Must the set of lines through the origin on which a nonconstant entire function is bounded be finite?

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.)

• 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? Sep 1, 2013 at 10:15
• 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. Sep 1, 2013 at 20:14
• Your intuition is totally wrong: $e^z$ is bounded on all directions in the left half-plane. May 27, 2017 at 8:13

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 $$G(z)=\int_0^\infty e^{i\pi t}t^{-t}\cosh(t\sqrt{z})dt\int_0^\infty e^{i\pi t}t^{-t}\cos(t\sqrt{z})dt .$$

• 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. 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 $$\frac{\pi}{2}< \arg z<\frac{3\pi}{2}.$$

• 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. Jun 27, 2010 at 23:07
• 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. Jun 27, 2010 at 23:18
• 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. Jun 28, 2010 at 12:03
• 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. Jun 28, 2010 at 12:20