$B _{ell ^{2}} ^{+}$ with the norm $Vertvert cdot Vertvert _{sqrt{2}}$ don't have normal structure











up vote
0
down vote

favorite












Consider the space $ell ^{2}$ with the standard norm
begin{align*}
Vert x Vert _{2} = left( sum _{i =1} ^{infty} x _{i} ^{2} right) ^{1/2}
end{align*}
and we define the equivalent norm
begin{align*}
Vertvert x Vertvert _{sqrt{2}}= max{ Vert x Vert _{2}, sqrt{2}Vert x Vert _{infty} } mbox{.}
end{align*}



Let's define the positive part of the unit ball
begin{align*}
B _{ell ^{2}} ^{+} = lbrace x in ell ^{2}: ; Vert x Vert _{2} leqslant 1, ; x _{i} geqslant 0 rbrace mbox{.}
end{align*}
I want show that $B _{ell ^{2}} ^{+}$ with the norm $Vertvert cdot Vertvert _{sqrt{2}}$ don't have normal structure, and for show that, I should show that diam$(B _{ell ^{2}} ^{+})$ = $r _{x}(B _{ell ^{2}} ^{+})$. I showed that diam$(B _{ell ^{2}} ^{+}) = 1$, but I don't know why $r _{x}(B _{ell ^{2}} ^{+}) =1$. What element in $B _{ell ^{2}} ^{+}$ can take for prove that $r _{x}(B _{ell ^{2}} ^{+}) =1$?






functional-analysis fixedpoints







share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    Consider the space $ell ^{2}$ with the standard norm
    begin{align*}
    Vert x Vert _{2} = left( sum _{i =1} ^{infty} x _{i} ^{2} right) ^{1/2}
    end{align*}
    and we define the equivalent norm
    begin{align*}
    Vertvert x Vertvert _{sqrt{2}}= max{ Vert x Vert _{2}, sqrt{2}Vert x Vert _{infty} } mbox{.}
    end{align*}



    Let's define the positive part of the unit ball
    begin{align*}
    B _{ell ^{2}} ^{+} = lbrace x in ell ^{2}: ; Vert x Vert _{2} leqslant 1, ; x _{i} geqslant 0 rbrace mbox{.}
    end{align*}
    I want show that $B _{ell ^{2}} ^{+}$ with the norm $Vertvert cdot Vertvert _{sqrt{2}}$ don't have normal structure, and for show that, I should show that diam$(B _{ell ^{2}} ^{+})$ = $r _{x}(B _{ell ^{2}} ^{+})$. I showed that diam$(B _{ell ^{2}} ^{+}) = 1$, but I don't know why $r _{x}(B _{ell ^{2}} ^{+}) =1$. What element in $B _{ell ^{2}} ^{+}$ can take for prove that $r _{x}(B _{ell ^{2}} ^{+}) =1$?






    functional-analysis fixedpoints







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Consider the space $ell ^{2}$ with the standard norm
      begin{align*}
      Vert x Vert _{2} = left( sum _{i =1} ^{infty} x _{i} ^{2} right) ^{1/2}
      end{align*}
      and we define the equivalent norm
      begin{align*}
      Vertvert x Vertvert _{sqrt{2}}= max{ Vert x Vert _{2}, sqrt{2}Vert x Vert _{infty} } mbox{.}
      end{align*}



      Let's define the positive part of the unit ball
      begin{align*}
      B _{ell ^{2}} ^{+} = lbrace x in ell ^{2}: ; Vert x Vert _{2} leqslant 1, ; x _{i} geqslant 0 rbrace mbox{.}
      end{align*}
      I want show that $B _{ell ^{2}} ^{+}$ with the norm $Vertvert cdot Vertvert _{sqrt{2}}$ don't have normal structure, and for show that, I should show that diam$(B _{ell ^{2}} ^{+})$ = $r _{x}(B _{ell ^{2}} ^{+})$. I showed that diam$(B _{ell ^{2}} ^{+}) = 1$, but I don't know why $r _{x}(B _{ell ^{2}} ^{+}) =1$. What element in $B _{ell ^{2}} ^{+}$ can take for prove that $r _{x}(B _{ell ^{2}} ^{+}) =1$?






      functional-analysis fixedpoints







      share|cite|improve this question















      Consider the space $ell ^{2}$ with the standard norm
      begin{align*}
      Vert x Vert _{2} = left( sum _{i =1} ^{infty} x _{i} ^{2} right) ^{1/2}
      end{align*}
      and we define the equivalent norm
      begin{align*}
      Vertvert x Vertvert _{sqrt{2}}= max{ Vert x Vert _{2}, sqrt{2}Vert x Vert _{infty} } mbox{.}
      end{align*}



