Mixing
Compactness in weak$^*$-topology on $l_1(mathbb{N})$
$begingroup$
Let $K$ denote the closed unit ball of $l_1(mathbb{N})$ (considered as a vector space over $mathbb{C}$).
Is $mathrm{co}(mathrm{Ext}(K))$ (the convex hull of its extreme points) compact in the weak$^∗$-topology on $l_1(mathbb{N})$? Justify your answer.
(Here $l_1(mathbb{N})$ is identified as the dual space of $c_0(mathbb{N})$)
I know these two facts:
- $K$ is the norm closure of $mathrm{co}(mathrm{Ext}(K))$
- Banach-Alaoglu's theorem tells me that $K$ is compact in the weak$^*$-topology on $l_1(mathbb{N})$
functional-analysis compactness lp-spaces weak-topology
asked Jan 17 at 9:54
user289143user289143
916313
$endgroup$
add a comment |
$begingroup$
Let $K$ denote the closed unit ball of $l_1(mathbb{N})$ (considered as a vector space over $mathbb{C}$).
Is $mathrm{co}(mathrm{Ext}(K))$ (the convex hull of its extreme points) compact in the weak$^∗$-topology on $l_1(mathbb{N})$? Justify your answer.
(Here $l_1(mathbb{N})$ is identified as the dual space of $c_0(mathbb{N})$)
I know these two facts:
- $K$ is the norm closure of $mathrm{co}(mathrm{Ext}(K))$
- Banach-Alaoglu's theorem tells me that $K$ is compact in the weak$^*$-topology on $l_1(mathbb{N})$
functional-analysis compactness lp-spaces weak-topology
asked Jan 17 at 9:54
user289143user289143
916313
$endgroup$
$begingroup$
Does $text{co}(A)$ denote the convex hull of $A$?
$endgroup$
– BigbearZzz
Jan 17 at 10:00
$begingroup$
Yes, question edited.
$endgroup$
– user289143
Jan 17 at 10:01
$begingroup$
Hint: the extremal points of $K$ have at most one nonzero term.
$endgroup$
– Mindlack
Jan 17 at 10:09
$begingroup$
I know that $mathrm{Ext}(K)= { lambda e_n : lambda in mathbb{C},|lambda|=1, n in mathbb{N} }$
$endgroup$
– user289143
Jan 17 at 10:12
$begingroup$
@user289143 I didn't realize that you knew $mathrm{Ext}(K)= { lambda e_n : lambda in mathbb{C},|lambda|=1, n in mathbb{N} }$ so I included that in my answer as well. Anyway, I hope the rest helps.
$endgroup$
– BigbearZzz
Jan 17 at 10:43
add a comment |
$begingroup$
Let $K$ denote the closed unit ball of $l_1(mathbb{N})$ (considered as a vector space over $mathbb{C}$).
Is $mathrm{co}(mathrm{Ext}(K))$ (the convex hull of its extreme points) compact in the weak$^∗$-topology on $l_1(mathbb{N})$? Justify your answer.
(Here $l_1(mathbb{N})$ is identified as the dual space of $c_0(mathbb{N})$)
I know these two facts:
- $K$ is the norm closure of $mathrm{co}(mathrm{Ext}(K))$
- Banach-Alaoglu's theorem tells me that $K$ is compact in the weak$^*$-topology on $l_1(mathbb{N})$
functional-analysis compactness lp-spaces weak-topology
asked Jan 17 at 9:54
user289143user289143
916313
$endgroup$
Let $K$ denote the closed unit ball of $l_1(mathbb{N})$ (considered as a vector space over $mathbb{C}$).
Is $mathrm{co}(mathrm{Ext}(K))$ (the convex hull of its extreme points) compact in the weak$^∗$-topology on $l_1(mathbb{N})$? Justify your answer.
