Estimation of the Hermite Polynomials using Plancherel-Rotach asymptotics

Question

Suppose $H_n(x)$ is a Hermite Polynomial such that $$\int_{\mathbb{R}} H_n(x) H_m(x) e^{-x^2} dx = \delta_{m,n}.$$ I want to show for $ \phi_n(x) = H_n(x)e^{-\frac{X^2}{2}}$ $$ \left( \phi_n(x)\phi_{n-1}(y) - \phi_{n-1}(x) \phi_n(y) \right)^2 \leq \frac{C}{n},$$ for all $x,y$ and $n$, some constant $C$ (doesn't depend on $x,y,n$). I have tried to use Plancherel-Rotach estimates, but I haven't succeeded. I wonder if you have ever seen somewhere a similar result before.

Answer 1

I have no idea if anyone still has interest in this question, but this claim is false.

It is known (for example, in here) that one may find $x_N$ such that $\phi_N(x_N)\ge CN^{-1/6}$ for some $C>0$.

Additionally, by the Plancherel-Rotach Asymptotics, we have that: $$\phi_{2N}(0)=\sqrt{\frac{2}{\pi}}\frac{1}{(2N)^{1/4}}(1+O(N^{-1}))$$ and the otherhand we have that $\phi_{2N+1}(0)=0$.

Thus we see that: $$(\phi_{2N}(0)\phi_{2N-1}(x_{2N-1})-\phi_{2N}(x_{2N-1})\phi_{2N-1}(0))^2=(\phi_{2N}(0)\phi_{2N-1}(x_{2N-1}))^2$$ $$\ge (\sqrt{\frac{2}{\pi}}\frac{1}{2^{1/4}}C)^2N^{-10/12}(1+O(N^{-1}))$$

Which shows that your desired bound cannot hold.