Mixing
Convergence in Lebesgue-Bochner Space $L^{infty}(0,T,L^1(Gamma))$
$begingroup$
Let $Gamma$ be a compact $C^2$ manifold and suppose that $f_n$ is a non negative sequence of functions such that ${vert vert f_n vert vert}_{L^{infty}(0,T,L^1(Gamma))} le C$
I am interested in deducing convergence of $f_n$
MY ATTEMPTS:
- Since $L^1$ is not reflexive, $L^{infty}(0,T,L^1(Gamma))$ is also not reflexive and thus from the boundedness of $f_n$ I can't obtain a weak convergent subsequence.
- After that, I wondered if I could have a weak-* convergence so I thought the Banach-Alaoglu theorem. But again, this didn't work because I couldn't find the Banach and separable space whose dual is $L^{infty}(0,T,L^1(Gamma))$
2nd EDIT: I just came up with the following idea for which I need also verification:
Consider the space of continuous functions with compact support on $Gamma$, i.e $C_c(Gamma)$. Since $Gamma$ is compact, we know that $C_c(Gamma)$ is also Banach and separable. Its dual space is the space of (signed) Radon
measures on $Gamma$ with finite mass which is denoted by $mathcal M(Gamma)$.
If $L^{infty}(0,T,mathcal M(Gamma))$ is contained in the dual space of $L^1(0,T,mathcal C_c(Gamma))$ then by Banach-Alaoglu theorem a weak-* convergent subsequence is obtained.
However I'm not completely sure if the duality argument that I used holds.
At this point I've been stuck. I would really appreciate any help or even hints.
Thanks in advance!
functional-analysis analysis convergence bochner-spaces
asked Jan 23 at 11:39
kaithkolesidoukaithkolesidou
1,086512
$endgroup$
|
show 11 more comments
$begingroup$
Let $Gamma$ be a compact $C^2$ manifold and suppose that $f_n$ is a non negative sequence of functions such that ${vert vert f_n vert vert}_{L^{infty}(0,T,L^1(Gamma))} le C$
I am interested in deducing convergence of $f_n$
MY ATTEMPTS:
- Since $L^1$ is not reflexive, $L^{infty}(0,T,L^1(Gamma))$ is also not reflexive and thus from the boundedness of $f_n$ I can't obtain a weak convergent subsequence.
- After that, I wondered if I could have a weak-* convergence so I thought the Banach-Alaoglu theorem. But again, this didn't work because I couldn't find the Banach and separable space whose dual is $L^{infty}(0,T,L^1(Gamma))$
2nd EDIT: I just came up with the following idea for which I need also verification:
Consider the space of continuous functions with compact support on $Gamma$, i.e $C_c(Gamma)$. Since $Gamma$ is compact, we know that $C_c(Gamma)$ is also Banach and separable. Its dual space is the space of (signed) Radon
measures on $Gamma$ with finite mass which is denoted by $mathcal M(Gamma)$.
If $L^{infty}(0,T,mathcal M(Gamma))$ is contained in the dual space of $L^1(0,T,mathcal C_c(Gamma))$ then by Banach-Alaoglu theorem a weak-* convergent subsequence is obtained.
However I'm not completely sure if the duality argument that I used holds.
At this point I've been stuck. I would really appreciate any help or even hints.
Thanks in advance!
functional-analysis analysis convergence bochner-spaces
asked Jan 23 at 11:39
kaithkolesidoukaithkolesidou
1,086512
$endgroup$
$begingroup$
If it were $L^infty(Omega)$, where $Omega$ is a $sigma$-finite measure space, then $L^infty$-boundedness would give you the existence of a subsequence that converges weakly-$star$. Surely there is a version of this result for the spaces you are working with. (I see that you added this in your edit. Try looking in these lecture notes of John Hunter:
$endgroup$
– Giuseppe Negro
Jan 23 at 12:24
$begingroup$
@GiuseppeNegro I just edited my question with a possible solution (?)...Could you please tell me your opinion?
$endgroup$
– kaithkolesidou
Jan 23 at 12:26
$begingroup$
math.stackexchange.com/a/1717522/8157
$endgroup$
– Giuseppe Negro
Jan 23 at 12:27
$begingroup$
I think that what you did is OK.
