If an entire function is bounded for all $z \in \mathbb{C}$, than it's a constant by Liouville's theorem. Of course an entire function can be bounded on lines through the origin $z=r \exp(i \phi), \phi= \text{const.}, r \in \mathbb{R}$ without being constant (e.g. $\cos(z^n)$ is bounded on $n$ lines).

What is the maximum cardinality of the set of "directions" $\phi$ for which an entire function can be bounded without being constant?

From intuition I would expect only finitely many directions. Is this correct?

(Picard's second theorem says that in any open set containing $\infty$ every value with possibly a single exception is taken infinitely often by an entire non-constant function. Here I'm asking a somehow "orthogonal" question, looking for lines through $\infty$ where an entire non-constant function is bounded.)

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    $\begingroup$ This is a very interesting question -- the accepted answer startles and amazes me -- but I find the title hard to read and understand. (The way I read it, an answer would be: "Certainly yes -- e.g. $z^2$ is not bounded on any line through the origin.") Would(n't) something like "Must the set of lines through the origin on which a nonconstant entire function is bounded be finite?" be better? $\endgroup$ Sep 1, 2013 at 10:15
  • $\begingroup$ Many thanks for your comment. Actually, I was not happy with the formulation of my question, but couldn't find a better one (I'm not a native English speaker). I will change the question according to your suggestion. $\endgroup$ Sep 1, 2013 at 20:14
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    $\begingroup$ Your intuition is totally wrong: $e^z$ is bounded on all directions in the left half-plane. $\endgroup$ May 27, 2017 at 8:13

2 Answers 2


Newman gave an example in 1976 of a non-constant entire function bounded on each line through the origin in "An entire function bounded in every direction".

I like the second sentence of the article:

This is exactly what is needed to confuse students who have just struggled to comprehend the meaning of Liouville's theorem.

Armitage gave examples in 2007 of non-constant entire functions that go to zero in every direction in "Entire functions that tend to zero on every line". For this I have only seen the MR review. (If you don't have MathSciNet access, the link should still give you the publication information to find the article.)

Update: I just decided to take a look at the Armitage paper, and the introduction was enlightening:

Although every bounded entire (holomorphic) function on $\mathbb{C}$ is constant (Liouville’s theorem), it has been known for more than a hundred years that there exist nonconstant entire functions $f$ such that $f(z) → 0$ as $z →∞$ along every line through 0 (see, for example, Lindelöf’s book [10, pp. 119–122] of 1905). And it has been known for more than eighty years that such functions can tend to 0 along any line whatsoever (see Mittag-Leffler [11], Grandjot [8], and Bohr [4]). Further references to related work are given in Burckel’s review [5] of Newman’s note [12]. Entire functions with radial decay are used by Beardon and Minda [3] and Ullrich [14] in studies of pointwise convergent sequences of entire functions.

Armitage goes on to mention that Mittag-Leffler and Grandjot also gave explicit constructions, but states, "The examples given in what follows may nevertheless be of some interest because of their comparative simplicity." The examples are $$F(z)=\exp\left(-\int_0^\infty t^{-t}\cosh(tz^2)dt\right) - \exp\left(-\int_0^\infty t^{-t}\cosh(2tz^2)dt\right)$$ and $$G(z)=\int_0^\infty e^{i\pi t}t^{-t}\cosh(t\sqrt{z})dt\int_0^\infty e^{i\pi t}t^{-t}\cos(t\sqrt{z})dt .$$

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    $\begingroup$ If an entire function is bounded on every line through the origin, then to every point on the unit circle there is assigned a bound, a nonnegative number that is the maximum absolute value of the function on the corresponding line. That function on the unit circle must therefore have a discontinuity. $\endgroup$ Nov 24, 2020 at 21:45

The Mittag-Leffler function $E_{\alpha,1}$, $\alpha>0$, is bounded in the sector $$\frac{\alpha\pi}{2}< \arg z<2\pi-\frac{\alpha\pi}{2}.$$

In particular, $e^z=E_{1,1}(z)$ is bounded in $$\frac{\pi}{2}< \arg z<\frac{3\pi}{2}.$$

  • $\begingroup$ Though Andrey, I think one of the points of the question is that the restriction of the function be bounded on both the positive and negative directions of the angular sector. $\endgroup$ Jun 27, 2010 at 23:07
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    $\begingroup$ That's OK, just take $z\mapsto e^{z^2}$ and consider $\pi/4<\arg z<3\pi/4$. +1: This answers the cardinality question much more simply. $\endgroup$ Jun 27, 2010 at 23:18
  • $\begingroup$ Thanks for the comments. @Willie Wong: In any case, $E_{\alpha,1}$ is bounded everywhere except for a small sector when $\alpha$ is small. This allows for a continuum of complete lines where the function is bounded. $\endgroup$ Jun 28, 2010 at 12:03
  • $\begingroup$ Ah! Good points both from Jonas and Andrey. That cleared things up for me. I don't remember my Mittag-Leffler function well, and was under the mis-impression that $\alpha$ has to be integral. Sorry about the noise. $\endgroup$ Jun 28, 2010 at 12:20

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