If every absolutely convergent series is convergent in $X$, then $X$ is a Banach space.
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Here is my try..
By hypothesis, the convergence of $sum_{i=1}^{infty}|x_n|$ in $mathbb{R}$ $implies$ the convergence of $sum_{i=1}^{infty} x_n$ in $X$.
Pick a Cauchy sequence ${ f_n}$ in $X$. Then $forall$ $epsilon > 0$, $exists$ $n>m>N$ s.t $|f_n - f_m |< epsilon$.
and $| |f_n| - |f_m| | leq |f_n - f_m |< epsilon$ that is ${|f_n|}$ is a Cauchy sequence in $mathbb{R}$ thus it converges in $mathbb{R}$. Hence it is bounded (i.e. $exists k>0$ s.t | |$f_n$| | $leq k$, $forall$ $n=1,2,..$).
$implies S_n=sum_{i=1}^{n}|f_n| leq nk$ but also the sequence ${S_n}$ is monotonically increasing thus ${S_n}$ converges in $mathbb{R}$ $implies sum_{i=1}^{infty} f_n$ converges in $X$ by hypothesis thus the sequence $f_n rightarrow 0$ in $X$. Thus $X$ is a Banach space.
real-analysis functional-analysis proof-verification banach-spaces
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up vote
1
down vote
favorite
Here is my try..
By hypothesis, the convergence of $sum_{i=1}^{infty}|x_n|$ in $mathbb{R}$ $implies$ the convergence of $sum_{i=1}^{infty} x_n$ in $X$.
Pick a Cauchy sequence ${ f_n}$ in $X$. Then $forall$ $epsilon > 0$, $exists$ $n>m>N$ s.t $|f_n - f_m |< epsilon$.
and $| |f_n| - |f_m| | leq |f_n - f_m |< epsilon$ that is ${|f_n|}$ is a Cauchy sequence in $mathbb{R}$ thus it converges in $mathbb{R}$. Hence it is bounded (i.e. $exists k>0$ s.t | |$f_n$| | $leq k$, $forall$ $n=1,2,..$).
$implies S_n=sum_{i=1}^{n}|f_n| leq nk$ but also the sequence ${S_n}$ is monotonically increasing thus ${S_n}$ converges in $mathbb{R}$ $implies sum_{i=1}^{infty} f_n$ converges in $X$ by hypothesis thus the sequence $f_n rightarrow 0$ in $X$. Thus $X$ is a Banach space.
real-analysis functional-analysis proof-verification banach-spaces
Take a look at the definition of boundedness and then maybe reconsider your assertion that $S_n$ is bounded. ;)
– MaoWao
Nov 16 at 23:46
Oh yes, $S_n$ can't be bounded as the set of natural numbers is not bounded above. Thank you!
– HybridAlien
Nov 17 at 0:05
If you type |f| instead of ||f|| it looks like $|f|$ instead of $||f||$.
– DanielWainfleet
Nov 17 at 13:57
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here is my try..
By hypothesis, the convergence of $sum_{i=1}^{infty}|x_n|$ in $mathbb{R}$ $implies$ the convergence of $sum_{i=1}^{infty} x_n$ in $X$.
Pick a Cauchy sequence ${ f_n}$ in $X$. Then $forall$ $epsilon > 0$, $exists$ $n>m>N$ s.t $|f_n - f_m |< epsilon$.
and $| |f_n| - |f_m| | leq |f_n - f_m |< epsilon$ that is ${|f_n|}$ is a Cauchy sequence in $mathbb{R}$ thus it converges in $mathbb{R}$. Hence it is bounded (i.e. $exists k>0$ s.t | |$f_n$| | $leq k$, $forall$ $n=1,2,..$).
$implies S_n=sum_{i=1}^{n}|f_n| leq nk$ but also the sequence ${S_n}$ is monotonically increasing thus ${S_n}$ converges in $mathbb{R}$ $implies sum_{i=1}^{infty} f_n$ converges in $X$ by hypothesis thus the sequence $f_n rightarrow 0$ in $X$. Thus $X$ is a Banach space.
real-analysis functional-analysis proof-verification banach-spaces
Here is my try..
By hypothesis, the convergence of $sum_{i=1}^{infty}|x_n|$ in $mathbb{R}$ $implies$ the convergence of $sum_{i=1}^{infty} x_n$ in $X$.
Pick a Cauchy sequence ${ f_n}$ in $X$. Then $forall$ $epsilon > 0$, $exists$ $n>m>N$ s.t $|f_n - f_m |< epsilon$.
and $| |f_n| - |f_m| | leq |f_n - f_m |< epsilon$ that is ${|f_n|}$ is a Cauchy sequence in $mathbb{R}$ thus it converges in $mathbb{R}$. Hence it is bounded (i.e. $exists k>0$ s.t | |$f_n$| | $leq k$, $forall$ $n=1,2,..$).
