Decreasing and complete monotonicity of functions defined by derivatives of completely monotonic function involving trigamma function

Hong-Ping Yin; Ling-Xiong Han; Feng Qi

doi:10.1515/dema-2024-0041

Article Open Access

Decreasing and complete monotonicity of functions defined by derivatives of completely monotonic function involving trigamma function

Hong-Ping Yin , Ling-Xiong Han and Feng Qi

Published/Copyright: August 13, 2024

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From the journal Demonstratio Mathematica Volume 57 Issue 1

Abstract

In this study, using convolution theorem of the Laplace transforms, a monotonicity rule for the ratio of two Laplace transforms, Bernstein’s theorem for completely monotonic functions, and other analytic techniques, the authors verify decreasing property of a ratio between three derivatives of a function involving trigamma function and find the necessary and sufficient conditions for a function defined by three derivatives of a function involving trigamma function to be completely monotonic. These results confirm previous guesses posed by Qi and generalize the corresponding known conclusions.

Keywords: decreasing monotonicity; complete monotonicity; completely monotonic function; trigamma function; derivative; ratio; convolution theorem; inequality; monotonicity rule; Laplace transform; Bernstein’s theorem; exponential function guess

MSC 2010: 33B15; 26A48; 26D07; 33B10; 44A10; 44A35

1 Introduction

In the literature [1, Section 6.4], the function

Γ ( z ) = ∫ 0 ∞ t z − 1 e − t d t , ℜ ( z ) > 0

and its logarithmic derivative ψ ( z ) = [ ln Γ ( z ) ] ′ = Γ ′ ( z ) Γ ( z ) are called Euler’s gamma function and digamma function, respectively. Further, the functions ψ ′ ( z ) , ψ ″ ( z ) , ψ ‴ ( z ) , and ψ ( 4 ) ( z ) are known as the trigamma, tetragamma, pentagamma, and hexagamma functions, respectively. All the derivatives ψ ( k ) ( z ) for k ≥ 0 are known as polygamma functions.

Recall from Chapter XIII in [2], Chapter 1 in [3], and Chapter IV in [4] that if a function f ( t ) on an interval I has derivatives of all orders on I and satisfies inequalities ( − 1 ) n f ( n ) ( t ) ≥ 0 for t ∈ I and n ∈ { 0 } ∪ N , then we call f ( t ) a completely monotonic function on I . There have been plenty of literature dedicated to the study and applications of completely monotonic functions, logarithmically completely monotonic functions, and completely monotonic degrees.

Let

Φ ( x ) = x ψ ′ ( x ) − 1 = x ψ ′ ( x ) − 1 x , x ∈ ( 0 , ∞ ) .

Lemma 2 in [5] states that the function ( − 1 ) k Φ ( k ) ( x ) for k ≥ 0 is completely monotonic on ( 0 , ∞ ) . The completely monotonic function Φ ( x ) and its derivatives have been initially investigated by Qi and his coauthors in a series of papers such as [5–7]. This topic was reviewed and surveyed in the article [8].

In [6, Theorem 4.1] and [7, Theorem 4], Qi turned out the following necessary and sufficient conditions and double inequality:

if and only if α ≥ 2 , the function H α ( x ) = Φ ′ ( x ) + α Φ 2 ( x ) is completely monotonic on ( 0 , ∞ ) ;
if and only if α ≤ 1 , the function − H α ( x ) is completely monotonic on ( 0 , ∞ ) ;
the double inequality − 2 < Φ ′ ( x ) Φ 2 ( x ) < − 1 is valid and sharp in the sense that the lower and upper bounds − 2 and − 1 cannot be replaced by any bigger and smaller ones, respectively.

In [6, Theorem 1.1], Qi found the following necessary and sufficient conditions and limits:

if and only if β ≥ 2 , the function H β ( x ) = Φ ′ ( x ) Φ β ( x ) is decreasing on ( 0 , ∞ ) , with the limits
lim x → 0 + H β ( x ) = − 1 , β = 2 ; 0 , β > 2 and lim x → ∞ H β ( x ) = − 2 , β = 2 ; − ∞ , β > 2 ;
if β ≤ 1 , the function H β ( x ) is increasing on ( 0 , ∞ ) , with the limits
H β ( x ) → − ∞ , x → 0 + ; 0 , x → ∞ .

