# $L^\infty$ bound of $x^m \psi_n(x)$ where $\psi_n$ is a Hermite function and $m,n \in \mathbb{N}$ - extension from Cramer's inequality

For each $$n \in \mathbb{N}$$, the Hermite function $$\psi_n : \mathbb{R} \to \mathbb{R}$$ is a Schwartz function defined by $$$$\psi_n(x):=(-1)^n(2^n n!\sqrt{\pi})^{-1/2} e^{x^2/2} \frac{d^n}{dx^n}e^{-x^2}$$$$ according to the Wikipedia article : https://en.wikipedia.org/wiki/Hermite_polynomials#Differential-operator_representation

Moreover, it is written there that each Hermite function satisfies the $$L^\infty$$ bound $$$$\lvert \psi_n(x) \rvert \leq \pi^{-1/4} \text{ for all } x \in \mathbb{R}, n \in \mathbb{N}$$$$ which is known as the Cramer's inequality.

Now, I am trying to find a bound of $$$$\sup_{x \in \mathbb{R}} \lvert x^m \psi_n(x) \rvert$$$$ for any given $$n,m \in \mathbb{N}$$.

More specifically, I am curious about the asymptotic behavior of this supremum as $$n \to \infty$$ while $$m$$ is fixed. Would it remain bounded or diverge?

• It seems like there should a reasonable bound for each $m$ and $n$, coming from the expansion $x^{m}\psi_{n}(x)=\sum_{j}c_{j}\psi_{n+m-j}(x)$.
– Buzz
Commented Jul 17, 2023 at 0:21

$$\newcommand\ep\varepsilon$$According to the 6th display in this section, for any fixed $$\ep>0$$ we have $$\psi_n(x)=(2/\pi)^{1/4}(\pi n)^{-1/4}(\sin t)^{-1/2}s_n(x)$$ if $$n\to\infty$$, $$x=\sqrt{2n+1}\,\cos t$$, and $$t\in[\ep,\pi-\ep]$$, and $$s_n(x):=\sin\Big(\frac{3\pi}4+\frac{2n+1}4\,(\sin2t-2t)\Big)+O(1/n).$$
Fix now any $$t\in(0,\pi/2)$$ such that $$\sin2t-2t$$ is irrational, and let $$x_n:=\sqrt{2n+1}\,\cos t$$. Then $$\limsup_n s_n(x_n)=1$$ and hence $$\limsup_n x_n^m\, s_n(x_n)=\infty$$ for each integer $$m\ge1$$.
So, for each integer $$m\ge1$$, $$$$\sup_{x\in\mathbb R}|x^m\,\psi_n(x)|\text{ is unbounded in n}.$$$$
Using Szegő, formula (8.22.14) (say with $$t=0$$), we see that, moreover, for each real $$m>1/12$$, $$$$\lim_n\sup_{x\in\mathbb R}|x|^m\,|\psi_n(x)|=\infty.$$$$