(Here $l_1(mathbb{N})$ is identified as the dual space of $c_0(mathbb{N})$)
I know these two facts:
- $K$ is the norm closure of $mathrm{co}(mathrm{Ext}(K))$
- Banach-Alaoglu's theorem tells me that $K$ is compact in the weak$^*$-topology on $l_1(mathbb{N})$
functional-analysis compactness lp-spaces weak-topology
functional-analysis compactness lp-spaces weak-topology
asked Jan 17 at 9:54
user289143user289143
916313
asked Jan 17 at 9:54
user289143user289143
916313
edited Jan 17 at 10:01
user289143
asked Jan 17 at 9:54
user289143user289143
916313
asked Jan 17 at 9:54
user289143user289143
916313
asked Jan 17 at 9:54
user289143user289143
916313
916313
$begingroup$
Does $text{co}(A)$ denote the convex hull of $A$?
$endgroup$
– BigbearZzz
Jan 17 at 10:00
$begingroup$
Yes, question edited.
$endgroup$
– user289143
Jan 17 at 10:01
$begingroup$
Hint: the extremal points of $K$ have at most one nonzero term.
$endgroup$
– Mindlack
Jan 17 at 10:09
$begingroup$
I know that $mathrm{Ext}(K)= { lambda e_n : lambda in mathbb{C},|lambda|=1, n in mathbb{N} }$
$endgroup$
– user289143
Jan 17 at 10:12
$begingroup$
@user289143 I didn't realize that you knew $mathrm{Ext}(K)= { lambda e_n : lambda in mathbb{C},|lambda|=1, n in mathbb{N} }$ so I included that in my answer as well. Anyway, I hope the rest helps.
$endgroup$
– BigbearZzz
Jan 17 at 10:43
add a comment |
$begingroup$
Does $text{co}(A)$ denote the convex hull of $A$?
$endgroup$
– BigbearZzz
Jan 17 at 10:00
$begingroup$
Yes, question edited.
$endgroup$
– user289143
Jan 17 at 10:01
$begingroup$
Hint: the extremal points of $K$ have at most one nonzero term.
$endgroup$
– Mindlack
Jan 17 at 10:09
$begingroup$
I know that $mathrm{Ext}(K)= { lambda e_n : lambda in mathbb{C},|lambda|=1, n in mathbb{N} }$
$endgroup$
– user289143
Jan 17 at 10:12
$begingroup$
@user289143 I didn't realize that you knew $mathrm{Ext}(K)= { lambda e_n : lambda in mathbb{C},|lambda|=1, n in mathbb{N} }$ so I included that in my answer as well. Anyway, I hope the rest helps.
$endgroup$
– BigbearZzz
Jan 17 at 10:43
$begingroup$
Does $text{co}(A)$ denote the convex hull of $A$?
$endgroup$
– BigbearZzz
Jan 17 at 10:00
$begingroup$
Does $text{co}(A)$ denote the convex hull of $A$?
$endgroup$
– BigbearZzz
Jan 17 at 10:00
$begingroup$
Yes, question edited.
$endgroup$
– user289143
Jan 17 at 10:01
$begingroup$
Yes, question edited.
$endgroup$
– user289143
Jan 17 at 10:01
$begingroup$
Hint: the extremal points of $K$ have at most one nonzero term.
$endgroup$
– Mindlack
Jan 17 at 10:09
$begingroup$
Hint: the extremal points of $K$ have at most one nonzero term.
$endgroup$
– Mindlack
Jan 17 at 10:09
$begingroup$
I know that $mathrm{Ext}(K)= { lambda e_n : lambda in mathbb{C},|lambda|=1, n in mathbb{N} }$
$endgroup$
– user289143
Jan 17 at 10:12
$begingroup$
I know that $mathrm{Ext}(K)= { lambda e_n : lambda in mathbb{C},|lambda|=1, n in mathbb{N} }$
$endgroup$
– user289143
Jan 17 at 10:12
$begingroup$
@user289143 I didn't realize that you knew $mathrm{Ext}(K)= { lambda e_n : lambda in mathbb{C},|lambda|=1, n in mathbb{N} }$ so I included that in my answer as well. Anyway, I hope the rest helps.
$endgroup$
– BigbearZzz
Jan 17 at 10:43
$begingroup$
@user289143 I didn't realize that you knew $mathrm{Ext}(K)= { lambda e_n : lambda in mathbb{C},|lambda|=1, n in mathbb{N} }$ so I included that in my answer as well. Anyway, I hope the rest helps.
