Abstract

We prove some monotonicity properties of functions involving gamma and q-gamma functions.

1. Introduction

The π‘ž-gamma function is defined for positive real numbers π‘₯ and π‘žβ‰ 1 byΞ“π‘ž(π‘₯)=(1βˆ’π‘ž)∞1βˆ’π‘₯𝑛=01βˆ’π‘žπ‘›+11βˆ’π‘žπ‘›+π‘₯Ξ“,0<π‘ž<1,π‘ž(π‘₯)=(π‘žβˆ’1)1βˆ’π‘₯π‘žβˆž(1/2)π‘₯(π‘₯βˆ’1)𝑛=01βˆ’π‘žβˆ’(𝑛+1)1βˆ’π‘žβˆ’(𝑛+π‘₯),π‘ž>1.(1.1) We note here according to [1] that the limit of Ξ“π‘ž(π‘₯) as π‘žβ†’1βˆ’ gives back the well-known Euler's gamma function: limπ‘žβ†’1βˆ’Ξ“π‘žξ€œ(π‘₯)=Ξ“(π‘₯)=∞0𝑑π‘₯π‘’βˆ’π‘‘π‘‘π‘‘π‘‘.(1.2) Note also that from the definition we have for positive π‘₯ and 0<π‘ž<1, Ξ“1/π‘ž(π‘₯)=π‘ž(π‘₯βˆ’1)(1βˆ’π‘₯/2)Ξ“π‘ž(π‘₯),(1.3) we see that limπ‘žβ†’1Ξ“π‘ž(π‘₯)=Ξ“(π‘₯). For historical remarks on gamma and π‘ž-gamma functions, we refer the reader to [1–3].

There exists an extensive and rich literature on inequalities for the gamma and π‘ž-gamma functions. For some developments in this area, we refer the reader to the articles in [2–7] and the references therein. Many of these inequalities follow from the monotonicity properties of functions which are closely related to Ξ“ (resp., Ξ“π‘ž) and its logarithmic derivative πœ“ (resp. πœ“π‘ž). Here we recall that a function 𝑓(π‘₯) is said to be completely monotonic on (π‘Ž,𝑏) if it has derivatives of all orders and (βˆ’1)π‘˜π‘“(π‘˜)(π‘₯)β‰₯0, π‘₯∈(π‘Ž,𝑏), π‘˜β‰₯0 and 𝑓(π‘₯) is said to be strictly completely monotonic on (π‘Ž,𝑏) if (βˆ’1)π‘˜π‘“(π‘˜)(π‘₯)>0, π‘₯∈(π‘Ž,𝑏), π‘˜β‰₯0. Lemma 2.3 below asserts that 𝑓(π‘₯)=π‘’βˆ’β„Ž(π‘₯) is completely monotonic on an interval if β„Žβ€² is. Following [8], we call such functions 𝑓(π‘₯) logarithmically completely monotonic. For similar notions of logarithmically completely monotonic functions and the history of the notions, we refer the reader to the articles in [9–12].

We note here that limπ‘žβ†’1πœ“π‘ž(π‘₯)=πœ“(π‘₯) (see [13]) and that πœ“β€² and πœ“ξ…žπ‘ž are completely monotonic functions on (0,+∞) (see [5, 14]). Thus, one expects to deduce results on gamma and π‘ž-gamma functions from properties of (logarithmically) completely monotonic functions, by applying them to functions related to πœ“β€² or πœ“ξ…žπ‘ž. It is our goal in this paper to obtain some results on gamma and π‘ž-gamma functions via this approach. As an example, we recall the following result of Bustoz and Ismail (the case π‘ž=1) as well as Ismail and Muldoon.

Theorem 1.1 (see [15, Theorem  3], [4, Theorem  2.5]). Let π‘Ž+1β‰₯𝑏>π‘Ž, 𝛼=max(βˆ’π‘Ž,βˆ’π‘) and for π‘ž>0 and π‘žβ‰ 1, define π‘”π‘žξ‚΅(π‘₯;π‘Ž,𝑏,𝑐)=1βˆ’π‘žπ‘₯+𝑐1βˆ’π‘žπ‘Žβˆ’π‘Ξ“π‘ž(π‘₯+𝑏)Ξ“π‘ž(π‘₯+π‘Ž),π‘₯>𝛼.(1.4) For π‘ž=1, let 𝑔1(π‘₯;π‘Ž,𝑏,𝑐)=limπ‘žβ†’1π‘”π‘ž(π‘₯;π‘Ž,𝑏,𝑐). Then π‘”π‘ž(π‘₯;π‘Ž,𝑏,𝑐) is logarithmically completely monotonic on (𝛼,+∞) if 𝑐≀(π‘Ž+π‘βˆ’1)/2 and 1/π‘”π‘ž(π‘₯;π‘Ž,𝑏,𝑐) is logarithmically completely monotonic on (𝛼,+∞) if 𝑐β‰₯π‘Ž.

It follows immediately from the above theorem that for 0<π‘ž<1 and 0<𝑠<1, one has [4] for π‘₯>0: ξ‚΅1βˆ’π‘žπ‘₯+𝑠/2ξ‚Ά1βˆ’π‘ž1βˆ’π‘ <Ξ“π‘ž(π‘₯+1)Ξ“π‘ž<ξ‚΅(π‘₯+𝑠)1βˆ’π‘žπ‘₯+𝑠1βˆ’π‘ž1βˆ’π‘ .(1.5) Alzer later [3] determined the best values 𝑒(π‘ž,𝑠),𝑣(π‘ž,𝑠) such that the inequalitiesξ‚΅1βˆ’π‘žπ‘₯+𝑒(π‘ž,𝑠)ξ‚Ά1βˆ’π‘ž1βˆ’π‘ <Ξ“π‘ž(π‘₯+1)Ξ“π‘ž<ξ‚΅(π‘₯+𝑠)1βˆ’π‘žπ‘₯+𝑣(π‘ž,𝑠)ξ‚Ά1βˆ’π‘ž1βˆ’π‘ (1.6) hold for π‘ž>0,0<𝑠<1,π‘₯>0 to be⎧βŽͺ⎨βŽͺβŽ©π‘’(π‘ž,𝑠)=ln((π‘žπ‘ βˆ’π‘ž)/(1βˆ’π‘ )(1βˆ’π‘ž))𝑠lnπ‘ž,0<π‘ž<1,2ξ‚€,π‘ž>1,𝑣(π‘ž,𝑠)=ln1βˆ’(1βˆ’π‘ž)Ξ“π‘ž1/(π‘ βˆ’1)(𝑠).lnπ‘ž(1.7)

Motivated by the above results, we will show in Section 3 that the function π‘”π‘ž(π‘₯;𝑠,1,𝑒(π‘ž,𝑠)) as defined in Theorem 1.1 is logarithmically completely monotonic on (0,+∞) for 0<π‘ž<1. This will enable us to deduce the left-hand side inequality of Alzer's result above for 0<π‘ž<1. The derivatives of πœ“(π‘₯) are known as polygamma functions and in Section 4 we will prove some inequalities involving the polygamma functions.

