On 2013-04-24, Dan <email@example.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 firstname.lastname@example.org Phone: (765)494-6054 FAX: (765)494-0558