For k ∈ { 0 } ∪ N and λ k , μ k ∈ R , let

J k , λ k ( x ) = Φ ( 2 k + 1 ) ( x ) + λ k [ Φ ( k ) ( x ) ] 2 and J k , μ k ( x ) = Φ ( 2 k + 1 ) ( x ) [ ( − 1 ) k Φ ( k ) ( x ) ] μ k

on ( 0 , ∞ ) . In [5, Theorems 1 and 2], Qi presented the following necessary and sufficient conditions, limits, and double inequality:

if and only if λ k ≥ ( 2 k + 2 ) ! k ! ( k + 1 ) ! , the function J k , λ k ( x ) is completely monotonic on ( 0 , ∞ ) ;
if and only if λ k ≤ 1 2 ( 2 k + 2 ) ! k ! ( k + 1 ) ! , the function − J k , λ k ( x ) is completely monotonic on ( 0 , ∞ ) ;
if and only if μ k ≥ 2 , the function J k , μ k ( x ) is decreasing on ( 0 , ∞ ) , with the limits
lim x → 0 + J k , μ k ( x ) = − 1 2 ( 2 k + 2 ) ! k ! ( k + 1 ) ! , μ k = 2 ; 0 , μ k > 2
and
lim x → ∞ J k , μ k ( x ) = − ( 2 k + 2 ) ! k ! ( k + 1 ) ! , μ k = 2 ; − ∞ , μ k > 2 ;
if μ k ≤ 1 , the function J k , μ k ( x ) is increasing on ( 0 , ∞ ) , with the limits
J k , μ k ( x ) → − ∞ , x → 0 + ; 0 , x → ∞ ;
the double inequality
− ( 2 k + 2 ) ! k ! ( k + 1 ) ! < Φ ( 2 k + 1 ) ( x ) [ Φ ( k ) ( x ) ] 2 < − 1 2 ( 2 k + 2 ) ! k ! ( k + 1 ) !
is valid on ( 0 , ∞ ) and sharp in the sense that the lower and upper bounds cannot be replaced by any larger and smaller numbers, respectively.

For k ≥ m ≥ 0 , let

J k , m ( x ) = Φ ( 2 k + 2 ) ( x ) Φ ( k − m ) ( x ) Φ ( k + m + 1 ) ( x )

on ( 0 , ∞ ) . In [5, Remark 3], Qi guessed that the function J k , m ( x ) for k ≥ m ≥ 0 should be decreasing on ( 0 , ∞ ) and that the double inequality

(1) − 2 ( 2 k + 2 ) ! k ! ( k + 1 ) ! < J k , 0 ( x ) < − ( 2 k + 2 ) ! k ! ( k + 1 ) !

for k ≥ 0 should be valid on ( 0 , ∞ ) and sharp in the sense that the lower and upper bounds cannot be replaced by any larger and smaller numbers, respectively.

For m , n ∈ { 0 } ∪ N and ω m , n ∈ R , let

(2) Y m , n ( x ) = Φ ( m + n + 1 ) ( x ) Φ ( m ) ( x ) Φ ( n ) ( x )

and

(3) Y m , n ; ω m , n ( x ) = Φ ( m + n + 1 ) ( x ) + ω m , n Φ ( m ) ( x ) Φ ( n ) ( x ) .

It is clear that

Y m , n ( x ) = Y n , m ( x ) , Y m , n ; ω m , n ( x ) = Y n , m ; ω n , m ( x ) , Y k − m , k + m + 1 ( x ) = J k , m ( x ) , Y k , k ; ω k , k ( x ) = J k , ω k , k ( x ) , Y 0 , 0 ; ω 0 , 0 ( x ) = H ω 0 , 0 ( x ) .

In this study, we will prove decreasing property of the function Y m , n ( x ) and find necessary and sufficient conditions on ω m , n for ± ( − 1 ) m + n + 1 Y m , n ; ω m , n ( x ) to be completely monotonic on ( 0 , ∞ ) . These results confirm the above guesses and generalize corresponding ones in [5–7] mentioned above.

2 Lemmas

The following lemmas are necessary in this study.

