Functional inequality on $\mathbb{Z}^d$ 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 $L \rightarrow \infty$?
 A: 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
$$
\int_{\mathbb{R}^d} \Big ({\frac{1}{|x|+1}}  \Big )^{d+1} \,dx
 = c_d \int_0^1 \frac{r^{d-1}}{(r+1)^{d+1}}\,dr < \infty
$$

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. 
