Use of Holder inequality in gradient estimate for harmonic function.
$begingroup$
While reading the book "Elliptic Partial Differential Equations" by Han and Lin, I failed to understand the proof of the interior gradient estimate for harmonic functions. The theorem says that if $u$ is harmonic in $B_1 (subset mathbb{R}^n)$, then $ sup_{B_{1/2}}{|Du|} le csup_{partial B_1}{|u|}$ for some positive constant $c=c(n).$ The proof illustrated in the book begins with choosing a cut-off function $phi = eta^2$ for some $eta in C_{0}^{1}(B_1)$ with $eta = 1$ in $B_{1/2}.$ Directly calculating, one can show $Delta(eta^2|Du|^2) = 2etaDeltaeta|Du|^2 + 2|Deta|^2|Du|^2 + 8etasum_{i, j =1}^{n}{D_{i}eta D_{j}u D_{ij}u} + 2eta^2sum_{i,j=1}^{n}(D_{ij}u)^2.$ What I do not know is the following inequality: $Delta(eta^2|Du|^2) ge (2etaDeltaeta-6|Deta|^2)|Du|^2.$ The book explains that "Holder inequality" is used here, but the only thing that I know regarding to Holder inequality is what appears in the usual real analysis book in the chapter of $L^p$ spaces (inequality for integral). I guess by the form of the equation, we need to say something about the third term: $8etasum_{i, j =1}^{n}{D_{i}eta D_{j}u D_{ij}u}.$ To be specific, obtaining the inequality $8etasum_{i, j =1}^{n}{D_{i}eta D_{j}u D_{ij}u} ge -8|Deta|^2|Du|^2$ will end the proof.
Can anyone help me to see how Holder inequality could be used to prove the claimed inequality? Thanks in advance.
pde harmonic-functions holder-inequality
asked Jan 11 at 13:05
EuduardoEuduardo
1838
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add a comment |
$begingroup$
While reading the book "Elliptic Partial Differential Equations" by Han and Lin, I failed to understand the proof of the interior gradient estimate for harmonic functions. The theorem says that if $u$ is harmonic in $B_1 (subset mathbb{R}^n)$, then $ sup_{B_{1/2}}{|Du|} le csup_{partial B_1}{|u|}$ for some positive constant $c=c(n).$ The proof illustrated in the book begins with choosing a cut-off function $phi = eta^2$ for some $eta in C_{0}^{1}(B_1)$ with $eta = 1$ in $B_{1/2}.$ Directly calculating, one can show $Delta(eta^2|Du|^2) = 2etaDeltaeta|Du|^2 + 2|Deta|^2|Du|^2 + 8etasum_{i, j =1}^{n}{D_{i}eta D_{j}u D_{ij}u} + 2eta^2sum_{i,j=1}^{n}(D_{ij}u)^2.$ What I do not know is the following inequality: $Delta(eta^2|Du|^2) ge (2etaDeltaeta-6|Deta|^2)|Du|^2.$ The book explains that "Holder inequality" is used here, but the only thing that I know regarding to Holder inequality is what appears in the usual real analysis book in the chapter of $L^p$ spaces (inequality for integral). I guess by the form of the equation, we need to say something about the third term: $8etasum_{i, j =1}^{n}{D_{i}eta D_{j}u D_{ij}u}.$ To be specific, obtaining the inequality $8etasum_{i, j =1}^{n}{D_{i}eta D_{j}u D_{ij}u} ge -8|Deta|^2|Du|^2$ will end the proof.
Can anyone help me to see how Holder inequality could be used to prove the claimed inequality? Thanks in advance.
pde harmonic-functions holder-inequality
asked Jan 11 at 13:05
EuduardoEuduardo
1838
$endgroup$
add a comment |
$begingroup$
While reading the book "Elliptic Partial Differential Equations" by Han and Lin, I failed to understand the proof of the interior gradient estimate for harmonic functions. The theorem says that if $u$ is harmonic in $B_1 (subset mathbb{R}^n)$, then $ sup_{B_{1/2}}{|Du|} le csup_{partial B_1}{|u|}$ for some positive constant $c=c(n).$ The proof illustrated in the book begins with choosing a cut-off function $phi = eta^2$ for some $eta in C_{0}^{1}(B_1)$ with $eta = 1$ in $B_{1/2}.$ Directly calculating, one can show $Delta(eta^2|Du|^2) = 2etaDeltaeta|Du|^2 + 2|Deta|^2|Du|^2 + 8etasum_{i, j =1}^{n}{D_{i}eta D_{j}u D_{ij}u} + 2eta^2sum_{i,j=1}^{n}(D_{ij}u)^2.$ What I do not know is the following inequality: $Delta(eta^2|Du|^2) ge (2etaDeltaeta-6|Deta|^2)|Du|^2.$ The book explains that "Holder inequality" is used here, but the only thing that I know regarding to Holder inequality is what appears in the usual real analysis book in the chapter of $L^p$ spaces (inequality for integral). I guess by the form of the equation, we need to say something about the third term: $8etasum_{i, j =1}^{n}{D_{i}eta D_{j}u D_{ij}u}.$ To be specific, obtaining the inequality $8etasum_{i, j =1}^{n}{D_{i}eta D_{j}u D_{ij}u} ge -8|Deta|^2|Du|^2$ will end the proof.
