Open linear subspace of a Hilbert space.












4












$begingroup$


Does there exist any open linear (vector) subspace of a Hilbert space? I could not think of any example.



Actually, I was reading the book by Simmons, there almost in every theorem it assumed that "If M is a closed linear subspace".It seemed natural to me to think about subspaces which are not closed. I have an got an example which is not closed:
Take the Hilbert space H = C[0,1], with L^2 norm and the subspace set of all polynomials, it is not closed because it's closure is H and not open can be found here Set of all polynomials on [0, 1/2] is not open in C[0, 1/2]. Then I asked myself an example of to think of an open set. But I could lead myself nowhere, as I am not familiar with infinite dimensional vector space. Not closed does not necessarily mean open.






functional-analysis







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Linear span of a ball = ?
    $endgroup$
    – metamorphy
    Jan 2 at 5:08






  • 1




    $begingroup$
    @metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say?
    $endgroup$
    – epsilon_delta
    Jan 2 at 5:18








  • 1




    $begingroup$
    I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite).
    $endgroup$
    – metamorphy
    Jan 2 at 5:21






  • 2




    $begingroup$
    Sorry, please disregard my close vote; I misread the question.
    $endgroup$
    – angryavian
    Jan 2 at 5:27






  • 1




    $begingroup$
    Nice question. Glad you didn't give up on it. :) +1.
    $endgroup$
    – DisintegratingByParts
    Jan 2 at 6:02


















4












$begingroup$


Does there exist any open linear (vector) subspace of a Hilbert space? I could not think of any example.



Actually, I was reading the book by Simmons, there almost in every theorem it assumed that "If M is a closed linear subspace".It seemed natural to me to think about subspaces which are not closed. I have an got an example which is not closed:
Take the Hilbert space H = C[0,1], with L^2 norm and the subspace set of all polynomials, it is not closed because it's closure is H and not open can be found here Set of all polynomials on [0, 1/2] is not open in C[0, 1/2]. Then I asked myself an example of to think of an open set. But I could lead myself nowhere, as I am not familiar with infinite dimensional vector space. Not closed does not necessarily mean open.






functional-analysis







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Linear span of a ball = ?
    $endgroup$
    – metamorphy
    Jan 2 at 5:08






  • 1




    $begingroup$
    @metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say?
    $endgroup$
    – epsilon_delta
    Jan 2 at 5:18








  • 1




    $begingroup$
    I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite).
    $endgroup$
    – metamorphy
    Jan 2 at 5:21






  • 2




    $begingroup$
    Sorry, please disregard my close vote; I misread the question.
    $endgroup$
    – angryavian
    Jan 2 at 5:27






  • 1




    $begingroup$
    Nice question. Glad you didn't give up on it. :) +1.
    $endgroup$
    – DisintegratingByParts
    Jan 2 at 6:02
















4












4








4


1



$begingroup$


Does there exist any open linear (vector) subspace of a Hilbert space? I could not think of any example.



Actually, I was reading the book by Simmons, there almost in every theorem it assumed that "If M is a closed linear subspace".It seemed natural to me to think about subspaces which are not closed. I have an got an example which is not closed:
Take the Hilbert space H = C[0,1], with L^2 norm and the subspace set of all polynomials, it is not closed because it's closure is H and not open can be found here Set of all polynomials on [0, 1/2] is not open in C[0, 1/2]. Then I asked myself an example of to think of an open set. But I could lead myself nowhere, as I am not familiar with infinite dimensional vector space. Not closed does not necessarily mean open.






functional-analysis







share|cite|improve this question











$endgroup$




Does there exist any open linear (vector) subspace of a Hilbert space? I could not think of any example.



Actually, I was reading the book by Simmons, there almost in every theorem it assumed that "If M is a closed linear subspace".It seemed natural to me to think about subspaces which are not closed. I have an got an example which is not closed:
Take the Hilbert space H = C[0,1], with L^2 norm and the subspace set of all polynomials, it is not closed because it's closure is H and not open can be found here Set of all polynomials on [0, 1/2] is not open in C[0, 1/2]. Then I asked myself an example of to think of an open set. But I could lead myself nowhere, as I am not familiar with infinite dimensional vector space. Not closed does not necessarily mean open.






