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Extra info for Spectral geometry: direct and inverse problems
Example text
43) that p nt (zi , zi+1 ) ≥ c . 41), we obtain pt (x, y) ≥ ... B(x1 ,r) n = = ≥ ≥ cn dμ(zn−1 ) . . dμ(z1 ) V (x, r) V (x1 , r) . . V (xn−1 , r) B(xn−1 ,r) cn c ≥c V (x, r) V x, (t/n)1/β cn V x, t1/β cn nα /β c ≥c 1/β 1/β V x, t V x, (t/n) V x, t1/β c exp (−Cn) V x, t1/β ⎛ 1 ⎞ β−1 β c d (x, y) ⎠. 27). 6. 5, we have E (u) all u ∈ L2 (M ). Consequently, F = W β/2,2 . Eβ (u) for Proof. 2 we have E (u) ≥ cEβ (u). 5), we obtain E (u) ≤ CEβ (u), which finishes the proof. G. Aronson, Non-negative solutions of linear parabolic equations Ann.
C below), amenability of the groups BA(X) for all finite sets X (therefore, amenability of all groups generated by bounded automata) would follow from amenability just of all the Mother groups M(X). It is worth noting that the groups generated by bounded automata form a subclass of the class of contracting self-similar groups (see [BN03, Nek05]). It is still an open question whether all contracting groups are amenable. However, Nekrashevych [Nek08] recently established a weaker property: contracting groups contain no free groups with ≥ 2 generators.
9. Given two points x, y ∈ M , a chain connecting x and y is any finite sequence n {xk }k=0 of points in M such that x0 = x, xn = y. We say that a metric space satisfies the chain condition if there is a constant C > 0 such that for any positive integer n and for all x, y ∈ M there is a chain {xk }nk=0 connecting x and y, such that d (x, y) for all k = 0, 1, . . , n − 1. 25) d (xk , xk+1 ) ≤ C n For example, the geodesic distance on any length space satisfies the chain condition. On the other hand, the combinatorial distance on a graph does not satisfy it.