# Finding $f(x)$ of functional equation

I would appreciate if somebody could help me with the following problem:

Q: Find all conti-function $f(x)~ (x>0)$

$$xf(x^2)=f(x)$$

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please considering accept an answer. –  Ma Ming Apr 30 '14 at 23:11

Since $x^2f(x^2)=xf(x)$, let $g(x):=x(f)$, then $g(x^2)=g(x)$.

It follows that $g(x)=\lim_{n\to \infty} g(\sqrt[2^n]{x})=g(1):=C$. So $f(x)=C/x$.

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$xf(x^2)=f(x)$

$x^2f(x^2)=xf(x)$

Let $g(x)=xf(x)$ ,

Then $g(x^2)=g(x)$

Let $\begin{cases}x_1=\log_2x\\g_1(x_1)=g(x)\end{cases}$ ,

Then $g_1(2x_1)=g_1(x_1)$

Let $\begin{cases}x_2=\log_2x_1\\g_2(x_2)=g_1(x_1)\end{cases}$ ,

Then $g_2(x_2+1)=g_2(x_2)$

$g_2(x_2)=\Theta(x_2)$ , where $\Theta(x_2)$ is an arbitrary periodic function with unit period

$g(x)=\Theta(\log_2\log_2x)$ , where $\Theta(x)$ is an arbitrary periodic function with unit period

$\therefore f(x)=\dfrac{\Theta(\log_2\log_2x)}{x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period

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