2. Lemmas

The following lemma lists some facts about πœ“(π‘₯) and πœ“π‘ž(π‘₯). These can be found, for example, in [3, equation (2.7)] and [5, equations  (1.2)–(1.5)].

Lemma 2.1. For π‘₯>0, πœ“π‘ž(π‘₯)=βˆ’ln(1βˆ’π‘ž)+lnπ‘žβˆžξ“π‘›=1π‘žπ‘›π‘₯1βˆ’π‘žπ‘›,0<π‘ž<1,(2.1)(βˆ’1)𝑛+1πœ“(𝑛)ξ€œ(π‘₯)=∞0π‘’βˆ’π‘₯𝑑𝑑𝑛1βˆ’π‘’βˆ’π‘‘π‘‘π‘‘=𝑛!βˆžξ“π‘˜=01(π‘₯+π‘˜)𝑛+1,𝑛β‰₯1,(2.2)πœ“(𝑛)(π‘₯+1)=πœ“(𝑛)(π‘₯)+(βˆ’1)𝑛𝑛!π‘₯𝑛+1,𝑛β‰₯0,(2.3)1πœ“(π‘₯)=lnπ‘₯βˆ’βˆ’12π‘₯12π‘₯2ξ‚€1+𝑂π‘₯3,π‘₯⟢+∞,(2.4)(βˆ’1)𝑛+1πœ“(𝑛)(π‘₯)=(π‘›βˆ’1)!π‘₯𝑛+𝑛!2π‘₯𝑛+1ξ‚€1+𝑂π‘₯𝑛+2,𝑛β‰₯1,π‘₯⟢+∞.(2.5)

The set of all the completely monotonic functions on an interval is equipped with a ring structure with the usual addition and multiplication of functions. The next two simple lemmas can also be used to construct new (logarithmically) completely monotonic functions; part of Lemma 2.2 is contained in [4, Lemma  1.3].

Lemma 2.2. If 𝑓(π‘₯) is completely monotonic on some interval (π‘Ž,𝑏), then so is 𝑓(π‘₯)βˆ’π‘“(π‘₯+𝑐) on (π‘Ž,𝑏)∩(π‘Žβˆ’π‘,π‘βˆ’π‘) for any 𝑐>0. Consequently, if 𝑓(π‘₯) is logarithmically completely monotonic on some interval (π‘Ž,𝑏), then so is 𝑓(π‘₯)/𝑓(π‘₯+𝑐) on (π‘Ž,𝑏)∩(π‘Žβˆ’π‘,π‘βˆ’π‘) for any 𝑐>0.

Lemma 2.3 (see [15, Lemma  2.1]). If 𝑓′(π‘₯) is completely monotonic on an interval, then exp(βˆ’π‘“(π‘₯)) is also completely monotonic on the same interval.

Lemma 2.4. Let π‘Žπ‘– and 𝑏𝑖(𝑖=1,…,𝑛) be real numbers such that 0<π‘Ž1β‰€β‹―β‰€π‘Žπ‘›, 0<𝑏1≀⋯≀𝑏𝑛, and βˆ‘π‘˜π‘–=1π‘Žπ‘–β‰€βˆ‘π‘˜π‘–=1𝑏𝑖 for π‘˜=1,…,𝑛. If the function 𝑓(π‘₯) is decreasing and convex on (0,+∞), then 𝑛𝑖=1𝑓𝑏𝑖≀𝑛𝑖=1π‘“ξ€·π‘Žπ‘–ξ€Έ.(2.6) If βˆ‘π‘›π‘–=1π‘Žπ‘–=βˆ‘π‘›i=1𝑏𝑖, then one only needs 𝑓(π‘₯) to be convex for the above inequality to hold.

The above lemma, except the last part, is a special case of Lemma  2 in [2]. This lemma follows from the theory of majorization, for example, see the discussions in [16].

Lemma 2.5 (Hadamard's inequality). Let 𝑓(π‘₯) be a convex function on [π‘Ž,𝑏], then π‘“ξ‚€π‘Ž+𝑏2≀1ξ€œπ‘βˆ’π‘Žπ‘π‘Žπ‘“(π‘₯)𝑑π‘₯≀𝑓(π‘Ž)+𝑓(𝑏)2.(2.7)

Lemma 2.6. For real numbers 0<𝑠,π‘ž<1 and any integer 𝑛β‰₯1, ξ‚΅π‘žπ‘ βˆ’π‘žξ‚Ά(1βˆ’π‘ )(1βˆ’π‘ž)𝑛β‰₯π‘žπ‘›π‘ βˆ’π‘žπ‘›(1βˆ’π‘ )(1βˆ’π‘žπ‘›).(2.8)

Proof. We can recast the above inequality as 𝐸(βˆ’1,π‘ βˆ’1;π‘ž,1)β‰₯𝐸(βˆ’π‘›,𝑛(π‘ βˆ’1);π‘ž,1),(2.9) where for positive numbers π‘₯,𝑦 and real numbers π‘Ÿ,𝑠 with π‘Ÿπ‘ (π‘Ÿβˆ’π‘ )(π‘₯βˆ’π‘¦)β‰ 0, we define ξ‚΅π‘ŸπΈ(π‘Ÿ,𝑠;π‘₯,𝑦)=π‘ β‹…π‘¦π‘ βˆ’π‘₯π‘ π‘¦π‘Ÿβˆ’π‘₯π‘Ÿξ‚Ά1/(π‘ βˆ’π‘Ÿ).(2.10) The assertion of the lemma now follows from the fact that 𝐸(π‘Ÿ,𝑠;π‘₯,𝑦) is increasing in both π‘₯ and 𝑦 and in both π‘Ÿ and 𝑠 (see, e.g., Theorem  1 in [17] for a proof of this).

Lemma 2.7. Let π‘š>𝑛β‰₯1 be two integers, then for any fixed constant 0<𝑐<1, the function π‘Ž(𝑑;π‘š,𝑛,𝑐)=π‘‘π‘šβˆ’π‘›+π‘‘π‘›βˆ’π‘(1+π‘‘π‘š)(2.11) has exactly one root when 𝑑β‰₯1.

