# Determining a lower bound on the Hausdorff dimension of a set

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).

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In my answer I am talking about a case where the Hausdorff dimension can be computed exactly - I had missed the point that you were asking for a lower bound (and in the title, to boot!). Which must mean you have something more complicated in mind. Perhaps you could give us a hint of what sort of sets you are thinking of? Otherwise, the question seems rather extraordinarily general. – Harald Hanche-Olsen Mar 3 '10 at 21:38
P.S. someone who can edit questions might fix the spelling of "dimension" in the subject ... useful for future searches, say. – Gerald Edgar Mar 4 '10 at 18:25

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.

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 $s$.

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No, citing your own book should be fine so long as it's relevant, which I assume it is. If someone disagrees, let them say so on meta and leave a note to that effect here. – Harald Hanche-Olsen Mar 3 '10 at 23:19
I hope that no one would object to citing your own book. One of the strengths of Math Overflow is that we have world-class mathematicians answering questions here. Why would we want to tie their hands in their own areas of expertise? – Tom Church Mar 4 '10 at 0:38

There are lower bound based on potential theoretic methods. Still, you need to have a measure $\mu$ supported on your set $E$. Define the $s$-energy of $\mu$ as

$I_s(\mu) = \iint |x-y|^{-s} \mu({\rm d} x) \mu({\rm d} y)$

If $I_s(\mu) < \infty$, then $\dim_{\rm H} (E) \geq s$. [Theorem 4.13, Falconer Fractal Geometry 2nd Ed.] IIRC, the proof follows from the density-based lower bound that Gerald gave. But if your measure allows easy estimate of potential, then it might be more convenient.

Moreover, you have the following characterization of Hausdorff dimension:

$\dim_{\rm H} (E) = \inf \{ s \geq 0 : C_s(E) = 0 \} = \sup \{s \geq 0 : C_s(E) > 0\}$,

where $C_s(E)$ is the $s$-capacity of $E$, defined as

$C_s(E) = \sup\{I_s(\mu)^{-1}: \mu \text{ is a probability measure supported on }E \}$

There are also lower bounds based on the Fourier transform of $\mu$. See Sec. 4.4 of Falconer's book.

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For one family of fractals there is a fairly straightforward method: If the set $G$ is the fixed point set of a number of similitudes $S_1,\ldots,S_n$ with contraction factors $s_1,\ldots,s_n$ then the Hausdorff dimension $D$ satisfies $$\sum_{i=1}^n s_i^D=1.$$ A bit more explanation may be in order: Each $S_i$ is an affine map on the form $S_i=a_i+s_iR_i$ where $R_i$ is a rotation (mirror symmetries allowed). These produce a map $S$ on the set of nonempty compact subsets of $\mathbb{R}^d$, say, given by $$S(F)=\bigcup_{i=1}^n S_i(F).$$ This map is a contraction in the Hausdorff metric on the space, and $G$ is the unique fixed point of the map.

There is one caveat: The sets $S_i(G)$ must be mutually disjoint, or almost so. For example the Sierpinski triangle: It is generated by three similitudes each mapping a given triangle to one half as big, so we should get $3\cdot(1/2)^D=1$. But the three subtriangles meet in three different points; however, these points are corners, and this counts as negligible.

I apologize for not having a reference handy. I learned this from a handwritten note which likewise lacked references, and I haven't had the fortitude to go and chase one down. If you can read Norwegian, there is a proof in my small note here. The main idea of the proof is fairly simple, however: It's a question of covering $G$ by small balls and seeing how these covers scale under $S_i$.

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I realize that this does not quite answer the question, but I am leaving the answer up anyway, since it does answer a closely related question after all. If someone has a literature reference for the result I mention, I'd be happy to learn about it. – Harald Hanche-Olsen Mar 3 '10 at 21:39
I think the requirement "must be mutually disjoint, or almost so" is actually "must have mutually disjoint interiors". And there is also a method for extending this to when $G$ is the fixed point of an uncountable number of what you call similitudes. – Mark Bell Mar 3 '10 at 21:54
Well, most fractals have empty interiors to begin with, so “mutually disjoint interiors” is a somewhat vacuous concept. The important bit is, when you try to cover $G$ with small balls in an optimal way, a negligible proportion of those balls manage to intersect more than one of the sub-pieces. – Harald Hanche-Olsen Mar 3 '10 at 23:18
The assumption that $S_i(G)$'s are disjoint is known as the strong separation condition. One weaker condition that often suffices is called (Moran's) open set condition: there exists non-empty bounded open set $U$ such that $\cup_{i=1}^n n S_i(U) \subset U$ with the union disjoint. See e.g. Falconer's Techniques in fractal geometry, p. 35. – mr.gondolier Mar 4 '10 at 19:41

Finding good lower bounds of the Hausdorff dimension of the attractors to dissipative dynamical systems is a very difficult problem.

In the early 1970s Arnold explicitly stated it as an important open problem in case of the 2D Navier-Stokes equations in a bounded domain (see the compendium of Arnold's problems); in various forms the question probably goes back to Kolmogorov's seminar in the 1950s. If atrractor's dimension can be shown to grow indefinitely along the Reynolds number Re, this can be interpreted as a manifestation of the turbulence of the fluid flow.

The upper estimates of attractors' Hausdorff dimension are much easier. For instance, the best known result for the 2D Navier-Stokes equations with the Dirichlet boundary conditions in the domain $\Omega$ says that

$$\dim_H {\rm Attr}\leq \frac{|\Omega|}{\pi \nu^2}\ \|f\|_{L^2}$$

(here $\nu$ is the viscosity and $f$ is the external force). There are no satisfactory lower bounds of the Hausdorff dimension in this case. The situation is slightly better in case of the Navier-Stokes equations on a torus: good lower bounds are known but only for very specific forces $f$.

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