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What is the predual of $L^1$
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Is there a nice characterization of the predual of $L^1$? So, what does the space $X$ look like, such that $X^*=L^1$, where the star denotes the dual of a Banach space. How do you start to find such preduals in general?
For some context, it is well known that given a measure space $(S, Sigma, mu)$, $L^p := L^p(S, mu)$ is a Banach space for $pin (1,infty)$ and that $L^p cong (L^q)^*$ where $q$ is the Holder conjugate of $p$, that is $frac 1p + frac 1q =1$. It is also known that $L^1$ is the predual of $L^infty$. This leaves the above questions as the only remaining case.
When $S$ is (for example) finite of course the question is moot. If you like one can consider only very simple measure space, like $[0,1]$ with the Lebesgue measure.
functional-analysis banach-spaces lp-spaces
edited Jul 31 at 15:58
John Ma
37.5k93669
asked Apr 27 '12 at 14:54
math
1,39211639
add a comment |Â
up vote
12
down vote
favorite
Is there a nice characterization of the predual of $L^1$? So, what does the space $X$ look like, such that $X^*=L^1$, where the star denotes the dual of a Banach space. How do you start to find such preduals in general?
For some context, it is well known that given a measure space $(S, Sigma, mu)$, $L^p := L^p(S, mu)$ is a Banach space for $pin (1,infty)$ and that $L^p cong (L^q)^*$ where $q$ is the Holder conjugate of $p$, that is $frac 1p + frac 1q =1$. It is also known that $L^1$ is the predual of $L^infty$. This leaves the above questions as the only remaining case.
When $S$ is (for example) finite of course the question is moot. If you like one can consider only very simple measure space, like $[0,1]$ with the Lebesgue measure.
functional-analysis banach-spaces lp-spaces
edited Jul 31 at 15:58
John Ma
37.5k93669
asked Apr 27 '12 at 14:54
math
1,39211639
4
Note also that the question what is "the" predual does not make sense in general. The space $ell^1$ has many non-isomorphic preduals, for example $C(K)$ for $K$ countable and compact.
– t.b.
Apr 27 '12 at 18:03
@ t.b: Thanks for pointing this out! To be honest, since I didn't know predual of $L^1$ I didn't think there could be various.
– math
Apr 28 '12 at 15:50
I think $K$ also needs to be Hausdorff no?
– Christian Bueno
Mar 9 '15 at 16:21
add a comment |Â
up vote
12
down vote
favorite
up vote
12
down vote
favorite
Is there a nice characterization of the predual of $L^1$? So, what does the space $X$ look like, such that $X^*=L^1$, where the star denotes the dual of a Banach space. How do you start to find such preduals in general?
For some context, it is well known that given a measure space $(S, Sigma, mu)$, $L^p := L^p(S, mu)$ is a Banach space for $pin (1,infty)$ and that $L^p cong (L^q)^*$ where $q$ is the Holder conjugate of $p$, that is $frac 1p + frac 1q =1$. It is also known that $L^1$ is the predual of $L^infty$. This leaves the above questions as the only remaining case.
When $S$ is (for example) finite of course the question is moot. If you like one can consider only very simple measure space, like $[0,1]$ with the Lebesgue measure.
functional-analysis banach-spaces lp-spaces
edited Jul 31 at 15:58
John Ma
37.5k93669
asked Apr 27 '12 at 14:54
math
1,39211639
Is there a nice characterization of the predual of $L^1$? So, what does the space $X$ look like, such that $X^*=L^1$, where the star denotes the dual of a Banach space. How do you start to find such preduals in general?
For some context, it is well known that given a measure space $(S, Sigma, mu)$, $L^p := L^p(S, mu)$ is a Banach space for $pin (1,infty)$ and that $L^p cong (L^q)^*$ where $q$ is the Holder conjugate of $p$, that is $frac 1p + frac 1q =1$. It is also known that $L^1$ is the predual of $L^infty$. This leaves the above questions as the only remaining case.
