Summability of double partial derivatives of $frac{1}{|x|}$ in dimension $3$












1












$begingroup$


I know the fact that its laplacian is equal to the Dirac delta function. However, is it true that its partial derivatives of order 2 belong to $L^{1}(mathbb{R}^3)$? And how should I show that, in case this is true? I had thought of calculating derivatives in standard coordinates and then pass to spherical coordinates to see if something got simpler, but I am not sure this would work.






real-analysis analysis partial-derivative lp-spaces laplacian







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I know the fact that its laplacian is equal to the Dirac delta function. However, is it true that its partial derivatives of order 2 belong to $L^{1}(mathbb{R}^3)$? And how should I show that, in case this is true? I had thought of calculating derivatives in standard coordinates and then pass to spherical coordinates to see if something got simpler, but I am not sure this would work.






    real-analysis analysis partial-derivative lp-spaces laplacian







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I know the fact that its laplacian is equal to the Dirac delta function. However, is it true that its partial derivatives of order 2 belong to $L^{1}(mathbb{R}^3)$? And how should I show that, in case this is true? I had thought of calculating derivatives in standard coordinates and then pass to spherical coordinates to see if something got simpler, but I am not sure this would work.






      real-analysis analysis partial-derivative lp-spaces laplacian







      share|cite|improve this question









      $endgroup$




      I know the fact that its laplacian is equal to the Dirac delta function. However, is it true that its partial derivatives of order 2 belong to $L^{1}(mathbb{R}^3)$? And how should I show that, in case this is true? I had thought of calculating derivatives in standard coordinates and then pass to spherical coordinates to see if something got simpler, but I am not sure this would work.






      real-analysis analysis partial-derivative lp-spaces laplacian




      real-analysis analysis partial-derivative lp-spaces laplacian






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 29 '18 at 20:57









      tommy1996qtommy1996q

      591414




      591414






















          1 Answer
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          $begingroup$

          The derivatives of order 2 of $u(x)=frac{1}{|x|}$does not belong to $L^1$. The sum $left(partial^2_1+ partial^2_2+partial^2_3 right) u= delta_0 notin L^1$ so at least one of them is not in $L^1$ (and by symmetry none of them).



          As you mentioned an other option is to compute these derivatives:
          $$partial_i u =frac{x_i}{|x|^3}$$
          so:
          $$partial_{ij}^2 u = begin{cases}frac{1}{|x|^3}-frac{3x_i^2} {|x^5|} text{ if } i=j \
          -3frac{x_i x_j}{|x|^5} text{ if } i neq j end{cases}$$
          but none of these functions are in $L^1$. For example using spherical coordinates you obtain
          $$iiint_{mathbb{R}^3} frac{x_1 x_2}{|x|^5} dx_1d_x2d_x3 =int_0^infty int_0^{2 pi} int_0^pi frac{r^2 sin(theta)^2 |cos(phi) sin(phi)|}{r^5} r^2 sin(theta) dtheta dphi dr=int_0^infty frac{1}{r} left(int_0^{2pi}int_0^pi sin(theta)^3 frac{1}{2} |sin(2phi)| dtheta dphi right) dr=+infty$$.



          The same behavior emerges in all cases: a singularity like $frac{1}{r}$ which is not integrable.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            You are absolutely right, I ecplained it wrong. What I wanted to show is that the convolution of those derivatives with a function belonging to $L^1$ belongs to $L^1$, so the Dirac delta doesn’t bother me.
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:09








          • 1




            $begingroup$
            Can I show that those derivatives behave like a properly rescaled Dirac delta? Like the double derivative with respect to only 1 variable behave like a third of the laplacian (by simmetry this should hold)
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:11






          • 1




            $begingroup$
            The singularity should appear in the case of the laplacian as well. So shouldn’t there be a way to get a properly rescaled Dirac delta eith the double detivatives as well?
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:14






          • 1




            $begingroup$
            @tommy1996q It is very interesting question :-). The $1/r$ singularity is not too bad as you are in a weak $L^1$ space. But unfortunately Young inequality for $L^p/L^{q,infty}$ spaces does not hold for $p=q=1$. So you are in a way in a "critical" case so it is maybe possible to have such property but one must use some properties of $1/|x|$ to do so.
            $endgroup$
            – Delta-u
            Nov 30 '18 at 14:20












          • $begingroup$
            Well, if $p=1$ is a problem, feel free to take higher values. The key point here is I want to see how to work with those derivatives in a useful way, like it is done for the laplacian
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:33











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          1












          $begingroup$

          The derivatives of order 2 of $u(x)=frac{1}{|x|}$does not belong to $L^1$. The sum $left(partial^2_1+ partial^2_2+partial^2_3 right) u= delta_0 notin L^1$ so at least one of them is not in $L^1$ (and by symmetry none of them).



