I asked this on [Math.SE](https://math.stackexchange.com/q/2443677/415941) 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).
> 
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> 


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?



The $n$-th derivative has $n$ zeros. Can such a function be unbounded?
As suggested by Mateusz Kwaśnicki, the function $f : x \mapsto (1+x^2)^{s}$ is bell-shaped and unbounded for any $s \in (0,\frac{1}{2})$.


It is easy to see that $f^{(n)}(x) = P\_n(x) (1+x^2)^{s-n}$ where $P\_n$ is a polynomial of degree $\leq n$. Actually
$$
P\_{n+1}(x) = (1+x^2) P\_n'(x) - 2(n-s)xP\_n(x).
$$
Let $a\_n$ be the coefficient of $x^n$ in $P\_n$. Then $a\_{n+1} = n a\_n - 2(n-s)a\_n = (2s-n) a\_n$. In particular $(-1)^{n-1} a\_n > 0$ for $n \geq 1$, so that $P\_n$ has degree exactly $n$ (this is where we use $s < \frac{1}{2}$).


One first checks that $f'$ has a single zero.
We then assume that $f^{(n)}$ has exactly $n$ distinct simple zeroes $x\_n < x\_{n-1} < \dots < x\_1$ for some $n \geq 1$. Then
$$
(-1)^{n-1} f^{(n+1)}(x\_1) = (-1)^{n-1} P\_n'(x\_1) (1+x\_1^2)^{s-n} > 0
$$
since $(-1)^{n-1} P\_n(x) > 0$ for $x > x\_1$. But $(-1)^{n-1} f^{(n+1)}(x) < 0$ for large $x$ so $f^{(n+1)}$ has a zero in $(x\_1,+\infty)$. Similarly $f^{(n+1)}$ has a zero in $(-\infty ,x\_n)$. Together with the zeroes of $f^{(n+1)}$ between the $x\_i$'s, we get $\geq n+1$ distinct zeroes (and therefore exactly $n+1$ zeroes).


