
Re: AskanAnalysis problem
Posted:
Apr 24, 2013 12:54 PM


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. > > 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...
Maybe if you restrict it such that f:[0,1] > [0,1]? Just a hunch,not sure.

