Classes of completely monotone and Bernstein functions deﬁned by convexity properties of their spectral measures

: We were interested in Bernstein and L´evy measures having certain convexity-type properties. The convexity-type properties were an extension of the harmonic convexity property considered in [9]. We characterized the corresponding completely monotone and Bernstein functions. We hope this paper can aid with understanding the analogous properties and open questions presented in [8, 9]

Completely monotone (CM) functions find many applications in analysis and probability and an excellent introduction into their properties can be found in the monographs [7,11].A function g : (0, ∞) → [0, ∞) is called Bernstein, if it is C ∞ and satisfies (−1) n−1 g (n) (x) ≥ 0, for all x > 0 and n ∈ N.
We see from the definition that if g is a Bernstein function (BF), then g is CM.These two classes of functions have classic integral representations, which are useful for our developments.
Theorem 1 (Bernstein).A function f is CM, if and only if it can be expressed as a Laplace transform where µ is a Radon measure on [0, ∞), such that the integral converges for all x > 0.
The measure µ in the Bernstein representation will be called the Bernstein measure of f .
The measure ν in this representation is usually called the Lévy measure of the Bernstein function g, and (a, b, ν) is called the Lévy triplet of g.The constants a and b ≥ 0 are called the killing rate and the drift term respectively.
Recently, research has focused on different subclasses of CM and BFs.In [8], the authors investigated CM and BFs with measures that satisfy certain convexity properties.A measure µ on [0, ∞) is called harmonically convex if x → xµ[0, x] is a convex function on (0, ∞).A measure ν on (0, ∞) is said to have harmonically concave tail if x → xν(x, ∞) is a concave function on (0, ∞).Among the main results in [8] are the following: Theorem 3.For any CM function f and a number α ∈ (0, 2/3], there exists a unique harmonically convex measure µ α on [0, ∞), such that Theorem 4. For any Bernstein function g and a number α ∈ (0, 2/3], there exists a unique triplet (a, b, ν α ), such that where a, b ≥ 0 are constants, and ν α is a measure on (0, ∞) with harmonically concave tail.The measure ν α satisfies the integrability condition One of the open problems formulated in [8] was to find the largest possible value of r for which Theorems 3 and 4 hold for all values of α in the interval (0, r].This question was successfully answered in [2].It was shown there, see [2,Theorem 6.3], that Theorems 3 and 4 hold for all α ∈ (0, α * ], where x ≈ 0.717461058844... and α * is the largest value for which Theorems 3 and 4 hold.Theorem 3 suggests that it is natural to consider the set, denoted H CM , of all BFs h, such that the composition f • h is a CM function having a harmonically convex measure for any CM function f .Analogously, Theorem 4 suggests to consider the set, denoted H BF , of all BFs h, such that the composition g • h is a BF with measure that has harmonically concave tail for any BF g.In this way, the results in [2] show that {x α : α ∈ [0, α * ]} ⊂ H CM ∩H BF , so the latter two sets are non-empty.These two sets of functions have surprising properties, see Section 7 in [8]: (1) We have ) is a BF for all t ≥ 0; (3) For any Bernstein function f and g ∈ H BF , one has f • g ∈ H BF ; (4) The set H BF is closed with respect to point-wise convergence.
Apart from these properties, very little is known about the set H BF .Is it a convex set?What are its generators?(A function f ∈ H BF is called a generator for the class H BF if it cannot be represented as a composition g • h for some non-affine BF g and some h ∈ H BF .)A characterization of the BFs, having Lévy measure with harmonically concave tail was proven in [8,Lemma 6.1].It states that a BF g has Lévy measure ν with harmonically concave tail if and only if g(x) − xg (x) is a BF.The feat in [2] was accomplished by relaxing this property and considering the class BF s of all BFs g, such that sg(x) − xg (x) is Bernstein, for some s > 0.Then, the set H BF was extended to BF * s , the later being the set of all BFs g, such that 1 − e −tg(x) ∈ BF s , for all t > 0. See Definition 1.4 in [2], where for technical reasons the killing rate and the drift term are removed from g.In particular, [2,Theorem 6.3], shows that The latter is related to a problem on the unimodality of reciprocal positive stable distributions raised by Simon in [10].
In the current work, we hope to shed more light into these classes of CM and BFs, by relaxing the notion of harmonic convexity, see Definition 1.For a value of a parameter β ∈ [0, 1], we say that a function . Thus, we consider Bernstein measures that are β-convex and Lévy measures with β-concave tail, see Definition 2. The main results may be succinctly summarized as follows and they parallel those in [8,9].
