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

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- Jun 19th 2010, 07:08 AMChandru1Continuous 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 AMAckbeet
Is f also differentiable?

- Jun 19th 2010, 07:36 AMChandru1
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 AMBruno 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 AMJG89
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 PMChandru1hi
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.