If the closed unit ball of Banach space has at least one extreme point, must the Banach space the be a dual...
up vote
4
down vote
favorite
Let $X$ be a Banach space.
By Banach-Alaoglu and Krein-Milman Theorems, one can show that if $X$ is a dual space, then $X$ must have at least one extreme point of the closed unit ball.
I am interested in its converse.
More precisely,
Question: Let $X$ be a Banach space.
If the closed unit ball of $X$ has at least one extreme point, must $X$ be a dual space?
I feel that the statement above is negative.
However, I could not produce a counterexample.
In fact, the only Banach spaces which I know that are not dual spaces are $c_0$ and $C_0(mathbb{R})$ (the latter set is the collection of all real-valued continuous function vanishing at infinity) because both sets have no extreme point.
fa.functional-analysis banach-spaces convexity duality extreme-points
asked Dec 2 at 7:06
Idonknow
213312
|
show 1 more comment
up vote
4
down vote
favorite
Let $X$ be a Banach space.
By Banach-Alaoglu and Krein-Milman Theorems, one can show that if $X$ is a dual space, then $X$ must have at least one extreme point of the closed unit ball.
I am interested in its converse.
More precisely,
Question: Let $X$ be a Banach space.
If the closed unit ball of $X$ has at least one extreme point, must $X$ be a dual space?
I feel that the statement above is negative.
However, I could not produce a counterexample.
In fact, the only Banach spaces which I know that are not dual spaces are $c_0$ and $C_0(mathbb{R})$ (the latter set is the collection of all real-valued continuous function vanishing at infinity) because both sets have no extreme point.
fa.functional-analysis banach-spaces convexity duality extreme-points
asked Dec 2 at 7:06
Idonknow
213312
3
When you say that "$X$ has at least one extreme point" do you mean that the closed unit ball of $X$ has at least on extreme point?
– Martin Sleziak
Dec 2 at 7:11
I have added the tag (extreme-points), since it seems to me a good fit to the question. There exists also (krein-milman-theorem) tag, but that one would probably be a stretch.
– Martin Sleziak
Dec 2 at 7:13
4
This post on Mathematics site seems to be about the same question: Krein-Milman and dual spaces.
– Martin Sleziak
Dec 2 at 7:20
@MartinSleziak Yes, I mean the closed unit ball of $X.$
– Idonknow
Dec 2 at 7:42
1
@TarasBanakh: There are infinite compact $K$ for which $C(K)$ is a dual space: these are precisely the hyperstonean $K$, e.g., $betamathbb{N}$. (On the other hand there are non-dual $C(K)$ for which the unit ball is the norm-closed convex hull of its extreme points, e.g. $alphamathbb{N}$. These are precisely the totally disconected $K$.)
– Dirk Werner
Dec 2 at 22:04
|
show 1 more comment
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $X$ be a Banach space.
By Banach-Alaoglu and Krein-Milman Theorems, one can show that if $X$ is a dual space, then $X$ must have at least one extreme point of the closed unit ball.
I am interested in its converse.
More precisely,
Question: Let $X$ be a Banach space.
If the closed unit ball of $X$ has at least one extreme point, must $X$ be a dual space?
I feel that the statement above is negative.
However, I could not produce a counterexample.
In fact, the only Banach spaces which I know that are not dual spaces are $c_0$ and $C_0(mathbb{R})$ (the latter set is the collection of all real-valued continuous function vanishing at infinity) because both sets have no extreme point.
fa.functional-analysis banach-spaces convexity duality extreme-points
asked Dec 2 at 7:06
Idonknow
213312
Let $X$ be a Banach space.
By Banach-Alaoglu and Krein-Milman Theorems, one can show that if $X$ is a dual space, then $X$ must have at least one extreme point of the closed unit ball.
I am interested in its converse.
More precisely,
Question: Let $X$ be a Banach space.
If the closed unit ball of $X$ has at least one extreme point, must $X$ be a dual space?
I feel that the statement above is negative.
