New York Journal of Mathematics

New York J. Math. 13 (2007) 175–198.

Rényi dimension and Gaussian filtering

Terry A. Loring

Abstract. Consider the partition function S_{μ}^{q}(ϵ) associated in the theory of Rényi dimension to a finite Borel measure μ on Euclidean d-space. This partition function S_{μ}^{q}(ϵ) is the sum of the q-th powers of the measure applied to a partition of d-space into d-cubes of width ϵ. We further Guérin’s investigation of the relation between this partition function and the Lebesgue L^{p} norm (L^{q} norm) of the convolution of μ against an approximate identity of Gaussians. We prove a Lipschitz-type estimate on the partition function. This bound on the partition function leads to results regarding the computation of Rényi dimension. It also shows that the partition function is of O-regular variation.

We find situations where one can or cannot replace the partition function by a discrete version. We discover that the slopes of the least-square best fit linear approximations to the partition function cannot always be used to calculate upper and lower Rényi dimension.

Contents

1. Introduction 176 2. Norms after convolution 178 3. Bounding the partition function 180 4. Application to discrete limits 183 5. Examples 184 6. Rényi dimension of convolutions 187 7. Best fit slopes 188 8. More examples 194 9. Modified Rényi dimensions 196 References 198

Received November 19, 2006. Revised July 6, 2007. Mathematics Subject Classification. 28A80, 28A78. Key words and phrases. Rényi dimension, fractal, regular variation, least squares, Laplacian

pyramid, convolution, Gaussian, Matuszewska indices. This work was supported in part by DARPA Contract N00014-03-1-0900.

175^{ISSN 1076-9803/07}

1. Introduction

The Rényi dimensions of a finite Borel measure μ on ℝ^{d} are derived from slopes of long secants of the log-log plot of the function

ϵ ↦ S_{μ}^{q}(ϵ) = ∑_{k∈ℤ^{d}} μ(ϵk + ϵ𝕀)^{q},

where

𝕀 = [0, 1) × [0, 1) × ⋯ × [0, 1).

There are exceptions for q = 0, 1. In this paper we only address the cases 0 < q < 1 and 1 < q < ∞. In this introduction we wish to avoid convergence issues, so let us also assume that μ has bounded support.

For any x_{0}, we set

D_{q}^{±}(μ) = lim_{x→−∞} _{inf}^{sup} \frac{1}{q − 1} \frac{ln (S_{μ}^{q}(e^{x})) − ln (S_{μ}^{q}(e^{x_{0}}))}{x − x_{0}}.

The constant terms are irrelevant, so this is usually written as

D_{q}^{±}(μ) = lim_{x→−∞} _{inf}^{sup} \frac{1}{q − 1} \frac{ln (S_{μ}^{q}(e^{x}))}{x}

or

D_{q}^{±}(μ) = lim_{ϵ→0} _{inf}^{sup} \frac{1}{q − 1} \frac{ln (∑_{k∈ℤ^{d}} μ(ϵk + ϵ𝕀)^{q})}{ln(ϵ)}.

Moving an exponent inside the log gives

D_{q}^{±}(μ) = lim_{ϵ→0} _{inf}^{sup} \frac{q}{q − 1} \frac{ln (∥k ↦ μ(ϵk + ϵ𝕀)∥_{q})}{ln(ϵ)}.

This shows a relationship between convolution, L^{p}-norms and Rényi dimension, because

(χ|_{(−ϵ𝕀)} ∗ μ)(x) = μ(x + ϵ𝕀).

Here we have used χ|_{(−ϵ𝕀)} to denote the characteristic function of −ϵ𝕀.

Guérin ([7]) showed a more general relation between convolutions, L^{p}-norms and Rényi dimensions. He showed that for many choices of a scalar-valued function g on ℝ^{d}, if

1 < q < ∞,

and if we set

g_{ϵ}(x) = ϵ^{−d}g(ϵ^{−1}x),

then

D_{q}^{±}(μ) = lim_{ϵ→0} _{inf}^{sup} \frac{1}{q − 1} \frac{ln (ϵ^{d(q−1)} ∥g_{ϵ} ∗ μ∥_{q}^{q})}{ln(ϵ)},

or

(1)  D_{q}^{±}(μ) = d + lim_{ϵ→0} _{inf}^{sup} \frac{q}{q − 1} \frac{ln (∥g_{ϵ} ∗ μ∥_{q})}{ln(ϵ)}.

Guérin allowed g from a large class of complex-valued functions. A technical improvement on Guérin’s result is given in Section 2, with additional restrictions on g but allowing 0 < q < 1. For a given μ, and a “nice” function g ≥ 0, we establish a uniform bound on difference

ln (ϵ^{d(q−1)} ∥g_{ϵ} ∗ μ∥_{q}^{q}) − ln (S_{μ}^{q}(ϵ)) .

This estimate allows us to analyze sequences

\frac{ln (S_{μ}^{q}(ϵ_{n}))}{ln(ϵ_{n})}

by looking instead at

\frac{ln (∥g_{ϵ_{n}} ∗ μ∥_{q})}{ln(ϵ_{n})}.

This will be advantageous if we choose g properly.

Most importantly, we wish to let g be a standard Gaussian on ℝ^{d}. As we have the convention

g_{ϵ}(x) = ϵ^{−d}g(ϵ^{−1}x),

the semigroup rule ends up as

g_{ϵ} ∗ g_{η} = g\sqrt{ϵ^{2}+η^{2}}.

We find that ∥g_{ϵ} ∗ μ∥_{q} gives us information not apparent in the sums S_{μ}^{q}(ϵ). Specifically, we find constants A and B so that

|ln (S_{μ}^{q}(e^{x})) − ln (S_{μ}^{q}(e^{y}))| ≤ A + B|x − y|.

It follows that S_{μ}^{q}(x^{−1}) is of O-regular variation.

Generalizing a result of Riedi ([16]), we show that

\frac{1}{q − 1} \frac{ln (S_{μ}^{q}(ϵ_{n}))}{ln(ϵ_{n})}

can be used to calculate D_{q}^{±}(μ), even if ϵ_{n} converges to zero somewhat faster than geometrically. The specific requirement is that

lim_{n→∞} \frac{ln(ϵ_{n+1})}{ln(ϵ_{n})} = 1.

Examples are exhibited that show that this result is in some sense the best possible.

We also give some new estimates on the Rényi dimensions of a convolution μ ∗ ν in terms of the Rényi dimensions of μ and ν.

In Sections 7 and 8 we consider some of the changes that occur when one replaces

\frac{ln (S_{μ}^{q}(e^{x})) − ln (S_{μ}^{q}(e^{x_{0}}))}{x − x_{0}}

by the slope of a least-squares best fit line over [x, x_{0}] to the function

t ↦ ln (S_{μ}^{q}(e^{t})) .

We exhibit an example where these least-squares slopes do not determine the upper Rényi dimension.

The examples we give have features that occur only on scales that grow doubly exponentially. In the final section we suggest an alteration of Rényi dimension that better detects the aberrant nature of these examples.

2. Norms after convolution

Here follows our main technical result, Lemma 2.3. Our initial interest here was in the context of image analysis, where convolution by scaled Gaussians is common. For example, see [4].

For 1 < q < ∞, we have an easy finite bound on the partition function

S_{μ}^{q}(ϵ) = ∑_{k∈ℤ^{d}} μ(ϵk + ϵ𝕀)^{q},

specifically

S_{μ}^{q}(ϵ) ≤ ( ∑_{k∈ℤ^{d}} μ(ϵk + ϵ𝕀))^{q}
= μ(ℝ^{d})^{q}.

