Mixing
Limit Question in Vectorspace with induced norm
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I found this question in a Collection of a Real Analysis Exercises.And i'm having struggles to understand the problem.
Let X be a Vectorspace together with a scalarproduct $left< .,. right> :Xtimes Xrightarrow mathbb Rquad $ and the norm $left| sqrt { left< x,x right> } right| $. If ${ x }_{ n }epsilon X$ and $lim _{ nrightarrow infty }{ left| { x }_{ n } right| } =rho $
than
we have $lim _{ n,mrightarrow infty }{ supquad left| { x }_{ n }-{ x }_{ m } right| } =0$ or $lim _{ n,mrightarrow infty }{ infquad left| { x }_{ n }+{ x }_{ m } right| } <2rho $
I dont get how the assumtions that are given should leado to those restults which is making it ahrd to begin the proof. Could somehown help me to understand the situation and maybe give a hint on how to start the proof?
real-analysis
asked 2 days ago
johnka
264111
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up vote
0
down vote
favorite
I found this question in a Collection of a Real Analysis Exercises.And i'm having struggles to understand the problem.
Let X be a Vectorspace together with a scalarproduct $left< .,. right> :Xtimes Xrightarrow mathbb Rquad $ and the norm $left| sqrt { left< x,x right> } right| $. If ${ x }_{ n }epsilon X$ and $lim _{ nrightarrow infty }{ left| { x }_{ n } right| } =rho $
than
we have $lim _{ n,mrightarrow infty }{ supquad left| { x }_{ n }-{ x }_{ m } right| } =0$ or $lim _{ n,mrightarrow infty }{ infquad left| { x }_{ n }+{ x }_{ m } right| } <2rho $
I dont get how the assumtions that are given should leado to those restults which is making it ahrd to begin the proof. Could somehown help me to understand the situation and maybe give a hint on how to start the proof?
real-analysis
asked 2 days ago
johnka
264111
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I found this question in a Collection of a Real Analysis Exercises.And i'm having struggles to understand the problem.
Let X be a Vectorspace together with a scalarproduct $left< .,. right> :Xtimes Xrightarrow mathbb Rquad $ and the norm $left| sqrt { left< x,x right> } right| $. If ${ x }_{ n }epsilon X$ and $lim _{ nrightarrow infty }{ left| { x }_{ n } right| } =rho $
than
we have $lim _{ n,mrightarrow infty }{ supquad left| { x }_{ n }-{ x }_{ m } right| } =0$ or $lim _{ n,mrightarrow infty }{ infquad left| { x }_{ n }+{ x }_{ m } right| } <2rho $
I dont get how the assumtions that are given should leado to those restults which is making it ahrd to begin the proof. Could somehown help me to understand the situation and maybe give a hint on how to start the proof?
real-analysis
asked 2 days ago
johnka
264111
I found this question in a Collection of a Real Analysis Exercises.And i'm having struggles to understand the problem.
Let X be a Vectorspace together with a scalarproduct $left< .,. right> :Xtimes Xrightarrow mathbb Rquad $ and the norm $left| sqrt { left< x,x right> } right| $. If ${ x }_{ n }epsilon X$ and $lim _{ nrightarrow infty }{ left| { x }_{ n } right| } =rho $
than
we have $lim _{ n,mrightarrow infty }{ supquad left| { x }_{ n }-{ x }_{ m } right| } =0$ or $lim _{ n,mrightarrow infty }{ infquad left| { x }_{ n }+{ x }_{ m } right| } <2rho $
I dont get how the assumtions that are given should leado to those restults which is making it ahrd to begin the proof. Could somehown help me to understand the situation and maybe give a hint on how to start the proof?
real-analysis
real-analysis
asked 2 days ago
johnka
264111
asked 2 days ago
johnka
264111
edited 2 days ago
asked 2 days ago
johnka
264111
asked 2 days ago
johnka
264111
asked 2 days ago
johnka
264111
264111
add a comment |
add a comment |
1 Answer
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You have typed $epsilon$ instead of $rho$. The result is an easy consequence of parallelogram identity: $|x_n-y_n|^{2}=2|x_n|^{2}+2|y_n|^{2}-|x_n+y_n|^{2}$. Suppose $lim inf |x_n+y_n| geq 2rho$. Then the inequality gives $lim sup|x_n-y_n|^{2} leq 2rho ^{2} +2rho^{2}-4rho^{2}=0$ so $|x_n-y_n| to 0$.
answered 2 days ago
Kavi Rama Murthy
40.8k31751
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
You have typed $epsilon$ instead of $rho$. The result is an easy consequence of parallelogram identity: $|x_n-y_n|^{2}=2|x_n|^{2}+2|y_n|^{2}-|x_n+y_n|^{2}$. Suppose $lim inf |x_n+y_n| geq 2rho$. Then the inequality gives $lim sup|x_n-y_n|^{2} leq 2rho ^{2} +2rho^{2}-4rho^{2}=0$ so $|x_n-y_n| to 0$.
answered 2 days ago
Kavi Rama Murthy
40.8k31751
add a comment |
up vote
1
down vote
accepted
You have typed $epsilon$ instead of $rho$. The result is an easy consequence of parallelogram identity: $|x_n-y_n|^{2}=2|x_n|^{2}+2|y_n|^{2}-|x_n+y_n|^{2}$. Suppose $lim inf |x_n+y_n| geq 2rho$. Then the inequality gives $lim sup|x_n-y_n|^{2} leq 2rho ^{2} +2rho^{2}-4rho^{2}=0$ so $|x_n-y_n| to 0$.
answered 2 days ago
Kavi Rama Murthy
40.8k31751
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have typed $epsilon$ instead of $rho$. The result is an easy consequence of parallelogram identity: $|x_n-y_n|^{2}=2|x_n|^{2}+2|y_n|^{2}-|x_n+y_n|^{2}$. Suppose $lim inf |x_n+y_n| geq 2rho$. Then the inequality gives $lim sup|x_n-y_n|^{2} leq 2rho ^{2} +2rho^{2}-4rho^{2}=0$ so $|x_n-y_n| to 0$.
answered 2 days ago
Kavi Rama Murthy
40.8k31751
You have typed $epsilon$ instead of $rho$. The result is an easy consequence of parallelogram identity: $|x_n-y_n|^{2}=2|x_n|^{2}+2|y_n|^{2}-|x_n+y_n|^{2}$. Suppose $lim inf |x_n+y_n| geq 2rho$. Then the inequality gives $lim sup|x_n-y_n|^{2} leq 2rho ^{2} +2rho^{2}-4rho^{2}=0$ so $|x_n-y_n| to 0$.
answered 2 days ago
Kavi Rama Murthy
40.8k31751
answered 2 days ago
Kavi Rama Murthy
40.8k31751
answered 2 days ago
Kavi Rama Murthy
40.8k31751
answered 2 days ago
Kavi Rama Murthy
40.8k31751
40.8k31751
add a comment |
add a comment |
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