When $S$ is (for example) finite of course the question is moot. If you like one can consider only very simple measure space, like $[0,1]$ with the Lebesgue measure.
functional-analysis banach-spaces lp-spaces
edited Jul 31 at 15:58
John Ma
37.5k93669
asked Apr 27 '12 at 14:54
math
1,39211639
edited Jul 31 at 15:58
John Ma
37.5k93669
edited Jul 31 at 15:58
John Ma
37.5k93669
edited Jul 31 at 15:58
John Ma
37.5k93669
37.5k93669
asked Apr 27 '12 at 14:54
math
1,39211639
asked Apr 27 '12 at 14:54
math
1,39211639
asked Apr 27 '12 at 14:54
math
1,39211639
1,39211639
4
Note also that the question what is "the" predual does not make sense in general. The space $ell^1$ has many non-isomorphic preduals, for example $C(K)$ for $K$ countable and compact.
– t.b.
Apr 27 '12 at 18:03
@ t.b: Thanks for pointing this out! To be honest, since I didn't know predual of $L^1$ I didn't think there could be various.
– math
Apr 28 '12 at 15:50
I think $K$ also needs to be Hausdorff no?
– Christian Bueno
Mar 9 '15 at 16:21
add a comment |Â
4
Note also that the question what is "the" predual does not make sense in general. The space $ell^1$ has many non-isomorphic preduals, for example $C(K)$ for $K$ countable and compact.
– t.b.
Apr 27 '12 at 18:03
@ t.b: Thanks for pointing this out! To be honest, since I didn't know predual of $L^1$ I didn't think there could be various.
– math
Apr 28 '12 at 15:50
I think $K$ also needs to be Hausdorff no?
– Christian Bueno
Mar 9 '15 at 16:21
4
4
Note also that the question what is "the" predual does not make sense in general. The space $ell^1$ has many non-isomorphic preduals, for example $C(K)$ for $K$ countable and compact.
– t.b.
Apr 27 '12 at 18:03
Note also that the question what is "the" predual does not make sense in general. The space $ell^1$ has many non-isomorphic preduals, for example $C(K)$ for $K$ countable and compact.
– t.b.
Apr 27 '12 at 18:03
@ t.b: Thanks for pointing this out! To be honest, since I didn't know predual of $L^1$ I didn't think there could be various.
– math
Apr 28 '12 at 15:50
@ t.b: Thanks for pointing this out! To be honest, since I didn't know predual of $L^1$ I didn't think there could be various.
– math
Apr 28 '12 at 15:50
I think $K$ also needs to be Hausdorff no?
– Christian Bueno
Mar 9 '15 at 16:21
I think $K$ also needs to be Hausdorff no?
– Christian Bueno
Mar 9 '15 at 16:21
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
15
down vote
accepted
In fact, $L_1[0,1]$ has no pre-dual. More is true: $L_1$ cannot be embedded is a separable dual space. See, e.g., Theorem 6.3.7 in Kalton and Albiac's Topics in in Banach Space Theory.
answered Apr 27 '12 at 15:18
David Mitra
61.8k694158
4
@Math: $ell^1=L^1(mathbbN)$ (with counting measure) famously has predual $c_0$, the subspace of $L^infty(mathbbN)$ consisting of sequences converging to $0$.
– Chris Eagle
Apr 27 '12 at 15:38
2
@math I should point out that it is not hard to show that $L_1[0,1]$ is not isometric to a dual space using an extreme point argument and the Krein-Milman Theorem.
– David Mitra
Apr 27 '12 at 15:55
2
@math A result due to Johnson and Zippen: Let $X$ be a separable $L_1(mu)$ predual. Then there is a subspace $Y$ of $C(Delta)$, where $Delta$ is the Cantor set, such that $X$ is isometric to $C(Delta)/Y$. Here is the reference: W.B. Johnson and M. Zippen, Every separable predual of an $L_1$-space is a quotient of $C(Delta)$. Israel J. Math 16 (1973), 198-202. See also here.
