Why a spectral resolution should be left-continuous?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
3
down vote

favorite












The usual definition I can find (almost) everywhere on the literature of a spectral resolution on a Hilbert space $mathscrH$ is this: a family of orthogonal projections $(E_t)_t in mathbbR$ such that



  1. $t leq s implies E_t leq E_s$

  2. $lim_t rightarrow -infty E_t = 0 , quad lim_t rightarrow +infty E_t$

  3. $lim_t searrow s E_t = E_s$,

where the limit is always taken in the strong topology sense. I wonder why one would ask the property 3 of a spectral resolution and not left-continuity. I've heard that right-continuity should actually be a better choice for a measure-theoretical reason: we can always take the limit of a measure of a sequence of increasing measurable sets and that doesn't work always if the sequence is decreasing, but I'm not convinced of that.










share|cite|improve this question



























    up vote
    3
    down vote

    favorite












    The usual definition I can find (almost) everywhere on the literature of a spectral resolution on a Hilbert space $mathscrH$ is this: a family of orthogonal projections $(E_t)_t in mathbbR$ such that



    1. $t leq s implies E_t leq E_s$

    2. $lim_t rightarrow -infty E_t = 0 , quad lim_t rightarrow +infty E_t$

    3. $lim_t searrow s E_t = E_s$,

    where the limit is always taken in the strong topology sense. I wonder why one would ask the property 3 of a spectral resolution and not left-continuity. I've heard that right-continuity should actually be a better choice for a measure-theoretical reason: we can always take the limit of a measure of a sequence of increasing measurable sets and that doesn't work always if the sequence is decreasing, but I'm not convinced of that.










    share|cite|improve this question

























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      The usual definition I can find (almost) everywhere on the literature of a spectral resolution on a Hilbert space $mathscrH$ is this: a family of orthogonal projections $(E_t)_t in mathbbR$ such that



      1. $t leq s implies E_t leq E_s$

      2. $lim_t rightarrow -infty E_t = 0 , quad lim_t rightarrow +infty E_t$

      3. $lim_t searrow s E_t = E_s$,

      where the limit is always taken in the strong topology sense. I wonder why one would ask the property 3 of a spectral resolution and not left-continuity. I've heard that right-continuity should actually be a better choice for a measure-theoretical reason: we can always take the limit of a measure of a sequence of increasing measurable sets and that doesn't work always if the sequence is decreasing, but I'm not convinced of that.










      share|cite|improve this question















      The usual definition I can find (almost) everywhere on the literature of a spectral resolution on a Hilbert space $mathscrH$ is this: a family of orthogonal projections $(E_t)_t in mathbbR$ such that



      1. $t leq s implies E_t leq E_s$

      2. $lim_t rightarrow -infty E_t = 0 , quad lim_t rightarrow +infty E_t$

      3. $lim_t searrow s E_t = E_s$,

      where the limit is always taken in the strong topology sense. I wonder why one would ask the property 3 of a spectral resolution and not left-continuity. I've heard that right-continuity should actually be a better choice for a measure-theoretical reason: we can always take the limit of a measure of a sequence of increasing measurable sets and that doesn't work always if the sequence is decreasing, but I'm not convinced of that.







      functional-analysis spectral-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 30 at 1:59

























      asked Aug 30 at 1:48









      B. Chinaski

      462211




      462211




















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          2
          down vote



          accepted










          It doesn't make a difference whether you require left-continuity or right-continuity. The two definitions are equivalent, up to a reparameterization. Namely, if $(E_t)$ satisfies the definition with right-continuity and you define $F_t=limlimits_tnearrow s E_s$ then $(F_t)$ will satisfy the definition with left-continuity. Conversely, if $(F_t)$ satisfies the definition with right-continuity, then $E_t=limlimits_tsearrow sF_s$ satisfies the definition with left-continuity, and these two operations are inverse to each other.



          The idea here is that $E_t$ represents the "measure" of $(-infty,t]$, while $F_t$ represents the "measure" of $(-infty,t)$. This is equivalent data, since a measure on all Borel sets can be recovered from either one. (Note that the issue you allude to with descending sequences of measurable sets is not relevant, since that's only an issue for measures that could be infinite due to not being able to subtract $infty-infty$, whereas our measures are projection-valued and we can always subtract projections.)






          share|cite|improve this answer




















          • I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
            – B. Chinaski
            Aug 30 at 3:06










          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: false,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          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%2f2899014%2fwhy-a-spectral-resolution-should-be-left-continuous%23new-answer', 'question_page');

          );

          Post as a guest






























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          It doesn't make a difference whether you require left-continuity or right-continuity. The two definitions are equivalent, up to a reparameterization. Namely, if $(E_t)$ satisfies the definition with right-continuity and you define $F_t=limlimits_tnearrow s E_s$ then $(F_t)$ will satisfy the definition with left-continuity. Conversely, if $(F_t)$ satisfies the definition with right-continuity, then $E_t=limlimits_tsearrow sF_s$ satisfies the definition with left-continuity, and these two operations are inverse to each other.



