1. ## Trouble with Integration

f:R ---> R is a continuous function and f(x) = f(2x) is true for all real numbers
If f(1)=3 then find the value of
$\displaystyle \int_{-1}^{1}f[f(x)] dx$

2. ## Re: Trouble with Integration

Originally Posted by grgrsanjay
f:R ---> R is a continuous function and f(x) = f(2x) is true for all real numbers If f(1)=3 then find the value of $\displaystyle \int_{-1}^{1}f[f(x)] dx$
Necessary condition: Taking into account that $\displaystyle g(x)=3$ is continuous in $\displaystyle \mathbb{R}$ and satisfies $\displaystyle g(x)=g(2x)\;\wedge \;g(1)=3$ for all $\displaystyle x\in \mathbb{R}$ then, if the solution does not depend on $\displaystyle f$ we necessarily have $\displaystyle \int_{-1}^{1}f[f(x)] \;dx=\int_{-1}^{1}3 \;dx=6$ .

3. ## Re: Trouble with Integration

We can show that f is necessarily constant.

Since $\displaystyle f(x) = f(2x)$ for all x,
$\displaystyle f(x/2) = f(x)$ .....(just substitute x/2 for x)
$\displaystyle f(x/(2^2)) = f(x)$ ..... similarly
...
$\displaystyle f(x/(2^n)) = f(x)$

Now let $\displaystyle n \to \infty$ and apply the continuity of f.

4. ## Re: Trouble with Integration

Another way: the solutions of the functional equation $\displaystyle f(2x)=f(x)$ are $\displaystyle f(x)=\phi(\log x)$ where $\displaystyle \phi$ is any periodic function with period $\displaystyle \log 2$ . According to the hypothesis, necessarily $\displaystyle f(x)=3$ .