MichelFest: Minkowski & Hausdorff Dimensions

I left off the last post on An Introduction to Fractal Geometry discussing the issue of dimension for fractals. Namely, using the defined box dimension, we saw that the Cantor Set and the Sierpinski Carpet have non-integer dimension. We also saw some issues with the Box Dimension, specifically with density, and thus will seek different notions of dimension in this post, in the form of Minkowski Dimension and Hausdorff Dimension.

As promised, this discussion will be mathematically rigorous. In this post, we’ll first discuss Minkowski Dimension and Minkowski Content, and then Hausdorff Measure and Hausdorff Dimension.


I. Minkowski Dimension and Minkowski Content

 

Let us prove this, to gain some insight into the matter. In particular, let’s prove the upper and lower Box and Minkowski Dimensions are equal.

In the remainder of this post, let |A|_n denote the n-dimensional Lebesgue measure of some Lebesgue measurable set A \subset \mathbb{R}^{n}. Let A^\epsilon denote the set

A^\epsilon := \{x \in \mathbb{R}^n : d(x,A) < \epsilon \}.

In a similar way to how we defined the box dimension, we define the upper Minkowski Dimension of A, \text{dim}^{*}_{M}(A), as

\text{dim}^{*}_{M}(A) : = n-\limsup \limits_{\epsilon \to 0^{+}} \frac{\log(|A^\epsilon|_n)}{\log(\epsilon)},

and we define the lower Minkowski Dimension of A,\text{dim}_{*M}(A), as

\text{dim}_{*M}(A) : = n-\liminf \limits_{\epsilon \to 0^{+}} \frac{\log(|A^\epsilon|_n)}{\log(\epsilon)},

if these are equal, we call their common value \text{dim}_{M}(A) the Minkowski Dimension of A.

At first this definition may seem a bit, but not to worry, it’s actually equivalent to the Box Dimension.

Let us prove this, to gain some insight into the matter. In particular, let’s prove the upper and lower Box and Minkowski Dimensions are equal. Instead of letting N_\epsilon(A) denote the number of closed boxes needed to cover A, we shall use open balls. These can easily be seen to give equivalent definitions of the Box Dimension.

Note first that |A^{\epsilon}|_n \leq c_n (2\epsilon)^{n}N_\epsilon(A), where c_n is the volume of the n-dimensional ball. Namely, if we have a cover of A by \epsilon balls, then surely, using the same balls with twice the radius, they will cover A^{\epsilon}. Taking the log of both sides (log monotonicity) and dividing by -\log(\epsilon) yields

\frac{\log(|A^{\epsilon}|_n)}{-\log(\epsilon)} \leq \frac{\log(2^nc_n) + n \log(\epsilon) + \log(N_{\epsilon}(A))}{-\log(\epsilon)}.

Taking the lim sup as \epsilon \to 0^{+}, adding n to the lhs, we have

n- \limsup \limits_{\epsilon \to 0^{+}} \frac{\log(|A^\epsilon|_n)}{\log(\epsilon)} \leq \text{dim}_{M}(A).

For the reverse inequality, use the observation that c_n\epsilon^n N'_\epsilon(A) \leq |A^{\epsilon}|_n where N^{'}_\epsilon(A) is the maximal number of disjoint open balls with centre in A, known as a packing. Clearly these cover at most A^{\epsilon}. Now, to compare N' and N, observe that N_{2\epsilon} \leq N^{'}_{\epsilon} since given a maximal packing, if we double the radius of the balls, we will surely cover the entire set. To see this, we need only note that each point not in the packing can not be more than \epsilon away than all members of the packing, for if it were this would surely contradict the maximality.

Ergo, to prove the equivalence of the dimensions, we just need to prove the reverse inequality, which is done in the same fashion as the first, since c_n\epsilon^n N_{2\epsilon}(A) \leq |A^{\epsilon}|_n .

This proves the equivalence of the Box Dimension and the Minkowski Dimension.

