Why is $L³$ continuously embedded into $H^{-1}$, the dual of $H¹_0$?
Why is $L³$ continuously embedded into $H^{-1}$, the dual of $H_0^1$?
In this article https://drive.google.com/open?id=0B0MDtuRPAebZQ3NvRFVQekJ4VEpTMldiRWxmbU9GLVM4MDNF the authors claim, in the proof of Proposition 4, that $L^3(Omega)$ is continuously embedded into $H^{-1}(Omega)$, the dual of $H_0^1(Omega)$, being $Omega in mathbb{R}^3$ a bounded open set.
The injections $H_0^1 subset L^2 subset H^{-1}$ are well known, but I am not able to prove the claim for $L^3$.
Any references, hints or suggestions will be the most appreciated.
pde hilbert-spaces sobolev-spaces
asked Dec 7 at 15:17
Danilo Gregorin
626413
add a comment |
Why is $L³$ continuously embedded into $H^{-1}$, the dual of $H_0^1$?
In this article https://drive.google.com/open?id=0B0MDtuRPAebZQ3NvRFVQekJ4VEpTMldiRWxmbU9GLVM4MDNF the authors claim, in the proof of Proposition 4, that $L^3(Omega)$ is continuously embedded into $H^{-1}(Omega)$, the dual of $H_0^1(Omega)$, being $Omega in mathbb{R}^3$ a bounded open set.
The injections $H_0^1 subset L^2 subset H^{-1}$ are well known, but I am not able to prove the claim for $L^3$.
Any references, hints or suggestions will be the most appreciated.
pde hilbert-spaces sobolev-spaces
asked Dec 7 at 15:17
Danilo Gregorin
626413
add a comment |
Why is $L³$ continuously embedded into $H^{-1}$, the dual of $H_0^1$?
In this article https://drive.google.com/open?id=0B0MDtuRPAebZQ3NvRFVQekJ4VEpTMldiRWxmbU9GLVM4MDNF the authors claim, in the proof of Proposition 4, that $L^3(Omega)$ is continuously embedded into $H^{-1}(Omega)$, the dual of $H_0^1(Omega)$, being $Omega in mathbb{R}^3$ a bounded open set.
The injections $H_0^1 subset L^2 subset H^{-1}$ are well known, but I am not able to prove the claim for $L^3$.
Any references, hints or suggestions will be the most appreciated.
pde hilbert-spaces sobolev-spaces
asked Dec 7 at 15:17
Danilo Gregorin
626413
Why is $L³$ continuously embedded into $H^{-1}$, the dual of $H_0^1$?
In this article https://drive.google.com/open?id=0B0MDtuRPAebZQ3NvRFVQekJ4VEpTMldiRWxmbU9GLVM4MDNF the authors claim, in the proof of Proposition 4, that $L^3(Omega)$ is continuously embedded into $H^{-1}(Omega)$, the dual of $H_0^1(Omega)$, being $Omega in mathbb{R}^3$ a bounded open set.
The injections $H_0^1 subset L^2 subset H^{-1}$ are well known, but I am not able to prove the claim for $L^3$.
Any references, hints or suggestions will be the most appreciated.
pde hilbert-spaces sobolev-spaces
pde hilbert-spaces sobolev-spaces
asked Dec 7 at 15:17
Danilo Gregorin
626413
asked Dec 7 at 15:17
Danilo Gregorin
626413
asked Dec 7 at 15:17
Danilo Gregorin
626413
asked Dec 7 at 15:17
Danilo Gregorin
626413
asked Dec 7 at 15:17
Danilo Gregorin
626413
626413
add a comment |
add a comment |
2 Answers
2
active
oldest
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By Sobolev, $H^1_0 subset L^{2^*}=L^6$, because $frac1{2^*}=frac12-frac13=frac16$.
So $L^3subset L^{6/5} = (L^6)^* subset (H^1_0)^* = H^{-1}$.
This shows that the lowest exponent that you can embed is $6/5$, and $3$ is much more than needed.
