Prove that $C_0$ is Banach.












1












$begingroup$


Let $x^n in C_0$ is Cauchy.



$rightarrow$ For $epsilon> 0$ there is N such that $n,m ge N $ $$ lVert x^m -x^nrVert< frac {epsilon} {2} $$
we know that for every k $$lvert x^m_k -x^n_krvert le sup_{ige 1} lvert x^m_i -x^n_irvert < frac {epsilon} {2} $$
So $x^n_k$ is Cauchy in $mathbb R$ which is Banach so $$x^n_k rightarrow x_k in mathbb R or lvert x^n_k -x_krvert le frac {epsilon} {2} $$
by this we can say that there is an N, $nge N$ such that $lVert x^n -xrVert = sup_{ige 1} lvert x^n_i -x_i lvert < frac {epsilon} {2} $ $ $



which means $x^n rightarrow x$



Since $x^m in c_0$,
there is some $N'$ such that $|x_i^m| < {1 over 2 } epsilon$ for $i ge N' (m=N)$



Now to show that $xin C_0$



$$lvert x_irvert le lvert x^m_i-x_irvert+ lvert x^m_irvert <frac {epsilon} {2}+frac {epsilon} {2}=epsilon for ige N' $$



This gives us $x_irightarrow 0$ for $ige N'$



So $xin C_0$



Is this Correct?






functional-analysis







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $x^n in C_0$ is Cauchy.



    $rightarrow$ For $epsilon> 0$ there is N such that $n,m ge N $ $$ lVert x^m -x^nrVert< frac {epsilon} {2} $$
    we know that for every k $$lvert x^m_k -x^n_krvert le sup_{ige 1} lvert x^m_i -x^n_irvert < frac {epsilon} {2} $$
    So $x^n_k$ is Cauchy in $mathbb R$ which is Banach so $$x^n_k rightarrow x_k in mathbb R or lvert x^n_k -x_krvert le frac {epsilon} {2} $$
    by this we can say that there is an N, $nge N$ such that $lVert x^n -xrVert = sup_{ige 1} lvert x^n_i -x_i lvert < frac {epsilon} {2} $ $ $



    which means $x^n rightarrow x$



    Since $x^m in c_0$,
    there is some $N'$ such that $|x_i^m| < {1 over 2 } epsilon$ for $i ge N' (m=N)$



    Now to show that $xin C_0$



    $$lvert x_irvert le lvert x^m_i-x_irvert+ lvert x^m_irvert <frac {epsilon} {2}+frac {epsilon} {2}=epsilon for ige N' $$



    This gives us $x_irightarrow 0$ for $ige N'$



    So $xin C_0$



    Is this Correct?






    functional-analysis







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $x^n in C_0$ is Cauchy.



      $rightarrow$ For $epsilon> 0$ there is N such that $n,m ge N $ $$ lVert x^m -x^nrVert< frac {epsilon} {2} $$
      we know that for every k $$lvert x^m_k -x^n_krvert le sup_{ige 1} lvert x^m_i -x^n_irvert < frac {epsilon} {2} $$
      So $x^n_k$ is Cauchy in $mathbb R$ which is Banach so $$x^n_k rightarrow x_k in mathbb R or lvert x^n_k -x_krvert le frac {epsilon} {2} $$
      by this we can say that there is an N, $nge N$ such that $lVert x^n -xrVert = sup_{ige 1} lvert x^n_i -x_i lvert < frac {epsilon} {2} $ $ $



      which means $x^n rightarrow x$



      Since $x^m in c_0$,
      there is some $N'$ such that $|x_i^m| < {1 over 2 } epsilon$ for $i ge N' (m=N)$



      Now to show that $xin C_0$



      $$lvert x_irvert le lvert x^m_i-x_irvert+ lvert x^m_irvert <frac {epsilon} {2}+frac {epsilon} {2}=epsilon for ige N' $$



      This gives us $x_irightarrow 0$ for $ige N'$



      So $xin C_0$



      Is this Correct?






      functional-analysis







      share|cite|improve this question











      $endgroup$




      Let $x^n in C_0$ is Cauchy.



      $rightarrow$ For $epsilon> 0$ there is N such that $n,m ge N $ $$ lVert x^m -x^nrVert< frac {epsilon} {2} $$
      we know that for every k $$lvert x^m_k -x^n_krvert le sup_{ige 1} lvert x^m_i -x^n_irvert < frac {epsilon} {2} $$
      So $x^n_k$ is Cauchy in $mathbb R$ which is Banach so $$x^n_k rightarrow x_k in mathbb R or lvert x^n_k -x_krvert le frac {epsilon} {2} $$
      by this we can say that there is an N, $nge N$ such that $lVert x^n -xrVert = sup_{ige 1} lvert x^n_i -x_i lvert < frac {epsilon} {2} $ $ $



      which means $x^n rightarrow x$



      Since $x^m in c_0$,
      there is some $N'$ such that $|x_i^m| < {1 over 2 } epsilon$ for $i ge N' (m=N)$



      Now to show that $xin C_0$



      $$lvert x_irvert le lvert x^m_i-x_irvert+ lvert x^m_irvert <frac {epsilon} {2}+frac {epsilon} {2}=epsilon for ige N' $$



      This gives us $x_irightarrow 0$ for $ige N'$



      So $xin C_0$



      Is this Correct?






      functional-analysis




      functional-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 6 at 6:26







      Hitman

















      asked Jan 5 at 21:53









      HitmanHitman

      1749




      1749






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          The question has been modified based on this answer. The answer below is for the first version of the question.
          First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
          Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you please check again? I have edited the question.
            $endgroup$
            – Hitman
            Jan 6 at 0:19










