Proof of Lemma 2 in Jürgen Moser's Proof of the Nash / De Giorgi / Moser Theorem












1












$begingroup$


I am currently studying Jürgen Moser's "A New Proof of de Giorgi's Theorem Concerning the Regularity Problem for Elliptic Differential Equations" (which can be found here: https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160130308).



In the Proof of the second Lemma there is one step in between two lines which I cannot follow (p.466, just before equation (22)), in spite trying to understand it for hours now. I will briefly sum up the problem:
Let $|x|< rho$ and $|xi|=1$, $w in L^2(B_{rho}(0))$. ($dsigma $ is the surface measure)



Then
$$ int_{0}^{rho} r^{n-1} dr int_{|xi|=1} 1 dsigma int^r_0 |w_x (x+ t xi)| dt$$
$$leq int_{rleq 2 rho} r^{n-1} dr int^r_0 frac{w_x (z)}{|x-z|^{n-1}} dt$$
holds.



So I tried myself and got to this point, using polar coordinates:
$$ int_{|xi|=1} 1 dsigma int^r_0 |w_x (x+ t xi)| d t = int_{B_{r}(0)} frac{|w_x (x+ |z| xi)|}{|z|^{n-1}} d z$$



I further though if I can find a $xi$ which minimizes $|w_x (x+ |z| xi)|$, then that would yield
$$|w_x (x+ |z| xi)| leq |w_x (x+ |z| frac{z}{|z|})|=|w_x (x+ z)|$$



From here on though, I don't get any further - I tried substituting in a clever way, but for me it didn't work out. I am in particular confused about the term $int_{rleq 2 rho} r^{n-1} dr$ regarding the larger set over which is integrated.
Thank you for any help in advance!






real-analysis ordinary-differential-equations pde elliptic-equations







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am currently studying Jürgen Moser's "A New Proof of de Giorgi's Theorem Concerning the Regularity Problem for Elliptic Differential Equations" (which can be found here: https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160130308).



    In the Proof of the second Lemma there is one step in between two lines which I cannot follow (p.466, just before equation (22)), in spite trying to understand it for hours now. I will briefly sum up the problem:
    Let $|x|< rho$ and $|xi|=1$, $w in L^2(B_{rho}(0))$. ($dsigma $ is the surface measure)



    Then
    $$ int_{0}^{rho} r^{n-1} dr int_{|xi|=1} 1 dsigma int^r_0 |w_x (x+ t xi)| dt$$
    $$leq int_{rleq 2 rho} r^{n-1} dr int^r_0 frac{w_x (z)}{|x-z|^{n-1}} dt$$
    holds.



    So I tried myself and got to this point, using polar coordinates:
    $$ int_{|xi|=1} 1 dsigma int^r_0 |w_x (x+ t xi)| d t = int_{B_{r}(0)} frac{|w_x (x+ |z| xi)|}{|z|^{n-1}} d z$$



    I further though if I can find a $xi$ which minimizes $|w_x (x+ |z| xi)|$, then that would yield
    $$|w_x (x+ |z| xi)| leq |w_x (x+ |z| frac{z}{|z|})|=|w_x (x+ z)|$$



    From here on though, I don't get any further - I tried substituting in a clever way, but for me it didn't work out. I am in particular confused about the term $int_{rleq 2 rho} r^{n-1} dr$ regarding the larger set over which is integrated.
    Thank you for any help in advance!






    real-analysis ordinary-differential-equations pde elliptic-equations







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I am currently studying Jürgen Moser's "A New Proof of de Giorgi's Theorem Concerning the Regularity Problem for Elliptic Differential Equations" (which can be found here: https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160130308).



      In the Proof of the second Lemma there is one step in between two lines which I cannot follow (p.466, just before equation (22)), in spite trying to understand it for hours now. I will briefly sum up the problem:
      Let $|x|< rho$ and $|xi|=1$, $w in L^2(B_{rho}(0))$. ($dsigma $ is the surface measure)



      Then
      $$ int_{0}^{rho} r^{n-1} dr int_{|xi|=1} 1 dsigma int^r_0 |w_x (x+ t xi)| dt$$
      $$leq int_{rleq 2 rho} r^{n-1} dr int^r_0 frac{w_x (z)}{|x-z|^{n-1}} dt$$
      holds.



      So I tried myself and got to this point, using polar coordinates:
      $$ int_{|xi|=1} 1 dsigma int^r_0 |w_x (x+ t xi)| d t = int_{B_{r}(0)} frac{|w_x (x+ |z| xi)|}{|z|^{n-1}} d z$$



      I further though if I can find a $xi$ which minimizes $|w_x (x+ |z| xi)|$, then that would yield
      $$|w_x (x+ |z| xi)| leq |w_x (x+ |z| frac{z}{|z|})|=|w_x (x+ z)|$$



      From here on though, I don't get any further - I tried substituting in a clever way, but for me it didn't work out. I am in particular confused about the term $int_{rleq 2 rho} r^{n-1} dr$ regarding the larger set over which is integrated.
      Thank you for any help in advance!






      real-analysis ordinary-differential-equations pde elliptic-equations







      share|cite|improve this question









      $endgroup$




      I am currently studying Jürgen Moser's "A New Proof of de Giorgi's Theorem Concerning the Regularity Problem for Elliptic Differential Equations" (which can be found here: https://onlinelibrary.wiley.com/doi/abs/10.1002/cpa.3160130308).