Lemma 1

(Convolution theorem for the Laplace transforms [4, pp. 91–92]) Let f k ( t ) for k = 1 , 2 be piecewise continuous in arbitrary finite intervals included in ( 0 , ∞ ) . If there exist some constants M k > 0 and c k ≥ 0 such that ∣ f k ( t ) ∣ ≤ M k e c k t for k = 1 , 2 , then

∫ 0 ∞ ∫ 0 t f 1 ( u ) f 2 ( t − u ) d u e − s t d t = ∫ 0 ∞ f 1 ( u ) e − s u d u ∫ 0 ∞ f 2 ( v ) e − s v d v .

Lemma 2

[9, Lemma 4] Let the functions A ( t ) and B ( t ) ≠ 0 be defined on ( 0 , ∞ ) such that their Laplace transforms ∫ 0 ∞ A ( t ) e − x t d t and ∫ 0 ∞ B ( t ) e − x t d t exist. If the ratio A ( t ) B ( t ) is increasing, then the ratio ∫ 0 ∞ A ( t ) e − x t d t ∫ 0 ∞ B ( t ) e − x t d t is decreasing on ( 0 , ∞ ) .

Lemma 3

Let x , y ∈ R such that 0 < 2 x < y .

When y > 2 x > 2 2 + 1 ln 2 = 6.885390 … , the function
F ( x , y ) = 2 1 x − 1 y − x + 1 2 2 y − x y − x − 2 x x − 2 y − x − 2 x ( y − x ) x
is positive.
For k , m ∈ N such that 6 ≤ 2 m < k , the sequence F ( m , k ) is positive.

Proof

The function F ( x , y ) can be rearranged as

F ( x , y ) = 2 ( y − 2 x ) + 2 x − 1 [ 2 − y + x + ( x − 2 ) 2 y − 2 x ] x ( y − x ) .

Therefore, it suffices to prove 2 − y + x + ( x − 2 ) 2 y − 2 x > 0 , that is,

(4) 2 y − 2 x > y − x − 2 x − 2 .

Replacing y − 2 x by t in (4) leads to

(5) 2 t > t + x − 2 x − 2 = 1 + t x − 2

for t > 0 and x > 2 . Inequality (5) can be reformulated as x > 2 + t 2 t − 1 . Since the function t 2 t − 1 is decreasing from ( 0 , ∞ ) onto 0 , 1 ln 2 , it is sufficient for x > 2 + 1 ln 2 = 3.442695 … .

Repeating those arguments before Inequality (4) hints us that, for proving F ( m , k ) > 0 , it is sufficient to show

2 k − 2 m > k − m − 2 m − 2 = 1 + k − 2 m m − 2 ,

which can be rewritten as

(6) k − 2 m 2 k − 2 m − 1 < m − 2 .

Since t 2 t − 1 is decreasing in t ∈ ( 0 , ∞ ) and k − 2 m ≥ 1 , the largest value of the left-hand side in inequality (6) is 1 2 1 − 1 = 1 which means that the strict inequality (6) is valid for all m ≥ 4 . As a result, the sequence F ( m , k ) is positive for all m ≥ 4 .

When m = 3 , sequence F ( 3 , k ) is

F ( 3 , k ) = 2 k − 32 k + 128 48 ( k − 3 ) = 2 5 [ 2 k − 5 − ( k − 4 ) ] 48 ( k − 3 ) ,

which is positive for all k > 2 ⋅ 3 = 6 . The proof of Lemma 3 is complete.□

Lemma 4

Let

h ( t ) = e t ( e t − 1 − t ) ( e t − 1 ) 2 , t ≠ 0 ; 1 2 , t = 0

on ( − ∞ , ∞ ) . Then, for any fixed s ∈ ( 0 , 1 ) , the ratio h ( s t ) h s ( t ) is increasing in t from ( 0 , ∞ ) onto 1 2 1 − s , 1 .

Proof

It is easy to see that

lim t → 0 h ( s t ) h s ( t ) = lim t → 0 h ( s t ) lim t → 0 h s ( t ) = 1 2 1 2 s = 1 2 1 − s

and

lim t → ∞ H s ( t ) = lim t → ∞ h ( s t ) lim t → 0 h s ( t ) = 1 1 s = 1 .

