Local dimension of fractal measure associated with the $(0, 1, a)$ - problem: the case $a = 6$

Abstract

Let X1,X2, ... be a sequence of independent, identically distributed(i.i.d) random variables each taking values 0, 1, a with equal probability 1/3. Let μ be the probability measure induced by S = 􀀟 ∞n =1 3−nXn. Let α(s) (resp.α(s), α(s)) denote the local dimension (resp. lower, upper local dimension) of s ∈ supp μ, and let α = sup{α(s) : s ∈ supp μ}; α = inf{α(s) : s ∈ supp μ} E = {α : α(s) = α for some s ∈ supp μ}. In the case a = 3, E = [2/3, 1], see [6]. It was hoped that this result holds true with a = 3k, for any k ∈ N. We prove that it is not the case. In fact, our result shows that for k = 2(a = 6), α = 1, α = 1 − log(1+√5)−log 2 2 log 3 ≈ 0.78099 and E = [1 − log(1+√5)−log 2 2 log 3 , 1].Let X1,X2, ... be a sequence of independent, identically distributed(i.i.d) random variables each taking values 0, 1, a with equal probability 1/3. Let μ be the probability measure induced by S = 􀀟 ∞n =1 3−nXn. Let α(s) (resp.α(s), α(s)) denote the local dimension (resp. lower, upper local dimension) of s ∈ supp μ, and let α = sup{α(s) : s ∈ supp μ}; α = inf{α(s) : s ∈ supp μ} E = {α : α(s) = α for some s ∈ supp μ}. In the case a = 3, E = [2/3, 1], see [6]. It was hoped that this result holds true with a = 3k, for any k ∈ N. We prove that it is not the case. In fact, our result shows that for k = 2(a = 6), α = 1, α = 1 − log(1+√5)−log 2 2 log 3 ≈ 0.78099 and E = [1 − log(1+√5)−log 2 2 log 3 , 1].