$endgroup$
– Giuseppe Negro
Jan 23 at 12:28
$begingroup$
@GiuseppeNegro My only doubt is if $mathcal M(Gamma)$ is a separable space in order for the duality argument to be valid. I saw in your answer in that post you recommend a book in evolution equations. I'll chech it now. Thanks a lot
$endgroup$
– kaithkolesidou
Jan 23 at 12:32
|
show 11 more comments
$begingroup$
Let $Gamma$ be a compact $C^2$ manifold and suppose that $f_n$ is a non negative sequence of functions such that ${vert vert f_n vert vert}_{L^{infty}(0,T,L^1(Gamma))} le C$
I am interested in deducing convergence of $f_n$
MY ATTEMPTS:
- Since $L^1$ is not reflexive, $L^{infty}(0,T,L^1(Gamma))$ is also not reflexive and thus from the boundedness of $f_n$ I can't obtain a weak convergent subsequence.
- After that, I wondered if I could have a weak-* convergence so I thought the Banach-Alaoglu theorem. But again, this didn't work because I couldn't find the Banach and separable space whose dual is $L^{infty}(0,T,L^1(Gamma))$
2nd EDIT: I just came up with the following idea for which I need also verification:
Consider the space of continuous functions with compact support on $Gamma$, i.e $C_c(Gamma)$. Since $Gamma$ is compact, we know that $C_c(Gamma)$ is also Banach and separable. Its dual space is the space of (signed) Radon
measures on $Gamma$ with finite mass which is denoted by $mathcal M(Gamma)$.
If $L^{infty}(0,T,mathcal M(Gamma))$ is contained in the dual space of $L^1(0,T,mathcal C_c(Gamma))$ then by Banach-Alaoglu theorem a weak-* convergent subsequence is obtained.
However I'm not completely sure if the duality argument that I used holds.
At this point I've been stuck. I would really appreciate any help or even hints.
Thanks in advance!
functional-analysis analysis convergence bochner-spaces
asked Jan 23 at 11:39
kaithkolesidoukaithkolesidou
1,086512
$endgroup$
Let $Gamma$ be a compact $C^2$ manifold and suppose that $f_n$ is a non negative sequence of functions such that ${vert vert f_n vert vert}_{L^{infty}(0,T,L^1(Gamma))} le C$
I am interested in deducing convergence of $f_n$
MY ATTEMPTS:
- Since $L^1$ is not reflexive, $L^{infty}(0,T,L^1(Gamma))$ is also not reflexive and thus from the boundedness of $f_n$ I can't obtain a weak convergent subsequence.
- After that, I wondered if I could have a weak-* convergence so I thought the Banach-Alaoglu theorem. But again, this didn't work because I couldn't find the Banach and separable space whose dual is $L^{infty}(0,T,L^1(Gamma))$
2nd EDIT: I just came up with the following idea for which I need also verification:
Consider the space of continuous functions with compact support on $Gamma$, i.e $C_c(Gamma)$. Since $Gamma$ is compact, we know that $C_c(Gamma)$ is also Banach and separable. Its dual space is the space of (signed) Radon
measures on $Gamma$ with finite mass which is denoted by $mathcal M(Gamma)$.
If $L^{infty}(0,T,mathcal M(Gamma))$ is contained in the dual space of $L^1(0,T,mathcal C_c(Gamma))$ then by Banach-Alaoglu theorem a weak-* convergent subsequence is obtained.
However I'm not completely sure if the duality argument that I used holds.
At this point I've been stuck. I would really appreciate any help or even hints.
Thanks in advance!
functional-analysis analysis convergence bochner-spaces
functional-analysis analysis convergence bochner-spaces
asked Jan 23 at 11:39
kaithkolesidoukaithkolesidou
1,086512
asked Jan 23 at 11:39
kaithkolesidoukaithkolesidou
1,086512
edited Jan 23 at 14:52
kaithkolesidou
asked Jan 23 at 11:39
kaithkolesidoukaithkolesidou
1,086512
asked Jan 23 at 11:39
kaithkolesidoukaithkolesidou
1,086512
asked Jan 23 at 11:39
kaithkolesidoukaithkolesidou
1,086512
1,086512
$begingroup$
If it were $L^infty(Omega)$, where $Omega$ is a $sigma$-finite measure space, then $L^infty$-boundedness would give you the existence of a subsequence that converges weakly-$star$. Surely there is a version of this result for the spaces you are working with. (I see that you added this in your edit. Try looking in these lecture notes of John Hunter:
$endgroup$
– Giuseppe Negro
Jan 23 at 12:24
$begingroup$
@GiuseppeNegro I just edited my question with a possible solution (?)...Could you please tell me your opinion?
