Do non-continuous Sobolev maps pull back closed forms to weakly closed forms?












2















$newcommand{R}{mathbb R}$
$newcommand{N}{mathbb N}$
$newcommand{de}{delta}$
$newcommand{sig}{sigma}$
$newcommand{Average}[1]{leftlangle#1rightrangle} $
$newcommand{IP}[2]{Average{#1,#2}}$



Let $n,d in mathbb{N}$, and let $Omega subseteq R^n$ be open.
Let $f in W^{1,k}(Omega,R^d)$. Let $omega in Omega^k(R^d)$ be a smooth closed $k$-form; Is it true that $f^*omega$ is weakly closed? i.e. does
$$int_{Omega} IP{f^* omega}{de sig}=0$$
hold for every compactly-supported $k+1$-form $sigma in Omega^{k+1}(Omega)$?



I have read that this should be true, but I am only able to show this in two special cases:




  1. The form $omega$ is constant.


  2. $f$ is continuous.



Does this hold for non-continuous Sobolev maps in general?




Here is the problem as I see it: We approximate $f$ via smooth functions; suppose that $f_n in C^{infty}(Omega,R^d)$ satisfy $f_n to f$ in $W^{1,k}(Omega,R^d)$. Then
$$
int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{d f_n^* omega}{ sig}=0.
$$

Now, we need to justify the passage to the limit $int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}$.



If $omega$ is constant, that is $omega_q= alpha$ independently of $q in R^d$, where $alpha$ is a fixed element in $bigwedge^k (R^d)^* $, then we have
$$
|f^* omega-f_n^* omega| le |alpha| , left|bigwedge^{k} df-bigwedge^{k} df_nright|_{op},
$$

thus
$$
left|int_{Omega} IP{f^* omega}{de sig}- IP{ f_n^* omega}{de sig}right| le |alpha| |de sig|_{sup} int_{Omega} left|bigwedge^{k} df-bigwedge^{k} df_nright|.
$$

and The RHS tends to zero since Sobolev approximation lifts to exterior powers.



When $omega$ is not constant, we have a problem that the point of evaluation "moves with the function" that pulls back, that is
$$
begin{split}
& |f^* omega-f_n^* omega|(p)= \
&left|omega_{f(p)} circ bigwedge^{k} df_p -omega_{f_n(p)} circ bigwedge^{k} (df_n)_p right| le \
&left|big(omega_{f(p)}-omega_{f_n(p)}) circ bigwedge^{k} df_p +omega_{f_n(p)} circ big( bigwedge^{k} df_p-bigwedge^{k} (df_n)_p) right| le \
&|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right|
end{split}
$$



So, we we need to estimate $omega_{f(p)}-omega_{f_n(p)}$. When $f$ is continuous, we can take $f_n$ which converges uniformly to $f$, and thus we can continue the estimate:



$$
begin{split}
&|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
&|Tomega|_{sup} , cdot , |f(p)-f_n(p)| , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
&|Tomega|_{sup} , cdot , |f-f_n|_{sup} , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| stackrel{L^1}{to} 0,
%&|alpha circ brk{bigwedge^{k} df_p-bigwedge^{k} (df_n)_p} | le |alpha| cdot |bigwedge^{k} df_p-bigwedge^{k} (df_n)_p|_{op}.
end{split}
$$



So, does the preservation of closedness hold in general (for non-continuous maps) or is there a counter-example?










share|cite|improve this question



























    2















    $newcommand{R}{mathbb R}$
    $newcommand{N}{mathbb N}$
    $newcommand{de}{delta}$
    $newcommand{sig}{sigma}$
    $newcommand{Average}[1]{leftlangle#1rightrangle} $
    $newcommand{IP}[2]{Average{#1,#2}}$



    Let $n,d in mathbb{N}$, and let $Omega subseteq R^n$ be open.
    Let $f in W^{1,k}(Omega,R^d)$. Let $omega in Omega^k(R^d)$ be a smooth closed $k$-form; Is it true that $f^*omega$ is weakly closed? i.e. does
    $$int_{Omega} IP{f^* omega}{de sig}=0$$
    hold for every compactly-supported $k+1$-form $sigma in Omega^{k+1}(Omega)$?



