$\textbf{Definition 1.}$ Let $\{F_n\}_{n \in \mathbb{N}}$ be a sequence of finite subsets of $\mathbb{Z}^d$. We say that $\{F_n\}_{n \in \mathbb{N}}$ is a Folner sequence if for every $v \in \mathbb{Z}^d$, $\lim_{n \rightarrow +\infty} \frac{|(F_n + v) \Delta F_n|}{|F_n|} = 0$.
$\textbf{Definition 2.}$ Let $\{F_n\}_{n \in \mathbb{N}}$ be a sequence of finite subsets of $\mathbb{Z}^d$. We say that $\{F_n\}_{n \in \mathbb{N}}$ is a Folner sequence if $\lim_{n \rightarrow +\infty} \frac{|\partial F_n|}{|F_n|} = 0$, where $\partial F = \{v \in \mathbb{Z}^d : v \notin F, d(v, F) = 1\}$.
What I did was the following:
$(1 \implies 2)$ First of all, note that as $\{F_n\}_{n \in \mathbb{N}}$ is a Folner sequence in the sense of definition $1$, we have that $\lim_{n \rightarrow +\infty} \frac{|(F_n \pm e_i) \Delta F_n|}{|F_n|} = 0$, for all $i \in \{1, \dots, d\}$. Thus, $\lim_{n \rightarrow +\infty} \frac{|(F_n \pm e_i) \setminus F_n|}{|F_n|} = 0$, for all $i \in \{1, \dots, d\}$.
Now, note that $|\partial F_n| \leq |(F_n + e_1) \setminus F_n| + \dots + |(F_n + e_d) \setminus F_n| + |(F_n - e_1) \setminus F_n)| + \dots + |(F_n - e_d) \setminus F_n|$. The, by the remark made above, we have that
$\lim_{n \rightarrow +\infty} \frac{|\partial F_n|}{|F_n|} = 0$.
Now, I have two questions:
1) Is it clear (and correct) the argument above?
2) I'm having some trouble to show that $2 \implies 1$. Could someone help me?
Thanks in advance!
Given $v$, pick a sequence $v_0=0,v_1,\ldots, v_m=v$ such that $d(v_i,v_{i+1})=1$.
Now note that a point $x\in(F_n+v)\mathrel{\Delta} F_n$ is either a point $x\in F_n$ such that $x-v\notin F_n$ or vice versa. Thus in the sequence $x-v_0,x-v_1,\ldots, x-v_m$, we encounter a switch form $\in F_n$ to $\notin F_n$ or vice versa. Thus to every point in $(F_n+v)\mathrel{\Delta} F_n$, we can associate an index $i\in\{0,\ldots, m\}$, a point in $\delta F_n$, and an additional "bit" whether $x\in F_n$ or $x\in F_n+v$. We conclude $$|(F_n+v)\mathrel{\Delta} F_n|\le |\delta F_n|\cdot (m+1)\cdot 2.$$