      Let's define the positive part of the unit ball
      begin{align*}
      B _{ell ^{2}} ^{+} = lbrace x in ell ^{2}: ; Vert x Vert _{2} leqslant 1, ; x _{i} geqslant 0 rbrace mbox{.}
      end{align*}
      I want show that $B _{ell ^{2}} ^{+}$ with the norm $Vertvert cdot Vertvert _{sqrt{2}}$ don't have normal structure, and for show that, I should show that diam$(B _{ell ^{2}} ^{+})$ = $r _{x}(B _{ell ^{2}} ^{+})$. I showed that diam$(B _{ell ^{2}} ^{+}) = 1$, but I don't know why $r _{x}(B _{ell ^{2}} ^{+}) =1$. What element in $B _{ell ^{2}} ^{+}$ can take for prove that $r _{x}(B _{ell ^{2}} ^{+}) =1$?






      functional-analysis fixedpoints




      functional-analysis fixedpoints






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 20 at 2:09

























      asked Nov 19 at 21:44









      Amanda Gael

      155




      155






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          You can take $x_i=2^{-i/2}$, and then the distance of this element from zero is precisely $1$ in your norm, because the sum of the squares is $1$ and the supremum is $1/sqrt{2}$.






          share|cite|improve this answer





















          • Thanks for help me :)
            – Amanda Gael
            Nov 20 at 2:11










          • the sequence $ y$ with each $ y_ {i} = 2 ^ {- 1/2} $ if it belongs to the ball, but the norm of $ x - and $ has a different result: how $Vertvert x - y Vertvert _{sqrt{2}}= max { Vert x - y Vert _{2}, sqrt{2} Vert x - y Vert _{infty} }$, so we have that
            – Amanda Gael
            Dec 4 at 1:35












          • begin{align*} Vert x - y Vert _{2} &= left( sum _{i = 1} ^{infty} (x - y) right) ^{1/2} \ &= left( sum _{i = 1} ^{infty} x _{i} ^{2} - 2 sum _{i = 1} ^{infty} frac{x _{i}}{2 ^{i/2}} + sum _{i = 1} ^{infty} frac{1}{2 ^{i/2}} right) ^{1/2} end{align*}
            – Amanda Gael
            Dec 4 at 2:02











          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',
          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%2f3005566%2fb-ell-2-with-the-norm-vert-vert-cdot-vert-vert-sqrt2-do%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








          up vote
          1
          down vote



          accepted










          You can take $x_i=2^{-i/2}$, and then the distance of this element from zero is precisely $1$ in your norm, because the sum of the squares is $1$ and the supremum is $1/sqrt{2}$.






          share|cite|improve this answer





















          • Thanks for help me :)
            – Amanda Gael
            Nov 20 at 2:11










          • the sequence $ y$ with each $ y_ {i} = 2 ^ {- 1/2} $ if it belongs to the ball, but the norm of $ x - and $ has a different result: how $Vertvert x - y Vertvert _{sqrt{2}}= max { Vert x - y Vert _{2}, sqrt{2} Vert x - y Vert _{infty} }$, so we have that
            – Amanda Gael
            Dec 4 at 1:35












          • begin{align*} Vert x - y Vert _{2} &= left( sum _{i = 1} ^{infty} (x - y) right) ^{1/2} \ &= left( sum _{i = 1} ^{infty} x _{i} ^{2} - 2 sum _{i = 1} ^{infty} frac{x _{i}}{2 ^{i/2}} + sum _{i = 1} ^{infty} frac{1}{2 ^{i/2}} right) ^{1/2} end{align*}
            – Amanda Gael
            Dec 4 at 2:02















          up vote
          1
          down vote



          accepted










          You can take $x_i=2^{-i/2}$, and then the distance of this element from zero is precisely $1$ in your norm, because the sum of the squares is $1$ and the supremum is $1/sqrt{2}$.






          share|cite|improve this answer





















          • Thanks for help me :)
            – Amanda Gael
            Nov 20 at 2:11










          • the sequence $ y$ with each $ y_ {i} = 2 ^ {- 1/2} $ if it belongs to the ball, but the norm of $ x - and $ has a different result: how $Vertvert x - y Vertvert _{sqrt{2}}= max { Vert x - y Vert _{2}, sqrt{2} Vert x - y Vert _{infty} }$, so we have that
            – Amanda Gael
            Dec 4 at 1:35












          • begin{align*} Vert x - y Vert _{2} &= left( sum _{i = 1} ^{infty} (x - y) right) ^{1/2} \ &= left( sum _{i = 1} ^{infty} x _{i} ^{2} - 2 sum _{i = 1} ^{infty} frac{x _{i}}{2 ^{i/2}} + sum _{i = 1} ^{infty} frac{1}{2 ^{i/2}} right) ^{1/2} end{align*}
            – Amanda Gael
            Dec 4 at 2:02













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          You can take $x_i=2^{-i/2}$, and then the distance of this element from zero is precisely $1$ in your norm, because the sum of the squares is $1$ and the supremum is $1/sqrt{2}$.