Direct differentiating and expanding to power series give

d d t h ( s t ) h s ( t ) = − s e ( 1 + s ) t ( t − 2 ) e ( 1 + 2 s ) t + ( t + 2 ) e 2 s t + ( 2 − s t ) e ( 2 + s ) t + 4 ( s − 1 ) t e ( 1 + s ) t − ( 2 s t 2 + 3 s t + 2 ) e s t − ( s t + 2 ) e 2 t + ( 2 s t 2 + 3 t + 2 ) e t + ( s − 1 ) t ( e t − 1 ) 3 ( e s t − 1 ) 3 h s + 1 ( t ) = s e ( 1 + s ) t ∑ k = 7 ∞ ( 3 k + 2 ) s k + 2 ( 2 s + 1 ) k + k s ( s + 2 ) k − 1 + 2 k 2 s k − 1 + 4 k ( s + 1 ) k − 1 + 2 k ( 1 + 2 k − 2 ) s + 2 k + 1 − [ 2 k + 1 s k + 2 ( s + 2 ) k + 4 k s ( s + 1 ) k − 1 + 2 k ( 1 + 2 k − 2 ) s k − 1 + k ( 2 s + 1 ) k − 1 + 2 k 2 s + 3 k + 2 ] t k k ! ( e t − 1 ) 3 ( e s t − 1 ) 3 h s + 1 ( t ) = s e ( 1 + s ) t ∑ k = 7 ∞ ∑ m = 1 k − 1 k 2 k − m k − 1 m − 1 + 4 k k − 1 m + 2 m + 1 k m s m + 2 k ( 1 − k + 2 k − 2 ) ( s − s k − 1 ) − ∑ m = 1 k − 1 k 2 m k − 1 m + 4 k k − 1 m − 1 + 2 k − m + 1 k m s m t k k ! ( e t − 1 ) 3 ( e s t − 1 ) 3 h s + 1 ( t ) = s e ( 1 + s ) t ∑ k = 7 ∞ ∑ m = 1 k − 1 k 2 k − m k − 1 m − 1 + 4 k k − 1 m + 2 m + 1 k m ( s m − s k − m ) + 2 k ( 1 − k + 2 k − 2 ) ( s − s k − 1 ) t k k ! ( e t − 1 ) 3 ( e s t − 1 ) 3 h s + 1 ( t ) = s e ( 1 + s ) t ∑ k = 7 ∞ ∑ 3 ≤ m < k 2 k 2 k − m k − 1 m − 1 − k 2 m k − 1 k − m − 1 + 4 k k − 1 m − 4 k k − 1 k − m + 2 m + 1 k m − 2 k − m + 1 k k − m ( s m − s k − m ) t k k ! ( e t − 1 ) 3 ( e s t − 1 ) 3 h s + 1 ( t ) = s e ( 1 + s ) t ∑ k = 7 ∞ 2 ∑ 3 ≤ m < k 2 k ! ( m − 1 ) ! ( k − m − 1 ) ! 2 1 m − 1 k − m + 1 2 2 k − m k − m − 2 m m − 2 k − m − 2 m ( k − m ) m ( s m − s k − m ) t k k ! ( e t − 1 ) 3 ( e s t − 1 ) 3 h s + 1 ( t ) = s 4 e ( 1 + s ) t 1 − s 36 t 7 + 1 − s 2 45 t 8 + 22 ( 1 − s 3 ) + 15 ( s − s 2 ) 2160 t 9 + 52 ( 1 − s 4 ) + 63 ( s − s 3 ) 15120 t 10 + 285 ( 1 − s 5 ) + 470 ( s − s 4 ) + 238 ( s 2 − s 3 ) 302400 t 11 + … ( e t − 1 ) 3 ( e s t − 1 ) 3 h s + 1 ( t ) .

Utilizing Lemma 3 reveals that the derivative d d t h ( s t ) h s ( t ) is positive for s ∈ ( 0 , 1 ) and t > 0 . Consequently, for s ∈ ( 0 , 1 ) , the ratio h ( s t ) h s ( t ) is increasing in t > 0 . The proof of Lemma 4 is complete.□

Lemma 5

[5, Lemma 2] For k ≥ 0 , the function ( − 1 ) k Φ ( k ) ( x ) is completely monotonic on ( 0 , ∞ ) , with the limits

(7) ( − 1 ) k x k + 1 Φ ( k ) ( x ) → k ! , x → 0 + ; k ! 2 , x → ∞ .