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 2 at 20:40







epsilon_delta

















asked Jan 2 at 5:02









epsilon_deltaepsilon_delta

286




286








  • 3




    $begingroup$
    Linear span of a ball = ?
    $endgroup$
    – metamorphy
    Jan 2 at 5:08






  • 1




    $begingroup$
    @metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say?
    $endgroup$
    – epsilon_delta
    Jan 2 at 5:18








  • 1




    $begingroup$
    I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite).
    $endgroup$
    – metamorphy
    Jan 2 at 5:21






  • 2




    $begingroup$
    Sorry, please disregard my close vote; I misread the question.
    $endgroup$
    – angryavian
    Jan 2 at 5:27






  • 1




    $begingroup$
    Nice question. Glad you didn't give up on it. :) +1.
    $endgroup$
    – DisintegratingByParts
    Jan 2 at 6:02
















  • 3




    $begingroup$
    Linear span of a ball = ?
    $endgroup$
    – metamorphy
    Jan 2 at 5:08






  • 1




    $begingroup$
    @metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say?
    $endgroup$
    – epsilon_delta
    Jan 2 at 5:18








  • 1




    $begingroup$
    I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite).
    $endgroup$
    – metamorphy
    Jan 2 at 5:21






  • 2




    $begingroup$
    Sorry, please disregard my close vote; I misread the question.
    $endgroup$
    – angryavian
    Jan 2 at 5:27






  • 1




    $begingroup$
    Nice question. Glad you didn't give up on it. :) +1.
    $endgroup$
    – DisintegratingByParts
    Jan 2 at 6:02










3




3




$begingroup$
Linear span of a ball = ?
$endgroup$
– metamorphy
Jan 2 at 5:08




$begingroup$
Linear span of a ball = ?
$endgroup$
– metamorphy
Jan 2 at 5:08




1




1




$begingroup$
@metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say?
$endgroup$
– epsilon_delta
Jan 2 at 5:18






$begingroup$
@metamorphy If you take a ball in R^n, and take an open ball around a point. then it contains at n linearly independent vectors, span of those vectors give you whole R^n. We know whole space is an open set but my point was in asking a proper subspace. Or what else you wanted to say?
$endgroup$
– epsilon_delta
Jan 2 at 5:18






1




1




$begingroup$
I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite).
$endgroup$
– metamorphy
Jan 2 at 5:21




$begingroup$
I'm telling you that there's no proper subspace. A nonempty open set contains some ball, the linear span of which is the whole space (this is true for any dimensions, including infinite).
$endgroup$
– metamorphy
Jan 2 at 5:21




2




2




$begingroup$
Sorry, please disregard my close vote; I misread the question.
$endgroup$
– angryavian
Jan 2 at 5:27




$begingroup$
Sorry, please disregard my close vote; I misread the question.
$endgroup$
– angryavian
Jan 2 at 5:27




1




1




$begingroup$
Nice question. Glad you didn't give up on it. :) +1.
$endgroup$
– DisintegratingByParts
Jan 2 at 6:02






$begingroup$
Nice question. Glad you didn't give up on it. :) +1.
$endgroup$
– DisintegratingByParts
Jan 2 at 6:02












2 Answers
2






active

oldest

votes


















2












$begingroup$

No.



