On a theorem in multi-parameter potential theory

We prove a theorem on additive Levy processes and give applications

By reviewing the whole process of the proof of Theorem 1.1 of [1] given by Khoshnevisan, Xiao and Zhong, our Theorem 1.1 certainly follows if we instead prove the following statement: Let X be any additive Lévy process in IR d .
Clearly, all we have to do is to complete Eq. (5.11) of [1] without bothering ourselves with Condition (1.3) of [1]. Since δ 0 is the only probability measure on F = {0}, letting η → 0, k → ∞, and ε → 0 and using the integrability condition Here, ± is merely a symbol for each possible arrangement of the minus signs; e.g., X 1 − X 2 + X 3 , X 1 − X 2 − X 3 , X 1 + X 2 + X 3 and so on. Let Ψ ± be the Lévy exponent for X ± t . Since −X j has Lévy exponent Ψ j , E Ψ ± (µ) = E Ψ (µ) for all X ± t and where the first summation is taken over the collection of all the X ± t . On the other hand, remains unchanged for all X ± t as long as µ is an N −fold product measure on IR N + . Proposition 10.3 of [1] and Theorem 2.1 of [1] together state that for any additive Lévy process X, where λ r is the restriction of the Lebesgue measure λ N in IR N to [0, r] N and k 1 , k 2 ∈ (0, ∞) are two constants depending only on r, N, d, π. Note that λ r is an N −fold product measure on IR N + . Thus, there exists a constant c 2 ∈ (0, ∞) depending only on N and r such that for all X ± t . Since |1 + z| = |1 +z| where z is a complex number, IR d N j=1 |1 + Ψ ± j (ξ)| −1 dξ < +∞ as well. Therefore, by (1.4), (1.3) follows, so does the theorem. 2

The Range of An Additive Lévy Process
As the first application, we use Theorem 1.1 to compute dim H X(IR N + ). Here, dim H denotes the Hausdorff dimension. To begin, we introduce the standard d-parameter additive α-stable Lévy process in IR d for α ∈ (0, 1) : that is, the S j are independent standard α-stable Lévy processes in IR d with the common Lévy exponent |ξ| α .
Theorem 2.1 Let X be any N -parameter additive Lévy process in IR d with Lévy exponent (Ψ 1 , · · · , Ψ N ). Then Proof Let C β denote the Riesz capacity. By Theorem 7.2 of [1], for all β ∈ (0, d) and Note that S 1−β/d + X is a (d + N, d)−additive Lévy process. Thus, by Theorem 1.1 and the fact that β < d and Re 1 1+Ψ j (ξ) ∈ (0, 1], we have for all β ∈ (0, d), Thanks to the Frostman theorem, it remains to show that C β (X(IR N + )) > 0 is a trivial event. Let E β denote the Riesz energy. By Plancherel's theorem, given any β ∈ (0, d), there is a constant holds for all probability measures ν in IR d . Consider the 1-killing occupation measure Clearly, O is a probability measure supported on X(IR N + ). It is easy to verify that It follows from (2.4) that

The Set of k-Multiple Points
First, we mention a q-potential density criterion: Let X be an additive Lévy process and assume that X has an a.e. positive q-potential density on IR d for some q ≥ 0. Then for all Borel sets F ⊂ IR d , The argument is elementary but crucially hinges on the property: can be replaced by X for all b ∈ IR N + ; moreover, the second condition "a.e. positive on IR d " is absolutely necessary for the direction ⇐= in (2.5); see for example Proposition 6.2 of [1].
Let X 1 , · · · , X k be k independent Lévy process in IR d . Define Z is a k-parameter additive Lévy process taking values in IR d(k−1) .
Assume that Z has an a.e. positive q-potential density for some q ≥ 0. [A special case is that if for each j = 1, · · · , k, X j has a one-potential density u 1 j > 0, λ d -a.e., then Z has an a.e. positive 1-potential density on IR d(k−1) .] Then Proof For any IR d -valued random variable X and ξ 1 , In particular, the Lévy process (X j , −X j ) has Lévy exponent Ψ j (ξ 1 − ξ 2 ). It follows that the corresponding integral in (1.1) for Z equals Since Z has an a.e. positive q-potential density, by (2.5) (2.6) now follows from Theorem 1.1. 2 For each β ∈ (0, d) and S 1−β/d independent of X 1 , · · · , X k , define Z S,β is a k + d parameter additive Lévy process taking values in IR dk .
Assume that for each β ∈ (0, d), Z S,β has an a.e. positive q-potential density on IR dk for some q ≥ 0. (q might depend on β.) [A special case is that if for each j = 1, · · · , k, X j has a onepotential density u 1 j > 0, λ d -a.e., then Z S,β has an a.e. positive 1-potential density on IR dk for all (2.7) Proof According to the argument, Eq. (4.96)-(4.102), in Proof of Theorem 3.2. of Khoshnevisan, Shieh, and Xiao [2], it suffices to show that for all β ∈ (0, d) and S 1−β/d independent of X 1 , · · · , X k , Similarly, the corresponding integral in (1.1) for Z S,β equals with ξ k = 0. Since Z S,β has an a.e. positive q-potential density, by (2.5) and Theorem 1.1 Finally, use the cyclic transformation: Let X be a Lévy process in IR d . Fix any path X t (ω). A point x ω ∈ IR d is said to be a k-multiple point of X(ω) if there exist k distinct times t 1 , t 2 , · · · , t k such that X t 1 (ω) = X t 2 (ω) = · · · = X t k (ω) = x ω . Denote by E ω k the set of k-multiple points of X(ω). It is well known that E k can be identified with k j=1 X j ((0, ∞)) where the X j are i.i.d. copies of X. Thus, Theorem 2.2 and Theorem 2.3 imply the next theorem.
Theorem 2.4 Let (X, Ψ) be any Lévy process in IR d . Assume that X has a one-potential density u 1 > 0, λ d -a.e. Let E k be the k-multiple-point set of X. Then (2.10)

