Properties of Bernstein functions of several complex variables

ISSN 0001-4346, Mathematical Notes, 2013, Vol. 93, No. 2, pp. 257–265. © Pleiades Publishing, Ltd., 2013.Original Russian Text © A. R. Mirotin, 2013, published in Matematicheskie Zametki, 2013, Vol. 93, No. 2, pp. 216–226.

Properties of Bernstein Functionsof Several Complex Variables

A. R. Mirotin*

Gomel State UniversityReceived August 20, 2009; in final form, May 25, 2012

Abstract—A multidimensional generalization of the class of Bernstein functions is introduced andthe properties of functions belonging to this class are studied. In particular, a new proof of theintegral representation of Bernstein functions of several variables is given. Examples are considered.

DOI: 10.1134/S0001434613010288

Keywords: Bernstein function of several variables, absolutely monotone function, integralrepresentation, Markov process.

1. INTRODUCTIONBernstein functions of one variable arise in mathematical analysis, potential theory (see [1, pp. 141–

142]), probability theory (see [2, Chap. XIII, Sec. 7], [3]) as well as in operator theory (see, forexample, [4]). The class of Bernstein functions of several variables was introduced by the author in [5] andwas used in [5]–[7] to construct the functional calculus of generators of multiparameter semigroups ofoperators. These functions also arise, in a natural way, in the theory of Markov processes. In this paper,we establish a number of properties of functions of this class, including their integral representation,present new methods for the construction of Bernstein functions of one and two variables, and alsoindicate the relationship of this class with the theory of Markov processes.

It should be noted that, in the literature, we often encounter the class B of (positive) Bernsteinfunctions of one variable whose theory is identical to the theory (introduced below) of the class T1 ofnegative Bernstein functions, because elements of the latter class are obtained from functions of class Bby reflection with respect to the origin.

In what follows, all the measures are assumed to be Radon and the inequalities between vectors arecoordinate-wise. For vectors a, b ∈ C

n, the expression a · b denotes∑n

i=1 aibi.

2. MAIN PROPERTIES OF FUNCTIONS OF CLASS Tn

Definition 1 ([5]). We say that a nonpositive function from C∞((−∞; 0)n) belongs to the class Tn ifall of its partial derivatives of first order are absolutely monotone (a function from C∞((−∞; 0)n) is saidto be absolutely monotone if it is nonnegative together with its partial derivatives of all orders).

The last condition on the function ψ ∈ Tn is equivalent to the inequality ∂αψ ≥ 0 for any multi-indexα �= 0. Functions of class Tn are also called negative Bernstein functions of n variables.

Since the function ψ ∈ Tn is monotone increasing in each variable separately, the inequality x ≤ y,where x, y ∈ (−∞; 0)n, implies ψ(x) ≤ ψ(y). Obviously, Tn is a cone under pointwise addition offunctions and multiplication by a scalar. It is also invariant under translations by vectors from (0;∞)n.Moreover, in view of the monotonicity property noted above, the function ψ(s− c)−ψ(−c) belongs to Tn

together with ψ (c ∈ (0;∞)n).The following theorem was announced in [5] and proved in [7] by using a nontrivial result from [1] (for

the one-dimensional case see, for example, [8, pp. 256–258]). A direct proof of the theorem is presentedbelow. The other statements contained in the present paper are consequences of this theorem (furtherthe expression s → −0 means that s1 → −0, . . . , sn → −0).

*E-mail: [email protected]

257

258 MIROTIN

Theorem 1. Let m and n be natural numbers. For the function ψ : (−∞; 0)n → (−∞; 0], thefollowing assertions are equivalent:

1) ψ ∈ Tn;

2) for s ∈ (−∞; 0)n,

ψ(s) = c0 + c1 · s +ˆ

Rn+\{0}

(es·u − 1) dμ(u), (2.1)

where

c0 = ψ(−0) := lims→−0

ψ(s), c1 = ∇ψ(−∞) ∈ Rn+,

and the positive measure μ on Rn+ \ {0} is uniquely defined and possesses the properties,

indicated in [7, Lemma 3.1];

3) for any function f absolutely monotone on (−∞; 0)m, the composite function (s, r) →f(ψ(s), r) is absolutely monotone on (−∞; 0)m+n−1, s ∈ (−∞; 0)n, r ∈ (−∞; 0)m−1.

