Theorem in Adams Sobolev spaces book requires $uin L^p(Omega) cap L^r(Omega)$ but we only have $u in C^infty$...
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Let $Omega$ be a (open) domain in $mathbb{R}^n$. In Theorem 4.19 of Adams book on Sobolev spaces he makes use of Theorem 2.11 (An Interpolation Inequality). Theorem 2.11 requires that if we have $1le p < q < r$ and $uin L^p(Omega) cap L^r(Omega)$, then we have that $uin L^q(Omega)$ and
$$
||u||_q le ||u||_p^theta ||u||_r^{1-theta},
$$
for $0 < theta < 1$.
However in Theorem 4.19 we only have $uin C^infty(Omega)$ so how can he apply Theorem 2.11?
I don't know if it makes any difference, but he also states that $u$ and all its derivatives are extended by zero outside $Omega$ in Theorem 4.19. Is it this extension by zero that lets him have know that $uin L^p(Omega) cap L^r(Omega)$ and thus apply Theorem 2.11?
functional-analysis pde sobolev-spaces lp-spaces
asked Nov 29 '18 at 9:16
sonicboomsonicboom
3,67082652
$endgroup$
add a comment |
$begingroup$
Let $Omega$ be a (open) domain in $mathbb{R}^n$. In Theorem 4.19 of Adams book on Sobolev spaces he makes use of Theorem 2.11 (An Interpolation Inequality). Theorem 2.11 requires that if we have $1le p < q < r$ and $uin L^p(Omega) cap L^r(Omega)$, then we have that $uin L^q(Omega)$ and
$$
||u||_q le ||u||_p^theta ||u||_r^{1-theta},
$$
for $0 < theta < 1$.
However in Theorem 4.19 we only have $uin C^infty(Omega)$ so how can he apply Theorem 2.11?
I don't know if it makes any difference, but he also states that $u$ and all its derivatives are extended by zero outside $Omega$ in Theorem 4.19. Is it this extension by zero that lets him have know that $uin L^p(Omega) cap L^r(Omega)$ and thus apply Theorem 2.11?
functional-analysis pde sobolev-spaces lp-spaces
asked Nov 29 '18 at 9:16
sonicboomsonicboom
3,67082652
$endgroup$
1
$begingroup$
You might say that the inequality is obviously true (but useless) if $unotin L^p$ or $unotin L^q$ since the right-hand side is infinite. You can't conclude that $uin L^q$ of course. That with the extension outside of $Omega$ can't be an important fact since after all $Omega=mathbb{R}^n$ is allowed.
$endgroup$
– Olivier Moschetta
Nov 29 '18 at 9:20
$begingroup$
Actually I see how it all works out now from the rest of the theorem so nevermind.
$endgroup$
– sonicboom
Nov 29 '18 at 9:52
2
$begingroup$
@sonicboom then maybe you should post an answer, so that this question does not appear as unanswered
$endgroup$
– supinf
Nov 29 '18 at 10:31
add a comment |
$begingroup$
Let $Omega$ be a (open) domain in $mathbb{R}^n$. In Theorem 4.19 of Adams book on Sobolev spaces he makes use of Theorem 2.11 (An Interpolation Inequality). Theorem 2.11 requires that if we have $1le p < q < r$ and $uin L^p(Omega) cap L^r(Omega)$, then we have that $uin L^q(Omega)$ and
$$
||u||_q le ||u||_p^theta ||u||_r^{1-theta},
$$
for $0 < theta < 1$.
However in Theorem 4.19 we only have $uin C^infty(Omega)$ so how can he apply Theorem 2.11?
I don't know if it makes any difference, but he also states that $u$ and all its derivatives are extended by zero outside $Omega$ in Theorem 4.19. Is it this extension by zero that lets him have know that $uin L^p(Omega) cap L^r(Omega)$ and thus apply Theorem 2.11?
functional-analysis pde sobolev-spaces lp-spaces
asked Nov 29 '18 at 9:16
sonicboomsonicboom
3,67082652
$endgroup$
Let $Omega$ be a (open) domain in $mathbb{R}^n$. In Theorem 4.19 of Adams book on Sobolev spaces he makes use of Theorem 2.11 (An Interpolation Inequality). Theorem 2.11 requires that if we have $1le p < q < r$ and $uin L^p(Omega) cap L^r(Omega)$, then we have that $uin L^q(Omega)$ and
$$
||u||_q le ||u||_p^theta ||u||_r^{1-theta},
$$
for $0 < theta < 1$.
However in Theorem 4.19 we only have $uin C^infty(Omega)$ so how can he apply Theorem 2.11?
I don't know if it makes any difference, but he also states that $u$ and all its derivatives are extended by zero outside $Omega$ in Theorem 4.19. Is it this extension by zero that lets him have know that $uin L^p(Omega) cap L^r(Omega)$ and thus apply Theorem 2.11?
functional-analysis pde sobolev-spaces lp-spaces
functional-analysis pde sobolev-spaces lp-spaces
asked Nov 29 '18 at 9:16
sonicboomsonicboom
3,67082652
asked Nov 29 '18 at 9:16
sonicboomsonicboom
3,67082652
asked Nov 29 '18 at 9:16
sonicboomsonicboom
3,67082652
asked Nov 29 '18 at 9:16
sonicboomsonicboom
3,67082652
asked Nov 29 '18 at 9:16
sonicboomsonicboom
3,67082652
3,67082652
1
$begingroup$
You might say that the inequality is obviously true (but useless) if $unotin L^p$ or $unotin L^q$ since the right-hand side is infinite. You can't conclude that $uin L^q$ of course. That with the extension outside of $Omega$ can't be an important fact since after all $Omega=mathbb{R}^n$ is allowed.
