The unit ball of $l^{1}$ is closed and has empty interior in $l^{2}$
$begingroup$
I know the definitions:
We say the sequence $(a_{n})$ belongs to $l^p$ if
$$left(sum_{n=1}^{infty}|a_n|^pright)^{frac 1p}<infty. $$
So, the unit ball of $l^1$, and I will denote this by $B_{1},$ is the set of all sequences $(a_n)$ satisfying $$ sum_n |a_ n|leq1.$$
To prove that $B_1$ has empty interior I think I need use the Baire's theorem, for $l^p$ is complete. So, how can I write $B_1=bigcup_{ninmathbb{N}}A_n$, with $int,A_n=emptyset$?
general-topology metric-spaces
asked Dec 12 '18 at 10:49
Mateus RochaMateus Rocha
817117
$endgroup$
add a comment |
$begingroup$
I know the definitions:
We say the sequence $(a_{n})$ belongs to $l^p$ if
$$left(sum_{n=1}^{infty}|a_n|^pright)^{frac 1p}<infty. $$
So, the unit ball of $l^1$, and I will denote this by $B_{1},$ is the set of all sequences $(a_n)$ satisfying $$ sum_n |a_ n|leq1.$$
To prove that $B_1$ has empty interior I think I need use the Baire's theorem, for $l^p$ is complete. So, how can I write $B_1=bigcup_{ninmathbb{N}}A_n$, with $int,A_n=emptyset$?
general-topology metric-spaces
asked Dec 12 '18 at 10:49
Mateus RochaMateus Rocha
817117
$endgroup$
$begingroup$
First, it should be $sum_n{|a_n|}$. Second, what you describe (with equality) is normally called the unit sphere: en.wikipedia.org/wiki/Unit_sphere; the unit ball would be $sum_n{|a_n|} le 1$. For the problem, I'd say a very direct approach to "find a $l^1$ near sequence with a norm $neq 1$" seems much easier.
$endgroup$
– Ingix
Dec 12 '18 at 10:56
$begingroup$
Well, as you can see from the answer below, you should have edited your question to use sphere and keep the "=", instead of of keeping ball and editing your equality to inequality. As it stands now, using the usual definition of a (closed) unit ball, it of course has an interior (all the sequences with norm < 1).
$endgroup$
– Ingix
Dec 12 '18 at 11:21
$begingroup$
Kavi's answer is a good place to look. The useful facts to be proved are: the unit ball $B_1$ of $l^1$ is closed in $l^2$ in the the $l^2$ topology, and has empty interior in $l^2$ in the $l^2$ topology. For the second one, you can indeed use the Baire category theorem.
$endgroup$
– GEdgar
Dec 12 '18 at 12:53
add a comment |
$begingroup$
I know the definitions:
We say the sequence $(a_{n})$ belongs to $l^p$ if
$$left(sum_{n=1}^{infty}|a_n|^pright)^{frac 1p}<infty. $$
So, the unit ball of $l^1$, and I will denote this by $B_{1},$ is the set of all sequences $(a_n)$ satisfying $$ sum_n |a_ n|leq1.$$
To prove that $B_1$ has empty interior I think I need use the Baire's theorem, for $l^p$ is complete. So, how can I write $B_1=bigcup_{ninmathbb{N}}A_n$, with $int,A_n=emptyset$?
general-topology metric-spaces
asked Dec 12 '18 at 10:49
Mateus RochaMateus Rocha
817117
$endgroup$
I know the definitions:
We say the sequence $(a_{n})$ belongs to $l^p$ if
$$left(sum_{n=1}^{infty}|a_n|^pright)^{frac 1p}<infty. $$
So, the unit ball of $l^1$, and I will denote this by $B_{1},$ is the set of all sequences $(a_n)$ satisfying $$ sum_n |a_ n|leq1.$$
To prove that $B_1$ has empty interior I think I need use the Baire's theorem, for $l^p$ is complete. So, how can I write $B_1=bigcup_{ninmathbb{N}}A_n$, with $int,A_n=emptyset$?
general-topology metric-spaces
general-topology metric-spaces
asked Dec 12 '18 at 10:49
Mateus RochaMateus Rocha
817117
asked Dec 12 '18 at 10:49
Mateus RochaMateus Rocha
817117
edited Dec 12 '18 at 11:00
Mateus Rocha
asked Dec 12 '18 at 10:49
Mateus RochaMateus Rocha
817117
asked Dec 12 '18 at 10:49
Mateus RochaMateus Rocha
817117
asked Dec 12 '18 at 10:49
Mateus RochaMateus Rocha
817117
817117
$begingroup$
First, it should be $sum_n{|a_n|}$. Second, what you describe (with equality) is normally called the unit sphere: en.wikipedia.org/wiki/Unit_sphere; the unit ball would be $sum_n{|a_n|} le 1$. For the problem, I'd say a very direct approach to "find a $l^1$ near sequence with a norm $neq 1$" seems much easier.
