# Convolutions and Toeplitz Operators

Let be $d>0$ an integer number and consider the Cartesian product $\mathbb Z^d$ as metric space, with the distance between $x,y\in\mathbb Z^d$ given by $\|x-y\|_1=\sum_{j=0}^d|x_j-y_j|$.

Let be $g:[0,\infty)\to\mathbb [0,\infty)$ a function having the two following properties:

1) $\sum_{z\in \mathbb Z^d}g(\|z\|_1)$ is convergent;

2) there is a positive constant $K\in \mathbb R$ (which depends only on $g$) such that for any $x,y\in\mathbb Z^d$, we have
$$\sum_{z\in\mathbb Z^d}g(\|x-z\|_1)g(\|z-y\|_1)\leq K g(\|x-y\|_1),$$

Question: Can we determine lower bounds for the ratio decay of $g(\|x-y\|_1)$ when $\|x-y\|_1$ goes to infinity ?

Examples:

Ex1: For any $\varepsilon>0$ $$g(\|z\|_1)=\frac{1}{1+\|z\|_1^{d+\varepsilon}}$$ has the properties 1 and 2.

For the other hand, $$g(\|z\|_1)=e^{-r\|z\|_1},$$ where $r>0$, breaks the property 2.

Edit:

I added the Toeplitz operator tag, because of the asymptotic behavior for $g$ (in terms of the lower bounds) could be obtained thinking $g(\|x-y\|_1)$ as matrix elements of a Toeplitz operator $A:L^p(\mathbb Z^d,2^{\mathbb Z^d},\sharp)\to L^p(\mathbb Z^d,2^{\mathbb Z^d},\sharp)$. In fact, in this point of view, we ask for lower bounds for the entries of a Toeplitz operator satisfying $(A^2)_{xy}\leq K A_{xy}$, where $(A^2)_{xy}$ is the $xy$ element of the matrix $A^2$.

Remark: The issues pointed out by Thomas Kragh in the comments were fixed by not considering $g$ as a function of space $\mathbb Z^d$ and requiring it to be positive.

Any reference or help, even for partial answer is very welcome.

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What is $x-z$ when $z \in \mathbb{Z}^d$ and $x\in \mathbb{Z}$? and if $d\neq 2$ what is $g(x,y)$? If you mean $x,y\in \mathbb{Z}^d$ then again what is $g(x,y)$ for any $d$? – Thomas Kragh Apr 27 '10 at 8:23
Sorry for the mistypes. I edited the question correcting it. So $g(x,y)$ should be $g(x-y)$ and $x$ and $y$ in fact, belongs to $\mathbb Z^d$. – Leandro Apr 27 '10 at 8:43
but now 2 implies a stronger 2 where $K=0$. Indeed $\sum g(x-z)g(z-y) = \sum g(x+y-z)g(z)$ which for fixed $x+y$ can have $x-y$ going to infinity, so the left hand side is always zero. – Thomas Kragh Apr 27 '10 at 9:17