$endgroup$
– BigbearZzz
Jan 17 at 10:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Claim 1: The extreme points of the (closed) unit ball $K=B_{l^1}$ of $l^1(Bbb N)$ is precisely the set $E={ alpha e_i:iinBbb N, |alpha|=1 }$ where
$e_i=(0,0,dots,0,1,0,0,dots)$, consisting of $0$ for all but the $i^{text{th}}$ coordinate.
Proof: If $e_i = (1-lambda)x+lambda y$ for some $x,yin B_{l^1}$, then $(1-lambda)x_i+lambda y_i=1$ and $|x_i|,|y_i|le 1$. From these we deduce that $x_i=y_i=1$ and hence all other entries of $x,y$ are zeroes. Thus $x_i=y_i=e_i$, similar for $alpha e_i$.
On the other hand, any point $xin B_{l^1}backslash E$ must have a coordinate such that $|x_i|<1$. We can deduce from this that $x$ is not an extreme point (why?). QED
Having known that, it is easy to see that for any $xin text{co}(E)$, $x$ can have only finitely many nonzero terms, so $text{co}(E) subsetneq B_{l^1}$.
On the other hand, Krein-Milman theorem states that $overline{text{co}} ^*(E)=B_{l^1}$ so $text{co}(E)$ is not a weak$^*$ closed set.
A weak$^*$ compact set must be weak$^*$ closed because the topology is Hausdorff, hence $text{co}(E)=text{co}(text{Ext}(K))$ is not weak$^*$ compact.
Edit: If you haven't learned the Krein-Milman theorem for a locally convex space yet, then you can use following line of reasoning, given that you knew $overline{text{co}}(E)=B_{l^1}$:
We clearly have
$$
B_{l^1} = overline{text{co}}(E) subset overline{text{co}} ^*(E)
$$
since weak$^*$ topology is coarser than the norm topology. Also, the reverse inclusion
$$
overline{text{co}} ^*(E) subset B_{l^1}
$$
comes from the fact that $text{co}(E) subset B_{l^1}$ and Banach-Alaoglu ($B_{l^1}$ is weak$^*$ compact and hence weak$^*$ closed). This shows that
$$overline{text{co}} ^*(E)=B_{l^1}.$$
answered Jan 17 at 10:23
BigbearZzzBigbearZzz
8,76121652
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076792%2fcompactness-in-weak-topology-on-l-1-mathbbn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Claim 1: The extreme points of the (closed) unit ball $K=B_{l^1}$ of $l^1(Bbb N)$ is precisely the set $E={ alpha e_i:iinBbb N, |alpha|=1 }$ where
$e_i=(0,0,dots,0,1,0,0,dots)$, consisting of $0$ for all but the $i^{text{th}}$ coordinate.
Proof: If $e_i = (1-lambda)x+lambda y$ for some $x,yin B_{l^1}$, then $(1-lambda)x_i+lambda y_i=1$ and $|x_i|,|y_i|le 1$. From these we deduce that $x_i=y_i=1$ and hence all other entries of $x,y$ are zeroes. Thus $x_i=y_i=e_i$, similar for $alpha e_i$.
On the other hand, any point $xin B_{l^1}backslash E$ must have a coordinate such that $|x_i|<1$. We can deduce from this that $x$ is not an extreme point (why?). QED
Having known that, it is easy to see that for any $xin text{co}(E)$, $x$ can have only finitely many nonzero terms, so $text{co}(E) subsetneq B_{l^1}$.
On the other hand, Krein-Milman theorem states that $overline{text{co}} ^*(E)=B_{l^1}$ so $text{co}(E)$ is not a weak$^*$ closed set.
A weak$^*$ compact set must be weak$^*$ closed because the topology is Hausdorff, hence $text{co}(E)=text{co}(text{Ext}(K))$ is not weak$^*$ compact.
