Timeline for $f\in (W^{1,p}(\Omega)\cap C(\Omega) \cap L^{\infty}(\Omega))\setminus C(\bar{\Omega})$, $f=0$ on $\partial \Omega$ imply $f\in W^{1,p}_{0}(\Omega)$?
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| Mar 11, 2018 at 14:07 | comment | added | Jeff | Let us continue this discussion in chat. | |
| Mar 11, 2018 at 14:07 | comment | added | Jeff | I already explained this to you; no, but you can modify it with a bump function so it is $W^{1,p}$ and equal to $\text{sign}(x_1)$ away from $B(0,\epsilon)$. | |
| Mar 11, 2018 at 9:13 | comment | added | Medo | @Jeff. Okay. Let us now connect the two half balls by the small ball $B(0; \epsilon )$ as you suggested. Will $x\mapsto sing (x_{1})$ still be a $W^{1,p}$ function on the connected domain ? | |
| Mar 11, 2018 at 5:24 | comment | added | Jeff | No, what I said is correct. The whole domain is made up of two connected components, each of which has a Lipschitz boundary, but the union does not have a Lipschitz boundary. To have a Lipschitz boundary, you need to be able to touch the boundary with a cone from the exterior, which is impossible along a slit. | |
| Mar 10, 2018 at 21:56 | comment | added | Medo | @Jeff. I am confused. On Jan 16 at 22:40, you commented "Either half has a Lipschitz boundary and the trace is well-defined for either half considered separately". Did you mean to say neither "Neither half has a Lipschitz boundary" ? | |
| Mar 10, 2018 at 19:06 | comment | added | Jeff | The previous domains are not Lipschitz. You need to review the definition of a Lipschitz domain. | |
| Mar 10, 2018 at 15:36 | comment | added | Medo | @Jeff. So, we still do not have an example where $f\in W^{1,p}(\Omega_{i})$, $i=1,2$, $\Omega_{1}$ is a domain not Lipschitz and $f$ does not have trace on $\partial \Omega_{1}$, while $\Omega_{2}$ is a Lipschitz domain and $f$ does have trace on $\Omega_{2}$. This will show that Lipschitz condition on the boundary is necessary for a $W^{1,p}$ function to have trace. Notice that the domain in your previous examples are Lipschitz. Thanks | |
| Jan 17, 2018 at 13:49 | comment | added | Jeff | Union $\Omega$ with a small ball $B^0(0,\epsilon)$ to make it connected, and multiply $f$ by a smooth bump function $\phi\in C^\infty(\mathbb{R}^n)$ satisfying $\phi=0$ in $B(0,2\epsilon)$ and $\phi=1$ for $|x|>3\epsilon$. The new function will agree with $\text{sgn}(x_1)$ for $|x|>3\epsilon$. | |
| Jan 17, 2018 at 10:10 | comment | added | Medo | @Jeff. Thanks a lot. I did learn a lot from your comments. One last concern though. How modify $\Omega$ (the unit ball minus $\{x_{1}=0\}$) to be connected, but still have $f=sgn(x_{1})$, $x\in \Omega$, a $W^{1,p}(\Omega)$ function without a trace, somewhere on the boundary) because the boundary is not Lipschitz? Thanks again for your patience | |
| Jan 16, 2018 at 22:40 | comment | added | Jeff | Either half has a Lipschitz boundary and the trace is well-defined for either half considered separately. But $\Omega$ as a whole does not have a Lipschtiz boundary. (Take $x\in \partial \Omega$ with $x_1=0$. For $\partial \Omega$ to be Lipschitz, there must exist $r>0$ such that $\Omega \cap B(x,r) = \{x \, : \, \gamma(x) > 0\}$ for a Lipschitz function $\gamma$, which is clearly not true). | |
| Jan 16, 2018 at 22:00 | comment | added | Medo | @Jeff. Okay. You are right it. It is a $W^{1,p}$ function on that $\Omega$. Now update $\Omega$ to be only one of the two halves, say the half with $x_{1}>0$. This is a domain that is not Lipschitz right? Why does not it have trace $=1$ on $x_{1}=0$ ? | |
| Jan 16, 2018 at 19:30 | comment | added | Jeff | I shouldn't even say "weak" derivative; in this case $f \in C^\infty(\bar{\Omega})$ (in the sense in Evans book, which means $f$ and all its derivatives are uniformly continuous on $\Omega$). | |
| Jan 16, 2018 at 19:23 | comment | added | Jeff | Yes it is $W^{1,p}$. Its weak derivative is identically zero in $\Omega$. Note in the definition of weak derivative you test with functions $\phi \in C^\infty_c(\Omega)$; in particular, $\phi$ vanishes in a neighborhood of the set where $\{x_1=0\}$. | |
| Jan 16, 2018 at 18:24 | comment | added | Medo | @Jeff. Once again, this $f$ is not a $W^{1,p}$ function on the given $\Omega$. Would you like to show/refer to an example that shows there is a minimal smoothness requirement on the boundary necessary for the existence (not $L^{p}$ boundedness) of the trace of a Sobolev function? | |
| Jan 16, 2018 at 13:35 | comment | added | Jeff | @NateEldredge gave a good answer to the only sensible (and non-trivial) interpretation of this question. | |
