Let $H\_k(x)$ be (probabilists' or physicists', does not matter for this question) Hermite polynomials.


It is well-known that all the gaps between consecutive roots of $H\_k(x)$ are at least a multiple of $1 / \sqrt{k}$ (see, e.g., Szego, "Orthogonal Polynomials", page 130).


Is the same known to be true if we consider the roots of $H\_k(x)$ and $H\_{k+1}(x)$ together? Equivalently, is it true that the zeros of $H\_{k+1}(x)$ are roughly half-way between its respective local minima and maxima?


Looking at the numerical values for $k \leq 100$, the desired bound seems to be true.



Gaps between roots of consecutive Hermite polynomials
The answers are yes and yes.


Consider the Hermite Gauss functions : $\psi\_n(x)=e^{-x^2/2}H\_n(x) $, we have two properties:
$$\psi\_n''(x)+(2n+1-x^2)\psi\_n(x)=0 $$
and 
$$\psi\_{n+1}=\psi\_n'(x) +x\psi\_n(x) $$
For $n$ large, we take a lenght $L>0$ which is very small (for example $L=1/\log{n}$). On any $[x\_0-L/2,x\_0+L/2]\subset ]- \sqrt{2n},\sqrt{2n}[$, $\psi\_n$ is rountly the solution of $$\psi\_n''(x)+(2n+1-x\_0^2)\psi\_n(x)=0 $$ and therefore $\psi\_n(x)=A\sin(\omega\_0 x+\alpha)$ on $[x\_0-L/2,x\_0+L/2]$ with $\omega\_0=\sqrt{2n+1-x\_0^2}$ and $\psi\_{n+1}(x)=-\omega\_0 A \cos(\omega\_0 x+\alpha)+x\_0A\sin(\omega\_0 x+\alpha)$. 
In particular the zeros of $\psi\_n$ are placed periodically every $\pi/\omega\_0$ and the zeros of $\psi\_{n+1}$ are just shifted by $\omega\_0 (\pi/2 + \arctan(x\_0/\omega\_0))$


For $x\_0=\pm \sqrt{2n}$, the $\sin$ has to be replaced by the Airy function with a scale $1/\sqrt{n}$. And the distribution of its zeros and the zeros of its derivative are the same (with the scale $1/\sqrt{n}$) <https://en.wikipedia.org/wiki/Airy_function> 