With this notion established, we will now give ourselves some notion of size using this dimension. Namely, we’re going to essentially skew the Lebesgue Measure of some set by a small amount, depending on the Minkowski Dimension. Let \partial A denote the topological boundary of A.

We define the lower Minkowski Content of a set $latexA&s=2$ to be

\mathcal{M}_{*}(A) = \liminf \limits_{\epsilon \to 0^{+}} \epsilon^{-(n-\text{dim}_M)}|(\partial A)^{\epsilon}|_n

and define the upper Minkowski Content as

\mathcal{M}^{*}(A) = \limsup \limits_{\epsilon \to 0^{+}} \epsilon^{-(n-\text{dim}_M)}|(\partial A)^{\epsilon}|_n

if these two values are equal, we say the set A is Minkowski Measurable, and denote their common value by \mathcal{M}(A), called the Minkowski Content. So as I said, all we’ve essentially done is taken the Lebesgue measure and skewed it by a constant depending on the dimension of the set.

The two most important properties of the Minkowski Content and Minkowski Dimension are:

  1. If \mathcal{M}(A) exists, 0<\mathcal{M}(A)<\infty.
  2. All other properties of \text{dim}_B(A) are immediately inherited by \text{dim}_M(A) by the proven equivalence.

Notice the particular importance of (1.). When we set out, we were looking for some way to assign size to a set that is non-zero and non-infinite. We have done so with this Minkowski content, provided we measure the set with respect to the Minkowski Dimension. Though, the Minkowski Dimension thus inherits the issues of the Box Dimension by (2.).

We’re going to define the Minkowski Dimension in another way in a bit. The reason we’re going to do this is quite intriguing from a theoretical construction. Recall how the above was constructed; namely, we defined the Minkowski Dimension and then said hey this is the thing that makes the Minkowski Content finite! Well after we define the Hausdorff Measure and Dimension, we’re going to see how we can get the Minkowski Dimension from the Minkowski Content.

 


 

II. Hausdorff Measure and Dimension

 

 

We’ll start off with the definition of the Hausdorff Measure (which truly is a measure most of the time!), and then will discuss its uses and implications. Again, let A \subset \mathbb{R}^n .

By a \delta-cover of a set A, we mean an at most countable collection of sets \mathcal{U} = \{U_i \}_{i=1}^{\infty} such that A is covered by their union, and \text{diam}(U_i) < \delta for all i , the usual diameter of a set. Define

\mathscr{H}_{\delta}^{s}(A) := \inf \{ \sum \limits_{i} (\text{diam}(U_i))^s : \mathcal{U} \, \text{is a} \, \delta- \, \text{cover} \} .

Note that as \delta \to 0^{+} , \mathscr{H}_{\delta}^{s} is montonically non-decreasing. To see this, note that the collection of all \delta-covers decreases in size as \delta decreases, and so the infimum must be non-decreasing. Ergo, the limit as \delta tends to 0 exists, and we define the s – dimensional Hausdorff Measure as

\mathscr{H}^{s} (A) = \lim \limits_{\delta \to 0^{+}} \mathscr{H}_{\delta}^{s}(A) .

Let us present some properties of the Hausdorff Measure, most without proof, but whose proofs can be found in Fractal Geometry, Falconer.

  1. \mathscr{H}^{s} (\emptyset) = 0
  2. If A \subset B , \mathscr{H}^{s} (A) \leq \mathscr{H}^{s} (B).
  3. If s \in \mathbb{N} ,   \mathscr{H}^{s} (A) = c^{-1}_n |A|_n.
  4. \mathscr{H}^{s} (A) is countably subadditive.
  5. \forall s \in [0,n] \mathscr{H}^{s} is a measure.

 

We prove only (4.), (1.) and (2.) are easily shown, and (3.) appears in [Falc].

Let \epsilon >0 . Let A = \bigcup \limits_{i=1}^{\infty} . Let \mathcal{U}_i be \delta covers such that

\sum \limits_{U \in \mathcal{U}_i} \text{diam}(U)^s < \mathscr{H}_\delta^{s} (A_i) + \frac{\epsilon}{2^i} .

