Mixing
Folland Chapter 6 Problem 20b
Suppose $f_n in L^infty(X,M,mu)$ for $n=1,2,dots$, $sup_n ||f_n||_infty < infty$, and $f_n to f$ a.e.. Suppose $mu$ is $sigma$-finite. We must show $f_n to f$ in the weak* topology (viewing $L^infty$ as a subspace of the dual of $L^1$).
My question is why we need $mu$ to be $sigma$-finite. What is wrong with the following proof? We are asked to show that for each $g in L^1$, that $int f_ng to int fg$. But doesn't this follow directly from Dominated convergence theorem since $f_ng to fg$ a.e. and $|f_ng| le Mg in L^1$ for each $n$, where $M := sup_n ||f_n||_infty < infty$.
real-analysis analysis measure-theory
asked Apr 22 '17 at 1:24
mathworker21
8,6521928
add a comment |
Suppose $f_n in L^infty(X,M,mu)$ for $n=1,2,dots$, $sup_n ||f_n||_infty < infty$, and $f_n to f$ a.e.. Suppose $mu$ is $sigma$-finite. We must show $f_n to f$ in the weak* topology (viewing $L^infty$ as a subspace of the dual of $L^1$).
My question is why we need $mu$ to be $sigma$-finite. What is wrong with the following proof? We are asked to show that for each $g in L^1$, that $int f_ng to int fg$. But doesn't this follow directly from Dominated convergence theorem since $f_ng to fg$ a.e. and $|f_ng| le Mg in L^1$ for each $n$, where $M := sup_n ||f_n||_infty < infty$.
real-analysis analysis measure-theory
asked Apr 22 '17 at 1:24
mathworker21
8,6521928
add a comment |
Suppose $f_n in L^infty(X,M,mu)$ for $n=1,2,dots$, $sup_n ||f_n||_infty < infty$, and $f_n to f$ a.e.. Suppose $mu$ is $sigma$-finite. We must show $f_n to f$ in the weak* topology (viewing $L^infty$ as a subspace of the dual of $L^1$).
My question is why we need $mu$ to be $sigma$-finite. What is wrong with the following proof? We are asked to show that for each $g in L^1$, that $int f_ng to int fg$. But doesn't this follow directly from Dominated convergence theorem since $f_ng to fg$ a.e. and $|f_ng| le Mg in L^1$ for each $n$, where $M := sup_n ||f_n||_infty < infty$.
real-analysis analysis measure-theory
asked Apr 22 '17 at 1:24
mathworker21
8,6521928
Suppose $f_n in L^infty(X,M,mu)$ for $n=1,2,dots$, $sup_n ||f_n||_infty < infty$, and $f_n to f$ a.e.. Suppose $mu$ is $sigma$-finite. We must show $f_n to f$ in the weak* topology (viewing $L^infty$ as a subspace of the dual of $L^1$).
My question is why we need $mu$ to be $sigma$-finite. What is wrong with the following proof? We are asked to show that for each $g in L^1$, that $int f_ng to int fg$. But doesn't this follow directly from Dominated convergence theorem since $f_ng to fg$ a.e. and $|f_ng| le Mg in L^1$ for each $n$, where $M := sup_n ||f_n||_infty < infty$.
real-analysis analysis measure-theory
real-analysis analysis measure-theory
asked Apr 22 '17 at 1:24
mathworker21
8,6521928
asked Apr 22 '17 at 1:24
mathworker21
8,6521928
edited Nov 20 '18 at 22:26
asked Apr 22 '17 at 1:24
mathworker21
8,6521928
asked Apr 22 '17 at 1:24
mathworker21
8,6521928
asked Apr 22 '17 at 1:24
mathworker21
8,6521928
8,6521928
add a comment |
add a comment |
1 Answer
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If $mu$ is not $sigma$ finite, then you cannot conclude that the dual of $L^1$ is $L^infty$ but only that $L^infty$ is contained in the dual of $L^1$. Anyway your proof works.
answered Apr 22 '17 at 2:35
Gio67
12.5k1626
Why do we need the dual of $L^1$ to be $L^infty$? We just need $L^infty$ to be a subspace.
