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This is an open problem that I learned from Thomas Simon. I will completely understand if the question is judged as non-research level (and it is indeed not related to my research), but I believe a solution would result in some nice, publishable mathematics. The point of this post is to popularise the problem.

We need two definitions.

Definition 1: We say that a function $f$ is bell-shaped on $\mathbb{R}$ if it is infinitely smooth, converges to zero at $\pm \infty$ and the $n$-th derivative of $f$ has $n$ zeroes in $I$ (counting multiplicity).

Definition 2: We say that the derivative is interlacing for $f$ on $\mathbb{R}$ if $f$ is infinitely smooth and for every $n \geqslant 0$ the zeroes of $f^{(n)}$ and $f^{(n+1)}$ interlace (counting multiplicity).

When we say that the zeroes of $f$ and $g$ interlace, we mean that each connected component of $\{x \in I : f(x) \ne 0\}$ contains exactly one zero of $g$, except perhaps for unbounded components, which are allowed to contain no zero of $g$ — at least when all zeroes of $f$ and $g$ are simple. The definition in the general case is somewhat more involved, but hopefully clear.

Problem: Describe all bell-shaped functions. Describe all functions $f$ such that the derivative is interlacing for $f$.

Natural modifications are allowed. For example:

  • $f$ can be assumed to be analytic or entire;
  • the problem can be restricted to a finite or semi-infinite interval.

By description we mean essentially arbitrary condition that is easier to check than the definition. Ideally, one could hope for a result similar to Bernstein's characterisation of completely monotone functions as Laplace transforms, which can be re-phrased as follows: if $f^{(n)}$ has no zeroes on $\mathbb{R}$, then $f$ is the Laplace transform of either a measure on $[0, \infty)$ (a completely monotone function), or a measure on $(-\infty, 0]$ (a totally monotone function). However, the answer for bell-shaped functions is likely much more complicated.


Easy observations:

  • It is clear that $f$ is bell-shaped if and only if it has no zero, it converges to zero at $\pm \infty$ and the derivative is interlacing for $f$.
  • By Rolle's theorem, if $f$ has no zeroes and $f$ converges to zero at $\pm \infty$, then $f^{(n)}$ has at least $n$ zeroes.
  • The function $\exp(-x^2)$ is bell-shaped: its $n$-th derivative is $P_n(x) \exp(-x^2)$ for a polynomial $P_n$ of degree $n$, so it has no more than $n$ zeroes.
  • Similarly, the function $f(x) = (1 + x^2)^{-p}$ is bell-shaped for $p > 0$ (and the derivative is interlacing for $f$ also when $p \in (-\tfrac12, 0)$): the $n$-th derivative of $f$ is of the form $P_n(x) = (1 + x^2)^{-p-n}$ for a polynomial $P_n$ of degree $n$.
  • In the same vein, $f(x) = x^{-p} \exp(-x^{-1})$ is bell-shaped on $(0, \infty)$ for any $p > 0$: $f^{(n)}(x) = P_n(x) x^{-p - 2n} \exp(-x^{-1})$ for a polynomial $P_n$ of degree $n$.
  • The derivative is interlacing for a polynomial if and only if it only has real roots.
  • More generally, the derivative is interlacing for any locally uniform limit of polynomials with only real roots. This class includes $\sin x$, $\exp(x)$, $\exp(-x^2)$, some Bessel functions and many more; see Chapter 5 in Steven Fisk's book.

This topic appeared at least in probability literature, in an erroneous article by W. Gawronski, followed by two articles by T. Simon on stable laws and passage times (with W. Jedidi).


EDIT:

Having read through Widder and Hirschman's book and other references pointed out in Alexandre Eremenko's answer (which was really helpful!), I was able to extend Thomas Simon's work and prove that certain functions are bell-shaped. The preprint is available at arXiv; all comments welcome.

