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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
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
>
>
> ---------------------------------
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