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Re: [harmonic] L^/infinity functions

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  • 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 1 of 7 , Feb 13, 2007
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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 2 of 7 , Feb 13, 2007
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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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      • 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 3 of 7 , Feb 14, 2007
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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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                                                                             Albert Schweitzer
        • 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 4 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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