Suppose f is CM with measure µ and define Then, as shown in Table 1, we have the following characterization of β-convexity (β-concavity) of the measure µ: Similarly, suppose g is Bernstein with Lévy triplet (a, b, ν).Define Then, as shown in Table 2, we have the following characterization of β-convexity (β-concavity) of the tail of the measure ν: The paper is organized as followed: Section 2 introduces the background concepts, notions, and some useful preliminary results.In Section 3, we characterize the CM functions having a β-convex (β-concave) measure and the BFs having a Lévy measure with β-convex (β-concave) tail.Section 4 contains several corollaries from the results in Section 3. Finally, the Appendix collects several classical properties of the Lebesgue-Stieltjes integral that are difficult to find in the formulation that we need.

β-convexity and β-concavity
The function h is concave if the opposite inequality holds.If h is twice differentiable in an open interval I, then h is convex on I if and only if its second order derivative is non-negative on I. Convex functions are continuous (in fact locally Lipschitz) on the interior of their domain.The directional derivatives exist (both left and right, in the extended sense) for every x ∈ I.The right directional derivative, denoted h + (x), is right-continuous, while the left directional derivative, denoted h − (x), is left-continuous.When h is convex, then both h + (x) and h − (x) are non-decreasing functions in x, see [6,Theorem 24.1].Moreover, for any x, y in the interior of I we have see [6, Corollary 24.2.1] for details.In addition, if h is convex and y > x, then We consider β ∈ [0, 1] in the following content without further notice.A function h is 0-convex if it is convex; and it is 1-convex precisely when h(1/x) is convex.The latter equivalence follows from Lemma 2.2 in [5], that we state for completeness.
When h(1/x) is convex, we say that h is harmonically convex, since it satisfies the inequality for every x, y > 0. Such functions are also called reciprocally convex in [5].Thus, β-convexity connects the notions of convexity and harmonic/reciprocal convexity.
The following equivalence is an immediate consequence from Lemma 1.
If h : (0, ∞) → R is β-convex, then the directional derivatives of h(x) exist for all x > 0.More precisely, it can be shown that The cumulative distribution function for measure µ on [0, ∞) is denoted by while the tail of measure ν on (0, ∞) is denoted by ν(x) := ν(x, ∞).
Note that ν(x) is non-increasing and a right-continuous function.
The next examples illustrate this concept.

Inverse formula for the Laplace transformation
A CM function f uniquely determines its Bernstein measure µ.Indeed, for all k ∈ N and t > 0, define the operator The following is an inversion formula for the Laplace-Stieltjes integrals, see [11, Chapter VII, Theorems 6a and 7a].
Theorem 5 (Inversion formula).Suppose f is a CM function with measure µ. a) For every t > 0, lim If measure µ has density u(t), then for every t > 0 in the Lebesgue set of u(t), Note that the Lebesgue set of a function contains its points of continuity.If t > 0 is a point of continuity of F µ , the right-hand side of (2.4) becomes F µ (t) − F µ (0).
In order to deal with the higher order derivatives in the inversion formula we need the following identity which can be proved by induction (see also [1,Lemma 2.7.12] for a more general case).For any integer k ≥ 0 and a C k+1 function r on (0, ∞), the following identity holds: (2.5)

Limiting properties for measures with β-convexity type properties
Suppose f is CM function with measure µ.An integration by parts in (1.1) leads to for any x > 0. A direct consequence of the definition of a Bernstein function is µ({0}) = f (∞).With the latter, one can also represent f as where − f (t) is CM.As a non-increasing function, that is integrable at infinity, is o(1/t) as t approaches infinity, we obtain for any k ≥ 1.
Proof.The proof uses the inequality Indeed, by (2.6), we have The fact that µ has no mass at zero implies F µ (0) = lim x→∞ f (x)=0, and (2.8) follows.
for any x > 0.
Proof.As a product of two CM functions, f (x)/x is also CM and so is −( f (x)/x) .By (2.6), we have where in the penultimate equality we used Lemma 5.This shows that the last integral has to be finite.Since µ is β-convex (or β-concave), using (2.1), we obtain Therefore, c has to be zero.The first limit in (2.9) follows.
To see the second limit, for a fixed > 0, use (2.2) to bound Then, using (2.6) one can see that the last expression converges to zero as t approaches infinity.If µ is β-concave, then t β F µ (t) is concave and non-decreasing.Thus, t β F µ (t) + is non-increasing and non-negative.Notice (0,1) ) as t approaches zero by Lemma 8.The first limit in (2.9) follows from here.
