Suppose that $f:\mathbb{C}\rightarrow \mathbb{C}$ is a non-constant entire function. By Liouville's theorem, we know that $f$ must take on arbitrarily large values. However Liouville doesn't say anything about what this large set must look like. In particular is it possible that the large values of $f$ are concentrated on a set of small measure?

More precisely, does there exist a non-constant entire function $f$ such that $\lambda(\{x: |f(x)|>1 \})<\infty$? Here $\lambda$ denotes the $2$-dimensional Lebesgue measure.

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    $\begingroup$ Consider $g(z) = \exp\big(1 -f(z)\big)$. We have that $|g| = \exp\big(1 -\Re (f)\big)$. It follows that \begin{align} \left\{z\in\Bbb C\mid |g(z)| < 1 \right\} &= \left\{z\in\Bbb C\mid \exp\big(1-\Re (f(z))\big)< 1 \right\} \\&= \left\{z\in\Bbb C\mid 1-\Re\big(f(z)\big)< 0 \right\} \\&= \left\{z\in\Bbb C\mid \Re\big(f(z)\big)> 1 \right\} \subset \left\{z\in\Bbb C\mid |f(z)| > 1 \right\} \end{align} So $\lambda\big(\left\{z\in\Bbb C\mid |g(z)| < 1 \right\}\big)<\infty$. In other words, if such a function exists for large values, it does for small values too. $\endgroup$ – Fimpellizieri Jul 30 '18 at 19:32
  • $\begingroup$ Certainly there are functions $g$ with $\lambda(\{z\in \mathbb{C} : |g(z)| < 1\})<\infty.$ For instance take $g(z) = z.$ $\endgroup$ – William Swartworth Jul 30 '18 at 19:40
  • $\begingroup$ In fact, any non-constant polynomial works for $g$. (Are there other such functions?) $\endgroup$ – William Swartworth Jul 30 '18 at 19:48
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    $\begingroup$ You might be interested to know about the existence of a neighboring theory, "value distribution theory"/"Nevanlinna theory". Thus is devoted more to how many points in the domain go to a single point in the codomain, though. I couldn't see how to apply any of the main results to your actual question. $\endgroup$ – user98602 Aug 2 '18 at 11:30
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    $\begingroup$ To say something more precise: Consider the function $N_f(r) = \frac{1}{2\pi} \int_0^{2\pi} \text{max}(0, \log(|f(re^{i\theta}|)) d\theta$. This is precisely a measure of how large $f$ gets, the integral supported on the region of the circle of radius $r$ where $|f| > 1$. One simple theorem of Nevanlinna theory is that if $f$ is entire and $N_f(r)$ is bounded, then $f$ is constant; if $N_f = O(\log r)$, then $f$ is polynomial. So $N_f \to \infty$, and at a decent pace. But you could imagine this happening so that $|f| > 1$ only near small patches of the north and south pole of $S^1(r)$... $\endgroup$ – user98602 Aug 2 '18 at 19:44

This set can have finite measure. See, for example, MR0537357 Golʹdberg, A. A. Sets on which the modulus of an entire function has a lower bound, Sibirsk. Mat. Zh. 20 (1979), no. 3, 512–518, 691. (There is an English translation: Siberian Math. Journal, 20 (1980) 360-364).


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