# Continuous function

• Jun 19th 2010, 07:08 AM
Chandru1
Continuous function
f is a continuous function from R->R and f satisfies f(2x+1)=f(x), then prove that f is constant.
• Jun 19th 2010, 07:14 AM
Ackbeet
Is f also differentiable?
• Jun 19th 2010, 07:36 AM
Chandru1
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.
• Jun 19th 2010, 07:40 AM
Bruno J.
It doesn't need to be differentiable!

Hint : show that $\displaystyle f$ is constant on a dense subset of $\displaystyle \mathbb{R}$.
• Jun 19th 2010, 10:00 AM
JG89
If it's of any help, f is differentiable at x = 1 with derivative 0, and f(1) = f(0).

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

And $\displaystyle f(0) = f(1 + 2*0) = f(1)$.
• Jun 19th 2010, 06:57 PM
Chandru1
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.