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Let $W^{1,p}$ be the Sobolev space of $L^p$ functions with $L^p$ first derivatives. Let $W^{1,p}_0$ be the closure of the test functions in $W^{1,p}$. I am not explicitly writing the domain of the functions because I expect it won't matter, but call it $\Omega$ if you wish. What I will need is $\Gamma$ to be the boundary. I know we can define an operator $\gamma_0:W^{1,p}\to L^p(\Gamma)$ such that it is a bounded linear operator and its restriction to smooth functions is the restriction operator $u\mapsto u|_\Gamma$. I have been told the following holds:

$$W^{1,p}_0=\ker\gamma_0.$$

I can easily see $\subseteq$: if $u\in W^{1,p}_0$, then by definition we have test functions $u_n$ converging to $u$ in $W^{1,p}$, but test functions belong to the kernel and $\gamma_0$ is continuous, hence:

$$\gamma_0u=\lim_{n\to\infty}\gamma_0u_n=0.$$

But what about the converse? Assume $\gamma_0u=0$. How do I prove this implies $u$ is the limit of test functions? I wasn't able to find that on the internet, and the teacher decided to omit the proof, so here I am asking for a proof. How do I proceed? I know that I can find smooth functions $u_n\to u$ in $W^{1,p}$ (e.g. convolutions with mollifiers, which are not necessarily test functions, I mean the convolutions, the mollifiers are) since smooth functions on $\overline\Omega$ are dens i $W^{1,p}$, but how do I show they are (at least eventually) with zero trace? I can only see that their traces converge to 0 in $W^{1,p}$ by continuity of $\gamma_0$…

Calvin Khor
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MickG
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    Suppose $u_n$ are smooth and converge to $u$ in $W^{1,p}$. Define $A_n={ x \in \Omega : d(x,\partial \Omega) \geq 1/n }$ and similarly $B_n={ x \in \Omega : d(x,\partial \Omega) \geq 2/n }$. Introduce $v_n$ which are smooth, equal to $u_n$ on $B_n$, and $0$ outside $A_n$. Argue that if $u$ is in fact in $W^{1,p}_0$ (not $W^{1,p}$ as a whole) then $v_n$ converges to $u$ as well. The details are not actually that easy, but this is the idea. – Ian Jan 10 '16 at 00:34
  • Sorry, in the previous comment I meant to say "...if $u$ in fact has trace zero ...". That is, my suggestion was for the trace zero implies $W^{1,p}_0$ side of the argument. The other side is easy, as noted in the OP and the accepted answer. – Ian Jan 10 '16 at 12:16
  • @Ian : so the technical details you are talking about is that for any compacts $B_n \subset A_n$ (strictly included) there is a smooth approximation of $1_{B_n}$, being exactly the identity on $B_n$, and being exactly zero outside of $A_n$ ? I know how to construct this on $\mathbb{R}$, but on $\mathbb{R}^n$ I'm not sure what is the argument (even if it is "obvious" that it exists) – reuns May 08 '16 at 23:20
  • @user1952009 Yes; as I recall the proof is a standard mollification argument, where you convolve $1_{B_n}$ with a mollifier whose support has diameter $1/n$ or something along these lines. If you know how to do it on $\mathbb{R}$ it should be essentially the same argument, the main point is that convolution only increases the support of the result by a small amount if the function you are convolving with has a small support. In the accepted answer they do essentially the same thing but proceed by straightening the boundary first so that the construction can be done explicitly. – Ian May 08 '16 at 23:21

1 Answers1

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In Evans book (Chapter 5) we can find a proof for bounded $\Omega$ with $C^1$ boundary. (The author says to omit it in a first reading.) Here it is in full:

THEOREM 2 (Trace-zero functions in $W^{1,p}$). Assume $U$ is bounded and $\partial U$ is $C^1$. Suppose furthermore that $u\in W^{1,p}(U)$. Then $$u\in W^{1,p}_0(U)\ \ \ \textit{if and only if}\ \ \ Tu=0\ \, \textit{on}\,\ \partial U.\tag4$$

Proof$^*$.