Can anyone help me to see how Holder inequality could be used to prove the claimed inequality? Thanks in advance.
pde harmonic-functions holder-inequality
asked Jan 11 at 13:05
EuduardoEuduardo
1838
$endgroup$
While reading the book "Elliptic Partial Differential Equations" by Han and Lin, I failed to understand the proof of the interior gradient estimate for harmonic functions. The theorem says that if $u$ is harmonic in $B_1 (subset mathbb{R}^n)$, then $ sup_{B_{1/2}}{|Du|} le csup_{partial B_1}{|u|}$ for some positive constant $c=c(n).$ The proof illustrated in the book begins with choosing a cut-off function $phi = eta^2$ for some $eta in C_{0}^{1}(B_1)$ with $eta = 1$ in $B_{1/2}.$ Directly calculating, one can show $Delta(eta^2|Du|^2) = 2etaDeltaeta|Du|^2 + 2|Deta|^2|Du|^2 + 8etasum_{i, j =1}^{n}{D_{i}eta D_{j}u D_{ij}u} + 2eta^2sum_{i,j=1}^{n}(D_{ij}u)^2.$ What I do not know is the following inequality: $Delta(eta^2|Du|^2) ge (2etaDeltaeta-6|Deta|^2)|Du|^2.$ The book explains that "Holder inequality" is used here, but the only thing that I know regarding to Holder inequality is what appears in the usual real analysis book in the chapter of $L^p$ spaces (inequality for integral). I guess by the form of the equation, we need to say something about the third term: $8etasum_{i, j =1}^{n}{D_{i}eta D_{j}u D_{ij}u}.$ To be specific, obtaining the inequality $8etasum_{i, j =1}^{n}{D_{i}eta D_{j}u D_{ij}u} ge -8|Deta|^2|Du|^2$ will end the proof.
Can anyone help me to see how Holder inequality could be used to prove the claimed inequality? Thanks in advance.
pde harmonic-functions holder-inequality
pde harmonic-functions holder-inequality
asked Jan 11 at 13:05
EuduardoEuduardo
1838
asked Jan 11 at 13:05
EuduardoEuduardo
1838
edited Jan 11 at 13:21
Euduardo
asked Jan 11 at 13:05
EuduardoEuduardo
1838
asked Jan 11 at 13:05
EuduardoEuduardo
1838
asked Jan 11 at 13:05
EuduardoEuduardo
1838
1838
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From googling, I managed to end up at these MIT notes that expand slightly on the proof (but don't use Holder's). The trick is as follows. Note that it is enough to show
$$ 8 eta sum_{ij} D_i eta D_j u D_{ij} u + 2eta^2 sum_{ij} (D_{ij}u)^2 ge - 8 |Deta |^2 |Du|^2$$
Note that in each term, $eta$ and $D_{ij}u$ appear in equal powers. So define
$$ X_{ij} = eta D_{ij} u ,quad Y_{ij} = D_i eta D_j u$$
Then the above inequality is equivalent to
$$ sum_{ij} 4X_{ij}Y_{ij} + 4Y_{ij}^2 + X_{ij}^2 ge 0$$
but of course this LHS is nothing but
$$ sum_{ij} 2X_{ij}(2Y_{ij}) + (2Y_{ij})^2 + X_{ij}^2 = sum_{ij} (2Y_{ij}+X_{ij})^2$$
which is clearly non-negative, so the original inequality that we wanted to prove is also true.
For vectors, the inequality $|a+b|^2ge 0$ can be used to prove Cauchy-Schwarz ($sum_i a_ib_i le |a||b|$) so I can imagine there is a version of this proof that uses Cauchy-Schwarz...but I don't see it.
