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






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oldest

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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











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1 Answer
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1 Answer
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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


















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