How to find the function $f$ given $f(f(x)) = xf(x)$?

Question

I was wondering if there is a continuous function such that $f(f(x)) = xf(x)$ for every positive number $x$.

Answer 1

Sure. $$ f(x) = x^{\frac{1 + \sqrt 5}{2}} $$

Answer 2

This is not yet a full answer for the proof, but possibly it is a good step to one. I also think that the problem is not more than a standard exercise in some textbook, but since there is not yet a more qualified answer here, I'll do some naive try so far...

To save notation, let#s write the h'th iterate $\underset{h \text{ times }}{\underbrace {f(...f(f(x)))}}$ as $x_h$ and its p'th power as $x_h^p$ where we understand, that the superscript gets evalauted after the subscript.

Then we can state the sequence:
$$ x = x_{-2} \cdot x_{-1} \\ x = x_{-4} \cdot x_{-3}^2 \cdot x_{-2} \\ x = x_{-6} \cdot x_{-5}^3 \cdot x_{-4}^3 \cdot x_{-3} \\ x = x_{-8} \cdot x_{-7}^4 \cdot x_{-6}^6 \cdot x_{-5}^4\cdot x_{-4} \\ \cdots $$ We observe, that the exponents are the binomial coefficients if powers of 2 $(=(1+1))$ are expanded. Now the idea is, to hope, that we can introduce a limit and that we can assume, that in the limit the difference between the iterates become insignificant below some epsilon, such that we can write $$ x = \lim (x_{-2h})^{2^h} $$ If we assume, that $x_{-2h}<x_{-h}$ then we can even write $$ (x_{-2h})^{2^h}< x < (x_{-h})^{2^h} $$ or $$ (x_{-2h+1})^{2^h}< f(x) < (x_{-h+1})^{2^h} $$ and then $$ (x_{-\infty} + \epsilon_1)^{2^h}< f(x) < (x_{-\infty} + \epsilon_2)^{2^h} $$ and then from a vanishing difference $\epsilon_1 - \epsilon_2 $ deduce, that the h'th iterate of f is necessarily of the form of the h'th iterate of a power $ax^b$ with some fixed a and b . Here I'm stuck because I've not much experience with the formal handling of such limts, but perhaps this is an intuitive path where one can proceed further...

Answer 3

In fact this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2315.pdf.

Let $\begin{cases}x=u(t)\\f=u(t+1)\end{cases}$ ,

Then $u(t+2)=u(t)u(t+1)$

Let $u(t)=e^{v(t)}$ ,

Then $e^{v(t+2)}=e^{v(t)}e^{v(t+1)}$

$e^{v(t+2)}=e^{v(t)+v(t+1)}$

$v(t+2)=v(t)+v(t+1)+2n\pi i$ , $\forall n\in\mathbb{Z}$

$v(t+2)-v(t+1)-v(t)=2n\pi i$ , $\forall n\in\mathbb{Z}$

Let $v(t)=v_c(t)+A$ ,

Then $v_c(t+2)+A-(v_c(t+1)+A)-(v_c(t)+A)=2n\pi i$

$v_c(t+2)-v_c(t+1)-v_c(t)-A=2n\pi i$

$\therefore A=-2n\pi i$

For $v_c(t+2)-v_c(t+1)-v_c(t)=0$ ,

$v_c(t)=C_1(t)\left(\dfrac{1+\sqrt{5}}{2}\right)^t+C_2(t)\left(\dfrac{1-\sqrt{5}}{2}\right)^t$ , where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period

$\therefore v(t)=C_1(t)\left(\dfrac{1+\sqrt{5}}{2}\right)^t+C_2(t)\left(\dfrac{1-\sqrt{5}}{2}\right)^t-2n\pi i$ , $\forall n\in\mathbb{Z}$ , where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period

Hence $u(t)=e^{C_1(t)\left(\frac{1+\sqrt{5}}{2}\right)^t+C_2(t)\left(\frac{1-\sqrt{5}}{2}\right)^t-2n\pi i}$ , $\forall n\in\mathbb{Z}$ , where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period

$u(t)=e^{C_1(t)\left(\frac{1+\sqrt{5}}{2}\right)^t}e^{C_2(t)\left(\frac{1-\sqrt{5}}{2}\right)^t}$ , where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period

$\therefore\begin{cases}x=e^{C_1(t)\left(\frac{1+\sqrt{5}}{2}\right)^t}e^{C_2(t)\left(\frac{1-\sqrt{5}}{2}\right)^t}\\f=e^{C_1(t)\left(\frac{1+\sqrt{5}}{2}\right)^{t+1}}e^{C_2(t)\left(\frac{1-\sqrt{5}}{2}\right)^{t+1}}\end{cases}$ , where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period