Let $a_{n,m}:\mathbb{Z}^2\to\mathbb{R}_{\geq0}$ be a double sequence of positive numbers. I'm trying to understand something about decay conditions for this sequence.
Are the following two conditions equivalent?
Condition 1 There are two positive constants $\alpha,\beta>0$ such that $\sum_{n,m}a_{n,m}e^{+\alpha|n-m|-\beta|m|}$ is finite.
Condition 2 There are three positive constants $C,\gamma,\delta>0$ such that $a_{n,m}\leq C e^{-\gamma|n-m|+\delta|m|}$ for all $n,m$.
I think they are, as I detail below, but I would like to know that I'm not missing something:
1 to 2 Let $\gamma,\delta>0$. Because all terms are positive, we have $$ a_{n,m}e^{+\gamma|n-m|-\delta|m|} \leq \sum_{n',m'}a_{n',m'}e^{+\gamma|n'-m'|-\delta|m'|} < \infty $$ where the last inequality follows if we choose $\gamma:=\alpha,\delta:=\beta$. Since this last RHS is finite for all $n,m$, we define that as $C$ and multiply by $e^{-\gamma|n-m|+\delta|m|}$ to get the condition.
2 to 1 We have for $\alpha,\beta>0$, $$\sum_{n,m}a_{n,m}e^{+\alpha|n-m|-\beta|m|}\leq\sum_{n,m}C e^{-\gamma|n-m|+\delta|m|+\alpha|n-m|-\beta|m|} $$so pick $\alpha:=\frac{1}{2}\gamma,\beta:=2\delta$ to get the the RHS is finite, if we first sum over $n$ (shifting by $m$ by translation invariance of $\mathbb{Z}$) and then over $m$.
It is not obviuous to me that $\sum_{m,n} e^{-a|n-m|-b|m|} <\infty$ for all positive numbers a and b. However, using the inequality $|n-m| \geq |n|-|m|$ we can see that the series converges if $b>a$. So if you make a better choice of $\alpha$ and $\beta$ in the proof of 2) implies 1) it works fine. You can take $\alpha =\frac {\gamma} n$ and $\beta =n \delta$ with $n>1+\frac \gamma \delta$.