Let $N$ be a normed space and



$M subsetneq N tag 1$



a proper subspace. Then $M$ contains no nonempty open set. For if



$emptyset ne U subset M tag 2$



were open, with



$M ni m in U, tag 3$



we could find $rho > 0$ such that the open ball



$B(m, rho) subset U; tag 4$



then picking any



$0 ne v in N setminus M tag 5$



the vector



$m + alpha (v - m) in B(m, rho) tag 6$



if $0 ne alpha in Bbb R$ is sufficiently small, since



$Vert (m + alpha (v - m)) - m Vert = Vert alpha (v - m) Vert = vert alpha vert Vert v - m Vert < rho tag 7$



for



$vert alpha vert < dfrac{rho}{Vert v - m Vert}; tag 8$



but then



$m + alpha(v - m) in M, tag 9$



whence



$alpha(v - m) = m + alpha(v - m) - m in M, tag{10}$



whence



$v - m in M, tag{11}$



whence



$v = v - m + m in M, tag{12}$



in contradiction to (5); therefore no $B(m, rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    If $M leq mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 in B_r(0) subset M$, but for every (none-zero) vector $v in mathcal{H}$, we have
    $$ frac{r}{2Vert v Vert} v in B_r(0) subset M $$
    But M is a linear subspace so $ v in M $. Thus the only open subspaces of $ mathcal{H} $ are $ mathcal{H} $ itself.






    share|cite|improve this answer











    $endgroup$














      Your Answer








      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%2f3059150%2fopen-linear-subspace-of-a-hilbert-space%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      No.



      Let $N$ be a normed space and



      $M subsetneq N tag 1$



      a proper subspace. Then $M$ contains no nonempty open set. For if



      $emptyset ne U subset M tag 2$



      were open, with



      $M ni m in U, tag 3$



      we could find $rho > 0$ such that the open ball



      $B(m, rho) subset U; tag 4$



      then picking any



      $0 ne v in N setminus M tag 5$



      the vector



      $m + alpha (v - m) in B(m, rho) tag 6$



      if $0 ne alpha in Bbb R$ is sufficiently small, since



      $Vert (m + alpha (v - m)) - m Vert = Vert alpha (v - m) Vert = vert alpha vert Vert v - m Vert < rho tag 7$



      for



      $vert alpha vert < dfrac{rho}{Vert v - m Vert}; tag 8$



      but then



      $m + alpha(v - m) in M, tag 9$



      whence



      $alpha(v - m) = m + alpha(v - m) - m in M, tag{10}$



      whence



      $v - m in M, tag{11}$



      whence



      $v = v - m + m in M, tag{12}$



      in contradiction to (5); therefore no $B(m, rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        No.



        Let $N$ be a normed space and



        $M subsetneq N tag 1$



        a proper subspace. Then $M$ contains no nonempty open set. For if



        $emptyset ne U subset M tag 2$



        were open, with



        $M ni m in U, tag 3$



        we could find $rho > 0$ such that the open ball



        $B(m, rho) subset U; tag 4$



        then picking any



        $0 ne v in N setminus M tag 5$



        the vector



        $m + alpha (v - m) in B(m, rho) tag 6$



        if $0 ne alpha in Bbb R$ is sufficiently small, since



        $Vert (m + alpha (v - m)) - m Vert = Vert alpha (v - m) Vert = vert alpha vert Vert v - m Vert < rho tag 7$



        for



        $vert alpha vert < dfrac{rho}{Vert v - m Vert}; tag 8$



        but then



        $m + alpha(v - m) in M, tag 9$



        whence



        $alpha(v - m) = m + alpha(v - m) - m in M, tag{10}$



        whence



        $v - m in M, tag{11}$



        whence



        $v = v - m + m in M, tag{12}$



        in contradiction to (5); therefore no $B(m, rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          No.