However, I could not produce a counterexample.
In fact, the only Banach spaces which I know that are not dual spaces are $c_0$ and $C_0(mathbb{R})$ (the latter set is the collection of all real-valued continuous function vanishing at infinity) because both sets have no extreme point.
fa.functional-analysis banach-spaces convexity duality extreme-points
fa.functional-analysis banach-spaces convexity duality extreme-points
asked Dec 2 at 7:06
Idonknow
213312
asked Dec 2 at 7:06
Idonknow
213312
edited Dec 2 at 23:09
asked Dec 2 at 7:06
Idonknow
213312
asked Dec 2 at 7:06
Idonknow
213312
asked Dec 2 at 7:06
Idonknow
213312
213312
3
When you say that "$X$ has at least one extreme point" do you mean that the closed unit ball of $X$ has at least on extreme point?
– Martin Sleziak
Dec 2 at 7:11
I have added the tag (extreme-points), since it seems to me a good fit to the question. There exists also (krein-milman-theorem) tag, but that one would probably be a stretch.
– Martin Sleziak
Dec 2 at 7:13
4
This post on Mathematics site seems to be about the same question: Krein-Milman and dual spaces.
– Martin Sleziak
Dec 2 at 7:20
@MartinSleziak Yes, I mean the closed unit ball of $X.$
– Idonknow
Dec 2 at 7:42
1
@TarasBanakh: There are infinite compact $K$ for which $C(K)$ is a dual space: these are precisely the hyperstonean $K$, e.g., $betamathbb{N}$. (On the other hand there are non-dual $C(K)$ for which the unit ball is the norm-closed convex hull of its extreme points, e.g. $alphamathbb{N}$. These are precisely the totally disconected $K$.)
– Dirk Werner
Dec 2 at 22:04
|
show 1 more comment
3
When you say that "$X$ has at least one extreme point" do you mean that the closed unit ball of $X$ has at least on extreme point?
– Martin Sleziak
Dec 2 at 7:11
I have added the tag (extreme-points), since it seems to me a good fit to the question. There exists also (krein-milman-theorem) tag, but that one would probably be a stretch.
– Martin Sleziak
Dec 2 at 7:13
4
This post on Mathematics site seems to be about the same question: Krein-Milman and dual spaces.
– Martin Sleziak
Dec 2 at 7:20
@MartinSleziak Yes, I mean the closed unit ball of $X.$
– Idonknow
Dec 2 at 7:42
1
@TarasBanakh: There are infinite compact $K$ for which $C(K)$ is a dual space: these are precisely the hyperstonean $K$, e.g., $betamathbb{N}$. (On the other hand there are non-dual $C(K)$ for which the unit ball is the norm-closed convex hull of its extreme points, e.g. $alphamathbb{N}$. These are precisely the totally disconected $K$.)
– Dirk Werner
Dec 2 at 22:04
3
3
When you say that "$X$ has at least one extreme point" do you mean that the closed unit ball of $X$ has at least on extreme point?
– Martin Sleziak
Dec 2 at 7:11
When you say that "$X$ has at least one extreme point" do you mean that the closed unit ball of $X$ has at least on extreme point?
– Martin Sleziak
Dec 2 at 7:11
I have added the tag (extreme-points), since it seems to me a good fit to the question. There exists also (krein-milman-theorem) tag, but that one would probably be a stretch.
– Martin Sleziak
Dec 2 at 7:13
I have added the tag (extreme-points), since it seems to me a good fit to the question. There exists also (krein-milman-theorem) tag, but that one would probably be a stretch.
– Martin Sleziak
Dec 2 at 7:13
4
4
This post on Mathematics site seems to be about the same question: Krein-Milman and dual spaces.
– Martin Sleziak
Dec 2 at 7:20
This post on Mathematics site seems to be about the same question: Krein-Milman and dual spaces.
