Functional inequality on $\mathbb{Z}^d$

Question

Let $B_L = \{ x \in \mathbb{Z}^d : |x| < L\}$ be a box of side length $L \in \mathbb{N}$, where $| \cdot|$ is the $L_{\infty}$ norm and and assume $d \geq 3$. Let for any $L \in \mathbb{N}$ , $f_L : \mathbb{Z}^d \rightarrow \mathbb{R}$ be a function. Assume that the functions $f_L$ are such that there exists a constant $C < \infty$ such that for any $L$, $$ \sum_{x \in {B_L}} \sum_{y \in {B_L}} f_L(x-y) \Big ({\frac{1}{|x|+1}} \Big )^{d-2} \Big ({\frac{1}{|y|+1}} \Big )^{d-2} < C. $$ Does this imply that for any $x \in \mathbb{Z}^d$, $$ f_L(x) \rightarrow 0 $$ as $L \rightarrow \infty$?

Answer 1

Doesn't seem so. Let $f_L$ be $1$ at the origin $(0,0,\dots, 0)$ and $f_L=0$ everywhere else. Then $$ \sum_{x \in {B_L}} \sum_{y \in {B_L}} f_L(x-y) \Big ({\frac{1}{|x|+1}} \Big )^{d-2} \Big ({\frac{1}{|y|+1}} \Big )^{d-2} = \sum_{x \in {B_L}} \Big ({\frac{1}{|x|+1}} \Big )^{2d-4} $$ When $d\ge 5$, we have $2d-4\ge d+1$, and the sum $$ \sum_{x \in \mathbb{Z}^d} \Big ({\frac{1}{|x|+1}} \Big )^{d+1} $$ converges by comparison to $$ \int_{\mathbb{R}^d} \Big ({\frac{1}{|x|+1}} \Big )^{d+1} \,dx = c_d \int_0^1 \frac{r^{d-1}}{(r+1)^{d+1}}\,dr < \infty $$

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For $d=3,4$ one has to obtain more cancellation in the sum. For example, let $f_L=1$ at $(0,0,\dots,0)$ and $f_L=-1$ at $(1, 0, \dots, 0)$, and $f_L=0$ elsewhere. Then the sum $$ \sum_{x \in {B_L}} \sum_{y \in {B_L}} f_L(x-y) \Big ({\frac{1}{|x|+1}} \Big )^{d-2} \Big ({\frac{1}{|y|+1}} \Big )^{d-2} $$ splits into terms that nearly cancel each other: specifically, it is $$\sum_{x \in {B_L}} \Big ({\frac{1}{|x|+1}} \Big )^{d-2} \left[\Big ({\frac{1}{|x|+1}}\Big )^{d-2} - \Big ({\frac{1}{|x-e_1|+1}} \Big )^{d-2}\right] $$ where $e_1$ is the first standard basis vector. The expression in square brackets is $O(1/|x|^{d-1})$, so the sum is controlled by $$\sum_{x \in {B_L}} \Big ({\frac{1}{|x|+1}} \Big )^{2d-3}$$ which for $d=4$ yields a convergent series again: $2d-3 = 5$ is enough decay.

For $d=3$ one can get higher order cancellation using the values $1, -2, 1$ instead of $1, -1$, etc.