For the second limit, note that t 1−β e −xt is a decreasing function for large enough t.Thus, t 1−β t β F µ (t) + e −xt is decreasing and we can apply Lemma 7 to the second integral in (2.10).
(2.13) (For more details about (2.12) and (2.13), refer to (3.3) and (3.6) in [7].)Integration by parts, using the facts that (2.15) Proof.Without loss of generality, we can assume a = b = 0.As g is a Bernstein function, g(x)/x is CM, and so is −(g(x)/x) .By (2.12) and (2.13), we obtain where we used Lemma 6 in the penultimate equality and Lemma 5 in the last equality.Now, applying Lemma 6 again, we conclude that This shows that the last integral has to be finite for all x > 0. Since ν has β-convex (or β-concave) tail, using Corollary 1 and (2.1), we obtain Denote h(t) := (t 1−β ν(1/t)) + .Applying Lemma 6 again, we have (0,∞) Next, observe that for all x > 0, we have (0,1) (2.17) If ν has β-convex tail, then t 1−β ν(1/t) is convex and non-decreasing, hence h is non-decreasing and non-negative.By (2.2), for an > 0, we obtain Analogously to (2.11), one can see that the right-hand side converges to 0 as t approaches zero, showing the first limit in (2.15).Now, h(1/t) is non-increasing and non-negative, and so is t −β h(1/t).As a non-increasing function which is integrable near zero, see (2.17), is o(1/u) as u approaches zero, we have The second limit in (2.15) follows.
If ν has β-concave tail, then t 1−β ν(1/t) is concave and non-decreasing, hence h is non-increasing and non-negative.The function t β−2 e −x/t is also non-increasing for t close to zero.Hence, from (2.16) and Lemma 7, we conclude that t β−2 h(t)e −x/t is o(1/t) as t approaches zero.This shows the first limit in (2.15).
Next, h(1/u) is non-decreasing and non-negative, and so is Therefore, c has to be zero and the second limit in (2.15) follows.
By replacing t with 1/t, the previous lemma can be reformulated as follows.

Implications of the β-convexity properties of the measures µ and ν
In this section, we characterize CM and BFs with β-convexity properties on their measures.In Theorem 6, we consider CM functions with β-convex and β-concave measures.In Theorem 7, BFs whose measures have β-concave tail and β-convex tail are considered.These results extend the characterizations in [8,9].The boundary cases of both Theorems 6 and 7, when β ∈ {0, 1}, are explored in Corollaries 7 to 14. Theorem 6. Suppose f is a CM function with measure µ.Consider the function Proof.Notice F can be rewritten as Thus, without loss of generality, we can assume µ({0}) = 0. a) For sufficiency, suppose F is CM.Anticipating the use of the inversion formula in Theorem 5, define where L k is the operator defined in (2.3).We claim that for every k ≥ 2, the function t → t β G k (t) is convex on the positive reals.Indeed, , and using (2.5), we have , therefore, by Lemma 2, we obtain (0,t] . So, we have Putting everything together and after a trivial calculation, using that (x f (x) ) , we obtain As F is CM, we know that (−1) k−1 F (k−1) (x) ≥ 0, for all x > 0 and k ≥ 2, which implies (t β G k (t)) ≥ 0. This concludes the claim.Let F µ be continuous at x, y > 0 and at the convex combination (1 − α)x + αy, where α ∈ [0, 1].Then, (2.4) shows that lim We will show that F µ is continuous on (0, ∞), thus completing the proof.
Recall that F µ is a right-continuous, non-decreasing function.Let u > 0 be a jump point for F µ , that is, F µ (u−) < F µ (u).Let {y n } be a sequence where F µ is continuous, that decreases and converges to u. (Recall that the points of discontinuity of F µ is countable.)Choose a sequence {x n } of points of continuity of F µ , that converges to u from the left.Synchronized with {x n }, choose a sequence {α n } ⊂ [1/4, 3/4], such that the convex combination (1 − α n )x n + α n y n is to the right of u and is a point of continuity of F µ .By compactness, we may assume that {α n } converges to an α ∈ [1/4, 3/4].So we have lim k→∞ G k (t) = F µ (t) for every t ∈ {x n , y n , (1 − α n )x n + α n y n , n ∈ N}.The convexity of the functions t β G k (t), in the limit gives for each n ∈ N. Letting n approach infinity, the right-continuity of F µ , shows that Using that α 1, and u > 0, gives F µ (u−) ≥ F µ (u), which is a contradiction.Therefore, F µ is continuous on (0, ∞).Now we show necessity.Suppose µ is a β-convex measure, we prove that F is CM.First, by (2.6) and [4, Theorem A.5.2], we have To simplify the notation, denote It is not difficult to see that both are well defined.With that notation, we can rewrite By (2.6) and Lemma 3, using Fubini's theorem, we have and (Note that the above equations also hold if µ is β-concave.)To summarize, we have Therefore, it can be shown that The convexity of t β F µ (t) implies that (t β F µ (t)) + is right-continuous and non-decreasing.Thus, the last integral is the Laplace transform of t 2−β d t β F µ (t) + on (0, ∞).By the Bernstein representation theorem, F(x) is CM.