Proof. We have π‘Žβ€²(𝑑;π‘š,𝑛,𝑐)=π‘‘π‘šβˆ’1((π‘šβˆ’π‘›)π‘‘βˆ’π‘›+π‘›π‘‘π‘›βˆ’π‘šβˆ’π‘π‘š).(2.12) The function 𝑑↦(π‘šβˆ’π‘›)π‘‘βˆ’π‘›+π‘›π‘‘π‘›βˆ’π‘šβˆ’π‘π‘š is clearly decreasing when 𝑑β‰₯1. By considering the cases 𝑑=1 and 𝑑→+∞ we conclude that π‘Žβ€²(𝑑;π‘š,𝑛,𝑐) has exactly one root when 𝑑β‰₯1. It follows from this and Cauchy's mean value theorem that π‘Ž(𝑑;π‘š,𝑛,𝑐) has at most two roots when 𝑑β‰₯1. This combined with the observation that π‘Ž(1;π‘š,𝑛,𝑐)>0 and lim𝑑→+βˆžπ‘Ž(𝑑;π‘š,𝑛,𝑐)<0 yields the desired conclusion.

Lemma 2.8. For 𝑑β‰₯𝑠β‰₯0, one has 𝑠1βˆ’π‘’βˆ’π‘ β‹…π‘‘βˆ’π‘ 1βˆ’π‘’βˆ’(π‘‘βˆ’π‘ )β‰₯𝑑1βˆ’π‘’βˆ’π‘‘.(2.13)

Proof. We write 𝑓π‘₯(π‘₯)=1βˆ’π‘’βˆ’π‘₯(2.14) and we observe that ξ‚΅π‘“ξ…ž(𝑑)𝑓(𝑑)ξ…ž=π‘’βˆ’π‘‘ξ‚€π‘‘2βˆ’ξ€·π‘’π‘‘/2βˆ’π‘’βˆ’π‘‘/2ξ€Έ2𝑑2(1βˆ’π‘’βˆ’π‘‘)2≀0,(2.15) where the last inequality follows from 𝑒𝑑/2βˆ’π‘’βˆ’π‘‘/2β‰₯𝑑,𝑑β‰₯0.(2.16) The above inequality follows from the fact that the derivative of sinh(𝑑)βˆ’π‘‘ is cosh(𝑑)βˆ’1>0 for 𝑑>0 and that sinh(0)βˆ’0=0. We now deduce that 𝑓′(π‘‘βˆ’π‘ )βˆ’π‘“(π‘‘βˆ’π‘ )𝑓′(𝑑)𝑓(𝑑)β‰₯0,(2.17) for 𝑑β‰₯𝑠β‰₯0. This implies that the function 𝑑↦ln𝑓(π‘‘βˆ’π‘ )βˆ’ln𝑓(𝑑) is increasing for 𝑑>𝑠. Thus we get ln𝑓(𝑠)+ln𝑓(π‘‘βˆ’π‘ )βˆ’ln𝑓(𝑑)β‰₯lim𝑑→𝑠+(ln𝑓(𝑠)+ln𝑓(π‘‘βˆ’π‘ )βˆ’ln𝑓(𝑑))=0,(2.18) which is the desired result.

3. Main Results

Theorem 3.1. Let π‘Žπ‘– and 𝑏𝑖(𝑖=1,…,𝑛) be real numbers such that 0β‰€π‘Ž1β‰€β‹―β‰€π‘Žπ‘›, 0≀𝑏1≀⋯≀𝑏𝑛, and βˆ‘π‘˜π‘–=1π‘Žπ‘–β‰€βˆ‘π‘˜π‘–=1𝑏𝑖 for π‘˜=1,…,𝑛. If π‘“ξ…žξ…ž(π‘₯) is completely monotonic on (0,+∞), then exp𝑛𝑖=1𝑓π‘₯+π‘Žπ‘–ξ€Έξ€·βˆ’π‘“π‘₯+𝑏𝑖ξƒͺξ€Έξ€Έ(3.1) is logarithmically completely monotonic on (0,+∞).

Proof. It suffices to show that βˆ’π‘›ξ“π‘–=1𝑓′π‘₯+π‘Žπ‘–ξ€Έξ€·βˆ’π‘“β€²π‘₯+𝑏𝑖(3.2) is completely monotonic on (0,+∞) or for π‘˜β‰₯1, (βˆ’1)π‘˜π‘›ξ“π‘–=1𝑓(π‘˜)ξ€·π‘₯+π‘Žπ‘–ξ€Έβ‰₯(βˆ’1)π‘˜π‘›ξ“π‘–=1𝑓(π‘˜)ξ€·π‘₯+𝑏𝑖.(3.3) By Lemma 2.4, it suffices to show that (βˆ’1)π‘˜π‘“(π‘˜)(π‘₯) is decreasing and convex on (0,+∞) or equivalently, (βˆ’1)π‘˜π‘“(π‘˜+1)(π‘₯)≀0 and (βˆ’1)π‘˜π‘“(π‘˜+2)(π‘₯)β‰₯0 for π‘˜β‰₯1. The last two inequalities hold since we assume that π‘“ξ…žξ…ž(π‘₯) is completely monotonic on (0,+∞). This completes the proof.

As a direct consequence of Theorem 3.1, we now generalize a result of Alzer [2, Theorem  10].

Corollary 3.2. Let π‘Žπ‘– and 𝑏𝑖 (𝑖=1,…,𝑛) be real numbers such that 0β‰€π‘Ž1β‰€β‹―β‰€π‘Žπ‘›, 0≀𝑏1≀⋯≀𝑏𝑛, and βˆ‘π‘˜π‘–=1π‘Žπ‘–β‰€βˆ‘π‘˜π‘–=1𝑏𝑖 for π‘˜=1,…,𝑛. Then, π‘₯βŸΌπ‘›ξ‘i=1Ξ“π‘žξ€·π‘₯+π‘Žπ‘–ξ€ΈΞ“π‘žξ€·π‘₯+𝑏𝑖(3.4) is logarithmically completely monotonic on (0,+∞).

Proof. Apply Theorem 3.1 to 𝑓(π‘₯)=lnΞ“π‘ž(π‘₯) and note that π‘“ξ…žξ…ž(π‘₯)=πœ“ξ…žπ‘ž(π‘₯) is completely monotonic on (0,+∞) and this completes the proof.

Theorem 3.3. Let π‘“ξ…žξ…ž(π‘₯) be completely monotonic on (0,+∞), then for 0≀𝑠≀1, the functions ξ‚€βˆ’ξ‚€ξ‚€π‘₯⟼exp𝑓(π‘₯+1)βˆ’π‘“(π‘₯+𝑠)βˆ’(1βˆ’π‘ )𝑓′π‘₯+1+𝑠2,π‘₯⟼exp𝑓(π‘₯+1)βˆ’π‘“(π‘₯+𝑠)βˆ’(1βˆ’π‘ )2ξ‚Ά(𝑓′(π‘₯+1)+𝑓′(π‘₯+𝑠))(3.5) are logarithmically completely monotonic on (0,+∞).