Proof. The implication 1 ⇒ 3 is immediately verified by differentiation.

To prove the implication 3 ⇒ 1, it suffices to consider the case f(x) = evx1 , v > 0.

The implication 2 ⇒ 1 is proved by differentiation under the sign of the integral in (2.1).

Let us prove the implication 1 ⇒ 2. By the multidimensional version of the Bernstein–Widdertheorem, the absolutely monotone function ∂ψ/∂si, i = 1, . . . , n, is an n-dimensional Laplace transformof the positive measure σi, i.e., it has the integral representation

∂ψ(s)∂si

=ˆ

Rn+

es·u dσi(u).

Consider the sets Ui = {u ∈ Rn+ | ui > 0} and define the Radon measure μi on Ui by the ruleˆ

Ui

ϕdμi =ˆ

Ui

ϕ(u)1ui

dσi(u),

where ϕ is a continuous function with compact support contained in Ui. Note that the sets Ui constitutean open covering of the space R

n+ \ {0}. In addition, the measures μi are coordinated in the sense that,

for any pair (i, j), their restrictions to Ui ∩ Uj are the same:

μi|(Ui∩Uj) = μj|(Ui∩Uj), i.e.,1ui

dσi(u) =1uj

dσj(u) in Ui ∩ Uj .

Indeed, ui dσj(u) = uj dσi(u), because both sides are representing measures for the mixed derivative∂2ψ/(∂si∂sj) = ∂2ψ/(∂sj∂si). Therefore, (see, for example, [9, Chap. 3, Sec. 2, Proposition 1]), thereexists a positive measure μ on R

n+ \ {0} such that μ|Ui = μi.

Suppose for the time being that ∂ψ(−0)/∂si �= ∞ for all i. Let us show then that the functionu → es·u − 1 is μ-integrable for any s ∈ (−∞; 0)n. For δ > 0, consider the sets Ei = {u ∈ R

n+ | ui ≥ δ}.

Since the following inequality holds for all i:Êi

es·u dμ =ˆ

Ei

es·u 1ui

dσi(u) ≤ 1δ

Êi

es·u dσi(u) ≤ 1δ

∂ψ(s)∂si

,

it follows that the function u → es·u is μ-integrable on the setn⋃

j=1

Ej = Rn+ \ [0; δ)n.

MATHEMATICAL NOTES Vol. 93 No. 2 2013

PROPERTIES OF BERNSTEIN FUNCTIONS OF SEVERAL COMPLEX VARIABLES 259

Setting successively s1 → −0, . . . , sn → −0 in this inequality, we obtainÊj

dμ ≤ δ−1 ∂ψ(−0)∂si

�= ∞ for all i

and, therefore, the function u → es·u − 1 is also μ-integrable on Rn+ \ [0; δ)n. Further, the function

u → (−s) · u is μ-integrable on [0; δ)n \ {0}, because the measures uj dμ(u) = dσj(u) are finite on thisset. Then the relation 1 − es·u ∼ (−s) · u, u → 0, shows that the function u → es·u − 1 is μ-integrableon the set [0; δ)n \ {0} for a sufficiently small δ > 0. Finally, we find that this function is μ-integrable onthe set R

n+ \ {0}.

Therefore, for all i,

∂ψ(s)∂si

=ˆ

Rn+

es·u dσi(u) =∂

∂si

(ˆR

n+\{0}

(es·u − 1) dμ(u) + ci1si

)

,

where ci1 = σi({0}) = ∂ψ(−∞)/∂si, whence we have (2.1).

