Does there exist an entire function $f$ such that $f(z)+f(z^2)=z^3$ for all $z \in \mathbb{C}$?
This problem showed up on a qualifying exam; these exams enjoy including problems that utilize Liouville's theorem.
Method 1: Show $f$ is bounded and use Liouville
Here, I would suppose $f$ is entire and non-constant. If I can show that $f$ is bounded, I will have a contradiction which means that $f$ is constant.
Method 2: Use Taylor Expansion Since $f$ is entire, $f$ has a Taylor series expansion about $z=0$. Thus,
$$f(z) = f(0) + f'(0)z + \frac{f''(0)}{2}z^2 + \dots \tag{1}$$
By hypothesis,
$$f(z) = z^3-f(z^2) \tag{2}$$
But now, we can write
$$f(z^2) = f(0) + f'(0)z^2 + \frac{f''(0)}{2}z^4 + \dots \tag{3}$$
Then, rewriting $(2)$, we get
$$f(0) + f'(0)z + \frac{f''(0)}{2}z^2 + \dots = z^3 – (f(0) + f'(0)z^2 + \frac{f''(0)}{2}z^4 + \dots) \tag{4}$$
I seek to derive a contradiction here but I experience no such luck. I'm suspecting that the first method of showing $f$ is bounded might be a better bet. Any ideas?
Best Answer
Expanding @Empy2's comment, suppose $f$ is an entire function that solves the given functional equation. Then by applying the equation twice,
$$ f(z) = z^3 - f(z^2) = z^3 - z^6 + f(z^4). $$
Recursively applying this relation, we get
$$ f(z) = \sum_{k=0}^{2N-1} (-1)^k z^{3\cdot 2^k} + f(z^{4^N}). \tag{1}\label{e:1}$$
Also note that plugging $z = 0$ to the functional equation gives $f(0) = 0$. Hence, as $N \to \infty$, the right-hand side of the above identity converges to
$$ f(z) = \sum_{k=0}^{\infty} (-1)^k z^{3\cdot 2^k}. $$
So by the uniqueness of power series, the right-hand side must be the power series expansion of $f(z)$ about $z = 0$. However, it is clear that the right-hand side has radius of convergence $R = 1$, a contradiction! For example,
$$ f(e^{2 \pi i/9}) = \sum_{k=0}^{\infty} (-1)^k e^{2^{k+1}\pi i/3} = \sum_{k=0}^{\infty} \frac{(-1)^{k+1} + i\sqrt{3}}{2} $$
and this diverges.
Alternative solution. The last line suggests another solution: Plugging $z = e^{2\pi i/9}$ and $N = 6$ to $\eqref{e:1}$, we get
\begin{align*} f(e^{2\pi i/9}) &= \sum_{k=0}^{11} (-1)^k e^{2^{k+1}\pi i/3} + f(z^{2^{13}\pi i/9}) \\ &= 6 i\sqrt{3} + f(e^{2\pi i/9}), \end{align*}
a contradiction!