Proof. We may assume 0≀𝑠<1. We will prove the first assertion and the second one can be shown similarly. It suffices to show that 𝑓′(π‘₯+1)βˆ’π‘“β€²(π‘₯+𝑠)βˆ’(1βˆ’π‘ )π‘“ξ…žξ…žξ‚€π‘₯+1+𝑠2(3.6) is completely monotonic on (0,+∞) or for π‘˜β‰₯1, 1ξ€œ1βˆ’π‘ π‘₯+1π‘₯+𝑠(βˆ’1)π‘˜+1𝑓(π‘˜+1)(𝑑)𝑑𝑑β‰₯(βˆ’1)π‘˜+1𝑓(π‘˜+1)ξ‚€π‘₯+1+𝑠2.(3.7) The last inequality holds by Lemma 2.5 and our assumption that π‘“ξ…žξ…ž(π‘₯) is completely monotonic on (0,+∞). This completes the proof.

Corollary 3.4. For 0≀𝑠≀1, the functions Ξ“π‘₯βŸΌπ‘ž(π‘₯+𝑠)Ξ“π‘žξ‚€(π‘₯+1)exp(1βˆ’π‘ )πœ“π‘žξ‚€π‘₯+1+𝑠2,Γπ‘₯βŸΌπ‘ž(π‘₯+1)Ξ“π‘žξ‚΅βˆ’(π‘₯+𝑠)exp(1βˆ’π‘ )2ξ€·πœ“π‘ž(π‘₯+1)+πœ“π‘žξ€Έξ‚Ά(π‘₯+𝑠)(3.8) are logarithmically completely monotonic on (0,+∞).

Proof. Apply Theorem 3.3 to 𝑓(π‘₯)=lnΞ“π‘ž(π‘₯) and note that π‘“ξ…žξ…ž(π‘₯)=πœ“ξ…žπ‘ž(π‘₯) is completely monotonic on (0,+∞) and this completes the proof.

By applying Lemma 2.5 to 𝑓(π‘₯)=βˆ’πœ“π‘ž(π‘₯), we obtain the following theorem.

Theorem 3.5. For positive π‘₯ and 0≀𝑠≀1, ξ‚΅exp(1βˆ’π‘ )2ξ€·πœ“π‘ž(π‘₯+1)+πœ“π‘žξ€Έξ‚Άβ‰€Ξ“(π‘₯+𝑠)π‘ž(π‘₯+1)Ξ“π‘žξ‚€(π‘₯+𝑠)≀exp(1βˆ’π‘ )πœ“π‘žξ‚€π‘₯+1+𝑠2.(3.9)

The upper bound in Theorem 3.5 is due to Ismail and Muldoon [4]. Our proof here is similar to that of Corollary  3 in [18].

Our next result refines the left-hand side inequality of (1.6) for 0<π‘ž<1.

Theorem 3.6. Let 0<𝑠<1 and 0<π‘ž<1. Let 𝑒(π‘ž,𝑠) be defined by (1.7) and let the function π‘”π‘ž(π‘₯;𝑠,1,𝑒(π‘ž,𝑠)) be defined as in Theorem 1.1. Then π‘”π‘ž(π‘₯;𝑠,1,𝑒(π‘ž,𝑠)) is logarithmically completely monotonic on (0,+∞).

Proof. Define β„Žπ‘ž(π‘₯)=βˆ’lnπ‘”π‘ž(π‘₯;𝑠,1,𝑒(π‘ž,𝑠))=βˆ’lnΞ“π‘ž(π‘₯+1)+lnΞ“π‘žξ‚΅(π‘₯+𝑠)+(1βˆ’π‘ )ln1βˆ’π‘žπ‘₯+𝑒(π‘ž,𝑠)ξ‚Ά.1βˆ’π‘ž(3.10) It suffices to show that β„Žξ…žπ‘ž(π‘₯) is completely monotonic on (0,+∞). We have β„Žξ…žπ‘ž(π‘₯)=βˆ’πœ“π‘ž(π‘₯+1)+πœ“π‘žπ‘ž(π‘₯+𝑠)βˆ’(1βˆ’π‘ )lnπ‘žπ‘₯+𝑒(π‘ž,𝑠)1βˆ’π‘žπ‘₯+𝑒(π‘ž,𝑠).(3.11) Using the expression (2.1), we can rewrite β„Žξ…žπ‘ž(π‘₯) as β„Žξ…žπ‘žξƒ©(π‘₯)=βˆ’lnπ‘žβˆžξ“π‘›=1π‘žπ‘›π‘₯(π‘žπ‘›βˆ’π‘žπ‘›π‘ )1βˆ’π‘žπ‘›π‘ž+(1βˆ’π‘ )π‘₯+𝑒(π‘ž,𝑠)1βˆ’π‘žπ‘₯+𝑒(π‘ž,𝑠)ξƒͺ,0<π‘ž<1.(3.12) Expanding (1βˆ’π‘žπ‘₯+𝑒(π‘ž,𝑠))βˆ’1, we may further rewrite β„Žξ…žπ‘ž(π‘₯) as β„Žξ…žπ‘ž(π‘₯)=βˆ’lnπ‘žβˆžξ“π‘›=1π‘žπ‘›π‘₯1βˆ’π‘žπ‘›π‘€π‘ž,𝑛(𝑠),0<π‘ž<1,(3.13) where π‘€π‘ž,𝑛(𝑐)=π‘žπ‘›βˆ’π‘žπ‘›π‘ +(1βˆ’π‘ )π‘žπ‘›π‘’(π‘ž,𝑠)(1βˆ’π‘žπ‘›).(3.14) In order for β„Žξ…žπ‘ž(π‘₯) to be completely monotonic on (0,+∞), it suffices to show π‘€π‘ž,𝑛(𝑠)β‰₯0 for 0<𝑠<1. This is just Lemma 2.6 and this completes the proof.

Theorem 3.6 implies that π‘”π‘ž(π‘₯;𝑠,1,𝑒(π‘ž,𝑠))>limπ‘₯β†’+βˆžπ‘”π‘ž(π‘₯;𝑠,1,𝑒(π‘ž,𝑠))=1, where the limit can be easily evaluated using (1.1) and we recover the left-hand side inequality of (1.6) for 0<π‘ž<1.