  1. Suppose first $u\in W_0^{1,p}(U).$ Then by definition there exist functions $u_m\in C_c^\infty(U)$ such that $$u_m\to u\quad{\rm in}\ W^{1,p}(U).$$ As $Tu_m=0$ on $\partial U\ (m=1,...)$ and $T:W^{1,p}(U)\to L^p (\partial U)$ is a bounded linear operator, we deduce $Tu=0$ on $\partial U$.

  2. The converse statement is more difficult. Let us assume that $$\tag{5} Tu=0\quad{\rm on}\ \partial U.$$ Using partitions of unity and flattening out $\partial U$ as usual, we may as well assume $$\begin{cases}u\in W^{1,p}(\Bbb R^n _+),\ \ u \ {\rm has\, compact\, support\, in\ }\overline{{ \Bbb R}}^n_+, \\ \qquad\ Tu=0\ {\rm on }\ \partial \Bbb R^n_+=\Bbb R^{n-1}. \end{cases}\tag6$$ Then since $Tu=0$ on $\Bbb R^{n-1}$, there exist functions $u_m\in C^1(\overline{ \Bbb R}^n_+)$ such that $$u_m\to u\ \ \ \ {\rm in} \ \, W^{1,p}(\Bbb R^n_+)\tag7$$ and $$Tu_m=u_m|_{\Bbb R^{n-1}}\to0\ \ \ \ {\rm in}\ L^p(\Bbb R^{n-1}).$$ Now if $x'\in\Bbb R^{n-1}$, $x_n\geq0$, we have $$|u_m(x',x_n)|\leq|u_m(x',0)|+\int_0^{x_n}|u_{m,x_n}(x',t)|\, dt.$$ Thus $$\begin{aligned}&\int_{\Bbb R^{n-1}}|u_m(x',x_n)|^p \, dx' \\ \leq &\; C\left(\int_{\Bbb R^{n -1}}|u_m(x',0)|^p\, dx' + x_n^{p-1}\int_0^{x_n}\int_{\Bbb R^{n-1}}|Du_m(x',t)|^p dx'\, dt\right)\end{aligned}.$$ Letting $m\to\infty$ and recalling (7), (8), we deduce: $$\int_{\Bbb R^{n-1}}|u(x',x_n)|^p dx'\leq Cx^{p-1}_n \int_0^{x_n}\int_{\Bbb R^{n-1}}|Du|^p dx'dt\tag9$$ for a.e. $x_n>0$.

  3. Next let $\zeta\in C^\infty(\Bbb R)$ satisfy $$\zeta\equiv1\ {\rm on}\ [0,1],\ \zeta\equiv0\ {\rm on}\ {\Bbb R}-[0,2],\ \ \ 0\leq \zeta\leq1,$$ and write $$\begin{cases}\zeta_m(x):=\zeta(mx_n)\ \ \ \ (x\in\Bbb R^n_+) \\ w_m:=u(x)(1-\zeta_m).\end{cases}$$ Then $$\begin{cases}w_{m,x_n}=u_{x_n}(1-\zeta_m)-mu\zeta'\\ D_{x'}w_m=D_{x'}u(1-\zeta_m).\end{cases}$$ Consequently $$\begin{align}\int_{\Bbb R^n_+}|Dw_m-Du|^p \, dx&\leq C\int_{\Bbb R^n_+}|\zeta_m|^p|Du|^p\, dx \\ & \qquad\ +Cm^p\int_0^{2/m}\int_{\Bbb R^{n-1}}|u|^p\, dx'dt\\ &=:A+B. \end{align}$$ Now $$\tag{11}A\to0\quad{\rm as}\ m\to\infty,$$ since $\zeta_m\neq0$ only if $0\leq x_n\leq 2/m$. To estimate the term $B$, we utilize inequality (9): $$\tag{12}\begin{align}B&\leq Cm^p\left(\int_0^{2/m}t^{p-1}dt\right)\left(\int_0^{2/m}\int_{\Bbb R^{n-1}}|Du|^p dx'dx_n\right)\\ &\leq C\int_0^{2/m}\int_{\Bbb R^{n-1}}|Du|^p dx'dx_n\to 0\quad{\rm as}\ m\to 0.\end{align}$$ Employing (10)-(12), we deduce $Dw_m\to Du$ in $L^p(\Bbb R^n_+)$. Since clearly $w_m\to u$ in $L^p (\Bbb R^n_+)$, we conclude $$w_m\to u\quad{\rm in}\ W^{1,p}(\Bbb R^n _+).$$ But $w_m=0$ if $0<x_n<1/m$. We can therefore mollify the $w_m$ to produce functions $u_m\in C_c^\infty(\Bbb R^n _+)$ such that $u_m\to u$ in $W^{1,p}(\Bbb R^n_+)$. Hence $u\in W^{1,p}_0(\Bbb R^n _+)$. $\tag*{$\square$}$

