Let $C\_\lambda$ be the classical Cantor set associated to a real number $0<\lambda<\frac{1}{2}$, as defined for example in the book of K. J. Falconer The geometry of fractal sets. I recall briefly the construction. Starting with the unit interval, we remote from the center of the interval an interval of length $1-2\lambda$. After $n$ steps, we obtain $2^n$ intervals equally spaced out, each one of length $\lambda^n$. For the Hausdorff measure $\mathscr{H}^s$ of dimension $s$, we use the function $g(S)=\text{diam}(S)^s$ for the Caratheodory construction and we define the upper density of a set $A\subset \mathbb{R}^n$, at a point $x\in\mathbb{R}^n$ by
$$
\Theta^{\*s}(A,x)=\limsup\_{r\rightarrow 0}\frac{\mathscr{H}^s(A\cap B(x,r))}{(2r)^s}
$$
and for the lower density $\Theta\_\*^s$ we replace $\limsup$ by $\liminf$. We can prove that $\dim C\_\lambda=\frac{\log 2}{\log(1/\lambda)}=s\_\lambda$, and even $\mathscr{H}^{s\_\lambda}(C\_\lambda)=1$, and I wanted to obtain precise estimates for the upper and lower density of $C\_\lambda$. In fact I think that I managed to prove that for all $x\in C\_\lambda$, we have
$$
2^{-(s\_\lambda+1)}\leq \Theta\_\*^{s\_\lambda}(C\_\lambda,x)\leq 2^{-s\_\lambda}\leq \Theta^{\*s\_\lambda}(C\_\lambda,x)\leq 1.
$$
The problem is that I know that there is a lower bound more precise for $\Theta^{\*s\_\lambda}(C\_\lambda,x)$, namely, a constant $c>2^{-s\_\lambda}$, such that $\Theta^{\*s\_\lambda}(C\_\lambda,x)\geq c$ for all $x\in C\_\lambda$. I would like to get a confirmation of the validity of the calculated estimates, and I would really appreciate a hint for the last lower bound. 



Precise density estimates for Cantor sets
**Upper densities**


In the following, I freely use the well known fact that $s\_\lambda$-Hausdorff measure gives mass $2^{-k}$ to all intervals that make up the stage $k$ in the construction of $C\_\lambda$.


I don't think it is correct that $\Theta^{\* s\_\lambda}(C\_\lambda,x)\ge c$ for all $x\in C\_\lambda$ and some $c>2^{-s\_\lambda}$.


For example, let $\lambda=1/3$ so that $C=C\_\lambda$ is the middle-thirds Cantor set. Write $s=s\_{1/3}=\tfrac{\log 2}{\log 3}$, and
$$
D(x,r) = \frac{\mathcal{H}^{s}(B(x,r)\cap C)}{(2r)^s}
$$


By self-similarity, $D(0,r)=D(0,r/3)$, so 
$$
\Theta^{\*s}(C,0) = \max\{ D(0,r): r\in [1/3,1]\}.
$$
Now if $r\in [1/3,2/3]$, then $\mathcal{H}^s(B(0,r))=1/2=(1/3)^s$ so $D(0,r)\le 2^{-s}$. Otherwise, write $r=2/3+t$ for some $t\in [0,1/3]$. Then
$$
\mathcal{H}^s(B(0,r)\cap C) = \mathcal{H}^s([0,1/3]\cap C)+\mathcal{H}^s([2/3,2/3+t]\cap C)\le \tfrac{1}{2}+t^{-s},
$$
whence 
$$
D(0,r) \le \frac{\tfrac{1}{2}+t^{s}}{2^s (2/3+t)^s} \le 2^{-s},
$$
from elementary calculus. Hence $\Theta^{\*s}(C,0)=2^{-s}$. There's nothing special about $\lambda=1/3$ here. Also, although the point $0$ is special, the same holds for any ternary point.




---


**Lower densities**


The lower density problem was essentially solved, for more general self-similar sets, in [**Feng, De-Jun**. Exact packing measure of linear Cantor sets. *Math. Nachr.* 248/249 (2003), 102--109]. In Theorem 1.1, a formula is given for the infimum of
$$
\frac{\mathcal{H}^{s\_\lambda}(B(x,r)\cap C\_\lambda)}{(2r)^s}
$$
over all $x\in C\_\lambda$ such that $B(x,r)\subset [0,1]$. Note that the example after Theorem 1.1 are exactly the central Cantor sets (with $\lambda=(1-\beta)/2$). It seems this value is strictly larger than $2^{-s\_\lambda-1}$ for all $\lambda\in (0,1/2)$.


In Theorem 2.1, Feng shows that $\Theta\_\*^{s\_\lambda}(C\_\lambda,x)$ equals this infimum for almost all $x$. Clearly, this infimum equals the smallest possible value of $\Theta\_\*^{s\_\lambda}(C\_\lambda,x)$, at least if we exclude $x=0,1$ (and if $\lambda\le 1/3$ this restriction is not necessary, as any extreme point of a construction interval will have the same density as $0$ and $1$).


**Edit** (after Paul-Benjamin's comment): to give an upper bound for the lower density at every point of $C\_\lambda$, I consider again the case $\lambda=1/3$ for concreteness, although a similar argument should work for any $\lambda$. Suppose first that $x$ is such that it belongs to infinitely many "left intervals" and infinitely many "right intervals" in the construction (or in other words the ternary expansion of $x$ has infinitely many zeros and twos). Then there are infinitely many $k$ such that the distance from $x$ to the boundary of the level $k$ interval of the construction of $C$ containing $x$ is at least $(2/9) 3^{-k}$. Let $r=r\_k=3^{-k}(1+2/9)$. Then the ball $B(x,r)$ intersects only one interval of level $k$ in the construction, so that 
$$
\frac{\mathcal{H}^s(B(x,r)\cap C)}{(2r)^s} = \frac{1}{2^s(11/9)^s} < 2^{-s}.
$$
Otherwise, without loss of generality, from some generation on $x$ is always on the left interval of the construction. In that case, $B(x, 2 3^{-k})$ meets only one interval of the construction for all sufficiently large $k$ and we get an even smaller upper bound for the lower density. Hence $b=2^{-s} (11/9)^{-s}$ works (I'm not claiming this is optimal, though it might be).


