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

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David C. Ullrich

Posts: 3,016
Registered: 12/13/04
Re: Ask-an-Analysis problem
Posted: Apr 24, 2013 7:16 PM
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On Wed, 24 Apr 2013 17:52:09 +0000 (UTC), Herman Rubin
<hrubin@skew.stat.purdue.edu> wrote:

>On 2013-04-24, Dan <dan.ms.chaos@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.


??? Since log_2 is 1-1, allowing h to be any function
here says that f can be 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.





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