Sorry if this question has been asked already but I didn't find it. Given a symmetric Toeplitz matrix of the form
$\left[\begin{array}{llll} a_0 & a_1 & \dots & a_n\\ a_1 & a_0 & \dots & a_{n-1}\\ \vdots& & & \\ a_n & a_{n-1} & \dots & a_0\\ \end{array} \right]$
Suppose we have the relation $a_0\geq a_1\geq\dots\geq a_n$. A simple $3 \times 3$ example will show you that the inverse need not have this property (even in terms of absolute value). But does it at least have the property that away from the main diagonal, the difference decreases? That is, do we have $|a_1-a_0|\geq|a_2-a_1|\geq\dots\geq|a_n-a_{n-1}|$?
After some experimenting, it seems like we might be able to expect this, but I am wondering if this is known already as I know very little about Toeplitz forms? A followup question I have is if this is true, then is it "easy" to see that it should remain true for bi-infinite Toeplitz matrices of the same form?
The actual matrix I am interested in, for what it's worth, is $A_\lambda = (e^{-\lambda|j-k|^2})_{j,k\in\mathbb{Z}}$.