Space of sequences such thtat $sum_{n=0}^{infty}2^n|a_n|<+infty$












-1














Consider the space of sequences of real numbers ${vec{a_i}}$ where each $vec{a_i}={a_{i_n}}$ is such that $sum_{n=0}^{infty}2^n|a_{i_n}|<+infty$. Then how could we better describe the space? Typically, does the space have countable linear basis,is it complete with respect to the norm $|vec{a_i}|=sum_n2^n|a_{i_n}|$,whether it is finite dimensional?



One example that comes to mind is of the sequence $a_n=frac1{2^{2n}}$. Hence, I think the space is countable. How should I proceed further. Is any representation like Riesz representation work here? Thanks beforehand.






sequences-and-series functional-analysis banach-spaces







share|cite|improve this question




















  • 1




    Is it assumed that $a_n>0$? And if $a_n=1/(2^n)$ don't you get the (divergent) series of all $1$'s?
    – coffeemath
    Dec 13 '18 at 9:29








  • 1




    I don't understand the question. What do you actually want?
    – BigbearZzz
    Dec 13 '18 at 9:38






  • 1




    Your expression for $||cdot||$ is not a norm...
    – BigbearZzz
    Dec 13 '18 at 9:41






  • 1




    @vidyarthi because this expression can be negative whereas a norm can't
    – igortsts
    Dec 13 '18 at 9:42








  • 1




    Tip: if banach space is infinite-dimensional, it's hamel basis is necessarily uncountable
    – igortsts
    Dec 13 '18 at 9:47
















-1














Consider the space of sequences of real numbers ${vec{a_i}}$ where each $vec{a_i}={a_{i_n}}$ is such that $sum_{n=0}^{infty}2^n|a_{i_n}|<+infty$. Then how could we better describe the space? Typically, does the space have countable linear basis,is it complete with respect to the norm $|vec{a_i}|=sum_n2^n|a_{i_n}|$,whether it is finite dimensional?



One example that comes to mind is of the sequence $a_n=frac1{2^{2n}}$. Hence, I think the space is countable. How should I proceed further. Is any representation like Riesz representation work here? Thanks beforehand.






sequences-and-series functional-analysis banach-spaces







share|cite|improve this question




















  • 1




    Is it assumed that $a_n>0$? And if $a_n=1/(2^n)$ don't you get the (divergent) series of all $1$'s?
    – coffeemath
    Dec 13 '18 at 9:29








  • 1




    I don't understand the question. What do you actually want?
    – BigbearZzz
    Dec 13 '18 at 9:38






  • 1




    Your expression for $||cdot||$ is not a norm...
    – BigbearZzz
    Dec 13 '18 at 9:41






  • 1




    @vidyarthi because this expression can be negative whereas a norm can't
    – igortsts
    Dec 13 '18 at 9:42








  • 1




    Tip: if banach space is infinite-dimensional, it's hamel basis is necessarily uncountable
    – igortsts
    Dec 13 '18 at 9:47














-1












-1








-1


2





Consider the space of sequences of real numbers ${vec{a_i}}$ where each $vec{a_i}={a_{i_n}}$ is such that $sum_{n=0}^{infty}2^n|a_{i_n}|<+infty$. Then how could we better describe the space? Typically, does the space have countable linear basis,is it complete with respect to the norm $|vec{a_i}|=sum_n2^n|a_{i_n}|$,whether it is finite dimensional?



One example that comes to mind is of the sequence $a_n=frac1{2^{2n}}$. Hence, I think the space is countable. How should I proceed further. Is any representation like Riesz representation work here? Thanks beforehand.






sequences-and-series functional-analysis banach-spaces







share|cite|improve this question















Consider the space of sequences of real numbers ${vec{a_i}}$ where each $vec{a_i}={a_{i_n}}$ is such that $sum_{n=0}^{infty}2^n|a_{i_n}|<+infty$. Then how could we better describe the space? Typically, does the space have countable linear basis,is it complete with respect to the norm $|vec{a_i}|=sum_n2^n|a_{i_n}|$,whether it is finite dimensional?



