Reformulation of amenability and growth rate of a group in terms of general metric spaces. Updates: Changed a bit the definition to include infinite dimensional Banach spaces; Included questions 0 and subquestion. Improved (I hope) the presentation.
The growth rate of a group, being a metric property, should have some reformulation just in terms of metric spaces. Also the amenability, thinking of Folner's characterization (which basically means that we can find arbitrarily big set with very small boundary), should have some reformulation in terms of metric space (there are indeed some). Now, before proving that convex-like structures in the sense of Nate Brown are embeddable into Banach spaces and solving (even too easily) all my problems, I have spent some time playing with some strange metric spaces, catching a general property that seems to do some job. 
Let $(X,d)$ be a metric space. If $x\in X$ and $R>0$, I denote with $B(x,R)$ the open ball and with $S(x,R)$ the sphere around $x$ with radius $R$.

Definition: Let $p_R(x)$ be the number (possibly infinite) of open, continuous and injective maps $f:B(x,R)\rightarrow X$ such that 
  
  
*
  
*$f(B(x,R))\subseteq S(x,R)$
  
*the images $f(B(x,R))$ are mutually disjoint
  
  
  I say that $(X,d)$ has the property SB (Small Boundary) if for all $x\in X$ one has $\sum_{R>0}p_R(x)<\infty$.

Examples: Banach spaces have this property. The metric linear space $(\mathbb R,d)$, with $d(x,y)=min(1,|x-y|)$ does not have SB (indeed $p_1(0)=\infty$).
Now let $G$ be a finitely generated group with a fixed symmetric set of generators and consider the word distance with respect to this generating set. Let $x=e$, being $e$ the identity element in $G$. Besides the questions number 0: does $p_R(e)$ depend on $S$? (maybe); and  does the $\sup$ of the definition depend on $S$? (.... sometimes I believe in God ...)  I guess that property SB should have some relation with the amenability and with the growh rate of $G$. For instance it seems to me quite evident that Subexponential growth implies SB. A very natural question now would be

Question 1: Does amenability imply SB?

It seems to me that one property that should be true for groups and that would make some thing easier is

Subquestion: Is $p_R(e)$ non-decreasing in $R$?

another interesting question would be so,e converse of question 1:

Question 2: Does SB for metric spaces (probably with some extra-condition) implies the existence of some good measure?

I have no more ideas at the moment, but still think that would be nice to find some (maybe not this one) unifying concept.
Thanks in advance for any comments,
Valerio1.
 A: Some thoughts...

Definition:
  Let $(P,\leq)$ be a partially ordered set. A finite subset
  $C=(c_1,c_2,...c_k)$ of $P$ is called complete chain if
  
  
*
  
*$C$ is a chain
  
*If $p\in P$ is such that $c_i\leq p\leq c_{i+1}$, then
  $p=c_i$, or $p=c_{i+1}$
  
  
  The number $k$ is called length of the chain.

If $P=[0,\infty)$ with the standard order, then any complete chain
is formed by just one point. If $P=\mathbb N$ with the standard
order, then any finite subset of consecutive numbers is a complete
chain.
Let $(X,d)$ be a metric space, $A\subseteq X$ an open and bounded subset and $k$ be a positive integer.

Definition: The discrete $k$-neighborhood of $A$ is the set $N_k(A)$ of elements $x\in X$ such that the set $d(x,a)$, $a\in A$, is a complete chain of length at most $k$ inside the set $P=d(y,A)$, $y\in X$, with the usual order.

Let $A$ be fixed, for any positive integer $n$, let $k(n)$ be the smallest positive integer $k$ such that there is a continuous, open and injective map $f:X\rightarrow X$
such that $f(A)\subseteq N_k(A)\setminus A$.
Now let $s(k(n))$ be the greatest positive integer (possibly
infinite) $s$ such that there are continuous, open and injective functions $f_1,...f_s: X\rightarrow X$
such that


*

*$f_i(A)\subseteq N_k(A)\setminus A$

*$f_i(A)\cap f_j(A)=\emptyset$, for all $i\neq j$



Definition: I say that the metric space $(X,d)$ has the small boundary property
  (property SB) if, for all open and bounded subset $A$ of $X$, whenever
  $$
\sum_{n\in\mathbb N}s(k(n))=\infty
$$
  then $k(n)$ is not bounded.

It should be clear that (even infinitely dimensional) Banach
spaces have the property SB. On the other hand, the metric linear
space $(\mathbb R,d)$, with $d(x,y)=\min(1,|x-y|)$ does not have
property SB, since $s(k(1))=\infty$, for $A$ equal to the open unit ball about $0$ - notice that one has $k(1)=1$.
The following result shows that this could be the right way.

Fact: Let $G$ be a finitely generated group, let $S$ be a symmetric
  generating set and let $d_S$ be the word distance with respect to
  $S$. Then $G$ is amenable if and only if $(G,d_S)$ has the property
  SB.

Proof: Recall the doubling condition: $G$ verifies the doubling condition if there is $k>0$ such that $|N_k(F)|\geq2|F|$, for all finite subset $F$ of $G$, where here $N_k(F)$ is the set of elements $g$ of $G$ such that $d_S(g,F)\leq k$. It is known that the doubling condition is equivalent for a group to be non-amenable. On the other hand, it seems to me that in this case the property SB reduces to the negation of the doubling condition.
Comments? New ideas? Some application for metric spaces?
Bibliography: the fact that amenability is equivalent to the doubling condition is proved in Theorem 32 of
T. Ceccherini–Silberstein, R.I. Grigorchuk and P. de la Harpe: Amenability and paradoxical decompositions for pseudogroups and for discrete metric spaces. Proc. Steklov Inst. Math. 224 (1999),
57-97