For 0 < q < 1, it is possible to have S_{μ}^{q}(ϵ) = ∞.

Definition 2.1. A finite Borel measure μ on ℝ^{d} is q-finite if S_{μ}^{q}(1) < ∞. Notice that if μ has bounded support then μ is q-finite for all 0 < q < 1.

Barbaroux, Germinet, and Tcheremchantsev have the following result implicitly in [1].

Lemma 2.2. Let μ be a finite Borel measure on ℝ^{d}. For any 0 < q < 1, the following are equivalent:

(a) μ is q-finite.

(b) There exists ϵ > 0 for which S_{μ}^{q}(ϵ) < ∞.

(c) For all ϵ > 0 it is true that S_{μ}^{q}(ϵ) < ∞.

Proof. In Section 3 of [1] it is shown that if μ is q-finite then S_{μ}^{q}(ϵ) < ∞ for small ϵ. Also, it is shown that if μ is not q-finite then S_{μ}^{q}(ϵ) = ∞ for small ϵ. A rescaling argument show that if S_{μ}^{q}(ϵ) is ever finite then it is finite for all small ϵ, while if it is ever infinite, it is infinite for all small ϵ. Therefore, the partition function is either finite for all ϵ or infinite for all ϵ. □

The proof of the following borrows from the methods in [1].

Lemma 2.3. Suppose g is a real-valued rapidly decreasing function on ℝ^{d} that is nonnegative, bounded, and nonzero at 0. Let

g_{ϵ}(x) = ϵ^{−d}g(ϵ^{−1}x).

Suppose μ is a finite Borel measure on ℝ^{d}. If 1 < q < ∞, or if 0 < q < 1 and μ is q-finite, then there exists a constant 1 < C < ∞ so that for all positive ϵ,

(2)  C^{−1} ≤ \frac{ϵ^{d(q−1)} ∥g_{ϵ} ∗ μ∥_{q}^{q}}{S_{μ}^{q}(ϵ)} ≤ C.

Here the L^{q} norm is with respect to Lebesgue measure. Therefore

D_{q}^{±}(μ) = d + lim_{ϵ→0} _{inf}^{sup} \frac{q}{q − 1} \frac{ln (∥g_{ϵ} ∗ μ∥_{q})}{ln(ϵ)}.

Proof. We will use m to denote Lebesgue measure, to keep it straight from μ. Let us denote the open unit rectangle at the origin by 𝔻, so

𝔻 = (−1, 1) × (−1, 1) × ⋯ × (−1, 1).

Recall 𝕀 is the product of d copies of [0, 1).

Let us denote by μ^{(ϵ)} the sequences over ℤ^{d} given by

μ_{n}^{(ϵ)} = μ(ϵn + ϵ𝕀).

Thus

S_{μ}^{q}(ϵ) = ∑_{k∈ℤ^{d}} μ(ϵk + ϵ𝕀)^{q} = ∥μ^{(ϵ)}∥_{q}^{q}

(the norm here is on l^{q}(ℤ^{d})).

An obvious rescaling reduces this theorem to the special case where

inf{g(x) | x ∈ 𝔻} > 0,

so let us make this assumption. We compute

(3) ∥g_{ϵ} ∗ μ∥_{q}^{q} = ϵ^{−qd} ∑_{j∈ℤ^{d}} ∫_{ϵj+ϵ𝕀} ( ∑_{k∈ℤ^{d}} ∫_{ϵk+ϵ𝕀} g(ϵ^{−1}(x − y)) dμ(y))^{q} dm(x).

If x ∈ ϵj + ϵ𝕀 and y ∈ ϵk + ϵ𝕀 then

ϵ^{−1}(x − y) ∈ (j − k) + 𝔻.

Let us define γ and Γ as sequences over ℤ^{d}, by

γ_{n} = inf{g(x) | x ∈ n + 𝔻}

and

Γ_{n} = sup{g(x) | x ∈ n + 𝔻}.

These give us bounds on the g(ϵ^{−1}(x − y)) term inside integrals in (3). For an upper bound,

∥g_{ϵ} ∗ μ∥_{q}^{q} ≤ ϵ^{−qd} ∑_{j∈ℤ^{d}} ( ∑_{k∈ℤ^{d}} Γ_{j−k}μ(ϵk + ϵ𝕀))^{q} ϵ^{d}

= ϵ^{d(1−q)} ∥Γ ∗ μ^{(ϵ)}∥_{q}^{q} .

For a lower bound,

∥g_{ϵ} ∗ μ∥_{q}^{q} ≥ ϵ^{d(1−q)} ∥γ ∗ μ^{(ϵ)}∥_{q}^{q} .

First assume 1 < q < ∞. The assumption that g is nonzero on 𝔻 is here used to obtain ∥γ∥_{q} > 0. From the rapidly decreasing assumption we obtain ∥Γ∥_{1} < ∞. Since

∥g_{ϵ} ∗ μ∥_{q} ≤ ϵ^{d \frac{(1−q)}{q}} ∥Γ ∗ μ^{(ϵ)}∥_{q}
≤ ϵ^{d \frac{(1−q)}{q}} ∥Γ∥_{1} ∥μ^{(ϵ)}∥_{q}

and

∥g_{ϵ} ∗ μ∥_{q} ≥ ϵ^{d \frac{(1−q)}{q}} ∥γ ∗ μ^{(ϵ)}∥_{q}
≥ ϵ^{d \frac{(1−q)}{q}} ∥γ∥_{q} ∥μ^{(ϵ)}∥_{q} ,

we may take

C = max (∥Γ∥_{1}^{q} , ∥γ∥_{q}^{−q}) .

Now assume 0 < q < 1. The assumptions on g tell us ∥γ∥_{1} > 0 and ∥Γ∥_{q} < ∞. In this case we may take

C = max (∥Γ∥_{q}^{q} , ∥γ∥_{1}^{−q}) .  □

3. Bounding the partition function

In this section we use Lemma 2.3 to establish bounds on

ln (S_{μ}^{q}(e^{x+h})) − ln (S_{μ}^{q}(e^{x}))

that are of first-order in h and hold for all x. Recall that the partition function for μ is

S_{μ}^{q}(ϵ) = ∑_{k∈ℤ^{d}} μ(ϵk + ϵ𝕀)^{q}.

Much of this section is familiar, but one conclusion seems novel: S_{μ}^{q}(x^{−1}) is almost decreasing. By this we mean

sup_{x<y} \frac{S_{μ}^{q}(y^{−1})}{S_{μ}^{q}(x^{−1})} < ∞.

(See [3].)

In this section, g denotes a standard Gaussian, and for ϵ > 0 we set

g_{ϵ}(x) = ϵ^{−d}g(ϵ^{−1}x).

Let us also adopt the notation

I_{μ}^{q}(ϵ) = ∥g_{ϵ} ∗ μ∥_{q}^{q} .

Lemma 3.1. Suppose μ is a finite Borel measure. If 1 < q < ∞ then

ϵ_{1} ≤ ϵ_{2} ⟹ ∥g_{ϵ_{2}} ∗ μ∥_{q} ≤ ∥g_{ϵ_{1}} ∗ μ∥_{q} .

