Find all entire functions that satisfy the following equality

Question

For $n\geq 2$ I need to find all entire functions $f:\mathbb{C} \rightarrow \mathbb{C}$ Such that $f(z^n)=f(z)^n$ for all complex numbers $z$. I've tried expanding it's series at 0 and have found some relations with the coefficients, but none seem to finish the problem. I couldn't conclude using Liouville's theorem neither.. My gut feeling tells me that the only the functions of the form $z^k$ will work, but I am a bit stuck.

Answer 1

Perhaps there is a more elegant and self-contained way of doing this, but I think we can use normal numbers to our advantage. I'll focus on the specific case of $n=2$, but similar arguments should work for other natural values of $n$. This is not a complete proof, but I think it's a good starting point.

Let $\omega$ be a normal number in base $2$, and let $f$ be an entire function satisfying $f(z^2) = f(z)^2$ for all $z\in\mathbb C$. Because $\omega$ is normal, we have that the numbers taking the form $$a_n = \exp\Big(2\pi i \omega \cdot 2^n\Big)$$ are dense in the unit circle. Further, we have that $$f(a_n) = f(a_0)^{2^n}$$ By the identity principle, since the $a_n$ have an accumulation point in $\mathbb C$ (because they are dense in $\mathbb S^1$) this means that $f$ is uniquely determined by its image of $a_0$.

Note that if $|f(a_0)| > 1$, then $f$ is unbounded on the unit circle, which is impossible for an entire function, so this cannot be. On the other hand, if $|f(a_0)| < 1$, then $f$ gets arbitrarily close to zero in arbitrary neighborhoods of any point on the unit circle, which is not possible except for the zero function $f=0$. So aside from the zero function (a trivial case which you did not mention in your question), we only need to consider cases when $|f(a_0)|=1$.

From here, we just need to consider the case when $f(a_0)=e^{2\pi i\lambda}$. When $\lambda$ is an integer multiple of $\omega$, then we have one of the power functions you mentioned (and when it is a nonnegative integer multiple, we have an entire power function).

Maybe someone else can take it from here? Some casework on the ratio $\lambda/\omega$ might allow us to narrow down the values that $\lambda$ can take, and thereby classify the possible functions satisfying this equation.

Answer 2

Let $n\geq 2$ and consider an entire function $f:\mathbb{C}\to\mathbb{C}$ such that $f(z^n)=f(z)^n$. We consider first the case where $f=g$ where $g$ is a function such that $g(0)\neq 0$.

We want to differentiate the expression $g(z^n)$, $k$ times. For $k\geq 1$, $$\frac{d^k}{dz^k}g(z^n)=\sum_{j=1}^kg^{(j)}(z^n)h_j(z)$$ for some entire functions $h_j$ where $h_k(0)=0$. We proceed by induction. For $k=1$ we have $$\frac{d}{dz}g(z^n)=g'(z^n)nz^{n-1}.$$ We thus have $h_0(z)=0$ and $h_1(z)=nz^{n-1}$. Note that $h_1(0)=0$. Assume now that it holds for $k=l$. This gives us $$\frac{d^l}{dz^l}g(z^n)=\sum_{j=1}^lg^{(j)}(z^n)h_j(z)$$ and thus $$\frac{d^{l+1}}{dz^{l+1}}g(z^n)=\frac d{dz}\sum_{j=1}^lg^{(j)}(z^n)h_j(z)=\sum_{j=1}^l\frac d{dz}g^{(j)}(z^n)h_j(z)=$$$$\sum_{j=1}^lg^{(j+1)}(z^n)nz^{n-1}h_j(z)+g^{(j)}(z^n)h_j'(z)=\sum_{j=1}^{l+1}g^{(j)}(z^n)\tilde{h}_j(z).$$ Note again that $\tilde{h}_{l+1}(0)=n\cdot0^{n-1}h_{l}(0)=0$ which proves the statement above.

We now try to differentiate $g(z)^n$, $k$ times. For $k\geq1$, $$\frac{d^k}{dz^k}g(z)^n=\sum_{j=1}^kg^{(j)}(z)h_j(z)$$ for some entire functions $h_j$ where $h_k(0)\neq0$. For $k=1$ we have $$\frac{d}{dz}g(z)^n=g'(z)ng(z)^{n-1}.$$ Here $h_1(0)=ng(0)^{n-1}\neq0$ by the assumption on $g$. Assume now that it holds for $k=l$, we then have $$\frac{d^l}{dz^l}g(z)^n=\sum_{j=1}^lg^{(j)}(z)h_j(z)$$ where $h_l(0)\neq0$ and it follows that $$\frac{d^{l+1}}{dz^{l+1}}g(z)^n=\frac d{dz}\sum_{j=1}^lg^{(j)}(z)h_j(z)=\sum_{j=1}^l\frac d{dz}g^{(j)}(z)h_j(z)=$$$$\sum_{j=1}^lg^{(j+1)}(z)h_j(z)+g^{(j)}(z)h_j'(z)=\sum_{j=1}^{l+1}g^{(j)}(z)\tilde{h}_j(z)$$ where $\tilde{h}_{l+1}(0)=h_l(0)\neq0$. This proves the second induction.