Finally, if ∂ψ(−0)/∂si = ∞ for some i, then, for ε > 0, we consider the vector ε := (ε, . . . , ε) and thefunction ψε(s) := ψ(s − ε) from Tn for which ∂ψε(−0)/∂si �= ∞ for all i. Since

∂ψε(s)∂si

=ˆ

Rn+

e(s−ε)·u dσi(u) =ˆ

Rn+

es·ue−ε·u dσi(u)

is the Laplace transform of the measure dσεi (u) = e−ε·u dσi(u) and

ψε(0) = ψ(−ε), σεi ({0}) = σi({0}),

it follows that the integral representation (2.1) for ψε is of the form

ψ(s − ε) = ψ(−ε) + c1 · s +ˆ

Rn+\{0}

(es·u − 1)e−ε·u dμ(u).

To complete the proof of formula (2.1), it remains to let ε tend to +0 and invoke a theorem of B. Levi.Finally, the uniqueness of the measure μ is a consequence of the uniqueness theorem for the Laplace

transform, because differentiation in (2.1) under the sign of the integral shows that ∂ψ/∂si is the Laplacetransform of the measure ui dμ(u).

Functions belonging to the class T :=⋃∞

n=1 Tn remain in this class under composition of functions.More precisely, the following corollary, showing that T is an operad in the sense of [10], is valid.

Corollary 1. If ψ1(s) ∈ Tn and ψ2(r) ∈ Tm, then the function ψ(r, ′s) := ψ1(ψ2(r), ′s) belongs toTm+n−1 (here ′s stands for (s2, . . . , sn)). In particular, the class T1 is stable under composition.

Proof. It follows from Theorem 1 that, for any absolutely monotone function f on (−∞; 0), the functionf1(s) := f(ψ1(s)) is absolutely monotone. Similarly, the function f(ψ(r, ′s)) = f1(ψ2(r), ′s) is alsoabsolutely monotone, and again, by Theorem 1, the function ψ(r, ′s) belongs to the class Tm+n−1.

The properties of the representing measure μ are expressed by the following statement.

Lemma 1 ([7]). For the measure μ from the integral representation (2.1) and a sufficiently smallδ > 0, the functions u → uj are μ-integrable on [0; δ)n \ {0}, j = 1, . . . , n, and μ(Rn

+ \ [0; δ)n) < ∞.

Remark 1. Each function ψ ∈ Tn can be extended by the formula

ψ(z) = c0 + c1 · z +ˆ

Rn+\{0}

(ez·u − 1) dμ(u) (2.2)

to a function holomorphic in the domain

{Re z < 0} := {z ∈ Cn : Re zj < 0, j = 1, . . . , n}

and continuous in its closure (the absolute convergence of the integral in (2.2) was proved in [7]). Theclass of the functions arising from such an analytic continuation will be denoted, as before, by Tn.


260 MIROTIN

In what follows, Sθ is the closure of the sector{

ζ ∈ C :π

2+

θ

2< arg ζ <

3π2

− θ

2

}

, θ ∈ [0;π].

Theorem 2. Any function from Tn which is not identically zero maps the domain {Re z < 0} intothe half-plane {Re ζ < 0}, and the cone Sn

θ into the sector Sθ.

Proof. Setting z = s + iy in (2.2), we find that

Reψ(z) = c0 + c1 · s +ˆ

Rn+\{0}

(es·u cos(y · u) − 1) dμ(u). (2.3)

Since each summand on the right-hand side of (2.3) is nonpositive, it follows that Re ψ(z) ≤ 0 forall z with Re z < 0; further, the equality is attained only if each summand on the right-hand side of (2.3)is zero. The case μ �= 0 is of interest. In this case, the equality Re ψ(z) = 0 implies that c0 = c1 = 0and es·u0 cos(y · u0) = 1 for some u0 ∈ R

n+ \ {0}. The last equality is impossible, because s < 0, which

proves the first assertion of the theorem.To prove the second assertion, we take z = s + iy ∈ Sn

θ , z �= 0, i.e.,

sj < 0 and |yj|/(−sj) ≤ cot(

θ

2

)

, j = 1, . . . , n.