The class of functions that I was able to handle includes density functions of infinitely divisible distributions with Lévy measure of the form $\nu(x) dx$, where $x \nu(x)$ and $x \nu(-x)$ are completely monotone on $(0, \infty)$. Such functions are called extended generalised gamma convolutions (EGGC) by L. Bondesson, and it includes all stable distributions (thus correcting an error in W. Gawronski's article).

Of course this does not answer the question about characterisation of all bell-shaped functions: this remains an open problem.

However, I am not aware of any bell-shaped function outside this class. Is there any?

It allows one to show that certain simple functions are bell-shaped. For example, if I did not make a mistake, $f(x) = \dfrac{1}{(1 + x^2)(4 + x^2)}$ is an EGGC, so it is bell-shaped. Out of curiosity: is there any elementary way to prove that this particular function $f$ is bell-shaped?

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    $\begingroup$ This is a nice question. The survey that I know on the topic is Polya, On the zeros of derivatives of a function and its analytic character, 1942, but it is a little bit out of date. $\endgroup$ – Alexandre Eremenko Oct 3 '17 at 20:49
  • $\begingroup$ @AlexandreEremenko: I have to admit I am completely ignorant in this area, and so I was not aware of Pólya's article. Thanks for the reference: even though it is outdated, it is still a very good read! $\endgroup$ – Mateusz Kwaśnicki Oct 4 '17 at 21:35
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This is a set of comments (the problem of completely describing the bell-shaped functions is probably too hard and its complete solution will probably be more than just a "publishable paper":-)).

  1. In the definition of bell-shaped it is better to say that ALL derivatives tend to $0$ as $x\to\pm\infty$. This makes more sense, and most authors define it like this.

  2. The large class of examples of bell-shaped functions is given by the so-called Polya frequency functions (Laplace transforms of reciprocals of Laguerre Polya functions), see Hirschman and Widder, Convolution Transform, Ch. IV, sect. 5.1. It seems reasonable to conjecture that all bell-shaped functions are of this kind.

  3. Assumption that the bell-shaped function is entire is a restrictive one. There are bell-shaped functions which are not entire, $1/(1+x^2)$ for example. If the function is assumed to be entire, AND of the form $g(x)e^{-x^2}$, where $g$ is a real canonical product of genus $1$, then it has derivative-interlacing property only if $g$ has all zeros real. This follows from a deep result of Haseo Ki and Young-One Kim, Duke Math J. 104 (2000), no. 1, 45–73. As a corollary, we obtain that $e^{-z^2}$ is the only bell-shaped function of this form.

  4. Hirschman (Proc. AMS, 1, (1950). 63–65.) proved the conjecture of Schoenberg that there are no bell shaped functions with compact support. But there are bell shaped functions which are zero on a ray.

EDIT. My conjecture in 2 is certainly wrong: according to the theorem of T. Simon (second reference in the question text), all positive stable densities are bell-shaped, but they are not Polya frequency functions.

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  • $\begingroup$ Thanks for these comments and references! I plan to write a follow-up with some more specific conjectures, and your comment no. 3 already answers the very first one. Regarding comment no. 1, if $f$ converges to zero at infinity and $f'$ is ultimately monotone, then $f'$ converges to zero too, so in fact all derivatives of a bell-shaped function automatically go to zero at $\pm\infty$. But you are right in that it is perhaps more natural to say this straight away. $\endgroup$ – Mateusz Kwaśnicki Oct 4 '17 at 21:41
  • $\begingroup$ Regarding the edit, an even simpler counterexample is $(1+x^2)^{-s}$, which is not log-concave, and hence it is not a Pólya frequency function. $\endgroup$ – Mateusz Kwaśnicki Oct 14 '17 at 21:37
  • $\begingroup$ @Mateusz Kwasnicki: Yes. All this shows that this class of bell-shaped functions is large, and that there is no plausible conjecture how to describe all of it. $\endgroup$ – Alexandre Eremenko Oct 14 '17 at 22:24

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