Extremas of derivatives of Gaussian Consider the family of functions
$$ \phi_n(x) = \frac{d^n}{dx^n} e^{-\frac{1}{2}x^2} = (-1)^n He_n(x) e^{-\frac{1}{2}x^2} $$
Where $He_n(x)$ is the $n$-th (probabilists') Hermite polynomial. From plotting these function for a few $n$ I have come to the conclusion that the absolute values of the extrema seem to be always decreasing away from the center; that is:
Conjecure: If |x|<|y| with $\phi_n'(x) = 0 = \phi_n'(y)$ then $|\phi_n(x)|>|\phi_n(y)|$
I would really like to prove this but I've been stuck on it for a while now. So far I have tried to work with some of the special properties of the Hermite polynomials


*

*recursion relation $He_{n+1}(x) = xHe_n(x) -He_n'(x)$

*Appell property: $ He_n'(x) = nHe_{n-1}(x)$


These allow us to rewrite
$$ \phi_n'(x_0) = 0  \iff He_{n+1}(x_0) = 0  \iff x_0 H_n(x_0) = nHe_{n-1}(x_0)$$
Here note that $He_n$ is an even/odd polynomial function depending on whether or not $n$ is even/odd. Thus $He_{n+1}(0)=0 \implies He_{n-1}(0)=0$ for $n\ge3$, so we can divide by $x_0$ even if it is $0$  if we treat the right hand side as a limit.
$$He_n(x_0) = \frac{n}{x_0}He_{n-1}(x_0)$$
Now to prove the conjecture we would need to show that for given $|x_0| < |y_0|$, $\phi_n'(x_0) = 0 = \phi_n'(y_0)$ holds:
$$ |(-1)^n He_n(x_0) e^{-\frac{1}{2}x_0^2}| > |(-1)^n He_n(y_0) e^{-\frac{1}{2}y_0^2}|
$$
Which by the above is equivalent to
$$ \frac{|y_0|}{|x_0|}\frac{|He_{n-1}(x_0)|}{|He_{n-1}(y_0)|} e^{\frac{1}{2}(y_0^2 - x_0^2)} > 1 $$
However here the problem is of course that we don't know what $He_{n-1}(x_0)$ is; so the question is if it is possible to bound  $\frac{|He_{n-1}(x_0)|}{|He_{n-1}(y_0)|}$ given that  $He_{n+1}(x_0) = 0 = He_{n+1}(y_0)  $. But maybe this is the wrong approach to the problem altogether... Any help and ideas are greatly appreciated.
EDIT: I just found a very interesting source: in Gabor Szegö's book Orthogonal Polynomials Theorem 7.6.3 claims that the local maxima of the function $|H_n(x)|e^{-\frac{1}{2}x^2}$ form an increasing sequence for $x\ge 0$
But mind you that he is using the physicists Hermite polynomials instead. The conversion between the two is given by
$$ H_n(x) = \sqrt{2}^n He_n(\sqrt{2}x)  $$
I.e. the roots/extrema of $H_n$ are closer to the center than those of $He_n(x)$ which might just make the difference on whether or not the sequence is increasing/decreasing due to the fast decay of the exponential factor $e^{-\frac{1}{2}x^2}$.
Sadly the proof of the book is very much like spaghetti code and I am not seeing right now how to adapt it.
 A: Alright I managed to decode the the proof in Szegö's book and apply it to the probabilists' Hermite polynomials. Hidden in a footnote on page 166, there is a very neat theorem due to Sonin, Butlewski and Pólya:
Assume that $y(x)$ solves $$(g(x)y')'+ f(x)y = 0$$ where $f,g$ are positive, continuously differentiable functions. Then the local maxima of $|y|$ form an increasing/decreasing sequence for $x\ge 0$ if $f(x)g(x)$ is decreasing/increasing.
Proof multiplying by  $2g(x)y'$ gives 
$$D(g(x)y')^2 + f(x)g(x)Dy^2 = 0 
\quad\text{i.e. } Dy^2 = -\frac{1}{f(x)g(x)}D(g(x)y')^2 $$
Now consider the function $$\psi(x) = y(x)^2 + \frac{1}{f(x)g(x)}(g(x)y')^2$$ Evidently, $\psi(x) = y(x)^2$ when $y'(x)=0$, so in particular for every maxima of $|y|$. Moreover for inc/dec $f(x)g(x)$, it is a dec/inc function itself since:
$$ D\psi(x) = -\frac{(f(x)g(x))'}{(f(x)g(x))^2}(g(x)y'(x))^2$$
which proofs the theorem.
Now this applies to the function $He_n(x)e^{-\frac{1}{2}x^2}$ since it satisfies the equation
$$ (e^{+\frac{1}{2}x^2} y')' + (n+1)e^{+\frac{1}{2}x^2} y = 0 $$
All of this raises an interesting side question: since the sequence of maxima of the function in question $|p(x)| e^{-\frac{1}{2}x^2}$ is increasing for the physicists Hermite polynomials $p=H_n$, and decreasing for the probabilists Hermite polynomials $p=He_n$, and they are scaled versions of each other, one is only left to wonder if there is some scaled inbetween version such that the sequence of maxima is constant. This should at least be possible for $n\le 4$.
If anyone has some more insight to offer I am more than happy to award the bounty since it's going to expire anyway.
