I'm seeking for a function $f: \mathbb{C} → \mathbb{C}$ such that:

  • $f$ is complex analytic

  • $f$ has no singularities

  • $f$ is real-valued over the reals

  • $f$ is bounded over the reals

  • $f'$ is positive real over the reals

  • $f''$ is positive (resp. negative) real over the negative (resp. positive) reals

In other words, I'm looking for a sigmoid-like real function that extends to an entire, complex analytic function.

It can be readily seen that some basic functions fail. For example, "the" sigmoid function $f(z) = (1 + \exp(-z))^{-1}$, has singularities on $\pi i + \langle 2\pi i \rangle$. For another example, if $f(z) = \arctan z$, this function is multivalued, and no branch cut admits an analytic function.

Due to Liouville's Theorem, such function cannot be bounded on the entire complex plane. Yet this question requires it to be bounded on the reals. Does this looser requirement enable a solution to be there?


1 Answer 1


One example is $$ f(z) = e^{-e^{-z}} \, . $$ For $x \in \Bbb R$ is $0 < f(x) < 1$, $$ f'(x) = e^{-e^{-x}} e^{-x} $$ is strictly positive, and $$ f''(x) = e^{-e^{-x}} e^{-2x} (1-e^x) $$ is positive for $x < 0$ and negative for $x > 0$.

enter image description here

$g(z) = f(z)-f(-z)$ has the same properties and is point-symmetric with respect to the origin:

enter image description here

More “entire sigmoid functions” can be constructed as $$ f(z) = \int_0^z h(t) \, dt $$ where $h$ is entire, positive on the real axis, increasing on $(-\infty, 0)$, decreasing on $(0, \infty)$, with $\int_{-\infty}^\infty h(t) \, dt < \infty$. One such function is the Gauss error function $$ \operatorname{erf}(z) = \frac{2}{\sqrt \pi}\int_0^z e^{-t^2} \, dt \, . $$

  • $\begingroup$ A bit unrelated with the problem at hand, but why does the order need to be at least $2$? Bounded over the reals imply only order at least $1$ (see $\sin$) and not sure that monotonic requires order $2$ either $\endgroup$
    – Conrad
    Feb 25, 2022 at 5:42
  • $\begingroup$ @Conrad: You are right, that was wrong! I'll remove that part again, thanks. $\endgroup$
    – Martin R
    Feb 25, 2022 at 6:22
  • $\begingroup$ I was thinking if one can construct an order $1$ function as here - I see no reason why not but on the other hand cannot come up with an easy example - $\endgroup$
    – Conrad
    Feb 25, 2022 at 6:41
  • $\begingroup$ @Conrad: Just in case that you are interested: math.stackexchange.com/q/4390816/42969. $\endgroup$
    – Martin R
    Feb 25, 2022 at 10:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .