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

Question

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 $m \ge 1$.

=============================

Any help will be appreciated! Thanks!

Answer 1

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.

Answer 2

HINT: any smooth function $u_n$ with compact support approximating $1$ well will be close to $1$ on most of $\Omega$ and close to $0$ elsewhere. This implies that there will always be a region where the approximating function has a big gradient. Try to derive a contradiction by showing that the norm of $1 - u_n$ stays bounded away frow $0$ thanks to the contribution from the gradient part of the norm.