Is f also differentiable?
How do we prove it for dense subsets. The only way i can think of is prove this true for rationals since they are dense in R.
But i have an other idea. Let g(x)=f(x-1)=> g(2x)=f(2x-1)=f(x-1)=g(x). But by doing this repeatedly we can have g(x)->g(0). Does this work.