Asympotic behaviour of $\sum_{z \in \mathbb{Z}^d} (1+ \| x-z\|)^{2-d} (1+ \| y-z\|)^{2-d}$

Question

I want to look at the asymptotic behaviour of a sum in $\mathbb{Z}^d$. We have two points $x,y \in \mathbb{Z}^d$ and consider the sum $$ \sum_{z \in \mathbb{Z}^d} (1+ || x-z ||)^{2-d} (1+ || y-z ||)^{2-d} $$
As all norms on $\mathbb{R}^n$ are equivalent, the concrete specification of the norm doesn't matter. I know this should diverge for $d \leq 4$ and behave like $||x-y||^{4-d}$ else but I'm unsure how to tackle this problem. First, I thought one could approximate it using the integral over $\mathbb{R}^n$ which would work if we had just one factor in the sum. Then maybe a clever rearrangement of the summands could help but haven't found one yet.

Answer 1

Here is a guideline on how not to prove this. For an elegant proof see the other answer (the answer was deleted... I have reproduced it at the end of this post).

Let's take $\Vert \cdot \Vert$ to mean the usual Euclidean norm. I'll assume that $2-d\leq 0$ and $\Vert x \Vert\geq 42$ for the rest of my answer. Before doing any estimate, I'd like to get rid of one of the variables. Shifting yields

\begin{align*} \sum_{z\in \mathbb{Z}^d} (1+\Vert x-z \Vert)^{2-d} (1+\Vert y-z \Vert)^{2-d} =\sum_{z\in \mathbb{Z}^d} (1+\Vert z \Vert)^{2-d} (1+\Vert (x-y)-z \Vert)^{2-d}. \end{align*} Thus, it is enough to consider the case $y=0$, which I will do from now on. We split our series into three parts and estimate them individually. \begin{align*} \sum_{z\in \mathbb{Z}^d} (1+\Vert z \Vert)^{2-d} (1+\Vert x-z \Vert)^{2-d} &= \sum_{z\in \mathbb{Z}^d \ : \ \Vert z\Vert < \Vert x \Vert/2} (1+\Vert z \Vert)^{2-d} (1+\Vert x-z \Vert)^{2-d} \\ &\qquad+\sum_{z\in \mathbb{Z}^d \ : \ \Vert x \Vert/2 \leq \Vert z\Vert \leq 2\Vert x \Vert} (1+\Vert z \Vert)^{2-d} (1+\Vert x-z \Vert)^{2-d} \\ &\qquad+\sum_{z\in \mathbb{Z}^d \ : \ \Vert z\Vert > 2\Vert x \Vert} (1+\Vert z \Vert)^{2-d} (1+\Vert x-z \Vert)^{2-d} \\ &=: \Xi_1+\Xi_2+\Xi_3, \end{align*} where $\Xi_i$ represent the $i$th term in the previous expression.

For the first term we note that for $\Vert z \Vert< \Vert x \Vert/2$ we have $$ \Vert x - z \Vert \geq \Vert x \Vert - \Vert z \Vert > \Vert x \Vert/2. $$ This implies \begin{align*} \Xi_1 \leq& (\Vert x \Vert/2)^{2-d} \sum_{z\in \mathbb{Z}^d \ : \ \Vert z\Vert < \Vert x \Vert/2} (1+\Vert z \Vert)^{2-d} \\ &\leq (\Vert x \Vert/2)^{2-d} \int_{\Vert z \Vert \leq \Vert x \Vert} \frac{1}{(\frac{1}{2}+\Vert z \Vert)^{d-2}} \\ &= \frac{C_d}{\Vert x \Vert^{d-2}} \int_0^{\Vert x \Vert} \frac{r^{d-1}}{(\frac{1}{2}+r)^{d-2}} dr \\ &\leq \frac{\widetilde{C}_d}{\Vert x \Vert^{d-4}}. \end{align*} Here I used $C_d, \widetilde{C}_d$ to denote constants depending only on $d$ and which we do not need to care about.

Next we deal with $\Xi_3$. We note that for $\Vert z \Vert \geq 2 \Vert x \Vert$ we have $$ \Vert z - x \Vert \vert \geq \Vert z \Vert - \Vert x \Vert \geq \Vert z \Vert/2. $$ Thus, we get \begin{align*} \Xi_3 &\leq 2^{d-2} \sum_{z\in \mathbb{Z}^d \ : \ \Vert z \Vert \geq 2\Vert x \Vert} \Vert z \Vert^{4-2d} \\ &\leq 2^{d-2} \int_{\Vert z \Vert \geq \Vert x \Vert} \Vert z \Vert^{4-2d} dz \\ &\leq C_d \int_{\Vert x \Vert}^\infty r^{3-d} dr \\ &\leq \widetilde{C}_d \Vert x \Vert^{4-d}. \end{align*} Finally we take care of $\Xi_2$. This term is somewhat similar to $\Xi_1$. We note that for $\Vert z \Vert \leq 2\Vert x \Vert$ we get $\Vert x-z\Vert \leq 3\Vert x \Vert$. With this observation we get \begin{align*} \Xi_2 &\leq 2^{d-2} \Vert x \Vert^{2-d} \sum_{z\in \mathbb{Z}^d \ : \ \Vert x \Vert/2 \leq \Vert z \Vert \leq 2 \Vert x \Vert} (1+\Vert x - z \Vert)^{2-d} \\ &\leq 2^{d-2} \Vert x \Vert^{2-d} \sum_{z\in \mathbb{Z}^d \ : \ \Vert z \Vert \leq 3 \Vert x \Vert} (1+\Vert z \Vert)^{2-d} \\ &\leq C_d \Vert x \Vert^{4-d}, \end{align*} where I have skipped passing to the integral and then to radial coordinates (it's the same as for $\Xi_1$).