Edit: If you haven't learned the Krein-Milman theorem for a locally convex space yet, then you can use following line of reasoning, given that you knew $overline{text{co}}(E)=B_{l^1}$:
We clearly have
$$
B_{l^1} = overline{text{co}}(E) subset overline{text{co}} ^*(E)
$$
since weak$^*$ topology is coarser than the norm topology. Also, the reverse inclusion
$$
overline{text{co}} ^*(E) subset B_{l^1}
$$
comes from the fact that $text{co}(E) subset B_{l^1}$ and Banach-Alaoglu ($B_{l^1}$ is weak$^*$ compact and hence weak$^*$ closed). This shows that
$$overline{text{co}} ^*(E)=B_{l^1}.$$
answered Jan 17 at 10:23
BigbearZzzBigbearZzz
8,76121652
$endgroup$
add a comment |
$begingroup$
Claim 1: The extreme points of the (closed) unit ball $K=B_{l^1}$ of $l^1(Bbb N)$ is precisely the set $E={ alpha e_i:iinBbb N, |alpha|=1 }$ where
$e_i=(0,0,dots,0,1,0,0,dots)$, consisting of $0$ for all but the $i^{text{th}}$ coordinate.
Proof: If $e_i = (1-lambda)x+lambda y$ for some $x,yin B_{l^1}$, then $(1-lambda)x_i+lambda y_i=1$ and $|x_i|,|y_i|le 1$. From these we deduce that $x_i=y_i=1$ and hence all other entries of $x,y$ are zeroes. Thus $x_i=y_i=e_i$, similar for $alpha e_i$.
On the other hand, any point $xin B_{l^1}backslash E$ must have a coordinate such that $|x_i|<1$. We can deduce from this that $x$ is not an extreme point (why?). QED
Having known that, it is easy to see that for any $xin text{co}(E)$, $x$ can have only finitely many nonzero terms, so $text{co}(E) subsetneq B_{l^1}$.
On the other hand, Krein-Milman theorem states that $overline{text{co}} ^*(E)=B_{l^1}$ so $text{co}(E)$ is not a weak$^*$ closed set.
A weak$^*$ compact set must be weak$^*$ closed because the topology is Hausdorff, hence $text{co}(E)=text{co}(text{Ext}(K))$ is not weak$^*$ compact.
Edit: If you haven't learned the Krein-Milman theorem for a locally convex space yet, then you can use following line of reasoning, given that you knew $overline{text{co}}(E)=B_{l^1}$:
We clearly have
$$
B_{l^1} = overline{text{co}}(E) subset overline{text{co}} ^*(E)
$$
since weak$^*$ topology is coarser than the norm topology. Also, the reverse inclusion
$$
overline{text{co}} ^*(E) subset B_{l^1}
$$
comes from the fact that $text{co}(E) subset B_{l^1}$ and Banach-Alaoglu ($B_{l^1}$ is weak$^*$ compact and hence weak$^*$ closed). This shows that
$$overline{text{co}} ^*(E)=B_{l^1}.$$
answered Jan 17 at 10:23
BigbearZzzBigbearZzz
8,76121652
$endgroup$
add a comment |
$begingroup$
Claim 1: The extreme points of the (closed) unit ball $K=B_{l^1}$ of $l^1(Bbb N)$ is precisely the set $E={ alpha e_i:iinBbb N, |alpha|=1 }$ where
$e_i=(0,0,dots,0,1,0,0,dots)$, consisting of $0$ for all but the $i^{text{th}}$ coordinate.
Proof: If $e_i = (1-lambda)x+lambda y$ for some $x,yin B_{l^1}$, then $(1-lambda)x_i+lambda y_i=1$ and $|x_i|,|y_i|le 1$. From these we deduce that $x_i=y_i=1$ and hence all other entries of $x,y$ are zeroes. Thus $x_i=y_i=e_i$, similar for $alpha e_i$.
On the other hand, any point $xin B_{l^1}backslash E$ must have a coordinate such that $|x_i|<1$. We can deduce from this that $x$ is not an extreme point (why?). QED
Having known that, it is easy to see that for any $xin text{co}(E)$, $x$ can have only finitely many nonzero terms, so $text{co}(E) subsetneq B_{l^1}$.
On the other hand, Krein-Milman theorem states that $overline{text{co}} ^*(E)=B_{l^1}$ so $text{co}(E)$ is not a weak$^*$ closed set.
A weak$^*$ compact set must be weak$^*$ closed because the topology is Hausdorff, hence $text{co}(E)=text{co}(text{Ext}(K))$ is not weak$^*$ compact.
