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Let $\Omega\subset\mathbb{R}^N$ be a bounded domain and $p\in (1,\infty)$. It is know that there exists a unique bounded surjective linear map $T: W^{1,p}(\Omega)\to W^{1-1/p,p}(\partial\Omega)$ with the property that $$Tu=u_{|\partial\Omega},\ \forall\ u\in C^\infty(\overline{\Omega})$$

It is also know that there exists a bounded linear map $\ell : W^{1-1/p,p}(\partial\Omega) \to W^{1,p}(\Omega)$ which is a right inverse for $T$. This map is sometimes called the lift map (see Necas Theorem 5.7). In proving the existence of $\ell$, Necas explicit constructs the map $\ell$.

I want to avoid this construction and use an argument from Functional Analysis, to wit, Theorem 2.12 of Brezis. To use it, I have to prove that $W_0^{1,p}(\Omega)$ is complemented in $W^{1,p}(\Omega)$, however I am stuck here. How would I go about this?

Tomás
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1 Answers1

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Partial answer: let's consider the simplest case: let $p=2$. Fix a $g\in H^{1/2}(\partial \Omega)$. For any $v \in H^1_0(\Omega)$, if there is some $w\in H^1(\Omega)$, the trace of $w$ is $g$, then $$ w\in H^1_0(\Omega)^{\perp}\iff \langle w,v\rangle_{H^1}:=\int_{\Omega} (\nabla w \cdot \nabla v + wv) = 0 \quad \text{for all } v \in H^1_0(\Omega). $$ This is equivalent to say that $w$ is the weak solution to $$ \begin{cases} -\Delta w + w = 0 &\text{in } \Omega, \\ w = g\in H^{1/2}(\partial \Omega) &\text{on }\partial \Omega. \end{cases}\tag{1} $$ This roughly tells us that the orthogonal complement of $H^1_0(\Omega)$ should be functions like $w$.

Such $w$ is constructed similar to an extension from a $g\in H^{1/2}(\partial \Omega)$. Due to problem (1)'s well-posedness, we can define a linear operator based on (1): $$L: H^{1/2}(\partial \Omega)\to H^1(\Omega),\; g\mapsto Lg:= w.$$ We also have $\|Lg\|_{H^1(\Omega)} \leq c\|g\|_{H^{1/2}(\partial \Omega)}$. This result says the boundedness of $L$. The argument can be found in the two additional lemmas in the updates below.

Also notice that if $g=0$, then $w=0$, this implies the intersection is $\{0\}$. Hence for any $u \in H^1(\Omega)$, let the trace operator be $T$ $$ u = \underbrace{L(Tu)}_{\in (H^1_0(\Omega))^{\perp}} + \underbrace{\big(u - L(Tu)\big)}_{\in H^1_0(\Omega)}. $$ This argument can be easily extended to $p\geq 2$ case for $W^{1,p}\subset W^{1,2}$.

Some updates:

Lemma 1: Suppose $\Omega$ has whatever regularity is needed. Let $u\in H^1(\Omega)$ be the weak solution of$$ \begin{cases} -\Delta u + u = f &\text{in } \Omega, \\ u = g\in H^{1/2}(\partial \Omega) &\text{on }\partial \Omega. \end{cases} $$ There is a $v\in H^1(\Omega)$ such that $u-v \in H^1_0(\Omega)$, then $$ \|u\|_{H^1(\Omega)}\leq C\Big(\|f\|_{H^{-1}(\Omega)} + \|v\|_{H^{1}(\Omega)}\Big).\tag{2} $$

Proof: Use $u-v\in H^1_0$ as the test function: $$ \int_{\Omega} \nabla u\cdot\nabla(u-v) + \int_{\Omega}u(u-v) = \int_{\Omega}f(u-v), $$ which is $$ \|u\|_{H^1(\Omega)} := \int_{\Omega} |\nabla u|^2 +\int_{\Omega}u^2 = \int_{\Omega}f(u-v) + \int_{\Omega}\nabla u\cdot\nabla v + \int_{\Omega}uv \\ \leq \frac{1}{\epsilon} \|f\|_{H^{-1}(\Omega)} \epsilon \|u-v\|_{H^1(\Omega)} + \frac{1}{2}\int_{\Omega} \big(|\nabla u|^2+|\nabla v|^2\big) + \frac{1}{2}\int_{\Omega}(u^2+v^2). $$ Then use Minkowski inequality again on $\|u-v\|_{H^1(\Omega)}$, choose $\epsilon$ small enought such that the $\|u\|_{H^1(\Omega)}$ term can be absorbed by the left. Also notice the choice of $\epsilon$ is irrelevant of $v$. The estimate in (2) will follow.

