Does anyone know of a good method for finding a lower bound of the Hausdorff dimension of a set $G$?


The only method I could find is to find an $\alpha$-Hölder function $f \colon G \to H$ then $\dim\_H(G) \geq \alpha \dim\_H(\operatorname{im}(f))$. Choosing $f$ cleverly will mean that $\operatorname{im}(f)$ will be a set whose Hausdorff dimension is already known (or at least a lower bound for it is known).



Determining a lower bound on the Hausdorff dimension of a set
Upper bound for the Hausdorff dimension is often easy, from the definition.


Lower bound can be harder. One method can be used if you have a measure on your set. Even better, a measure that naturally fits with the structure of the set. Then lower bounds for the Hausdorff dimension come from density computations for that measure.


(Would citing my own book here be considered crass?)


Packing dimension may be opposite. The lower bound is easy from the definition, but the upper bound harder. Again a density with respeact to a measure can help with this upper bound.


**added March 4**


This density theorem is found in: G. Edgar, *Integral, Probability, and Fractal Measures* (Springer 1998) Theorem 1.5.14, p. 52.


*Definitions* ... Let $E \subseteq \mathbb{R}^n$ be a Borel set, let $\mu$ be a nonzero measure
on $E$, let $s>0$ be a real number. Write
$$
 B\_r(x) = \{y \in E \colon |y-x|\le r\}
$$
for a closed ball. The **upper** $s$-**density** of $\mu$ at a point
$x \in \mathbb{R}^n$ is
$$
 \overline{D}^s\_\mu(x) = \limsup\_{r \to 0} \frac{\mu(B\_r(x))}{(2r)^s} .
$$


A consequence of *Theorem* 1.5.14 is then: If
$\sup\_{x \in E} \overline{D}^s\_\mu(x) < \infty$, then the Hausdorff dimension
of $E$ is at least $s$.