If 0 < q < 1 and μ is q-finite, then

ϵ_{1} ≤ ϵ_{2} ⟹ ∥g_{ϵ_{2}} ∗ μ∥_{q} ≥ ∥g_{ϵ_{1}} ∗ μ∥_{q} .

Proof. If 1 < q < ∞ then

∥g_{η} ∗ μ∥_{q} = ∥g\sqrt{η^{2}−ϵ^{2}} ∗ g_{ϵ} ∗ μ∥_{q}
≤ ∥g\sqrt{η^{2}−ϵ^{2}} ∥_{1} ∥g_{ϵ} ∗ μ∥_{q}
= ∥g_{ϵ} ∗ μ∥_{q} .

For 0 < q < 1 the middle step becomes

∥g\sqrt{η^{2}−ϵ^{2}} ∗ g_{ϵ} ∗ μ∥_{q} ≥ ∥g\sqrt{η^{2}−ϵ^{2}} ∥_{1} ∥g_{ϵ} ∗ μ∥_{q} .  □

Lemma 3.2. Suppose μ is a finite Borel measure and that n is a natural number. For 1 < q < ∞, if ϵ > 0 then

0 ≤ ln (S_{μ}^{q}(ϵ)) − ln (S_{μ}^{q}(2^{−n}ϵ)) ≤ n ln (2^{d(q−1)}) .

For 0 < q < 1, if μ is q-finite and ϵ > 0 then

n ln (2^{d(q−1)}) ≤ ln (S_{μ}^{q}(ϵ)) − ln (S_{μ}^{q}(2^{−n}ϵ)) ≤ 0.

Proof. Suppose 1 < q < ∞. Given a disjoint union of Borel sets,

E_{1} ∪ ⋯ ∪ E_{2^{d}} = F,

we have the estimates

2^{d(1−q)}μ(F)^{q} ≤ ∑_{j=1}^{2^{d}} μ(E_{j})^{q} ≤ μ(F)^{q}.

Therefore

ln (S_{μ}^{q}(ϵ)) + ln(2^{d(1−q)}) ≤ ln (S_{μ}^{q}(2^{−1}ϵ)) ≤ ln (S_{μ}^{q}(ϵ)) .

For 0 < q < 1 the inequalities are all easily reversed. □

Lemma 3.3. Suppose μ is a finite Borel measure. If 1 < q < ∞, or if 0 < q < 1, and μ is q-finite, then there is a constant E so that

| ln(ϵ) − ln(η)| ≤ ln(2)

implies

|ln (S_{μ}^{q}(ϵ)) − ln (S_{μ}^{q}(η))| ≤ E.

Proof. We may assume

2^{−1}ϵ ≤ η ≤ ϵ.

By Lemma 2.3, there is a constant C so that for all ρ > 0,

|d(q − 1) ln(ρ) + ln (I_{μ}^{q}(ρ)) − ln (S_{μ}^{q}(ρ))| ≤ C.

If we set

Q = d(q − 1) ln (2^{− \frac{1}{2}} ϵ)

and

D = C + d|q − 1| ln (2^{\frac{1}{2}} )

then

2^{−1}ϵ ≤ ρ ≤ ϵ ⟹ |Q + ln (I_{μ}^{q}(ρ)) − ln (S_{μ}^{q}(ρ))| ≤ D.

Now assume 1 < q < ∞. Lemma 3.1 gives us

ln (I_{μ}^{q}(ϵ)) ≤ ln (I_{μ}^{q}(η)) ≤ ln (I_{μ}^{q}(2^{−1}ϵ))

and Lemma 3.2 gives us

ln (S_{μ}^{q}(2^{−1}ϵ)) ≤ ln (S_{μ}^{q}(ϵ)) .

182 Terry A. Loring We put this information together as

ln (S_{μ}^{q}(ϵ)) − 2D ≤ Q + ln (I_{μ}^{q}(ϵ)) − D
≤ Q + ln (I_{μ}^{q}(η)) − D
≤ ln (S_{μ}^{q}(η))
≤ Q + ln (I_{μ}^{q}(η)) + D
≤ Q + ln (I_{μ}^{q}(2^{−1}ϵ)) + D
≤ ln (S_{μ}^{q}(2^{−1}ϵ)) + 2D
≤ ln (S_{μ}^{q}(ϵ)) + 2D

For 0 < q < 1 the proof is similar. □ Theorem 3.4. Suppose μ is a finite Borel measure on ℝ^{d}. Let B = d(q − 1). For 1 < q < ∞, there is constant A so that

0 < η < ϵ

implies

−A ≤ ln (S_{μ}^{q}(ϵ)) − ln (S_{μ}^{q}(η)) ≤ A + B(ln(ϵ) − ln(η)).

For 0 < q < 1, if μ is q-finite then there is constant A so that

0 < η < ϵ

implies

−A + B(ln(ϵ) − ln(η)) ≤ ln (S_{μ}^{q}(ϵ)) − ln (S_{μ}^{q}(η)) ≤ A.

Proof. Assume first that 1 < q < ∞. For some natural number n,

2^{−n−1}ϵ ≤ η ≤ 2^{−n}ϵ.

This means

n ≤ \frac{1}{ln(2)}(ln(ϵ) − ln(η)) + 1

and so by the last three lemmas, ln (S_{μ}^{q}(ϵ)) − B(ln(ϵ) − ln(η)) − B ln(2) − E

≤ ln (S_{μ}^{q}(ϵ)) − nB ln(2) − E
≤ ln (S_{μ}^{q}(2^{−n}ϵ)) − E
≤ S_{μ}^{q}(η)
≤ ln (S_{μ}^{q}(2^{−n}ϵ)) + E
≤ ln (S_{μ}^{q}(ϵ)) + E
≤ ln (S_{μ}^{q}(ϵ)) + B ln(2) + E

so we can set

A = B ln(2) + E.

For 0 < q < 1 the proof is similar. □

4. Application to discrete limits

Riedi [15, 16] shows that

(4)  D_{p}^{±}(μ) = lim_{n→∞} _{inf}^{sup} \frac{1}{q − 1} \frac{ln (S_{μ}^{q}(ϵ_{n}))}{ln(ϵ_{n})}

for ϵ_{n} ↘ 0, so long as

lim sup_{n→∞}(ln(ϵ_{n}) − ln(ϵ_{n+1})) < ∞.

Indeed, he works with all p ∈ ℝ, and shows that grids other than

{ϵk + ϵ𝕀 | k ∈ ℤ^{d}}

can be used. What concerns us here is that Riedi showed that (4) is valid for a geometric series. We can go further, to allow sequences such a ϵ_{n} = b^{−n^{r}}.

Lemma 4.1. Suppose μ is a finite Borel measure on ℝ^{d}, and q ≠ 1 is a positive number. If 0 < q < 1 then also suppose μ is q-finite. If ϵ_{n} ↘ 0 and η_{n} ↘ 0 with

lim_{n→∞} \frac{ln(η_{n})}{ln(ϵ_{n})} = 1,

then

lim_{n→∞} |\frac{ln (S_{μ}^{q}(η_{n}))}{ln(η_{n})} − \frac{ln (S_{μ}^{q}(ϵ_{n}))}{ln(ϵ_{n})}| = 0.

