Analysis Problem of the Day 41

Today’s problem appeared as Problem 11 on the UCLA Spring 2021 Analysis Qual:

Problem 11: For an entire function $f,$ define $f^{[n]}(z)$ to be the composition of $f$ with itself $n$ times, with $f^{[1]}(z)=f(z).$

a) Show that if $f^{[n]}$ is a polynomial for some $n,$ then $f$ is a polynomial.

b) Show that $f^{[n]}(z) \not = e^z$ for any $n \geq 2.$

Solution: a) We proceed by contrapositive. By the Great Picard theorem, if $f$ is not a polynomial, $f$ has an essential singularity at $\infty$ and therefore, the image $f(U_0)$ of a neighborhood $U_0$ of $\infty$ attains all except at most one value of $\mathbb{C}$ infinitely many times. In particular, $f(U_0) \supset U_1,$ where $U_1$ is a neighborhood of infinity. Inductively, if $f^{[n+1]}(U_0) \supset f(U_n),$ where $U_n$ is a neighborhood infinity, implying that $f^{[n+1]}(U_0)$ is dense in $\mathbb{C}.$ But the image of a polynomial in a sufficiently small neighborhood of infinity is not dense in $\mathbb{C}$ (since $p(z) \to \infty$ as $z \to \infty$ for any polynomial $p$ ), so $f^{[n]}$ cannot be a polynomial. Thus, if $f^{[n]}$ is a polynomial, so is $f.$

b) Note that if $f(\mathbb{C})=\mathbb{C},$ then $f^{[n]}(\mathbb{C}) = \mathbb{C},$ and since $e^z$ omits 0 in its image, it follows that $f$ has to omit a value – namely, also 0. Thus, $f(z) = e^{g(z)}$ for some entire $g.$ It follows that for $n \geq 2,$ $f^{[n]}(z) = e^{g(e^{h(z)})}$ for some entire $h.$ Thus, if $f^{[n]}(z)=e^z,$ taking complex logarithms on both sides yields $g(e^{h(z)})+2\pi i n =z$ for some $n \in \mathbb{Z}.$ But by Great Picard again, there exist $z_1,z_2$ such that $h(z_1)-h(z_2)=2\pi i,$ i.e. $e^{h(z)}$ is not injective. Thus $g(e^{h(z)})$ is not injective, which is a contradiction. Thus, such an $f$ cannot exist.

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