    I have read that this should be true, but I am only able to show this in two special cases:




    1. The form $omega$ is constant.


    2. $f$ is continuous.



    Does this hold for non-continuous Sobolev maps in general?




    Here is the problem as I see it: We approximate $f$ via smooth functions; suppose that $f_n in C^{infty}(Omega,R^d)$ satisfy $f_n to f$ in $W^{1,k}(Omega,R^d)$. Then
    $$
    int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{d f_n^* omega}{ sig}=0.
    $$

    Now, we need to justify the passage to the limit $int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}$.



    If $omega$ is constant, that is $omega_q= alpha$ independently of $q in R^d$, where $alpha$ is a fixed element in $bigwedge^k (R^d)^* $, then we have
    $$
    |f^* omega-f_n^* omega| le |alpha| , left|bigwedge^{k} df-bigwedge^{k} df_nright|_{op},
    $$

    thus
    $$
    left|int_{Omega} IP{f^* omega}{de sig}- IP{ f_n^* omega}{de sig}right| le |alpha| |de sig|_{sup} int_{Omega} left|bigwedge^{k} df-bigwedge^{k} df_nright|.
    $$

    and The RHS tends to zero since Sobolev approximation lifts to exterior powers.



    When $omega$ is not constant, we have a problem that the point of evaluation "moves with the function" that pulls back, that is
    $$
    begin{split}
    & |f^* omega-f_n^* omega|(p)= \
    &left|omega_{f(p)} circ bigwedge^{k} df_p -omega_{f_n(p)} circ bigwedge^{k} (df_n)_p right| le \
    &left|big(omega_{f(p)}-omega_{f_n(p)}) circ bigwedge^{k} df_p +omega_{f_n(p)} circ big( bigwedge^{k} df_p-bigwedge^{k} (df_n)_p) right| le \
    &|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right|
    end{split}
    $$



    So, we we need to estimate $omega_{f(p)}-omega_{f_n(p)}$. When $f$ is continuous, we can take $f_n$ which converges uniformly to $f$, and thus we can continue the estimate:



    $$
    begin{split}
    &|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
    &|Tomega|_{sup} , cdot , |f(p)-f_n(p)| , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
    &|Tomega|_{sup} , cdot , |f-f_n|_{sup} , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| stackrel{L^1}{to} 0,
    %&|alpha circ brk{bigwedge^{k} df_p-bigwedge^{k} (df_n)_p} | le |alpha| cdot |bigwedge^{k} df_p-bigwedge^{k} (df_n)_p|_{op}.
    end{split}
    $$



    So, does the preservation of closedness hold in general (for non-continuous maps) or is there a counter-example?










    share|cite|improve this question

























      2












      2








      2


      1






      $newcommand{R}{mathbb R}$
      $newcommand{N}{mathbb N}$
      $newcommand{de}{delta}$
      $newcommand{sig}{sigma}$
      $newcommand{Average}[1]{leftlangle#1rightrangle} $
      $newcommand{IP}[2]{Average{#1,#2}}$



      Let $n,d in mathbb{N}$, and let $Omega subseteq R^n$ be open.
      Let $f in W^{1,k}(Omega,R^d)$. Let $omega in Omega^k(R^d)$ be a smooth closed $k$-form; Is it true that $f^*omega$ is weakly closed? i.e. does
      $$int_{Omega} IP{f^* omega}{de sig}=0$$
      hold for every compactly-supported $k+1$-form $sigma in Omega^{k+1}(Omega)$?



      I have read that this should be true, but I am only able to show this in two special cases:




      1. The form $omega$ is constant.


      2. $f$ is continuous.



      Does this hold for non-continuous Sobolev maps in general?