| Jan 16, 2018 at 13:26 | comment | added | Jeff | @Medo Take $\Omega = B^0(0,1)\setminus \{x_1=0\}$ and set $f=1$ when $x_1>0$ and $f=-1$ when $x_1<0$. What is the trace of $f$ on the portion of $\partial \Omega$ where $x_1=0$? You can construct similar examples on connected domains. | |
| Jan 16, 2018 at 13:25 | comment | added | Jeff | @Medo The punctured point is part of the boundary... | |
| Jan 16, 2018 at 9:13 | comment | added | Medo | @Nate. However $\sin(1/x)∉W^{1,p}((0,1))$ for any $ p≥1$. Or is that your point ? | |
| Jan 16, 2018 at 9:12 | comment | added | Medo | @Jeff. Could you give an example for the necessity of the boundary being Lipschitz for the mere existence of a trace of a $W^{1,p}(\Omega)$ function on $\partial \Omega$? | |
| Jan 16, 2018 at 9:10 | comment | added | Medo | @Jeff. I do not understand why would we want to know the trace at a puncture point. When I said the trace of a function $f\in C(\Omega)\cap L^{\infty}(\Omega)$, the assumption that $f\in W^{1,p}(\Omega)$ should be understood, since we are talking about traces. | |
| Jan 16, 2018 at 0:00 | comment | added | Jeff | The trace is not defined for arbitrary domains; you need some boundary regularity. Consider the punctured plane $\Omega = \mathbb{R}^n\setminus \{0\}$. Then the trace of a function is pointwise evaluation at $x=0$, which is not well-defined unless $p>n$. The unboundedness of $\Omega$ is not essential in the example. | |
| Jan 15, 2018 at 23:56 | comment | added | Jeff | Yes, but the equivalence between zero trace and $W^{1,p}_0$ requires Lipschitz boundary. | |
| Jan 15, 2018 at 23:21 | comment | added | Nate Eldredge | @Medo: Let $\Omega=(0,1) \subset \mathbb{R}^1$. Let $f(x) = \sin(1/x)$ which is in $C(\Omega) \cap L^\infty(\Omega)$. How will you define its trace at $x=0 \in \partial \Omega$? | |
| Jan 15, 2018 at 20:01 | history | edited | Medo | CC BY-SA 3.0 |
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| Jan 15, 2018 at 20:00 | comment | added | Medo | Please correct me if I am wrong. Let $\Omega$ be a domain. (no smoothness conditions on $\Omega$). The trace of a function $f\in C(\Omega) \cap L^{\infty} (\Omega) $ exists in the pointwise sense. | |
| Jan 15, 2018 at 19:48 | comment | added | Medo | @Jeff. Sorry the for the earlier confusing version of the question. Now you see I meant $f=0$ on $\partial \Omega$ in the pointwise sense, e.g. $f=0$ on $\Omega^{c}$. You said (provided the boundary is Lipschitz). But this is necessary for the $L^{p}$ boundedness of the trace. I believe it is not required to define the trace. Example: $f\in C^{\infty}_{c}(\Omega)$ then $f\in W^{1,p}_{0}(\Omega)$ no matter how rough the boundary is. | |
| Jan 15, 2018 at 19:41 | comment | added | Medo | @LSpice. I rephrased the question. It was a bit confusing. It turns out to be very simple. Sorry about that. | |
| Jan 15, 2018 at 19:39 | history | edited | Medo | CC BY-SA 3.0 |
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| Jan 15, 2018 at 19:34 | history | edited | Medo | CC BY-SA 3.0 |
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| Jan 15, 2018 at 4:34 | comment | added | Jeff | The question is really what do you mean by $f=0$ on $\partial \Omega$? Do you mean the trace of $f$ is zero? If so, then $f\in W^{1,p}_0$ (provided boundary is Lipschitz)? If not, then you should clarify what $f=0$ on $\partial \Omega$ means. | |
| Jan 14, 2018 at 20:39 | comment | added | Fan Zheng | I guess the problem is that the title and the text have different interpretations. | |
| Jan 14, 2018 at 15:55 | comment | added | LSpice | You currently have two opposite answers to this question, @Nate's saying 'yes' and yours saying 'no'. Do you disagree with Nate's answer? Or is it just that you are taking the question to mean "must $f$ lie in $W_0^{1, p}(\Omega)$" and Nate is taking it to mean "can $f$ lie in $W_0^{1, p}(\Omega)$"? (You have used 'is' instead of 'must' or 'can', which, I think, is ambiguous.) | |
| Jan 14, 2018 at 8:49 | review | First posts | |||
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| Jan 14, 2018 at 8:22 | answer | added | Medo | timeline score: 0 | |
| Jan 13, 2018 at 18:45 | answer | added | Nate Eldredge | timeline score: 2 | |
| Jan 13, 2018 at 18:12 | history | edited | Medo | CC BY-SA 3.0 |
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| Jan 13, 2018 at 11:32 | history | edited | Medo | CC BY-SA 3.0 |
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| Jan 13, 2018 at 11:15 | history | asked | Medo | CC BY-SA 3.0 |