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I'd like to have some hints for a problem I bumped into some times ago but I was not able to solve (even if I think the most is done...).


Before the problem, let me recall some definitions. Let $\Omega \subset \mathbb{R}^N$ be a nonempty open set (n.o.s., for short) and consider the Sobolev space $W^{1,2}(\Omega)$ (which is Hilbert's with its natural inner product $\langle u,v\rangle := \int_\Omega u\ v+\nabla u\cdot \nabla v$).

$\Omega$ is called an extension domain (for $W^{1,2}$) iff there exists a function $P:W^{1,2}(\Omega) \to W^{1,2}(\mathbb{R}^N)$ (usually called extension operator) s.t.:

  1. $P$ is linear;
  2. $Pu\Big|_\Omega = u$ a.e. in $\Omega$ for all $u\in W^{1,2}(\Omega)$;
  3. there exists a constant $C=C(\Omega)$ s.t. $\lVert Pu\rVert_{L^2(\mathbb{R}^N)} \leq C\ \lVert u\rVert_{L^2(\Omega)}$ and $\lVert Pu\rVert_{W^{1,2}(\mathbb{R}^N)} \leq C\ \lVert u\rVert_{W^{1,2}(\Omega)}$ for all $u\in W^{1,2}(\Omega)$.

For example, if $\Omega$ is a n.o.s. and $\partial \Omega$ is sufficiently smooth (e.g. Lipschitz) then $\Omega$ is an extension domain (cfr. Brezis, 9.2).


And here we go.

Let $\Omega \subset \mathbb{R}^N$ be a n.o.s. and assume that for each $u\in W^{1,2}(\Omega)$ there exist some $U\in W^{1,2}(\mathbb{R}^N)$ s.t. $U\Big|_\Omega =u$ a.e.

Prove that $\Omega$ is an extension domain.

Moreover, the following hint was given in the lecture:

Use the Hilbert space structure of $W^{1,2}$.

Actually, the problem here lies in finding a good candidate to be the extension operator $P: W^{1,2}(\Omega) \to W^{1,2}(\mathbb{R}^N)$. In order to find such a $P$, I thought to proceed as follows.

For each $u\in W^{1,2}(\Omega)$ I set: $$K_u := \left\{ U \in W^{1,2}(\mathbb{R}^N) |\quad U\Big|_\Omega =u \text{ a.e.}\right\} \; ;$$ it is easily seen that $K_u$ is a nonempty closed convex set in $W^{1,2}(\mathbb{R}^N)$, hence there exists a unique $\tilde{u} \in K_u$ having minimal norm and it is characterized by the variational inequality: $$\tag{V}\forall U\in K_u,\qquad \langle \tilde{u} ,\tilde{u} -U\rangle \leq 0\; .$$ Now, $\tilde{u}$ seems to be a good candidate to be the extension of $u$ I'm looking for, therefore I set: $$P:W^{1,2}(\Omega) \ni u \mapsto \tilde{u} \in W^{1,2}(\mathbb{R}^N)$$ and I have to show that such a $P$ satisfies properties 1, 2, 3.

1. $P$ is linear. First of all, I note that $K_{\alpha u} =\alpha\ K_u$; thus $P(\alpha u) =\widetilde{\alpha\ u} =\alpha\ \tilde{u} =\alpha\ Pu$ follows from (V) and $P$ is homogenous. It remains to prove that $P$ is additive, i.e. that I get $P(u_1+u_2)=Pu_1+Pu_2$ for all $u_1,u_2\in W^{1,2}(\Omega)$. For $W\in K_{u+v}$ I have $W-\tilde{v} \in K_u$, $W-\tilde{u} \in K_v$ therefore: $$\langle \tilde{u} +\tilde{v}, \tilde{u} +\tilde{v}-W\rangle =\langle \tilde{u} , \tilde{u} - (W -\tilde{v})\rangle +\langle \tilde{v} , \tilde{v} - (W -\tilde{u})\rangle \leq 0 \; ,$$ thus $P(u+v) =\widetilde{u+v} =\tilde{u} +\tilde{v} =Pu+Pv$.

2. $Pu\Big|_\Omega = u$ almost everywhere. This comes from $Pu=\tilde{u} \in K_u$.

But now I'm in trouble in proving 3... Any hints?

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Maybe 3. follow from the closed graph theorem. –  Davide Giraudo Jan 6 '12 at 20:08

1 Answer 1

That is, indeed, a Baire category game. You have a linear mapping $P$ from one Hilbert space to another. Note that the pre-image of every closed ball centered at the origin is closed (indeed, if $u_j\to u$, and $\|Pu_j\|$ are bounded by $R$, then some subsequence of $Pu_j$ converges to something weakly but that something restricted to $\Omega$ must be $u$ and its norm is at most $R$, so $\|Pu\|\le R$ by the construction. Now, the union of those pre-images is the entire space, so one of them has an inner point, which is enough to conclude that $P$ is a bounded linear mapping.

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