– Martin Sleziak
Dec 2 at 7:20
@MartinSleziak Yes, I mean the closed unit ball of $X.$
– Idonknow
Dec 2 at 7:42
@MartinSleziak Yes, I mean the closed unit ball of $X.$
– Idonknow
Dec 2 at 7:42
1
1
@TarasBanakh: There are infinite compact $K$ for which $C(K)$ is a dual space: these are precisely the hyperstonean $K$, e.g., $betamathbb{N}$. (On the other hand there are non-dual $C(K)$ for which the unit ball is the norm-closed convex hull of its extreme points, e.g. $alphamathbb{N}$. These are precisely the totally disconected $K$.)
– Dirk Werner
Dec 2 at 22:04
@TarasBanakh: There are infinite compact $K$ for which $C(K)$ is a dual space: these are precisely the hyperstonean $K$, e.g., $betamathbb{N}$. (On the other hand there are non-dual $C(K)$ for which the unit ball is the norm-closed convex hull of its extreme points, e.g. $alphamathbb{N}$. These are precisely the totally disconected $K$.)
– Dirk Werner
Dec 2 at 22:04
|
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
8
down vote
accepted
Every separable Banach space $X$ can be equivalently renormed so that every point in the unit sphere is an extreme point: Take an injective bounded linear operator $T$ from $X$ into $ell_2$ and use $|x| := |x|_X + |Tx|_2$. Of course, there are many separable Banach spaces that are not isomorphic to a separable conjugate space, including (as Dirk pointed out) those that fail the Radon Nikodym property.
answered Dec 2 at 22:25
Bill Johnson
24.1k369116
add a comment |
up vote
7
down vote
No. Let $X$ and $Y$ be Banach spaces, and set $Z=Xoplus Y$, with $||(x,y)||:=|x|+|y|$. Assume that $x$ is a extreme point of $X$ with $|x|=1$. Then $(x,0)$ becomes an extreme point of $Z$; indeed, if
$$(x,0)=frac12(a,y)+frac12(b,z)$$
for $(a,y),(b,z)$ in the unit ball of $Z$, we then have $a=x=b$, since $x$ is an extreme point, but then $1=|x|=|a|leq||(a,y)||leq 1$, so $y=0$, and analogously, $z=0$.
So, $L^2(mathbb R)oplus L^1(mathbb R)$, is not a dual space, but its unit ball has extreme points.
answered Dec 2 at 9:33
Meisam Soleimani Malekan
1,1071311
1
How can we prove that $L^2(mathbb R)oplus L^1(mathbb R)$ is not a dual space?
– Idonknow
Dec 2 at 14:56
1
@Idonknow This is because, $Xoplus Y$ is dual iff both $X$ and $Y$ are dual.
– Meisam Soleimani Malekan
Dec 2 at 16:03
3
@MeisamSoleimaniMalekan: I am not absolutely positive about your previous comment: Write $C[0,1]^*$ as $L_1[0,1] oplus_1 Y$ with $Y=$ singular measures w.r.t. the Lebesgue measure. So a dual can have a non-dual $ell_1$-direct summand. My argument: If $L_2oplus L_1$ were a dual, it would, being separable, have the RNP, and hence $L_1$ would have the RNP, which it doesn't.
– Dirk Werner
Dec 2 at 22:03
@DirkWerner Thanks for the comment, you are right!
– Meisam Soleimani Malekan
Dec 3 at 3:01
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
Every separable Banach space $X$ can be equivalently renormed so that every point in the unit sphere is an extreme point: Take an injective bounded linear operator $T$ from $X$ into $ell_2$ and use $|x| := |x|_X + |Tx|_2$. Of course, there are many separable Banach spaces that are not isomorphic to a separable conjugate space, including (as Dirk pointed out) those that fail the Radon Nikodym property.
answered Dec 2 at 22:25
Bill Johnson
24.1k369116
add a comment |
up vote
8
down vote
accepted
Every separable Banach space $X$ can be equivalently renormed so that every point in the unit sphere is an extreme point: Take an injective bounded linear operator $T$ from $X$ into $ell_2$ and use $|x| := |x|_X + |Tx|_2$. Of course, there are many separable Banach spaces that are not isomorphic to a separable conjugate space, including (as Dirk pointed out) those that fail the Radon Nikodym property.