For 𝛼β‰₯0, let𝑓𝛼1(π‘₯)=βˆ’lnΞ“(π‘₯)+π‘₯βˆ’21lnπ‘₯βˆ’π‘₯+12πœ“β€²(π‘₯+𝛼).(3.15) Alzer [19, Theorem  1] showed that for fixed 0<𝑠<1, the function exp(βˆ’(𝑓𝛼(π‘₯+𝑠)βˆ’π‘“π›Ό(π‘₯+1))) is strictly completely monotonic on (0,+∞) if and only if 𝛼β‰₯1/2 and the function exp(𝑓𝛼(π‘₯+𝑠)βˆ’π‘“π›Ό(π‘₯+1)) is strictly completely monotonic on (0,+∞) if and only if 𝛼=0. In view of Lemmas  2.2 and  2.3, Alzer's result would follow if one can show that π‘“ξ…žπ›Ό(π‘₯) is strictly completely monotonic on (0,+∞) if and only if 𝛼β‰₯1/2 and that βˆ’π‘“ξ…žπ›Ό(π‘₯) is strictly completely monotonic on (0,+∞) if and only if 𝛼=0. We now establish the above assertions in the following.

Proposition 3.7. For 𝛼β‰₯0, let 𝑓𝛼(π‘₯) be defined as in (3.15). Then π‘“ξ…žπ›Ό(π‘₯) is strictly completely monotonic on (0,+∞) if and only if 𝛼β‰₯1/2 and βˆ’π‘“ξ…žπ›Ό(π‘₯) is strictly completely monotonic on (0,+∞) if and only if 𝛼=0.

Proof. As πœ“β€²(π‘₯) is completely monotonic on (0,+∞), we may just focus on the case 𝛼=1/2 or 0. We have π‘“ξ…žπ›Ό1(π‘₯)=βˆ’πœ“(π‘₯)+lnπ‘₯βˆ’+12π‘₯πœ“12ξ…žξ…ž(π‘₯+𝛼),π‘“π›Όξ…žξ…ž(π‘₯)=βˆ’πœ“ξ…ž1(π‘₯)+π‘₯+12π‘₯2+1πœ“12(3)(π‘₯+𝛼).(3.16) Using the asymptotic expressions (2.4) and (2.5), we see that limπ‘₯β†’+βˆžπ‘“π›Ό(𝑛)(π‘₯)=0 for any integer 𝑛β‰₯0. It is then easy to see that the β€œif” part of the assertions of the proposition will follow if we can show that π‘“ξ…žξ…ž1/2(π‘₯+1)βˆ’π‘“ξ…žξ…ž1/2(π‘₯) is strictly completely monotonic on (0,+∞) and 𝑓0ξ…žξ…ž(π‘₯)βˆ’π‘“0ξ…žξ…ž(π‘₯+1) is completely monotonic on (0,+∞). Using (2.3), it is easy to see that π‘“ξ…žξ…ž1/2(π‘₯+1)βˆ’π‘“ξ…žξ…ž1/2π‘₯(π‘₯)=2+π‘₯+1/84π‘₯2(1+π‘₯)2(π‘₯+1/2)4=14(1+π‘₯)2(π‘₯+1/2)4ξ‚€11+π‘₯+18π‘₯2.(3.17) It is easy to see from this and the ring structure of completely monotonic functions on (0,+∞) that π‘“ξ…žξ…ž1/2(π‘₯+1)βˆ’π‘“ξ…žξ…ž1/2(π‘₯) is strictly completely monotonic on (0,+∞). Similarly, one shows that 𝑓0ξ…žξ…ž(π‘₯)βˆ’π‘“0ξ…žξ…ž(π‘₯+1) is completely monotonic on (0,+∞).
To prove the β€œonly if” part of the assertions of the proposition, we note that π‘“π›Όξ…žξ…ž(π‘₯+1)βˆ’π‘“π›Όξ…žξ…ž(π‘₯)=(4π›Όβˆ’2)π‘₯3+ξ€·6𝛼2ξ€Έπ‘₯βˆ’12+4𝛼3π‘₯+𝛼42π‘₯2(1+π‘₯)2(π‘₯+𝛼)4,(3.18) from which we see easily by considering the case π‘₯β†’+∞ that π‘“π›Όξ…žξ…ž(π‘₯) fails to be increasing if 𝛼<1/2. Similarly, by considering the case π‘₯β†’0+ that π‘“π›Όξ…žξ…ž(π‘₯) fails to be decreasing if 𝛼>0 and this completes the proof.

4. Some Inequalities Involving Polygamma Functions

For integers 𝑝β‰₯π‘šβ‰₯𝑛β‰₯π‘žβ‰₯0 and any real number 𝑐, we define 𝐹𝑝,π‘š,𝑛,π‘ž(π‘₯;𝑐)=(βˆ’1)π‘š+π‘›πœ“(π‘š)(π‘₯)πœ“(𝑛)(π‘₯)βˆ’π‘(βˆ’1)𝑝+π‘žπœ“(𝑝)(π‘₯)πœ“(π‘ž)(π‘₯).(4.1) Here we set πœ“(0)(π‘₯)=βˆ’1 for convenience.

For 𝑛β‰₯2, a result of Alzer and Wells [20, Theorem  2.1] asserts that the function 𝐹𝑛+1,𝑛,𝑛,π‘›βˆ’1(π‘₯;𝑐) is completely monotonic on (0,+∞) if and only if 𝑐≀(π‘›βˆ’1)/𝑛 and βˆ’πΉπ‘›+1,𝑛,𝑛,π‘›βˆ’1(π‘₯;𝑐) is completely monotonic on (0,+∞) if and only if 𝑐β‰₯𝑛/(𝑛+1).

We denote 𝑐𝑝,π‘š,𝑛,π‘ž=(π‘šβˆ’1)!(π‘›βˆ’1)!(π‘βˆ’1)!(π‘žβˆ’1)!,π‘žβ‰₯1;𝑐𝑝,π‘š,𝑛,0=(π‘šβˆ’1)!(π‘›βˆ’1)!(,π‘‘π‘βˆ’1)!𝑝,π‘š,𝑛,π‘ž=π‘š!𝑛!.𝑝!π‘ž!(4.2) and note that 0<𝑐𝑝,π‘š,𝑛,π‘ž,𝑑𝑝,π‘š,𝑛,π‘ž<1 when 𝑝+π‘ž=π‘š+𝑛,𝑝>π‘š. We now extend the result of Alzer and Wells to the following.

Theorem 4.1. Let 𝑝>π‘šβ‰₯𝑛>π‘žβ‰₯0 be integers satisfying π‘š+𝑛=𝑝+π‘ž. The function 𝐹𝑝,π‘š,𝑛,π‘ž(π‘₯;𝑐𝑝,π‘š,𝑛,π‘ž) is completely monotonic on (0,+∞). The function βˆ’πΉπ‘,π‘š,𝑛,π‘ž(π‘₯;𝑑𝑝,π‘š,𝑛,π‘ž) is also completely monotonic on (0,+∞) when π‘ž>0.

