I asked this on Math.SE some days ago, but without any success. For some application I need a formal definition of bell-shaped function. So I had the following idea:

Definition. A $C^\infty$-function $f:\Bbb R\to\Bbb R$ should be called bell-shaped if for all $n\geq 1$ the $n$-th derivative has exactly $n$ zeros (counted with multiplicity).

E.g. this works for $\exp(-x^2)$. As this is my try to formalize an informal term, I have no way to check if this is correct. However, I tried to prove that such functions must be bounded (for me, bell-shaped functions are always bounded). But I had no success.

Question: Is a bell-shaped function (in the sense above) always bounded?

I asked this on Math.SE some days ago, but without any success. For some application I need a formal definition of bell-shaped function. So I had the following idea:

Definition. A $C^\infty$-function $f:\Bbb R\to\Bbb R$ should be called bell-shaped if for all $n\geq 1$ the $n$-th derivative has exactly $n$ zeros (counted with multiplicity).

E.g. this works for $\exp(-x^2)$. As this is my try to formalize an informal term, I have no way to check if this is correct. However, I tried to prove that such functions must be bounded (for me, intuitively bell-shaped functions are always bounded). But I had no success.

Question: Is a bell-shaped function (in the sense above) always bounded?