Is it v1,v2,v3,.. converging sequence of space $mathbb{E}$? [closed]
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$(mathbb{E}, Vert{.}Vert)$ - Banach space over scalars $mathbb{R}$ and if $v_1, v_2, v_3,..$ it's the sequence of this space that with $forall vinmathbb{E}$ a sequence $ Vert v - v_1Vert,Vert v - v_2Vert, Vert v - v_3Vert,...$ is real converging sequence. Is it necessary that $v_1, v_2, v_3,..$ converging sequence of space $mathbb{E}$? Why?
sequences-and-series functional-analysis convergence banach-spaces
asked Nov 22 at 19:37
Bambeil
75
closed as off-topic by Did, amWhy, José Carlos Santos, Leucippus, Cesareo Nov 23 at 1:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-2
down vote
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$(mathbb{E}, Vert{.}Vert)$ - Banach space over scalars $mathbb{R}$ and if $v_1, v_2, v_3,..$ it's the sequence of this space that with $forall vinmathbb{E}$ a sequence $ Vert v - v_1Vert,Vert v - v_2Vert, Vert v - v_3Vert,...$ is real converging sequence. Is it necessary that $v_1, v_2, v_3,..$ converging sequence of space $mathbb{E}$? Why?
sequences-and-series functional-analysis convergence banach-spaces
asked Nov 22 at 19:37
Bambeil
75
closed as off-topic by Did, amWhy, José Carlos Santos, Leucippus, Cesareo Nov 23 at 1:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
What are your thoughts on the problem?
– MisterRiemann
Nov 22 at 19:39
I think it isn't necessary that converging sequence and it's enough to show a counterexample, but I failed with it. @MisterRiemann
– Bambeil
Nov 22 at 20:08
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
$(mathbb{E}, Vert{.}Vert)$ - Banach space over scalars $mathbb{R}$ and if $v_1, v_2, v_3,..$ it's the sequence of this space that with $forall vinmathbb{E}$ a sequence $ Vert v - v_1Vert,Vert v - v_2Vert, Vert v - v_3Vert,...$ is real converging sequence. Is it necessary that $v_1, v_2, v_3,..$ converging sequence of space $mathbb{E}$? Why?
sequences-and-series functional-analysis convergence banach-spaces
asked Nov 22 at 19:37
Bambeil
75
$(mathbb{E}, Vert{.}Vert)$ - Banach space over scalars $mathbb{R}$ and if $v_1, v_2, v_3,..$ it's the sequence of this space that with $forall vinmathbb{E}$ a sequence $ Vert v - v_1Vert,Vert v - v_2Vert, Vert v - v_3Vert,...$ is real converging sequence. Is it necessary that $v_1, v_2, v_3,..$ converging sequence of space $mathbb{E}$? Why?
sequences-and-series functional-analysis convergence banach-spaces
sequences-and-series functional-analysis convergence banach-spaces
asked Nov 22 at 19:37
Bambeil
75
asked Nov 22 at 19:37
Bambeil
75
edited Nov 22 at 22:50
asked Nov 22 at 19:37
Bambeil
75
asked Nov 22 at 19:37
Bambeil
75
asked Nov 22 at 19:37
Bambeil
75
75
closed as off-topic by Did, amWhy, José Carlos Santos, Leucippus, Cesareo Nov 23 at 1:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Did, amWhy, José Carlos Santos, Leucippus, Cesareo Nov 23 at 1:10
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Leucippus, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
What are your thoughts on the problem?
– MisterRiemann
Nov 22 at 19:39
I think it isn't necessary that converging sequence and it's enough to show a counterexample, but I failed with it. @MisterRiemann
– Bambeil
Nov 22 at 20:08
add a comment |
What are your thoughts on the problem?
– MisterRiemann
Nov 22 at 19:39
I think it isn't necessary that converging sequence and it's enough to show a counterexample, but I failed with it. @MisterRiemann
– Bambeil
Nov 22 at 20:08
What are your thoughts on the problem?
– MisterRiemann
Nov 22 at 19:39
What are your thoughts on the problem?
– MisterRiemann
Nov 22 at 19:39
I think it isn't necessary that converging sequence and it's enough to show a counterexample, but I failed with it. @MisterRiemann
– Bambeil
Nov 22 at 20:08
I think it isn't necessary that converging sequence and it's enough to show a counterexample, but I failed with it. @MisterRiemann
– Bambeil
Nov 22 at 20:08
add a comment |
2 Answers
2
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0
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accepted
Take your favourite two elements $u, v_1 in mathbb E$ and define for $n > 1$
$$ v_n = v_1+nu. $$
Then the sequence $Vert v - v_1 Vert, Vert v_1 - v_2 Vert, Vert v_2 - v_3 Vert$ is constantly equal to $Vert u Vert$ (except perhaps for the first term), and hence convergent. But
$$ Vert v_n Vert = Vert v_1 + nu Vert geq nVert u Vert - Vert v_1 Vert, $$
so that $(v_n)_{n=1}^infty$ is unbounded and hence cannot converge.
answered Nov 22 at 20:13
MisterRiemann
5,7031624
Thanks, but look, I find a mistake in that my problem and now corrected it. Will this answer stay correct? @MisterRiemann
– Bambeil
Nov 22 at 22:54
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0
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Take $mathbb{E}=mathbb{R}$ and $v_n=sum_{i=1}^n 1/i$. Then $|v_n-v_{n-1}|=1/nrightarrow 0$. But ${v_n}$ is not a convergent sequence.
