Solving the functional equation $f(2x) = 4f(x)^2$ for a continuous $f$ from $\Bbb{R}$ to $\Bbb{R}$

Question

I found an interesting functional equation:

$f: \mathbb{R}\to \mathbb{R}$ is continuous $$f(2x) = 4f(x)^2$$

I've found the following solutions:

$f(x) = 0$, and $f(x) = 1/4 a^x$ where $a$ is an arbitrary non-negative real number.

I can't prove uniqueness but I've noticed the following:

1. If $f(x)$ is a solution then $f(x)a^x$ is also a solution.

2. Using induction you can prove that if the solution is unique on an arbitrarily small open interval $(-\epsilon,\epsilon)$, it must be unique over the whole real line.

I also wondered about an extension of this problem:

$f: \mathbb{R}\to\mathbb{R}$ is continuous

$$f(2x) = 4f(x)^2 + c$$

for arbitrary $c$ element of $\mathbb{R}$.

Any ideas?

Answer 1

As Ivan Neretin says in the comments solutions are extremely non-unique. Here is a large class of solutions, using the fact that $g(x) = 4x^2$ is invertible if we restrict to the non-negative reals. Somewhat more generally let's consider

$$f(2x) = g(f(x))$$

where $g$ is invertible on the non-negative reals. We can choose the values of $f$ on the interval $[1, 2]$ to be any non-negative continuous function such that $f(2) = g(f(1))$. This determines the value of $f$ on any interval $[2^k, 2^{k+1}], k \in \mathbb{Z}$ since we must have

$$f(2^k x) = g^k(f(x)), x \in [1, 2]$$

where $g^k$ denotes the $k$-fold composition. The condition that $f(2) = g(f(1))$ guarantees $f(2^k) = g(f(2^{k-1}))$ which guarantees continuity on the positive reals.

To get continuity at $0$ we need the limit $L = \lim_{k \to - \infty} g^k(f(1))$ to exist, and then we can do the same thing for the negative reals, where the values of $f$ are determined by its values on the interval $[-2, -1]$ subject to the condition that $f(-2) = g(f(-1))$, and the limit $\lim_{k \to - \infty} g^k(f(-1))$ must exist and equal $L$. $L$ must be a fixed point of $g$, so if $g$ has no fixed points there are no solutions. And $f(1)$ and $f(-1)$ must be in the basin of attraction of $L$ with respect to the action of $g^{-1}$.

Edit: Sorry, the condition for continuity at $0$ is much stronger than this, I'm being silly. We actually need $f(0) = L = \lim_{k \to - \infty} g^k(f(x))$ to exist for every positive $x$, or equivalently for every $x \in [1, 2]$. So every value of $f$ must be in the basin of attraction of the fixed point $L$.

$g(x) = 4x^2$ has fixed points $x = 0, \frac{1}{4}$, so those are the possible values of $L = f(0)$ in this case. With respect to the action of $g^{-1}(x) = \sqrt{ \frac{x}{4} }$, the basin of attraction of $\frac{1}{4}$ is $(0, \infty)$, so if $f(x) > 0$ for $x > 0$ then $f(0) = \frac{1}{4}$ and if $f(x)$ ever takes the value $0$ then $f(0) = 0$ which gives $f(x) = 0$ identically.

For $g(x) = 4x^2 + c$ the analysis gets more annoying due to lack of invertibility. The fixed point equation $4x^2 - x + c = 0$ has discriminant $\Delta = 1 - 16c$ so there are no fixed points for $c > \frac{1}{16}$.