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Let $T \in\left(0,1\right)$, $n\in\mathbb{N}$ and $e_n = [1,\ldots,1]\in\mathbb{R}^n$. Consider the covariance matrix $\mathfrak{A}_n = \left[sinc\left(\frac{T\left(r-s\right)}{n}\right)\right]^n_{r,s=1}\in\mathbb{R}^{n\times n}$ in which $sinc(x) = \frac{\sin\left(x\right)}{x}$. How can we prove that for large enough $n$, there is a universal constant $M > 0$ such that

\begin{equation} e^T_n\mathfrak{A}^{-1}_ne_n \leq M \end{equation}

So, the sum of the entries of $\mathfrak{A}^{-1}_n$ is bounded. My simulation shows that the above inequality is valid. I've tried to use the following result to overcome the lack of knowledge about $\mathfrak{A}^{-1}_n$.

For any $w\in\mathbb{R}^n$ and any positive definite (invertible) matrix, $\Sigma\in\mathbb{R}^{n\times n}$,

\begin{equation*} w^T \Sigma^{-1}w = \frac{1}{\min_{v^Tw = 1} v^T\Sigma v} \end{equation*}

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  • $\begingroup$ How to see that it is covariance matrix? Is it a correlation matrix? $\endgroup$
    – M. Lin
    Dec 3, 2014 at 21:14
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    $\begingroup$ The function $sinc\left(Tt\right)$ has a non-negative Fourier transform. So to prove the positive definiteness, you just need to write the entries of $\mathfrak{A}_n$ in terms of their Fourier transform. You can even replace the sinc function with any function with non-negative Fourier transform. $\endgroup$
    – Student
    Dec 4, 2014 at 4:23
  • $\begingroup$ Did you find any proof for that? Or an approximation for $M$ $\endgroup$ Aug 23, 2021 at 13:53

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