Pedro
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  • It might be a good idea to insert screenshots of the couple of pages of proof, for the sake of the post's self-containment. Here are my screenshots: 1, 2, 3 and 4. Also, does this proof extend to products of $N$ intervals in $\mathbb{R}^N$? I'm asking because in my course almost everything was proved for two cases: $C^1$ with bounded frontier and product of intervals. – MickG Jan 10 '16 at 11:51
  • Just accepted. If you agree to the screenshots, I used your answer to upload them so I need but type an edit summary, click save, and the screenshots will be in the answer, so just say aye, and I shall save :). – MickG Jan 10 '16 at 11:59
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    @MickG I don't know the answer for your question on the generalization of the proof. Yes, the screenshots are good. I agree. – Pedro Jan 10 '16 at 12:07
  • Added 'em screenshots :). – MickG Jan 10 '16 at 12:10
  • @MickG The last screenshot is repeated (links 3 and 4 in your comment are the same). – Pedro Jan 10 '16 at 12:17
  • Will fix after lunch. – MickG Jan 10 '16 at 12:18
  • Fixed it. ${}{}{}$ – MickG Jan 10 '16 at 13:27
  • Why does (7) imply $\int|u_m(x′,x_n)|^p \rightarrow \int|u(x′,x_n)|^p$? – MaudPieTheRocktorate Aug 29 '18 at 03:40
  • Could anyone explain me why the inequality between (7) and (9) holds true? –  Apr 01 '20 at 14:14
  • @C.Bishop $\int_a^b f'(t);dt=f(b)-f(a)$ and thus $|f(b)|= |f(a)+ \int_a^b f'(t);dt|$. – Pedro Apr 01 '20 at 15:54
  • I mean the one immediately above the (9), but i think i have done by using holder inequality. Thank you @Pedro –  Apr 01 '20 at 16:25
  • Sorry, one last question. How about the inequality above the (11)? Could i have a hint? –  Apr 01 '20 at 17:06
  • @C.Bishop You are right, the one immediately above (9) is obtained by Hölder: \begin{align} |u_m(x',x_n)|&\leq|u_m(x',0)|+\int_0^{x_n}|Du_{m}(x',t)|, dt\ &\leq |u_m(x',0)|+\left(\int_0^{x_n} 1^{p/(p-1)}, dt\right)^{(p-1)/p}\left(\int_0^{x_n}|Du_{m}(x',t)|^p, dt\right)^{1/p}\ &=|u_m(x',0)|+x_n^{(p-1)/p}\left(\int_0^{x_n}|Du_{m}(x',t)|^p, dt\right)^{1/p} \end{align} Taking the power $p$ the result follows (from this). – Pedro Apr 01 '20 at 19:23
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    @C.Bishop From the one above (11), note that (by this) $$Dw_{m}=(1-\zeta_m)Du-uD(1-\zeta_m) =(1-\zeta_m)Du-mu\zeta'(mx_n)$$ and thus $$|Dw_m-Du|^p=|\zeta_mDu-mu\zeta'|^p\leq C|\zeta_m|^p|Du|^p+Cm^p|u|^p|\zeta'(mx_n)|^p.$$ Therefore, $$\int_{\mathbb R^n_+}|Dw_m-Du|^p,dx\leq C\int_{\mathbb R^n_+}|\zeta_m|^p|Du|^p,dx+Cm^p\int_0^{x_n}\int_{\mathbb R^{n-1}}|u(x',t)|^p|\zeta'(mt)|^p,dx'dt$$ and the result follows (because $\zeta=0$ outside $[0,2]$). – Pedro Apr 01 '20 at 19:23
