Integral Inequality with L-2 Norm
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On page 135 of The Mathematical Theory of Finite Element Methods (Brenner and Scott), I encountered the following inequality:
$left | int_{Gamma} overline{v} - v , ds right | leq |Gamma |^{1/2} || overline{v} - v ||_{L^2(Gamma)},$.
The mean $overline{v}$ is defined as
$overline{v} = frac{1}{text{meas}(Omega)} int_{Omega} v(x) , dx ,$
and $Gamma = partial Omega$.
The inequality is part of a proof that the bilinear form for the Poisson equation with Dirichlet boundary conditions is coercive.
My question is: could I replace the $overline{v}-v$ on both sides of the inequality with some (more general) function? If so, what would the restrictions on the function be? I'm not sure where the inequality comes from.
functional-analysis integral-inequality
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rainbowgigi
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On page 135 of The Mathematical Theory of Finite Element Methods (Brenner and Scott), I encountered the following inequality:
$left | int_{Gamma} overline{v} - v , ds right | leq |Gamma |^{1/2} || overline{v} - v ||_{L^2(Gamma)},$.
The mean $overline{v}$ is defined as
$overline{v} = frac{1}{text{meas}(Omega)} int_{Omega} v(x) , dx ,$
and $Gamma = partial Omega$.
The inequality is part of a proof that the bilinear form for the Poisson equation with Dirichlet boundary conditions is coercive.
My question is: could I replace the $overline{v}-v$ on both sides of the inequality with some (more general) function? If so, what would the restrictions on the function be? I'm not sure where the inequality comes from.
functional-analysis integral-inequality
asked yesterday
rainbowgigi
62
New contributor
rainbowgigi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
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up vote
1
down vote
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On page 135 of The Mathematical Theory of Finite Element Methods (Brenner and Scott), I encountered the following inequality:
$left | int_{Gamma} overline{v} - v , ds right | leq |Gamma |^{1/2} || overline{v} - v ||_{L^2(Gamma)},$.
The mean $overline{v}$ is defined as
$overline{v} = frac{1}{text{meas}(Omega)} int_{Omega} v(x) , dx ,$
and $Gamma = partial Omega$.
The inequality is part of a proof that the bilinear form for the Poisson equation with Dirichlet boundary conditions is coercive.
My question is: could I replace the $overline{v}-v$ on both sides of the inequality with some (more general) function? If so, what would the restrictions on the function be? I'm not sure where the inequality comes from.
functional-analysis integral-inequality
asked yesterday
rainbowgigi
62
New contributor
rainbowgigi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
On page 135 of The Mathematical Theory of Finite Element Methods (Brenner and Scott), I encountered the following inequality:
$left | int_{Gamma} overline{v} - v , ds right | leq |Gamma |^{1/2} || overline{v} - v ||_{L^2(Gamma)},$.
The mean $overline{v}$ is defined as
$overline{v} = frac{1}{text{meas}(Omega)} int_{Omega} v(x) , dx ,$
and $Gamma = partial Omega$.
The inequality is part of a proof that the bilinear form for the Poisson equation with Dirichlet boundary conditions is coercive.
My question is: could I replace the $overline{v}-v$ on both sides of the inequality with some (more general) function? If so, what would the restrictions on the function be? I'm not sure where the inequality comes from.
functional-analysis integral-inequality
functional-analysis integral-inequality
asked yesterday
rainbowgigi
62
New contributor
rainbowgigi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
rainbowgigi
62
New contributor
rainbowgigi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
asked yesterday
rainbowgigi
62
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rainbowgigi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked yesterday
rainbowgigi
62
asked yesterday
rainbowgigi
62
62
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rainbowgigi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 Answer
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This is just Hoelder inequality. Let me denote $w:=bar v -v$. Then
$$
|int_Gamma w ds| le int_Gamma |w|cdot 1 ds le |1|_{L^2(Gamma)} |w|_{L^2(Gamma)} = |Gamma|^{1/2}|w|_{L^2(Gamma)}.
$$
This works if $win L^2(Gamma)$ and $|Gamma|<+infty$.
answered yesterday
daw
23.6k1544
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This is just Hoelder inequality. Let me denote $w:=bar v -v$. Then
$$
|int_Gamma w ds| le int_Gamma |w|cdot 1 ds le |1|_{L^2(Gamma)} |w|_{L^2(Gamma)} = |Gamma|^{1/2}|w|_{L^2(Gamma)}.
$$
This works if $win L^2(Gamma)$ and $|Gamma|<+infty$.
answered yesterday
daw
23.6k1544
add a comment |
up vote
1
down vote
This is just Hoelder inequality. Let me denote $w:=bar v -v$. Then
$$
|int_Gamma w ds| le int_Gamma |w|cdot 1 ds le |1|_{L^2(Gamma)} |w|_{L^2(Gamma)} = |Gamma|^{1/2}|w|_{L^2(Gamma)}.
$$
This works if $win L^2(Gamma)$ and $|Gamma|<+infty$.
answered yesterday
daw
23.6k1544
add a comment |
up vote
1
down vote
up vote
1
down vote
This is just Hoelder inequality. Let me denote $w:=bar v -v$. Then
$$
|int_Gamma w ds| le int_Gamma |w|cdot 1 ds le |1|_{L^2(Gamma)} |w|_{L^2(Gamma)} = |Gamma|^{1/2}|w|_{L^2(Gamma)}.
$$
This works if $win L^2(Gamma)$ and $|Gamma|<+infty$.
answered yesterday
daw
23.6k1544
This is just Hoelder inequality. Let me denote $w:=bar v -v$. Then
$$
|int_Gamma w ds| le int_Gamma |w|cdot 1 ds le |1|_{L^2(Gamma)} |w|_{L^2(Gamma)} = |Gamma|^{1/2}|w|_{L^2(Gamma)}.
$$
This works if $win L^2(Gamma)$ and $|Gamma|<+infty$.
answered yesterday
daw
23.6k1544
answered yesterday
daw
23.6k1544
answered yesterday
daw
23.6k1544
answered yesterday
daw
23.6k1544
23.6k1544
add a comment |
add a comment |
rainbowgigi is a new contributor. Be nice, and check out our Code of Conduct.
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