$endgroup$
– kaithkolesidou
Jan 23 at 12:26
$begingroup$
math.stackexchange.com/a/1717522/8157
$endgroup$
– Giuseppe Negro
Jan 23 at 12:27
$begingroup$
I think that what you did is OK.
$endgroup$
– Giuseppe Negro
Jan 23 at 12:28
$begingroup$
@GiuseppeNegro My only doubt is if $mathcal M(Gamma)$ is a separable space in order for the duality argument to be valid. I saw in your answer in that post you recommend a book in evolution equations. I'll chech it now. Thanks a lot
$endgroup$
– kaithkolesidou
Jan 23 at 12:32
|
show 11 more comments
$begingroup$
If it were $L^infty(Omega)$, where $Omega$ is a $sigma$-finite measure space, then $L^infty$-boundedness would give you the existence of a subsequence that converges weakly-$star$. Surely there is a version of this result for the spaces you are working with. (I see that you added this in your edit. Try looking in these lecture notes of John Hunter:
$endgroup$
– Giuseppe Negro
Jan 23 at 12:24
$begingroup$
@GiuseppeNegro I just edited my question with a possible solution (?)...Could you please tell me your opinion?
$endgroup$
– kaithkolesidou
Jan 23 at 12:26
$begingroup$
math.stackexchange.com/a/1717522/8157
$endgroup$
– Giuseppe Negro
Jan 23 at 12:27
$begingroup$
I think that what you did is OK.
$endgroup$
– Giuseppe Negro
Jan 23 at 12:28
$begingroup$
@GiuseppeNegro My only doubt is if $mathcal M(Gamma)$ is a separable space in order for the duality argument to be valid. I saw in your answer in that post you recommend a book in evolution equations. I'll chech it now. Thanks a lot
$endgroup$
– kaithkolesidou
Jan 23 at 12:32
$begingroup$
If it were $L^infty(Omega)$, where $Omega$ is a $sigma$-finite measure space, then $L^infty$-boundedness would give you the existence of a subsequence that converges weakly-$star$. Surely there is a version of this result for the spaces you are working with. (I see that you added this in your edit. Try looking in these lecture notes of John Hunter:
$endgroup$
– Giuseppe Negro
Jan 23 at 12:24
$begingroup$
If it were $L^infty(Omega)$, where $Omega$ is a $sigma$-finite measure space, then $L^infty$-boundedness would give you the existence of a subsequence that converges weakly-$star$. Surely there is a version of this result for the spaces you are working with. (I see that you added this in your edit. Try looking in these lecture notes of John Hunter:
$endgroup$
– Giuseppe Negro
Jan 23 at 12:24
$begingroup$
@GiuseppeNegro I just edited my question with a possible solution (?)...Could you please tell me your opinion?
$endgroup$
– kaithkolesidou
Jan 23 at 12:26
$begingroup$
@GiuseppeNegro I just edited my question with a possible solution (?)...Could you please tell me your opinion?
$endgroup$
– kaithkolesidou
Jan 23 at 12:26
$begingroup$
math.stackexchange.com/a/1717522/8157
$endgroup$
– Giuseppe Negro
Jan 23 at 12:27
$begingroup$
math.stackexchange.com/a/1717522/8157
$endgroup$
– Giuseppe Negro
Jan 23 at 12:27
$begingroup$
I think that what you did is OK.
$endgroup$
– Giuseppe Negro
Jan 23 at 12:28
$begingroup$
I think that what you did is OK.
$endgroup$
– Giuseppe Negro
Jan 23 at 12:28
$begingroup$
@GiuseppeNegro My only doubt is if $mathcal M(Gamma)$ is a separable space in order for the duality argument to be valid. I saw in your answer in that post you recommend a book in evolution equations. I'll chech it now. Thanks a lot
$endgroup$
– kaithkolesidou
Jan 23 at 12:32
$begingroup$
@GiuseppeNegro My only doubt is if $mathcal M(Gamma)$ is a separable space in order for the duality argument to be valid. I saw in your answer in that post you recommend a book in evolution equations. I'll chech it now. Thanks a lot
$endgroup$
– kaithkolesidou
Jan 23 at 12:32
|
show 11 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The dual space of $L^1(0,T; C(Gamma))$ is $L^infty_w(0,T;mathcal M(Gamma))$ and this space consists of weak-$*$ measurable functions, see, e.g., Theorem 10.1.16 in "Handbook of applied analysis" by Papageorgiou and Kyritsi-Yiallourou.
answered Jan 23 at 18:58
gerwgerw
19.6k11334
$endgroup$
$begingroup$
Exactly what I was looking for! Thanks a lot for your help!