          • $begingroup$
            Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
            $endgroup$
            – Kavi Rama Murthy
            Jan 6 at 0:38










          • $begingroup$
            I think now it is okay.
            $endgroup$
            – Hitman
            Jan 6 at 0:59










          • $begingroup$
            what do you say?
            $endgroup$
            – Hitman
            Jan 6 at 1:00






          • 1




            $begingroup$
            In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
            $endgroup$
            – Kavi Rama Murthy
            Jan 6 at 4:47











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063237%2fprove-that-c-0-is-banach%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          The question has been modified based on this answer. The answer below is for the first version of the question.
          First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
          Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you please check again? I have edited the question.
            $endgroup$
            – Hitman
            Jan 6 at 0:19










          • $begingroup$
            Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
            $endgroup$
            – Kavi Rama Murthy
            Jan 6 at 0:38










          • $begingroup$
            I think now it is okay.
            $endgroup$
            – Hitman
            Jan 6 at 0:59










          • $begingroup$
            what do you say?
            $endgroup$
            – Hitman
            Jan 6 at 1:00






          • 1




            $begingroup$
            In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
            $endgroup$
            – Kavi Rama Murthy
            Jan 6 at 4:47
















          1












          $begingroup$

          The question has been modified based on this answer. The answer below is for the first version of the question.
          First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
          Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could you please check again? I have edited the question.
            $endgroup$
            – Hitman
            Jan 6 at 0:19










          • $begingroup$
            Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
            $endgroup$
            – Kavi Rama Murthy
            Jan 6 at 0:38










          • $begingroup$
            I think now it is okay.
            $endgroup$
            – Hitman
            Jan 6 at 0:59










          • $begingroup$
            what do you say?
            $endgroup$
            – Hitman
            Jan 6 at 1:00






          • 1




            $begingroup$
            In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
            $endgroup$
            – Kavi Rama Murthy
            Jan 6 at 4:47














          1












          1








          1





          $begingroup$

          The question has been modified based on this answer. The answer below is for the first version of the question.
          First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
          Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.






          share|cite|improve this answer











          $endgroup$



          The question has been modified based on this answer. The answer below is for the first version of the question.
          First correction: $|a_n|<epsilon /2$ for all $n$ and $a_n to a$ does not imply $|a|<epsilon /2$. It implies $|a|leq epsilon /2$.
          Second correction: in the last part $|x|$ has no meaning for a sequence $x$. Use coordinates. You should write $|x_i| leq |x_i^{n}|+|x_i^{n}-x_i|$. Then you should make the argument more rigorous as follows. First choose $n$ such that $|x_i^{n}-x_i|<epsilon /2$ for all $i$. Fixing this $n$ choose $k$ such that $|x_i^{n}|<epsilon /2$ for all $i geq k$. Now you get $|x_i| <epsilon$ for all $i geq k$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 0:39

























          answered Jan 5 at 23:53









          Kavi Rama MurthyKavi Rama Murthy

          65.4k42766




          65.4k42766












          • $begingroup$
            Could you please check again? I have edited the question.
            $endgroup$
            – Hitman
            Jan 6 at 0:19










          • $begingroup$
            Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
            $endgroup$
            – Kavi Rama Murthy
            Jan 6 at 0:38










          • $begingroup$
            I think now it is okay.
            $endgroup$
            – Hitman
            Jan 6 at 0:59










          • $begingroup$
            what do you say?
            $endgroup$
            – Hitman
            Jan 6 at 1:00






          • 1




            $begingroup$
            In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
            $endgroup$
            – Kavi Rama Murthy
            Jan 6 at 4:47


















          • $begingroup$
            Could you please check again? I have edited the question.
            $endgroup$
            – Hitman
            Jan 6 at 0:19










          • $begingroup$
            Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
            $endgroup$
            – Kavi Rama Murthy
            Jan 6 at 0:38










          • $begingroup$
            I think now it is okay.
            $endgroup$
            – Hitman
            Jan 6 at 0:59










          • $begingroup$
            what do you say?
            $endgroup$
            – Hitman
            Jan 6 at 1:00






          • 1




            $begingroup$
            In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
            $endgroup$
            – Kavi Rama Murthy
            Jan 6 at 4:47
















          $begingroup$
          Could you please check again? I have edited the question.
          $endgroup$
          – Hitman
          Jan 6 at 0:19




          $begingroup$
          Could you please check again? I have edited the question.
          $endgroup$
          – Hitman
          Jan 6 at 0:19












          $begingroup$
          Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
          $endgroup$
          – Kavi Rama Murthy
          Jan 6 at 0:38




          $begingroup$
          Your last part is still not rigorous. There are two variables $i$ and $n$ and you have to state clearly that you first choose $n$ and then the final inequality becomes true for all $i$ sufficiently large.
          $endgroup$
          – Kavi Rama Murthy
          Jan 6 at 0:38












          $begingroup$
          I think now it is okay.
          $endgroup$
          – Hitman
          Jan 6 at 0:59




          $begingroup$
          I think now it is okay.
          $endgroup$
          – Hitman
          Jan 6 at 0:59












          $begingroup$
          what do you say?
          $endgroup$
          – Hitman
          Jan 6 at 1:00




          $begingroup$
          what do you say?
          $endgroup$
          – Hitman
          Jan 6 at 1:00




          1




          1




          $begingroup$
          In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
          $endgroup$
          – Kavi Rama Murthy
          Jan 6 at 4:47




          $begingroup$
          In the last part you did not specify $m$. If you take $m=N$ your proof will be correct.
          $endgroup$
          – Kavi Rama Murthy
          Jan 6 at 4:47


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063237%2fprove-that-c-0-is-banach%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Ellipse (mathématiques)

          Image
          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
          Read more

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
          Read more