Q1: Let $p\geq 1$, and let $f\in W^{1,p}(\Omega)\cap C(\Omega)$. Assume also $f\in L^{\infty}(\Omega)$ and $f=0$ on $\partial \Omega$. Is it true that $f\in W^{1,p}_{0}(\Omega)$ even if $f\notin C(\bar{\Omega})$ ?
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2$\begingroup$ 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.) $\endgroup$– LSpiceCommented Jan 14, 2018 at 15:55
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$\begingroup$ I guess the problem is that the title and the text have different interpretations. $\endgroup$– Fan ZhengCommented Jan 14, 2018 at 20:39
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$\begingroup$ 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. $\endgroup$– JeffCommented Jan 15, 2018 at 4:34
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$\begingroup$ @LSpice. I rephrased the question. It was a bit confusing. It turns out to be very simple. Sorry about that. $\endgroup$– MedoCommented Jan 15, 2018 at 19:41
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$\begingroup$ @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. $\endgroup$– MedoCommented Jan 15, 2018 at 19:48
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2 Answers
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(The question got modified and this may not answer it anymore.)
It is possible to have a function $f$ which is $C^\infty$ and bounded on $\Omega$ and is in $W^{1,p}_0(\Omega)$, yet does not have any continuous extension to $\overline{\Omega}$; in particular, setting $f = 0$ on $\partial \Omega$ does not result in a continuous function.
Let $\Omega$ be some domain in $\mathbb{R}^d$, $d \ge 2$, and let $\varphi$ be some smooth bump function supported inside the unit ball and with $\varphi(0) = 1$. Let $\varphi_\epsilon(x) = \varphi(x/\epsilon)$. With a change of variables, you can compute that $\|\varphi_\epsilon\|_{W^{1,p}} \approx \epsilon^{d-1}$.
Now choose a sequence of disjoint open balls $B_n$ inside $\Omega$ of radius $r_n$, where $\sum r_n < \infty$, whose centers $x_n$ converge to some $x \in \partial \Omega$. Let $g_n(x) = \varphi_{r_n}(x-x_n)$, which is compactly supported inside $B_n$ and has $g_n(x_n)=1$. Then the sum $f = \sum_n g_n$ converges in $W^{1,1}_0(\Omega)$, but $f$ is greater than $1/2$ everywhere on some neighborhood of $\{x_1, x_2, \dots\}$, so $f$ cannot be continuous up to the boundary, even after modification on a null set. Moreover, $f$ is continuous (even $C^\infty$) inside $\Omega$, since the sum is locally finite inside $\Omega$.
answered Jan 13, 2018 at 18:45
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$\begingroup$ Thanks a lot Nate. I am a little confused though. You are saying $f \in W^{1,1}_{0}(\Omega)$ yet there can be a subset of the boundary with positive Lebesgue surface measure on which $f\neq 0$. Does not the latter exclude $f$ from $W^{1,1}_{0}(\Omega)$ ? $\endgroup$– MedoCommented Jan 13, 2018 at 19:39
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$\begingroup$ @Medo: No, that's not true. $f$ is zero everywhere on the boundary. Note that every $g_n$ is compactly supported inside $\Omega$. $\endgroup$ Commented Jan 14, 2018 at 4:42
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$\begingroup$ Thanks a gain. What do you mean by "so f does not vanish continuously at the boundary" ? $\endgroup$– MedoCommented Jan 14, 2018 at 8:17
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$\begingroup$ @Medo: Poor phrasing. I meant that although $f$ does vanish on the boundary, it cannot be continuous up to the boundary. $\endgroup$ Commented Jan 14, 2018 at 15:16
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The answer is No.
There can be $f\in W^{1,p}(\Omega)\cap C(\Omega) \cap L^{\infty}(\Omega)$ such that $f=0$ on $\partial \Omega$ but $f\notin W^{1,p}_{0}(\Omega)$.
If $f\notin C({\bar{\Omega}})$, then its trace on $\partial \Omega$ can be different from zero, and in that case $f\notin W^{1,p}_{0}(\Omega)$.
Example: simply the characteristic function $\chi_{\Omega}$
The trace of a function $f \in C(\Omega)$ at $x\in \partial \Omega$ is the limit of $f$ when we move from the interior of $\Omega$ toward $x$. If $f\in C(\Omega) \cap L^{\infty}(\Omega)$ then its trace exists. If $f\in C({\bar{\Omega}})$ then the trace (the limit) at any $x\in\partial \Omega$ coincides with $f(x)$. So, assuming $f=0$ on $\partial \Omega$ and continuity on $\Omega$ are not enough.
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$\begingroup$ There is a lot wrong with this answer (and question). Functions in $W^{1,p}(\Omega)$ are equivalence classes of functions defined on the open set $\Omega$. Aside from the fact that $\partial \Omega$ has measure zero, the values of $f$ on $\partial \Omega$ are not even part of the definition of a Sobolev space function. There is no sensible way to say that the characteristic function of $\Omega$ as an element of $W^{1,p}(\Omega)$ is zero on the boundary. This is why the notion of trace was invented (and trace is not exactly the limit of $f$ as you move to the boundary; it is a bit more subtle). $\endgroup$– JeffCommented Jan 16, 2018 at 3:33
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$\begingroup$ Also, a function in $C(\Omega)\cap L^\infty(\Omega)$ need not have a trace; just consider $\sin(1/x)$ on $\Omega=(0,1)$. $\endgroup$– JeffCommented Jan 16, 2018 at 3:34