On the lacunary function $f(z) = \frac{r + f(-1)}{\sqrt 5}$

Question

Consider the lacunary series

$$ f(z) = \sum_{n=1}^{\infty} a_n z^n $$

with radius $1$ and natural boundary at the unit circle.
The $a_n$ are real and strict positive, and also

$$\sum_{n=1}^{\infty} a_n = 1 $$

and the boundary thus bounded.
Moreover

$$\sum_{n=1}^{\infty} n^k a_n < \infty $$

for every integer $k$.
Hence the sequence $1/a_n$ is subexponential but superpolynomial, and it follows the boundary value of $f$ on the natural boundary is bounded and even infinitely differentiable ($C^{\infty}$).
Let $r$ be a complex number strictly within the unit circle, i.e. $|r| < 1$.

Question. Is it true that the identity

$$f(z) = \frac{r + f(-1)}{\sqrt 5}$$

holds for every $f$ and every $r$?

Answer 1

Let's suppose without restriction in generality that $f(1) = 1$ and $f(-1) = -1$, and moreover notice that

• $f$ only has 2 fixed points being $+1,-1$, and
• all iterates of $f$ converge to them or to $0$.

Now consider

$$g(z) = f(f(f(z)))=f^{\circ 3}(z).$$

Clearly since $f$ is lacunary, $g(z) - f(z) \neq 0$ within radius $1$.
Also $g(z)$ maps $z$ to smaller $z$ near the boundary, therefore $g(z)$ is more or less a contraction.

But $g$ is of the same type of $f$.

Therefore for a sufficiently large number of iterations of $g$ we have a counterexample.