          As you mentioned an other option is to compute these derivatives:
          $$partial_i u =frac{x_i}{|x|^3}$$
          so:
          $$partial_{ij}^2 u = begin{cases}frac{1}{|x|^3}-frac{3x_i^2} {|x^5|} text{ if } i=j \
          -3frac{x_i x_j}{|x|^5} text{ if } i neq j end{cases}$$
          but none of these functions are in $L^1$. For example using spherical coordinates you obtain
          $$iiint_{mathbb{R}^3} frac{x_1 x_2}{|x|^5} dx_1d_x2d_x3 =int_0^infty int_0^{2 pi} int_0^pi frac{r^2 sin(theta)^2 |cos(phi) sin(phi)|}{r^5} r^2 sin(theta) dtheta dphi dr=int_0^infty frac{1}{r} left(int_0^{2pi}int_0^pi sin(theta)^3 frac{1}{2} |sin(2phi)| dtheta dphi right) dr=+infty$$.



          The same behavior emerges in all cases: a singularity like $frac{1}{r}$ which is not integrable.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            You are absolutely right, I ecplained it wrong. What I wanted to show is that the convolution of those derivatives with a function belonging to $L^1$ belongs to $L^1$, so the Dirac delta doesn’t bother me.
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:09








          • 1




            $begingroup$
            Can I show that those derivatives behave like a properly rescaled Dirac delta? Like the double derivative with respect to only 1 variable behave like a third of the laplacian (by simmetry this should hold)
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:11






          • 1




            $begingroup$
            The singularity should appear in the case of the laplacian as well. So shouldn’t there be a way to get a properly rescaled Dirac delta eith the double detivatives as well?
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:14






          • 1




            $begingroup$
            @tommy1996q It is very interesting question :-). The $1/r$ singularity is not too bad as you are in a weak $L^1$ space. But unfortunately Young inequality for $L^p/L^{q,infty}$ spaces does not hold for $p=q=1$. So you are in a way in a "critical" case so it is maybe possible to have such property but one must use some properties of $1/|x|$ to do so.
            $endgroup$
            – Delta-u
            Nov 30 '18 at 14:20












          • $begingroup$
            Well, if $p=1$ is a problem, feel free to take higher values. The key point here is I want to see how to work with those derivatives in a useful way, like it is done for the laplacian
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:33
















          1












          $begingroup$

          The derivatives of order 2 of $u(x)=frac{1}{|x|}$does not belong to $L^1$. The sum $left(partial^2_1+ partial^2_2+partial^2_3 right) u= delta_0 notin L^1$ so at least one of them is not in $L^1$ (and by symmetry none of them).



          As you mentioned an other option is to compute these derivatives:
          $$partial_i u =frac{x_i}{|x|^3}$$
          so:
          $$partial_{ij}^2 u = begin{cases}frac{1}{|x|^3}-frac{3x_i^2} {|x^5|} text{ if } i=j \
          -3frac{x_i x_j}{|x|^5} text{ if } i neq j end{cases}$$
          but none of these functions are in $L^1$. For example using spherical coordinates you obtain
          $$iiint_{mathbb{R}^3} frac{x_1 x_2}{|x|^5} dx_1d_x2d_x3 =int_0^infty int_0^{2 pi} int_0^pi frac{r^2 sin(theta)^2 |cos(phi) sin(phi)|}{r^5} r^2 sin(theta) dtheta dphi dr=int_0^infty frac{1}{r} left(int_0^{2pi}int_0^pi sin(theta)^3 frac{1}{2} |sin(2phi)| dtheta dphi right) dr=+infty$$.



          The same behavior emerges in all cases: a singularity like $frac{1}{r}$ which is not integrable.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            You are absolutely right, I ecplained it wrong. What I wanted to show is that the convolution of those derivatives with a function belonging to $L^1$ belongs to $L^1$, so the Dirac delta doesn’t bother me.
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:09








          • 1




            $begingroup$
            Can I show that those derivatives behave like a properly rescaled Dirac delta? Like the double derivative with respect to only 1 variable behave like a third of the laplacian (by simmetry this should hold)
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:11






          • 1




            $begingroup$
            The singularity should appear in the case of the laplacian as well. So shouldn’t there be a way to get a properly rescaled Dirac delta eith the double detivatives as well?
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:14