answered Dec 2 at 22:25
Bill Johnson
24.1k369116
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
Every separable Banach space $X$ can be equivalently renormed so that every point in the unit sphere is an extreme point: Take an injective bounded linear operator $T$ from $X$ into $ell_2$ and use $|x| := |x|_X + |Tx|_2$. Of course, there are many separable Banach spaces that are not isomorphic to a separable conjugate space, including (as Dirk pointed out) those that fail the Radon Nikodym property.
answered Dec 2 at 22:25
Bill Johnson
24.1k369116
Every separable Banach space $X$ can be equivalently renormed so that every point in the unit sphere is an extreme point: Take an injective bounded linear operator $T$ from $X$ into $ell_2$ and use $|x| := |x|_X + |Tx|_2$. Of course, there are many separable Banach spaces that are not isomorphic to a separable conjugate space, including (as Dirk pointed out) those that fail the Radon Nikodym property.
answered Dec 2 at 22:25
Bill Johnson
24.1k369116
answered Dec 2 at 22:25
Bill Johnson
24.1k369116
answered Dec 2 at 22:25
Bill Johnson
24.1k369116
answered Dec 2 at 22:25
Bill Johnson
24.1k369116
24.1k369116
add a comment |
add a comment |
up vote
7
down vote
No. Let $X$ and $Y$ be Banach spaces, and set $Z=Xoplus Y$, with $||(x,y)||:=|x|+|y|$. Assume that $x$ is a extreme point of $X$ with $|x|=1$. Then $(x,0)$ becomes an extreme point of $Z$; indeed, if
$$(x,0)=frac12(a,y)+frac12(b,z)$$
for $(a,y),(b,z)$ in the unit ball of $Z$, we then have $a=x=b$, since $x$ is an extreme point, but then $1=|x|=|a|leq||(a,y)||leq 1$, so $y=0$, and analogously, $z=0$.
So, $L^2(mathbb R)oplus L^1(mathbb R)$, is not a dual space, but its unit ball has extreme points.
answered Dec 2 at 9:33
Meisam Soleimani Malekan
1,1071311
1
How can we prove that $L^2(mathbb R)oplus L^1(mathbb R)$ is not a dual space?
– Idonknow
Dec 2 at 14:56
1
@Idonknow This is because, $Xoplus Y$ is dual iff both $X$ and $Y$ are dual.
– Meisam Soleimani Malekan
Dec 2 at 16:03
3
@MeisamSoleimaniMalekan: I am not absolutely positive about your previous comment: Write $C[0,1]^*$ as $L_1[0,1] oplus_1 Y$ with $Y=$ singular measures w.r.t. the Lebesgue measure. So a dual can have a non-dual $ell_1$-direct summand. My argument: If $L_2oplus L_1$ were a dual, it would, being separable, have the RNP, and hence $L_1$ would have the RNP, which it doesn't.
– Dirk Werner
Dec 2 at 22:03
@DirkWerner Thanks for the comment, you are right!
– Meisam Soleimani Malekan
Dec 3 at 3:01
add a comment |
up vote
7
down vote
No. Let $X$ and $Y$ be Banach spaces, and set $Z=Xoplus Y$, with $||(x,y)||:=|x|+|y|$. Assume that $x$ is a extreme point of $X$ with $|x|=1$. Then $(x,0)$ becomes an extreme point of $Z$; indeed, if
$$(x,0)=frac12(a,y)+frac12(b,z)$$
for $(a,y),(b,z)$ in the unit ball of $Z$, we then have $a=x=b$, since $x$ is an extreme point, but then $1=|x|=|a|leq||(a,y)||leq 1$, so $y=0$, and analogously, $z=0$.