One example that comes to mind is of the sequence $a_n=frac1{2^{2n}}$. Hence, I think the space is countable. How should I proceed further. Is any representation like Riesz representation work here? Thanks beforehand.






sequences-and-series functional-analysis banach-spaces




sequences-and-series functional-analysis banach-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 9:50







vidyarthi

















asked Dec 13 '18 at 9:27









vidyarthividyarthi

2,9341832




2,9341832








  • 1




    Is it assumed that $a_n>0$? And if $a_n=1/(2^n)$ don't you get the (divergent) series of all $1$'s?
    – coffeemath
    Dec 13 '18 at 9:29








  • 1




    I don't understand the question. What do you actually want?
    – BigbearZzz
    Dec 13 '18 at 9:38






  • 1




    Your expression for $||cdot||$ is not a norm...
    – BigbearZzz
    Dec 13 '18 at 9:41






  • 1




    @vidyarthi because this expression can be negative whereas a norm can't
    – igortsts
    Dec 13 '18 at 9:42








  • 1




    Tip: if banach space is infinite-dimensional, it's hamel basis is necessarily uncountable
    – igortsts
    Dec 13 '18 at 9:47














  • 1




    Is it assumed that $a_n>0$? And if $a_n=1/(2^n)$ don't you get the (divergent) series of all $1$'s?
    – coffeemath
    Dec 13 '18 at 9:29








  • 1




    I don't understand the question. What do you actually want?
    – BigbearZzz
    Dec 13 '18 at 9:38






  • 1




    Your expression for $||cdot||$ is not a norm...
    – BigbearZzz
    Dec 13 '18 at 9:41






  • 1




    @vidyarthi because this expression can be negative whereas a norm can't
    – igortsts
    Dec 13 '18 at 9:42








  • 1




    Tip: if banach space is infinite-dimensional, it's hamel basis is necessarily uncountable
    – igortsts
    Dec 13 '18 at 9:47








1




1




Is it assumed that $a_n>0$? And if $a_n=1/(2^n)$ don't you get the (divergent) series of all $1$'s?
– coffeemath
Dec 13 '18 at 9:29






Is it assumed that $a_n>0$? And if $a_n=1/(2^n)$ don't you get the (divergent) series of all $1$'s?
– coffeemath
Dec 13 '18 at 9:29






1




1




I don't understand the question. What do you actually want?
– BigbearZzz
Dec 13 '18 at 9:38




I don't understand the question. What do you actually want?
– BigbearZzz
Dec 13 '18 at 9:38




1




1




Your expression for $||cdot||$ is not a norm...
– BigbearZzz
Dec 13 '18 at 9:41




Your expression for $||cdot||$ is not a norm...
– BigbearZzz
Dec 13 '18 at 9:41




1




1




@vidyarthi because this expression can be negative whereas a norm can't
– igortsts
Dec 13 '18 at 9:42






@vidyarthi because this expression can be negative whereas a norm can't
– igortsts
Dec 13 '18 at 9:42






1




1




Tip: if banach space is infinite-dimensional, it's hamel basis is necessarily uncountable
– igortsts
Dec 13 '18 at 9:47




Tip: if banach space is infinite-dimensional, it's hamel basis is necessarily uncountable
– igortsts
Dec 13 '18 at 9:47










1 Answer
1






active

oldest

votes


















0














${(a_n):sum 2^{n} |a_n| <infty}$ with $|(a_n)|= {sum 2^{n} |a_n| }$ is separable infinite dimensional Banach space. In fact it just an $L^{1}(mu)$ for a suitable measure $mu$. [ Without absolute value you don't even get a norm as pointed out in the comments]. Define a measure $mu$ on the power set of $mathbb N$ by $mu {n}=2^{n}$. Then $int fdmu =sum 2^{n} f(n)$ for any $f:mathbb N to mathbb R$ whcih is $mu$ integrable. The identification of $f in L^{1}(mu)$ with the sequence $(f(n))$ shows that our space is nothing but $L^{1}(mu)$. Note also that $(L^{1}(mu))^{*}=L^{infty}(mu)=ell ^{infty}$.






share|cite|improve this answer























  • so it is an infinite dimensional banach space and has no countable basis, right?
    – vidyarthi
    Dec 13 '18 at 9:47






  • 1




    @KaviRamaMurthy Are you sure? It looks more like $L^1(mu)$ with weighted atomic Borel measure to me.
    – BigbearZzz
    Dec 13 '18 at 9:51