Lemma 6

(Bernstein’s theorem [4, p. 161, Theorem 12b]) A function f ( x ) is completely monotonic on ( 0 , ∞ ) if and only if

(8) f ( x ) = ∫ 0 ∞ e − x t d σ ( t ) , x ∈ ( 0 , ∞ ) ,

where σ ( s ) is non-decreasing and the integral in (8) converges for x ∈ ( 0 , ∞ ) .

3 Decreasing property

In this section, we prove that the function Y m , n ( x ) defined in (2) is decreasing.

Theorem 1

For m , n ∈ { 0 } ∪ N , the function Y m , n ( x ) defined in (2) is decreasing in x from ( 0 , ∞ ) onto the interval − 2 ( m + n + 1 ) ! m ! n ! , − ( m + n + 1 ) ! m ! n ! . Consequently, for m , n ∈ { 0 } ∪ N , the double inequality

(9) − 2 ( m + n + 1 ) ! m ! n ! < Y m , n ( x ) < − ( m + n + 1 ) ! m ! n !

is valid on ( 0 , ∞ ) and sharp in the sense that the lower and upper bounds cannot be replaced by any larger and smaller numbers, respectively.

Proof

In the proof of [7, Theorem 4], Qi established that

(10) Φ ( x ) = ∫ 0 ∞ h ( t ) e − x t d t .

Then, the ratio Y m , n ( x ) can be rewritten as

Y m , n ( x ) = − ∫ 0 ∞ t m + n + 1 h ( t ) e − x t d t ∫ 0 ∞ t m h ( t ) e − x t d t ∫ 0 ∞ t n h ( t ) e − x t d t = − ∫ 0 ∞ t m + n + 1 h ( t ) e − x t d t ∫ 0 ∞ ∫ 0 t u m ( t − u ) n h ( u ) h ( t − u ) d u e − x t d t ,

where we used Lemma 1. Based on Lemma 2, in order to prove decreasing property of Y m , n ( x ) , it suffices to show that the ratio

(11) Y m , n ( t ) = t m + n + 1 h ( t ) ∫ 0 t u m ( t − u ) n h ( u ) h ( t − u ) d u

is decreasing in t ∈ ( 0 , ∞ ) . By changing the variable u = 1 + v 2 t , the denominator of Y m , n ( t ) becomes

t 2 m + n + 1 ∫ − 1 1 ( 1 + v ) m ( 1 − v ) n h 1 + v 2 t h 1 − v 2 t d v .

Accordingly, we obtain

1 Y m , n ( t ) = ∫ − 1 1 ( 1 + v ) m ( 1 − v ) n h 1 + v 2 t h 1 − v 2 t d v 2 m + n + 1 h ( t )

(12) = 1 2 m + n + 1 ∫ − 1 1 ( 1 + v ) m ( 1 − v ) n h 1 + v 2 t h 1 − v 2 t h ( t ) d v = 1 2 m + n + 1 ∫ − 1 1 ( 1 + v ) m ( 1 − v ) n h ( s t ) h s ( t ) h ( ( 1 − s ) t ) h 1 − s ( t ) d v ,

where s = 1 + v 2 ∈ ( 0 , 1 ) . From Lemma 4, we find that the function h ( s t ) h s ( t ) h ( ( 1 − s ) t ) h 1 − s ( t ) is increasing in t ∈ ( 0 , ∞ ) for any fixed s ∈ ( 0 , 1 ) . Hence, the function Y m , n ( t ) is decreasing on ( 0 , ∞ ) . Therefore, the function Y m , n ( x ) for m , n ∈ { 0 } ∪ N is decreasing on ( 0 , ∞ ) .

Making use of the limits in (7) in Lemma 5 yields

Y m , n ( x ) = − ( − 1 ) m + n + 1 x m + n + 2 Φ ( m + n + 1 ) ( x ) [ ( − 1 ) m x m + 1 Φ ( m ) ( x ) ] [ ( − 1 ) n x k + m + 2 Φ ( n ) ( x ) ] → − ( m + n + 1 ) ! m ! n ! , x → 0 + ; − 2 ( m + n + 1 ) ! m ! n ! , x → ∞ .

The proof of Theorem 1 is complete.□

4 Necessary and sufficient conditions of complete monotonicity

In this section, we discover necessary and sufficient conditions on ω m , n for the function ± ( − 1 ) m + n + 1 Y m , n ; ω m , n ( x ) defined in (3) to be completely monotonic.

