On Wed, 24 Apr 2013 17:52:09 +0000 (UTC), Herman Rubin <firstname.lastname@example.org> wrote:
>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.
??? 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.