$C^k_b$ with sup-norm not complete Let $C^n_b=\{ f : I\rightarrow \mathbb{C}: f~n\textrm{-times continuously differentiable and } \|f\|_{n,\infty} < \infty\}$, where $\emptyset\neq I\subseteq\mathbb{R}$ denotes an open interval and $\|f\|_{n,\infty}=\Sigma_{k=0}^n\|f^{(k)}\|_\infty$ be the function space of our concern. Equipped with the supremum norm this should not be complete.
Does anyone have a (classic) example of a Cauchy sequence for proving the noncompletness?
 A: The idea is to consider sequence of continuous functions that converge to the function of class $C([a,b])\setminus C^1([a,b])$. Then one needs to integrate this functions $n$ times and prove that the result is Cauchy sequence without limits in $C^n([a,b])$. It will not have the limit because $n$-th derivatives will converge to non-$C^1([a,b])$ function. 
For simplicity wee consider case $[a,b]=[-1,1]$. Define functions
$$
y_k(t)=\int_0^t\arctan(ks)ds
$$
One can check that
$$
\lim\limits_{k\to\infty}y_k(t)=\frac{\pi}{2}|t|\notin C^1([-1,1])
$$
One can check that $n$-th antideriveative of $y_k$ is
$$
x_k(t)=\int_0^t\arctan(ks)\frac{(t-s)^n}{n!}ds
$$
You can show that
$$
|x_k^{(r)}(t)-x_l^{(r)}(t)|\leq M\int_0^{|t|}|\arctan (ks)-\arctan(ls)|ds\tag{1}
$$
for any $r=0,\ldots,n$ $k,l\in \mathbb{N}$ and $t\in[-1,1]$. Here
$$
M=\max\limits_{r=0,\ldots,n}\max\limits_{(s,t)\in[-1,1]^2}\frac{d}{dt}\frac{(t-s)^m}{m!}
$$
Using $(1)$ you can show that $\{x_n:n\in\mathbb{N}\}\subset C^n([-1,1])$ is a Cauchy sequence. It is not convergent by arguments given above.