Proof. Let A and B be the constants from Theorem 3.4, and let

A_{0} = A + |ln(S_{μ}^{p}(1)|

so that

|ln (S_{μ}^{q}(ϵ))| ≤ A_{0} + C| ln(ϵ)|

for all ϵ. We may assume ϵ_{n} ≤ e^{−1} and η_{n} ≤ e^{−1}, in which case

|\frac{ln (S_{μ}^{q}(η_{n}))}{ln(η_{n})} − \frac{ln (S_{μ}^{q}(ϵ_{n}))}{ln(ϵ_{n})}|

≤ \frac{|ln (S_{μ}^{q}(η_{n})) − ln (S_{μ}^{q}(ϵ_{n}))|}{− ln(η_{n})} + \frac{|ln (S_{μ}^{q}(ϵ_{n}))|}{− ln(ϵ_{n})} |\frac{ln(ϵ_{n})}{ln(η_{n})} − 1|

≤ \frac{A + C| ln(η_{n}) − ln(ϵ_{n})|}{− ln(η_{n})} + \frac{A_{0} − C ln(ϵ_{n})}{− ln(ϵ_{n})} |\frac{ln(ϵ_{n})}{ln(η_{n})} − 1|

≤ \frac{−A}{ln(η_{n})} + C |1 − \frac{ln(ϵ_{n})}{ln(η_{n})}| + (A_{0} + C) |\frac{ln(ϵ_{n})}{ln(η_{n})} − 1|

→ 0. □

Theorem 4.2. Suppose μ is a finite Borel measure on ℝ^{d} and q ≠ 1 is a positive number. If 0 < q < 1 then also suppose μ is q-finite. If ϵ_{n} ↘ 0 and

lim_{n→∞} \frac{ln(ϵ_{n+1})}{ln(ϵ_{n})} = 1,

then

D_{q}^{±}(μ) = \frac{1}{q − 1} lim_{n→∞} _{inf}^{sup} \frac{ln (S_{μ}^{q}(ϵ_{n}))}{ln(ϵ_{n})}.

Moreover, the sequence

(5)  n ↦ \frac{ln (S_{μ}^{q}(ϵ_{n}))}{ln(ϵ_{n})}

and the net

(6)  ϵ ↦ \frac{ln (S_{μ}^{q}(ϵ))}{ln(ϵ)}

have the same accumulation points.

Proof. Suppose x is an accumulation point of the net in (6). This means there is a sequence η_{n} ↘ 0 so that

lim_{n→∞} \frac{ln (S_{μ}^{q}(η_{n}))}{ln(η_{n})} = x.

Let the sequence k_{n} be defined so that

ϵ_{k_{n}+1} ≤ η_{n} ≤ ϵ_{k_{n}}.

Of course, k_{n} ↘ 0 and

lim_{n→∞} \begin{matrix}ln & (η_{n}) \\ ln & (ϵ_{k_{n}}) \end{matrix} = 1.

The last lemma tells us

lim_{n→∞} \frac{ln (S_{μ}^{q}(ϵ_{k_{n}}))}{ln(ϵ_{k_{n}})} = lim_{n→∞} \frac{ln (S_{μ}^{q}(η_{n}))}{ln(η_{n})} = x.

Thus x is also an accumulation point of the sequence in (5). □

5. Examples

It is possible to use simple recursive definitions to create a Borel measure μ, with support in [0, 1], so that the partition function

S_{μ}^{q}(ϵ) = ∑_{k∈ℤ} μ(ϵk + ϵ𝕀)^{q}

behaves almost any way we would like. However, we do need to respect Lemma 3.2.

Lemma 5.1. Suppose

0 ≤ a_{n} ≤ 1

for n ≥ 1. If q ≠ 1 is a positive real number, there is a Borel probability measure μ on [0, 1] for which

ln (S_{μ}^{q}(2^{−n})) = (1 − q) ln(2) ∑_{j=1}^{n} a_{n}.

Moreover, the net

ϵ ↦ \frac{1}{q − 1} \frac{ln (S_{μ}^{q}(ϵ))}{ln(ϵ)}

has the same accumulation points as the sequence

n ↦ \frac{1}{n} ∑_{j=1}^{n} a_{j}.

Proof. First let’s define F, the cumulative distribution function. We start with

F (0) = 0 and F (1) = 1.

Choose ω_{n} in [0, \frac{1}{2}] so that

ln (ω_{n}^{q} + (1 − ω_{n})^{q}) = (1 − q) ln(2)a_{n}.

Define F on the dyadic rationals between 0 and 1 by

F (\frac{2k + 1}{2^{n+1}}) = ω_{n}F (\frac{k}{2^{n}} ) + (1 − ω_{n})F (\frac{k + 1}{2^{n}}) .

Thus F is nondecreasing on the dyadic rationals; set it to 0 on dyadics less than 0 and to 1 on dyadics greater than 1. For any n and any k with 0 < k < 2^{−n} we have

F (\frac{k}{2^{n}} ) − F (\frac{k}{2^{n}} − \frac{1}{2^{n+1}} ) = ω_{n} (F (\frac{k}{2^{n}} ) − F (\frac{k}{2^{n}} − \frac{1}{2^{n}} )) .

Since we chose ω_{n} to the left of \frac{1}{2}, we have

F (\frac{k}{2^{n}} ) − F (\frac{k}{2^{n}} − \frac{1}{2^{n+1}} ) ≤ \frac{1}{2} (F (\frac{k}{2^{n}} ) − F (\frac{k}{2^{n}} − \frac{1}{2^{n}} )) .

Therefore, if m ≥ 1,

F (\frac{k}{2^{n}} ) − F (\frac{k}{2^{n}} − \frac{1}{2^{n+m}} ) ≤ (\frac{1}{2})^{m} (F (\frac{k}{2^{n}} ) − F (\frac{k}{2^{n}} − \frac{1}{2^{n}} )) .

Since F is bounded between 0 and 1, we have for any dyadic rational r,

\frac{k}{2^{n}} − \frac{1}{2^{n+m}} ≤ r < \frac{k}{2^{n}} ⟹ F (\frac{k}{2^{n}} ) − F (r) ≤ (\frac{1}{2})^{m} .

Since F is nondecreasing, this says

F (s) = sup_{s>r∈ℤ[\frac{1}{2}]} F (r)

for all s in ℤ[\frac{1}{2}].

Let us extend F to ℝ by

F (x) = sup_{x>r∈ℤ[\frac{1}{2}]} F (r).

It is routine to verify that F is left continuous and nondecreasing.

Since F is nondecreasing and left continuous, we have an associated measure μ which satisfies

μ([a, b)) = F (b) − F (a).

By design,

μ ([\frac{2k}{2^{n+1}} , \frac{2k + 1}{2^{n+1}})) = (1 − ω_{n})μ ([\frac{k}{2^{n}} , \frac{k + 1}{2^{n}}))

and

μ ([\frac{2k + 1}{2^{n+1}}, \frac{2k + 2}{2^{n+1}})) = ω_{n}μ ([\frac{k}{2^{n}} , \frac{k + 1}{2^{n}})) .

This tells us

S_{μ}^{q}(2^{−n−1}) = (ω_{n}^{q} + (1 − ω_{n})^{q})S_{μ}^{q}(2^{−n})

and so also

ln (S_{μ}^{q}(2^{−n−1})) = (1 − q) ln(2)a_{n} + ln (S_{μ}^{q}(2^{−n})) .

Since

S_{μ}^{q}(1) = ∥μ∥_{1} = 1,

induction gives us

ln (S_{μ}^{q}(2^{−n})) = (1 − q) ln(2) ∑_{j=1}^{n} a_{j}.