Theorem 2

For m , n ∈ { 0 } ∪ N and ω m , n ∈ R ,

if and only if ω m , n ≤ ( m + n + 1 ) ! m ! n ! , the function ( − 1 ) m + n + 1 Y m , n ; ω m , n ( x ) is completely monotonic on ( 0 , ∞ ) ;
if and only if ω m , n ≥ 2 ( m + n + 1 ) ! m ! n ! , the function ( − 1 ) m + n Y m , n ; ω m , n ( x ) is completely monotonic on ( 0 , ∞ ) ;
the double inequality (9) is valid on ( 0 , ∞ ) and sharp in the sense that the lower and upper bounds cannot be replaced by any larger and smaller numbers, respectively.

Proof

As done in the proof of Theorem 1, by virtue of the integral representation (10) and Lemma 1, we acquire

( − 1 ) m + n + 1 Y m , n ; ω m , n ( x ) = ∫ 0 ∞ t m + n + 1 h ( t ) e − x t d t − ω m , n ∫ 0 ∞ t m h ( t ) e − x t d t ∫ 0 ∞ t n h ( t ) e − x t d t = ∫ 0 ∞ t m + n + 1 h ( t ) − ω m , n ∫ 0 t u m ( t − u ) n h ( u ) h ( t − u ) d u e − x t d t = ∫ 0 ∞ 1 − ω m , n Y m , n ( t ) t m + n + 1 h ( t ) e − x t d t ,

where Y m , n ( t ) is defined by (11) and it has been proved in the proof of Theorem 1 to be decreasing on ( 0 , ∞ ) . From Lemma 4, we conclude that the function h ( s t ) h s ( t ) h ( ( 1 − s ) t ) h 1 − s ( t ) is increasing in t from ( 0 , ∞ ) onto 1 2 , 1 . Accordingly, by virtue of (12), we arrive at the sharp inequalities

1 2 m + n + 2 ∫ − 1 1 ( 1 + v ) m ( 1 − v ) n d v < 1 Y m , n ( t ) < 1 2 m + n + 1 ∫ − 1 1 ( 1 + v ) m ( 1 − v ) n d v .

Since

∫ − 1 1 ( 1 + v ) m ( 1 − v ) n d v = ∫ 0 1 [ ( 1 + v ) m ( 1 − v ) n + ( 1 − v ) m ( 1 + v ) n ] d v = 2 m + n + 1 B ( m + 1 , n + 1 ) = 2 m + n + 1 m ! n ! ( m + n + 1 ) ! ,

where we used the formula

∫ 0 1 [ ( 1 + x ) μ − 1 ( 1 − x ) ν − 1 + ( 1 + x ) ν − 1 ( 1 − x ) μ − 1 ] d x = 2 μ + ν − 1 B ( μ , ν ) = 2 μ + ν − 1 Γ ( μ ) Γ ( ν ) Γ ( μ + ν )

for ℜ ( μ ) , ℜ ( ν ) > 0 in [10, p. 321, 3.214], the double inequality

1 2 m ! n ! ( m + n + 1 ) ! < 1 Y m , n ( t ) < m ! n ! ( m + n + 1 ) !

is valid and sharp on ( 0 , ∞ ) . Consequently, by virtue of Lemma 6, if and only if ω m , n ≤ ( m + n + 1 ) ! m ! n ! , the function ( − 1 ) m + n + 1 Y m , n ; ω m , n ( x ) is completely monotonic on ( 0 , ∞ ) ; if and only if ω m , n ≥ 2 ( m + n + 1 ) ! m ! n ! , the function ( − 1 ) m + n Y m , n ; ω m , n ( x ) is completely monotonic on ( 0 , ∞ ) .

The double inequality (9) follows from complete monotonicity of the functions ± ( − 1 ) m + n + 1 Y m , n ; ω m , n ( x ) . The proof of the sharpness of the double inequality (9) is the same as done in the proof of Theorem 1. The proof of Theorem 2 is complete.□

5 Remarks

In this section, we list several remarks related to our main results and their proofs in this study.

Remark 1

Lemma 4 in this study generalizes a conclusion in [6, Lemma 2.3], which reads that the function h ( 2 t ) h 2 ( t ) is decreasing from ( 0 , ∞ ) onto ( 1 , 2 ) .