      In the Proof of the second Lemma there is one step in between two lines which I cannot follow (p.466, just before equation (22)), in spite trying to understand it for hours now. I will briefly sum up the problem:
      Let $|x|< rho$ and $|xi|=1$, $w in L^2(B_{rho}(0))$. ($dsigma $ is the surface measure)



      Then
      $$ int_{0}^{rho} r^{n-1} dr int_{|xi|=1} 1 dsigma int^r_0 |w_x (x+ t xi)| dt$$
      $$leq int_{rleq 2 rho} r^{n-1} dr int^r_0 frac{w_x (z)}{|x-z|^{n-1}} dt$$
      holds.



      So I tried myself and got to this point, using polar coordinates:
      $$ int_{|xi|=1} 1 dsigma int^r_0 |w_x (x+ t xi)| d t = int_{B_{r}(0)} frac{|w_x (x+ |z| xi)|}{|z|^{n-1}} d z$$



      I further though if I can find a $xi$ which minimizes $|w_x (x+ |z| xi)|$, then that would yield
      $$|w_x (x+ |z| xi)| leq |w_x (x+ |z| frac{z}{|z|})|=|w_x (x+ z)|$$



      From here on though, I don't get any further - I tried substituting in a clever way, but for me it didn't work out. I am in particular confused about the term $int_{rleq 2 rho} r^{n-1} dr$ regarding the larger set over which is integrated.
      Thank you for any help in advance!






      real-analysis ordinary-differential-equations pde elliptic-equations




      real-analysis ordinary-differential-equations pde elliptic-equations






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 5 '18 at 14:07









      MaxMax

      586




      586






















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          $begingroup$

          Recall the standard polar coordinate representation on balls,
          $$ int_{B_{r}(x)} f(z) ,mathrm{d}z = int_0^{r} int_{S^{n-1}} t^{n-1} f(x + txi), mathrm{d}xi,mathrm{d}t.$$
          If we apply this with $|f(z)|/|x-z|^{n-1}$ instead of $f,$ noting that $t = |x-z|$ we get,
          $$ int_{S^{n-1}} int_0^r |f(x+txi)| ,mathrm{d}t,mathrm{d}xi = int_{B_r(x)} frac{|f(z)|}{|x-z|^{n-1}} ,mathrm{d}z. $$
          Multiplying by $r^{n-1}$ on both sides and integrating on $(0,rho)$ we get,
          $$ int_0^{rho} r^{n-1} int_{S^{n-1}} int_0^r |f(x+txi)| ,mathrm{d}t,mathrm{d}xi,mathrm{d}r = int_0^{rho} r^{n-1} int_{B_r(x)} frac{|f(z)|}{|x-z|^{n-1}} ,mathrm{d}z,mathrm{d}r. $$



          This should be sufficient for your purposes, assuming there is a typo in the paper and the $w_x$ should be $|w_x|$ instead. I also think the $2rho$ is also a typo, as in the next line it's no longer there.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot for the answer! Still took me a while to come back and look at it again, but now I figured it out and I have to say that your presentation is way clearer than the one in the paper. :-)
            $endgroup$
            – Max
            Jan 9 at 16:34











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          active

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          active

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          1












          $begingroup$

          Recall the standard polar coordinate representation on balls,
          $$ int_{B_{r}(x)} f(z) ,mathrm{d}z = int_0^{r} int_{S^{n-1}} t^{n-1} f(x + txi), mathrm{d}xi,mathrm{d}t.$$
          If we apply this with $|f(z)|/|x-z|^{n-1}$ instead of $f,$ noting that $t = |x-z|$ we get,
          $$ int_{S^{n-1}} int_0^r |f(x+txi)| ,mathrm{d}t,mathrm{d}xi = int_{B_r(x)} frac{|f(z)|}{|x-z|^{n-1}} ,mathrm{d}z. $$
          Multiplying by $r^{n-1}$ on both sides and integrating on $(0,rho)$ we get,
          $$ int_0^{rho} r^{n-1} int_{S^{n-1}} int_0^r |f(x+txi)| ,mathrm{d}t,mathrm{d}xi,mathrm{d}r = int_0^{rho} r^{n-1} int_{B_r(x)} frac{|f(z)|}{|x-z|^{n-1}} ,mathrm{d}z,mathrm{d}r. $$



          This should be sufficient for your purposes, assuming there is a typo in the paper and the $w_x$ should be $|w_x|$ instead. I also think the $2rho$ is also a typo, as in the next line it's no longer there.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot for the answer! Still took me a while to come back and look at it again, but now I figured it out and I have to say that your presentation is way clearer than the one in the paper. :-)
            $endgroup$
            – Max
            Jan 9 at 16:34
















