1. Continuous function

f is a continuous function from R->R and f satisfies f(2x+1)=f(x), then prove that f is constant.

2. Is f also differentiable?

3. Hi--

No f is not differentiable. Ok, suppose you have an counterexample then please do provide me with one or else, prove the result with differentiability holding true. As far as i know, i dont think f is differntiable.

4. It doesn't need to be differentiable!

Hint : show that $f$ is constant on a dense subset of $\mathbb{R}$.

5. If it's of any help, f is differentiable at x = 1 with derivative 0, and f(1) = f(0).

$\frac{f(1+2h) - f(1)}{2h} = 0$ for all non-zero h.

And $f(0) = f(1 + 2*0) = f(1)$.

6. hi

Hi--

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.