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

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

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$f(x) = 0$ for $\forall x$ :) – Patrick Li Sep 29 '12 at 5:14
You must have $f(1)=0$ or $1$. – Henry Sep 29 '12 at 5:57
Did you mean to require $f(x)$ to be positive as well? – Hurkyl Sep 30 '12 at 13:45
At the invertible places, the equation also reads $f(x)=x\cdot f^{-1}(x)\$. – NikolajK Sep 30 '12 at 21:18

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

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How to get that? – Patrick Li Sep 29 '12 at 5:16
try $x^\beta$ and solve for $\beta$ – Will Jagy Sep 29 '12 at 5:25
That's a clever idea. How about functions with other forms? – Patrick Li Sep 29 '12 at 5:38
Is there a proof that leads to f(x) = x^{\frac{1 + \sqrt 5}{2}} – Geokal Sep 29 '12 at 5:41
@Geokal: You can easily prove $f(x) = x^{\frac{1 + \sqrt 5}{2}}$. You cannot prove it is the only solution as there is at least one more which is similar but with a different coefficient. – Henry Sep 29 '12 at 5:51

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...

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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

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While a neat approach, there's still a fairly large gap between the solution to the problem you solved and the solution to the problem asked. This gap includes checking which of your solutions make $f$ a well-defined partial function of $x$, which solutions are real, and checking how (and if!) different solutions can be patched together to define an $f$ as a total function of $x$. (Also, the OP hasn't specified whether $f(x)$ is allowed to be non-positive which makes a difference, since he doesn't require the equation to hold for nonpositive $x$) – Hurkyl Sep 30 '12 at 21:09
@ Hurkyl:If we require f(x) to be positive ,as you said above, then are things easier ? – Geokal Oct 1 '12 at 18:50
@Geokal: It means the functional equation holds for every element of the image of $f$. If $f(x)$ is allowed to be negative, but also allow $f(f(x)) \neq x f(x)$ for negative $x$, things become trickier. – Hurkyl Oct 1 '12 at 19:16