I have a set of points in $\mathbb{R}^2$, of the form:

$\Bigg\{\left(\frac{a}{\ell^2},\frac{b}{\ell^3}\right): \ell \in \mathbb{N}^+\Bigg\}$

where $a$ and $b$ are some real positive numbers.

I am interested to know the box dimension of this set. Is there a simple way to determine this analytically?


I believe the answer is $1/3$, independent of the values of $a$ and $b$.

To see this, we first show that the box counting dimension of the set $$E = \{1,1/4,1/9,\ldots,1/n^2,\ldots\}$$ is $1/3$. The computations below very much mimic those of example 3.5 on page 45 of Falconer's Fractal Geometry.

To this end, let $\varepsilon>0$ and choose $N$ to be the unique natural number such that $$\frac{1}{N^2} - \frac{1}{(N+1)^2}<\varepsilon\leq\frac{1}{(N-1)^2}-\frac{1}{N^2}.$$ Now, since $\varepsilon \leq 1/(N-1)^2-1/N^2$, we need at least $N-1$ sets of diameter $\varepsilon$ to cover $E$ - one for each of the numbers $1,1/4,\ldots,1/(N-1)^2$. Thus, $N_{\varepsilon}(E) \geq N-1$. And since $$\varepsilon > \frac{1}{N^2} - \frac{1}{(N+1)^2} > \frac{1}{(N+1)^3},$$ $1/\varepsilon<(N+1)^3$. Thus, $$\frac{\log(N_{\varepsilon}(E))}{\log(1/\varepsilon)} > \frac{\log(N-1)}{\log((N+1)^3)} \to \frac{1}{3}.$$

This much shows that $$\liminf_{\varepsilon\to0^+}\frac{\log(N_{\varepsilon}(E))}{\log(1/\varepsilon)} \geq \frac{1}{3}.$$

To show that $$\limsup_{\varepsilon\to0^+}\frac{\log(N_{\varepsilon}(E))}{\log(1/\varepsilon)} \leq \frac{1}{3},$$ we use the fact that, since $\varepsilon>1/(N+1)^3$, all the points of $E$ in $[0,1/(N+1)^2]$ may be covered with $N+1$ sets of size $\varepsilon$. That leaves us with only $N$ more points so that $E$ may be covered with $2N+1$ sets of size $\varepsilon$ so that $N_{\varepsilon}(E)\geq 2N+1$. Also using the fact that $$\varepsilon\leq\frac{1}{(N-1)^2}-\frac{1}{N^2}< \frac{6}{N^3}$$ for $n>2$, we get that $$\frac{\log(N_{\varepsilon}(E))}{\log(1/\varepsilon)} < \frac{\log(2N+1)}{\log((N+1)^3/6)} \to \frac{1}{3}.$$

Now, it's just a couple of steps to get from here to your set of interest - one small step and one not so small.

First, if we define $$E_a = \{a,a/4,a/9,\ldots,a/n^2,\ldots\}$$ for a positive number $a$, then we again have a set of box dimension $1/3$ because box dimension is preserved under similarity transformations.

Now, let's denote your set by $E_{a,b}$. The projection of your set onto the $x$-axis does not increase distance. Thus, it cannot increase box dimension so that the lower box dimension of $E_{a,b}$ is at least $1/3$. On the other hand, $E_{a,b}$ is the image of $E_a$ under the function $F:\mathbb R\to\mathbb R^2$ defined by $F(x) = (x,b(x/a)^{3/2})$. Since we're mapping to the graph of a differentiable function, this is a Lipschitz map and, therefore, also does not increase dimension. So the upper box dimension of $E_{a,b}$ is at most $1/3$. Taking these together, we a well defined box dimension of $1/3$.


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