Now, \mathcal{U} = \bigcup \limits_{i=1}^{\infty} is a \delta -cover for A . Thus,

\mathscr{H}_\delta^{s} (A) \leq \sum \limits_{i=1}^{\infty} \sum \limits_{U \in \mathcal{U}_i} \text{diam}(U)^s < \sum \limits_{i=1}^{\infty} \mathscr{H}_\delta^{s} + \epsilon .

Let \epsilon \to 0 and then \delta \to 0^{+}  to conclude countable subadditivity.

One can actually note that the value s for which 0 < \mathscr{H}^{s} < \infty is unique. To see this, suppose 0< \mathscr{H}^{s}(A) < \infty. If t>s , then if \mathcal{U} is a \delta cover that, then, since all such members of the cover have diameter less than \delta ,

\mathscr{H}^{t}_\delta \leq \sum \limits_{U \in \mathcal{U}} (\text{diam}(U))^t \leq  \delta^{t-s} \sum \limits_{U \in \mathcal{U}} (\text{diam}(U))^s

and if s>t, in the same manner,

\delta^{t-s} \sum \limits_{U \in \mathcal{U}} (\text{diam}(U))^s \leq  \sum \limits_{U \in \mathcal{U}} (\text{diam}(U))^t

so taking the limit as \delta \to 0 in both cases, we see that \mathscr{H}^{t}(A) = 0 or \mathscr{H}^{t}(A) = \infty

Ergo, the graph of \mathscr{H}^{s} with respect to s looks like

Screen Shot 2016-06-24 at 10.29.10 AM.

Thus there is one unique value at which the measure seems to give a good notion of size for some set. But as we said before, the idea is to take a measure in the same dimension as the object. This motivates the Hausdorff Dimension to be defined as this unique value. Namely, the Hasudorff Dimension of  a set A is given as

\text{dim}_H(A) : = \inf \{s \geq 0 : \mathscr{H}^{s}(A) = 0 \} = \sup \{s \geq 0 : \mathscr{H}^{s} (A) = \infty \}.

We therefore have done the exact opposite as we did in the case of Minkowski Dimension. In the case of Minkowski Dimension, we defined the dimension, and showed it gave a finite measure in the case of Minkowski Content. In the case of Hausdorff Dimension, we defined a plausible measure and then gave the dimension as that value which let the measure be finite.

It should therefore be noted that we may do the same in the case of Minkowski Dimension. If we define the Minkowski Dimension as

\text{dim}_M(A) : = \inf \{s \geq 0 : \mathcal{M}^{s}(A) = 0 \} = \sup \{s \geq 0 : \mathcal{M}^{s} (A) = \infty \},

we get an equivalent definition of the Minkowski Dimension, and this equivalence is proven on pages 78-79 of Geometry of Sets and Measures in Euclidean Spaces by Mattila.

The Hausdorff Measure behaves fairly well, as indicated by such propositions as

If f is a Hölder Continuous function, meaning

|f(x)-f(y)| \leq c |x-y|^{\alpha},

for c, \alpha >0 , then

\mathscr{H}^{\frac{s}{\alpha}}(f(F)) \leq c^{\frac{s}{\alpha}} \mathscr{H}^{s}(F).

whose proof is also in [Falc].

Let us finish with a final remark on the intuition on the Hausdorff Dimension compared to the Box/Minkowski Dimension.

In particular, we have that

\text{dim}_H \leq \text{dim}_{*B} \leq \text{dim}^{*}_B.

Intuitively,

\text{dim}_H = \, \, \text{number of cubes} \, \leq \epsilon

\text{dim}_B = \, \, \text{number of cubes} \, = \epsilon

which is the intuitive reason why the Hausdorff Dimension is smaller than the Box Dimension, because it can be more refined.


Originally I was going to discuss Iterated Function Systems and Moran’s Theorem in this post, but they’re extensive topics that deserve their own post, so that’ll be next. We’ve so far seen how to define dimension and size for fractals, but not where they come from or how to construct them. This will be the topic of the next post.

Cheers,

J.T.

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