The Packing Measure of the Trajectory of a One-Dimensional Symmetric Cauchy Process

Abstract

Let 𝑋_𝑡={𝑋(𝑡),𝑡≥0} be a one-dimensional symmetric Cauchy process. We prove that, for any measure function, 𝜑,𝜑−𝑝(𝑋[0,𝜏]) is zero or infinite, where 𝜑−𝑝(𝐸) is the 𝜑-packing measure of 𝐸, thus solving a problem posed by Rezakhanlou and Taylor in 1988.

1. Introduction

Let 𝑋_𝑡={𝑋(𝑡),𝑡≥0} be a strictly stable Levy process taking values in 𝑅^𝑛 (𝑛-dimensional Euclidean space) of index 𝛼∈(0,2], that is, a Markov process with stationary independent increment whose characteristic function is given by 𝐸𝑒^𝑖^{𝑢,𝑋𝑡}=𝑒^{−𝑡𝜓𝛼}^𝑢.(1.1)

Here, 𝑢 and 𝑋_𝑡 are points in 𝑅^𝑛, (𝑢,𝑥) is the ordinary inner product in 𝑅^𝑛, and ‖𝑥‖²=(𝑥,𝑥). The Levy exponent 𝜓_𝛼(𝑢) is of the form 𝜓_𝛼𝑢=||𝑢||^𝛼_𝑆𝑛𝑤_𝛼𝑢,𝑦𝜇𝑑𝑦,(1.2)where 𝑤_𝛼𝑢,𝑦=1−𝑖sgn𝑢,𝑦tan𝜋𝛼2|||‖‖𝑢𝑢‖‖,𝑦|||^𝛼𝑤if𝛼≠1,₁𝑢,𝑦=|||𝑢‖‖𝑢‖‖,𝑦|||+2𝜋𝑖𝑢,𝑦log|||𝑢,𝑦|||.(1.3)𝜇(𝑑𝑦) is an arbitrary finite measure on the unit sphere 𝑆_𝑛 in 𝑅^𝑛, not supported on a diametrical plane. If in (1.2) 𝜇 is the uniform distribution on 𝑆_𝑛, 𝑋_𝑡 is called the isotropic stable Levy process with index 𝛼. In this case, 𝜓_𝛼(𝑢)=𝜆|𝑢|^𝛼 for some 𝜆>0. When 𝛼=1, 𝜇 must also have the origin as its center of mass, that is, _𝑆𝑛𝑦𝜇𝑑𝑦=0,(1.4)and the resulting process is the symmetric Cauchy process.

If ∫_𝑆𝑛𝑦𝜇(𝑑𝑦)≠0 for 𝛼=1, we have the strictly asymmetric Cauchy process. When 𝛼=2, we obtain the standard Brownian motion.

We assume that our process has been defined so that the strong Markov property is valid and all sample paths are right continuous and have left limits everywhere.

It is well known that the sample paths 𝑋_𝑡 of strictly stable Levy processes determine trajectories in 𝑅^𝑛 that are random fractal sets.

We are interested in the range of the processes, that is, the random set 𝑅_𝜏 generated by 𝑋_𝑡 and defined by 𝑅_𝜏=𝑋0,𝜏=𝑥∈𝑅^𝑛∶𝑥=𝑋𝑡forsome𝑡∈0,𝜏.(1.5)

The Hausdorff and packing measures serve as useful tools for analyzing fine properties of Levy processes.

The problem of determining the exact Hausdorff measure of the range of those processes for 𝛼∈(0,2] has been completely solved. See, for example, [1].

The study of the exact packing measure of the range of a stochastic process has a more recent history, starting with the work of Taylor and Tricot [2].

The packing measure of the trajectory was found in [2] by Taylor and Tricot for transient Brownian motion. The corresponding problem for the range of strictly stable processes, 𝛼<𝑛, was solved by Taylor [3].

Further results on the asymmetric Cauchy process and subordinators have been established by Rezakhanlou and Taylor [4] and Fristedt and Taylor [5], respectively.

For the critical cases, 𝛼=𝑛, the only known result is due to Le Gall and Taylor [6]. They proved that if 𝑋(𝑡) is a planar Brownian motion, 𝛼=𝑛=2, 𝜑−𝑝[𝑋([0,𝑡])] is either zero or infinite for any measure function 𝜑. Hence, the packing measure problem of the symmetric Cauchy process on the line remained open.

The main objective of this paper is to show that for 𝛼=𝑛=1, a similar result to that of planner Brownian motion holds for the packing measure of the range of a one-dimensional symmetric Cauchy process with different criteria on 𝜑.

2. Preliminaries

In this section, we start by recalling the definition and properties of packing measure and packing dimension introduced by Taylor and Tricot [2].