– mathworker21
Apr 22 '17 at 7:20
yes, it is enough to be a subspace. Functions in $L^infty$ are in the dual of $L^1$ even if the measure is not $sigma$ finite. So you are perfectly right.
– Gio67
Apr 22 '17 at 11:20
Thanks. Also, even in finite measure spaces, it need not be the case that $(L^1)^* = L^infty$.
– mathworker21
Apr 22 '17 at 18:46
Actually the necessary and sufficient conditions for the dual of $L^1$ to be $L^infty$ is that the measure be localizable and have the finite subset property. A sigma-finite or a finite measure satisfies both
– Gio67
Apr 23 '17 at 3:47
Ah yes, I was thinking the other way around. My apologies
– mathworker21
Apr 23 '17 at 4:45
add a comment |
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1 Answer
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1 Answer
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oldest
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If $mu$ is not $sigma$ finite, then you cannot conclude that the dual of $L^1$ is $L^infty$ but only that $L^infty$ is contained in the dual of $L^1$. Anyway your proof works.
answered Apr 22 '17 at 2:35
Gio67
12.5k1626
Why do we need the dual of $L^1$ to be $L^infty$? We just need $L^infty$ to be a subspace.
– mathworker21
Apr 22 '17 at 7:20
yes, it is enough to be a subspace. Functions in $L^infty$ are in the dual of $L^1$ even if the measure is not $sigma$ finite. So you are perfectly right.
– Gio67
Apr 22 '17 at 11:20
Thanks. Also, even in finite measure spaces, it need not be the case that $(L^1)^* = L^infty$.
– mathworker21
Apr 22 '17 at 18:46
Actually the necessary and sufficient conditions for the dual of $L^1$ to be $L^infty$ is that the measure be localizable and have the finite subset property. A sigma-finite or a finite measure satisfies both
– Gio67
Apr 23 '17 at 3:47
Ah yes, I was thinking the other way around. My apologies
– mathworker21
Apr 23 '17 at 4:45
add a comment |
If $mu$ is not $sigma$ finite, then you cannot conclude that the dual of $L^1$ is $L^infty$ but only that $L^infty$ is contained in the dual of $L^1$. Anyway your proof works.
answered Apr 22 '17 at 2:35
Gio67
12.5k1626
Why do we need the dual of $L^1$ to be $L^infty$? We just need $L^infty$ to be a subspace.
– mathworker21
Apr 22 '17 at 7:20
yes, it is enough to be a subspace. Functions in $L^infty$ are in the dual of $L^1$ even if the measure is not $sigma$ finite. So you are perfectly right.
– Gio67
Apr 22 '17 at 11:20
Thanks. Also, even in finite measure spaces, it need not be the case that $(L^1)^* = L^infty$.
– mathworker21
Apr 22 '17 at 18:46
Actually the necessary and sufficient conditions for the dual of $L^1$ to be $L^infty$ is that the measure be localizable and have the finite subset property. A sigma-finite or a finite measure satisfies both
– Gio67
Apr 23 '17 at 3:47
Ah yes, I was thinking the other way around. My apologies
– mathworker21
Apr 23 '17 at 4:45
add a comment |
If $mu$ is not $sigma$ finite, then you cannot conclude that the dual of $L^1$ is $L^infty$ but only that $L^infty$ is contained in the dual of $L^1$. Anyway your proof works.
answered Apr 22 '17 at 2:35
Gio67
12.5k1626
If $mu$ is not $sigma$ finite, then you cannot conclude that the dual of $L^1$ is $L^infty$ but only that $L^infty$ is contained in the dual of $L^1$. Anyway your proof works.
answered Apr 22 '17 at 2:35
Gio67
12.5k1626
edited Apr 22 '17 at 11:19
answered Apr 22 '17 at 2:35
Gio67
12.5k1626
answered Apr 22 '17 at 2:35
Gio67
12.5k1626
answered Apr 22 '17 at 2:35
Gio67
12.5k1626
12.5k1626
Why do we need the dual of $L^1$ to be $L^infty$? We just need $L^infty$ to be a subspace.
