# Functional equation f(x)=f(x^2)

1. Jan 14, 2006

### cliowa

Hello everybody
I'm given a continuous function f (from the real numbers to the real numbers) which I know obeys the following functional equation:
$$f(x)=f(x^{2})$$
How can I proof that this function is constant?
I started out like this: Looking at a number x in $$[0,1[$$ I said to myself that:
$$f(x)=f(x^{2})=f(x^{4})=f(x^{16})=...\rightarrow f(0).$$.
As this relation holds for all x in [0,1[ (looking at $$x^{2^{n}} (n \rightarrow \infty$$), it should be true that f(x)=f(-x)=f(0) (the minus because of symmetry, the square). I can also conclude that f(0)=f(1), for f is continuous. But how do I proceed? I would like to show that this function is constant for all x in the real numbers?
Any help will be appreciated.
Regards...Cliowa

Last edited: Jan 14, 2006
2. Jan 14, 2006

### shmoe

If x is a positive real, you also have $$f(x)=f(\sqrt{x})$$. Use this for x>1.

3. Jan 14, 2006

### cliowa

Oh my gosh, I should have gotten that one on my own.
Now it seems clear. Thanks alot, shmoe.
Best regards
Cliowa

4. Jan 15, 2006

### benorin

If $$f:\mathbb{R}\rightarrow\mathbb{R}$$ is a continuous function such that $$f(x)=f(x^{2}),\forall x\in\mathbb{R}$$, then

i. f is an even function since $$f(x)=f(x^{2})\Rightarrow f(-x)=f((-x)^{2})=f(x^{2})=f(x)$$

ii. $$f(x)=f(x^{2^{n}}),\forall n\in\mathbb{Z}$$ which can be proved inductively, or simply reasoned much as you have above:

for positive integer n,

$$f(x)=f(x^{2})=f(x^{4})=f(x^{16})=\cdots =f(x^{2^{n}})$$

and for negative integer n=-k,

$$f(x)=f(\sqrt{x})=f(\sqrt{\sqrt{x}})=f(\sqrt{\sqrt{\sqrt{x}}})=\cdots =f(x^{2^{-k}})$$