In addition, this book (A Basic Course in Partial Differential Equations) by Han expands on the proof in a different way, using the Cauchy inequality (i.e. Young's inequality for products, a.k.a. the "rob-Peter-to-pay-Paul" inequality).
answered Jan 11 at 14:58
Calvin KhorCalvin Khor
12.3k21438
$endgroup$
$begingroup$
Thanks Khor! I've learned a lot reading your solution.
$endgroup$
– Euduardo
Jan 12 at 2:57
$begingroup$
@Euduardo You're welcome :)
$endgroup$
– Calvin Khor
Jan 12 at 10:32
add a comment |
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$begingroup$
From googling, I managed to end up at these MIT notes that expand slightly on the proof (but don't use Holder's). The trick is as follows. Note that it is enough to show
$$ 8 eta sum_{ij} D_i eta D_j u D_{ij} u + 2eta^2 sum_{ij} (D_{ij}u)^2 ge - 8 |Deta |^2 |Du|^2$$
Note that in each term, $eta$ and $D_{ij}u$ appear in equal powers. So define
$$ X_{ij} = eta D_{ij} u ,quad Y_{ij} = D_i eta D_j u$$
Then the above inequality is equivalent to
$$ sum_{ij} 4X_{ij}Y_{ij} + 4Y_{ij}^2 + X_{ij}^2 ge 0$$
but of course this LHS is nothing but
$$ sum_{ij} 2X_{ij}(2Y_{ij}) + (2Y_{ij})^2 + X_{ij}^2 = sum_{ij} (2Y_{ij}+X_{ij})^2$$
which is clearly non-negative, so the original inequality that we wanted to prove is also true.
For vectors, the inequality $|a+b|^2ge 0$ can be used to prove Cauchy-Schwarz ($sum_i a_ib_i le |a||b|$) so I can imagine there is a version of this proof that uses Cauchy-Schwarz...but I don't see it.
In addition, this book (A Basic Course in Partial Differential Equations) by Han expands on the proof in a different way, using the Cauchy inequality (i.e. Young's inequality for products, a.k.a. the "rob-Peter-to-pay-Paul" inequality).
answered Jan 11 at 14:58
Calvin KhorCalvin Khor
12.3k21438
$endgroup$
$begingroup$
Thanks Khor! I've learned a lot reading your solution.
$endgroup$
– Euduardo
Jan 12 at 2:57
$begingroup$
@Euduardo You're welcome :)
$endgroup$
– Calvin Khor
Jan 12 at 10:32
add a comment |
$begingroup$
From googling, I managed to end up at these MIT notes that expand slightly on the proof (but don't use Holder's). The trick is as follows. Note that it is enough to show
$$ 8 eta sum_{ij} D_i eta D_j u D_{ij} u + 2eta^2 sum_{ij} (D_{ij}u)^2 ge - 8 |Deta |^2 |Du|^2$$
Note that in each term, $eta$ and $D_{ij}u$ appear in equal powers. So define
$$ X_{ij} = eta D_{ij} u ,quad Y_{ij} = D_i eta D_j u$$
Then the above inequality is equivalent to
$$ sum_{ij} 4X_{ij}Y_{ij} + 4Y_{ij}^2 + X_{ij}^2 ge 0$$
but of course this LHS is nothing but
$$ sum_{ij} 2X_{ij}(2Y_{ij}) + (2Y_{ij})^2 + X_{ij}^2 = sum_{ij} (2Y_{ij}+X_{ij})^2$$
which is clearly non-negative, so the original inequality that we wanted to prove is also true.
For vectors, the inequality $|a+b|^2ge 0$ can be used to prove Cauchy-Schwarz ($sum_i a_ib_i le |a||b|$) so I can imagine there is a version of this proof that uses Cauchy-Schwarz...but I don't see it.
In addition, this book (A Basic Course in Partial Differential Equations) by Han expands on the proof in a different way, using the Cauchy inequality (i.e. Young's inequality for products, a.k.a. the "rob-Peter-to-pay-Paul" inequality).
answered Jan 11 at 14:58
Calvin KhorCalvin Khor
12.3k21438
$endgroup$
$begingroup$
Thanks Khor! I've learned a lot reading your solution.
