I'll state the theorem I am posing up front, and then explain why I think this theorem appears to be true. I am asking if anyone can prove it, or knows references to where it is proved. Please, forgive me if it is trivial, or if a counter example is trivial too.

If $\phi(s,z) : \mathbb{C}_{\Re(s)>0} \times \mathbb{C} \to \mathbb{C}$ is holomorphic in both variables, and $$\phi(s_0,\phi(s_1,z)) = \phi(s_0 + s_1,z)$$ then necessarily $\frac{\partial^2}{\partial z^2} \phi(s,z) = 0$, so that $$\phi(s,z) = e^{qs}(z-z_0) + z_0$$ for $q,z_0 \in \mathbb{C}$ or $$\phi(s,z) = z+cs$$ for $c \in \mathbb{C}$

I ask this question because

A) I've never encountered a counter example in an extensive study of semigroups.

B) I can prove on a whole bunch of occasions if $\phi(a,z) = f(z)$; for $\Re(a) > 0$, for specific $f$; implies $\phi(s,z)$ cannot exist.

Examples include $f = \sin,\cos,\exp,\exp(p(z)), p(\exp(z))$ where $p$ is an arbitrary polynomial. If $f$ has a super attracting fixed point, no such $\phi$ exists. If $f^{\circ n}$ has fixed points $f$ doesn't have, then no such $\phi$ exists. In all these cases, no such solution exists for the orbits of $f$ either. I'm wondering if this is a universal trait. That, in some sense, the theorem above can be a kind of Liouville theorem.

Let's say that $\phi$ *satisfies* $f$, if $\phi(a,z) = f(z)$ for some $\Re(a) > 0$. To highlight the similarity to Liouville's theorem, the original theorem can be restated as

If $f:\mathbb{C} \to \mathbb{C}$ and some semigroup $\phi$ satisfies $f$, then $f= mz+b$ for some $m,b \in \mathbb{C}$.

Supposing the pair $f(\cdot)$ and $\phi(s,\cdot)$ don't have to map $\mathbb{C}$ to itself, that $\mathbb{C}$ is weakened to the unit disk $\mathbb{D}$, and $f(0) = 0$ with $|f'(0)| \neq 0,1$, then there always exists a $\phi$ *satisfying* $f$. It seems though, as soon as we lift from $\mathbb{D}$ (or any simply connected domain biholomorphic to $\mathbb{D}$) to $\mathbb{C}$ it fails.

I can show the result in a restricted form, which I also think is interesting

If $p:\mathbb{C} \to \mathbb{C}$ is a polynomial, and there exists a semigroup $\phi$ satisfying $p$ then $p(z) = mz+b$ for some $m,b \in \mathbb{C}$

My main avenue of approach for deriving a contradiction has been to consider the Weierstrass product. It follows that if $\phi(s_0,z_n) = z_n$ then $\phi(s,z_n) = z_n$, so $\phi$ has a countable list of fixed points invariant on our choice of $s$, allowing us to say

$$\phi(s,z) = z + e^{g(s,z)}\prod_{n=0}^{\infty}(1-\frac{z}{z_n})e^{-p_n(z/z_n)}$$

where $p_n(z) = \sum_{j=1}^n \frac{z^j}{j}$. I've been fiddling around with this as it seems like a smart idea. Since $\lim_{s\to 0}\phi(s,z) = z$, and we have a semigroup property, it seems reasonable to think we can show that for all $\epsilon >0$ there exists a $\delta > 0$ such that $|\phi(\delta,z)-z| < C_\delta e^{|z|^{\epsilon}}$ then by Hadamard

$$\sum_{n=0}^\infty \frac{1}{|z_n|} < \infty$$

which greatly reduces the candidates for $f$ that can be satisfied by some $\phi$—$f$'s fixed points have to be sufficiently spaced out, so to speak.

Overall, I'm lost on the general case, and seem to be only able to prove it in specific circumstances. I think it says something rather profound about multiplication and addition, that, for another reason, they are incredibly special. Help, comments, suggestions, edits, anything is welcome, thanks.

Brian A. Coomes, Polynomial Flows on $\mathbb{C}^n$, Trans. A.M.S. Vol. 320, No. 2 (Aug., 1990), pp. 493-506$\endgroup$ – YCor Feb 3 at 17:02