# A Class of Logarithmically Completely Monotonic Functions and Their Applications

Authors: Senlin Guo
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We study the recent investigations on a class of functions which are logarithmically completely monotonic. Two open problems are also presented.


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## Body

## 1. Introduction

Recall [1] that a positive function f is said to be logarithmically completely monotonic (LCM) on an open interval I if f has derivatives of all orders on I and for all n∈N∶={1,2,3,…},
(1)(-1)n[ln⁡f(x)](n)≥0.LCM functions are related to completely monotonic (CM) functions [2], strongly logarithmically completely monotonic (SLCM) functions [3], almost strongly completely monotonic (ASCM) functions [3], almost completely monotonic (ACM) functions [4], Laplace transforms, and Stieltjes transforms and have wide applications. It is evident that the set of SLCM functions is a nontrivial subset of the set of LCM functions, which is a nontrivial subset of the set of CM functions, and that the set of CM functions is a nontrivial subset of the set of ACM functions. It was established [3] that the set of SLCM functions is a nontrivial subset of the set of ASCM functions and that the set of SLCM functions on the interval (0,∞) is disjoint with the set of strongly completely monotonic (SCM) functions (see [5] for its definition) on the interval (0,∞).It is well known that the classical Euler gamma function is defined forx>0 by
(2)Γ(z)=∫0∞tx-1e-tdt.
The logarithmic derivative of Γ(z), denoted by
(3)ψ(z)=Γ′(z)Γ(z),
is called psi function, and ψ(k) for k∈N are called polygamma functions.Forα,γ∈R and β≥0, define
(4)fα,β,γ(x)∶=[exΓ(x+β)xx+β-α]γ,x∈(0,∞),
which is encountered in probability and statistics.Sincefα,β,γ(x)  (γ>0) is logarithmically completely monotonic if and only if fα,β,1(x) is logarithmically completely monotonic and fα,β,γ(x)  (γ<0) is logarithmically completely monotonic if and only if fα,β,-1(x) is logarithmically completely monotonic, we only need to study the logarithmically complete monotonicity of the function
(5)fα,β,±1(x)=[exΓ(x+β)xx+β-α]±1,x∈(0,∞).In [6, Theorem 3.2], it was proved that the function f1/2,0,1(x) is decreasing and logarithmically convex from (0,∞) onto (2π,∞) and that the function f1,0,1(x) is increasing and logarithmically concave from (0,∞) onto (1,∞).In [7, Theorem 1], for showing
(6)bb-1aa-1ea-b<Γ(b)Γ(a)<bb-1/2aa-1/2ea-b
for
(7)b>a>1,
monotonic properties of the functions ln⁡fα,0,1(x) and ln⁡fα,0,1(x) on the interval (1,∞) were obtained.In [8, Theorem 2], it was presented that the function fα,0,1(x) is decreasing on the interval (c,∞) for c≥0 if and only if
(8)α≤12
and increasing on the interval (c,∞) if and only if
(9)α≥{c[ln⁡c-ψ(c)]ifc>0,1ifc=0.In [9], after proving the logarithmically completely monotonic property of the functions f1/2,0,1(x) and f1,0,-1(x), in virtue of Jensen’s inequality for convex functions, the upper and lower bounds for the Gurland’s ratio were established: for positive numbers x and y, the inequality
(10)xx-1/2yy-1/2[(x+y)/2]x+y-1≤Γ(x)Γ(y)[Γ((x+y)/2)]2≤xx-1yy-1[(x+y)/2]x+y-2
holds true, where the middle term in (10) is called Gurland’s ratio [10].In [11] the authors proved the following result.Theorem 1 (see [11]).
If(11)2α≤1≤β,
then the function fα,β,1(x) is logarithmically completely monotonic on the interval (0,∞).The necessary and sufficient conditions for the functionsfα,0,1(x) and fα,0,-1(x) to be logarithmically completely monotonic on the interval (0,∞) were also given in [11].Using monotonic properties of the functionsf1/2,0,1(x) and f1,0,-1(x), the inequality (6) was extended (see [11, Remark 1]) from
(12)b>a>1
to
(13)b>a>0.In [12] the authors proved the following results.Theorem 2 (see [12]).
Ifβ>0 and α≤0, then the function fα,β,1(x) is logarithmically completely monotonic on the interval (0,∞).Theorem 3 (see [12]).
Forβ>0, a necessary condition for the function fα,β,1(x) to be logarithmically completely monotonic on the interval (0,∞) is that
(14)α≤min⁡{β,12}.Theorem 4 (see [12]).
Forβ≥1, a necessary and sufficient condition for the function fα,β,1(x) to be logarithmically completely monotonic on the interval (0,∞) is that
(15)α≤12.As direct consequences of the above results, the following Kečkić-Vasić-type inequality is deduced.Theorem 5 (see [12]).
Letx and y be positive numbers with x≠y.(1)
Forβ≥1, the following inequality
(16)I(x,y)>[(xy)α-βΓ(x+β)Γ(y+β)]1/(x-y)holds true if and only ifα≤1/2, where
