I'm seeking for a function $f: \mathbb{C} → \mathbb{C}$ such that:
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$f$ is complex analytic
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$f$ has no singularities
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$f$ is real-valued over the reals
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$f$ is bounded over the reals
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$f'$ is positive real over the reals
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$f''$ 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?
Best Answer
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 $x > 0$.
$g(z) = f(z)-f(-z)$ has the same properties and is point-symmetric with respect to the origin:
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 $$ \operatorname{erf}(z) = \frac{2}{\sqrt \pi}\int_0^z e^{-t^2} \, dt \, . $$