– David Mitra
Apr 27 '12 at 16:43
6
David's comment expanded. In a dual space, the unit ball has lots of extreme points (since, in the weak* topology, it is the closed convex hull of its extreme points: Krein-Milman theorem). But the unit ball of $L^1[0,1]$ has no extreme points at all! This shows $L^1[0,1]$ is not isometric to a dual space. The isomorphic theorem is harder.
– GEdgar
Apr 27 '12 at 17:44
2
@XanderHenderson I've added an answer for the Krein-Milman argument which should answer this question and the duplicate source: I'm afraid the more general argument in this answer is beyond me.
– B. Mehta
Jul 30 at 22:33
 |Â
show 4 more comments
up vote
6
down vote
Banach-Alaoglu-Bourbaki: For $X$ a Banach space, the closed unit ball of $X^*$ is weak*-compact.
Krein-Milman: For $X$ a locally convex topological vector space, and $K$ a compact convex subset, $K$ is the closed convex hull of its extreme points.
Suppose $X^* = L^1[0,1]$. By BAB, the closed unit ball of $L^1[0,1]$ is weak*-compact. In addition, weak* topologies are always locally convex (see, for instance, here) and $L^1[0,1]$ is clearly convex so Krein-Milman gives that it is the closed convex hull of its extreme points.
But the closed unit ball of $L^1[0,1]$ has no extreme points: Suppose $|f|_1 = alpha leq 1$, $f neq 0$. Then $H(s) = int_0^s |f(t)| , dt$ is a continuous function, with $H(0) = 0$ and $H(1) = alpha$. Thus there is some $s_0 in (0,1)$ with $H(s) = fracalpha2$, so set
$$g(t) = 2f chi_[0,s_0] quad h(t) = 2f chi_[s_0,1]$$
which satisfy $|g|_1 = |h|_1 = |f|_1 = alpha$ and $frac12(g + h) = f$, so $f$ is not an extreme point. Clearly $0$ is not an extreme point either, so the unit ball of $L^1[0,1]$ has no extreme points.
Thus we have a contradiction, and $L^1[0,1]$ has no pre-dual.
answered Jul 30 at 22:31
B. Mehta
11.6k21844
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
accepted
In fact, $L_1[0,1]$ has no pre-dual. More is true: $L_1$ cannot be embedded is a separable dual space. See, e.g., Theorem 6.3.7 in Kalton and Albiac's Topics in in Banach Space Theory.
answered Apr 27 '12 at 15:18
David Mitra
61.8k694158
4
@Math: $ell^1=L^1(mathbbN)$ (with counting measure) famously has predual $c_0$, the subspace of $L^infty(mathbbN)$ consisting of sequences converging to $0$.
– Chris Eagle
Apr 27 '12 at 15:38
2
@math I should point out that it is not hard to show that $L_1[0,1]$ is not isometric to a dual space using an extreme point argument and the Krein-Milman Theorem.
– David Mitra
Apr 27 '12 at 15:55
2
@math A result due to Johnson and Zippen: Let $X$ be a separable $L_1(mu)$ predual. Then there is a subspace $Y$ of $C(Delta)$, where $Delta$ is the Cantor set, such that $X$ is isometric to $C(Delta)/Y$. Here is the reference: W.B. Johnson and M. Zippen, Every separable predual of an $L_1$-space is a quotient of $C(Delta)$. Israel J. Math 16 (1973), 198-202. See also here.
– David Mitra
Apr 27 '12 at 16:43
6
David's comment expanded. In a dual space, the unit ball has lots of extreme points (since, in the weak* topology, it is the closed convex hull of its extreme points: Krein-Milman theorem). But the unit ball of $L^1[0,1]$ has no extreme points at all! This shows $L^1[0,1]$ is not isometric to a dual space. The isomorphic theorem is harder.
– GEdgar
Apr 27 '12 at 17:44
2
@XanderHenderson I've added an answer for the Krein-Milman argument which should answer this question and the duplicate source: I'm afraid the more general argument in this answer is beyond me.
– B. Mehta
Jul 30 at 22:33
 |Â
show 4 more comments
up vote
15
down vote
accepted
In fact, $L_1[0,1]$ has no pre-dual. More is true: $L_1$ cannot be embedded is a separable dual space. See, e.g., Theorem 6.3.7 in Kalton and Albiac's Topics in in Banach Space Theory.
answered Apr 27 '12 at 15:18
David Mitra
61.8k694158
4
@Math: $ell^1=L^1(mathbbN)$ (with counting measure) famously has predual $c_0$, the subspace of $L^infty(mathbbN)$ consisting of sequences converging to $0$.
– Chris Eagle
Apr 27 '12 at 15:38
2
@math I should point out that it is not hard to show that $L_1[0,1]$ is not isometric to a dual space using an extreme point argument and the Krein-Milman Theorem.
– David Mitra
Apr 27 '12 at 15:55
2
@math A result due to Johnson and Zippen: Let $X$ be a separable $L_1(mu)$ predual. Then there is a subspace $Y$ of $C(Delta)$, where $Delta$ is the Cantor set, such that $X$ is isometric to $C(Delta)/Y$. Here is the reference: W.B. Johnson and M. Zippen, Every separable predual of an $L_1$-space is a quotient of $C(Delta)$. Israel J. Math 16 (1973), 198-202. See also here.
– David Mitra
Apr 27 '12 at 16:43
6
David's comment expanded. In a dual space, the unit ball has lots of extreme points (since, in the weak* topology, it is the closed convex hull of its extreme points: Krein-Milman theorem). But the unit ball of $L^1[0,1]$ has no extreme points at all! This shows $L^1[0,1]$ is not isometric to a dual space. The isomorphic theorem is harder.
– GEdgar
Apr 27 '12 at 17:44
2
@XanderHenderson I've added an answer for the Krein-Milman argument which should answer this question and the duplicate source: I'm afraid the more general argument in this answer is beyond me.
– B. Mehta
Jul 30 at 22:33
 |Â
show 4 more comments
up vote
15
down vote
accepted
up vote
15
down vote
accepted
In fact, $L_1[0,1]$ has no pre-dual. More is true: $L_1$ cannot be embedded is a separable dual space. See, e.g., Theorem 6.3.7 in Kalton and Albiac's Topics in in Banach Space Theory.
answered Apr 27 '12 at 15:18
David Mitra
61.8k694158
In fact, $L_1[0,1]$ has no pre-dual. More is true: $L_1$ cannot be embedded is a separable dual space. See, e.g., Theorem 6.3.7 in Kalton and Albiac's Topics in in Banach Space Theory.
answered Apr 27 '12 at 15:18
David Mitra
61.8k694158
edited Apr 27 '12 at 15:24
answered Apr 27 '12 at 15:18
David Mitra
61.8k694158
answered Apr 27 '12 at 15:18
David Mitra
61.8k694158
answered Apr 27 '12 at 15:18
David Mitra
61.8k694158
61.8k694158
4
@Math: $ell^1=L^1(mathbbN)$ (with counting measure) famously has predual $c_0$, the subspace of $L^infty(mathbbN)$ consisting of sequences converging to $0$.
– Chris Eagle
Apr 27 '12 at 15:38
2
@math I should point out that it is not hard to show that $L_1[0,1]$ is not isometric to a dual space using an extreme point argument and the Krein-Milman Theorem.
– David Mitra
Apr 27 '12 at 15:55
2
@math A result due to Johnson and Zippen: Let $X$ be a separable $L_1(mu)$ predual. Then there is a subspace $Y$ of $C(Delta)$, where $Delta$ is the Cantor set, such that $X$ is isometric to $C(Delta)/Y$. Here is the reference: W.B. Johnson and M. Zippen, Every separable predual of an $L_1$-space is a quotient of $C(Delta)$. Israel J. Math 16 (1973), 198-202. See also here.