Now once again by induction we prove that $g^{(k)}(0)=0$ for all $k\geq1$. $g(z^n)=g(z)^n$ implies $g'(z^n)nz^{n-1}=ng(z)^{n-1}g'(z)$ and $0=0g'(0)=g(0)^{n-1}g'(0)$ where $g'(0)=0$ because $g(0)\neq0$. The statement thus holds for $k=1$. Assume now that $g^{(k)}(0)=0$ for $k\leq l$. By the two induction arguments above we get $$\sum_{j=1}^{l+1}g^{(j)}(z^n)h_j(z)=\frac{d^{l+1}}{dz^{l+1}}g(z^n)=\frac{d^{l+1}}{dz^{l+1}}g(z)^n=\sum_{j=1}^{l+1}g^{(j)}(z)\tilde{h}_j(z).$$ This gives $$\sum_{j=1}^{l+1}g^{(j)}(0)h_j(0)=\sum_{j=1}^{l+1}g^{(j)}(0)\tilde{h}_j(0)$$ which is the same as $$g^{({l+1})}(0)h_{l+1}(0)=g^{({l+1})}(0)\tilde{h}_{l+1}(0).$$ Note that $h_{l+1}(0)=0$ but $\tilde{h}_{l+1}(0)\neq 0$. We must thus have $g^{(l+1)}(0)=0$. We finally arrive with the conclusion that $g(z)=g(0)$ by a taylor expansion of the entire function $g$ and $g$ is thus constant. In particular $g(0)=g(0)^n$ and $g(0)=\zeta$ must be an $n-1$'th root of unity.

Suppose now instead that $f(0)=0$. Either $f(z)=0$ for all $z$ or $f$ has a root of some order $k$ at $0$, in this case $$g(z)=\frac{f(z)}{z^k}$$ is an entire function such that $g(0)=0$. It is not hard to show that $g$ satisfies the same functional equation and by the reasoning above $g(z)=\zeta$. We thus have $f(z)=\zeta z^k$ for an $n-1$'th root of unity $\zeta$. In conclusion we have shown that the only solutions are $f(z)=0$ or $f(z)=\zeta z^k$ for some $k\geq 0$.

Answer 3

This answer uses the Weierstrass factorization theorem.

First, prove that if $f$ is non zero, then $f$ has no non zero root. Consider, proving this for yourself before reading on.

Suppose that $f$ has a non-zero root $z_0$ [Edit: that is also not equal to 1], then ${z_0}^{1/n}$ is also a root. Iterating that argument, every member of the family ${z_0}^{1/n^p}$, for $p$ integer, is a root. That family has an accumulation point at 1 so $f(z)=0$ for all $z$ which contradicts the initial hypothesis.

[Edit: Now, consider the case where $f$ has a root that is equal to 1. Then, a $n$-th root of unity is also a root. As $n>1$, $f$ would have a root that is not equal to 1 nor zero which was proved to not be possible in the previous paragraph.]

Thus, according to the Weierstrass factorization theorem, if $f$ is non zero then,

$$f(z)=z^m e^{g(z)}$$

where $g(z)$ is an entire function

Thus, $g$ verifies

$$ng(z)=g(z^n)+2ik\pi$$

for some integer $k$

Taking the series expansion of $g$ with coefficients $a_l$, one finds that for $l>0$

$$a_l=na_{n l}$$

Iterating that relationship we have

$$a_l=n^p a_{ln^p}$$

for $p$ integer. $r a_r \rightarrow 0$ for $r\rightarrow\infty$ as the series representation of $g'(1)$ converges. Thus, $n^p a_{ln^p}$ also converges to $0$ for large $p$ and so $g$ is constant. It is then clear that

$$f(z)=z^m\Theta$$

where $$\Theta^n=\Theta$$

Answer 4

Focus on $\overline B(0,1)$ and its boundary $S^1$. If there was a point $z\in S^1$ such that $\vert f(z)\vert \gt 1$, then there is a $\vert z'\vert \lt 1$ where $\vert f(z')\vert \gt 1$ implying $f$ is unbounded in $\overline B(0,1)$, violating compactness.

If there is a point $z\in S^1$ such that $\vert f(z)\vert \lt 1$, then there is a $\delta$ neighborhood of $z$ where the image under $f$ still has modulus $\lt 1$ and for all primes $p$ large enough (also insisting $p\gt n$) there is a $p$th root of unity $\omega \neq 1$ within that $\delta$ neighborhood of $z$, i.e. $\vert f(\omega)\vert \lt 1$. We may denote the finite group of $p$th roots of unity as $G$ and the map $\phi(z)=z^n$ acts as an automorphism (e.g. by the Counting Formula). By finiteness of the group, there is some $g' \in G-\big\{1\big\}$ which has a minimum modulus $M\lt 1$ under the image of $f$. Then $\phi(g') \in G-\big\{1\big\}$ so $0\leq M \leq \big\vert f\big(\phi(g')\big)\big\vert = \big\vert\phi\big( f(g')\big)\big\vert =\phi\big(\vert f(g')\vert\big) = M^n \leq M \lt 1\implies M =0$; that is, $f$ maps a $p$th root of unity $\neq 1$ to $0$ for all $p$ large enough $\implies f =0$ since there are countably infinite number of primes and $S^1$ is compact.

It remains to consider the case of $f\big(S^1\big)\subseteq S^1$. $f$ must be non-zero on the punctured ball $B^*\big(0,1\big)$, otherwise it would have countably infinite zeros in $\overline B(0,1)$ and be identically zero. Let $f(0) = 0$ with multiplicity $k$ (a non-negative integer) and $F(z):=\frac{f(z)}{z^k}$ with $F(0)$ being the value implied by continuity and observe $F\big(S^1\big)\subseteq S^1$. Since $F$ has no zeros in $B(0,1)$ the max and min modulus theorems tell us it attains a max and min modulus on the boundary, i.e. $1\leq \big \vert F\big(B(0,1)\big)\big\vert \leq 1$ hence $F$ is constant by Max Modulus Theorem, and $F(1) = \lambda =F(1^n)=F(1)^n =\lambda^n$ i.e. $\lambda$ is an $n-1$th root of unity $\implies f(z) = \lambda \cdot z^k$.