Question on Sobolev extension onto boundary












3














Let $U subset mathbb R^3$ be an open, bounded and connected set with a $C^2-$regular boundary $partial U$. I'm trying to understand the following implication:




If $fin W^{2-1/2,2}(U)$ then $f{vert}_{partial U} in W^{1,2}(partial U)(*)$




So, I'm aware of this theorem:



General Trace Theorem: if $fin W^{1-1/p,p}(partial Omega)$, then there exists a function $f in W^{1,p}(Omega)$ such that $f{vert}_{partial Omega}=f$



QUESTION: Is the above theorem still valid if we replace $partial Omega$ with $U$ and $Omega$ with $partial U$ so that $(*)$ makes sense? If not, is there any other way to deduce $(*)$?



Any help is appreciated. Thanks in advance!



EDIT: regularity in $(*)$ fixed






functional-analysis pde sobolev-spaces trace regularity-theory-of-pdes







share|cite|improve this question
























  • You're asking about whether you can extend a function from the boundary into the domain?
    – Ian
    Nov 13 at 15:52










  • I think that you have the regularities reversed. Anyway, see here.
    – Giuseppe Negro
    Nov 13 at 15:55










  • @Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary...
    – kaithkolesidou
    Nov 13 at 15:56










  • The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning.
    – Ian
    Nov 13 at 15:57










  • @GiuseppeNegro What do you mean by reversed regularities?
    – kaithkolesidou
    Nov 13 at 16:00
















3














Let $U subset mathbb R^3$ be an open, bounded and connected set with a $C^2-$regular boundary $partial U$. I'm trying to understand the following implication:




If $fin W^{2-1/2,2}(U)$ then $f{vert}_{partial U} in W^{1,2}(partial U)(*)$




So, I'm aware of this theorem:



General Trace Theorem: if $fin W^{1-1/p,p}(partial Omega)$, then there exists a function $f in W^{1,p}(Omega)$ such that $f{vert}_{partial Omega}=f$



QUESTION: Is the above theorem still valid if we replace $partial Omega$ with $U$ and $Omega$ with $partial U$ so that $(*)$ makes sense? If not, is there any other way to deduce $(*)$?



Any help is appreciated. Thanks in advance!



EDIT: regularity in $(*)$ fixed






functional-analysis pde sobolev-spaces trace regularity-theory-of-pdes







share|cite|improve this question
























  • You're asking about whether you can extend a function from the boundary into the domain?
    – Ian
    Nov 13 at 15:52










  • I think that you have the regularities reversed. Anyway, see here.
    – Giuseppe Negro
    Nov 13 at 15:55










  • @Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary...
    – kaithkolesidou
    Nov 13 at 15:56










  • The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning.
    – Ian
    Nov 13 at 15:57










  • @GiuseppeNegro What do you mean by reversed regularities?
    – kaithkolesidou
    Nov 13 at 16:00














3












3








3


1





Let $U subset mathbb R^3$ be an open, bounded and connected set with a $C^2-$regular boundary $partial U$. I'm trying to understand the following implication:




If $fin W^{2-1/2,2}(U)$ then $f{vert}_{partial U} in W^{1,2}(partial U)(*)$




So, I'm aware of this theorem:



General Trace Theorem: if $fin W^{1-1/p,p}(partial Omega)$, then there exists a function $f in W^{1,p}(Omega)$ such that $f{vert}_{partial Omega}=f$



QUESTION: Is the above theorem still valid if we replace $partial Omega$ with $U$ and $Omega$ with $partial U$ so that $(*)$ makes sense? If not, is there any other way to deduce $(*)$?



Any help is appreciated. Thanks in advance!



EDIT: regularity in $(*)$ fixed






functional-analysis pde sobolev-spaces trace regularity-theory-of-pdes







share|cite|improve this question















Let $U subset mathbb R^3$ be an open, bounded and connected set with a $C^2-$regular boundary $partial U$. I'm trying to understand the following implication:




If $fin W^{2-1/2,2}(U)$ then $f{vert}_{partial U} in W^{1,2}(partial U)(*)$




So, I'm aware of this theorem:



General Trace Theorem: if $fin W^{1-1/p,p}(partial Omega)$, then there exists a function $f in W^{1,p}(Omega)$ such that $f{vert}_{partial Omega}=f$



QUESTION: Is the above theorem still valid if we replace $partial Omega$ with $U$ and $Omega$ with $partial U$ so that $(*)$ makes sense? If not, is there any other way to deduce $(*)$?



Any help is appreciated. Thanks in advance!



EDIT: regularity in $(*)$ fixed






functional-analysis pde sobolev-spaces trace regularity-theory-of-pdes




functional-analysis pde sobolev-spaces trace regularity-theory-of-pdes






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 13:25

























asked Nov 13 at 15:51









kaithkolesidou

959511




959511












  • You're asking about whether you can extend a function from the boundary into the domain?
    – Ian
    Nov 13 at 15:52










  • I think that you have the regularities reversed. Anyway, see here.
    – Giuseppe Negro
    Nov 13 at 15:55










  • @Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary...
    – kaithkolesidou
    Nov 13 at 15:56










  • The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning.
    – Ian
    Nov 13 at 15:57










  • @GiuseppeNegro What do you mean by reversed regularities?
    – kaithkolesidou
    Nov 13 at 16:00


















  • You're asking about whether you can extend a function from the boundary into the domain?
    – Ian
    Nov 13 at 15:52