With μ as constructed from a_{n} and q as indicated, Theorem 4.2 applies to give the final statement in the lemma. □

Theorem 5.2. Suppose 0 < q < 1 or 1 < q < ∞, and ϵ_{n} ↘ 0 with

lim inf_{n→∞} \frac{ln(ϵ_{n+1})}{ln(ϵ_{n})} > 1.

Then there is a Borel probability measure on [0, 1] so that

lim_{n→∞} \frac{ln (S_{μ}^{q}(ϵ_{n}))}{ln(ϵ_{n})}

exists but

lim_{ϵ→0} \frac{ln (S_{μ}^{q}(ϵ))}{ln(ϵ)}

does not.

Proof. By Lemma 4.1, it suffices to prove this in the special case where

ϵ_{n} = 4^{−k_{n}}

for some k_{n} ∈ ℕ. The hypothesis on the ϵ_{n} translates to the assumption that k_{n} is nondecreasing, with limit ∞, and that

lim inf_{n→∞} \frac{k_{n+1}}{k_{n}} > 1.

Select a subsequence k_{n_{l}} and some R > 1 so that

k_{n_{l}+1} ≥ Rk_{n_{l}}

for all l. Define a_{j} = \frac{1}{2} for all j except

2k_{n_{l}} < j ≤ (k_{n_{l}} + k_{n_{l}+1}) ⟹ a_{j} = 0

and

(k_{n_{l}} + k_{n_{l}+1}) < j ≤ 2k_{n_{l}+1} ⟹ a_{j} = 1

With μ as in Lemma 5.1,

\frac{1}{q − 1} \frac{ln (S_{μ}^{q}(ϵ_{n}))}{ln(ϵ_{n})} = \frac{1}{2k_{n}} ∑_{j=1}^{2k_{n}} a_{j} = \frac{1}{2}.

If

η_{n} = 2^{−(k_{n_{l}} +k_{n_{l}+1})}

then

\begin{aligned}\frac{1}{q − 1} \frac{ln (S_{μ}^{q}(η_{n}))}{ln(η_{n})} & = \frac{1}{k_{n_{l}} + k_{n_{l}+1}} ∑_{j=1}^{k_{n_{l}} +k_{n_{l}+1}} a_{n} \\ & = \frac{k_{n_{l}}}{k_{n_{l}} + k_{n_{l}+1}} \\ & > \frac{1}{2}. \end{aligned}  □

6. Rényi dimension of convolutions

Barbaroux, Germinet and Tcheremchantsev ([1]) establish bounds that relate D_{q}^{±}(μ ∗ ν) with D_{q}^{±}(μ) and D_{q}^{±}(ν), when q is positive and q ≠ 1. Here we establish related bounds.

Theorem 6.1. Suppose μ and ν are Borel measures on ℝ^{d}. If 1 < q < ∞ and \frac{1}{q} + 1 = \frac{1}{r} + \frac{1}{s} for some 1 < r, s < ∞, then

D_{q}^{−}(μ ∗ ν) ≥ \frac{q(r − 1)}{r(q − 1)}D_{r}^{−}(μ) + \frac{q(s − 1)}{s(q − 1)}D_{s}^{−}(ν).

If 0 < q < 1, and μ is q-finite, and \frac{1}{q} + 1 = \frac{1}{r} + \frac{1}{s} for some 0 < r, s < 1, then

D_{q}^{+}(μ ∗ ν) ≤ \frac{q(r − 1)}{r(q − 1)}D_{r}^{+}(μ) + \frac{q(s − 1)}{s(q − 1)}D_{s}^{+}(ν).

Proof. We again let g be the standard Gaussian.

Assume 1 < r, s, q < ∞ and \frac{1}{q} + 1 = \frac{1}{r} + \frac{1}{s}. By Young’s inequality (Theorem 1.2.12 in [6]),

d + \frac{q}{q − 1} \frac{ln(∥g_{\sqrt{2}ϵ} ∗ (μ ∗ ν)∥_{p})}{ln(\sqrt{2}ϵ)}

= d + \frac{q}{q − 1} \frac{ln(∥(g_{ϵ} ∗ μ) ∗ (g_{ϵ} ∗ ν)∥_{p})}{ln(ϵ)}

(7) ≥ d + \frac{q}{q − 1} (\frac{ln(∥g_{ϵ} ∗ μ∥_{r})}{ln(ϵ)} + \frac{ln(∥g_{ϵ} ∗ ν∥_{s})}{ln(ϵ)})

= \frac{q(r − 1)}{r(q − 1)} (d + \frac{r}{r − 1} \frac{ln ∥g_{ϵ} ∗ μ∥_{r}}{ln(ϵ)} ) + \frac{q(s − 1)}{s(q − 1)} (d + \frac{s}{s − 1} \frac{ln ∥g_{ϵ} ∗ ν∥_{s}}{ln(ϵ)} )

Notice that

0 < \frac{q(r − 1)}{r(q − 1)}, \frac{q(s − 1)}{s(q − 1)}

and

\frac{q(r − 1)}{r(q − 1)} + \frac{q(s − 1)}{s(q − 1)} = 1.

Now take lim inf of both sides of (7) and apply Guérin’s formula (1).

For 0 < q < 1, the inequality switches in (7), we apply lim sup to both sides, and then use Lemma 2.3. See [2] regarding Young’s inequality for 0 < q < 1. □

7. Best fit slopes

Rather than tracking long-term slopes

\frac{ln (S_{μ}^{q}(e^{x})) − ln (S_{μ}^{q}(e^{x_{0}}))}{x − x_{0}}

to determine a “dimension” for μ, one could look at slope information of the function

x ↦ ln (S_{μ}^{q}(e^{x})) .

in many ways. Here we consider the slopes of least-squares best fit lines.

This is not to advocate for or against using least-squares best fit lines for calculating fractal dimensions in practice. See [13, 12] for a discussion. See also [14].

Remark 7.1. Given a measurable function

ρ : [0, ∞) → ℝ

that is bounded on bounded intervals, the slope of closest line over [0, x] to ρ is

\frac{6}{x^{3}} ∫_{0}^{x}(2t − x)ρ(t) dt.

Here closest means with respect to the L^{2} norm using Lebesgue measure on [0, x].

Remark 7.2. Given a sequence

ρ : ℕ → ℝ,

and any positive real λ, the slope of the least-squares best fit line to

{(0, ρ_{0}), (λ, ρ_{1}), …, ((n − 1)λ, ρ_{n−1})}

is

\frac{6}{(n^{3} − n)λ} ∑_{k=0}^{n−1}(2k + 1 − n)ρ_{k}.

Definition 7.3. A function ρ : [0, ∞) → ℝ is nearly Lipschitz if ρ is measurable and there are finite constants A and B so that

|ρ(x) − ρ(y)| ≤ A + B|x − y|.

Lemma 7.4. If λ > 0 and ρ : [0, ∞) → ℝ is nearly Lipschitz, then the sequence

n ↦ \frac{6}{(nλ)^{3}} ∫_{0}^{nλ}(2t − nλ)ρ(t) dt

has the same accumulation points as the net

x ↦ \frac{6}{x^{3}} ∫_{0}^{x}(2t − x)ρ(t) dt.

Moreover, there is a constant C so that

1 ≤ x ≤ y ≤ x + λ

implies

|\frac{6}{x^{3}} ∫_{0}^{x}(2t − x)ρ(t) dt − \frac{6}{y^{3}} ∫_{0}^{y}(2t − y)ρ(t) dt| ≤ \frac{C}{x} .