– David Mitra
Apr 27 '12 at 16:43
6
David's comment expanded. In a dual space, the unit ball has lots of extreme points (since, in the weak* topology, it is the closed convex hull of its extreme points: Krein-Milman theorem). But the unit ball of $L^1[0,1]$ has no extreme points at all! This shows $L^1[0,1]$ is not isometric to a dual space. The isomorphic theorem is harder.
– GEdgar
Apr 27 '12 at 17:44
2
@XanderHenderson I've added an answer for the Krein-Milman argument which should answer this question and the duplicate source: I'm afraid the more general argument in this answer is beyond me.
– B. Mehta
Jul 30 at 22:33
 |Â
show 4 more comments
4
@Math: $ell^1=L^1(mathbbN)$ (with counting measure) famously has predual $c_0$, the subspace of $L^infty(mathbbN)$ consisting of sequences converging to $0$.
– Chris Eagle
Apr 27 '12 at 15:38
2
@math I should point out that it is not hard to show that $L_1[0,1]$ is not isometric to a dual space using an extreme point argument and the Krein-Milman Theorem.
– David Mitra
Apr 27 '12 at 15:55
2
@math A result due to Johnson and Zippen: Let $X$ be a separable $L_1(mu)$ predual. Then there is a subspace $Y$ of $C(Delta)$, where $Delta$ is the Cantor set, such that $X$ is isometric to $C(Delta)/Y$. Here is the reference: W.B. Johnson and M. Zippen, Every separable predual of an $L_1$-space is a quotient of $C(Delta)$. Israel J. Math 16 (1973), 198-202. See also here.
– David Mitra
Apr 27 '12 at 16:43
6
David's comment expanded. In a dual space, the unit ball has lots of extreme points (since, in the weak* topology, it is the closed convex hull of its extreme points: Krein-Milman theorem). But the unit ball of $L^1[0,1]$ has no extreme points at all! This shows $L^1[0,1]$ is not isometric to a dual space. The isomorphic theorem is harder.
– GEdgar
Apr 27 '12 at 17:44
2
@XanderHenderson I've added an answer for the Krein-Milman argument which should answer this question and the duplicate source: I'm afraid the more general argument in this answer is beyond me.
– B. Mehta
Jul 30 at 22:33
4
4
@Math: $ell^1=L^1(mathbbN)$ (with counting measure) famously has predual $c_0$, the subspace of $L^infty(mathbbN)$ consisting of sequences converging to $0$.
– Chris Eagle
Apr 27 '12 at 15:38
@Math: $ell^1=L^1(mathbbN)$ (with counting measure) famously has predual $c_0$, the subspace of $L^infty(mathbbN)$ consisting of sequences converging to $0$.
– Chris Eagle
Apr 27 '12 at 15:38
2
2
@math I should point out that it is not hard to show that $L_1[0,1]$ is not isometric to a dual space using an extreme point argument and the Krein-Milman Theorem.
– David Mitra
Apr 27 '12 at 15:55
@math I should point out that it is not hard to show that $L_1[0,1]$ is not isometric to a dual space using an extreme point argument and the Krein-Milman Theorem.
– David Mitra
Apr 27 '12 at 15:55
2
2
@math A result due to Johnson and Zippen: Let $X$ be a separable $L_1(mu)$ predual. Then there is a subspace $Y$ of $C(Delta)$, where $Delta$ is the Cantor set, such that $X$ is isometric to $C(Delta)/Y$. Here is the reference: W.B. Johnson and M. Zippen, Every separable predual of an $L_1$-space is a quotient of $C(Delta)$. Israel J. Math 16 (1973), 198-202. See also here.
– David Mitra
Apr 27 '12 at 16:43
@math A result due to Johnson and Zippen: Let $X$ be a separable $L_1(mu)$ predual. Then there is a subspace $Y$ of $C(Delta)$, where $Delta$ is the Cantor set, such that $X$ is isometric to $C(Delta)/Y$. Here is the reference: W.B. Johnson and M. Zippen, Every separable predual of an $L_1$-space is a quotient of $C(Delta)$. Israel J. Math 16 (1973), 198-202. See also here.
