On Wed, 24 Apr 2013 19:06:20 -0700, William Elliot <firstname.lastname@example.org> wrote:
>On Wed, 24 Apr 2013, email@example.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. > >Whoops, indeed a hypothesis was omitted. > >Assume for continuous f:[0,1] -> R here'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.
As has been pointed out several times already, there are trivial counterexamples, like f(x) = x^2.
> >> Forget what I said this morning. The question was either >> totally garbled or totally stupid to begin with. >> f(x) = x^2 is a counterexample. > >You could go to Ask-an-Algebraist forum at > at.yorku.ca/topology >to give your simple disproof directly to the primary source. >