In an answer here, several users reference entire functions which are bounded along many or every line. I have two questions tending in the opposite direction:

1) Is there an entire function (not a polynomial) which diverges along every path $\gamma:[0,1)\to\mathbb{C}$ such that $|\gamma(t)|\to\infty$ as $t\to1^-$?

2) Is there an entire function $f$ which satisfies the above conclusion in the right half plane (ie. diverges along any path approaching $\infty$ which is contained in the right half plane) but converges (say to $0$) along any path approaching $\infty$ which is contained in the left half plane?

  • $\begingroup$ Thanks very much to both Robert Isreal and Lasse Rempe-Gillan. Since they both provided answers to one of the problems, I awarded the solution to the first poster, Robert Israel. $\endgroup$ Aug 13, 2016 at 21:28

2 Answers 2


For (1), just take $f(z) = z$. (Or do you want to call that "converging to $\infty$"?)

For (2), the problem is what happens on the imaginary axis. Whether $f(it)$ diverges or converges to $0$ as $t \to +\infty$, you can take a continuous function $g: [0,\infty) \to (0,\infty)$ that goes to $0$ so rapidly that $f(it)$, $f(g(t) + it)$ and $f(-g(t) + it)$ have the same behaviour as $t \to +\infty$. Thus you can't have one diverging and the other converging to $0$.

EDIT: For the revised question, an example is $f(z) = z \sin(z)$. Note that on the boundary of the rectangle with corners $\pm (n+1/2) \pi\pm n i$ for positive integers $n$, $|f(z)| \ge n$. Since $\gamma(t)$ crosses infinitely many of those boundaries, $f(\gamma(t))$ diverges.

  • $\begingroup$ Good point on number one, I really meant for $f$ to have an essential singularity at $\infty$. I will make the change. $\endgroup$ Aug 11, 2016 at 16:00
  • $\begingroup$ Of course, absolutely right on second one as well. I will wait to accept the answer for the sake of the first question for now though. Thanks! $\endgroup$ Aug 11, 2016 at 16:06

To add a little bit more background in addition to Robert's answer:

The first thing to clarify is what you mean by "diverges". Like Robert, I think you mean "does not converge to a finite value". However, you can also ask your question with "diverges" in the sense of "diverges in the extended complex plane" (i.e., does not converge to infinity either).

Note that, if a transcendental entire function $f$ has a limiting value along a path to infinity, then this is called an asymptotic value. For example, $0$ is an asymptotic value of $z\mapsto e^z$.

So your question 1) can be asked, with the two variations above, as follows:

Is there a transcendental entire function having no finite asymptotic values? The answer is yes, for example take $\sin$ or $\cos$ (even more elementary examples in some sense than that of Robert). Furthermore, any transcendental entire function of order less than 1/2 has this property.

Is there a transcendental entire function not having infinity as an asymptotic value? The answer is negative: by a result of Iversen, there is always a path along which the function tends to infinity.

With respect to your question 2), as it is stated there is not much to add to Robert's answer. However, it may be worth noting that, if you are willing to be a bit more realistic with what you require on or near the boundaries, then almost anything is possible. A classical result that is very powerful is Arakelyan's approximation theorem. For example, it gives the following:

Observation. For any neighbourhood $U$ of the imaginary axis, there is an entire function $f$ with the following property. If $\gamma$ is a curve to infinity that does not intersect $U$, then $f$ tends to zero along $\gamma$ if $\gamma$ is in the left half-plane, and diverges along $\gamma$ if $\gamma$ is in the right half-plane.

Proof. By shrinking $U$, we can assume that $U$ is bounded by two Jordan curves through infinity, one in the left half-plane and one in the right half-plane. Then Arakelyan's theorem can be applied to $A:=\mathbb{C}\setminus U$; let $L$ and $R$ be the left and right components of this set. Define a function $g$ on $A$ by $g(z)=-\log z$ on $L$, and $g(z)=\log z$ on $B$ (taking any branch of the logarithm). By Arakelyan's theorem, there is an entire function $G$ such that $|G(z)-g(z)|\leq 1$ on $A$. Then $f(z):= e^{G(z)}$ is an entire function with the desired properties.

For some results on constructing entire functions with prescribed behaviour and good control everywhere in the complex plane, I refer to recent work of Chris Bishop.


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