$endgroup$
– Olivier Moschetta
Nov 29 '18 at 9:20
$begingroup$
Actually I see how it all works out now from the rest of the theorem so nevermind.
$endgroup$
– sonicboom
Nov 29 '18 at 9:52
2
$begingroup$
@sonicboom then maybe you should post an answer, so that this question does not appear as unanswered
$endgroup$
– supinf
Nov 29 '18 at 10:31
add a comment |
1
$begingroup$
You might say that the inequality is obviously true (but useless) if $unotin L^p$ or $unotin L^q$ since the right-hand side is infinite. You can't conclude that $uin L^q$ of course. That with the extension outside of $Omega$ can't be an important fact since after all $Omega=mathbb{R}^n$ is allowed.
$endgroup$
– Olivier Moschetta
Nov 29 '18 at 9:20
$begingroup$
Actually I see how it all works out now from the rest of the theorem so nevermind.
$endgroup$
– sonicboom
Nov 29 '18 at 9:52
2
$begingroup$
@sonicboom then maybe you should post an answer, so that this question does not appear as unanswered
$endgroup$
– supinf
Nov 29 '18 at 10:31
1
1
$begingroup$
You might say that the inequality is obviously true (but useless) if $unotin L^p$ or $unotin L^q$ since the right-hand side is infinite. You can't conclude that $uin L^q$ of course. That with the extension outside of $Omega$ can't be an important fact since after all $Omega=mathbb{R}^n$ is allowed.
$endgroup$
– Olivier Moschetta
Nov 29 '18 at 9:20
$begingroup$
You might say that the inequality is obviously true (but useless) if $unotin L^p$ or $unotin L^q$ since the right-hand side is infinite. You can't conclude that $uin L^q$ of course. That with the extension outside of $Omega$ can't be an important fact since after all $Omega=mathbb{R}^n$ is allowed.
$endgroup$
– Olivier Moschetta
Nov 29 '18 at 9:20
$begingroup$
Actually I see how it all works out now from the rest of the theorem so nevermind.
$endgroup$
– sonicboom
Nov 29 '18 at 9:52
$begingroup$
Actually I see how it all works out now from the rest of the theorem so nevermind.
$endgroup$
– sonicboom
Nov 29 '18 at 9:52
2
2
$begingroup$
@sonicboom then maybe you should post an answer, so that this question does not appear as unanswered
$endgroup$
– supinf
Nov 29 '18 at 10:31
$begingroup$
@sonicboom then maybe you should post an answer, so that this question does not appear as unanswered
$endgroup$
– supinf
Nov 29 '18 at 10:31
add a comment |
1 Answer
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When Adams states that $u in C^infty(Omega)$ in the opening paragraph of this theorem he doing so to show that we have inequality (13).
He then estimates an $L^p$ norm of $u$ and can make use of inequality (13) through the density of the $C^infty$ functions in $L^p$ spaces.
answered Nov 30 '18 at 17:49
sonicboomsonicboom
3,67082652
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add a comment |
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When Adams states that $u in C^infty(Omega)$ in the opening paragraph of this theorem he doing so to show that we have inequality (13).
He then estimates an $L^p$ norm of $u$ and can make use of inequality (13) through the density of the $C^infty$ functions in $L^p$ spaces.
answered Nov 30 '18 at 17:49
sonicboomsonicboom
3,67082652
$endgroup$
add a comment |
$begingroup$
When Adams states that $u in C^infty(Omega)$ in the opening paragraph of this theorem he doing so to show that we have inequality (13).
He then estimates an $L^p$ norm of $u$ and can make use of inequality (13) through the density of the $C^infty$ functions in $L^p$ spaces.
answered Nov 30 '18 at 17:49
sonicboomsonicboom
3,67082652
$endgroup$
add a comment |
$begingroup$
When Adams states that $u in C^infty(Omega)$ in the opening paragraph of this theorem he doing so to show that we have inequality (13).
He then estimates an $L^p$ norm of $u$ and can make use of inequality (13) through the density of the $C^infty$ functions in $L^p$ spaces.
answered Nov 30 '18 at 17:49
sonicboomsonicboom
3,67082652
$endgroup$
When Adams states that $u in C^infty(Omega)$ in the opening paragraph of this theorem he doing so to show that we have inequality (13).
He then estimates an $L^p$ norm of $u$ and can make use of inequality (13) through the density of the $C^infty$ functions in $L^p$ spaces.
answered Nov 30 '18 at 17:49
sonicboomsonicboom
3,67082652
answered Nov 30 '18 at 17:49
sonicboomsonicboom
3,67082652
answered Nov 30 '18 at 17:49
sonicboomsonicboom
3,67082652
answered Nov 30 '18 at 17:49
sonicboomsonicboom
3,67082652
3,67082652
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You might say that the inequality is obviously true (but useless) if $unotin L^p$ or $unotin L^q$ since the right-hand side is infinite. You can't conclude that $uin L^q$ of course. That with the extension outside of $Omega$ can't be an important fact since after all $Omega=mathbb{R}^n$ is allowed.
$endgroup$
– Olivier Moschetta
Nov 29 '18 at 9:20
$begingroup$
Actually I see how it all works out now from the rest of the theorem so nevermind.
$endgroup$
– sonicboom
Nov 29 '18 at 9:52
2
$begingroup$
@sonicboom then maybe you should post an answer, so that this question does not appear as unanswered
$endgroup$
– supinf
Nov 29 '18 at 10:31