– David Mitra
Apr 27 '12 at 16:43
6
6
David's comment expanded. In a dual space, the unit ball has lots of extreme points (since, in the weak* topology, it is the closed convex hull of its extreme points: Krein-Milman theorem). But the unit ball of $L^1[0,1]$ has no extreme points at all! This shows $L^1[0,1]$ is not isometric to a dual space. The isomorphic theorem is harder.
– GEdgar
Apr 27 '12 at 17:44
David's comment expanded. In a dual space, the unit ball has lots of extreme points (since, in the weak* topology, it is the closed convex hull of its extreme points: Krein-Milman theorem). But the unit ball of $L^1[0,1]$ has no extreme points at all! This shows $L^1[0,1]$ is not isometric to a dual space. The isomorphic theorem is harder.
– GEdgar
Apr 27 '12 at 17:44
2
2
@XanderHenderson I've added an answer for the Krein-Milman argument which should answer this question and the duplicate source: I'm afraid the more general argument in this answer is beyond me.
– B. Mehta
Jul 30 at 22:33
@XanderHenderson I've added an answer for the Krein-Milman argument which should answer this question and the duplicate source: I'm afraid the more general argument in this answer is beyond me.
– B. Mehta
Jul 30 at 22:33
 |Â
show 4 more comments
up vote
6
down vote
Banach-Alaoglu-Bourbaki: For $X$ a Banach space, the closed unit ball of $X^*$ is weak*-compact.
Krein-Milman: For $X$ a locally convex topological vector space, and $K$ a compact convex subset, $K$ is the closed convex hull of its extreme points.
Suppose $X^* = L^1[0,1]$. By BAB, the closed unit ball of $L^1[0,1]$ is weak*-compact. In addition, weak* topologies are always locally convex (see, for instance, here) and $L^1[0,1]$ is clearly convex so Krein-Milman gives that it is the closed convex hull of its extreme points.
But the closed unit ball of $L^1[0,1]$ has no extreme points: Suppose $|f|_1 = alpha leq 1$, $f neq 0$. Then $H(s) = int_0^s |f(t)| , dt$ is a continuous function, with $H(0) = 0$ and $H(1) = alpha$. Thus there is some $s_0 in (0,1)$ with $H(s) = fracalpha2$, so set
$$g(t) = 2f chi_[0,s_0] quad h(t) = 2f chi_[s_0,1]$$
which satisfy $|g|_1 = |h|_1 = |f|_1 = alpha$ and $frac12(g + h) = f$, so $f$ is not an extreme point. Clearly $0$ is not an extreme point either, so the unit ball of $L^1[0,1]$ has no extreme points.
Thus we have a contradiction, and $L^1[0,1]$ has no pre-dual.
answered Jul 30 at 22:31
B. Mehta
11.6k21844
add a comment |Â
up vote
6
down vote
Banach-Alaoglu-Bourbaki: For $X$ a Banach space, the closed unit ball of $X^*$ is weak*-compact.
Krein-Milman: For $X$ a locally convex topological vector space, and $K$ a compact convex subset, $K$ is the closed convex hull of its extreme points.
Suppose $X^* = L^1[0,1]$. By BAB, the closed unit ball of $L^1[0,1]$ is weak*-compact. In addition, weak* topologies are always locally convex (see, for instance, here) and $L^1[0,1]$ is clearly convex so Krein-Milman gives that it is the closed convex hull of its extreme points.
But the closed unit ball of $L^1[0,1]$ has no extreme points: Suppose $|f|_1 = alpha leq 1$, $f neq 0$. Then $H(s) = int_0^s |f(t)| , dt$ is a continuous function, with $H(0) = 0$ and $H(1) = alpha$. Thus there is some $s_0 in (0,1)$ with $H(s) = fracalpha2$, so set
$$g(t) = 2f chi_[0,s_0] quad h(t) = 2f chi_[s_0,1]$$
which satisfy $|g|_1 = |h|_1 = |f|_1 = alpha$ and $frac12(g + h) = f$, so $f$ is not an extreme point. Clearly $0$ is not an extreme point either, so the unit ball of $L^1[0,1]$ has no extreme points.