  • @BigbearZzz You are right. I have corrected my answer.
    – Kavi Rama Murthy
    Dec 13 '18 at 9:53










  • @BigbearZzz could you show what transformation would make it into $L^1(mu)$?
    – vidyarthi
    Dec 13 '18 at 9:54










  • @vidyarthi See the last part of my revised answer.
    – Kavi Rama Murthy
    Dec 13 '18 at 9:59











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037796%2fspace-of-sequences-such-thtat-sum-n-0-infty2na-n-infty%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









0














${(a_n):sum 2^{n} |a_n| <infty}$ with $|(a_n)|= {sum 2^{n} |a_n| }$ is separable infinite dimensional Banach space. In fact it just an $L^{1}(mu)$ for a suitable measure $mu$. [ Without absolute value you don't even get a norm as pointed out in the comments]. Define a measure $mu$ on the power set of $mathbb N$ by $mu {n}=2^{n}$. Then $int fdmu =sum 2^{n} f(n)$ for any $f:mathbb N to mathbb R$ whcih is $mu$ integrable. The identification of $f in L^{1}(mu)$ with the sequence $(f(n))$ shows that our space is nothing but $L^{1}(mu)$. Note also that $(L^{1}(mu))^{*}=L^{infty}(mu)=ell ^{infty}$.






share|cite|improve this answer























  • so it is an infinite dimensional banach space and has no countable basis, right?
    – vidyarthi
    Dec 13 '18 at 9:47






  • 1




    @KaviRamaMurthy Are you sure? It looks more like $L^1(mu)$ with weighted atomic Borel measure to me.
    – BigbearZzz
    Dec 13 '18 at 9:51










  • @BigbearZzz You are right. I have corrected my answer.
    – Kavi Rama Murthy
    Dec 13 '18 at 9:53










  • @BigbearZzz could you show what transformation would make it into $L^1(mu)$?
    – vidyarthi
    Dec 13 '18 at 9:54










  • @vidyarthi See the last part of my revised answer.
    – Kavi Rama Murthy
    Dec 13 '18 at 9:59
















0














${(a_n):sum 2^{n} |a_n| <infty}$ with $|(a_n)|= {sum 2^{n} |a_n| }$ is separable infinite dimensional Banach space. In fact it just an $L^{1}(mu)$ for a suitable measure $mu$. [ Without absolute value you don't even get a norm as pointed out in the comments]. Define a measure $mu$ on the power set of $mathbb N$ by $mu {n}=2^{n}$. Then $int fdmu =sum 2^{n} f(n)$ for any $f:mathbb N to mathbb R$ whcih is $mu$ integrable. The identification of $f in L^{1}(mu)$ with the sequence $(f(n))$ shows that our space is nothing but $L^{1}(mu)$. Note also that $(L^{1}(mu))^{*}=L^{infty}(mu)=ell ^{infty}$.






share|cite|improve this answer























  • so it is an infinite dimensional banach space and has no countable basis, right?
    – vidyarthi
    Dec 13 '18 at 9:47






  • 1




    @KaviRamaMurthy Are you sure? It looks more like $L^1(mu)$ with weighted atomic Borel measure to me.
    – BigbearZzz
    Dec 13 '18 at 9:51










  • @BigbearZzz You are right. I have corrected my answer.
    – Kavi Rama Murthy
    Dec 13 '18 at 9:53










  • @BigbearZzz could you show what transformation would make it into $L^1(mu)$?
    – vidyarthi
    Dec 13 '18 at 9:54










  • @vidyarthi See the last part of my revised answer.
    – Kavi Rama Murthy
    Dec 13 '18 at 9:59














0












0








0






${(a_n):sum 2^{n} |a_n| <infty}$ with $|(a_n)|= {sum 2^{n} |a_n| }$ is separable infinite dimensional Banach space. In fact it just an $L^{1}(mu)$ for a suitable measure $mu$. [ Without absolute value you don't even get a norm as pointed out in the comments]. Define a measure $mu$ on the power set of $mathbb N$ by $mu {n}=2^{n}$. Then $int fdmu =sum 2^{n} f(n)$ for any $f:mathbb N to mathbb R$ whcih is $mu$ integrable. The identification of $f in L^{1}(mu)$ with the sequence $(f(n))$ shows that our space is nothing but $L^{1}(mu)$. Note also that $(L^{1}(mu))^{*}=L^{infty}(mu)=ell ^{infty}$.