          The idea here is that $E_t$ represents the "measure" of $(-infty,t]$, while $F_t$ represents the "measure" of $(-infty,t)$. This is equivalent data, since a measure on all Borel sets can be recovered from either one. (Note that the issue you allude to with descending sequences of measurable sets is not relevant, since that's only an issue for measures that could be infinite due to not being able to subtract $infty-infty$, whereas our measures are projection-valued and we can always subtract projections.)






          share|cite|improve this answer




















          • I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
            – B. Chinaski
            Aug 30 at 3:06














          up vote
          2
          down vote



          accepted










          It doesn't make a difference whether you require left-continuity or right-continuity. The two definitions are equivalent, up to a reparameterization. Namely, if $(E_t)$ satisfies the definition with right-continuity and you define $F_t=limlimits_tnearrow s E_s$ then $(F_t)$ will satisfy the definition with left-continuity. Conversely, if $(F_t)$ satisfies the definition with right-continuity, then $E_t=limlimits_tsearrow sF_s$ satisfies the definition with left-continuity, and these two operations are inverse to each other.



          The idea here is that $E_t$ represents the "measure" of $(-infty,t]$, while $F_t$ represents the "measure" of $(-infty,t)$. This is equivalent data, since a measure on all Borel sets can be recovered from either one. (Note that the issue you allude to with descending sequences of measurable sets is not relevant, since that's only an issue for measures that could be infinite due to not being able to subtract $infty-infty$, whereas our measures are projection-valued and we can always subtract projections.)






          share|cite|improve this answer




















          • I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
            – B. Chinaski
            Aug 30 at 3:06












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          It doesn't make a difference whether you require left-continuity or right-continuity. The two definitions are equivalent, up to a reparameterization. Namely, if $(E_t)$ satisfies the definition with right-continuity and you define $F_t=limlimits_tnearrow s E_s$ then $(F_t)$ will satisfy the definition with left-continuity. Conversely, if $(F_t)$ satisfies the definition with right-continuity, then $E_t=limlimits_tsearrow sF_s$ satisfies the definition with left-continuity, and these two operations are inverse to each other.



          The idea here is that $E_t$ represents the "measure" of $(-infty,t]$, while $F_t$ represents the "measure" of $(-infty,t)$. This is equivalent data, since a measure on all Borel sets can be recovered from either one. (Note that the issue you allude to with descending sequences of measurable sets is not relevant, since that's only an issue for measures that could be infinite due to not being able to subtract $infty-infty$, whereas our measures are projection-valued and we can always subtract projections.)






          share|cite|improve this answer












          It doesn't make a difference whether you require left-continuity or right-continuity. The two definitions are equivalent, up to a reparameterization. Namely, if $(E_t)$ satisfies the definition with right-continuity and you define $F_t=limlimits_tnearrow s E_s$ then $(F_t)$ will satisfy the definition with left-continuity. Conversely, if $(F_t)$ satisfies the definition with right-continuity, then $E_t=limlimits_tsearrow sF_s$ satisfies the definition with left-continuity, and these two operations are inverse to each other.



          The idea here is that $E_t$ represents the "measure" of $(-infty,t]$, while $F_t$ represents the "measure" of $(-infty,t)$. This is equivalent data, since a measure on all Borel sets can be recovered from either one. (Note that the issue you allude to with descending sequences of measurable sets is not relevant, since that's only an issue for measures that could be infinite due to not being able to subtract $infty-infty$, whereas our measures are projection-valued and we can always subtract projections.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 30 at 2:48









          Eric Wofsey

          166k12195308




          166k12195308











          • I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
            – B. Chinaski
            Aug 30 at 3:06
















          • I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
            – B. Chinaski
            Aug 30 at 3:06















          I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
          – B. Chinaski
          Aug 30 at 3:06




          I thought you might end up on a issue like the one I mentioned when dealing with unbounded self-adjoint operators, for example... but what you said makes sense.
          – B. Chinaski
          Aug 30 at 3:06

















           

          draft saved


          draft discarded















































           


          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2899014%2fwhy-a-spectral-resolution-should-be-left-continuous%23new-answer', 'question_page');

          );

          Post as a guest













































































          這個網誌中的熱門文章

          tkz-euclide: tkzDrawCircle[R] not working

          How to combine Bézier curves to a surface?

          1st Magritte Awards