Remark 2

The function F ( x , y ) discussed in Lemma 3 can be reformulated as

F ( x , y ) = 1 x − 1 y − x 2 − 1 2 2 y − x y − x − 2 x x 1 y − x − 1 x − 2 y − x − 2 x ( y − x ) − x ,

in which the functions

2 y − x y − x − 2 x x 1 y − x − 1 x and 2 y − x − 2 x ( y − x ) − x

can be regarded as special means [11,12].

Let x , y ∈ R such that 0 < x < y 2 . Motivated by Lemma 3, we guess that

when 2 < x < y 2 , the function F ( x , y ) is positive;
when y > 4 and 0 < x < 2 , the function F ( x , y ) is negative.

Furthermore, one can discuss positivity and negativity of the function F ( x , y ) for all x , y satisfying 0 < x < y 2 .

Remark 3

When taking m = k and n = k + 1 , the double inequality (9) in Theorem 1 becomes the double inequality (1) guessed by the corresponding and third author in [5, Remark 3].

Remark 4

For m , n ∈ { 0 } ∪ N , direct differentiation gives

Y ′ ( x ) = Φ ( m + n + 2 ) ( x ) [ Φ ( m ) ( x ) Φ ( n ) ( x ) ] − Φ ( m + n + 1 ) ( x ) [ Φ ( m ) ( x ) Φ ( n ) ( x ) ] ′ [ Φ ( m ) ( x ) Φ ( n ) ( x ) ] 2

on ( 0 , ∞ ) . The decreasing monotonicity of Y m , n ( x ) in Theorem 1 implies that, for m , n ∈ { 0 } ∪ N , the inequality

Φ ( m + n + 1 ) ( x ) [ Φ ( m ) ( x ) Φ ( n ) ( x ) ] ′ > Φ ( m + n + 2 ) ( x ) [ Φ ( m ) ( x ) Φ ( n ) ( x ) ] ,

equivalently,

[ Φ ( m ) ( x ) Φ ( n ) ( x ) ] ′ Φ ( m ) ( x ) Φ ( n ) ( x ) < Φ ( m + n + 2 ) ( x ) Φ ( m + n + 1 ) ( x ) ,

is valid on ( 0 , ∞ ) .

We guess that, for m , n ∈ { 0 } ∪ N , the function

Φ ( m + n + 1 ) ( x ) [ Φ ( m ) ( x ) Φ ( n ) ( x ) ] ′ − Φ ( m + n + 2 ) ( x ) [ Φ ( m ) ( x ) Φ ( n ) ( x ) ]

is completely monotonic in x ∈ ( 0 , ∞ ) .

One can also consider necessary and sufficient conditions on Λ m , n ∈ R for m , n ∈ { 0 } ∪ N such that the function

Φ ( m + n + 1 ) ( x ) [ Φ ( m ) ( x ) Φ ( n ) ( x ) ] ′ − Λ m , n Φ ( m + n + 2 ) ( x ) [ Φ ( m ) ( x ) Φ ( n ) ( x ) ]

and its opposite are, respectively, completely monotonic on ( 0 , ∞ ) .

Remark 5

This work has two electronic preprints at the sites https://hal.archives-ouvertes.fr/hal-02998203v1 and https://doi.org/10.48550/arxiv.2405.19361.

hanlingxiong@outlook.com

# Dedicated to Professor Dr. Mourad E. H. Ismail at University of Central Florida.

Acknowledgements

The authors appreciate anonymous referees for their careful corrections, helpful suggestions, and valuable comments on the original version of this work.

Funding information: This work was partially supported by the National Natural Science Foundation of China (under Grant no. 12361013) and by the Natural Science Foundation of Inner Mongolia (under Grant nos. 2020LH01007 and 2024LHMS01014).
Author contributions: All authors contributed equally to the manuscript and read and approved the final manuscript.
Conflict of interest: The authors declare that they do not have any conflict of competing interests.
Ethical approval: The conducted research is not related to either human or animal use.
Data availability statement: Data sharing is not applicable to this article as no new data were created or analyzed in this study.

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Received: 2023-03-21

Revised: 2024-05-24

Accepted: 2024-06-07

Published Online: 2024-08-13

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/dema-2024-0041

Keywords for this article

decreasing monotonicity; complete monotonicity; completely monotonic function; trigamma function; derivative; ratio; convolution theorem; inequality; monotonicity rule; Laplace transform; Bernstein’s theorem; exponential function guess

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