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:
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On Sun, Mar 4, 2018 at 2:41 PM, Dan Asimov <dasimov(a)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.
> -----
> 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.
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