$endgroup$
– kaithkolesidou
Jan 24 at 10:11
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The dual space of $L^1(0,T; C(Gamma))$ is $L^infty_w(0,T;mathcal M(Gamma))$ and this space consists of weak-$*$ measurable functions, see, e.g., Theorem 10.1.16 in "Handbook of applied analysis" by Papageorgiou and Kyritsi-Yiallourou.
answered Jan 23 at 18:58
gerwgerw
19.6k11334
$endgroup$
$begingroup$
Exactly what I was looking for! Thanks a lot for your help!
$endgroup$
– kaithkolesidou
Jan 24 at 10:11
add a comment |
$begingroup$
The dual space of $L^1(0,T; C(Gamma))$ is $L^infty_w(0,T;mathcal M(Gamma))$ and this space consists of weak-$*$ measurable functions, see, e.g., Theorem 10.1.16 in "Handbook of applied analysis" by Papageorgiou and Kyritsi-Yiallourou.
answered Jan 23 at 18:58
gerwgerw
19.6k11334
$endgroup$
$begingroup$
Exactly what I was looking for! Thanks a lot for your help!
$endgroup$
– kaithkolesidou
Jan 24 at 10:11
add a comment |
$begingroup$
The dual space of $L^1(0,T; C(Gamma))$ is $L^infty_w(0,T;mathcal M(Gamma))$ and this space consists of weak-$*$ measurable functions, see, e.g., Theorem 10.1.16 in "Handbook of applied analysis" by Papageorgiou and Kyritsi-Yiallourou.
answered Jan 23 at 18:58
gerwgerw
19.6k11334
$endgroup$
The dual space of $L^1(0,T; C(Gamma))$ is $L^infty_w(0,T;mathcal M(Gamma))$ and this space consists of weak-$*$ measurable functions, see, e.g., Theorem 10.1.16 in "Handbook of applied analysis" by Papageorgiou and Kyritsi-Yiallourou.
answered Jan 23 at 18:58
gerwgerw
19.6k11334
answered Jan 23 at 18:58
gerwgerw
19.6k11334
answered Jan 23 at 18:58
gerwgerw
19.6k11334
answered Jan 23 at 18:58
gerwgerw
19.6k11334
19.6k11334
$begingroup$
Exactly what I was looking for! Thanks a lot for your help!
$endgroup$
– kaithkolesidou
Jan 24 at 10:11
add a comment |
$begingroup$
Exactly what I was looking for! Thanks a lot for your help!
$endgroup$
– kaithkolesidou
Jan 24 at 10:11
$begingroup$
Exactly what I was looking for! Thanks a lot for your help!
$endgroup$
– kaithkolesidou
Jan 24 at 10:11
$begingroup$
Exactly what I was looking for! Thanks a lot for your help!
$endgroup$
– kaithkolesidou
Jan 24 at 10:11
add a comment |
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$begingroup$
If it were $L^infty(Omega)$, where $Omega$ is a $sigma$-finite measure space, then $L^infty$-boundedness would give you the existence of a subsequence that converges weakly-$star$. Surely there is a version of this result for the spaces you are working with. (I see that you added this in your edit. Try looking in these lecture notes of John Hunter:
$endgroup$
– Giuseppe Negro
Jan 23 at 12:24
$begingroup$
@GiuseppeNegro I just edited my question with a possible solution (?)...Could you please tell me your opinion?
$endgroup$
– kaithkolesidou
Jan 23 at 12:26
$begingroup$
math.stackexchange.com/a/1717522/8157
$endgroup$
– Giuseppe Negro
Jan 23 at 12:27
$begingroup$
I think that what you did is OK.
$endgroup$
– Giuseppe Negro
Jan 23 at 12:28
$begingroup$
@GiuseppeNegro My only doubt is if $mathcal M(Gamma)$ is a separable space in order for the duality argument to be valid. I saw in your answer in that post you recommend a book in evolution equations. I'll chech it now. Thanks a lot
$endgroup$
– kaithkolesidou
Jan 23 at 12:32