 $}
\end{cases}$$
How to prove that $f_n$ is a Cauchy sequence with respect to $\lVert \cdot\rVert_1$?
If I use basic definition then I have to prove that $\lVert f_n – f_m\rVert_1 < \epsilon$ $\forall n, m > N$. But I am finding it difficult to prove this.
Please help me to understand how to prove that $f_n$ is Cauchy sequence in $C([0, 1])$.
Thanks
Best Answer
Let $m\leq n$ both natural numbers, then $$\|f_n-f_m\|_1 = \int_0^1 |f_n(x)-f_m(x)|\,\mathrm{d}x $$ $$ = \int_\frac{1}{2}^{\frac{1}{2}+\frac{1}{n}}(n-m)\left(x-\frac{1}{2}\right)\,\mathrm{d}x + \int_{\frac{1}{2}+\frac{1}{n}}^{\frac{1}{2}+\frac{1}{m}}\left(1-m\left(x-\frac{1}{2}\right)\right)\,\mathrm{d}x.$$
Now try to bound these integrals for $n,m\geq N$.