As the title suggests, I am trying to calculate the $s$-energy of the middle third Cantor set. I am reading Falconer's Fractal Geometry book, available here:
http://www.dm.uba.ar/materias/optativas/geometria_fractal/2006/1/Fractales/1.pdf
and this is an exercise (exercise 4.9 on page 45 of the pdf - 68 of the book).
First, we define the $s$-potential at a point $x\in\mathbb{R}^{n}$ as $$\phi_{s}(x)=\int\frac{\text{d}\mu(y)}{|x-y|^{s}}$$ and the define the $s$-energy as $$I_{s}(\mu)=\int\phi_{s}(x)\text{d}\mu(x)=\iint\frac{\text{d}\mu(x)\text{d}\mu(y)}{|x-y|^{s}}.$$
Now, let $F$ be the middle third Cantor set and let $\mu$ be the mass distribution on $F$ so that each $2^{k}$ $k$th level interval of length $3^{-k}$ has mass $2^{-k}$.
Estimate the $s$-energy of $\mu$ for $s<\log{2}/\log{3}$, and deduce that $\dim_{\text{H}}F\geq\log{2}/\log{3}$.
I get the feeling that this isn't an expecially hard exercise, but I'm stuck on where to start. Could someone help me please?
First, a general observation: if a finite measure $\mu$ on $\mathbb{R}$ has no atoms, then the diagonal $\{(x,x)\in\mathbb{R}^2\}$ has zero measure with respect to the product measure $\mu\times \mu$. To see why, partition $\mathbb R$ into $n$ intervals of measure $1/n$, and observe that the diagonal is covered by $n$ squares, each of which gets product measure $1/n^2$.
So we integrate over $x\ne y$; by symmetry, it suffices to integrate over $x<y$. Write $x,y$ in base-3 as
$$x = 0.\underset{n \text{ digits}}{\underbrace{\cdots}} 0\cdots , \quad y = 0.\underset{\text{same digits}}{\underbrace{\cdots}} 2\cdots $$ where $n$ is a nonnegative integer. Note that $|x-y|\ge 3^{-n-1}$. Also, the measure of all pairs $(x,y)$ as above is $$ 2^{n}(1/2)^{2(n+1)} = \frac14\cdot 2^{-n} $$ because there are $2^n$ choices of $n$ digits of $x$, and because fixing the first $(n+1)$ digits of a number restricts it to a subset of measure $(1/2)^{n+1}$.
Putting it all together, $$ \iint_{x<y} |x-y|^{-s}\,d\mu(x)\,d\mu(y) \le \sum_{n=0}^\infty 3^{s(n+1)} \frac14\cdot 2^{-n} = \frac{3^s}{4} \sum_{n=0}^\infty (3^s/2)^n $$ which converges when $3^s<2$.