Analysis Problem of the Day 39

Today’s problem appeared as Problem 7 on the UCLA Spring 2014 Analysis Qual:

Problem 7: Characterize all entire functions $f$ with $|f(z)|>0$ for $|z|$ large and $\limsup_{z \to \infty} \frac{|\log |f(z)||}{|z|}<\infty.$

Solution: Notice that the condition is equivalent to stating that $f$ has finitely many zeros and $|f(z)| \leq e^{C|z|},$ i.e. $f$ is of exponential type 1. By the Hadamard factorization theorem, it follows that $f(z) = z^m e^{az+b} \prod_{i=1}^n (1-\frac{z}{a_i}) = e^{az} p(z)$ for some constants $a,b \in \mathbb{C}$ and some nonzero polynomial $p.$

Alternative approach: Divide out by the zeros of $f$ and take the complex logarithm $\log f$ (which is well-defined since $f$ does not vanish). By assumption, $g=\log f$ is an entire function satisfying $|g(z)| \leq C|z|.$ By the polynomial variant of Liouville’s theorem, this implies that $g$ is at most a polynomial of degree $1,$ i.e. $g(z)=az+b.$ As above, this implies that $f(z)= e^{az} p(z).$

Remark: If the condition is replaced by $\limsup_{z \to \infty} \frac{|\log|f(z)||}{|z|^k}<\infty,$ it follows that $f(z)=p(z)e^{q(z)}$ where $p,q$ are polynomials and the degree of $q$ is at most $k.$

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