          Let $N$ be a normed space and



          $M subsetneq N tag 1$



          a proper subspace. Then $M$ contains no nonempty open set. For if



          $emptyset ne U subset M tag 2$



          were open, with



          $M ni m in U, tag 3$



          we could find $rho > 0$ such that the open ball



          $B(m, rho) subset U; tag 4$



          then picking any



          $0 ne v in N setminus M tag 5$



          the vector



          $m + alpha (v - m) in B(m, rho) tag 6$



          if $0 ne alpha in Bbb R$ is sufficiently small, since



          $Vert (m + alpha (v - m)) - m Vert = Vert alpha (v - m) Vert = vert alpha vert Vert v - m Vert < rho tag 7$



          for



          $vert alpha vert < dfrac{rho}{Vert v - m Vert}; tag 8$



          but then



          $m + alpha(v - m) in M, tag 9$



          whence



          $alpha(v - m) = m + alpha(v - m) - m in M, tag{10}$



          whence



          $v - m in M, tag{11}$



          whence



          $v = v - m + m in M, tag{12}$



          in contradiction to (5); therefore no $B(m, rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.






          share|cite|improve this answer









          $endgroup$



          No.



          Let $N$ be a normed space and



          $M subsetneq N tag 1$



          a proper subspace. Then $M$ contains no nonempty open set. For if



          $emptyset ne U subset M tag 2$



          were open, with



          $M ni m in U, tag 3$



          we could find $rho > 0$ such that the open ball



          $B(m, rho) subset U; tag 4$



          then picking any



          $0 ne v in N setminus M tag 5$



          the vector



          $m + alpha (v - m) in B(m, rho) tag 6$



          if $0 ne alpha in Bbb R$ is sufficiently small, since



          $Vert (m + alpha (v - m)) - m Vert = Vert alpha (v - m) Vert = vert alpha vert Vert v - m Vert < rho tag 7$



          for



          $vert alpha vert < dfrac{rho}{Vert v - m Vert}; tag 8$



          but then



          $m + alpha(v - m) in M, tag 9$



          whence



          $alpha(v - m) = m + alpha(v - m) - m in M, tag{10}$



          whence



          $v - m in M, tag{11}$



          whence



          $v = v - m + m in M, tag{12}$



          in contradiction to (5); therefore no $B(m, rho)$ as in (5) can exist, and $M$ cannot be open, since it contains no open set.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 6:15









          Robert LewisRobert Lewis

          49.1k23168




          49.1k23168























              3












              $begingroup$

              If $M leq mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 in B_r(0) subset M$, but for every (none-zero) vector $v in mathcal{H}$, we have
              $$ frac{r}{2Vert v Vert} v in B_r(0) subset M $$
              But M is a linear subspace so $ v in M $. Thus the only open subspaces of $ mathcal{H} $ are $ mathcal{H} $ itself.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                If $M leq mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 in B_r(0) subset M$, but for every (none-zero) vector $v in mathcal{H}$, we have
                $$ frac{r}{2Vert v Vert} v in B_r(0) subset M $$
                But M is a linear subspace so $ v in M $. Thus the only open subspaces of $ mathcal{H} $ are $ mathcal{H} $ itself.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  If $M leq mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 in B_r(0) subset M$, but for every (none-zero) vector $v in mathcal{H}$, we have
                  $$ frac{r}{2Vert v Vert} v in B_r(0) subset M $$
                  But M is a linear subspace so $ v in M $. Thus the only open subspaces of $ mathcal{H} $ are $ mathcal{H} $ itself.






                  share|cite|improve this answer











                  $endgroup$



                  If $M leq mathcal{H}$ a subspace of a Hilbert space (or generally any normed space) is open, then it contains a ball around the origin $0 in B_r(0) subset M$, but for every (none-zero) vector $v in mathcal{H}$, we have
                  $$ frac{r}{2Vert v Vert} v in B_r(0) subset M $$
                  But M is a linear subspace so $ v in M $. Thus the only open subspaces of $ mathcal{H} $ are $ mathcal{H} $ itself.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 2 at 7:02









                  Henno Brandsma

                  117k350128




                  117k350128










                  answered Jan 2 at 6:16









                  pitariverpitariver

                  469213




                  469213






























                      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%2f3059150%2fopen-linear-subspace-of-a-hilbert-space%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 change which sound is reproduced for terminal bell?

                      Image
                      3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
                      Read more

                      Can I use Tabulator js library in my java Spring + Thymeleaf project?

                      Image
                      0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
                      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