**Upload:** the general question has been answered in the negative. Now I am proposing a more specific question, which is actually the one that I am interested in.


I think that this is a quite natural question, but I was able neither to find anything in literature, nor to find an argument by myself.


Let $G=(V,E)$ be an infinite, locally finite, connected graph (maybe without loops and multiple edges) equipped with the shortest path metric. Given a finite subset $A$ of $V$, let $\partial A$ denote the set of vertex at distance $1$ from $A$. There are many definitions of isoperimetric constant. I am adopting the following one: $\iota(G)$ is the infimum of $\frac{|\partial A|}{|A|}$, when $A$ runs over the finite non-empty subsets of $V$.


Now, take an increasing and covering sequence $A\_n$ of finite non-empty subsets of $V$ and define the number $j$ to be the infimum of the $\frac{|\partial A\_n|}{|A\_n|}$'s.


The general question concerned possible relations between $j$ and $\iota(G)$. In particular, I was interesting in the following



> 
> **Original question:** Is it possible that $\iota(G)=0$ and $j>0$?
> 
> 
> 


It has trivially negative answer: take a tree with degree $\geq3$ and attach a copy of $\mathbb Z$ to one vertex.


The question that I am interested in is actually more specific. Fix a point $x$ and consider the sequence $A\_0=x$, $A\_1=\partial A\_0$, ... $A\_n=\partial A\_{n-1}$. Define $j(x)$ to be the infimum of the $\frac{|\partial A\_n|}{|A\_{n-1}|}$'s. Let $j$ be the infimum over $x$ of the $j(x)$'s. What about the following question?



> 
> **New question 1:** Is it possible that there are vertex $x$ and $y$ such that $j(x)\neq j(y)$?
> 
> 
> **New question 2:** Is it possible that $j=0$ and there exists $x$ such that $j(x)>0$?
> 
> 
> 


Basically I am looking for some local version of the isopetric constant...



How much local is the information encoded by the isoperimetric constant of a graph?
As far as I understand, you suppose that $\partial A$ is disjoint from $A$, and in the process you define $A\_n=A\_{n-1}\cup \partial A\_{n-1}$,


Under these assumptions, the answer to both questions is negative.


Consider a binary tree with a root vertex $a$ (that is, $a$ has degree 1, and each other vertex has degree 3). Next, for every positive integer $d$, identify all the vertices at distance $3d$ from $a$ (obtaining a vertex $v\_d$; the vertices $v\_d$ are different for differet $d$). Finally, take two disjoint copies of this graph and identify their root vertices (denote the resulting vertex by $a$ again).


Now, $j=j(a)=0$ since $|\partial A\_{3d-1}|=2$. On the other hand, let $b$ be one of the two neighbors of $a$. Then $|A\_n|\leq 2^{n+2}$, but $|\partial A\_n|\geq 2^{n-1}$, so $j(b)\geq 1/8$. Thus we have constructed a counterexample to both conjectures.


On the other hand, you may ask the same questions for a graph with a bounded degree. Then the answer to the second question is affirmative, though there still exist counterexamples to the first one.


Here is a counterexample in this case. Instead of glueing vertices, we will thin out a tree. First, take the same binary tree with a rot vertex $a$. Now, for some $n\_1$ large enough, delete some vertices of the $n\_1$th layer so that $|\partial A\_{n\_1-1}|\approx |A\_{n\_1-1}|/1000$ (we delete each vertex together with a subtree hanging on it). Note that now we have, say, $|\partial A\_{n\_1}|\geq |A\_{n\_1}|/800$. Next, take $n\_2$ much larger than
$n\_1$ and thin out $n\_2$th layer in the same manner, and so on. Finally, glue two copies of such tree at vertex $a$ again. The we have $j(a)=1/1000$ but $j(b)>1/1000$ if $b$ is a neighbor of $a$. 