Then|eu·s sin(u · y)|

1 − eu·s cos(u · y)=

|sin(u · y)|e−u·s − cos(u · y)

≤ |u · y|(1 − u · s) − 1

≤ cotθ

2.

Now relation (2.3), together with the equality

Im ψ(z) = c1 · y +ˆ

Rn+\{0}

es·u sin(y · u) dμ(u),

which also follows from (2.2), shows that

|Im ψ(z)| ≤ cot(

θ

2

)

(−Reψ(z)),

i.e., ψ(z) ∈ Sθ, which proves the assertion.

Definition 2. We say that a sequence of functions ψk from Tn converges in Tn to a function ψ if ψ isdefined on the set (−∞; 0)n and the functions ψk converge to ψ pointwise on this set.

The following result shows that the convergence for the class Tn described above is natural.

Theorem 3. Let the sequence of ψk converge to the function ψ in Tn. Then

1) ψ ∈ Tn;

2) the sequence of ψk converges to ψ uniformly on compact subsets of the domain {Re z < 0};

3) the sequence of ψk is uniformly bounded on compact subsets of the cone Snθ , θ ∈ (0;π).

Proof. As noted in the proof of Theorem 2, Re ψk(z) ≤ 0 for k = 1, 2, . . . , Re z ≤ 0, i.e., the functionsψk take values outside the half-plane {Re ζ > 0}. Therefore, by the multidimensional version of Montel’stheorem, the family (ψk) is normal, i.e., any one of its subsequences contains a subsequence uniformlyconvergent inside the domain {Re z < 0} to an analytic function or to infinity. However, the last caseis impossible in view of the pointwise convergence of ψk. The limits of all uniformly convergent (inside{Re z < 0}) subsequences of the sequence of ψk coincide with ψ on (−∞; 0)n and, therefore, by one ofthe forms of the uniqueness theorem, they coincide with each other on {Re z < 0} (see, for example, [11,



p. 32]). But, in that case, the sequence of ψk itself also converges uniformly inside {Re z < 0} to thefunction which is the analytic continuation of ψ. This continuation will also be denoted by ψ. Now, bythe Weierstrass theorem, for any multi-index α, as k → ∞, we have the convergence

∂αψk(s) → ∂αψ(s), s ∈ (−∞; 0)n,

and, therefore, ψ ∈ Tn, which proves the first two assertions of the theorem.To prove of the third assertion, we shall need the following statement.

Lemma 2. For any function ψ ∈ Tn with integral representation (2.2), the following inequalityholds:

Reψ(s + iy) − ψ(s) ≥ 2(Re ψ(r + iy) − c0 − c1 · r), (2.4)

where y ∈ Rn and r, s ∈ (−∞; 0)n.

Proof. In view of formula (2.3), we can write

Reψ(s + iy) = c0 + c1 · s +ˆ

Rn+\{0}

(es·u cos(y · u) − 1) dμ(u)

= c0 + c1 · s +ˆ

Rn+\{0}

(es·u − 1) cos(y · u) dμ(u) +ˆ

Rn+\{0}

(cos(y · u) − 1) dμ(u) ≤ 0.