$endgroup$
– Euduardo
Jan 12 at 2:57
$begingroup$
@Euduardo You're welcome :)
$endgroup$
– Calvin Khor
Jan 12 at 10:32
add a comment |
$begingroup$
From googling, I managed to end up at these MIT notes that expand slightly on the proof (but don't use Holder's). The trick is as follows. Note that it is enough to show
$$ 8 eta sum_{ij} D_i eta D_j u D_{ij} u + 2eta^2 sum_{ij} (D_{ij}u)^2 ge - 8 |Deta |^2 |Du|^2$$
Note that in each term, $eta$ and $D_{ij}u$ appear in equal powers. So define
$$ X_{ij} = eta D_{ij} u ,quad Y_{ij} = D_i eta D_j u$$
Then the above inequality is equivalent to
$$ sum_{ij} 4X_{ij}Y_{ij} + 4Y_{ij}^2 + X_{ij}^2 ge 0$$
but of course this LHS is nothing but
$$ sum_{ij} 2X_{ij}(2Y_{ij}) + (2Y_{ij})^2 + X_{ij}^2 = sum_{ij} (2Y_{ij}+X_{ij})^2$$
which is clearly non-negative, so the original inequality that we wanted to prove is also true.
For vectors, the inequality $|a+b|^2ge 0$ can be used to prove Cauchy-Schwarz ($sum_i a_ib_i le |a||b|$) so I can imagine there is a version of this proof that uses Cauchy-Schwarz...but I don't see it.
In addition, this book (A Basic Course in Partial Differential Equations) by Han expands on the proof in a different way, using the Cauchy inequality (i.e. Young's inequality for products, a.k.a. the "rob-Peter-to-pay-Paul" inequality).
answered Jan 11 at 14:58
Calvin KhorCalvin Khor
12.3k21438
$endgroup$
From googling, I managed to end up at these MIT notes that expand slightly on the proof (but don't use Holder's). The trick is as follows. Note that it is enough to show
$$ 8 eta sum_{ij} D_i eta D_j u D_{ij} u + 2eta^2 sum_{ij} (D_{ij}u)^2 ge - 8 |Deta |^2 |Du|^2$$
Note that in each term, $eta$ and $D_{ij}u$ appear in equal powers. So define
$$ X_{ij} = eta D_{ij} u ,quad Y_{ij} = D_i eta D_j u$$
Then the above inequality is equivalent to
$$ sum_{ij} 4X_{ij}Y_{ij} + 4Y_{ij}^2 + X_{ij}^2 ge 0$$
but of course this LHS is nothing but
$$ sum_{ij} 2X_{ij}(2Y_{ij}) + (2Y_{ij})^2 + X_{ij}^2 = sum_{ij} (2Y_{ij}+X_{ij})^2$$
which is clearly non-negative, so the original inequality that we wanted to prove is also true.
For vectors, the inequality $|a+b|^2ge 0$ can be used to prove Cauchy-Schwarz ($sum_i a_ib_i le |a||b|$) so I can imagine there is a version of this proof that uses Cauchy-Schwarz...but I don't see it.
In addition, this book (A Basic Course in Partial Differential Equations) by Han expands on the proof in a different way, using the Cauchy inequality (i.e. Young's inequality for products, a.k.a. the "rob-Peter-to-pay-Paul" inequality).
answered Jan 11 at 14:58
Calvin KhorCalvin Khor
12.3k21438
edited Jan 11 at 15:23
answered Jan 11 at 14:58
Calvin KhorCalvin Khor
12.3k21438
answered Jan 11 at 14:58
Calvin KhorCalvin Khor
12.3k21438
answered Jan 11 at 14:58
Calvin KhorCalvin Khor
12.3k21438
12.3k21438
$begingroup$
Thanks Khor! I've learned a lot reading your solution.
$endgroup$
– Euduardo
Jan 12 at 2:57
$begingroup$
@Euduardo You're welcome :)
$endgroup$
– Calvin Khor
Jan 12 at 10:32
add a comment |
$begingroup$
Thanks Khor! I've learned a lot reading your solution.
$endgroup$
– Euduardo
Jan 12 at 2:57
$begingroup$
@Euduardo You're welcome :)
$endgroup$
– Calvin Khor
Jan 12 at 10:32
$begingroup$
Thanks Khor! I've learned a lot reading your solution.
$endgroup$
– Euduardo
Jan 12 at 2:57
$begingroup$
Thanks Khor! I've learned a lot reading your solution.
$endgroup$
– Euduardo
Jan 12 at 2:57
$begingroup$
@Euduardo You're welcome :)
$endgroup$
– Calvin Khor
Jan 12 at 10:32
$begingroup$
@Euduardo You're welcome :)
$endgroup$
– Calvin Khor
Jan 12 at 10:32
add a comment |
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