Show that the function $u(x)=1 \in W^{m,\ 2}(\Omega)$, but not in $W_0^{m,\ 2}(\Omega)$ I have a problem:

For $\Omega$ be a domain in $\Bbb R^n$. Show that the function $u(x)=1 \in W^{m,\ 2}(\Omega)$, but not in $W_0^{m,\ 2}(\Omega)$, for all $m \ge 1$.

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Any help will be appreciated! Thanks!

First I would like to answer the first comment : It is only if you take the whole $\mathbb{R}^n$ that this two spaces coincide not at all for bounded domain. 
For the problem : I will do it on $\mathbb{R}$ and $m=1$ and $\Omega=(0,1)$ (easy to adapt). To show that $u$ belongs to $W^{m,2}(\Omega)$ you only need to show that the integration by parts formula (against function in $\mathcal{C}^\infty_c(\Omega)$) holds. Or just notice that $u\in \mathcal{C}^{\infty}(\Omega)$ theferore the classic (then the weak) derivatives exist and are $0$.
Now $W^{m,2}_0(\Omega)$ is by definition the closure of $\mathcal{C}^\infty_c(\Omega)$ with respect to the norm of $W^{m,2}(\Omega)$ :
$$\|\phi\|_{W^{1,2}(\Omega)}=\int|\phi|^2+\int|\phi^{\prime}|^2.$$ 
Now assume that $u$ belongs to $W^{m,2}_0(\Omega)$ then there exists a sequence $\phi_n$ in $\mathcal{C}^\infty_c(\Omega)$ such that. $\|\phi_n-u\|_{W^{2,1}(\Omega)}\rightarrow 0$ which by definition imply the followings
$$\int|\phi_n-1|^2\rightarrow 0\\ \int|\phi_n^{\prime}|^2\rightarrow 0\\
\phi_n(0)=\phi_n(1)=0$$ and   integration by 
parts of the quantity $\int(\phi_n-1)$ leads easily to a contradiction. Generally the only constant function which belongs to this space is $0$.
Roughly speaking $W^{m,2}_0(\Omega)$ are functions (with appropriate weak derivatives)  which vanishes in some sense at the boundary (rigorously the trace sense) and this is not the case for $u=1$. 
Note : if you are already familiar with the boundary trace of function in sobolev spaces it is obvious : $W^{m,p}_0(\Omega)$ are the function of $W^{m,p}(\Omega)$ for which the trace at the boundary is 0. But since $u$ is continuous the trace of $u$ on $\partial\Omega$ is only $u_{|_{\partial\Omega}}=1\neq 0$ 
and it is over.