Therefore,

|Re ψ(s + iy)| = |c0| + |c1 · s| +ˆ

Rn+\{0}

(1 − es·u)|cos(y · u)| dμ(u) +ˆ

Rn+\{0}

(1 − cos(y · u)) dμ(u)

≤ −c0 − c1 · s +ˆ

Rn+\{0}

(1 − es·u) dμ(u) +ˆ

Rn+\{0}

(1 − cos(y · u)) dμ(u)

= −ψ(s) +ˆ

Rn+\{0}

(1 − cos(y · u)) dμ(u). (2.5)

Since all the summands on the right-hand side and the integrand in (2.3) are nonpositive, we have

|Re ψ(r + iy)| = −c0 − c1 · r +ˆ

Rn+\{0}

(1 − er·u cos(y · u)) dμ(u). (2.6)

Note that, for t ∈ (0; 1], z ∈ [−1; 1], the inequality 1 − z ≤ 2(1 − tz) holds. Therefore,

1 − cos(y · u) ≤ 2(1 − er·u cos(y · u)),

and, in view of (2.6), we can writeˆR

n+\{0}

(1 − cos(y · u)) dμ(u) ≤ 2ˆ

Rn+\{0}

(1 − er·u cos(y · u)) dμ(u)

= 2(|Re ψ(r + iy)| + c0 + c1 · r).Therefore, inequality (2.5) implies that

|Re ψ(s + iy)| ≤ −ψ(s) + 2(|Re ψ(r + iy)| + c0 + c1 · r),i.e.,

−Reψ(s + iy) ≤ −ψ(s) + 2(−Re ψ(r + iy) + c0 + c1 · r),which concludes the proof of the lemma.

Proof of assertion 3 ). For some a < 0, any bounded set M from Snθ is contained in the set Δn, where

Δ = {ζ ∈ Sθ : Re ζ ≥ a}. For r = (a, . . . , a), by inequality (2.4), we have

|Re ψk(s + iy)| ≤ |ψk(s)| + 2|Re ψk(r + iy)| + ∇ψk(−∞) · r + 2ψk(0)≤ |ψk(s)| + 2|Re ψk(r + iy)| + 2ψk(0). (2.7)


262 MIROTIN

Further, since the sequence ψk(a, . . . , a), is convergent, it follows that it is bounded. Since thefunction s → ψk(s1, . . . , sn) increases in each variable, it follows that, for all s ≥ (a, . . . , a), we havethe inequality

ψk(a, . . . , a) ≤ ψk(s) ≤ 0,

i.e., the sequence of ψk is uniformly bounded on [a; 0]n. It follows from assertion 2) that the sequenceof ψk is uniformly bounded on the compact set K = {z ∈ Δn : Re zj = a, j = 1, . . . , n}. As provedabove, the first and third summands on the right-hand side of (2.7) are bounded if s + iy ∈ Δn, because,in that case, s ∈ [a; 0]n. The second summand is uniformly bounded, because r + iy ∈ K. Nowassertion 3) immediately follows from the inequality

|Im ψ(z)| ≤ cot(

θ

2

)

(−Re ψ(z))

estabished in the proof of Theorem 2.

3. EXAMPLES

The functions

cα − (c − s)α, 0 < α < 1, c ≥ 0, log b − log(b − s), acosh b − acosh(b − s), b ≥ 1,

belong to the class T1 (see, for example, [7] for the cases c = 0, b = 1; the general case follows fromthe remark after Definition 1). Let us present the theorem that allows us to construct new examples offunctions from this class.

Theorem 4. If the function w(−p) is the Laplace transform of a nonnegative nondecreasingpreimage and ψ ∈ T1, ψ′(−∞) = 0, then the function

ϕ(s) =ˆ 0

sψ(t)w(t) dt, s < 0, (3.1)

also belongs to T1 (under the assumption that the integral exists).

Proof. Obviously, ϕ(s) ≤ 0 for s < 0, because ψ(t) ≤ 0 and w(t) ≥ 0 for t < 0. Consider of two cases.