$endgroup$
– Ingix
Dec 12 '18 at 10:56
$begingroup$
Well, as you can see from the answer below, you should have edited your question to use sphere and keep the "=", instead of of keeping ball and editing your equality to inequality. As it stands now, using the usual definition of a (closed) unit ball, it of course has an interior (all the sequences with norm < 1).
$endgroup$
– Ingix
Dec 12 '18 at 11:21
$begingroup$
Kavi's answer is a good place to look. The useful facts to be proved are: the unit ball $B_1$ of $l^1$ is closed in $l^2$ in the the $l^2$ topology, and has empty interior in $l^2$ in the $l^2$ topology. For the second one, you can indeed use the Baire category theorem.
$endgroup$
– GEdgar
Dec 12 '18 at 12:53
add a comment |
$begingroup$
First, it should be $sum_n{|a_n|}$. Second, what you describe (with equality) is normally called the unit sphere: en.wikipedia.org/wiki/Unit_sphere; the unit ball would be $sum_n{|a_n|} le 1$. For the problem, I'd say a very direct approach to "find a $l^1$ near sequence with a norm $neq 1$" seems much easier.
$endgroup$
– Ingix
Dec 12 '18 at 10:56
$begingroup$
Well, as you can see from the answer below, you should have edited your question to use sphere and keep the "=", instead of of keeping ball and editing your equality to inequality. As it stands now, using the usual definition of a (closed) unit ball, it of course has an interior (all the sequences with norm < 1).
$endgroup$
– Ingix
Dec 12 '18 at 11:21
$begingroup$
Kavi's answer is a good place to look. The useful facts to be proved are: the unit ball $B_1$ of $l^1$ is closed in $l^2$ in the the $l^2$ topology, and has empty interior in $l^2$ in the $l^2$ topology. For the second one, you can indeed use the Baire category theorem.
$endgroup$
– GEdgar
Dec 12 '18 at 12:53
$begingroup$
First, it should be $sum_n{|a_n|}$. Second, what you describe (with equality) is normally called the unit sphere: en.wikipedia.org/wiki/Unit_sphere; the unit ball would be $sum_n{|a_n|} le 1$. For the problem, I'd say a very direct approach to "find a $l^1$ near sequence with a norm $neq 1$" seems much easier.
$endgroup$
– Ingix
Dec 12 '18 at 10:56
$begingroup$
First, it should be $sum_n{|a_n|}$. Second, what you describe (with equality) is normally called the unit sphere: en.wikipedia.org/wiki/Unit_sphere; the unit ball would be $sum_n{|a_n|} le 1$. For the problem, I'd say a very direct approach to "find a $l^1$ near sequence with a norm $neq 1$" seems much easier.
$endgroup$
– Ingix
Dec 12 '18 at 10:56
$begingroup$
Well, as you can see from the answer below, you should have edited your question to use sphere and keep the "=", instead of of keeping ball and editing your equality to inequality. As it stands now, using the usual definition of a (closed) unit ball, it of course has an interior (all the sequences with norm < 1).
$endgroup$
– Ingix
Dec 12 '18 at 11:21
$begingroup$
Well, as you can see from the answer below, you should have edited your question to use sphere and keep the "=", instead of of keeping ball and editing your equality to inequality. As it stands now, using the usual definition of a (closed) unit ball, it of course has an interior (all the sequences with norm < 1).
$endgroup$
– Ingix
Dec 12 '18 at 11:21
$begingroup$
Kavi's answer is a good place to look. The useful facts to be proved are: the unit ball $B_1$ of $l^1$ is closed in $l^2$ in the the $l^2$ topology, and has empty interior in $l^2$ in the $l^2$ topology. For the second one, you can indeed use the Baire category theorem.
$endgroup$
– GEdgar
Dec 12 '18 at 12:53
$begingroup$
Kavi's answer is a good place to look. The useful facts to be proved are: the unit ball $B_1$ of $l^1$ is closed in $l^2$ in the the $l^2$ topology, and has empty interior in $l^2$ in the $l^2$ topology. For the second one, you can indeed use the Baire category theorem.
$endgroup$
– GEdgar
Dec 12 '18 at 12:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I will answer the question in the title which says that the unit ball of $ell ^{1}$ has no interior in $ell ^{2}$. Suppose $(a_n)$ is an interior point. Then there exists $epsilon >0$ such that $sum |b_n-a_n|^{2} <epsilon$ implies $sum |b_n| leq 1$. Choose $N$ such that $sum_{n=N}^{infty} frac 1 {n^{2}} <epsilon$ an d $sum_{n=1}^{N-1} frac 1 n >1+sum |a_n|$. Let $b_n=a_n+frac 1 n$ for $n geq N$ and $b_n=a_n$ for $n <N$. Note that $sum |b_n-a_n|^{2} <epsilon$ but $sum |b_n| geq sum_{n=1}^{N-1} frac 1 n -sum_{n=1}^{N-1} |a_n|>1 $. So we have a contradiction.