          • 1




            $begingroup$
            @tommy1996q It is very interesting question :-). The $1/r$ singularity is not too bad as you are in a weak $L^1$ space. But unfortunately Young inequality for $L^p/L^{q,infty}$ spaces does not hold for $p=q=1$. So you are in a way in a "critical" case so it is maybe possible to have such property but one must use some properties of $1/|x|$ to do so.
            $endgroup$
            – Delta-u
            Nov 30 '18 at 14:20












          • $begingroup$
            Well, if $p=1$ is a problem, feel free to take higher values. The key point here is I want to see how to work with those derivatives in a useful way, like it is done for the laplacian
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:33














          1












          1








          1





          $begingroup$

          The derivatives of order 2 of $u(x)=frac{1}{|x|}$does not belong to $L^1$. The sum $left(partial^2_1+ partial^2_2+partial^2_3 right) u= delta_0 notin L^1$ so at least one of them is not in $L^1$ (and by symmetry none of them).



          As you mentioned an other option is to compute these derivatives:
          $$partial_i u =frac{x_i}{|x|^3}$$
          so:
          $$partial_{ij}^2 u = begin{cases}frac{1}{|x|^3}-frac{3x_i^2} {|x^5|} text{ if } i=j \
          -3frac{x_i x_j}{|x|^5} text{ if } i neq j end{cases}$$
          but none of these functions are in $L^1$. For example using spherical coordinates you obtain
          $$iiint_{mathbb{R}^3} frac{x_1 x_2}{|x|^5} dx_1d_x2d_x3 =int_0^infty int_0^{2 pi} int_0^pi frac{r^2 sin(theta)^2 |cos(phi) sin(phi)|}{r^5} r^2 sin(theta) dtheta dphi dr=int_0^infty frac{1}{r} left(int_0^{2pi}int_0^pi sin(theta)^3 frac{1}{2} |sin(2phi)| dtheta dphi right) dr=+infty$$.



          The same behavior emerges in all cases: a singularity like $frac{1}{r}$ which is not integrable.






          share|cite|improve this answer









          $endgroup$



          The derivatives of order 2 of $u(x)=frac{1}{|x|}$does not belong to $L^1$. The sum $left(partial^2_1+ partial^2_2+partial^2_3 right) u= delta_0 notin L^1$ so at least one of them is not in $L^1$ (and by symmetry none of them).



          As you mentioned an other option is to compute these derivatives:
          $$partial_i u =frac{x_i}{|x|^3}$$
          so:
          $$partial_{ij}^2 u = begin{cases}frac{1}{|x|^3}-frac{3x_i^2} {|x^5|} text{ if } i=j \
          -3frac{x_i x_j}{|x|^5} text{ if } i neq j end{cases}$$
          but none of these functions are in $L^1$. For example using spherical coordinates you obtain
          $$iiint_{mathbb{R}^3} frac{x_1 x_2}{|x|^5} dx_1d_x2d_x3 =int_0^infty int_0^{2 pi} int_0^pi frac{r^2 sin(theta)^2 |cos(phi) sin(phi)|}{r^5} r^2 sin(theta) dtheta dphi dr=int_0^infty frac{1}{r} left(int_0^{2pi}int_0^pi sin(theta)^3 frac{1}{2} |sin(2phi)| dtheta dphi right) dr=+infty$$.



          The same behavior emerges in all cases: a singularity like $frac{1}{r}$ which is not integrable.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 14:04









          Delta-uDelta-u

          5,3892618




          5,3892618








          • 1




            $begingroup$
            You are absolutely right, I ecplained it wrong. What I wanted to show is that the convolution of those derivatives with a function belonging to $L^1$ belongs to $L^1$, so the Dirac delta doesn’t bother me.
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:09








          • 1




            $begingroup$
            Can I show that those derivatives behave like a properly rescaled Dirac delta? Like the double derivative with respect to only 1 variable behave like a third of the laplacian (by simmetry this should hold)
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:11






          • 1




            $begingroup$
            The singularity should appear in the case of the laplacian as well. So shouldn’t there be a way to get a properly rescaled Dirac delta eith the double detivatives as well?
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:14






          • 1




            $begingroup$
            @tommy1996q It is very interesting question :-). The $1/r$ singularity is not too bad as you are in a weak $L^1$ space. But unfortunately Young inequality for $L^p/L^{q,infty}$ spaces does not hold for $p=q=1$. So you are in a way in a "critical" case so it is maybe possible to have such property but one must use some properties of $1/|x|$ to do so.
            $endgroup$
            – Delta-u
            Nov 30 '18 at 14:20












