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Re: [harmonic] L^/infinity functions

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  • wilson@cems.uvm.edu
    Quoting gnavada : It s false.
    Message 1 of 7 , Feb 13 9:43 AM
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      Quoting gnavada <gnavada@...>:

      It's false.

      > Hello,
      > I need a proof of the following fact: L^/infinity functions in R can
      > be made continuous by changing the values of functions on a set of
      > measure zero. Even a reference to the proof will help.
      > thanks
      > gowri
      >
      >
    • venku naidu
      Hai, I think that is not true in general. Particular you can visualise this by taking a step function on a bounded interval. bye.... wilson@cems.uvm.edu wrote:
      Message 2 of 7 , Feb 13 8:18 PM
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        Hai,
        I think that is not true in general. Particular you can visualise this by taking a step function on a bounded interval.
        bye....

        wilson@... wrote:
        Quoting gnavada <gnavada@yahoo. com>:

        It's false.

        > Hello,
        > I need a proof of the following fact: L^/infinity functions in R can
        > be made continuous by changing the values of functions on a set of
        > measure zero. Even a reference to the proof will help.
        > thanks
        > gowri
        >
        >




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      • Gowri Navada
        Hi Thanks. I got similar mails from others saying it is not true. I havnt checked a counterexample yet, but I ll do it. In page 192 of Rudin s Real and
        Message 3 of 7 , Feb 13 10:07 PM
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          Hi
          Thanks. I got similar mails from others saying it is
          not true. I havnt checked a counterexample yet, but
          I'll do it.
          In page 192 of Rudin's "Real and Complex Analysis", he
          uses a /beta which is in L^/infinity and then he says
          (after equation (5)) that by changing /beta(y)on a set
          of measure 0 we may assume it is continuous. If you
          have some time to spare can u pl. go through this pf
          and explain that step?
          Thank you very much
          regards,
          gowri

          --- venku naidu <venku_uoh@...> wrote:

          > Hai,
          > I think that is not true in general. Particular
          > you can visualise this by taking a step function on
          > a bounded interval.
          > bye....
          >
          > wilson@... wrote:
          > Quoting gnavada <gnavada@...>:
          >
          > It's false.
          >
          > > Hello,
          > > I need a proof of the following fact: L^/infinity
          > functions in R can
          > > be made continuous by changing the values of
          > functions on a set of
          > > measure zero. Even a reference to the proof will
          > help.
          > > thanks
          > > gowri
          > >
          > >
          >
          >
          >
          >
          >
          >
          > D.VENKU NAIDU,
          > CAUVERY HOSTEL,
          > ROOM NO:2045,
          > IIT,MADRAS,
          > CHENNAI-36
          >
          >
          > ---------------------------------
          > Now you can scan emails quickly with a reading
          > pane. Get the new Yahoo! Mail.




          ____________________________________________________________________________________
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          Get your game face on with the latest PS3 news and previews at Yahoo! Games.
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        • Bela A. Frigyik
          Hi Gowri, Look at equation (5) on pg. 192. We assumed, that phi(f) is not zero and that beta(y) is in L^ infinity. On the other hand, since the map f to f_y
          Message 4 of 7 , Feb 14 10:39 AM
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            Hi Gowri,

            Look at equation (5) on pg. 192. We assumed, that \phi(f) is not zero and that \beta(y) is in L^\infinity. On the other hand, since the map f \to f_y (translation by 'y') is continuous and the function \phi is a continuous functional we have a continuous function on the right hand side. Eq. (5) says that they are the same a.e. Change the value of  \beta(y) to (1/\phi(f)) \phi(f_y) for all 'y' at which eq. (5) doesn't hold. You end up with a new \beta function that agrees with the old one a.e. and agrees with (1/\phi(f)) \phi(f_y) everywhere. This latest function is continuous.

            Note, that the situation is special since the measure '\beta dm' comes from the Riesz representation theorem and definitely not arbitrary!

            Bela

            On 2/14/07, Gowri Navada <gnavada@...> wrote:

            Hi
            Thanks. I got similar mails from others saying it is
            not true. I havnt checked a counterexample yet, but
            I'll do it.
            In page 192 of Rudin's "Real and Complex Analysis", he
            uses a /beta which is in L^/infinity and then he says
            (after equation (5)) that by changing /beta(y)on a set
            of measure 0 we may assume it is continuous. If you
            have some time to spare can u pl. go through this pf
            and explain that step?
            Thank you very much
            regards,
            gowri

            --- venku naidu <venku_uoh@...> wrote:

            > Hai,
            > I think that is not true in general. Particular
            > you can visualise this by taking a step function on
            > a bounded interval.
            > bye....
            >
            > wilson@... wrote:
            > Quoting gnavada <gnavada@...>:
            >
            > It's false.
            >
            > > Hello,
            > > I need a proof of the following fact: L^/infinity
            > functions in R can
            > > be made continuous by changing the values of
            > functions on a set of
            > > measure zero. Even a reference to the proof will
            > help.
            > > thanks
            > > gowri
            > >
            > >
            >
            >
            >
            >
            >
            >
            > D.VENKU NAIDU,
            > CAUVERY HOSTEL,
            > ROOM NO:2045,
            > IIT,MADRAS,
            > CHENNAI-36
            >
            >
            > ---------------------------------
            > Now you can scan emails quickly with a reading
            > pane. Get the new Yahoo! Mail.

            __________________________________________________________
            Be a PS3 game guru.
            Get your game face on with the latest PS3 news and previews at Yahoo! Games.
            http://videogames.yahoo.com/platform?platform=120121




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          • zaher_hani
            Hi Gowri; I think the best you can do in approximating such function is Lusin s Theorem, which is sometimes very useful. Anyways, here I provide a counter
            Message 5 of 7 , Mar 3, 2007
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              Hi Gowri;

              I think the best you can do in approximating such function is Lusin's
              Theorem, which is sometimes very useful. Anyways, here I provide a
              counter example to your result (which I hope is right).

              There exists a sequence {En} of disjoint measurable sets on R such
              that
              (i) the measure of En is less than 2^n
              (ii)there exists a countable basis {In} of the topology of R then En
              intersects In in a set of positive measure for every n
              (note that you use this construction to solve exercise 8 of Rudin
              Chapter 2)

              Now consider the function f=sigma{(1/2)^n * (Characteristic function
              of En)/ (measure of En)}. Then since the En are disjoint f converges
              everywhere. It is easy to see that the L1 norm of f is finite, but its
              integral on any interval of R is infinite. This last fact proves that
              even if you change f on a set of measure 0, it will remain
              discontinuous.

              But you tell me: "f may not be in L^infinity". Then take the function
              g=exp{-f} then g is bounded and is continuous if and only if f is
              continuous (here we needed the fact that f is never infinite, which
              explains my choice of this particular f rather than more easier
              examples satisfying almost the same properties as f).

              g is in L^infinity (bounded). If I can change g on a set of measure 0
              to make it continuous then the same is true for f, which is a
              contradiction.

              Please check the above counter example and keep me in touch if
              something is wrong.
              Best Regards;
              Zaher Hani
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