b) The proof is very much analogous to the proof for part a), so we will only address the differences.For sufficiency, suppose −F is CM.Define G k (t) as (3.2).Without any further assumptions, analogously to (3.3), we have As −F is CM, we know that t β G k (t) is concave.Analogous proof by contradiction applies to verify the continuity of F µ .Therefore, the measure µ is β-concave, as G k (t) converges to F µ (t) for all t > 0 and G k is concave for all k ≥ 2.
For necessity, suppose µ is β-concave, we prove −F is CM.Using the notation A n and B m from part a), we have Thus, we obtain The concavity of t β F µ (t) implies that −(t β F µ (t)) + is right-continuous and non-decreasing.The last integral is the Laplace transform of t 2−β d − t β F µ (t) + on (0, ∞).Hence, −F is a CM function.
Theorem 7. Suppose g is BF with Lévy triplet (a, b, ν).Consider the function a) The measure ν has β-convex tail, if and only if G is CM.b) The measure ν has β-concave tail, if and only if −G is CM.
Proof.Without loss of generality, we can assume a = b = 0.By (2.12) we have where the operator L k is defined in (2.3).We claim that t β G k (t) is convex on (0, ∞) for every k ≥ 1.Notice that by (2.5), As G is CM, we know (−1) k G (k) (x) ≥ 0 for all x > 0 and k ≥ 1, which implies t β G k (t) is convex.This closes the claim.Let x and y belong to the interval (0, ∞) and let the function ν be continuous at these points as well as at the convex combination (1−α)x +αy, where α is in [0, 1].By Theorem 5, as k approaches infinity, the limit of G k equals ν(t) for any t taken from the set {x, y, (1 − α)x + αy}.Furthermore, since t β G k (t) is convex for all k ≥ 1, we have the inequality: To conclude the proof, we need to establish the continuity of ν on (0, ∞).
Recall that ν is non-increasing and right-continuous.Suppose there exists a point u > 0, where the function exhibits a jump, that is ν(u−) > ν(u).Let {y n } be a sequence where ν is continuous, that decreases and converges to u. Choose a sequence {x n } of points of continuity of ν that converges to u from the left.Together with it, choose a sequence {α n } in the interval [1/4, 3/4], such that the convex combination (1 − α n )x n + α n y n is to the left of u and is a point of continuity of ν.By compactness, we may assume that {α n } converges to an α in [1/4, 3/4].Thus, we can conclude that the limit of lim k→∞ G k (t) equals ν for every t in the set {x n , y n , (1 − α n )x n + α n y n , n ∈ N}.Given the convexity of the functions t β G k (t) for every k ≥ 1 and each n ∈ N, we obtain: Letting n approach infinity, the right-continuity of ν implies that Using that α 0 and u > 0, we arrive at ν(u−) ≤ ν(u), which is a contradiction.Therefore, ν is continuous on (0, ∞).Now we show necessity.Suppose measure ν has β-convex tail.As a result, function t β ν(t) is convex and s 1−β ν(1/s) is also convex by Corollary 1.We prove that G is CM.By (3.5) and change of variable Therefore, by [4, Theorem A.5.2], we have To simplify the notation, denote With these notations, we can rewrite By (2.13) and Lemma 4, we have and (Note that the above equations also hold if ν has β-concave tail.)To summarize, Therefore, it can be shown that As b) The proof is very much analogous to the proof for part a), so we will only address the difference.For sufficiency, suppose −G is CM.Define G k as (3.6).Without any further assumption, As −G is CM, we know that t β G k (t) is concave.Analogous proof by contradiction applies to verify the continuity of ν(t).As G k (t) converges to ν(t) for all t > 0, and as G k has β-concave tail for all k ≥ 1, the tail of ν is β-concave.