Let Φ be the class of functions: 𝜑∶0,𝛿→0,∞(2.1) which are right continuous and monotone increasing with 𝜑(0₊)=0 and for which there is a finite constant 𝑘>0 with 𝜑2𝑠𝜑𝑠≤𝑘for0<𝑠<𝛿2.(2.2)

The inequality (2.2) is a weak smoothness condition usually called a doubling property. A function 𝜑 in Φ is often called a measure function: 𝜑−𝑃𝐸=limsup_𝜀→0_𝑖𝜑2𝑟_𝑖∶𝐵𝑥_𝑖,𝑟_𝑖aredisjoint,𝑥^𝑖∈𝐸,𝑟_𝑖<𝜀,(2.3) where 𝐵(𝑥_𝑖,𝑟_𝑖) denotes the closure of the open ball 𝐵(𝑥_𝑖,𝑟_𝑖) which is  centered at 𝑥 and has radius 𝑟.

A sequence of closed balls satisfying the condition on the right side of (2.3) is called a 𝜀-packing of 𝐸.

𝜑−𝑃 is a 𝜑-packing premeasure. The 𝜑-packing measure on 𝑅^𝑛, denoted by 𝜑−𝑝, is obtained by defining 𝜑−𝑝𝐸=inf_𝑛𝜑−𝑃𝐸∶𝐸⊆_𝑛𝐸_𝑛.(2.4)It is proved in [2] that 𝜑−𝑝(𝐸) is a metric outer measure, and hence every Borel set in 𝑅^𝑛 is 𝜑−𝑝 measurable.

We can see that for any 𝐸⊂𝑅^𝑛, 𝜑−𝑝𝐸≤𝜑−𝑃𝐸.(2.5)This gives a way to determine the upper bound of 𝜑−𝑝(𝐸). Using the function 𝜑(𝑠)=𝑠^𝛼, 𝛼>0 gives the fractal index dim_𝑝𝐸=inf𝛼>0∶𝑠^𝛼−𝑝𝐸=0=sup𝛼>0∶𝑠^𝛼−𝑝𝐸=∞,(2.6)called the packing dimension of 𝐸.

In order to calculate the packing measure, we will use the following density theorem of Taylor and Tricot [2], which we will call Lemma 2.1.

Lemma 2.1. For a given 𝜑∈Φ, there exists a finite constant 𝑘>0 such that for any Borel measure 𝜇 on 𝑅^𝑛 with 0<‖𝜇‖=𝜇(𝑅^𝑛)<∞ and any Borel set 𝐸⊆𝑅^𝑛, 𝑘⁻¹𝜇𝐸inf_𝑥∈𝐸𝐷^𝜑−𝜇𝑥⁻¹≤𝜑−𝑝𝐸≤𝑘‖‖𝜇‖‖sup_𝑥∈𝐸𝐷^𝜑−𝜇𝑥⁻¹,(2.7)where 𝐷^𝜑−𝜇𝑥=liminf_𝑟↓0𝜇𝐵𝑥,𝑟𝜑2𝑟(2.8) is the lower 𝜑-density of 𝜇 at 𝑥.

One then uses the sample path 𝑋_𝑡 to define the random measure 𝜇𝐸=|||𝑡∈0,𝜏∶𝑋𝑡∈𝐸|||(2.9) known as the occupation measure of the trajectory; |⋅| denotes the Lebesgue measure.

This gives a Borel measure with 𝜇(𝑅)=‖𝑢‖=𝜏, and it is concentrated on 𝑋[0,𝜏] and spreads evenly on it.

If 𝑥=𝑋𝑡₀,0<𝑡⁰<𝜏,(2.10)then 𝜇𝐵𝑥,𝑟=^𝜏0𝐼_𝐵^𝑥,𝑟𝑋𝑡𝑑𝑡=𝑇𝑥,𝑟(2.11) is the sojourn time of 𝑋_𝑡 in the ball 𝐵(𝑥,𝑟) up to the time 𝜏. Define 𝜏=inf{𝑡>0∶|𝑥(𝑡)|>1}; then by a result in [7] about 𝜏 one has 𝐸⁰𝜏=1, where 𝐸⁰ is the associated expectation for the process started at 0. Denote ∫^∞0 by ∫₀₊. If 𝑥=0, one denotes 𝑇(𝑥,𝑟) by 𝑇(𝑟).

In [8], we exhibited a measure function 𝜑 satisfying the following criteria.

Theorem 2.2. Suppose 𝜑=𝑟ℎ(𝑟), where ℎ(𝑟) is a monotone nondecreasing function and 𝑇𝑟=^𝜏0𝐼_𝐵^0,𝑟𝑋𝑡𝑑𝑡;(2.12)then liminf_𝑟↓0𝑇𝑟𝜑𝑟=⎧⎪⎨⎪⎩0if⁰⁺ℎ𝑠∞otherw𝑠islne,1/𝑠=∞,(2.13) where 𝑋_𝑡 is a one-dimensional symmetric Cauchy process.