I'm assuming $|Omega|<infty$.
edited Dec 7 at 15:31
Strants
5,47221736
answered Dec 7 at 15:25
Federico
4,459512
add a comment |
The embedding $L^3subset H^{-1}$ follows from
the embeddings
$L^3subset L^2$
and
$L^2subset H^{-1}$
by composition of the embeddings.
answered Dec 7 at 15:26
supinf
5,9491027
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
By Sobolev, $H^1_0 subset L^{2^*}=L^6$, because $frac1{2^*}=frac12-frac13=frac16$.
So $L^3subset L^{6/5} = (L^6)^* subset (H^1_0)^* = H^{-1}$.
This shows that the lowest exponent that you can embed is $6/5$, and $3$ is much more than needed.
I'm assuming $|Omega|<infty$.
edited Dec 7 at 15:31
Strants
5,47221736
answered Dec 7 at 15:25
Federico
4,459512
add a comment |
By Sobolev, $H^1_0 subset L^{2^*}=L^6$, because $frac1{2^*}=frac12-frac13=frac16$.
So $L^3subset L^{6/5} = (L^6)^* subset (H^1_0)^* = H^{-1}$.
This shows that the lowest exponent that you can embed is $6/5$, and $3$ is much more than needed.
I'm assuming $|Omega|<infty$.
edited Dec 7 at 15:31
Strants
5,47221736
answered Dec 7 at 15:25
Federico
4,459512
add a comment |
By Sobolev, $H^1_0 subset L^{2^*}=L^6$, because $frac1{2^*}=frac12-frac13=frac16$.
So $L^3subset L^{6/5} = (L^6)^* subset (H^1_0)^* = H^{-1}$.
This shows that the lowest exponent that you can embed is $6/5$, and $3$ is much more than needed.
I'm assuming $|Omega|<infty$.
edited Dec 7 at 15:31
Strants
5,47221736
answered Dec 7 at 15:25
Federico
4,459512
By Sobolev, $H^1_0 subset L^{2^*}=L^6$, because $frac1{2^*}=frac12-frac13=frac16$.
So $L^3subset L^{6/5} = (L^6)^* subset (H^1_0)^* = H^{-1}$.
This shows that the lowest exponent that you can embed is $6/5$, and $3$ is much more than needed.
I'm assuming $|Omega|<infty$.
edited Dec 7 at 15:31
Strants
5,47221736
answered Dec 7 at 15:25
Federico
4,459512
edited Dec 7 at 15:31
Strants
5,47221736
edited Dec 7 at 15:31
Strants
5,47221736
edited Dec 7 at 15:31
Strants
5,47221736
5,47221736
answered Dec 7 at 15:25
Federico
4,459512
answered Dec 7 at 15:25
Federico
4,459512
answered Dec 7 at 15:25
Federico
4,459512
4,459512
add a comment |
add a comment |
The embedding $L^3subset H^{-1}$ follows from
the embeddings
$L^3subset L^2$
and
$L^2subset H^{-1}$
by composition of the embeddings.
answered Dec 7 at 15:26
supinf
5,9491027
add a comment |
The embedding $L^3subset H^{-1}$ follows from
the embeddings
$L^3subset L^2$
and
$L^2subset H^{-1}$
by composition of the embeddings.
answered Dec 7 at 15:26
supinf
5,9491027
add a comment |
The embedding $L^3subset H^{-1}$ follows from
the embeddings
$L^3subset L^2$
and
$L^2subset H^{-1}$
by composition of the embeddings.
answered Dec 7 at 15:26
supinf
5,9491027
The embedding $L^3subset H^{-1}$ follows from
the embeddings
$L^3subset L^2$
and
$L^2subset H^{-1}$
by composition of the embeddings.
answered Dec 7 at 15:26
supinf
5,9491027
answered Dec 7 at 15:26
supinf
5,9491027
answered Dec 7 at 15:26
supinf
5,9491027
answered Dec 7 at 15:26
supinf
5,9491027
5,9491027
add a comment |
add a comment |
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