share|cite|improve this answer














${(a_n):sum 2^{n} |a_n| <infty}$ with $|(a_n)|= {sum 2^{n} |a_n| }$ is separable infinite dimensional Banach space. In fact it just an $L^{1}(mu)$ for a suitable measure $mu$. [ Without absolute value you don't even get a norm as pointed out in the comments]. Define a measure $mu$ on the power set of $mathbb N$ by $mu {n}=2^{n}$. Then $int fdmu =sum 2^{n} f(n)$ for any $f:mathbb N to mathbb R$ whcih is $mu$ integrable. The identification of $f in L^{1}(mu)$ with the sequence $(f(n))$ shows that our space is nothing but $L^{1}(mu)$. Note also that $(L^{1}(mu))^{*}=L^{infty}(mu)=ell ^{infty}$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 13 '18 at 23:12

























answered Dec 13 '18 at 9:46









Kavi Rama MurthyKavi Rama Murthy

52.4k31955




52.4k31955












  • so it is an infinite dimensional banach space and has no countable basis, right?
    – vidyarthi
    Dec 13 '18 at 9:47






  • 1




    @KaviRamaMurthy Are you sure? It looks more like $L^1(mu)$ with weighted atomic Borel measure to me.
    – BigbearZzz
    Dec 13 '18 at 9:51










  • @BigbearZzz You are right. I have corrected my answer.
    – Kavi Rama Murthy
    Dec 13 '18 at 9:53










  • @BigbearZzz could you show what transformation would make it into $L^1(mu)$?
    – vidyarthi
    Dec 13 '18 at 9:54










  • @vidyarthi See the last part of my revised answer.
    – Kavi Rama Murthy
    Dec 13 '18 at 9:59


















  • so it is an infinite dimensional banach space and has no countable basis, right?
    – vidyarthi
    Dec 13 '18 at 9:47






  • 1




    @KaviRamaMurthy Are you sure? It looks more like $L^1(mu)$ with weighted atomic Borel measure to me.
    – BigbearZzz
    Dec 13 '18 at 9:51










  • @BigbearZzz You are right. I have corrected my answer.
    – Kavi Rama Murthy
    Dec 13 '18 at 9:53










  • @BigbearZzz could you show what transformation would make it into $L^1(mu)$?
    – vidyarthi
    Dec 13 '18 at 9:54










  • @vidyarthi See the last part of my revised answer.
    – Kavi Rama Murthy
    Dec 13 '18 at 9:59
















so it is an infinite dimensional banach space and has no countable basis, right?
– vidyarthi
Dec 13 '18 at 9:47




so it is an infinite dimensional banach space and has no countable basis, right?
– vidyarthi
Dec 13 '18 at 9:47




1




1




@KaviRamaMurthy Are you sure? It looks more like $L^1(mu)$ with weighted atomic Borel measure to me.
– BigbearZzz
Dec 13 '18 at 9:51




@KaviRamaMurthy Are you sure? It looks more like $L^1(mu)$ with weighted atomic Borel measure to me.
– BigbearZzz
Dec 13 '18 at 9:51












@BigbearZzz You are right. I have corrected my answer.
– Kavi Rama Murthy
Dec 13 '18 at 9:53




@BigbearZzz You are right. I have corrected my answer.
– Kavi Rama Murthy
Dec 13 '18 at 9:53












@BigbearZzz could you show what transformation would make it into $L^1(mu)$?
– vidyarthi
Dec 13 '18 at 9:54




@BigbearZzz could you show what transformation would make it into $L^1(mu)$?
– vidyarthi
Dec 13 '18 at 9:54












@vidyarthi See the last part of my revised answer.
– Kavi Rama Murthy
Dec 13 '18 at 9:59




@vidyarthi See the last part of my revised answer.
– Kavi Rama Murthy
Dec 13 '18 at 9:59


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3037796%2fspace-of-sequences-such-thtat-sum-n-0-infty2na-n-infty%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more