Edit: If you haven't learned the Krein-Milman theorem for a locally convex space yet, then you can use following line of reasoning, given that you knew $overline{text{co}}(E)=B_{l^1}$:
We clearly have
$$
B_{l^1} = overline{text{co}}(E) subset overline{text{co}} ^*(E)
$$
since weak$^*$ topology is coarser than the norm topology. Also, the reverse inclusion
$$
overline{text{co}} ^*(E) subset B_{l^1}
$$
comes from the fact that $text{co}(E) subset B_{l^1}$ and Banach-Alaoglu ($B_{l^1}$ is weak$^*$ compact and hence weak$^*$ closed). This shows that
$$overline{text{co}} ^*(E)=B_{l^1}.$$
answered Jan 17 at 10:23
BigbearZzzBigbearZzz
8,76121652
$endgroup$
Claim 1: The extreme points of the (closed) unit ball $K=B_{l^1}$ of $l^1(Bbb N)$ is precisely the set $E={ alpha e_i:iinBbb N, |alpha|=1 }$ where
$e_i=(0,0,dots,0,1,0,0,dots)$, consisting of $0$ for all but the $i^{text{th}}$ coordinate.
Proof: If $e_i = (1-lambda)x+lambda y$ for some $x,yin B_{l^1}$, then $(1-lambda)x_i+lambda y_i=1$ and $|x_i|,|y_i|le 1$. From these we deduce that $x_i=y_i=1$ and hence all other entries of $x,y$ are zeroes. Thus $x_i=y_i=e_i$, similar for $alpha e_i$.
On the other hand, any point $xin B_{l^1}backslash E$ must have a coordinate such that $|x_i|<1$. We can deduce from this that $x$ is not an extreme point (why?). QED
Having known that, it is easy to see that for any $xin text{co}(E)$, $x$ can have only finitely many nonzero terms, so $text{co}(E) subsetneq B_{l^1}$.
On the other hand, Krein-Milman theorem states that $overline{text{co}} ^*(E)=B_{l^1}$ so $text{co}(E)$ is not a weak$^*$ closed set.
A weak$^*$ compact set must be weak$^*$ closed because the topology is Hausdorff, hence $text{co}(E)=text{co}(text{Ext}(K))$ is not weak$^*$ compact.
Edit: If you haven't learned the Krein-Milman theorem for a locally convex space yet, then you can use following line of reasoning, given that you knew $overline{text{co}}(E)=B_{l^1}$:
We clearly have
$$
B_{l^1} = overline{text{co}}(E) subset overline{text{co}} ^*(E)
$$
since weak$^*$ topology is coarser than the norm topology. Also, the reverse inclusion
$$
overline{text{co}} ^*(E) subset B_{l^1}
$$
comes from the fact that $text{co}(E) subset B_{l^1}$ and Banach-Alaoglu ($B_{l^1}$ is weak$^*$ compact and hence weak$^*$ closed). This shows that
$$overline{text{co}} ^*(E)=B_{l^1}.$$
answered Jan 17 at 10:23
BigbearZzzBigbearZzz
8,76121652
edited Jan 17 at 10:37
answered Jan 17 at 10:23
BigbearZzzBigbearZzz
8,76121652
answered Jan 17 at 10:23
BigbearZzzBigbearZzz
8,76121652
answered Jan 17 at 10:23
BigbearZzzBigbearZzz
8,76121652
8,76121652
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076792%2fcompactness-in-weak-topology-on-l-1-mathbbn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Does $text{co}(A)$ denote the convex hull of $A$?
$endgroup$
– BigbearZzz
Jan 17 at 10:00
$begingroup$
Yes, question edited.
$endgroup$
– user289143
Jan 17 at 10:01
$begingroup$
Hint: the extremal points of $K$ have at most one nonzero term.
$endgroup$
– Mindlack
Jan 17 at 10:09
$begingroup$
I know that $mathrm{Ext}(K)= { lambda e_n : lambda in mathbb{C},|lambda|=1, n in mathbb{N} }$
$endgroup$
– user289143
Jan 17 at 10:12
$begingroup$
@user289143 I didn't realize that you knew $mathrm{Ext}(K)= { lambda e_n : lambda in mathbb{C},|lambda|=1, n in mathbb{N} }$ so I included that in my answer as well. Anyway, I hope the rest helps.
$endgroup$
– BigbearZzz
Jan 17 at 10:43