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I'm computing the lower bound for the Hausdorff dimension of a Cantor-like set; I've reduced it to computing $\lim_{k\rightarrow \infty, \delta \rightarrow 0^+}\frac{1}{(1+\delta)^{k(p-1)}}$, where $p$ is some prime. This limit depends on the relationship between $\delta, k$, otherwise it is undefined. (

I arrived to this limit by playing with an expression (for which no relation between $\delta, k$ was specified, but we eventually want to take the limit as $\delta$ tends to 0) $(*) \lim_{k\rightarrow \infty}\frac{\log(m_p*...*m_{kp})}{\log(n_{k+1})}=\lim_{k\rightarrow \infty}\frac{\log(n_0^{1+\delta}*...*n_0^{(1+\delta)^p})}{\log(n_0^{(1+\delta)^kp})}=\frac{(1+\delta)+...+(1+\delta)^p}{(1+\delta)^{kp}}$ I simplified this limit to get that it is $\ge \lim \frac{1}{(1+\delta)^{k(p-1)}}$

Note all the $\delta$s that appear are the same $\delta$. In the process of arriving at this limit, I never assumed there was a relationship between $\delta$ and $k$. Since I want the value of this limit to be a positive number less than $1$, I want to pick a specific relationship between $\delta$ and $k$ that will allow this to happen- which I am allowed to do. However, once I define a relationship $\delta_k=f(k)$, will this violate the steps I used to arrive to my limit expression if it involved $\delta, k$?

My instinct says yes because once I define $\delta_k=f(k)$, then each $\delta$ in the last equation of $(*)$ will be different, since it is coming from a different index $k$. Therefore, my steps need to be reconsidered.

On the other hand, all that was known originally that $\delta \rightarrow 0^+$, so why should it matter which path I choose for $\delta$ to approach zero? And therefore I should be able to do this without adjusting the steps used to get my limit lower bound expression.

Do I have to readjust my calculation and place a $\delta_k$ wherever there is a $\delta$ once I specify this relationship?

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up vote 3 down vote accepted

Bounding the Hausdorff dimension from below is hard. It is harder to estimate than the Minkowski dimension precisely because you cannot work with coverings on any particular scale. By definition the Hausdorff measure is a $\liminf $, and not just over all scales $\delta$ but also over all multiscale coverings.

The most common way to get a bound from below is to exhibit a positive measure $\mu$ which is supported on the set and satisfies $\mu (B)\le C (\mathrm {diam}\, B)^d$ for every ball $B$. This will imply that the Hausdorff dimension is at least $d$, by the (easy) converse of Frostman's lemma.

I don't claim to understand your attempted proof, but I'll be surprised if it can be easily made rigorous.

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This isn't my attempted proof, I just want to know if my steps $(*)$ need to be readjusted once I define a relationship between $\delta, k$ by making $\delta$ approach $0$ along the path $1/k$. – The Substitute Oct 1 '12 at 8:27

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