Rényi dimension and Gaussian filtering 189

Proof. Let A_{0} = A + |ρ(0)|, so that

|ρ(t)| ≤ A_{0} + Bt.

A change of variable shows

\frac{6}{x^{3}} ∫_{0}^{x}(2t − x)ρ(t) dt = 6 ∫_{0}^{1}(2t − 1)\frac{1}{x}ρ(xt) dt.

For x > 0 and y > 0,

|∫_{0}^{1}(2t − 1)\frac{1}{x}ρ(xt) dt − ∫_{0}^{1}(2t − 1)\frac{1}{y} ρ(yt) dt|

= |∫_{0}^{1}(2t − 1)\frac{1}{x} (ρ(xt) − ρ(yt)) dt + ∫_{0}^{1} (\frac{1}{x} − \frac{1}{y} ) (2t − 1)ρ(yt) dt|

≤ \frac{1}{x} ∫_{0}^{1} |2t − 1||ρ(xt) − ρ(yt)| dt + \frac{|x − y|}{xy} ∫_{0}^{1} |2t − 1||ρ(yt)| dt.

We estimate the first term via

∫_{0}^{1} |2t − 1||ρ(xt) − ρ(yt)| dt ≤ ∫_{0}^{1} |ρ(xt) − ρ(yt)| dt

≤ ∫_{0}^{1} A + B|x − y|t dt

= A + \frac{1}{2}B|x − y|.

For the second,

∫_{0}^{1} |2t − 1||ρ(yt)| dt ≤ ∫_{0}^{1} |ρ(yt)| dt
≤ ∫_{0}^{1} A_{0} + Byt dt

so

(8)  ∫_{0}^{1} |2t − 1||ρ(yt)| dt ≤ A_{0} + \frac{1}{2}By.

If we assume

x ≤ y ≤ x + λ

then

\begin{matrix}|∫_{0}^{1}(2t − 1)\frac{1}{x}ρ(xt) dt − ∫_{0}^{1}(2t − 1)\frac{1}{y} ρ(yt) dt| \\ ≤ \frac{1}{x} (A + \frac{1}{2}Bλ) + \frac{λ}{xy} (A_{0} + \frac{1}{2}By) \\ ≤ \frac{1}{x} (A + \frac{1}{2}Bλ) + \frac{λ}{x^{2}} A_{0} + \frac{1}{2} \frac{λ}{x}B \\ = (A + Bλ) \frac{1}{x} + (λA_{0}) \frac{1}{x^{2}} . \end{matrix}  □

190 Terry A. Loring Lemma 7.5. If λ > 0 and ρ : [0, ∞) → ℝ is nearly Lipschitz, then the sequence

n ↦ \frac{6}{(n^{3} − n)λ} ∑_{k=0}^{n−1}(2k + 1 − n)ρ(λk)

has the same accumulation points as the sequence

n ↦ \frac{6}{(nλ)^{3}} ∫_{0}^{nλ}(2t − nλ)ρ(t) dt.

Moreover, there is a constant C so that n ≥ 2 implies

|\frac{6}{(nλ)^{3}} ∫_{0}^{nλ}(2t − nλ)ρ(t) dt − \frac{6}{(n^{3} − n)λ} ∑_{k=0}^{n−1}(2k + 1 − n)ρ(λk)| ≤ \frac{C}{n} .

Proof. Let A_{0} = A + |ρ(0)|, so that

|ρ(t)| ≤ A_{0} + Bt.

Equation (8), with y = nλ, gives us the constant bound

|\frac{6}{(nλ)^{3}} ∫_{0}^{nλ}(2t − nλ)ρ(t) dt| = \frac{6}{nλ} |∫_{0}^{1}(2t − 1)ρ(nλt) dt|

≤ \frac{6}{nλ} (A_{0} + \frac{1}{2}Bnλ)

≤ 3 (\frac{A_{0}}{λ} + B)

as long as n ≥ 2. Also n ≥ 2 implies

|\frac{n^{3}}{n^{3} − n} − 1| ≤ \frac{2}{n^{2}} .

Therefore, it suffices to estimate the distance from

\frac{6}{(nλ)^{3}} ∫_{0}^{nλ}(2t − nλ)ρ(t) dt = \frac{6}{n^{3}λ} ∫_{0}^{n}(2t − n)ρ(λt) dt

to

\frac{6}{n^{3}λ} ∑_{k=0}^{n−1}(2k + 1 − n)ρ(λk).

For all n ≥ 2,

\begin{matrix}|\frac{6}{n^{3}λ} ∫_{0}^{n}(2t − n)ρ(λt) dt − \frac{6}{n^{3}λ} ∑_{k=1}^{n−1}(2k + 1 − n)ρ(λk)| \\ = \frac{6}{n^{3}λ} |∑_{k=0}^{n−1} ∫_{k}^{k+1}(2t − n)ρ(λt) dt − ∑_{k=0}^{n−1} ∫_{k}^{k+1}(2k + 1 − n)ρ(λk) dt| \\ ≤ \frac{6}{n^{3}λ} ∑_{k=0}^{n−1} ∫_{k}^{k+1} |(2t − n) − (2k + 1 − n)||ρ(λt)| dt \\ + \frac{6}{n^{3}λ} ∑_{k=0}^{n−1} ∫_{k}^{k+1} |2k + 1 − n||ρ(λt) − ρ(λk)| dt \\ ≤ \frac{6}{n^{3}λ} ∑_{k=0}^{n−1} ∫_{k}^{k+1} |ρ(λt)| + n|ρ(λt) − ρ(λk)| dt \\ ≤ \frac{6}{n^{3}λ} ∑_{k=0}^{n−1} ∫_{k}^{k+1} A_{0} + Bλt + n(A + Bλ(t − k)) dt \\ = (\frac{6A}{λ} + \frac{6|ρ(0)|}{λ} + 3B) \frac{1}{n^{2}} + (6B + \frac{6A}{λ}) \frac{1}{n}. \end{matrix}  □

Lemma 7.6. If λ > 0 and ρ : [0, ∞) → ℝ is nearly Lipschitz, then the sequence

n ↦ \frac{6}{(n^{3} − n)λ} ∑_{k=0}^{n−1}(2k + 1 − n)ρ(λk)

has the same accumulation points as the net

x ↦ \frac{6}{x^{3}} ∫_{0}^{x}(2t − x)ρ(t) dt.

Moreover, there are constants C and D so that

D ≤ nλ ≤ x ≤ (n + 1)λ

implies

|\frac{6}{x^{3}} ∫_{0}^{x}(2t − x)ρ(t) dt − \frac{6}{(n^{3} − n)λ} ∑_{k=0}^{n−1}(2k + 1 − n)ρ(λk)| ≤ \frac{C}{n} .

Proof. Let C_{1} and C_{2} be the constants from the last two lemmas. Suppose

max(1, 2λ) ≤ nλ ≤ x ≤ (n + 1)λ.

Let y = nλ. We have n ≥ 2 and

0 < y ≤ x ≤ y + λ.

