Loading ...
Sorry, an error occurred while loading the content.

L^/infinity functions

Expand Messages
  • gnavada
    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.
    Message 1 of 7 , Feb 12, 2007
    • 0 Attachment
      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
    • Maria Roginskaya
      What you want is not true. The fact you mean is that you can do it with a set of arbitrary small measure.
      Message 2 of 7 , Feb 13, 2007
      • 0 Attachment
        What you want is not true. The fact you mean is that you can do it with a
        set of arbitrary small measure.

        > 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
        >
        >
      • wilson@cems.uvm.edu
        Quoting gnavada : It s false.
        Message 3 of 7 , Feb 13, 2007
        • 0 Attachment
          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 4 of 7 , Feb 13, 2007
          • 0 Attachment
            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
            >
            >




            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.

          • 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 5 of 7 , Feb 13, 2007
            • 0 Attachment
              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
            • 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 6 of 7 , Feb 14, 2007
              • 0 Attachment
                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




                --
                "Az ember egy okos állat, amelyik úgy viselkedik, mint egy hülye."
                "Man is a clever animal who behaves like an imbecile."
                                                                                   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 7 of 7 , Mar 3, 2007
                • 0 Attachment
                  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
                Your message has been successfully submitted and would be delivered to recipients shortly.