Summability of double partial derivatives of $frac{1}{|x|}$ in dimension $3$
$begingroup$
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
asked Nov 29 '18 at 20:57
tommy1996qtommy1996q
591414
$endgroup$
add a comment |
$begingroup$
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
asked Nov 29 '18 at 20:57
tommy1996qtommy1996q
591414
$endgroup$
add a comment |
$begingroup$
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
asked Nov 29 '18 at 20:57
tommy1996qtommy1996q
591414
$endgroup$
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
asked Nov 29 '18 at 20:57
tommy1996qtommy1996q
591414
asked Nov 29 '18 at 20:57
tommy1996qtommy1996q
591414
asked Nov 29 '18 at 20:57
tommy1996qtommy1996q
591414
asked Nov 29 '18 at 20:57
tommy1996qtommy1996q
591414
asked Nov 29 '18 at 20:57
tommy1996qtommy1996q
591414
591414
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Nov 30 '18 at 14:04
Delta-uDelta-u
5,3892618
$endgroup$
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
|
show 2 more comments
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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.
answered Nov 30 '18 at 14:04
Delta-uDelta-u
5,3892618
$endgroup$
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
|
show 2 more comments
$begingroup$
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.
answered Nov 30 '18 at 14:04
Delta-uDelta-u
5,3892618
$endgroup$
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
|
show 2 more comments
$begingroup$
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.
answered Nov 30 '18 at 14:04
Delta-uDelta-u
5,3892618
$endgroup$
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.
answered Nov 30 '18 at 14:04
Delta-uDelta-u
5,3892618
answered Nov 30 '18 at 14:04
Delta-uDelta-u
5,3892618
answered Nov 30 '18 at 14:04
Delta-uDelta-u
5,3892618
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
|
show 2 more comments
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
|
show 2 more comments
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