For any 𝑡₀≥0, 𝑋(𝑡+𝑡₀)−𝑋(𝑡₀) is also a symmetric Cauchy process on the line since the finite-dimensional distribution of 𝑋(𝑡+𝑡₀)−𝑋(𝑡₀) is independent of 𝑡₀; see, for example, [1] for the strong Markov property of Cauchy processes.

The following corollary is then immediate.

Corollary 2.3. Let 𝑋_𝑡, 𝑡≥0, be a one-dimensional symmetric Cauchy process. Then, for any 𝑡₀≥0 with probability one, liminf_𝑟↓0𝑇𝑋𝑡₀,𝑟𝜑𝑟=⎧⎪⎨⎪⎩0if⁰⁺ℎ𝑠∞otherw𝑠islne,1/𝑠=∞,(2.14) where 𝜑 is as defined in Theorem 2.2.

One will also need an estimate for the small ball probability of the sojourn time 𝑇(𝑟), taken from [8, Theorem 3.1].

Lemma 2.4. Suppose 𝜑(𝑟)=𝑟ℎ(𝑟), where ℎ(𝑟) is a monotone increasing function. For 𝑇 defined in (2.12), then for any fixed constant 𝑐₁ and 𝑎_𝑘=𝜌^−𝑘, 𝜌>1, 𝑃𝑇𝑎_𝑘+1<𝑐¹𝜑𝑎_𝑘≤𝑐²ℎ𝑎_𝑘𝑘.(2.15)

In the next section, we will use the above results and some known techniques to calculate the packing measure of the trajectory of the one-dimensional symmetric Cauchy process.

3. The Measure of the Trajectory

In this section, we proceed to the main result.

Theorem 3.1. Let 𝑋(𝑡)={𝑋(𝑡):𝑡≥0} be a one-dimensional symmetric Cauchy process. If 𝜑(𝑟)=𝑟ℎ(𝑟), where ℎ is a nondecreasing function, then with probability one, 𝜑−𝑝𝑋0,𝜏=⎧⎪⎨⎪⎩0if₀₊ℎ𝑠∞otherw𝑠islne,1/𝑠<∞,(3.1) where 𝜑−𝑝(𝑋([0,𝜏])) is the 𝜑-packing measure of 𝑋([0,𝜏]).