Therefore

\begin{matrix}|\frac{6}{x^{3}} ∫_{0}^{x}(2t − x)ρ(t) dt − \frac{6}{(n^{3} − n)λ} ∑_{k=0}^{n−1}(2k + 1 − n)ρ(λk)| \\ ≤ |\frac{6}{x^{3}} ∫_{0}^{x}(2t − x)ρ(t) dt − \frac{6}{y^{3}} ∫_{0}^{y}(2t − y)ρ(t) dt| \\ + |\frac{6}{n^{3}λ^{3}} ∫_{0}^{nλ}(2t − nλ)ρ(t) dt − \frac{6}{(n^{3} − n)λ} ∑_{k=0}^{n−1}(2k + 1 − n)ρ(λk)| \\ ≤ \frac{C_{1}}{y} + \frac{C_{2}}{n} \\ = (\frac{C_{1}}{λ} + C_{2}) \frac{1}{n}. \end{matrix}  □

Theorem 7.7. Suppose μ is a finite Borel measure on ℝ^{d}, and 0 < q < 1 or 1 < q < ∞. If 0 < q < 1 then also suppose μ is q-finite. Suppose ν > 1. Let

S_{μ}^{q}(ϵ) = ∑_{k∈ℤ} μ(ϵk + ϵ𝕀)^{q}.

For x in [0, ∞), let m_{x} and b_{x} be the real numbers so that

t ↦ m_{x}t + b_{x}

is the linear function that minimizes

∫_{−x}^{0} (ln (S_{μ}^{q}(e^{t})) − (m_{x}t + b_{x}))^{2} dt.

(Here dt refers to Lebesgue measure on [−x, 0].) For a natural number n, let m̃_{n} and b̃_{n} the the real numbers so that

t ↦ m̃_{n}t + b̃_{n}

is the the least-squares best fit line to the pairs

{(ln(ν^{−k}), ln (S_{μ}^{q}(ν^{−k})))| k = 0, 1, …, n − 1} .

There are constants C and D so that

D ≤ n ln(ν) ≤ x ≤ (n + 1) ln(ν)

implies

|m_{x} − m̃_{n}| ≤ \frac{C}{n} .

In particular,

lim_{x→∞} _{inf}^{sup}m_{x} = lim_{n→∞} _{inf}^{sup}m̃_{n}.

Proof. By Theorem 3.4, there are constants A and B so that

|ln (S_{μ}^{q}(e^{−x})) − ln (S_{μ}^{q}(e^{−y}))| ≤ A + B |x − y|

for all x and y. Since ln (S_{μ}^{q}(e^{−x})) is Borel measurable, we see from Lemma 7.6 that there are constants C and D so that

D ≤ n ln(ν) ≤ x ≤ (n + 1) ln(ν)

Rényi dimension and Gaussian filtering 193

implies

\frac{6}{x^{3}} ∫_{0}^{x}(2t − x) ln (S_{μ}^{q} (e^{−t})) dt

is within \frac{C}{n} of

\frac{6}{(n^{3} − n) ln(ν)} ∑_{k=0}^{n−1}(2k + 1 − n) ln (S_{μ}^{q} (e^{−k ln(ν)})) .

That is,

(9)  \frac{6}{x^{3}} ∫_{0}^{x}(2t − x) ln (S_{μ}^{q} (e^{−t})) dt

is within \frac{C}{n} of

(10)  \frac{6}{(n^{3} − n) ln(ν)} ∑_{k=0}^{n−1}(2k + 1 − n) ln (S_{μ}^{q} (ν^{−k})) .

The quantity in (9) gives the slope of the best fit over [0, x] of

t ↦ ln (S_{μ}^{q}(e^{−t})) ,

and so

(11)  m_{x} = \frac{−6}{x^{3} ln(ν)} ∫_{0}^{x}(2t − x) ln (S_{μ}^{q}(e^{−t})) dt.

The quantity in (10) gives the slope of the best fit to

{(k ln(ν), ln (S_{μ}^{q}(ν^{−k})))| k = 0, 1, …, n − 1}

and so

(12) m̃_{n} = \frac{−6}{(n^{3} − n) ln(ν)} ∑_{k=0}^{n−1}(2k + 1 − n) ln (S_{μ}^{q}(ν^{−k})) .

We are done. □ Remark 7.8. It is interesting to note some alternative formulas:

(13)  m̃_{n} = \frac{∑_{k=0}^{n−1}(2k + 1 − n) ln (S_{μ}^{q}(ν^{−k}))}{∑_{k=0}^{n−1}(2k + 1 − n) ln (ν^{−k})};
(14) m̃_{n} = \frac{6}{(n^{3} − n) ln(ν)} ∑_{k=1}^{n−1} k(n − k) (ln (S_{μ}^{q}(ν^{−(k−1)}) − ln (S_{μ}^{q}(ν^{−k}))));
(15)  m̃_{n} = \frac{∑_{k=1}^{n−1} k(n − k) (ln (S_{μ}^{q}(ν^{−k})) − ln (S_{μ}^{q}(ν^{−(k−1)})))}{∑_{k=1}^{n−1} k(n − k) ln (ν^{−1})};
(16)  m_{x} = \frac{∫_{0}^{x}(2t − (x)) ln (S_{μ}^{q}(ν^{−t})) dt}{∫_{0}^{x}(2t − x) ln (ν^{−t}) dt}.

8. More examples

The slope of the least-squares best fit linear approximation to the partition function cannot always be used to determine the Renyi dimension. We show this by the following example. This example is far from what we hope to see in applications.

Lemma 8.1. For any 1 < q < ∞, there is a finite Borel measure μ on [0, 1] for which

lim sup_{x→−∞} \frac{ln (S_{μ}^{q}(e^{x}))}{x} < lim sup_{x→−∞} m_{x},

where

m_{x}t + b_{x} ≈ ln (S_{μ}^{q}(e^{t})) (−x ≤ t ≤ 0)

is the least-squares best fit line. More specifically, m_{x} and b_{x} minimize

∫_{−x}^{0} (ln (S_{μ}^{q}(e^{t})) − (m_{x}t + b_{x}))^{2} dt.

For any 0 < q < 1, there is a finite Borel measure μ on [0, 1] for which

lim inf_{x→−∞} \frac{ln (S_{μ}^{q}(e^{x}))}{x} > lim inf_{x→−∞} m_{x}.

Proof. We will use the μ associated with a sequence a_{n} and a choice of q, as in Lemma 5.1.

Let

a_{1} = \frac{30}{47}

and

a_{k} = \begin{cases}0 & if 48^{n} < k ≤ 12(48^{n}), any n ∈ ℕ \\ 1 & if 12(48^{n}) < k ≤ 36(48^{n}), any n ∈ ℕ \\ \frac{1}{2} & if 36(48^{n}) < k ≤ 48(48^{n}), any n ∈ ℕ. \end{cases}

When 1 < q < ∞,

lim sup_{ϵ→0} \frac{ln (S_{μ}^{q}(ϵ))}{ln(ϵ)} = lim sup_{n→0}(q − 1) ln(2)\frac{1}{n} ∑_{k=1}^{n} a_{k}

and

lim sup_{ϵ→0} m_{x} = lim sup_{n→0} \frac{6(q − 1) ln(2)}{n^{3}} ∑_{k=1}^{n−1} k(n − k)a_{k}.

Therefore we really only need to show that

lim sup_{n→∞} \frac{1}{n} ∑_{k=1}^{n} a_{k} < lim sup_{n→∞} \frac{6}{n^{3}} ∑_{k=1}^{n−1} k(n − k)a_{k}.

For q < 1 the (q − 1) reverses the inequalities and turns each lim inf into a lim sup, so the desired inequality reduces to the same thing.