answered Nov 22 at 19:58
John_Wick
1,134111
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Take your favourite two elements $u, v_1 in mathbb E$ and define for $n > 1$
$$ v_n = v_1+nu. $$
Then the sequence $Vert v - v_1 Vert, Vert v_1 - v_2 Vert, Vert v_2 - v_3 Vert$ is constantly equal to $Vert u Vert$ (except perhaps for the first term), and hence convergent. But
$$ Vert v_n Vert = Vert v_1 + nu Vert geq nVert u Vert - Vert v_1 Vert, $$
so that $(v_n)_{n=1}^infty$ is unbounded and hence cannot converge.
answered Nov 22 at 20:13
MisterRiemann
5,7031624
Thanks, but look, I find a mistake in that my problem and now corrected it. Will this answer stay correct? @MisterRiemann
– Bambeil
Nov 22 at 22:54
add a comment |
up vote
0
down vote
accepted
Take your favourite two elements $u, v_1 in mathbb E$ and define for $n > 1$
$$ v_n = v_1+nu. $$
Then the sequence $Vert v - v_1 Vert, Vert v_1 - v_2 Vert, Vert v_2 - v_3 Vert$ is constantly equal to $Vert u Vert$ (except perhaps for the first term), and hence convergent. But
$$ Vert v_n Vert = Vert v_1 + nu Vert geq nVert u Vert - Vert v_1 Vert, $$
so that $(v_n)_{n=1}^infty$ is unbounded and hence cannot converge.
answered Nov 22 at 20:13
MisterRiemann
5,7031624
Thanks, but look, I find a mistake in that my problem and now corrected it. Will this answer stay correct? @MisterRiemann
– Bambeil
Nov 22 at 22:54
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Take your favourite two elements $u, v_1 in mathbb E$ and define for $n > 1$
$$ v_n = v_1+nu. $$
Then the sequence $Vert v - v_1 Vert, Vert v_1 - v_2 Vert, Vert v_2 - v_3 Vert$ is constantly equal to $Vert u Vert$ (except perhaps for the first term), and hence convergent. But
$$ Vert v_n Vert = Vert v_1 + nu Vert geq nVert u Vert - Vert v_1 Vert, $$
so that $(v_n)_{n=1}^infty$ is unbounded and hence cannot converge.
answered Nov 22 at 20:13
MisterRiemann
5,7031624
Take your favourite two elements $u, v_1 in mathbb E$ and define for $n > 1$
$$ v_n = v_1+nu. $$
Then the sequence $Vert v - v_1 Vert, Vert v_1 - v_2 Vert, Vert v_2 - v_3 Vert$ is constantly equal to $Vert u Vert$ (except perhaps for the first term), and hence convergent. But
$$ Vert v_n Vert = Vert v_1 + nu Vert geq nVert u Vert - Vert v_1 Vert, $$
so that $(v_n)_{n=1}^infty$ is unbounded and hence cannot converge.
answered Nov 22 at 20:13
MisterRiemann
5,7031624
answered Nov 22 at 20:13
MisterRiemann
5,7031624
answered Nov 22 at 20:13
MisterRiemann
5,7031624
answered Nov 22 at 20:13
MisterRiemann
5,7031624
5,7031624
Thanks, but look, I find a mistake in that my problem and now corrected it. Will this answer stay correct? @MisterRiemann
– Bambeil
Nov 22 at 22:54
add a comment |
Thanks, but look, I find a mistake in that my problem and now corrected it. Will this answer stay correct? @MisterRiemann
– Bambeil
Nov 22 at 22:54
Thanks, but look, I find a mistake in that my problem and now corrected it. Will this answer stay correct? @MisterRiemann
– Bambeil
Nov 22 at 22:54
Thanks, but look, I find a mistake in that my problem and now corrected it. Will this answer stay correct? @MisterRiemann
– Bambeil
Nov 22 at 22:54
add a comment |
up vote
0
down vote
Take $mathbb{E}=mathbb{R}$ and $v_n=sum_{i=1}^n 1/i$. Then $|v_n-v_{n-1}|=1/nrightarrow 0$. But ${v_n}$ is not a convergent sequence.
answered Nov 22 at 19:58
John_Wick
1,134111
add a comment |
up vote
0
down vote
Take $mathbb{E}=mathbb{R}$ and $v_n=sum_{i=1}^n 1/i$. Then $|v_n-v_{n-1}|=1/nrightarrow 0$. But ${v_n}$ is not a convergent sequence.
answered Nov 22 at 19:58
John_Wick
1,134111
add a comment |
up vote
0
down vote
up vote
0
down vote
Take $mathbb{E}=mathbb{R}$ and $v_n=sum_{i=1}^n 1/i$. Then $|v_n-v_{n-1}|=1/nrightarrow 0$. But ${v_n}$ is not a convergent sequence.
answered Nov 22 at 19:58
John_Wick
1,134111
Take $mathbb{E}=mathbb{R}$ and $v_n=sum_{i=1}^n 1/i$. Then $|v_n-v_{n-1}|=1/nrightarrow 0$. But ${v_n}$ is not a convergent sequence.
answered Nov 22 at 19:58
John_Wick
1,134111
answered Nov 22 at 19:58
John_Wick
1,134111
answered Nov 22 at 19:58
John_Wick
1,134111
answered Nov 22 at 19:58
John_Wick
1,134111
1,134111
add a comment |
add a comment |
What are your thoughts on the problem?
– MisterRiemann
Nov 22 at 19:39
I think it isn't necessary that converging sequence and it's enough to show a counterexample, but I failed with it. @MisterRiemann
– Bambeil
Nov 22 at 20:08