As far as I understand, you suppose that $\partial A$ is disjoint from $A$, and in the process you define $A_n=A_{n-1}\cup \partial A_{n-1}$,
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 $a$. 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$. 1 As far as I understand, you suppose that$\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$a$. 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.