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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.

William Swartworth
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    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.

     – Fimpellizzeri Jul 30 '18 at 19:32
  • Certainly there are functions $g$ with $\lambda({z\in \mathbb{C} : |g(z)| < 1})<\infty.$ For instance take $g(z) = z.$ – William Swartworth Jul 30 '18 at 19:40
  • In fact, any non-constant polynomial works for $g$. (Are there other such functions?) – William Swartworth Jul 30 '18 at 19:48
  • Well, in truth I had hoped to get some mileage out of the fact that $g$ is the exponential of some entire function (which is not the case for polynomials); it's easy to modify that argument to show that $f$ (as in your question) can be assumed to be the exponential of some entire function.

    In neither case I've managed to go much further.

     – Fimpellizzeri Jul 30 '18 at 19:51
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    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. –  Aug 02 '18 at 11:30
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    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 $S^1(r)$... –  Aug 02 '18 at 19:44
  • where $|f|$ gets gigantic, and these small patches shrink as $r \to \infty$. I'm sure this doesn't happen, but a proof eludes me. I also feel like there should be some interplay between the small and large values, and feel like the Weierstrass factorization theorem might be useful, but was unable to get anything serious out of it. –  Aug 02 '18 at 19:45

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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).

Alexandre Eremenko
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