Proof. In order to apply the density Lemma 2.1, we have to calculate liminf_𝑟↓0𝜇𝐵𝑥,𝑟𝜑2𝑟.(3.2)But by Corollary 2.3, for each fixed 𝑡₀∈(0,𝜏) with probability one, liminf_𝑟↓0𝜇𝐵𝑋𝑡₀,𝑟𝜑𝑟=liminf^𝑟↓0𝑇𝑋𝑡₀,𝑟𝜑𝑟=0if₀₊ℎ𝑠𝑠ln(1/𝑠)=∞.(3.3)Then a Fubini argument gives |||𝑡∈0,𝜏∶liminf_𝑟↓0𝜇𝐵𝑋𝑡,𝑟𝜑𝑟=0a.s.|||=𝜏<∞(3.4) so that if 𝐸={𝑋(𝑡₀):𝑡₀∈(0,𝜏)}, then 𝐸⊆𝑋([0,𝜏]) and 𝜇(𝐸)=𝜏<∞ a.s. Using an application of the inequality of the density Lemma 2.1, we have 𝜑−𝑝𝐸=∞,(3.5)and thus 𝜑−𝑝𝑋([0,𝑡])=∞ with probability one if ∫₀₊(ℎ(𝑠)/𝑠ln(1/𝑠))=∞.
In order to prove the upper bound, we use density Lemma 2.1 in the other direction, as well as a “bad-point” argument similar to that in [3].
For each point 𝑥∈𝑅, let 𝑉_𝑘(𝑥) denote a semidyadic interval with length 2^−𝑘 whose complement is at distance 2^−𝑘−2 from a dyadic interval of length 2^−𝑘−2 which contains 𝑥.
Let Γ_𝐸=𝑉_𝑘𝑥∶𝑘=1,2,…,𝑥∈𝐸.(3.6)We use the intervals in Γ_𝐸 to replace the balls 𝐵(𝑥,𝑟) in (2.3) with length 2^−𝑘 replacing 2𝑟=diam𝐵(𝑥,𝑟).
This gives a new premeasure 𝜑−𝑃^𝑥𝑥(𝐸) comparable to 𝜑−𝑃 as follows. There exist positive finite constants 𝑘₁, 𝑘₂ such that, for all Borel sets 𝐸⊂𝑅, 𝑘₁𝜑−𝑃^𝑥𝑥𝐸≤𝜑−𝑃𝐸≤𝑘₂𝜑−𝑃^𝑥𝑥𝐸,(3.7) where 𝜑−𝑝^𝑥𝑥𝐸=inf_𝑖𝜑−𝑃^𝑥𝑥𝐸^𝑖∶𝐸⊂∪𝐸^𝑖.(3.8)For ^∞0+ℎ𝑠𝑠ln(1/𝑠)<∞,(3.9)let 𝐺=𝑡₀∈0,𝜏∶liminf𝜇𝐵𝑋𝑡⁰,𝑟𝜑2𝑟=∞(3.10) be the set of “good” points. A Fubini argument tells us that |𝐺|=𝜏<∞ a.s.; then using the density lemma in the other direction, we have 𝜑−𝑃𝑋𝐺=0.(3.11)Let 0,𝜏⧵𝐺=^∞_𝑖=1𝐺_𝑗,(3.12)where 𝐺_𝑗=𝑡∈0,𝜏∶liminf𝜇𝐵𝑋𝑡,𝑟𝜑2𝑟≤𝑗(3.13) is the set of “bad” points.
For 𝑡∈𝐺_𝑗, by monotonicity, we have for a positive constant 𝑐, 𝜇𝐵𝑋𝑡,2^−𝑘≤𝑐𝑗𝜑2^−𝑘,(3.14)for infinitely many 𝑘.
For fixed 𝑗, we can only get a contribution to 𝜑−𝑃^𝑥𝑥(𝑋(𝐺_𝑗)) from semidyadic intervals of length 2^−𝑘 if the dyadic interval of length 2^−𝑘−2 is entered by 𝑋(𝑡) at time 𝑡≤𝜏 and the restarted process leaves the interval of length 2^−𝑘−2 in less than 𝑗𝜑(2^−𝑘).
Thus, if 𝑆_𝑘 is a semidyadic interval of length 2^−𝑘, then 𝑆_𝑘 is bad if 𝑋(𝑠) enters inside dyadic interval of length 2^−𝑘−2 but spends less than 𝑗𝜑(2^−𝑘) in 𝑆_𝑘; otherwise it is “good”. Any 𝑡∈𝐺_𝑗 will be in infinitely many such bad 𝑆_𝑘.
The probability that 𝑆_𝑘 is bad given that it is entered is at most 𝑃𝑇2^−𝑘≤𝑐𝑗𝜑2^−𝑘≤𝑐ℎ2^−𝑘𝑘,(3.15) by Lemma 2.4.
Let 𝑁_𝑘(𝜏) be the number of intervals of length 2^−𝑘 that are entered by the time 𝜏, and let 𝐵_𝑘(𝜏) denote the number of those that are bad; then 𝐸𝐵_𝑘𝜏≤𝐸𝑁_𝑘𝜏ℎ2^−𝑘𝑘.(3.16)Leaving out the nonoverlapping requirement, we have, for a positive constant 𝑐₃, 𝐸𝜑−𝑝^𝑥𝑥(𝑋(𝐺_𝑗))≤𝑐₃∑^∞𝑘=𝑘₀𝐸𝐵_𝑘(𝜏)𝜑(2^−𝑘). Now, by [1, Lemma 4.1], 𝐸𝑁_𝑘(𝑠)≤𝑐₂2^𝑚, for a positive constant 𝑐₂.
Thus, using (3.16), we have 𝐸𝜑−𝑝^𝑥𝑥𝑋𝐺^𝑗≤𝑐^3∞_𝑘_=𝑘0ℎ2^−𝑘²𝑘→0a.s.as𝑘⁰→∞(3.17) since ∑((ℎ(2^−𝑘))²/𝑘)<∞ if ∑(ℎ(2^−𝑘)/𝑘)<∞ for ℎ(2^−𝑘) sufficiently small.
It follows that 𝜑−𝑝^𝑥𝑥𝑋(𝐺_𝑗)=0 a.s., and from (3.8), 𝜑−𝑝^𝑥𝑥𝑋(𝐺_𝑗)=0 a.s. So 𝜑−𝑝^𝑥𝑥𝑋(⋃^∞𝑗=1𝐺_𝑗)≤∑𝜑−𝑝^𝑥𝑥𝑋(𝐺_𝑗)=0.
By (3.12), 𝐺∪⋃^∞𝑗=1𝐺_𝑗=[0,𝜏], and therefore 𝜑−𝑝𝑋[0,𝜏]=0 if ∫₀₊(ℎ(𝑠)/𝑠ln(1/𝑠))𝑑𝑠<∞.
This completes the proof.