sums in Banach spaces
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Let $X$ be a Banach space and let $(x_n)$ be a sequence in $X$. One can define $sum_nx_n$ if the partial sum $s_n=sum_{k=1}^nx_n$ is convergent in $X$. Also a stronger version of sum can be defined like this: choose all finite subsets of $mathbb{N}$, let us call it $mathcal{F}$. Define a partial ordering $prec$ on $mathcal{F}$ by $F_1prec F_2$ if $F_1subseteq F_2$. Define a net $S$ on $mathcal{F}$ by $S_F=sum_{kin F}x_k$. Let us call $Nsum_nx_n$ exists if the net $(S_F)_{Finmathcal{F}}$ converges in $X$. It is clear that, Nsum_nx_n$ exists Rightarrow sum_nx_n$ exists, as sequence of partial sum is a subnet of the other one. Can someone give an example when $sum_nx_n$ exists but $Nsum_nx_n$ does not exists?
summation banach-spaces
asked Nov 22 at 19:08
Tanmoy Paul
1
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Let $X$ be a Banach space and let $(x_n)$ be a sequence in $X$. One can define $sum_nx_n$ if the partial sum $s_n=sum_{k=1}^nx_n$ is convergent in $X$. Also a stronger version of sum can be defined like this: choose all finite subsets of $mathbb{N}$, let us call it $mathcal{F}$. Define a partial ordering $prec$ on $mathcal{F}$ by $F_1prec F_2$ if $F_1subseteq F_2$. Define a net $S$ on $mathcal{F}$ by $S_F=sum_{kin F}x_k$. Let us call $Nsum_nx_n$ exists if the net $(S_F)_{Finmathcal{F}}$ converges in $X$. It is clear that, Nsum_nx_n$ exists Rightarrow sum_nx_n$ exists, as sequence of partial sum is a subnet of the other one. Can someone give an example when $sum_nx_n$ exists but $Nsum_nx_n$ does not exists?
summation banach-spaces
asked Nov 22 at 19:08
Tanmoy Paul
1
How about $x_n=(-1)^n 1/n$?
– John_Wick
Nov 22 at 19:18
Same difference between conditional and absolute convergence
– Federico
Nov 22 at 19:24
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $X$ be a Banach space and let $(x_n)$ be a sequence in $X$. One can define $sum_nx_n$ if the partial sum $s_n=sum_{k=1}^nx_n$ is convergent in $X$. Also a stronger version of sum can be defined like this: choose all finite subsets of $mathbb{N}$, let us call it $mathcal{F}$. Define a partial ordering $prec$ on $mathcal{F}$ by $F_1prec F_2$ if $F_1subseteq F_2$. Define a net $S$ on $mathcal{F}$ by $S_F=sum_{kin F}x_k$. Let us call $Nsum_nx_n$ exists if the net $(S_F)_{Finmathcal{F}}$ converges in $X$. It is clear that, Nsum_nx_n$ exists Rightarrow sum_nx_n$ exists, as sequence of partial sum is a subnet of the other one. Can someone give an example when $sum_nx_n$ exists but $Nsum_nx_n$ does not exists?
summation banach-spaces
asked Nov 22 at 19:08
Tanmoy Paul
1
Let $X$ be a Banach space and let $(x_n)$ be a sequence in $X$. One can define $sum_nx_n$ if the partial sum $s_n=sum_{k=1}^nx_n$ is convergent in $X$. Also a stronger version of sum can be defined like this: choose all finite subsets of $mathbb{N}$, let us call it $mathcal{F}$. Define a partial ordering $prec$ on $mathcal{F}$ by $F_1prec F_2$ if $F_1subseteq F_2$. Define a net $S$ on $mathcal{F}$ by $S_F=sum_{kin F}x_k$. Let us call $Nsum_nx_n$ exists if the net $(S_F)_{Finmathcal{F}}$ converges in $X$. It is clear that, Nsum_nx_n$ exists Rightarrow sum_nx_n$ exists, as sequence of partial sum is a subnet of the other one. Can someone give an example when $sum_nx_n$ exists but $Nsum_nx_n$ does not exists?
summation banach-spaces
summation banach-spaces
asked Nov 22 at 19:08
Tanmoy Paul
1
asked Nov 22 at 19:08
Tanmoy Paul
1
asked Nov 22 at 19:08
Tanmoy Paul
1
asked Nov 22 at 19:08
Tanmoy Paul
1
asked Nov 22 at 19:08
Tanmoy Paul
1
1
How about $x_n=(-1)^n 1/n$?
– John_Wick
Nov 22 at 19:18
Same difference between conditional and absolute convergence
– Federico
Nov 22 at 19:24
add a comment |
How about $x_n=(-1)^n 1/n$?
– John_Wick
Nov 22 at 19:18
Same difference between conditional and absolute convergence
– Federico
Nov 22 at 19:24
How about $x_n=(-1)^n 1/n$?
– John_Wick
Nov 22 at 19:18
How about $x_n=(-1)^n 1/n$?
– John_Wick
Nov 22 at 19:18
Same difference between conditional and absolute convergence
– Federico
Nov 22 at 19:24
Same difference between conditional and absolute convergence
– Federico
Nov 22 at 19:24
add a comment |
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How about $x_n=(-1)^n 1/n$?
– John_Wick
Nov 22 at 19:18
Same difference between conditional and absolute convergence
– Federico
Nov 22 at 19:24