Thus we have a contradiction, and $L^1[0,1]$ has no pre-dual.
answered Jul 30 at 22:31
B. Mehta
11.6k21844
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Banach-Alaoglu-Bourbaki: For $X$ a Banach space, the closed unit ball of $X^*$ is weak*-compact.
Krein-Milman: For $X$ a locally convex topological vector space, and $K$ a compact convex subset, $K$ is the closed convex hull of its extreme points.
Suppose $X^* = L^1[0,1]$. By BAB, the closed unit ball of $L^1[0,1]$ is weak*-compact. In addition, weak* topologies are always locally convex (see, for instance, here) and $L^1[0,1]$ is clearly convex so Krein-Milman gives that it is the closed convex hull of its extreme points.
But the closed unit ball of $L^1[0,1]$ has no extreme points: Suppose $|f|_1 = alpha leq 1$, $f neq 0$. Then $H(s) = int_0^s |f(t)| , dt$ is a continuous function, with $H(0) = 0$ and $H(1) = alpha$. Thus there is some $s_0 in (0,1)$ with $H(s) = fracalpha2$, so set
$$g(t) = 2f chi_[0,s_0] quad h(t) = 2f chi_[s_0,1]$$
which satisfy $|g|_1 = |h|_1 = |f|_1 = alpha$ and $frac12(g + h) = f$, so $f$ is not an extreme point. Clearly $0$ is not an extreme point either, so the unit ball of $L^1[0,1]$ has no extreme points.
Thus we have a contradiction, and $L^1[0,1]$ has no pre-dual.
answered Jul 30 at 22:31
B. Mehta
11.6k21844
Banach-Alaoglu-Bourbaki: For $X$ a Banach space, the closed unit ball of $X^*$ is weak*-compact.
Krein-Milman: For $X$ a locally convex topological vector space, and $K$ a compact convex subset, $K$ is the closed convex hull of its extreme points.
Suppose $X^* = L^1[0,1]$. By BAB, the closed unit ball of $L^1[0,1]$ is weak*-compact. In addition, weak* topologies are always locally convex (see, for instance, here) and $L^1[0,1]$ is clearly convex so Krein-Milman gives that it is the closed convex hull of its extreme points.
But the closed unit ball of $L^1[0,1]$ has no extreme points: Suppose $|f|_1 = alpha leq 1$, $f neq 0$. Then $H(s) = int_0^s |f(t)| , dt$ is a continuous function, with $H(0) = 0$ and $H(1) = alpha$. Thus there is some $s_0 in (0,1)$ with $H(s) = fracalpha2$, so set
$$g(t) = 2f chi_[0,s_0] quad h(t) = 2f chi_[s_0,1]$$
which satisfy $|g|_1 = |h|_1 = |f|_1 = alpha$ and $frac12(g + h) = f$, so $f$ is not an extreme point. Clearly $0$ is not an extreme point either, so the unit ball of $L^1[0,1]$ has no extreme points.
Thus we have a contradiction, and $L^1[0,1]$ has no pre-dual.
answered Jul 30 at 22:31
B. Mehta
11.6k21844
edited Jul 30 at 22:54
answered Jul 30 at 22:31
B. Mehta
11.6k21844
answered Jul 30 at 22:31
B. Mehta
11.6k21844
answered Jul 30 at 22:31
B. Mehta
11.6k21844
11.6k21844
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4
Note also that the question what is "the" predual does not make sense in general. The space $ell^1$ has many non-isomorphic preduals, for example $C(K)$ for $K$ countable and compact.
– t.b.
Apr 27 '12 at 18:03
@ t.b: Thanks for pointing this out! To be honest, since I didn't know predual of $L^1$ I didn't think there could be various.
– math
Apr 28 '12 at 15:50
I think $K$ also needs to be Hausdorff no?
– Christian Bueno
Mar 9 '15 at 16:21