1) Suppose that ψ(s) = eus − 1, where u ≥ 0. The function corresponding to ψ by formula (3.1)is denoted by ϕu. Let us verify that the function ϕ′

u(s) = −ψ(s)w(s) is absolutely monotone. Ifw(−p) = (Lf)(p), where f ≥ 0 is a nondecreasing pre-image (here and elsewhere, L is the Laplacetransform), then

ϕ′u(s) = −eusw(s) + w(s) =

ˆ ∞

0ersf(r) dr −

ˆ ∞

0e(u+r)sf(r) dr =

ˆ ∞

0ersf(r) dr

−ˆ ∞

uersf(r − u) dr =

ˆ u

0ersf(r) dr +

ˆ ∞

uers(f(r) − f(r − u)) dr

=ˆ ∞

0ersg(r) dr,

where g(r) = f(r) − f(r − u) ≥ 0, r ∈ R (in our case, f(s) = 0 for s < 0). Thus, ϕ′u(s) = (Lg)(−s),

s < 0, is an absolutely monotone function.

2) The function ψ ∈ T1 has the integral representation (2.1) in which c1 = 0. Then, by Fubini’stheorem, we have

ϕ(s) = c0

ˆ 0

sw(t) dt +

ˆ ∞

0

(ˆ 0

s(eut − 1)w(t) dt

)

dμ(u)

= c0

ˆ 0

sw(t) dt +

ˆ ∞

0ϕu(s) dμ(u) (3.2)

(Fubini’s theorem is applicable, because the integrand has constant sign in the domain of integration).



The first summand on the right-hand side of (3.2) belongs to T1, because it is nonpositive and thefunction

(

c0

ˆ 0

sw(t) dt

)′= (−c0)w(s)

is absolutely monotone.

Expressing the integral as the limit of integral sums, we see that the second summand in (3.2) is thelimit in T1 of the sequence of functions of the form (αj > 0)

σ(s) =m∑

j=1

ϕuj (s)αj ,

which belong to T1, as was proved above in case 1. By Theorem 3, we now see that the second summandin (3.2) also belongs to T1, which concludes the proof.

Example 1. The polylogarithmic function of order p ∈ N is of the form

Lip(z) =∞∑

n=1

zn

np, |z| < 1,

and can be analytically continued to the half-plane {Re z < 0}; here the following recurrence relation isvalid:

Lip(s) =ˆ 0

sLip−1(t)(−t−1) dt

(see, for example, [12]). As is easy to verify, Li1(s) = − log(1 − s) ∈ T1. In addition, −t−1 = (Lθ)(−t),where θ is the Heaviside function. Therefore, Theorem 4 and the recurrence relation show that Lip ∈ T1,whenever Lip−1 ∈ T1. By the induction principle, this means that Lip ∈ T1 for all p.

Functions from Tn can be constructed by using sums of functions from T1 (of various arguments) aswell as the composition operation. Let us point out another method for constructing functions from T2,which generalizes the first assertion of Theorem 11.1 from [7].

Theorem 5. Let the function ψ1 ∈ T1 have the integral representation

ψ1(s) = c0 +ˆ ∞

0(esu − 1) dμ1(u), with ω :=

ˆ ∞

0u dμ1(u) �= ∞.

Then the function

ψ(s1, s2) :=ψ1(s1) − ψ1(s2)

s1 − s2− ω

belongs to T2 and has the integral representation

ψ(s1, s2) =ˆ ∞

0

ˆ ∞

0(es1u1+s2u2 − 1) dμ(u1, u2),

where dμ(u1, u2) is the image of the measure dμ1(v) dw under the mapping

u1 =v + w

2, u2 =

v − w

2.

By the value of the function ψ for s1 = s2 we mean, as usual, its limit as s2 → s1.


264 MIROTIN

Proof. Let Δ be the angle in the (v,w) plane bounded by the bisectors of the first and fourth quadrants.Consider the double integral

12

¨Δ

(es1(v+w)/2+s2(v−w)/2 − 1) dμ1(v) dw =12

ˆ ∞

0dμ1(v)

ˆ v

−v(es1(v+w)/2+s2(v−w)/2 − 1) dw. (3.3)

In the case s1 �= s2, calculating the inner integral, we evaluate the right-hand side of (3.3) as

1s1 − s2

ˆ ∞

0(es1v − es2v) dμ1(v) − ω = ψ(s1, s2).

