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

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Herman Rubin

Posts: 399
Registered: 2/4/10
Re: Ask-an-Analysis problem
Posted: Apr 24, 2013 1:52 PM
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On 2013-04-24, Dan <> wrote:
> On Apr 24, 5:38 pm, wrote:
>> On Wed, 24 Apr 2013 02:19:43 -0700, William Elliot <>
>> 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 Phone: (765)494-6054 FAX: (765)494-0558

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