(17)I(a,b)=1e(bbaa)1/(b-a)(a>0,b>0,a≠b)is the identric or exponential mean.(2)
Forβ>0, the inequality (16) holds true also if α≤0.In [13], the following result was established.Theorem 6 (see [13]).
(
1
)
For β∈[0,1/2), if
(18)α≤β-e-4(1-β)2exp⁡(21-β),
then the function fα,β,1(x) is logarithmically completely monotonic on the interval (0,∞).
(
2
)
For β∈[1/2,1], if
(19)α≤min⁡{3β2-3β+1,12},
then the function fα,β,1(x) is logarithmically completely monotonic on the interval (0,∞).From Theorem6 we can directly obtain the following new result.Corollary 7.
(
1
)
For β∈[1/4,1/2], if
(20)α≤β-14,
then the function fα,β,1(x) is logarithmically completely monotonic on the interval (0,∞).
(
2
)
For β∈(1/2,3/4], if
(21)α≤β-13,
then the function fα,β,1(x) is logarithmically completely monotonic on the interval (0,∞).
(
3
)
For β∈(3/4,1], if
(22)α≤β-12,
then the function fα,β,1(x) is logarithmically completely monotonic on the interval (0,∞).A necessary and sufficient condition is obtained in [13] as follows.Theorem 8 (see [13]).
For(23)β∈{0}∪[12+36,∞),
a necessary and sufficient condition for the function fα,β,1(x) to be logarithmically completely monotonic on the interval (0,∞) is that
(24)α≤12.Regarding the logarithmically complete monotonicity for the functionfα,β,-1(x) and their applications. In [14], the authors proved the following results.Theorem 9 (see [14]).
If the functionfα,β,-1(x) is logarithmically completely monotonic on the interval (0,∞), then either
(25)β>0,α≥max⁡{β,12}
or
(26)β=0,α≥1.Theorem 10 (see [14]).
For(27)β≥12,
the necessary and sufficient condition for the function fα,β,-1(x) to be logarithmically completely monotonic on the interval (0,∞) is that
(28)α≥β.As first application, the following inequalities are derived by using logarithmically completely monotonic properties of the functionfα,β,±1(x) on the interval (0,∞).Theorem 11 (see [14]).
(
1
)
For k∈N, double inequalities
(29)ln⁡x-1x≤ψ(x)≤ln⁡x-12x,(k-1)!xk+k!2xk+1≤(-1)k+1ψ(k)(x)≤(k-1)!xk+k!xk+1
hold true on the interval (0,∞).
(
2
)
When β>0, inequalities
(30)ψ(x+β)≤ln⁡x+βx,(-1)kψ(k-1)(x+β)≥(k-2)!xk-1-β(k-1)!xk
hold true on the interval (0,∞) for k≥2.
(
3
)
When β≥1/2, inequalities
(31)ψ(x+β)≥ln⁡x,(-1)kψ(k-1)(x+β)≤(k-2)!xk-1
hold true on the interval (0,∞) for k≥2.
(
4
)
When β≥1, inequalities
(32)ψ(x+β)≤ln⁡x+β-1/2x,(-1)kψ(k-1)(x+β)≥(k-2)!xk-1-(β-1/2)(k-1)!xk
hold true on the interval (0,∞) for k≥2.As second application, the following inequalities are derived by using logarithmically convex properties of the functionfα,β,±1(x) on (0,∞).Theorem 12 (see [14]).
Letn∈N and
(33)xk>0(1≤k≤n).
Suppose also that
(34)∑k=1n‍pk=1(pk≥0).
If either
(35)β>0,α≤0
or
(36)β≥1,α≤12,
then
(37)∏k=1n‍[Γ(xk+β)]pkΓ(∑k=1n‍pkxk+β)≥∏k=1n‍xkpk(xk+β-α)(∑k=1n‍pkxk)∑k=1n‍pkxk+β-α.
If
(38)α≥β≥12,
then the inequality (37) reverses.As final application, the following inequality can be derived by using the decreasingly monotonic property of the functionfα,β,-1(x) on (0,∞).Theorem 13 (see [14]).
If(39)α≥β≥12,
then
(40)I(x,y)<[(xy)α-βΓ(x+β)Γ(y+β)]1/(x-y)
holds true for x,y∈(0,∞) with x≠y, where I(x,y), defined by (17), is the identric or exponential mean.The following results were shown in [15].Theorem 14 (see [15]).
For(41)β≥0,
a sufficient condition for the function fα,β,-1(x) to be logarithmically completely monotonic on the interval (0,∞) is that
(42)α≥max⁡{12,β,3β2-3β+1}.Remark 15.
From Theorems9 and 14 we see that the necessary and sufficient condition for the function fα,0,-1(x) to be logarithmically completely monotonic on the interval (0,∞) is that
(43)α≥1.
This result is Theorem 2 in [11]. Here we recovered it.Theorem 16 (see [15]).
Let(44)β∈[12-36,12].
Then the necessary and sufficient condition for the function fα,β,-1(x) to be logarithmically completely monotonic on the interval (0,∞) is that
(45)α≥12.The following results are applications of the above theorems.Theorem 17 (see [15]).
When(46)12-36≤β≤12,
the following inequalities
(47)ψ(x+β)≥ln⁡x-1/2-βx,(-1)kψ(k-1)(x+β)≤(k-2)!xk-1+(1/2-β)(k-1)!xkWWWWWWWWWW(k≥2)
hold true on the interval (0,∞).Theorem 18 (see [15]).
Letn∈N and
(48)xk>0(1≤k≤n).
Suppose also that
(49)∑k=1npk‍=1(pk≥0).
If
(50)0≤β≤12,α≥max⁡{12,3β2-3β+1},
then
(51)∏k=1n[Γ(xk+β)]pkΓ(∑k=1npkxk+β)≤∏k=1nxkpk(xk+β-α)(∑k=1npkxk)∑k=1npkxk+β-α.Theorem 19 (see [15]).
If(52)0≤β≤12,α≥max⁡{12,3β2-3β+1},
then
(53)I(x,y)<[(xy)α-βΓ(x+β)Γ(y+β)]1/(x-y)WWWWWiW(x>0;y>0;x≠y),
where in (53) I(x,y), defined by (17), is the identric or exponential mean.