Lemma 2: The weak solution $w$ of (2) satisfies the estimate $\|w\|_{H^1(\Omega)} \leq c\|g\|_{H^{1/2}(\partial \Omega)}$.

Proof: Choose any $v\in H^1_g(\Omega):=\{v\in H^1(\Omega): Tv = g\in H^{1/2}(\partial \Omega)\}$, i.e., the traces coincide. We have $w-v\in H^1_0$, hence by (2), we have $$ \|w\|_{H^1(\Omega)} \leq C\|v\|_{H^1(\Omega)} \quad\text{for all }v\in H^1_g(\Omega). \\ \implies c \|w\|_{H^1(\Omega)} \leq \inf_{v\in H^1_g(\Omega)}\|v\|_{H^1(\Omega)} =: \|g\|_{H^{1/2}(\partial \Omega)}. $$ The last step is just using an implicit way of introducing the half norm (even though proving this quotient like norm is equivalent to the usual fractional norm on boundary still needs a constructive proof......).

Shuhao Cao
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  • Because $W^{1.p}(\Omega)$ is reflexive, we know that there eixst $T:W^{-1,p'}(\Omega) \to W^{1,p}(\Omega)$ homeomorphism. What happens if we take $W_0^{1,p}(\Omega)^\perp\subset W^{-1,p'}(\Omega)$ and send it to $W^{1,p}(\Omega)$ by $T$. Is $T(W_0^{1,p}(\Omega)^\perp)$ a complment of $W_0^{1,p}(\Omega)$ in $W^{1,p}(\Omega)$? – Tomás Sep 05 '13 at 13:52
  • @Tomás If $W^{1,p}_0(\Omega)^{\perp} = {w\in W^{-1,p'}(\Omega): \langle u,w\rangle = 0,\text{ for any }u\in W^{1,p}(\Omega)}$, then no, there are multiple ways to construct the $T$ in your comment. To prove the direct sum of the complement, you have to pin down this map by a boundary value problem (working like an implicit constructed extension). For example, consider the BVP for $(-\Delta + cI)$ instead of $(-\Delta + I)$ in (1), you wouldn't get orthogonal complement. – Shuhao Cao Sep 05 '13 at 20:38
  • @Tomás Correct a typo, should be $u\in W^{1,p}_0(\Omega)$ in the definition. My point is that only knowing the existence of $T$ in your comment is not enough, we may ask question like following: for a fixed $u\in W^{1,p}(\Omega)$, is there a representation: $u = {u_0} + {Tw}$ for some $u_0\in W^{1,p}_0$, and $w\in W^{1,p}_0(\Omega)^{\perp}$? If this is true, then we have to have $\operatorname{Trace}({u}) = \operatorname{Trace}(Tw)$, this is another reason we should use boundary value problem to construct $T$. – Shuhao Cao Sep 05 '13 at 20:52
  • I see. Let me ask something about your answer: How do we know $L(H^{1/2}(\partial\Omega))$ is closed in $H^1(\Omega)$? – Tomás Sep 06 '13 at 20:53
  • Let me ask another question please: How do you prove the continuity of $L$, at least, what is the idea? – Tomás Sep 07 '13 at 18:56
  • @Tomás The continuity of $L$ follows from the the boundedness of $L$, and thus there holds the closedness of $L(H^{1/2}(\partial \Omega))$. – Shuhao Cao Sep 07 '13 at 19:38
  • Sorry, let me remake my question: How do you prove the boundedness of $L$? – Tomás Sep 07 '13 at 20:37
  • @Tomás Please see my updated answer. – Shuhao Cao Sep 08 '13 at 03:25
  • I will accpet your answer, because it was essential in the understanding of the problem. (+1) – Tomás Sep 08 '13 at 12:06
  • In the second sentence, I don't understand why $w$ is the weak solution to (1). There is no $g$ in the variational formulation. – timur Sep 09 '13 at 20:37
  • @timur Thank you so much for reminding me about the unclearness, I rephrased the statement, can you check it now? – Shuhao Cao Sep 09 '13 at 20:49