      Here is the problem as I see it: We approximate $f$ via smooth functions; suppose that $f_n in C^{infty}(Omega,R^d)$ satisfy $f_n to f$ in $W^{1,k}(Omega,R^d)$. Then
      $$
      int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{d f_n^* omega}{ sig}=0.
      $$

      Now, we need to justify the passage to the limit $int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}$.



      If $omega$ is constant, that is $omega_q= alpha$ independently of $q in R^d$, where $alpha$ is a fixed element in $bigwedge^k (R^d)^* $, then we have
      $$
      |f^* omega-f_n^* omega| le |alpha| , left|bigwedge^{k} df-bigwedge^{k} df_nright|_{op},
      $$

      thus
      $$
      left|int_{Omega} IP{f^* omega}{de sig}- IP{ f_n^* omega}{de sig}right| le |alpha| |de sig|_{sup} int_{Omega} left|bigwedge^{k} df-bigwedge^{k} df_nright|.
      $$

      and The RHS tends to zero since Sobolev approximation lifts to exterior powers.



      When $omega$ is not constant, we have a problem that the point of evaluation "moves with the function" that pulls back, that is
      $$
      begin{split}
      & |f^* omega-f_n^* omega|(p)= \
      &left|omega_{f(p)} circ bigwedge^{k} df_p -omega_{f_n(p)} circ bigwedge^{k} (df_n)_p right| le \
      &left|big(omega_{f(p)}-omega_{f_n(p)}) circ bigwedge^{k} df_p +omega_{f_n(p)} circ big( bigwedge^{k} df_p-bigwedge^{k} (df_n)_p) right| le \
      &|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right|
      end{split}
      $$



      So, we we need to estimate $omega_{f(p)}-omega_{f_n(p)}$. When $f$ is continuous, we can take $f_n$ which converges uniformly to $f$, and thus we can continue the estimate:



      $$
      begin{split}
      &|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
      &|Tomega|_{sup} , cdot , |f(p)-f_n(p)| , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
      &|Tomega|_{sup} , cdot , |f-f_n|_{sup} , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| stackrel{L^1}{to} 0,
      %&|alpha circ brk{bigwedge^{k} df_p-bigwedge^{k} (df_n)_p} | le |alpha| cdot |bigwedge^{k} df_p-bigwedge^{k} (df_n)_p|_{op}.
      end{split}
      $$



      So, does the preservation of closedness hold in general (for non-continuous maps) or is there a counter-example?










      share|cite|improve this question














      $newcommand{R}{mathbb R}$
      $newcommand{N}{mathbb N}$
      $newcommand{de}{delta}$
      $newcommand{sig}{sigma}$
      $newcommand{Average}[1]{leftlangle#1rightrangle} $
      $newcommand{IP}[2]{Average{#1,#2}}$



      Let $n,d in mathbb{N}$, and let $Omega subseteq R^n$ be open.
      Let $f in W^{1,k}(Omega,R^d)$. Let $omega in Omega^k(R^d)$ be a smooth closed $k$-form; Is it true that $f^*omega$ is weakly closed? i.e. does
      $$int_{Omega} IP{f^* omega}{de sig}=0$$
      hold for every compactly-supported $k+1$-form $sigma in Omega^{k+1}(Omega)$?



      I have read that this should be true, but I am only able to show this in two special cases:




      1. The form $omega$ is constant.


      2. $f$ is continuous.



      Does this hold for non-continuous Sobolev maps in general?




      Here is the problem as I see it: We approximate $f$ via smooth functions; suppose that $f_n in C^{infty}(Omega,R^d)$ satisfy $f_n to f$ in $W^{1,k}(Omega,R^d)$. Then
      $$
      int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{d f_n^* omega}{ sig}=0.
      $$

      Now, we need to justify the passage to the limit $int_{Omega} IP{f^* omega}{de sig}=lim_{n to infty} int_{Omega} IP{ f_n^* omega}{de sig}$.



