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 $12\lambda$. 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.
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 middlethirds 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 selfsimilarity, $D(0,r)=D(0,r/3)$, 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 selfsimilar sets, in [Feng, DeJun. Exact packing measure of linear Cantor sets. Math. Nachr. 248/249 (2003), 102109]. 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 $2^{s_\lambda1}$ 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 PaulBenjamin'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).

$\begingroup$ Thank you very much for your answer Mr. Shmerkin. But there is still something's missing. In the exercise, I had to determine bounds $0<a<b<c<d<\infty$ such that $a\leq\Theta_*\leq b<c\leq \Theta^*\leq d$. As far as I can see, we can take $a=2^{(s+1)},c=2^{s},d=1$, but I don't know how to get $b$ such that $a<b<c$, because in the article you mentionned, there is an upper bound for the lower density almost everywhere, but I'm supposed to get an estimate everywhere. $\endgroup$ – PaulBenjamin Nov 24 '14 at 10:44

$\begingroup$ I think that ingeneral, using the ShannonMcMillanBreiman theorem, one can get a bound for "most points", although a bound for every point is much more delicate, and I don't think it there will be one in general (as the exceptional sets for the ergodic theorem in a Bernoulli system is rather large). $\endgroup$ – Asaf Nov 24 '14 at 16:02

2$\begingroup$ @Asaf  this has very little to do with ShannonMcMillanBreiman or ergodic theory in general. Besides giving only a.e. results, SMB or the ergodic theorem give information about local dimension (i.e. asymptotic behavior of $\log \mu(B(x,r))/\log r$), not about densities (i.e. asymptotic behavior of $\mu(B(x,r))/(2r)^s$). But for this simple construction one can still find sharp bounds valid at every point by direct geometric considerations. $\endgroup$ – Pablo Shmerkin Nov 24 '14 at 22:42