Added: For some reasons the other answer was deleted by its author. Let me reproduce it here for people who cannot see the deleted answer. This is verbatim the deleted (up to fixed typos).

Take the infinite norm $\left\|\cdot \right\| = \left\|\cdot \right\|_\infty$, since for all $n\in \mathbb Z^d$, $u\in \left[n-\frac12\times \boldsymbol 1, n + \frac12\times\boldsymbol 1\right]$,

$$\left\|u \right\| \le \left\|n \right\| + \left\|u - n\right\| \le \left\|n \right\| + \frac12$$

and

$$\left\|u - x\right\| \le \left\|n - x\right\| + \left\|u - n\right\| \le \left\|x - n\right\| + \frac12$$

Hence, with $v=x/\vert x \vert$ one gets

\begin{align} \sum_{n\in \mathbb Z^d} \frac1{\left(1 + \left\|n\right\|\right)^{d-2}\left(1 + \left\|x-n\right\|\right)^{d-2}} &\le \sum_{n\in \mathbb Z^d} \int_{\left[n-\frac12\times \boldsymbol 1, n + \frac12\times\boldsymbol 1\right]} \frac1{\left(\frac12 + \left\|u\right\|\right)^{d-2}\left(\frac12 + \left\|x - u\right\|\right)^{d-2}} \mathrm d u\\ &=\int_{\mathbb R^d} \frac1{\left(\frac12 + \left\|u\right\|\right)^{d-2}\left(\frac12 + \left\|x - u\right\|\right)^{d-2}}\mathrm d u & \left(s := \frac1{\left\|x\right\|}u\right)\\ &= \int_{\mathbb R^d}\frac1{\left(\frac12 + \left\|\left\|x\right\|s\right\|\right)^{d-2}\left(\frac12 + \left\|x - \left\|x\right\|s\right\|\right)^{d-2}}\left\|x\right\|^d\mathrm d s\\ &= \frac1{\left\|x\right\|^{d-4}}\int_{\mathbb R^d} \frac1{\left(\frac1{2\left\|x\right\|} + \left\|s\right\|\right)^{d-2}\left(\frac1{2\left\|x\right\|} + \left\|v - s\right\|\right)^{d-2}}\mathrm d s\\ &\le \frac1{\left\|x\right\|^{d-4}}\int_{\mathbb R^d} \frac1{\left\|s\right\|^{d-2} \left\|v - s\right\|^{d-2}}\mathrm d s \end{align}

Finally, with the substitution $t=Us$ where $U\in \text{Mat}_{d\times d}(\mathbb{R})$ is a rotation such that $Ue_1=v$ (with $e_1$ being the first standard basis vector) we obtain \begin{align*} \sum_{n\in \mathbb Z^d} \frac1{\left(1 + \left\|n\right\|\right)^{d-2}\left(1 + \left\|x-n\right\|\right)^{d-2}} \leq \frac1{\left\|x\right\|^{d-4}}\int_{\mathbb R^d} \frac1{\left\|t\right\|^{d-2} \left\|e_1 - t\right\|^{d-2}}\mathrm d t. \end{align*} Now the integral does not depend on $x$ anymore and is finite. To see that the integral is finite, we can estimate it as

\begin{align*} \int_{\mathbb{R}^d} \frac{1}{\Vert t \Vert^{d-2} \Vert t-e_1\Vert^{d-2}} dt &= \int_{\mathbb{R}^d \setminus (B(0,1/2) \cup B(e_1,1/2))} \frac{1}{\Vert t \Vert^{d-2} \Vert t-e_1\Vert^{d-2}} dt \\ &\quad + \int_{B(0,1/2) \cup B(e_1,1/2)} \frac{1}{\Vert t \Vert^{d-2} \Vert t-e_1\Vert^{d-2}} dt \\ &\leq \left( \int_{\Vert t \Vert \geq 1/2} \frac{1}{\Vert t \Vert^{2(d-2)}} dt \right)^{1/2} + 2 \cdot 2^{d-2}\int_{B(0,1/2)} \frac{1}{\Vert s \Vert^{d-2}}dt<\infty. \end{align*}