In the case s1 = s2, similar calculations yield the following value for the same part:ˆ ∞

0(es1v − 1)v dμ1(v) = ψ′

1(s1) − ω = ψ(s1, s1).

On the other hand, making the change of variables v = u1 + u2, w = u1 − u2, in the integral on theleft-hand side of (3.3), we find that it equalsˆ ∞

0

ˆ ∞

0(es1u1+s2u2 − 1) dμ(u1, u2),

which concludes the proof.

Example 2. As noted above, the function ψ1(s) = log b − log(b − s), b ≥ 1, belongs to the class T1. Itis easy to calculate that, in this case, ω = 1/b. Therefore, the function

1s1 − s2

logb − s2

b − s1− 1

b

belongs to T2.

In conclusion, let us show how functions from the class Tn arise in the theory of Markov processes(for n = 1, this was established by Phillips [13, Theorem 23.15.1]; see also [3, pp. 1339–1340]).Consider a homogeneous (in time and space) Markov process ξ in the phase space R

n+ (see, for

example, [13, Sec. 23.13]). By the Chapman–Kolmogorov equation, the corresponding family oftransition probabilities P (t, x), t > 0, x ∈ R

n+, generates the convolution semigroup of probability

measures Pt, t > 0, on Rn+. Let gt(s) := (LPt)(−s), where L denotes the n-dimensional Laplace

transform and s < 0, is the one-sided characteristic function for the distribution function P (t, · ).Obviously, gt is absolutely monotone. In addition, for any m ∈ N,

gmt(s) =ˆ

Rn+

es·u dPmt(u) =ˆ

Rn+

es·u dP ∗mt (u) = (gt(s))m,

where P ∗mt denotes the mth convolution power of the measure Pt. In the last equality, replacing t

by (k/m)t, we obtain grt(s) = (gt(s))r for any positive rational number r = k/m, k,m ∈ N. If weadditionally assume that the mapping t → Pt is continuous, then, for any a > 0, the relation gat(s) =(gt(s))a holds. Indeed, if the sequence of positive rational numbers rk converges to a, then, in view ofour continuity condition for

gat(s) = limk→∞

grkt(s) = (gt(s))a

(see, for example, [9, p. 529 (Russian transl.), Proposition 14]). In particular, gt(s) = (g1(s))t. Further,as noted above, the totally monotone function g1(−r) = (LP1)(r), satisfies the equality

g1(−r)1/m = g1/m(−r), m ∈ N, r ∈ (0;∞)n,

and hence its representing measure is an infinitely divisible distribution (i.e., g1(−r) is infinitely divisiblein the sense of [14]). Therefore, g1(−r) = eψ(−r), where ψ ∈ Tn (in the one-dimensional case, this iswell known; see, for example, [2, Chap. XIII, Sec. 7]; in the general case, this follows, for example,from Theorem 3 in [14]). By analogy with the one-dimensional case, the function ψ is naturally called



the Laplace exponent of the process ξ. Thus, the Laplace exponent of the Markov process underconsideration belongs to Tn.

It should be noted that the role of the Laplace exponent in the theory of Markov processes is quiteconsiderable. For example, arguing as in [13, Theorem 23.15.2], where the case n = 1 was studied, wecan show that the generator of the Markov semigroup corresponding to ξ has the form ψ(−∇), wherethe value of the Laplace exponent ψ at the set of operators −∇ = (−∂/∂s1, . . . ,−∂/∂sn) is regarded inthe sense of the calculus constructed in [7].

ACKNOWLEDGMENTS

This work was supported by the State Program for Basic Research of the Republic of Belarus (grantno. 20061473).

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Properties of Bernstein functions of several complex variables