Proof. We first prove the assertion for 𝐹𝑝,π‘š,𝑛,π‘ž(π‘₯;𝑐𝑝,π‘š,𝑛,π‘ž) with π‘žβ‰₯1. The proof here uses the method in [20]. Using the integral representation (2.2) for (βˆ’1)𝑛+1πœ“(𝑛)(π‘₯) and using βˆ— for the Laplace convolution, we get 𝐹𝑝,π‘š,𝑛,π‘žξ€·π‘₯;𝑐𝑝,π‘š,𝑛,π‘žξ€Έ=ξ€œβˆž0π‘’βˆ’π‘₯𝑑𝑔(𝑑)𝑑𝑑,(4.3) where 𝑑𝑔(𝑑)=π‘š1βˆ’π‘’βˆ’π‘‘βˆ—π‘‘π‘›1βˆ’π‘’βˆ’π‘‘βˆ’π‘π‘,π‘š,𝑛,π‘žπ‘‘π‘1βˆ’π‘’βˆ’π‘‘βˆ—π‘‘π‘ž1βˆ’π‘’βˆ’π‘‘=ξ€œπ‘‘0ξ€·(π‘‘βˆ’π‘ )π‘šπ‘ π‘›βˆ’π‘π‘,π‘š,𝑛,π‘ž(π‘‘βˆ’π‘ )π‘π‘ π‘žξ€Έβ„Ž(π‘‘βˆ’π‘ )β„Ž(𝑠)𝑑𝑠,(4.4) with 1β„Ž(𝑠)=1βˆ’π‘’βˆ’π‘ .(4.5) It suffices to show that 𝑔(𝑑)β‰₯0. By change of variables 𝑠→𝑑𝑠 we can recast it as 𝑔(𝑑)=π‘‘π‘š+𝑛+1ξ€œ10ξ€·(1βˆ’π‘ )π‘šπ‘ π‘›βˆ’π‘π‘,π‘š,𝑛,π‘ž(1βˆ’π‘ )π‘π‘ π‘žξ€Έβ„Ž(𝑑(1βˆ’π‘ ))β„Ž(𝑑𝑠)𝑑𝑠.(4.6) We now break the above integral into two integrals, one from 0 to 1/2 and the other from 1/2 to 1. We make a further change of variable 𝑠→(1βˆ’π‘ )/2 for the first one and 𝑠→(1+𝑠)/2 for the second one. We now combine them to get 𝑑𝑔(𝑑)=2ξ‚π‘š+𝑛+1ξ€œ10π‘Žξ‚€1+𝑠1βˆ’π‘ ;π‘βˆ’π‘ž,π‘›βˆ’π‘ž,𝑐𝑝,π‘š,𝑛,π‘žξ‚ξ€·1βˆ’π‘ 2ξ€Έπ‘ž(1βˆ’π‘ )π‘βˆ’π‘žξ‚΅Γ—β„Žπ‘‘(1βˆ’π‘ )2ξ‚Άβ„Žξ‚΅π‘‘(1+𝑠)2𝑑𝑠,(4.7) where the function π‘Ž(𝑑;π‘š,𝑛,𝑐) is defined as in Lemma 2.7. Note that (1+𝑠)/(1βˆ’π‘ )β‰₯1 for 0≀𝑠<1, hence by Lemma 2.7, there is a unique number 0<𝑠0<1 such that π‘Žξ‚΅1+𝑠01βˆ’π‘ 0;π‘βˆ’π‘ž,π‘›βˆ’π‘ž,𝑐𝑝,π‘š,𝑛,π‘žξ‚Ά=0.(4.8) We further note it is shown in the proof of [20, Lemma  2.2] that the function ξ€·π‘ βŸΌ1βˆ’π‘ 2ξ€Έβ„Žξ‚΅π‘‘(1βˆ’π‘ )2ξ‚Άβ„Žξ‚΅π‘‘(1+𝑠)2ξ‚Ά(4.9) is a decreasing function on (0,1) so that for 0≀𝑠≀1, π‘Žξ‚€1+𝑠1βˆ’π‘ ;π‘βˆ’π‘ž,π‘›βˆ’π‘ž,𝑐𝑝,π‘š,𝑛,π‘žξ‚ξ€·1βˆ’π‘ 2ξ€Έπ‘ž(1βˆ’π‘ )π‘βˆ’π‘žβ„Žξ‚΅π‘‘(1βˆ’π‘ )2ξ‚Άβ„Žξ‚΅π‘‘(1+𝑠)2ξ‚Άξ‚€β‰₯π‘Ž1+𝑠1βˆ’π‘ ;π‘βˆ’π‘ž,π‘›βˆ’π‘ž,𝑐𝑝,π‘š,𝑛,π‘žξ‚ξ€·1βˆ’π‘ 2ξ€Έπ‘žβˆ’1(1βˆ’π‘ )π‘βˆ’π‘žξ€·1βˆ’π‘ 20ξ€Έξƒ©π‘‘ξ€·Γ—β„Ž1βˆ’π‘ 0ξ€Έ2ξƒͺβ„Žξƒ©π‘‘ξ€·1+𝑠0ξ€Έ2ξƒͺ.(4.10) Hence 𝑑𝑔(𝑑)β‰₯2ξ‚π‘š+𝑛+1ξ€·1βˆ’π‘ 20ξ€Έβ„Žξƒ©π‘‘ξ€·1βˆ’π‘ 0ξ€Έ2ξƒͺβ„Žξƒ©π‘‘ξ€·1+𝑠0ξ€Έ2ξƒͺβ‹…ξ€œ10π‘Žξ‚€1+𝑠1βˆ’π‘ ;π‘βˆ’π‘ž,π‘›βˆ’π‘ž,𝑐𝑝,π‘š,𝑛,π‘žξ‚ξ€·1βˆ’π‘ 2ξ€Έπ‘žβˆ’1(1βˆ’π‘ )π‘βˆ’π‘žπ‘‘π‘ .(4.11) Note that the integral above is (by reversing the process above on changing variables) 2π‘š+π‘›βˆ’1ξ€œ10ξ€·(1βˆ’π‘ )π‘šβˆ’1π‘ π‘›βˆ’1βˆ’π‘π‘,π‘š,𝑛,π‘ž(1βˆ’π‘ )π‘βˆ’1π‘ π‘žβˆ’1𝑑𝑠=0,(4.12) where the last step follows from the well-known beta function identity ξ€œπ΅(π‘₯,𝑦)=10𝑑π‘₯βˆ’1(1βˆ’π‘‘)π‘¦βˆ’1𝑑𝑑=Ξ“(π‘₯)Ξ“(𝑦)Ξ“(π‘₯+𝑦),π‘₯,𝑦>0,(4.13) and the well-known fact Ξ“(𝑛)=(π‘›βˆ’1)! for 𝑛β‰₯1.