      If $omega$ is constant, that is $omega_q= alpha$ independently of $q in R^d$, where $alpha$ is a fixed element in $bigwedge^k (R^d)^* $, then we have
      $$
      |f^* omega-f_n^* omega| le |alpha| , left|bigwedge^{k} df-bigwedge^{k} df_nright|_{op},
      $$

      thus
      $$
      left|int_{Omega} IP{f^* omega}{de sig}- IP{ f_n^* omega}{de sig}right| le |alpha| |de sig|_{sup} int_{Omega} left|bigwedge^{k} df-bigwedge^{k} df_nright|.
      $$

      and The RHS tends to zero since Sobolev approximation lifts to exterior powers.



      When $omega$ is not constant, we have a problem that the point of evaluation "moves with the function" that pulls back, that is
      $$
      begin{split}
      & |f^* omega-f_n^* omega|(p)= \
      &left|omega_{f(p)} circ bigwedge^{k} df_p -omega_{f_n(p)} circ bigwedge^{k} (df_n)_p right| le \
      &left|big(omega_{f(p)}-omega_{f_n(p)}) circ bigwedge^{k} df_p +omega_{f_n(p)} circ big( bigwedge^{k} df_p-bigwedge^{k} (df_n)_p) right| le \
      &|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right|
      end{split}
      $$



      So, we we need to estimate $omega_{f(p)}-omega_{f_n(p)}$. When $f$ is continuous, we can take $f_n$ which converges uniformly to $f$, and thus we can continue the estimate:



      $$
      begin{split}
      &|omega_{f(p)}-omega_{f_n(p)}| , , cdot , , left| bigwedge^{k} df_pright| +|omega_{f_n(p)} | , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
      &|Tomega|_{sup} , cdot , |f(p)-f_n(p)| , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| le \
      &|Tomega|_{sup} , cdot , |f-f_n|_{sup} , cdot , left| bigwedge^{k} df_pright| +|omega |_{sup} , , cdot , , left| bigwedge^{k} df_p-bigwedge^{k} (df_n)_p right| stackrel{L^1}{to} 0,
      %&|alpha circ brk{bigwedge^{k} df_p-bigwedge^{k} (df_n)_p} | le |alpha| cdot |bigwedge^{k} df_p-bigwedge^{k} (df_n)_p|_{op}.
      end{split}
      $$



      So, does the preservation of closedness hold in general (for non-continuous maps) or is there a counter-example?







      dg.differential-geometry ap.analysis-of-pdes sobolev-spaces regularity exterior-algebra






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      asked 2 hours ago









      Asaf ShacharAsaf Shachar

      2,2391746




      2,2391746






















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          This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf. The assumptions might be a bit different than yours, but it might be useful. I guess the assumptions in Lemma 4.1 are close to be optimal. I will expand my answer when I have time. Let me know if Lemma 4.1 suits you.






          share|cite|improve this answer

























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            This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf. The assumptions might be a bit different than yours, but it might be useful. I guess the assumptions in Lemma 4.1 are close to be optimal. I will expand my answer when I have time. Let me know if Lemma 4.1 suits you.






            share|cite|improve this answer






























              2














              This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf. The assumptions might be a bit different than yours, but it might be useful. I guess the assumptions in Lemma 4.1 are close to be optimal. I will expand my answer when I have time. Let me know if Lemma 4.1 suits you.






              share|cite|improve this answer




























                2












                2








                2







                This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf. The assumptions might be a bit different than yours, but it might be useful. I guess the assumptions in Lemma 4.1 are close to be optimal. I will expand my answer when I have time. Let me know if Lemma 4.1 suits you.






                share|cite|improve this answer















                This is I guess Lemma 4.1 in https://arxiv.org/pdf/1301.4978.pdf. The assumptions might be a bit different than yours, but it might be useful. I guess the assumptions in Lemma 4.1 are close to be optimal. I will expand my answer when I have time. Let me know if Lemma 4.1 suits you.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 42 mins ago









                Martin Sleziak

                2,95532028




                2,95532028










                answered 47 mins ago









                Piotr HajlaszPiotr Hajlasz

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                6,50442356






























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