          share|cite|improve this answer












          You can take $x_i=2^{-i/2}$, and then the distance of this element from zero is precisely $1$ in your norm, because the sum of the squares is $1$ and the supremum is $1/sqrt{2}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 at 22:09









          uniquesolution

          8,631823




          8,631823












          • Thanks for help me :)
            – Amanda Gael
            Nov 20 at 2:11










          • the sequence $ y$ with each $ y_ {i} = 2 ^ {- 1/2} $ if it belongs to the ball, but the norm of $ x - and $ has a different result: how $Vertvert x - y Vertvert _{sqrt{2}}= max { Vert x - y Vert _{2}, sqrt{2} Vert x - y Vert _{infty} }$, so we have that
            – Amanda Gael
            Dec 4 at 1:35












          • begin{align*} Vert x - y Vert _{2} &= left( sum _{i = 1} ^{infty} (x - y) right) ^{1/2} \ &= left( sum _{i = 1} ^{infty} x _{i} ^{2} - 2 sum _{i = 1} ^{infty} frac{x _{i}}{2 ^{i/2}} + sum _{i = 1} ^{infty} frac{1}{2 ^{i/2}} right) ^{1/2} end{align*}
            – Amanda Gael
            Dec 4 at 2:02


















          • Thanks for help me :)
            – Amanda Gael
            Nov 20 at 2:11










          • the sequence $ y$ with each $ y_ {i} = 2 ^ {- 1/2} $ if it belongs to the ball, but the norm of $ x - and $ has a different result: how $Vertvert x - y Vertvert _{sqrt{2}}= max { Vert x - y Vert _{2}, sqrt{2} Vert x - y Vert _{infty} }$, so we have that
            – Amanda Gael
            Dec 4 at 1:35












          • begin{align*} Vert x - y Vert _{2} &= left( sum _{i = 1} ^{infty} (x - y) right) ^{1/2} \ &= left( sum _{i = 1} ^{infty} x _{i} ^{2} - 2 sum _{i = 1} ^{infty} frac{x _{i}}{2 ^{i/2}} + sum _{i = 1} ^{infty} frac{1}{2 ^{i/2}} right) ^{1/2} end{align*}
            – Amanda Gael
            Dec 4 at 2:02
















          Thanks for help me :)
          – Amanda Gael
          Nov 20 at 2:11




          Thanks for help me :)
          – Amanda Gael
          Nov 20 at 2:11












          the sequence $ y$ with each $ y_ {i} = 2 ^ {- 1/2} $ if it belongs to the ball, but the norm of $ x - and $ has a different result: how $Vertvert x - y Vertvert _{sqrt{2}}= max { Vert x - y Vert _{2}, sqrt{2} Vert x - y Vert _{infty} }$, so we have that
          – Amanda Gael
          Dec 4 at 1:35






          the sequence $ y$ with each $ y_ {i} = 2 ^ {- 1/2} $ if it belongs to the ball, but the norm of $ x - and $ has a different result: how $Vertvert x - y Vertvert _{sqrt{2}}= max { Vert x - y Vert _{2}, sqrt{2} Vert x - y Vert _{infty} }$, so we have that
          – Amanda Gael
          Dec 4 at 1:35














          begin{align*} Vert x - y Vert _{2} &= left( sum _{i = 1} ^{infty} (x - y) right) ^{1/2} \ &= left( sum _{i = 1} ^{infty} x _{i} ^{2} - 2 sum _{i = 1} ^{infty} frac{x _{i}}{2 ^{i/2}} + sum _{i = 1} ^{infty} frac{1}{2 ^{i/2}} right) ^{1/2} end{align*}
          – Amanda Gael
          Dec 4 at 2:02




          begin{align*} Vert x - y Vert _{2} &= left( sum _{i = 1} ^{infty} (x - y) right) ^{1/2} \ &= left( sum _{i = 1} ^{infty} x _{i} ^{2} - 2 sum _{i = 1} ^{infty} frac{x _{i}}{2 ^{i/2}} + sum _{i = 1} ^{infty} frac{1}{2 ^{i/2}} right) ^{1/2} end{align*}
          – Amanda Gael
          Dec 4 at 2:02


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3005566%2fb-ell-2-with-the-norm-vert-vert-cdot-vert-vert-sqrt2-do%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

          How to send String Array data to Server using php in android

          Image
          0 I have a String array which contains value string array = {6,37,38,837,7,...} and I have also used some string values. I want to send that all string values and string array to my database which is located at server side. When I am click on submit button the data will stored in table. On the PHP side I want to read this array as it is,other thing i will do at background means PHP side. See my below code then you will get it what I am says. Here is my code: public class Hotels extends Activity { // Declare Variables JSONArray jsonarray = null; private static final String TAG_ID = "id"; public static final String TAG_NAME = "name"; public static final String TAG_LOCATION = "location"; public static final String TAG_DESC = "description"; String f_date, l_date;...
          Read more

          Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

          Image
          up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
          Read more

          Is anime1.com a legal site for watching anime?

          Image
          up vote 4 down vote favorite 1 Other anime streaming sites are blocking my IP because I'm Iranian while anime1.com isn't. I wanted to know if it's legal to watch/stream anime from this site. Thanks. anime-production share | improve this question edited Nov 27 at 22:38 W. Are 1,599 2 23 asked Nov 27 at 20:36 mhnajafi 23 3 add a comment ...
          Read more