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Topic: Ask-an-Analysis problem
Replies: 10   Last Post: Apr 25, 2013 1:01 PM

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dan.ms.chaos@gmail.com

Posts: 409
Registered: 3/1/08
Re: Ask-an-Analysis problem
Posted: Apr 25, 2013 6:50 AM
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On Apr 24, 8:52 pm, Herman Rubin <hru...@skew.stat.purdue.edu> wrote:
> On 2013-04-24, Dan <dan.ms.ch...@gmail.com> wrote:
>

> > On Apr 24, 5:38 pm, dullr...@sprynet.com wrote:
> >> On Wed, 24 Apr 2013 02:19:43 -0700, William Elliot <ma...@panix.com>
> >> wrote:

> >> >Assume for f:[0,1] -> R that there's some c /= 0,1 with
> >> >for all x in [0,1/2], f(x) = c.f(2x).
> >> >Show there's some k with for all x in [0,1], f(x) = kx.

>
> For one thing, the theorem is false.  The most one can hope
> to get is that f(x)=kx^q for some k and q>0.  On the open
> interval, q can even be negative.
>

> >> Not true. It's well known that there exists a nowhere-
> >> continuous function f : R -> R such that
> >> f(x+y) = f(x) + f(y) for all x, y.
> >> Maybe you omitted a hypothesis? Seems like it may be
> >> true for continuous f...

>
> One does not need to go that far.  The general solution is
> f(x)=x^q*h(log_2(x)), where h is any function.  Continuous
> does not get farther.  However, all derivatives bounded
> will get q to be a positive integer and h constant.
>

> > Maybe if you restrict it such that f:[0,1] -> [0,1]?
> > Just a hunch,not sure.

>
> It can be seen from the above that this will not suffice.
> Even requiring that f be 1-1 will not suffice to get more
> than the general solution with limits on h, but not just
> the multiple of a power.
>
> --
> This address is for information only.  I do not claim that these views
> are those of the Statistics Department or of Purdue University.
> Herman Rubin, Department of Statistics, Purdue University
> hru...@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558


I was referring to the stronger set of conditions , namely f(x+y) =
f(x) + f(y) , and f:[0,1]->[0,1] , not the op's problem .Although this
might seem off topic .



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