So, $L^2(mathbb R)oplus L^1(mathbb R)$, is not a dual space, but its unit ball has extreme points.
answered Dec 2 at 9:33
Meisam Soleimani Malekan
1,1071311
1
How can we prove that $L^2(mathbb R)oplus L^1(mathbb R)$ is not a dual space?
– Idonknow
Dec 2 at 14:56
1
@Idonknow This is because, $Xoplus Y$ is dual iff both $X$ and $Y$ are dual.
– Meisam Soleimani Malekan
Dec 2 at 16:03
3
@MeisamSoleimaniMalekan: I am not absolutely positive about your previous comment: Write $C[0,1]^*$ as $L_1[0,1] oplus_1 Y$ with $Y=$ singular measures w.r.t. the Lebesgue measure. So a dual can have a non-dual $ell_1$-direct summand. My argument: If $L_2oplus L_1$ were a dual, it would, being separable, have the RNP, and hence $L_1$ would have the RNP, which it doesn't.
– Dirk Werner
Dec 2 at 22:03
@DirkWerner Thanks for the comment, you are right!
– Meisam Soleimani Malekan
Dec 3 at 3:01
add a comment |
up vote
7
down vote
up vote
7
down vote
No. Let $X$ and $Y$ be Banach spaces, and set $Z=Xoplus Y$, with $||(x,y)||:=|x|+|y|$. Assume that $x$ is a extreme point of $X$ with $|x|=1$. Then $(x,0)$ becomes an extreme point of $Z$; indeed, if
$$(x,0)=frac12(a,y)+frac12(b,z)$$
for $(a,y),(b,z)$ in the unit ball of $Z$, we then have $a=x=b$, since $x$ is an extreme point, but then $1=|x|=|a|leq||(a,y)||leq 1$, so $y=0$, and analogously, $z=0$.
So, $L^2(mathbb R)oplus L^1(mathbb R)$, is not a dual space, but its unit ball has extreme points.
answered Dec 2 at 9:33
Meisam Soleimani Malekan
1,1071311
No. Let $X$ and $Y$ be Banach spaces, and set $Z=Xoplus Y$, with $||(x,y)||:=|x|+|y|$. Assume that $x$ is a extreme point of $X$ with $|x|=1$. Then $(x,0)$ becomes an extreme point of $Z$; indeed, if
$$(x,0)=frac12(a,y)+frac12(b,z)$$
for $(a,y),(b,z)$ in the unit ball of $Z$, we then have $a=x=b$, since $x$ is an extreme point, but then $1=|x|=|a|leq||(a,y)||leq 1$, so $y=0$, and analogously, $z=0$.
So, $L^2(mathbb R)oplus L^1(mathbb R)$, is not a dual space, but its unit ball has extreme points.
answered Dec 2 at 9:33
Meisam Soleimani Malekan
1,1071311
edited Dec 2 at 10:24
answered Dec 2 at 9:33
Meisam Soleimani Malekan
1,1071311
answered Dec 2 at 9:33
Meisam Soleimani Malekan
1,1071311
answered Dec 2 at 9:33
Meisam Soleimani Malekan
1,1071311
1,1071311
1
How can we prove that $L^2(mathbb R)oplus L^1(mathbb R)$ is not a dual space?
– Idonknow
Dec 2 at 14:56
1
@Idonknow This is because, $Xoplus Y$ is dual iff both $X$ and $Y$ are dual.
– Meisam Soleimani Malekan
Dec 2 at 16:03
3
@MeisamSoleimaniMalekan: I am not absolutely positive about your previous comment: Write $C[0,1]^*$ as $L_1[0,1] oplus_1 Y$ with $Y=$ singular measures w.r.t. the Lebesgue measure. So a dual can have a non-dual $ell_1$-direct summand. My argument: If $L_2oplus L_1$ were a dual, it would, being separable, have the RNP, and hence $L_1$ would have the RNP, which it doesn't.
– Dirk Werner
Dec 2 at 22:03
@DirkWerner Thanks for the comment, you are right!