Now we prove the assertion for 𝐹𝑝,π‘š,𝑛,0(π‘₯;𝑐𝑝,π‘š,𝑛,0). In this case 𝑝=π‘š+𝑛 and we note that π‘π‘š+𝑛,π‘š,𝑛,0ξ€œ=𝐡(π‘š,𝑛)=10π‘ π‘šβˆ’1(1βˆ’π‘ )π‘›βˆ’1𝑑𝑠,(4.14) and we use this to write π‘π‘š+𝑛,π‘š,𝑛,0π‘‘π‘š+𝑛1βˆ’π‘’βˆ’π‘‘=ξ€œπ‘‘0π‘ π‘šβˆ’1(π‘‘βˆ’π‘ )π‘›βˆ’1𝑑1βˆ’π‘’βˆ’π‘‘π‘‘π‘ .(4.15) It follows that 𝐹𝑝,π‘š,𝑛,0ξ€·π‘₯;𝑐𝑝,π‘š,𝑛,0ξ€Έ=ξ€œβˆž0π‘’βˆ’π‘₯π‘‘ξ‚΅π‘‘π‘š1βˆ’π‘’βˆ’π‘‘βˆ—π‘‘π‘›1βˆ’π‘’βˆ’π‘‘βˆ’π‘π‘š+𝑛,π‘š,𝑛,0π‘‘π‘š+𝑛1βˆ’π‘’βˆ’π‘‘ξ‚Ά=ξ€œπ‘‘π‘‘βˆž0π‘’βˆ’π‘₯π‘‘ξ‚΅ξ€œπ‘‘0π‘ π‘šβˆ’1(π‘‘βˆ’π‘ )π‘›βˆ’1ξ‚€s1βˆ’π‘’βˆ’π‘ β‹…π‘‘βˆ’π‘ 1βˆ’π‘’βˆ’(π‘‘βˆ’π‘ )βˆ’π‘‘1βˆ’π‘’βˆ’π‘‘ξ‚ξ‚Άπ‘‘π‘ π‘‘π‘‘β‰₯0,(4.16) where the last inequality follows from Lemma 2.8.
It remains to show the assertion for βˆ’πΉπ‘,π‘š,𝑛,π‘ž(π‘₯;𝑑𝑝,π‘š,𝑛,π‘ž). In this case we use the series representation in (2.2) for (βˆ’1)𝑛+1πœ“(𝑛)(π‘₯) to get βˆ’πΉπ‘,π‘š,𝑛,π‘žξ€·π‘₯;𝑑𝑝,π‘š,𝑛,π‘žξ€Έ=π‘š!𝑛!ξƒ©ξƒ©βˆžξ“π‘–=01(π‘₯+𝑖)𝑝+1ξƒͺξƒ©βˆžξ“π‘—=01(π‘₯+𝑗)π‘ž+1ξƒͺβˆ’ξƒ©βˆžξ“π‘–=01(π‘₯+𝑖)π‘š+1ξƒͺξƒ©βˆžξ“π‘—=01(π‘₯+𝑗)𝑛+1.ξƒͺξƒͺ(4.17) We note the following Binet-Cauchy identity: 𝑛𝑖=0π‘Žπ‘–π‘π‘–ξƒͺ𝑛𝑖=0𝑏𝑖𝑑𝑖ξƒͺβˆ’ξƒ©π‘›ξ“π‘–=0π‘Žπ‘–π‘‘π‘–ξƒͺ𝑛𝑖=0𝑏𝑖𝑐𝑖ξƒͺ=0≀𝑖<π‘—β‰€π‘›ξ€·π‘Žπ‘–π‘π‘—βˆ’π‘Žπ‘—π‘π‘–π‘ξ€Έξ€·π‘–π‘‘π‘—βˆ’π‘π‘—π‘‘π‘–ξ€Έ.(4.18) We now apply the above identity with π‘Žπ‘–=1(π‘₯+𝑖)π‘š+1,𝑏𝑖=1(π‘₯+𝑖)π‘ž+1,𝑐𝑖=1(π‘₯+𝑖)π‘βˆ’π‘š,𝑑𝑖=1(4.19) to get βˆ’πΉπ‘,π‘š,𝑛,π‘žξ€·π‘₯;𝑑𝑝,π‘š,𝑛,π‘žξ€Έξ“=π‘š!𝑛!0≀𝑖<𝑗1(π‘₯+𝑖)π‘š+1(π‘₯+𝑗)π‘ž+1βˆ’1(π‘₯+𝑗)π‘š+1(π‘₯+𝑖)π‘ž+11ξ‚Άξ‚΅(π‘₯+𝑖)π‘βˆ’π‘šβˆ’1(π‘₯+𝑗)π‘βˆ’π‘šξ‚Ά.(4.20) We note that the second factor on the right-hand side above is completely monotonic on (0,+∞) and also that 1(π‘₯+𝑖)π‘š+1(π‘₯+𝑗)π‘ž+1βˆ’1(π‘₯+𝑗)π‘š+1(π‘₯+𝑖)π‘ž+1=1(π‘₯+𝑖)π‘ž+11(π‘₯+𝑗)π‘ž+1ξ‚΅1(π‘₯+𝑖)π‘šβˆ’π‘žβˆ’1(π‘₯+𝑗)π‘šβˆ’π‘žξ‚Ά.(4.21) Certainly each factor on the right-hand side above is completely monotonic on (0,+∞) and it follows from the ring structure of completely monotonic functions on (0,+∞) that the left-hand side expression in (4.21) is also completely monotonic on (0,+∞). Hence by the ring structure of completely monotonic functions on (0,+∞) again we deduce that βˆ’πΉπ‘,π‘š,𝑛,π‘ž(π‘₯;𝑑𝑝,π‘š,𝑛,π‘ž) is completely monotonic on (0,+∞).

