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Consider the (physicist's) Hermite polynomials $H_n(x)$ which are divided by $$\sqrt{\sqrt{\pi} 2^n n!}$$ for the purpose of normalization.

These are orthogonal with respect to the weight function $e^{-x^2}$ and $\int_{\mathbb{R}}H_n(x)^2\,e^{-x^2}\,dx=1$. Define the family of functions $$f_N(x,y)=e^{-\frac{N}2(x^2+y^2)}\sum_{\ell=0}^{N-1} H_\ell(x) H_\ell(y).$$ Using tools from Random Matrix Theory, the following when treated as the variance of the number of eigenvalues in an interval, it is proven positive. However, I would like to ask:

QUESTION. Is there a direct proof of the below inequality that does not involve RMT? $$\int_a^bf_N(x,x)\,dx-\int_a^b\int_a^bf_N(x,y)^2 \, dx \, dy>0 \qquad \text{for all $N\geq1$}$$

Remark. It turns out that the above inequality is in fact valid for other orthogonal polynomials.

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As you said, the inequality holds for the orthonormal polynomials $P_{k}$, $k\geq0$, of any positive measure $\mu$, with support $K$, a compact subset of $\mathbb{C}$. It follows from a basic result about the Christoffel-Darboux kernel $K_{N}(w,z)$ and the Christoffel function $\lambda_{N}(z)$, associated to $\mu$, respectively defined by $$ K_{N}(w,z)=\sum_{k=0}^{N-1}P_{k}(w)\overline{P_{k}(z)},\qquad \lambda_{N}(z)=\min_{\deg P\leq N-1,~P(z)=1}\int_{K}|P(w)|^{2}d\mu(w).$$ Theorem : One has $$ \lambda_{N}(z)=\frac{1}{\sum_{k=0}^{N-1}|P_{k}(z)|^{2}}=\frac{1}{K_{N}(z,z)},\qquad z\in\mathbb{C},$$ and the minimum is attained (uniquely) when $$ P(w)=\Pi_{N,z}(w) :=K_{N}(w,z)/K_{N}(z,z). $$ For completeness, here is the proof : let $P(w)=\sum_{k=0}^{N-1}\alpha_{k}P_{k}(w)$ be any polynomial of degree $\leq N-1$. Then $$ \lambda_{N}(z)^{-1}=\max_{\alpha_{k}} \frac{|\sum_{k=0}^{N-1}\alpha_{k}P_{k}(z)|^{2}} {\sum_{k=0}^{N-1}|\alpha_{k}|^{2}}\leq \sum_{k=0}^{N-1}|P_{k}(z)|^{2}, $$ and the Cauchy-Schwarz inequality becomes an equality when $\alpha_{k}=\overline{P_{k}(z)}$, $k=0,\ldots,n$, which also gives the expression for the minimizing polynomial $\Pi_{N,z}(w)$.

Making use of the above result, we have, for any subset $K_{0}$ of $K$, \begin{align*} \int_{K_{0}}\int_{K_{0}} |K_{N}(w,z)|^{2}d\mu(z)d\mu(w) & =\int_{K_{0}}K_{N}(z,z)^{2}\int_{K_{0}}|\Pi_{N,z}(w)|^{2}d\mu(w)d\mu(z) \\[5pt] & \leq\int_{K_{0}}K_{N}(z,z)^{2}\lambda_{N}(z)d\mu(z) =\int_{K_{0}}K_{N}(z,z)d\mu(z). \end{align*} By the way, I think there is a missing normalization of the variables $x$ and $y$ in the Hermite polynomials $H_{l}$, in your definition of $f_{N}(x,y)$.

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  • $\begingroup$ Trying to learn here, I think I got an idea of the proof but not sure how $\lambda_{N}(z)=\min_{\deg P\leq N-1,~P(z)=1}\int_{K}|P(w)|^{2}d\mu(w)$ translates into $\lambda_{N}(z)^{-1}=\max_{\alpha_{k}}\frac{|\sum_{k=0}^{N-1}\alpha_{k}P_{k}(z)|^{2}}{\sum_{k=0}^{N-1}|\alpha_{k}|^{2}}$ if it were argmin instead of min it would look a bit like linear regression ($\hat\beta=\mbox{argmin}||X-\beta Y||\rightarrow \hat\beta=(X^TX)^{-1}XY)$ but still can't make sense of how things go $\endgroup$
    – Dabed
    Commented May 27 at 18:17
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    $\begingroup$ @Dabed the denominator in the ratio is the L2-norm squared of the polynomial (with respect to $\mu$). The numerator accounts for the normalization $P(z)=1$. $\endgroup$
    – user111
    Commented May 27 at 18:58
  • $\begingroup$ Yes but (and I should probably not insist much more since is my who is lacking in understanding) is that I fail to see what happens in between the two expressions $\lambda_{N}(z)^{-1}=(\min_{\deg P\leq N-1 ,~P(z)=1}\int_{K}|P(w)|^{2}d\mu(w))^{-1} =...?...= \max_{\alpha_{k}}\frac{|\sum_{k=0}^{N-1}\alpha_{k}P_{k}(z)|^{2}}{\sum_{k=0}^{N-1}|\alpha_{k}|^{2}}$ $\endgroup$
    – Dabed
    Commented May 27 at 20:49

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