  • @Pedro thanks again. I agree with everything you wrote, but what happens to the term $\vert\zeta^{\prime}(mt)\vert^p$? I ask you because in the inequality it disappears. –  Apr 03 '20 at 08:02
  • @C.Bishop $\zeta'\leq \text{constant}$ because $\zeta$ is smooth (the $C$ in my previous comment is different from the one in the book) – Pedro Apr 03 '20 at 08:19
  • Oh, it's clear. Thank you again @Pedro –  Apr 03 '20 at 08:40
  • @Pedro I am really sorry to try your patience, but I have a last question. When the author starts the proof, he says that $\exists u_m\subset C^1(\overline{\mathbb{R}^n_{+}})$ such that $u_m\longrightarrow u$ in $W^{1, p}(\mathbb{R}^n_{+})$. It is true because $C^1(\overline{\mathbb{R}^n_{+}})$ is dense in $W^{1, p}(\mathbb{R}^n_{+})$, isn't it? –  Apr 03 '20 at 09:14
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    @C.Bishop $C^1(\overline{\mathbb R^n_+})$ is indeed dense in $W^{1,p}(\mathbb R^n_+)$. However, I think we should be able to justify this passage by using the tools available in the book (general density results were proved only for bounded domains). Here there is an idea (not sure if it is completely correct). Since $u$ has compact support in $\overline{\mathbb R^n_+}$, there exists $r>0$ such $u=0$ outside $\Omega:=B(0;r)\cap\mathbb R^n_+$. From Thm 3 in Sec. 5.3, there exists $z_m\in C^\infty(\Omega)$ such that $z_m\to u$ in $W^{1,p}(\Omega)$. From Thm 1 in Sec. 5.4... – Pedro Apr 03 '20 at 15:34
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    ... there exist extensions $\overline{u}=Eu,u_m=Ez_m\in W^{1,p}(\mathbb R^n)$ such that $$\begin{align}|u_m-u|{W^{1,p}(\mathbb R^n+)}&=|u_m-\overline{u}|{W^{1,p}(\mathbb R^n+)}\ &\leq |u_m-\overline{u}|{W^{1,p}(\mathbb R^n)}=|E(z_m-u)|{W^{1,p}(\mathbb R^n)}\leq C|z_m-u|_{W^{1,p}(\Omega)}\to 0.\end{align}$$ From the proof of the said thm, $u_m\in C^1_c(\mathbb R^n)$ (because $z_m$ is smooth) and thus $u_m\in C^1(\overline{\mathbb R^n_+})$. – Pedro Apr 03 '20 at 15:34
  • I'll need to ressurect this comment chain to ask one thing: How does the integral B transforms into (12) exactly? If I directly apply (9), I get $B \leq Cm^p\int_0^{2/m} x_n^{p-1} \int_0^{x_n} \int_{\mathbb{R}^{n-1}} |Du(x,t)|^p dx' \ dt \ dx_n$. – AspiringMathematician Apr 01 '21 at 22:02
  • @AspiringMathematician Changing notation, we conclude from (9) that $$\begin{aligned} B&=Cm^p\int_0^{2/m}\int_{\Bbb R^{n-1}}|u(x',t)|^p, dx'dt\ &\leq Cm^p\int_0^{2/m}\left(Ct^{p-1} \int_0^{t}\int_{\Bbb R^{n-1}}|Du(x',x_n)|^p dx'dx_n\right)dt \end{aligned}$$ Since $t$ is at most $m/2$, it follows that $$\begin{aligned} B&\leq Cm^p\int_0^{2/m}\left(Ct^{p-1} \int_0^{m/2}\int_{\Bbb R^{n-1}}|Du(x',x_n)|^p dx'dx_n\right)dt\ &=Cm^p\left(\int_0^{2/m}t^{p-1} dt\right)\left(\int_0^{m/2}\int_{\Bbb R^{n-1}}|Du(x',x_n)|^p dx'dx_n\right) \end{aligned}$$ – Pedro Apr 02 '21 at 01:28