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?






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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







    share|cite|improve this question


























      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







      share|cite|improve this question















      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






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      edited 2 days ago

























      asked 2 days ago









      johnka

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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 $|x_n-y_n| to 0$.






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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 $|x_n-y_n| to 0$.






            share|cite|improve this answer

























              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 $|x_n-y_n| to 0$.






              share|cite|improve this answer























                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 $|x_n-y_n| to 0$.






                share|cite|improve this answer












                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 $|x_n-y_n| to 0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Kavi Rama Murthy

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