I am especially interested in trees but realized the question makes sense for any graph. I do mean undirected graphs, so the line << E is a subset of V x V - diagonal Should have read << E is a symmetric subset of V x V - diagonal. But I am surprised to hear that A^2n(f) (or A^(2n+1)(f)) might converge for *all* f. Can't I make f increase so fast in terms of distance from some fixed point P that A^(2n)(f)(P), say, gets arbitrarily large as n —> oo ? —Dan Andy Latto wrote: ----- On Sun, Mar 4, 2018 at 2:41 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Let G be an infinite connected graph
G = (V, E)
where V is a countably infinite set and E is a subset of V x V - diagonal,
You don't say whether (x, y) can be in E while (y, x) is not. Are you asking about directed or undirected graphs? The subject line refers to trees: did you intend to also specify the graph is acyclic?
satisfying 1) and 2):
----- 1) Every edge e in E has 2 endpoints, and any 2 vertices v,w in V bound at most one edge.
2) If for any v in V, deg(v) denotes the number of edges v belongs to, then for all v we have
2 <= deg(v) < oo.
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Question: Which f in C approach a limit under iteration of A:
lim A^n(f) = f^ n—>oo
for some f^ in C ???
I don't have a proof yet, but I assuming we're talking about undirected graphs, I believe that this will converge for all f if G is not bipartite. If G is a bipartite graph, which includes all acyclic graphs, then I think that lim A^2n(f) = f^ n—>oo and lim A^(2n+1)(f) = f^ n—>oo Will always both exist, but will generally be different. -----