Positively A Function

Fix some $a > 0$ and let $n = \frac{f(a)}{a}$, so that $f(a) = na$. Then by definition of $f$, we must have $n \ge 0$. For $k \in \mathbb{N}$, we denote $k$ applications of $f$ to $a$ by writing $f^k (a)$ (i.e. $f^1 (a) = f(a)$, $f^2 (a) = f(f(a))$, etc.)

It's easy to show, by induction, that for all $k \in \mathbb{N}$, we have: $f^k (a) = \frac{1}{5} \left((-3)^k(2-n) + 2^k(n+3) \right) a$

Suppose that $0 \le n < 2$. By the Archimedean Principle, we can find some $k \in \mathbb{N}$ such that $k > log_{3/2} \left(\frac{n+3}{2-n} \right)$ and such that $k$ is odd. For this value of $k$, we have $(-3)^k(2-n) + 2^k(n+3) < 0$, so that $f^k (a) < 0$. This is a contradiction since the range of $f$ is positive numbers.

Now suppose that $n > 2$. By the Archimedean Principle, we can find some $k \in \mathbb{N}$ such that $k > log_{3/2} \left(\frac{n+3}{n-2} \right)$ and such that $k$ is even. For this value of $k$, we have $(-3)^k(2-n) + 2^k(n+3) < 0$, so that $f^k (a) < 0$. This is a contradiction since the range of $f$ is positive numbers.

Therefore, we must have $n = 2$, so $f(a) = 2a$.

Therefore, if we unfix $a$, we get that $f(x) = 2x$ for all $x > 0$.

Hint: Look at the tags.

We let $x=a_0$ and $f^n(x)=a_n$. Plugging in $x=f^{n}(x)$ gives $a_{n+2}=6a_n-a_{n+1}$.

The characteristic polynomial of this recurrence relation is $c^2+c-6=0$ which has roots $c=2,-3$. Thus we can represent $a_n=\lambda_1(2)^n+\lambda_2(-3)^n$

all we need to do is find a function that satisfies $f(\lambda_1(2)^{n-1}+\lambda_2(-3)^{n-1})=\lambda_1(2)^n+\lambda_2(-3)^n$.

Now note that $(-3)^n$ grows faster than $(2)^n$, no matter what constants $\lambda_1,\lambda_2$ we pick. However, since $(-3)^n$ oscillates between positive and negative, that would imply that when the function is applied to some positive $x$, then $f(x)$ is negative! Thus, we must have $\lambda_2=0$. Therefore, we want $f(\lambda_1(2)^{n-1})=\lambda_1(2)^{n}$ Subbing $x=\lambda_1(2^{n-1})$, we see that $\boxed{f(x)=2x}$ which is our only solution.

Does this work?

Looking at the functional equation, it is apparent that f(x) must be linear and that it cannot have a constant term. Thus f(x)=k*x. To find the values that k can assume, substitute in the functional equation: k^2=k-6; solutions are k=-3 or 2. Since the function is defined in Positive Real space, it has to be f(x)=2x.

F(x) equal to -3x

f(x)=-3x is a solution. I found it by solving the quadratic equation f^2+f-6=0. Do not ask me why :-)

I can say that there are no other solutions but I may assume that the function is a polynomial since from the given (somehow confusing logic) that f is a polynomial function. But the assumption...

I found 2 solutions: f(x) = 2x, and f(x) = -3x. :)