– Meisam Soleimani Malekan
Dec 3 at 3:01
add a comment |
1
How can we prove that $L^2(mathbb R)oplus L^1(mathbb R)$ is not a dual space?
– Idonknow
Dec 2 at 14:56
1
@Idonknow This is because, $Xoplus Y$ is dual iff both $X$ and $Y$ are dual.
– Meisam Soleimani Malekan
Dec 2 at 16:03
3
@MeisamSoleimaniMalekan: I am not absolutely positive about your previous comment: Write $C[0,1]^*$ as $L_1[0,1] oplus_1 Y$ with $Y=$ singular measures w.r.t. the Lebesgue measure. So a dual can have a non-dual $ell_1$-direct summand. My argument: If $L_2oplus L_1$ were a dual, it would, being separable, have the RNP, and hence $L_1$ would have the RNP, which it doesn't.
– Dirk Werner
Dec 2 at 22:03
@DirkWerner Thanks for the comment, you are right!
– Meisam Soleimani Malekan
Dec 3 at 3:01
1
1
How can we prove that $L^2(mathbb R)oplus L^1(mathbb R)$ is not a dual space?
– Idonknow
Dec 2 at 14:56
How can we prove that $L^2(mathbb R)oplus L^1(mathbb R)$ is not a dual space?
– Idonknow
Dec 2 at 14:56
1
1
@Idonknow This is because, $Xoplus Y$ is dual iff both $X$ and $Y$ are dual.
– Meisam Soleimani Malekan
Dec 2 at 16:03
@Idonknow This is because, $Xoplus Y$ is dual iff both $X$ and $Y$ are dual.
– Meisam Soleimani Malekan
Dec 2 at 16:03
3
3
@MeisamSoleimaniMalekan: I am not absolutely positive about your previous comment: Write $C[0,1]^*$ as $L_1[0,1] oplus_1 Y$ with $Y=$ singular measures w.r.t. the Lebesgue measure. So a dual can have a non-dual $ell_1$-direct summand. My argument: If $L_2oplus L_1$ were a dual, it would, being separable, have the RNP, and hence $L_1$ would have the RNP, which it doesn't.
– Dirk Werner
Dec 2 at 22:03
@MeisamSoleimaniMalekan: I am not absolutely positive about your previous comment: Write $C[0,1]^*$ as $L_1[0,1] oplus_1 Y$ with $Y=$ singular measures w.r.t. the Lebesgue measure. So a dual can have a non-dual $ell_1$-direct summand. My argument: If $L_2oplus L_1$ were a dual, it would, being separable, have the RNP, and hence $L_1$ would have the RNP, which it doesn't.
– Dirk Werner
Dec 2 at 22:03
@DirkWerner Thanks for the comment, you are right!
– Meisam Soleimani Malekan
Dec 3 at 3:01
@DirkWerner Thanks for the comment, you are right!
– Meisam Soleimani Malekan
Dec 3 at 3:01
add a comment |
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3
When you say that "$X$ has at least one extreme point" do you mean that the closed unit ball of $X$ has at least on extreme point?
– Martin Sleziak
Dec 2 at 7:11
I have added the tag (extreme-points), since it seems to me a good fit to the question. There exists also (krein-milman-theorem) tag, but that one would probably be a stretch.
– Martin Sleziak
Dec 2 at 7:13
4
This post on Mathematics site seems to be about the same question: Krein-Milman and dual spaces.
– Martin Sleziak
Dec 2 at 7:20
@MartinSleziak Yes, I mean the closed unit ball of $X.$
– Idonknow
Dec 2 at 7:42
1
@TarasBanakh: There are infinite compact $K$ for which $C(K)$ is a dual space: these are precisely the hyperstonean $K$, e.g., $betamathbb{N}$. (On the other hand there are non-dual $C(K)$ for which the unit ball is the norm-closed convex hull of its extreme points, e.g. $alphamathbb{N}$. These are precisely the totally disconected $K$.)
– Dirk Werner
Dec 2 at 22:04