We note here that when π‘ž=0, the function βˆ’πΉπ‘š+𝑛,π‘š,𝑛,0(π‘₯;π‘‘π‘š+𝑛,π‘š,𝑛,0) is not completely monotonic on (0,+∞) in general, as we observe for example that it follows from (2.5) that as π‘₯β†’+∞, βˆ’πΉ2,1,1,0(π‘₯;𝑑2,1,1,0)<0.

Corresponding to π‘š=𝑛=1,𝑝=2,π‘ž=0 in Theorem 4.1, it was shown in the proof of [5, equation (4.39)] and in [21, Lemma  1.1] the following special case (in fact with strict inequality):ξ€·πœ“ξ…žξ€Έ(π‘₯)2+πœ“ξ…žξ…ž(π‘₯)β‰₯0,π‘₯>0.(4.22) We note that inequality (4.22) follows from the limiting case 𝑐→0 of the following inequalities for any 0<𝑐<1 and π‘₯>0:1𝑐(πœ“(π‘₯+𝑐)βˆ’πœ“(π‘₯))2>πœ“β€²(π‘₯)βˆ’πœ“β€²(π‘₯+𝑐)>(πœ“(π‘₯+𝑐)βˆ’πœ“(π‘₯))2,(4.23) where the above inequalities reverse when 𝑐>1.

Inequalities (4.23) are special cases of Theorem  1 in [10] as one can relate the inequalities in (4.23) to the properties concerning Ξ”0,𝑐;1(π‘₯) or Ξ”0,𝑐;1/𝑐(π‘₯) defined in [10]. The left-hand side inequality of (4.23) was also established in the proof of [22, Theorem  1.1] and the right-hand side inequality in (4.23) was proven in [23, Lemma  7].

To end this paper, we prove a π‘ž-analogue to (4.23).

Theorem 4.2. Let 0<π‘ž<1 and 0<𝑐<1 be fixed. Then for any π‘₯>0, 1βˆ’π‘ž1βˆ’π‘žπ‘ξ€·πœ“π‘ž(π‘₯+𝑐)βˆ’πœ“π‘žξ€Έ(π‘₯)2>π‘žπ‘₯ξ€·πœ“ξ…žπ‘ž(π‘₯)βˆ’πœ“ξ…žπ‘žξ€Έ>ξ€·πœ“(π‘₯+𝑐)π‘ž(π‘₯+𝑐)βˆ’πœ“π‘žξ€Έ(π‘₯)2.(4.24) The above inequalities reverse when 𝑐>1.

Proof. We first prove the left-hand side inequality of (4.24). For this, we define 𝑓(π‘₯)=π‘žπ‘₯ξ€·πœ“ξ…žπ‘ž(π‘₯+𝑐)βˆ’πœ“ξ…žπ‘žξ€Έ+(π‘₯)1βˆ’π‘ž1βˆ’π‘žπ‘ξ€·πœ“π‘ž(π‘₯+𝑐)βˆ’πœ“π‘žξ€Έ(π‘₯)2.(4.25) Applying (2.1), we obtain 𝑓(π‘₯)=(lnπ‘ž)2βˆžξ“π‘›=1π‘›π‘ž(𝑛+1)π‘₯(π‘žπ‘›π‘βˆ’1)1βˆ’π‘žπ‘›+1βˆ’π‘ž1βˆ’π‘žπ‘ξƒ©lnπ‘žβˆžξ“π‘›=1π‘žπ‘›π‘₯(π‘žπ‘›π‘βˆ’1)1βˆ’π‘žπ‘›ξƒͺ2=(1βˆ’π‘ž)(lnπ‘ž)21βˆ’π‘žπ‘βˆžξ“π‘›=2π‘žπ‘›π‘₯𝑔𝑛(𝑐;π‘ž),(4.26) where 𝑔𝑛(𝑐;π‘ž)=π‘›βˆ’1ξ“π‘˜=11βˆ’π‘žπ‘˜π‘ξ€Έξ€·1βˆ’π‘ž(π‘›βˆ’π‘˜)𝑐1βˆ’π‘žπ‘˜ξ€Έξ€·1βˆ’π‘žπ‘›βˆ’π‘˜ξ€Έβˆ’(1βˆ’π‘žπ‘)ξ€·1βˆ’π‘ž(π‘›βˆ’1)𝑐(ξ€·1βˆ’π‘ž)1βˆ’π‘žπ‘›βˆ’1ξ€Έξƒͺ.(4.27) It suffices to show that 𝑔𝑛(𝑐;π‘ž)>0 for 0<𝑐<1 and 𝑔𝑛(𝑐;π‘ž)<0 for 𝑐>1 when 𝑛β‰₯2. For this, we let 𝑦=π‘žπ‘ so that 0<𝑦<1 and it suffices to show the function (1βˆ’π‘¦π‘˜)(1βˆ’π‘¦π‘›βˆ’π‘˜)/((1βˆ’π‘¦)(1βˆ’π‘¦π‘›βˆ’1)) is increasing for 0<𝑦<1 and 1β‰€π‘˜β‰€π‘›βˆ’1. On taking the logarithmic derivative of the above function, we see that it suffices to show that β„Ž(π‘˜;𝑦)+β„Ž(π‘›βˆ’π‘˜;𝑦)β‰€β„Ž(1;𝑦)+β„Ž(π‘›βˆ’1;𝑦), where π‘§β„Ž(𝑧;𝑦)=1βˆ’π‘¦π‘§.(4.28) We now regard β„Ž(𝑧;𝑦) as a function of 𝑧 and note that β„Žξ…žξ…ž(𝑧;𝑦)=(ln𝑦)𝑦𝑧(1βˆ’π‘¦π‘§)3𝑒(𝑦𝑧),𝑒(𝑑)=2βˆ’2𝑑+ln𝑑+𝑑ln𝑑.(4.29) It is easy to see that 𝑒′(𝑑)>0 for 0<𝑑<1 so that 𝑒(𝑑)<𝑒(1)=0 for 0<𝑑<1. It follows that β„Žξ…žξ…ž(𝑧;𝑦)>0. We then deduce from this and Lemma 2.4 that β„Ž(π‘˜;𝑦)+β„Ž(π‘›βˆ’π‘˜;𝑦)β‰€β„Ž(1;𝑦)+β„Ž(π‘›βˆ’1;𝑦) holds and this completes the proof for the left-hand side inequality of (4.24). For the right-hand side inequality of (4.24), one proceeds similarly to the above argument to see that it suffices to show the function (1βˆ’π‘¦π‘˜)(1βˆ’π‘¦π‘›βˆ’π‘˜)/(1βˆ’π‘¦π‘›βˆ’1) is decreasing for 0<𝑦<1 and 1β‰€π‘˜β‰€π‘›βˆ’1. This follows from the observation that both functions (1βˆ’π‘¦π‘˜)/(1βˆ’π‘¦π‘›βˆ’1) and 1βˆ’π‘¦π‘›βˆ’π‘˜ are decreasing and this completes the proof.

Acknowledgments

This work was partially carried out while the author was visiting the American Institute of Mathematics in fall 2005 and the Centre de Recherches MathΓ©matiques at the UniversitΓ© de MontrΓ©al in spring 2006. The author would like to thank both the American Institute of Mathematics and the Centre de Recherches MathΓ©matiques at the UniversitΓ© de MontrΓ©al for their generous support and hospitality. The author is also grateful to the referees for their helpful comments and suggestions.