$implies S_n=sum_{i=1}^{n}|f_n| leq nk$ but also the sequence ${S_n}$ is monotonically increasing thus ${S_n}$ converges in $mathbb{R}$ $implies sum_{i=1}^{infty} f_n$ converges in $X$ by hypothesis thus the sequence $f_n rightarrow 0$ in $X$. Thus $X$ is a Banach space.
real-analysis functional-analysis proof-verification banach-spaces
real-analysis functional-analysis proof-verification banach-spaces
edited Nov 17 at 14:58
asked Nov 16 at 23:34
HybridAlien
2008
2008
Take a look at the definition of boundedness and then maybe reconsider your assertion that $S_n$ is bounded. ;)
– MaoWao
Nov 16 at 23:46
Oh yes, $S_n$ can't be bounded as the set of natural numbers is not bounded above. Thank you!
– HybridAlien
Nov 17 at 0:05
If you type |f| instead of ||f|| it looks like $|f|$ instead of $||f||$.
– DanielWainfleet
Nov 17 at 13:57
add a comment |
Take a look at the definition of boundedness and then maybe reconsider your assertion that $S_n$ is bounded. ;)
– MaoWao
Nov 16 at 23:46
Oh yes, $S_n$ can't be bounded as the set of natural numbers is not bounded above. Thank you!
– HybridAlien
Nov 17 at 0:05
If you type |f| instead of ||f|| it looks like $|f|$ instead of $||f||$.
– DanielWainfleet
Nov 17 at 13:57
Take a look at the definition of boundedness and then maybe reconsider your assertion that $S_n$ is bounded. ;)
– MaoWao
Nov 16 at 23:46
Take a look at the definition of boundedness and then maybe reconsider your assertion that $S_n$ is bounded. ;)
– MaoWao
Nov 16 at 23:46
Oh yes, $S_n$ can't be bounded as the set of natural numbers is not bounded above. Thank you!
– HybridAlien
Nov 17 at 0:05
Oh yes, $S_n$ can't be bounded as the set of natural numbers is not bounded above. Thank you!
– HybridAlien
Nov 17 at 0:05
If you type |f| instead of ||f|| it looks like $|f|$ instead of $||f||$.
– DanielWainfleet
Nov 17 at 13:57
If you type |f| instead of ||f|| it looks like $|f|$ instead of $||f||$.
– DanielWainfleet
Nov 17 at 13:57
add a comment |
1 Answer
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If a subsequence of a Cauchy sequenec converges, so does the entire sequence. Choose subsequence $f_{n_k}$ such that $|f_{n_k}-f_{n_{k+1}}| < frac 1 {2^{k}}$. Then $sum_k (f_{n_k}-f_{n_{k+1}})$ converges absolutely and hence it converges. By writing down the partial sums show that $lim_k f_{n_k}$ exists. This proves convergence of ${f_{n_k}}$ hence that of $(f_n)$.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If a subsequence of a Cauchy sequenec converges, so does the entire sequence. Choose subsequence $f_{n_k}$ such that $|f_{n_k}-f_{n_{k+1}}| < frac 1 {2^{k}}$. Then $sum_k (f_{n_k}-f_{n_{k+1}})$ converges absolutely and hence it converges. By writing down the partial sums show that $lim_k f_{n_k}$ exists. This proves convergence of ${f_{n_k}}$ hence that of $(f_n)$.
add a comment |
up vote
1
down vote
accepted
If a subsequence of a Cauchy sequenec converges, so does the entire sequence. Choose subsequence $f_{n_k}$ such that $|f_{n_k}-f_{n_{k+1}}| < frac 1 {2^{k}}$. Then $sum_k (f_{n_k}-f_{n_{k+1}})$ converges absolutely and hence it converges. By writing down the partial sums show that $lim_k f_{n_k}$ exists. This proves convergence of ${f_{n_k}}$ hence that of $(f_n)$.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If a subsequence of a Cauchy sequenec converges, so does the entire sequence. Choose subsequence $f_{n_k}$ such that $|f_{n_k}-f_{n_{k+1}}| < frac 1 {2^{k}}$. Then $sum_k (f_{n_k}-f_{n_{k+1}})$ converges absolutely and hence it converges. By writing down the partial sums show that $lim_k f_{n_k}$ exists. This proves convergence of ${f_{n_k}}$ hence that of $(f_n)$.
If a subsequence of a Cauchy sequenec converges, so does the entire sequence. Choose subsequence $f_{n_k}$ such that $|f_{n_k}-f_{n_{k+1}}| < frac 1 {2^{k}}$. Then $sum_k (f_{n_k}-f_{n_{k+1}})$ converges absolutely and hence it converges. By writing down the partial sums show that $lim_k f_{n_k}$ exists. This proves convergence of ${f_{n_k}}$ hence that of $(f_n)$.
answered Nov 17 at 0:45
Kavi Rama Murthy
41.4k31751
41.4k31751
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Take a look at the definition of boundedness and then maybe reconsider your assertion that $S_n$ is bounded. ;)
– MaoWao
Nov 16 at 23:46
Oh yes, $S_n$ can't be bounded as the set of natural numbers is not bounded above. Thank you!
– HybridAlien
Nov 17 at 0:05
If you type |f| instead of ||f|| it looks like $|f|$ instead of $||f||$.
– DanielWainfleet
Nov 17 at 13:57