Why is it enough to consider limits as $\to 0$ through $\{_k\}$ such that $_{k+1} \ge c_k$ for some $0 < c < 1$ to find $\dim_B F$?

Question

To set the stage, let me recall the definition of the box-counting dimension of a set $F \subset \mathbb R^n$.

The lower and upper box-counting dimensions of a subset $F$ of $ℝ^n$ are given by $$\underline{\dim}_B F = \underline{\lim}_{\delta\to 0} \frac{\log N_\delta(F)}{-\log \delta}$$ $$\overline{\dim}_B F = \overline{\lim}_{\delta\to 0} \frac{\log N_\delta(F)}{-\log \delta}$$ and the box-counting dimension of $F$ by $$\dim_B F = \lim_{\delta\to 0} \frac{\log N_\delta(F)}{-\log \delta}$$ (if this limit exists), where $N_\delta(F)$ is the smallest number of sets of diameter at most $$ that cover $F$.

Now, the author says that to find $\dim_B F$, it is enough to consider limits as $\to 0$ through $\{_k\}$ such that $_{k+1} \ge c_k$ for some $0 < c < 1$, in particular $\delta_k = c^k$. I have trouble seeing why this is true. Following is the reasoning provided by the author:

Note that if $\delta_{k+1} \le \delta < \delta_k$, then with $N_(F)$ the least number of sets in a $$-cover of F, $$\frac{\log N_\delta(F)}{-\log\delta} \le \frac{\log N_{\delta_{k+1}}(F)}{-\log\delta_k} = \frac{\log N_{\delta_{k+1}}(F)}{-\log\delta_{k+1} + \log{(\delta_{k+1}/\delta_k)}} \le \frac{\log N_{\delta_{k+1}}(F)}{-\log\delta_{k+1} + \log c}$$ and so $$\overline{\lim_{\delta\to 0}} \frac{\log N_\delta(F)}{-\log\delta} \le \overline{\lim_{k\to\infty}} \frac{\log N_{\delta_k}(F)}{-\log\delta_k}$$ The opposite inequality is trivial; the case of lower limits may be dealt with in the same way.

1. The author says $0 < c < 1$, so that $\log c < 0$. I do not understand where the following implication comes from: $$\frac{\log N_\delta(F)}{-\log\delta} \le \frac{\log N_{\delta_{k+1}}(F)}{-\log\delta_{k+1} + \log c} \implies \overline{\lim_{\delta\to 0}} \frac{\log N_\delta(F)}{-\log\delta} \le \overline{\lim_{k\to\infty}} \frac{\log N_{\delta_k}(F)}{-\log\delta_k} $$

2. For the opposite direction (assuming the implication in (1) is correct), I would do $$\frac{\log N_\delta(F)}{-\log\delta} \ge \frac{\log N_{\delta_{k}}(F)}{-\log\delta_{k+1}} = \frac{\log N_{\delta_{k}}(F)}{-\log\delta_{k} + \log(\delta_k/\delta_{k+1})} \ge \frac{\log N_{\delta_{k}}(F)}{-\log\delta_{k} - \log c}$$ Taking limits (this needs more justification), $$\overline\lim_{\delta\to 0} \frac{\log N_\delta(F)}{-\log\delta} \ge \overline\lim_{k\to\infty} \frac{\log N_{\delta_k}(F)}{-\log\delta_k}$$ which is the required inequality. Combining this with the previous inequality, stated in the question, we get $$\overline\lim_{\delta\to 0} \frac{\log N_\delta(F)}{-\log\delta} = \overline\lim_{k\to\infty} \frac{\log N_{\delta_k}(F)}{-\log\delta_k}$$ A similar procedure could be adopted for the case of $\underline\lim_{\delta\to 0}$ and $\underline\lim_{k\to \infty}$ (just take lower limits in place of upper ones, everything else unchanged). Is this right?

---

References.

• Fractal Geometry by Kenneth Falconer.

Answer 1

Below is a proof I have for the lower box dimension, from some handwritten notes I wrote in January 1991. I don't have time now to check the details, other than a quick proof-reading of the expressions, so let me know if something seems amiss and I'll try to address your concerns at some later time.

Let $\delta > 0$ be given. Then there exists $k$ such that $\delta_{k+1} < \delta < \delta_{k}.$ Therefore,

$$\frac{\log N_\delta(F)}{-\log\delta} \;\; \leq \;\; \frac{\log N_{\delta_{k+1}}(F)}{-\log\delta} \;\; \leq \;\; \frac{\log N_{\delta_{k+1}}(F)}{-\log\delta_k} \;\; = \;\; \frac{\log N_{\delta_{k+1}}(F)}{-\log\left(\frac{\delta_k}{\delta_{k+1}} \cdot \delta_{k+1}\right)} $$

$$ = \;\; \frac{\log N_{\delta_{k+1}}(F)}{-\log\left(\frac{\delta_k}{\delta_{k+1}}\right) - \log \delta_{k+1}} \;\; \leq \;\; \frac{\log N_{\delta_{k+1}}(F)}{\log c - \log \delta_{k+1}}, $$

where for the last inequality note that

$$\frac{\delta_{k+1}}{\delta_k} \geq c \;\; \implies \;\; \log \frac{\delta_k}{\delta_{k+1}} \leq \log \frac{1}{c} = -\log c \;\; \implies \;\; \log c \leq -\log\left(\frac{\delta_k}{\delta_{k+1}}\right),$$

and so using $\log c$ in "the last inequality" results in a smaller denominator (hence, a larger value).

Therefore,

$$ \liminf_{\delta \rightarrow 0^+}\, \frac{\log N_{\delta}(F)}{-\log \delta} \;\; \leq \;\; \liminf_{k \rightarrow \infty}\, \frac{\log N_{\delta_{k+1}}(F)}{\log c - \log \delta_{k+1}} $$

$$= \;\; \liminf_{k \rightarrow \infty}\, \left[ \frac{\;\;\frac{\log N_{\delta_{k+1}}(F)}{-\log \delta_{k+1}}\;\;}{\frac{\log c}{-\log \delta_{k+1}} + 1} \right] \;\; = \;\; \liminf_{k \rightarrow \infty}\, \frac{\log N_{\delta_{k+1}}(F)}{-\log \delta_{k+1}} \;\; = \;\; \liminf_{k \rightarrow \infty}\, \frac{\log N_{\delta_k}(F)}{-\log \delta_k}, $$

where in the second to last equality above we are using the fact that $\frac{\log c}{-\log \delta_{k+1}} \rightarrow 0$ as $k \rightarrow \infty.$

Interchanging $k$ with $k+1$ in the above is easily seen to reverse each inequality, and so the other inequality also holds.