answered Dec 12 '18 at 12:02
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
$endgroup$
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%2f3036532%2fthe-unit-ball-of-l1-is-closed-and-has-empty-interior-in-l2%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
$begingroup$
I will answer the question in the title which says that the unit ball of $ell ^{1}$ has no interior in $ell ^{2}$. Suppose $(a_n)$ is an interior point. Then there exists $epsilon >0$ such that $sum |b_n-a_n|^{2} <epsilon$ implies $sum |b_n| leq 1$. Choose $N$ such that $sum_{n=N}^{infty} frac 1 {n^{2}} <epsilon$ an d $sum_{n=1}^{N-1} frac 1 n >1+sum |a_n|$. Let $b_n=a_n+frac 1 n$ for $n geq N$ and $b_n=a_n$ for $n <N$. Note that $sum |b_n-a_n|^{2} <epsilon$ but $sum |b_n| geq sum_{n=1}^{N-1} frac 1 n -sum_{n=1}^{N-1} |a_n|>1 $. So we have a contradiction.
answered Dec 12 '18 at 12:02
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
$endgroup$
add a comment |
$begingroup$
I will answer the question in the title which says that the unit ball of $ell ^{1}$ has no interior in $ell ^{2}$. Suppose $(a_n)$ is an interior point. Then there exists $epsilon >0$ such that $sum |b_n-a_n|^{2} <epsilon$ implies $sum |b_n| leq 1$. Choose $N$ such that $sum_{n=N}^{infty} frac 1 {n^{2}} <epsilon$ an d $sum_{n=1}^{N-1} frac 1 n >1+sum |a_n|$. Let $b_n=a_n+frac 1 n$ for $n geq N$ and $b_n=a_n$ for $n <N$. Note that $sum |b_n-a_n|^{2} <epsilon$ but $sum |b_n| geq sum_{n=1}^{N-1} frac 1 n -sum_{n=1}^{N-1} |a_n|>1 $. So we have a contradiction.
answered Dec 12 '18 at 12:02
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
$endgroup$
add a comment |
$begingroup$
I will answer the question in the title which says that the unit ball of $ell ^{1}$ has no interior in $ell ^{2}$. Suppose $(a_n)$ is an interior point. Then there exists $epsilon >0$ such that $sum |b_n-a_n|^{2} <epsilon$ implies $sum |b_n| leq 1$. Choose $N$ such that $sum_{n=N}^{infty} frac 1 {n^{2}} <epsilon$ an d $sum_{n=1}^{N-1} frac 1 n >1+sum |a_n|$. Let $b_n=a_n+frac 1 n$ for $n geq N$ and $b_n=a_n$ for $n <N$. Note that $sum |b_n-a_n|^{2} <epsilon$ but $sum |b_n| geq sum_{n=1}^{N-1} frac 1 n -sum_{n=1}^{N-1} |a_n|>1 $. So we have a contradiction.
answered Dec 12 '18 at 12:02
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
$endgroup$
I will answer the question in the title which says that the unit ball of $ell ^{1}$ has no interior in $ell ^{2}$. Suppose $(a_n)$ is an interior point. Then there exists $epsilon >0$ such that $sum |b_n-a_n|^{2} <epsilon$ implies $sum |b_n| leq 1$. Choose $N$ such that $sum_{n=N}^{infty} frac 1 {n^{2}} <epsilon$ an d $sum_{n=1}^{N-1} frac 1 n >1+sum |a_n|$. Let $b_n=a_n+frac 1 n$ for $n geq N$ and $b_n=a_n$ for $n <N$. Note that $sum |b_n-a_n|^{2} <epsilon$ but $sum |b_n| geq sum_{n=1}^{N-1} frac 1 n -sum_{n=1}^{N-1} |a_n|>1 $. So we have a contradiction.
answered Dec 12 '18 at 12:02
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
answered Dec 12 '18 at 12:02
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
answered Dec 12 '18 at 12:02
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
answered Dec 12 '18 at 12:02
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
57.9k42160
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%2f3036532%2fthe-unit-ball-of-l1-is-closed-and-has-empty-interior-in-l2%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
$begingroup$
First, it should be $sum_n{|a_n|}$. Second, what you describe (with equality) is normally called the unit sphere: en.wikipedia.org/wiki/Unit_sphere; the unit ball would be $sum_n{|a_n|} le 1$. For the problem, I'd say a very direct approach to "find a $l^1$ near sequence with a norm $neq 1$" seems much easier.
$endgroup$
– Ingix
Dec 12 '18 at 10:56
$begingroup$
Well, as you can see from the answer below, you should have edited your question to use sphere and keep the "=", instead of of keeping ball and editing your equality to inequality. As it stands now, using the usual definition of a (closed) unit ball, it of course has an interior (all the sequences with norm < 1).
$endgroup$
– Ingix
Dec 12 '18 at 11:21
$begingroup$
Kavi's answer is a good place to look. The useful facts to be proved are: the unit ball $B_1$ of $l^1$ is closed in $l^2$ in the the $l^2$ topology, and has empty interior in $l^2$ in the $l^2$ topology. For the second one, you can indeed use the Baire category theorem.
$endgroup$
– GEdgar
Dec 12 '18 at 12:53