  • I think that you have the regularities reversed. Anyway, see here.
    – Giuseppe Negro
    Nov 13 at 15:55










  • @Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary...
    – kaithkolesidou
    Nov 13 at 15:56










  • The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning.
    – Ian
    Nov 13 at 15:57










  • @GiuseppeNegro What do you mean by reversed regularities?
    – kaithkolesidou
    Nov 13 at 16:00
















You're asking about whether you can extend a function from the boundary into the domain?
– Ian
Nov 13 at 15:52




You're asking about whether you can extend a function from the boundary into the domain?
– Ian
Nov 13 at 15:52












I think that you have the regularities reversed. Anyway, see here.
– Giuseppe Negro
Nov 13 at 15:55




I think that you have the regularities reversed. Anyway, see here.
– Giuseppe Negro
Nov 13 at 15:55












@Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary...
– kaithkolesidou
Nov 13 at 15:56




@Ian no... I am trying to understand the implication in $(*)$. There the extension is from the domain to boundary...
– kaithkolesidou
Nov 13 at 15:56












The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning.
– Ian
Nov 13 at 15:57




The given theorem is about extending from the inside of the domain to the boundary. Switching the two around would amount to extending a function given on the boundary to the domain, unless I am misunderstanding your intended meaning.
– Ian
Nov 13 at 15:57












@GiuseppeNegro What do you mean by reversed regularities?
– kaithkolesidou
Nov 13 at 16:00




@GiuseppeNegro What do you mean by reversed regularities?
– kaithkolesidou
Nov 13 at 16:00










1 Answer
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$newcommand{R}{mathbb{R}}$
The implication (*) is invalid, and for obvious reasons. The trace $f|_{partial U}$ is kind of a restriction of $f$, so you cannot expect it to have more derivatives than $f$!



To have an example, choose your favorite function $g in W^{1/2,2}(R)$ that doesn't belong to $W^{1,2}(R)$. Then consider
begin{align*}
U &= { (x,y,z) in R^3 : z > 0 }, \
f(x,y,z) & = g(x,y).
end{align*}
Then $f$ is locally in $W^{1/2,2}$, but its restriction to the boundary $partial U = R^2 times {0}$ is simply $g$, which is not in $W^{1,2}$.



Of course, one can modify this example to have a bounded domain.






share|cite|improve this answer





















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    $newcommand{R}{mathbb{R}}$
    The implication (*) is invalid, and for obvious reasons. The trace $f|_{partial U}$ is kind of a restriction of $f$, so you cannot expect it to have more derivatives than $f$!



    To have an example, choose your favorite function $g in W^{1/2,2}(R)$ that doesn't belong to $W^{1,2}(R)$. Then consider
    begin{align*}
    U &= { (x,y,z) in R^3 : z > 0 }, \
    f(x,y,z) & = g(x,y).
    end{align*}
    Then $f$ is locally in $W^{1/2,2}$, but its restriction to the boundary $partial U = R^2 times {0}$ is simply $g$, which is not in $W^{1,2}$.



    Of course, one can modify this example to have a bounded domain.






    share|cite|improve this answer


























      0














      $newcommand{R}{mathbb{R}}$
      The implication (*) is invalid, and for obvious reasons. The trace $f|_{partial U}$ is kind of a restriction of $f$, so you cannot expect it to have more derivatives than $f$!



      To have an example, choose your favorite function $g in W^{1/2,2}(R)$ that doesn't belong to $W^{1,2}(R)$. Then consider
      begin{align*}
      U &= { (x,y,z) in R^3 : z > 0 }, \
      f(x,y,z) & = g(x,y).
      end{align*}
      Then $f$ is locally in $W^{1/2,2}$, but its restriction to the boundary $partial U = R^2 times {0}$ is simply $g$, which is not in $W^{1,2}$.



      Of course, one can modify this example to have a bounded domain.






      share|cite|improve this answer
























        0












        0








        0






        $newcommand{R}{mathbb{R}}$
        The implication (*) is invalid, and for obvious reasons. The trace $f|_{partial U}$ is kind of a restriction of $f$, so you cannot expect it to have more derivatives than $f$!



        To have an example, choose your favorite function $g in W^{1/2,2}(R)$ that doesn't belong to $W^{1,2}(R)$. Then consider
        begin{align*}
        U &= { (x,y,z) in R^3 : z > 0 }, \
        f(x,y,z) & = g(x,y).
        end{align*}
        Then $f$ is locally in $W^{1/2,2}$, but its restriction to the boundary $partial U = R^2 times {0}$ is simply $g$, which is not in $W^{1,2}$.



        Of course, one can modify this example to have a bounded domain.






        share|cite|improve this answer












        $newcommand{R}{mathbb{R}}$
        The implication (*) is invalid, and for obvious reasons. The trace $f|_{partial U}$ is kind of a restriction of $f$, so you cannot expect it to have more derivatives than $f$!



        To have an example, choose your favorite function $g in W^{1/2,2}(R)$ that doesn't belong to $W^{1,2}(R)$. Then consider
        begin{align*}
        U &= { (x,y,z) in R^3 : z > 0 }, \
        f(x,y,z) & = g(x,y).
        end{align*}
        Then $f$ is locally in $W^{1/2,2}$, but its restriction to the boundary $partial U = R^2 times {0}$ is simply $g$, which is not in $W^{1,2}$.



        Of course, one can modify this example to have a bounded domain.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 13 at 18:51









        Michał Miśkiewicz

        2,743616




        2,743616






























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