Define the function : $$ x \to z_n(x)=\frac{e^{-x^2/2}He_n(x)}{\sqrt{(n+1)!}} $$ where $He_n$ is the probabilistic Hermite polynomials I'd like to show (see figure below) that : $max_{x\geq0} |z_n(x)|>max_{x\geq0} |z_{n+1}(x)|$ or equivalently that : $max_{x\geq0} z_n(x)^2>max_{x\geq0} z_{n+1}(x)^2$
So far, I have shown that the maximum of $z_n^2$ occurs either at $x=0$ (if n is even) or at $x=x_{1,n+1}$ (if n is odd) where $x_{1,n+1}$ denotes the first positive zero of $He_{n+1}$. I know that the relative maximas of $z_n^2$ decrease and I have the following relations :
$$ z_{n}^{'}\left(x\right)=-\sqrt{n+2}z_{n+1}\left(x\right). $$ $$ \sqrt{n+3}z_{n+2}\left(x\right)=xz_{n+1}\left(x\right)-\frac{n+1}{\sqrt{n+2}}z_{n}\left(x\right). $$ $$ z_{n}^{''}\left(x\right)+xz_{n}^{'}\left(x\right)+\left(n+1\right)z_{n}\left(x\right)=0. $$
Any idea on how to proceed ?
