Consider $(\mathcal{C}([-1,1]), \Vert . \Vert_{\infty})$ (i.e. the set of continuous functions over $[-1,1]$ with the $\sup$-norm).
It is well-known to be a Banach space.
However, I'm stuck in this example, where a sequence that cannot converge seems to be a Cauchy sequence.
Any help in finding out my mistake is thankfully appreciated! :)
Consider the sequence of functions: $$ f_n(x)= \begin{cases} 1, & \text{if $x \in [-1,0]$} \\ 1-nx, & \text{if $x \in (0, \frac{1}{n})$} \\ 0, & \text{if $x \in (\frac{1}{n},1]$} \end{cases} \qquad \in \quad \mathcal{C}([-1,1]) $$ (Each $f_n$ is a composition of two constant functions with the segment line from the point $(0,1)$ to the point $(\frac{1}{n},0)$. Therefore, it is continuous).
We have:
- $\quad f_n \rightarrow f = \chi_{[-1,0]} \quad \text{pointwise as $n \rightarrow \infty$}$
- $\quad f \notin \mathcal{C}([-1,1])$
Therefore, $\{f_n\}$ CANNOT be a Cauchy sequence. Otherwise $\{f_n\}$ should converge to $f$ in $\Vert . \Vert_{\infty}$ - this since $\{f_n\} \in (\mathcal{C}([-1,1]), \Vert . \Vert_{\infty})$ and $(\mathcal{C}([-1,1]), \Vert . \Vert_{\infty})$ is a Banach space.
BUT
$$ \Vert f_{n+k}-f_n \Vert_{\infty} = 1 - \frac{n}{n+k} \quad \xrightarrow{n \rightarrow \infty} \quad 0 \qquad (*) $$ And $(*)$ proves that $\{f_n\}$ is actually a Cauchy sequence.
No, $(f_n)_{n\in\Bbb N}$ is not a Cauchy sequence. To see where your mistake lies, consider the sequence of real numbers $(\log n)_{n\in\Bbb N}$. It diverges, and therefore it is not a Cauchy sequance, right?! But, for any $k\in\Bbb N$,$$\lim_{n\to\infty}|\log(n+k)-\log n|=\lim_{n\to\infty}\log\left(1+\frac kn\right)=0.$$