To show the necessity, suppose that the tail of ν is β-concave, we prove −G is CM.Following the notation C n and D m in part a), we also have Thus, we obtain It defines a Radon measure and −G is CM by definition.

Corollaries
This section contains several corollaries of the main results.where r β (s) = s 2−β (s β F µ (s)) for almost all s > 0.
Proof.We omit the proof since it is almost identical to the one for the next corollary.One just needs to replace ν(s) with F µ (s) and ν(ds) with µ(ds).where r β (s) = s 2−β (s β ν(s)) for almost all s > 0.
Proof.Using Fubini's theorem, several times, function G can be rewritten as Comparing (4.3) with (2.6) we make the following observation: G (resp.−G) is CM precisely when ρ β (resp.−ρ β ) has a non-negative, non-decreasing density function r β .In that case, solving for ν(ds) in (4.4) shows that necessarily ν has a density functionand and without loss of generality we may assume that it is of the form m(s)/s β+1 , s > 0. Substituting it in (4.4), we obtain Using that s β ν(s) is convex (resp.concave), the derivative of r β exists almost everywhere and differentiating (4.5) gives An application of Fubini's theorem finally gives which completing the proof.
Corollary 5.The function G, defined in (3.4), can never be a Bernstein function.
Proof.We use the notation and representations from the proof of Corollary 4. Comparing (4.3) with (2.12) we observe that: G is a BF precisely when ρ β has a non-negative, non-increasing density function r β .Assuming the latter, then (4.7) shows that ν has β-concave tail.Then, Theorem 7, part b) shows that −G is CM, which is a contradiction.
A similar proof shows the next analogous corollary.
Corollary 6.The function F, defined in (3.1), can never be a BF.
Standard facts about BFs imply that if xG(x) is a BF, then G is completely monotone.The next corollaries deal with special cases of the main results.Some of them re-derive several of the results in [9].Corollary 7. Suppose f is CM with measure µ.Then, µ is harmonically convex precisely when f (x) − x f (x) is CM.
Proof.By Theorem 6 part a), with β = 1, µ is harmonically convex precisely when x f (x) is CM.We show this condition is equivalent to f is CM, it suffices to show its nonnegativity.This is trivial, because f (x) ≥ 0 and f (x) ≤ 0 for all x > 0.
Proof.By Theorem 6 part a), applied to the shifted function f (x) − µ({0}) with β = 0, the measure µ is convex precisely when 2 f (x) + x f (x) is CM.We show this condition is equivalent to The first inequality holds because f (x) − µ({0}) ≥ 0. For the second inequality, as 2 f (x) The second inequality follows from here.
Corollary 10.Suppose f is CM with measure µ.Then, µ is concave precisely when f (x) + x f (x) is CM.
Proof.Without loss of generality, we can assume µ has no mass at zero.By Theorem 6 part b), with β = 0, the measure µ is concave, if and only if −2x f (x) − x f (x) is CM.We show this condition is equivalent to f (x) + x f (x) being CM.If f (x) + x f (x) is CM, then is CM.Conversely, if −2x f (x) − x f (x) is CM, to see f (x) + x f (x) is CM, it suffices to show it is non-negative.As its derivative is non-positive, f (x) + x f (x) is non-increasing.By (2.7), we obtain lim x→∞ f (x) + x f (x) = 0.
So f (x) + x f (x) ≥ 0. This completes the proof.
Proof.By Theorem 7 part a), applied to the shifted function g(x) − a − bx with β = 1, the measure ν has harmonically convex tail, if and only if xg (x) is completely monotone.We show this condition is equivalent to g(x) = a + bx.If g(x) = a + bx, then xg (x) = 0, which is CM.Conversely, if xg (x) is CM, then so is x f (x)(1/x) = f (x), that is g (x) ≥ 0. Because g is a Bernstein function, g (x) ≤ 0. Thus, we obtain g (x) = 0, which implies g(x) = a + bx.we obtain that g(x) + xg (x) ≤ 0, and thus its anti-derivative xg(x) is non-increasing.Because g(x) approaches zero as x approaches zero, we know xg(x) ≤ 0. However, g is a Bernstein function, indicating g(x)= 0. This concludes the proof.

e
−xt ν(t) dt.(3.5) a) We show sufficiency first.Suppose G is CM.Anticipating the use of the inversion formula in Theorem 5, define
s) + is non-decreasing.It defines a Radon measure.One can see G is CM by definition.