Elemental computation?
I´m trying to understand the proof of Thm 4. on this paper
https://arxiv.org/pdf/1501.06828.pdf. At the end it claims that after elemental computations it can be shown that
$$vert zvert ^{4H+ alpha -d}int_{mathbb{S}^{d-1}}int_0^inftyfrac{r^{d-1-alpha}}{(vert zvert^2 + r^2)^{2H}}(1-cos(r
(thetacdottheta_z)))drsigma(dtheta)leq Cvert z vert ^ {2 wedge (4H+alpha-d) } bigg(logfrac{1}{vert z vert}bigg)^{beta}$$
where
$rho=1$ if $2=4H+alpha-d$ and $rho=0$ in other case- $alpha< d < 4H+alpha $
- $frac{1}{2}<H<1$
$sigma$ is the uniform measure on $mathbb{S}^{d-1}$ and- $theta_z=frac{z}{vert zvert}$
- c is a positive constant
I don´t understand why it follows after elemental computation , I tried bounding $1-cos(r (thetacdottheta_z))$ by 1, but this clearly does not work. For example in the case $4H+alpha-d<2$ we have that after the change of variables $u=frac{r}{vert zvert}$
$$vert zvert ^{4H+ alpha -d}int_{mathbb{S}^{d-1}}int_0^inftyfrac{r^{d-1-alpha}}{(vert zvert^2 + r^2)^{2H}}drsigma(dtheta)=int_{mathbb{S}^{d-1}}int_0^inftyfrac{u^{d-1-alpha}}{(1+ u^2)^{2H}}drsigma(dtheta)$$
Do you have any idea for proving this claim?
real-analysis
asked Nov 26 at 10:58
Adrián Hinojosa Calleja
937
add a comment |
I´m trying to understand the proof of Thm 4. on this paper
https://arxiv.org/pdf/1501.06828.pdf. At the end it claims that after elemental computations it can be shown that
$$vert zvert ^{4H+ alpha -d}int_{mathbb{S}^{d-1}}int_0^inftyfrac{r^{d-1-alpha}}{(vert zvert^2 + r^2)^{2H}}(1-cos(r
(thetacdottheta_z)))drsigma(dtheta)leq Cvert z vert ^ {2 wedge (4H+alpha-d) } bigg(logfrac{1}{vert z vert}bigg)^{beta}$$
where
$rho=1$ if $2=4H+alpha-d$ and $rho=0$ in other case- $alpha< d < 4H+alpha $
- $frac{1}{2}<H<1$
$sigma$ is the uniform measure on $mathbb{S}^{d-1}$ and- $theta_z=frac{z}{vert zvert}$
- c is a positive constant
I don´t understand why it follows after elemental computation , I tried bounding $1-cos(r (thetacdottheta_z))$ by 1, but this clearly does not work. For example in the case $4H+alpha-d<2$ we have that after the change of variables $u=frac{r}{vert zvert}$
$$vert zvert ^{4H+ alpha -d}int_{mathbb{S}^{d-1}}int_0^inftyfrac{r^{d-1-alpha}}{(vert zvert^2 + r^2)^{2H}}drsigma(dtheta)=int_{mathbb{S}^{d-1}}int_0^inftyfrac{u^{d-1-alpha}}{(1+ u^2)^{2H}}drsigma(dtheta)$$
Do you have any idea for proving this claim?
real-analysis
asked Nov 26 at 10:58
Adrián Hinojosa Calleja
937
add a comment |
I´m trying to understand the proof of Thm 4. on this paper
https://arxiv.org/pdf/1501.06828.pdf. At the end it claims that after elemental computations it can be shown that
$$vert zvert ^{4H+ alpha -d}int_{mathbb{S}^{d-1}}int_0^inftyfrac{r^{d-1-alpha}}{(vert zvert^2 + r^2)^{2H}}(1-cos(r
(thetacdottheta_z)))drsigma(dtheta)leq Cvert z vert ^ {2 wedge (4H+alpha-d) } bigg(logfrac{1}{vert z vert}bigg)^{beta}$$
where
$rho=1$ if $2=4H+alpha-d$ and $rho=0$ in other case- $alpha< d < 4H+alpha $
- $frac{1}{2}<H<1$
$sigma$ is the uniform measure on $mathbb{S}^{d-1}$ and- $theta_z=frac{z}{vert zvert}$
- c is a positive constant
I don´t understand why it follows after elemental computation , I tried bounding $1-cos(r (thetacdottheta_z))$ by 1, but this clearly does not work. For example in the case $4H+alpha-d<2$ we have that after the change of variables $u=frac{r}{vert zvert}$
$$vert zvert ^{4H+ alpha -d}int_{mathbb{S}^{d-1}}int_0^inftyfrac{r^{d-1-alpha}}{(vert zvert^2 + r^2)^{2H}}drsigma(dtheta)=int_{mathbb{S}^{d-1}}int_0^inftyfrac{u^{d-1-alpha}}{(1+ u^2)^{2H}}drsigma(dtheta)$$
Do you have any idea for proving this claim?
real-analysis
asked Nov 26 at 10:58
Adrián Hinojosa Calleja
937
I´m trying to understand the proof of Thm 4. on this paper
https://arxiv.org/pdf/1501.06828.pdf. At the end it claims that after elemental computations it can be shown that
$$vert zvert ^{4H+ alpha -d}int_{mathbb{S}^{d-1}}int_0^inftyfrac{r^{d-1-alpha}}{(vert zvert^2 + r^2)^{2H}}(1-cos(r
(thetacdottheta_z)))drsigma(dtheta)leq Cvert z vert ^ {2 wedge (4H+alpha-d) } bigg(logfrac{1}{vert z vert}bigg)^{beta}$$
where
$rho=1$ if $2=4H+alpha-d$ and $rho=0$ in other case- $alpha< d < 4H+alpha $
- $frac{1}{2}<H<1$
$sigma$ is the uniform measure on $mathbb{S}^{d-1}$ and- $theta_z=frac{z}{vert zvert}$
- c is a positive constant
I don´t understand why it follows after elemental computation , I tried bounding $1-cos(r (thetacdottheta_z))$ by 1, but this clearly does not work. For example in the case $4H+alpha-d<2$ we have that after the change of variables $u=frac{r}{vert zvert}$
$$vert zvert ^{4H+ alpha -d}int_{mathbb{S}^{d-1}}int_0^inftyfrac{r^{d-1-alpha}}{(vert zvert^2 + r^2)^{2H}}drsigma(dtheta)=int_{mathbb{S}^{d-1}}int_0^inftyfrac{u^{d-1-alpha}}{(1+ u^2)^{2H}}drsigma(dtheta)$$
Do you have any idea for proving this claim?
real-analysis
real-analysis
asked Nov 26 at 10:58
Adrián Hinojosa Calleja
937
asked Nov 26 at 10:58
Adrián Hinojosa Calleja
937
asked Nov 26 at 10:58
Adrián Hinojosa Calleja
937
asked Nov 26 at 10:58
Adrián Hinojosa Calleja
937
asked Nov 26 at 10:58
Adrián Hinojosa Calleja
937
937
add a comment |
add a comment |
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