# Brownian Motion on Nested Fractals by Tom Lindstrom By Tom Lindstrom

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Extra resources for Brownian Motion on Nested Fractals

Example text

6C- s is the C, then BROWNIAN MOTION ON NESTED FRACTALS 37 A Proof: Assume that with an n-cell another, s belongs to the n-complex D associated D. Since the walk moves from one N-neighbor to A s, „ must also be an element of k-1 D. But then sk_1€COD = COD by the Nesting Axiom. The last topic I shall address in this section is the question of how many n-cells an element in E may belong to. 11. 13 Proposition. , , where i i 4> €¥ i has belongs to another n-cell and F. y€F. By Symmetry, F . -•/••# 1 .

Z - 7 1 1+1 t . _,_l' >o, 1+1 if and only if A 7i =7t ^ for an odd number of indices i. ,s t. , t . , . l l+l l l+l To prove that p is continuous, we shall need the following simple lemma. 3 Lemma. ,p )6j) of basic transition probabilities is used in the construction of Bi. TOM LINDSTR0M -1 P r o o f : Let k = ( J - , m. ) . ,p is d e c r e a s i n g , r 1=. p. < ( . Z 1=1 and hence p, >k 1 1 1=1 whenever us that if we just choose F is connected to than N, and thus if l m. ,p ) € y. Proposition IV.

In all our standard examples the multiplicity is two, but for the modified Koch-curve in Figure 12 it is three. 15 Proposition. An element in E belongs to at most p n-complexes. Proof: According to the Nesting Axiom, an element x in E can only belong to more than one n-complex if it is an n-point. I shall prove by induction on p n-comlexes. For n=1 , n that an n-point belongs to at most this is just the definition of p (note that by the Nesting Axiom it doesn't matter whether we count cells or complexes).