Intersection of Two Independent Subordinators
Let X t , t ≥ 0 be a process with X 0 = 0, taking values in IR + . First, we ask this question: What is a condition on X such that for all sets F ⊂ (0, ∞), For subordinators, still the existence and positivity of a q-potential density (q ≥ 0) is the only known useful condition to this question.
Let σ be a subordinator. Take an independent copy σ − of −σ. We then define a processσ on IR byσ s = σ s for s ≥ 0 andσ s = σ − −s for s < 0. Note thatσ is a process of the property: σ t+b −σ b , t ≥ 0 (independent ofσ b ) can be replaced by σ for all b ∈ IR.
Let X t , t ≥ 0 be any process in IR d . Then the q-potnetial density is nothing but the density of the expected q-occupation measure with respected to the Lebesgue measure. (When q = 0, assume that the expected 0-occupation measure is finite on the balls.) Since the reference measure is Lebesgue, one can easily deduce that if u is a q-potential density of X, then u(−x) is a q-potential density of −X. Consequently, if we define X s = X s for s ≥ 0 and X s = X − −s for s < 0 where X − is an independent copy of −X, then u(x) + u(−x) is a q-potential density of X. Conversely, if X has a q-potential density, then it has to be the form u(x) + u(−x), where u is a q-potential density of X. If σ is a subordinator, after a little thought we can conclude thatσ has an a.e. positive q-potential density on IR if and only if σ has an a.e. positive q-potential density on IR + . Lemma 2.5 If a subordinator σ has an a.e. positive q-potential density for some q ≥ 0 on IR + , then for all Borel sets F ⊂ (0, ∞), Since σ((0, ∞)) σ((−∞, 0)) = ∅ or at most {0}, by looking at the law of σ − , it is clear that Assume that E{λ 1 (F −σ((0, ∞)))} > 0. Since F ⊂ F * , E{λ 1 (F * −σ((0, ∞)))} > 0. From the above discussion,σ has an a.e. positive q-potential density. Moreover,σ is a process of the property: σ t+b −σ b , t ≥ 0 (independent ofσ b ) can be replaced by σ for all b ∈ IR. It follows from the standard q-potential density argument that P (F * σ(IR\{0}) = ∅) > 0. The direction =⇒ in (2.11) is elementary since σ has a q-potential density. 2 Theorem 2.6 Let σ 1 and σ 2 be two independent subordinators having the Lévy exponents Ψ 1 and Ψ 2 , respectively. Assume that σ 1 has an a.e. positive q-potential density for some q ≥ 0 on IR + . Then Note that our result does not require any continuity condition on the q-potential density.
Proof By Lemma 2.5 and Theorem 1.1, The proof is therefore completed. 2

A Fourier Integral Problem
This part of content can be found in Section 6 of [1]. It is an independent Fourier integral problem. Neither computing the Hausdorff dimension nor proving the existence of 1-potential density needs the discussion below. [But this Fourier integral problem might be of novelty to those who want to replace the Lévy exponent by the 1-potential density.] Let X be an additive Lévy process. Here is the question. Suppose that K : IR d → [0, ∞] is a symmetric function with K(x) < ∞ for x = 0 that satisfies K ∈ L 1 and K(ξ) = k 1 hold for all probability measures µ in IR d ? Here, k 1 , k 2 ∈ (0, ∞) are two constants. Consider the function K in the following example. Define X j t j = −Y j −t j for t j < 0 and X j t j = X j t j for t j ≥ 0, where Y j is an independent copy of X j and the Y j are independent of each other and of X as well. Then X t = X 1 t 1 + X 2 t 2 + · · · + X N t N , t ∈ IR N is a random field on IR N . Assume that X has a 1-potential density K. So, K ∈ L 1 and a direct check verifies that K is symmetric. By the definition of K, K(ξ) = IR N e − N j=1 |t j | Ee iξ· Xt dt. Evaluating this integral quadrant by quadrant and using the identity for Re(z j ) ≥ 0 (where is taken over the 2 N permutations of conjugate) yield K(ξ) = k 1 N j=1 Re 1 1+Ψ j (ξ) > 0. If K ∈ L 1 (even though this case is less interesting), on one hand by Fubini, |μ(ξ)| 2 K(ξ)dξ = e −iξ·(x−y) K(ξ)dξµ(dx)µ(dy) and on the other hand by inversion (assuming the inversion holds everywhere by modification on a null set), K(x − y)µ(dx)µ(dy) = (2π) −d e −iξ·(x−y) K(ξ)dξµ(dx)µ(dy).
Thus, (2.13) holds automatically in this case. If K is continuous at 0 and K(0) < ∞, then K ∈ L 1 . This is a standard fact. Since K ∈ L 1 and K > 0, a bottom line condition needed to prove (2.13) is that K is continuous at 0 on [0, ∞]. This paper makes no attempt to solve the general case K(0) = ∞.
Remark Lemma 6.1 of [1] is not valid. (The authors of [1] looked like not having a clear idea how to prove a result of that sort.) The assumption that Re N j=1 1 1+Ψ j (ξ) > 0 cannot (by any means) justify either equation in (6.4) of [1]. Fortunately, Lemma 6.1 played no role in [1], because Theorem 7.2 of [1] is an immediate consequence of the well-known identity (2.4) of the present paper and Theorem 1.5 of [1]. Nevertheless [1] indeed showed that the 1-potential density of an isotropic stable additive process is comparable to the Riesz kernel at 0, and therefore the 1-potential density is continuous at 0 on [0, ∞].