## 2. Open Problems

### 2.1. Open Problem 1

From Theorem8 we have already known, for
(54)β∈{0}∪[12+36,∞),
a necessary and sufficient condition for the function fα,β,1(x) to be logarithmically completely monotonic on the interval (0,∞).For(55)β∈(0,12+36), what is a necessary and sufficient condition for the function fα,β,1(x) to be logarithmically completely monotonic on the interval (0,∞)?Already Known. Theorem 3 gave a necessary condition; Theorem 6 provided a sufficient condition.


### 2.2. Open Problem 2

From Remark15, Theorems 10 and 16 we have already known, for
(56)β∈{0}∪[12-36,∞),
a necessary and sufficient condition for the function fα,β,-1(x) to be logarithmically completely monotonic on the interval (0,∞).For(57)β∈(0,12-36),
what is a necessary and sufficient condition for the function fα,β,-1(x) to be logarithmically completely monotonic on the interval (0,∞)?Already Known. Theorem 9 gave a necessary condition; Theorem 14 provided a sufficient condition.


## 2.1. Open Problem 1

From Theorem8 we have already known, for
(54)β∈{0}∪[12+36,∞),
a necessary and sufficient condition for the function fα,β,1(x) to be logarithmically completely monotonic on the interval (0,∞).For(55)β∈(0,12+36), what is a necessary and sufficient condition for the function fα,β,1(x) to be logarithmically completely monotonic on the interval (0,∞)?Already Known. Theorem 3 gave a necessary condition; Theorem 6 provided a sufficient condition.


## 2.2. Open Problem 2

From Remark15, Theorems 10 and 16 we have already known, for
(56)β∈{0}∪[12-36,∞),
a necessary and sufficient condition for the function fα,β,-1(x) to be logarithmically completely monotonic on the interval (0,∞).For(57)β∈(0,12-36),
what is a necessary and sufficient condition for the function fα,β,-1(x) to be logarithmically completely monotonic on the interval (0,∞)?Already Known. Theorem 9 gave a necessary condition; Theorem 14 provided a sufficient condition.



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