          1












          $begingroup$

          Recall the standard polar coordinate representation on balls,
          $$ int_{B_{r}(x)} f(z) ,mathrm{d}z = int_0^{r} int_{S^{n-1}} t^{n-1} f(x + txi), mathrm{d}xi,mathrm{d}t.$$
          If we apply this with $|f(z)|/|x-z|^{n-1}$ instead of $f,$ noting that $t = |x-z|$ we get,
          $$ int_{S^{n-1}} int_0^r |f(x+txi)| ,mathrm{d}t,mathrm{d}xi = int_{B_r(x)} frac{|f(z)|}{|x-z|^{n-1}} ,mathrm{d}z. $$
          Multiplying by $r^{n-1}$ on both sides and integrating on $(0,rho)$ we get,
          $$ int_0^{rho} r^{n-1} int_{S^{n-1}} int_0^r |f(x+txi)| ,mathrm{d}t,mathrm{d}xi,mathrm{d}r = int_0^{rho} r^{n-1} int_{B_r(x)} frac{|f(z)|}{|x-z|^{n-1}} ,mathrm{d}z,mathrm{d}r. $$



          This should be sufficient for your purposes, assuming there is a typo in the paper and the $w_x$ should be $|w_x|$ instead. I also think the $2rho$ is also a typo, as in the next line it's no longer there.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot for the answer! Still took me a while to come back and look at it again, but now I figured it out and I have to say that your presentation is way clearer than the one in the paper. :-)
            $endgroup$
            – Max
            Jan 9 at 16:34














          1












          1








          1





          $begingroup$

          Recall the standard polar coordinate representation on balls,
          $$ int_{B_{r}(x)} f(z) ,mathrm{d}z = int_0^{r} int_{S^{n-1}} t^{n-1} f(x + txi), mathrm{d}xi,mathrm{d}t.$$
          If we apply this with $|f(z)|/|x-z|^{n-1}$ instead of $f,$ noting that $t = |x-z|$ we get,
          $$ int_{S^{n-1}} int_0^r |f(x+txi)| ,mathrm{d}t,mathrm{d}xi = int_{B_r(x)} frac{|f(z)|}{|x-z|^{n-1}} ,mathrm{d}z. $$
          Multiplying by $r^{n-1}$ on both sides and integrating on $(0,rho)$ we get,
          $$ int_0^{rho} r^{n-1} int_{S^{n-1}} int_0^r |f(x+txi)| ,mathrm{d}t,mathrm{d}xi,mathrm{d}r = int_0^{rho} r^{n-1} int_{B_r(x)} frac{|f(z)|}{|x-z|^{n-1}} ,mathrm{d}z,mathrm{d}r. $$



          This should be sufficient for your purposes, assuming there is a typo in the paper and the $w_x$ should be $|w_x|$ instead. I also think the $2rho$ is also a typo, as in the next line it's no longer there.






          share|cite|improve this answer









          $endgroup$



          Recall the standard polar coordinate representation on balls,
          $$ int_{B_{r}(x)} f(z) ,mathrm{d}z = int_0^{r} int_{S^{n-1}} t^{n-1} f(x + txi), mathrm{d}xi,mathrm{d}t.$$
          If we apply this with $|f(z)|/|x-z|^{n-1}$ instead of $f,$ noting that $t = |x-z|$ we get,
          $$ int_{S^{n-1}} int_0^r |f(x+txi)| ,mathrm{d}t,mathrm{d}xi = int_{B_r(x)} frac{|f(z)|}{|x-z|^{n-1}} ,mathrm{d}z. $$
          Multiplying by $r^{n-1}$ on both sides and integrating on $(0,rho)$ we get,
          $$ int_0^{rho} r^{n-1} int_{S^{n-1}} int_0^r |f(x+txi)| ,mathrm{d}t,mathrm{d}xi,mathrm{d}r = int_0^{rho} r^{n-1} int_{B_r(x)} frac{|f(z)|}{|x-z|^{n-1}} ,mathrm{d}z,mathrm{d}r. $$



          This should be sufficient for your purposes, assuming there is a typo in the paper and the $w_x$ should be $|w_x|$ instead. I also think the $2rho$ is also a typo, as in the next line it's no longer there.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 18:10









          ktoiktoi

          2,3961616




          2,3961616












          • $begingroup$
            Thanks a lot for the answer! Still took me a while to come back and look at it again, but now I figured it out and I have to say that your presentation is way clearer than the one in the paper. :-)
            $endgroup$
            – Max
            Jan 9 at 16:34


















          • $begingroup$
            Thanks a lot for the answer! Still took me a while to come back and look at it again, but now I figured it out and I have to say that your presentation is way clearer than the one in the paper. :-)
            $endgroup$
            – Max
            Jan 9 at 16:34
















          $begingroup$
          Thanks a lot for the answer! Still took me a while to come back and look at it again, but now I figured it out and I have to say that your presentation is way clearer than the one in the paper. :-)
          $endgroup$
          – Max
          Jan 9 at 16:34




          $begingroup$
          Thanks a lot for the answer! Still took me a while to come back and look at it again, but now I figured it out and I have to say that your presentation is way clearer than the one in the paper. :-)
          $endgroup$
          – Max
          Jan 9 at 16:34


















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