An example of a non-convergent Cauchy sequence in $C^\infty$ with the Sobolev norm I apologize in advance if this question is a duplicate, but I couldn't find an answer.
Studying Sobolev Spaces, I came across with the following proposition:

The Sobolev Space $H^m(\Omega)$ is the completion, with respect to the Sobolev norm $\|\cdot\|_{H^m}$, of the space $C^{\infty}(\overline{\Omega}).$

Now I'm trying to find a non convergent Cauchy sequence in $C^{\infty}(\overline{\Omega})$ with the norm $\|\cdot\|_{H^m}$ but without success.
Can anyone give me some suggestion/hint?
Thanks.

Since the dimension of $\Omega$ was not specified, let's use a one-dimensional domain, $(-1, 1)$. Given $m$, let $k$ be an odd integer greater than $2m$. Consider the sequence of $C^\infty$ smooth functions
$f_n(x) = (x^2+ 1/n)^{k/2}$.  It converges to $|x|^k$ which is not $C^\infty$ smooth, but is $C^m$ smooth. Moreover, the derivatives of $f_n$ of orders $0, \dots, m$ converge uniformly to the corresponding derivatives of $f$, which implies $\|f_n-f\|_{H^m}\to 0$, hence $\{f_n\}$ is a Cauchy sequence with respect to the $H^m$ norm. 
Let's justify the claim about the convergence of derivatives. By induction, 
$$
f_n^{(j)}(x) = P_{j}(x, 1/n)  (x^2+ 1/n)^{k/2-j},\quad j=0, \dots, m
$$
where $P_{j}$ is some polynomial of two variables. When $n\to \infty$, we have $P_{j}(x, 1/n)\to P_j(x, 0)$ and $(x^2+ 1/n)^{k/2-j} \to |x|^{k-2j}$ uniformly on $\Omega$.

Essentially the same example, $(|x|^2 + 1/n)^{k/2}$, can be used in higher dimensions.
