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 .