For all n,

\frac{1}{48^{n+1}} ∑_{k=1}^{48^{n+1}} a_{k} = \frac{1}{48^{n+1}} [(∑_{k=1}^{48^{n}} a_{k}) + 24(48^{n}) + 12(48^{n})\frac{1}{2}]

= \frac{1}{48} [\frac{1}{48^{n}} (∑_{k=1}^{48^{n}} a_{k}) + 30] .

Since

\frac{1}{48} [\frac{30}{47} + 30] = \frac{30}{47}

we have found

\frac{1}{48^{n}} ∑_{k=1}^{48^{n}} a_{k} = \frac{30}{47}

for all n. The terms of value 0 will cause the average to fall until index 12(48^{n}). At this point, the average will be

\frac{1}{12} \frac{30}{47} + \frac{11}{12}0 = \frac{5}{94}

The next 12(48^{n}) terms are of value 1, so the average rises to

\frac{1}{3} \frac{5}{94} + \frac{2}{3}1 = \frac{193}{282}.

Next the average falls, due the terms of value \frac{1}{2}, until it is back to \frac{30}{47}. Therefore,

lim sup_{n→∞} \frac{1}{n} ∑_{k=1}^{n} a_{k} = \frac{193}{282}.

We now need just a decent estimate on

lim sup_{n→∞} \frac{6(q − 1)}{n^{3}} ∑_{k=1}^{n−1} k(n − k)a_{k}.

Since a_{n} > 0 for all n, we find

lim sup_{n→∞} \frac{6}{n^{3}} ∑_{k=1}^{n−1} k(n − k)a_{k}

≥ lim sup_{n→∞} \frac{6}{(48^{n})^{3}} ∑_{k=1}^{48^{n}} k(48^{n} − k)a_{k}

> lim sup_{n→∞} 6\frac{∑_{k=12(48^{n−1})+1}^{36(48^{n−1})} k(48^{n} − k) + ∑_{k=36(48^{n−1})+1}^{48^{n}} k(48^{n} − k)\frac{1}{2}}{(48^{n})^{3}}

= lim_{n→∞} 6 ∑_{k=12(48^{n−1})+1}^{36(48^{n−1})} \frac{k}{48^{n}} (1 − \frac{k}{48^{n}} ) \frac{1}{48^{n}}

+ lim_{n→∞} 3 ∑_{k=36(48^{n−1})+1}^{48^{n}} \frac{k}{48^{n}} (1 − \frac{k}{48^{n}} ) \frac{1}{48^{n}}

= 6 ∫ \frac{3}{4} \frac{1}{4} t (1 − t) dt + 3 ∫ 1_{\frac{3}{4}} t (1 − t) dt

= 3 (\frac{1}{2}1^{2} − \frac{1}{3}1^{3}) + 3 (\frac{1}{2} (\frac{3}{4})^{2} − \frac{1}{3} (\frac{3}{4})^{3}) − 6 (\frac{1}{2} (\frac{1}{4})^{2} − \frac{1}{3} (\frac{1}{4})^{3})

= \frac{49}{64}.

Thus

lim sup_{ϵ→0} \frac{ln (S_{μ}^{q}(ϵ))}{ln(ϵ)} = \frac{193}{282}(q − 1) = \frac{6176}{9024}(q − 1)

and

lim sup_{ϵ→0} m_{x} > \frac{49}{64}(q − 1) = \frac{6909}{9024}(q − 1).  □

9. Modified Rényi dimensions

The theory of regular variation and its extensions ([3]) give many ways to measure how closely a function f behaves like various powers x^{c} near ∞. Regular variation forces f to behave like a single power x^{δ}.

More realistic classes are those of extended variation and O-regularly varying functions. Both classes allow f to behave like x^{c} for c in a range (α, β), but they differ on the meaning of “behave.” (This is a bit vague. See [3].) The extended real numbers α and β are called Karamata indices in the case where f is of extended variation. For the class of O-regularly varying functions, these are called the Matuszewska indices.

Guido and Isola ([8, 9, 10]) have used the Matuszewska indices to define a new local fractal dimension. Stern ([17]) has suggested generally that the theories of extended variation and O-regular variation be applied to global fractal dimensions.

The example in Section 8 is rather unnatural. It can be explained away if we use Matuszewska indices to describe the “slope at infinity” of the partition function.

We use the following as a working definition of the Matuszewska indices. It is equivalent to the standard definition, cf. pages 68–73 of [3].

Definition 9.1. Suppose

f : [0, ∞) → (0, ∞)

is any function. The upper Matuszewska index of f is

α(f) = inf \{α ∈ ℝ | ∃X, C s.t. y ≥ x ≥ X ⟹ f(y) ≤ C (\frac{y}{x})^{α} f(x)\} .

Here X are C are to be understood to be positive real numbers. The lower Matuszewska index of f is

β(f) = sup \{β ∈ ℝ | ∃X, C s.t. y ≥ x ≥ X ⟹ f(y) ≥ C (\frac{y}{x})^{β} f(x)\} .

It is easy to show that

β(f) ≤ lim inf_{x→∞} \frac{ln(f(x))}{ln(x)} ≤ lim sup_{x→∞} \frac{ln(f(x))}{ln(x)} ≤ α(f).

Again, see [3]. The middle numbers are the so-called orders of f. In an unfortunate clash of terminology, the “upper and lower Rényi dimensions of order q” are the upper and lower orders of

x ↦ (S_{μ}^{q}(x^{−1}))^{\frac{1}{1−q}} .

Perhaps it is better to refer to q as the index.

Definition 9.2. If μ is a finite measure, and if 0 < q < 1 or 1 < q < ∞, the upper and lower Matuszewska Dimensions of index q are the upper and lower Matuszewska indices of the function

x ↦ (S_{μ}^{q}(x^{−1}))^{\frac{1}{1−q}} ,

denoted D_{q}^{++}(μ) and D_{q}^{−−}(μ) respectively.

Theorem 9.3. Suppose μ is a finite Borel measure on ℝ^{d}. If 1 < q < ∞, or if 0 < q < 1 and μ is q-finite, then the partition function S_{μ}^{q}(x^{−1}) is of extended variation and

0 ≤ D_{q}^{−−}(μ) ≤ D_{q}^{−}(μ) ≤ D_{q}^{+}(μ) ≤ D_{q}^{++}(μ) ≤ d.

Proof. The second and fourth inequalities come from the general facts about order and Matuszewska indices. The middle is even more standard. The outer inequalities are really just restatements of those in Theorem 3.4.

Equivalently, these inequalities show that the upper and lower Matuszewska indices of S_{μ}^{q}(x^{−1}) are bounded between and 0 and 1 − q. Since S_{μ}^{q}(x^{−1}) is measurable, we can apply [3, Theorem 2.1.7] to conclude that S_{μ}^{q}(x^{−1}) is of extended variation. □

Remark 9.4. In the example of Section 8:

D_{q}^{−−}(μ) = 0,

D_{q}^{−}(μ) = \frac{5}{94},

D_{q}^{+}(μ) = \frac{193}{282},

198 Terry A. Loring

D_{q}^{++}(μ) = 1.

Thus the upper and lower Matuszewska dimensions dismiss μ from “being selfsimilar” more resoundingly than do the upper and lower Rényi dimensions.

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Department of Mathematics and Statistics, University of New Mexico, Albuquerque, NM 87131, USA loring@math.unm.edu http://www.math.unm.edu/˜ loring/

This paper is available via http://nyjm.albany.edu/j/2007/13-11.html.