          • $begingroup$
            Well, if $p=1$ is a problem, feel free to take higher values. The key point here is I want to see how to work with those derivatives in a useful way, like it is done for the laplacian
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:33














          • 1




            $begingroup$
            You are absolutely right, I ecplained it wrong. What I wanted to show is that the convolution of those derivatives with a function belonging to $L^1$ belongs to $L^1$, so the Dirac delta doesn’t bother me.
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:09








          • 1




            $begingroup$
            Can I show that those derivatives behave like a properly rescaled Dirac delta? Like the double derivative with respect to only 1 variable behave like a third of the laplacian (by simmetry this should hold)
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:11






          • 1




            $begingroup$
            The singularity should appear in the case of the laplacian as well. So shouldn’t there be a way to get a properly rescaled Dirac delta eith the double detivatives as well?
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:14






          • 1




            $begingroup$
            @tommy1996q It is very interesting question :-). The $1/r$ singularity is not too bad as you are in a weak $L^1$ space. But unfortunately Young inequality for $L^p/L^{q,infty}$ spaces does not hold for $p=q=1$. So you are in a way in a "critical" case so it is maybe possible to have such property but one must use some properties of $1/|x|$ to do so.
            $endgroup$
            – Delta-u
            Nov 30 '18 at 14:20












          • $begingroup$
            Well, if $p=1$ is a problem, feel free to take higher values. The key point here is I want to see how to work with those derivatives in a useful way, like it is done for the laplacian
            $endgroup$
            – tommy1996q
            Nov 30 '18 at 14:33








          1




          1




          $begingroup$
          You are absolutely right, I ecplained it wrong. What I wanted to show is that the convolution of those derivatives with a function belonging to $L^1$ belongs to $L^1$, so the Dirac delta doesn’t bother me.
          $endgroup$
          – tommy1996q
          Nov 30 '18 at 14:09






          $begingroup$
          You are absolutely right, I ecplained it wrong. What I wanted to show is that the convolution of those derivatives with a function belonging to $L^1$ belongs to $L^1$, so the Dirac delta doesn’t bother me.
          $endgroup$
          – tommy1996q
          Nov 30 '18 at 14:09






          1




          1




          $begingroup$
          Can I show that those derivatives behave like a properly rescaled Dirac delta? Like the double derivative with respect to only 1 variable behave like a third of the laplacian (by simmetry this should hold)
          $endgroup$
          – tommy1996q
          Nov 30 '18 at 14:11




          $begingroup$
          Can I show that those derivatives behave like a properly rescaled Dirac delta? Like the double derivative with respect to only 1 variable behave like a third of the laplacian (by simmetry this should hold)
          $endgroup$
          – tommy1996q
          Nov 30 '18 at 14:11




          1




          1




          $begingroup$
          The singularity should appear in the case of the laplacian as well. So shouldn’t there be a way to get a properly rescaled Dirac delta eith the double detivatives as well?
          $endgroup$
          – tommy1996q
          Nov 30 '18 at 14:14




          $begingroup$
          The singularity should appear in the case of the laplacian as well. So shouldn’t there be a way to get a properly rescaled Dirac delta eith the double detivatives as well?
          $endgroup$
          – tommy1996q
          Nov 30 '18 at 14:14




          1




          1




          $begingroup$
          @tommy1996q It is very interesting question :-). The $1/r$ singularity is not too bad as you are in a weak $L^1$ space. But unfortunately Young inequality for $L^p/L^{q,infty}$ spaces does not hold for $p=q=1$. So you are in a way in a "critical" case so it is maybe possible to have such property but one must use some properties of $1/|x|$ to do so.
          $endgroup$
          – Delta-u
          Nov 30 '18 at 14:20






          $begingroup$
          @tommy1996q It is very interesting question :-). The $1/r$ singularity is not too bad as you are in a weak $L^1$ space. But unfortunately Young inequality for $L^p/L^{q,infty}$ spaces does not hold for $p=q=1$. So you are in a way in a "critical" case so it is maybe possible to have such property but one must use some properties of $1/|x|$ to do so.
          $endgroup$
          – Delta-u
          Nov 30 '18 at 14:20














          $begingroup$
          Well, if $p=1$ is a problem, feel free to take higher values. The key point here is I want to see how to work with those derivatives in a useful way, like it is done for the laplacian
          $endgroup$
          – tommy1996q
          Nov 30 '18 at 14:33




          $begingroup$
          Well, if $p=1$ is a problem, feel free to take higher values. The key point here is I want to see how to work with those derivatives in a useful way, like it is done for the laplacian
          $endgroup$
          – tommy1996q
          Nov 30 '18 at 14:33


















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          Musa

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          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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