– mathworker21
Apr 22 '17 at 7:20
yes, it is enough to be a subspace. Functions in $L^infty$ are in the dual of $L^1$ even if the measure is not $sigma$ finite. So you are perfectly right.
– Gio67
Apr 22 '17 at 11:20
Thanks. Also, even in finite measure spaces, it need not be the case that $(L^1)^* = L^infty$.
– mathworker21
Apr 22 '17 at 18:46
Actually the necessary and sufficient conditions for the dual of $L^1$ to be $L^infty$ is that the measure be localizable and have the finite subset property. A sigma-finite or a finite measure satisfies both
– Gio67
Apr 23 '17 at 3:47
Ah yes, I was thinking the other way around. My apologies
– mathworker21
Apr 23 '17 at 4:45
add a comment |
Why do we need the dual of $L^1$ to be $L^infty$? We just need $L^infty$ to be a subspace.
– mathworker21
Apr 22 '17 at 7:20
yes, it is enough to be a subspace. Functions in $L^infty$ are in the dual of $L^1$ even if the measure is not $sigma$ finite. So you are perfectly right.
– Gio67
Apr 22 '17 at 11:20
Thanks. Also, even in finite measure spaces, it need not be the case that $(L^1)^* = L^infty$.
– mathworker21
Apr 22 '17 at 18:46
Actually the necessary and sufficient conditions for the dual of $L^1$ to be $L^infty$ is that the measure be localizable and have the finite subset property. A sigma-finite or a finite measure satisfies both
– Gio67
Apr 23 '17 at 3:47
Ah yes, I was thinking the other way around. My apologies
– mathworker21
Apr 23 '17 at 4:45
Why do we need the dual of $L^1$ to be $L^infty$? We just need $L^infty$ to be a subspace.
– mathworker21
Apr 22 '17 at 7:20
Why do we need the dual of $L^1$ to be $L^infty$? We just need $L^infty$ to be a subspace.
– mathworker21
Apr 22 '17 at 7:20
yes, it is enough to be a subspace. Functions in $L^infty$ are in the dual of $L^1$ even if the measure is not $sigma$ finite. So you are perfectly right.
– Gio67
Apr 22 '17 at 11:20
yes, it is enough to be a subspace. Functions in $L^infty$ are in the dual of $L^1$ even if the measure is not $sigma$ finite. So you are perfectly right.
– Gio67
Apr 22 '17 at 11:20
Thanks. Also, even in finite measure spaces, it need not be the case that $(L^1)^* = L^infty$.
– mathworker21
Apr 22 '17 at 18:46
Thanks. Also, even in finite measure spaces, it need not be the case that $(L^1)^* = L^infty$.
– mathworker21
Apr 22 '17 at 18:46
Actually the necessary and sufficient conditions for the dual of $L^1$ to be $L^infty$ is that the measure be localizable and have the finite subset property. A sigma-finite or a finite measure satisfies both
– Gio67
Apr 23 '17 at 3:47
Actually the necessary and sufficient conditions for the dual of $L^1$ to be $L^infty$ is that the measure be localizable and have the finite subset property. A sigma-finite or a finite measure satisfies both
– Gio67
Apr 23 '17 at 3:47
Ah yes, I was thinking the other way around. My apologies
– mathworker21
Apr 23 '17 at 4:45
Ah yes, I was thinking the other way around. My apologies
– mathworker21
Apr 23 '17 at 4:45
add a comment |
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