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This is related to another question I asked: Directional derivative in a Sobolev-like inequality

Suppose that $u \in C_{0}^{\infty}(\Omega)$. Show that the linear map $u \to u(x,0) \in C^{\infty}(\mathbb{R}^{n})$ has a unique continuous extension as a map from $W^{1,p}_{0}(\Omega)$ to $L^{p}(\partial \Omega_{+})$, where $\partial \Omega_{+} \equiv \{x;(x,0)\in\Omega\}$.

I know that you can define extensions from one Sobolev space to another. Is the reason why we can do it here from a Sobolev space to an $L^{p}$ space because $L^{p}$ spaces are contained inside Sobolev spaces? Also, I was thinking about using the BLT Theorem, but do I know that the map from $u$ to $u(x,0)$ is bounded?

I'm very confused, and not as well-versed in functional analysis as I should be, so any assistance/hints you could give me for this proof would be much appreciated!

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  • $\begingroup$ Note that it is the boundary which is the domain in the second space. Such results are called "trace theorems" in the Sobolev norm machinery. $\endgroup$ – hardmath Feb 11 '14 at 1:43
  • $\begingroup$ Yes, I've been spending the past 9 days teasing about them. All I have so far is this: Suppose $u_{j}\in C_{0}^{\infty}(\Omega)$ and let $u_{j}\to u$ in $W_{0}^{1,p}$. So, by applying the inequality from math.stackexchange.com/questions/665593/… , $||u_{j}-u_{k}||_{L^{P}(\partial \Omega)}\leq C||u_{j}-u_{k}||_{W^{1,p}}$ so that $\{u_{j}\}$ is Cauchy in $L^{p}(\partial \Omega)$. So, we define $u=\lim_{j \to \infty}u_{j}$, the limit taken in $L^{p}$. Only thing is, I don't think this is thorough enough or even takes into account the fact.. $\endgroup$ – ALannister Feb 20 '14 at 3:02
  • $\begingroup$ .that the map is supposed to take $u$ to $L^{p}(\partial \Omega_{+})$, and not just $L^{p}(\partial \Omega)$. I think I have all the basic machinery there in the proof, but I need help fleshing out the proof and making it more precise. If somebody could help me with this, it would be much appreciated (and rewarded by a bounty of 50 reputation points!). $\endgroup$ – ALannister Feb 20 '14 at 3:06
  • $\begingroup$ Could you please edit your question answering the following questions. What is $\Omega$? How is $u(x,0)\in C^\infty(\mathbb{R}^n)$ if $u\in C_0^\infty(\Omega)$? $\endgroup$ – Tomás Feb 20 '14 at 11:23
  • $\begingroup$ @Tomas, all of that information is given in the problem that this is related to. Follow the above link to read more. Basically, I'm using the same $\Omega$ as given in that problem. Think of that problem as part (a) and this one as part (b). $u(x,0)$ is u restricted to the boundary, which is meaningless in regular $L^{p}$ or Sobolev spaces, hence why we have to do something else - namely this extension. $\endgroup$ – ALannister Feb 20 '14 at 11:37
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$\Omega$ is bounded, so there is a K such that $\Omega \subset \mathbb{R}^n\times (-\infty,K)$. Then it's basically John's hint from the other question: For $u \in C_0^\infty(\Omega)$, we have

$$u(x,0) = - \int_0^K \frac{\partial u}{\partial x_{n+1}}(x,t)\,dt.$$

Hölder's inequality gives

$$\lvert u(x,0)\rvert \leqslant \left(\int_0^K \left\lvert \frac{\partial u}{\partial x_{n+1}}(x,t)\right\rvert^p\,dt\right)^{1/p}\cdot \left(\int_0^K 1^{p/(p-1)}\,dt\right)^{(p-1)/p},$$

and from that we obtain

$$\int_{\partial \Omega_+} \lvert u(x,0)\rvert^p\,dx \leqslant K^{p-1} \int_{\partial\Omega_+} \int_0^K \left\lvert \frac{\partial u}{\partial x_{n+1}}(x,t)\right\rvert^p\,dt\,dx \leqslant K^{p-1} \left\lVert \frac{\partial u}{\partial x_{n+1}}\right\rVert_{L^p(\Omega)}^p,$$

or, writing $\rho(u)(x) = u(x,0)$,

$$\lVert \rho(u)\rVert_{L^p(\partial \Omega_+)} \leqslant K^{1-\frac1p}\left\lVert \frac{\partial u}{\partial x_{n+1}}\right\rVert_{L^p(\Omega)} \leqslant K^{1-\frac1p}\lVert u\rVert_{W_0^{1,p}(\Omega)}.\tag{1}$$

The estimate $(1)$ shows that

$$\rho \colon C_0^{\infty}(\Omega) \to L^p(\partial\Omega_+);\quad \rho(u)(x) = u(x,0)$$

is continuous if we endow $C_0^\infty(\Omega)$ with the subspace topology induced by $W_0^{1,p}(\Omega)$. By the denseness of $C_0^\infty(\Omega)$ in $W_0^{1,p}(\Omega)$, there is a unique continuous linear $\tilde{\rho}\colon W_0^{1,p}(\Omega) \to L^p(\partial\Omega_+)$ that extends $\rho$.

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  • $\begingroup$ There is one thing that I can't understand. If $u\in C_0^\infty(\Omega)$ then $u(x,0)=0$ for $x\in\partial\Omega_+$, hence, $\tilde{\rho}=0$. Am I right? $\endgroup$ – Tomás Feb 20 '14 at 12:44
  • $\begingroup$ The notation is a bit misleading. $\Omega$ is not open in $\mathbb{R}^{n+1}$, but in $\overline{\mathbb{R}^{n+1}_+} = \{ (x,x_{n+1}) \in \mathbb{R}^{n+1} : x_{n+1} \geqslant 0\}$. Thus the compact support of $u$ can contain a nontrivial part of the hyperplane $\{x_{n+1} = 0\}$. $\endgroup$ – Daniel Fischer Feb 20 '14 at 12:48
  • $\begingroup$ You are right. Thank you. $\endgroup$ – Tomás Feb 20 '14 at 12:49
  • $\begingroup$ Thanks, Daniel! I have to take a closer look at this, and I might have questions. $\endgroup$ – ALannister Feb 21 '14 at 17:50

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