# Some Monotonicity Properties of Gamma andq-Gamma Functions

Authors: Peng Gao
Journal: ISRN Mathematical Analysis
 (2011)

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## Abstract

We prove some monotonicity properties of functions involving  gamma andq-gamma functions.


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## Body

## 1. Introduction

Theq-gamma function is defined for positive real numbers x and q≠1 by(1.1)Γq(x)=(1-q)1-x∏n=0∞1-qn+11-qn+x,0<q<1,Γq(x)=(q-1)1-xq(1/2)x(x-1)∏n=0∞1-q-(n+1)1-q-(n+x),q>1.
We note here according to [1] that the limit of Γq(x) as q→1- gives back the well-known Euler's gamma function: (1.2)lim⁡q→1-Γq(x)=Γ(x)=∫0∞txe-tdtt.
Note also that from the definition we have for positive x and 0<q<1, (1.3)Γ1/q(x)=q(x-1)(1-x/2)Γq(x),
we see that limq→1Γq(x)=Γ(x). For historical remarks on gamma and q-gamma functions, we refer the reader to [1–3].There exists an extensive and rich literature on inequalities for the gamma andq-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 Γ (resp., Γq) and its logarithmic derivative ψ (resp. ψq). Here we recall that a function f(x) is said to be completely monotonic on (a,b) if it has derivatives of all orders and (-1)kf(k)(x)≥0, x∈(a,b), k≥0 and f(x) is said to be strictly completely monotonic on (a,b) if (-1)kf(k)(x)>0, x∈(a,b),  k≥0. Lemma 2.3 below asserts that f(x)=e-h(x) is completely monotonic on an interval if h′ is. Following [8], we call such functions f(x) 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 thatlimq→1ψq(x)=ψ(x) (see [13]) and that ψ′ and ψq′ are completely monotonic functions on (0,+∞) (see [5, 14]). Thus, one expects to deduce results on gamma and q-gamma functions from properties of (logarithmically) completely monotonic functions, by applying them to functions related to ψ′ or ψq′. It is our goal in this paper to obtain some results on gamma and q-gamma functions via this approach. As an example, we recall the following result of Bustoz and Ismail (the case q=1) as well as Ismail and Muldoon.Theorem 1.1 (see [15, Theorem  3], [4, Theorem  2.5]).
Leta+1≥b>a, α=max(-a,-c) and for q>0 and q≠1, define
(1.4)gq(x;a,b,c)=(1-qx+c1-q)a-bΓq(x+b)Γq(x+a),x>α.
For q=1, let g1(x;a,b,c)=limq→1gq(x;a,b,c). Then gq(x;a,b,c) is logarithmically completely monotonic on (α,+∞) if c≤(a+b-1)/2 and 1/gq(x;a,b,c) is logarithmically completely monotonic on (α,+∞) if c≥a.It follows immediately from the above theorem that for0<q<1 and 0<s<1, one has [4] for x>0: (1.5)(1-qx+s/21-q)1-s<Γq(x+1)Γq(x+s)<(1-qx+s1-q)1-s.
Alzer later [3] determined the best values u(q,s),v(q,s) such that the inequalities(1.6)(1-qx+u(q,s)1-q)1-s<Γq(x+1)Γq(x+s)<(1-qx+v(q,s)1-q)1-s
hold for q>0,0<s<1,x>0 to be(1.7)u(q,s)={ln⁡((qs-q)/(1-s)(1-q))ln⁡q,0<q<1,s2,q>1,v(q,s)=ln⁡(1-(1-q)Γq1/(s-1)(s))ln⁡q.Motivated by the above results, we will show in Section3 that the function gq(x;s,1,u(q,s)) as defined in Theorem 1.1 is logarithmically completely monotonic on (0,+∞) for 0<q<1. This will enable us to deduce the left-hand side inequality of Alzer's result above for 0<q<1. The derivatives of ψ(x) 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 aboutψ(x) and ψq(x). These can be found, for example, in [3, equation (2.7)] and [5, equations  (1.2)–(1.5)].Lemma 2.1.
Forx>0,
(2.1)ψq(x)=-ln⁡(1-q)+ln⁡q∑n=1∞qnx1-qn,0<q<1,(2.2)(-1)n+1ψ(n)(x)=∫0∞e-xttn1-e-tdt=n!∑k=0∞1(x+k)n+1,n≥1,(2.3)ψ(n)(x+1)=ψ(n)(x)+(-1)nn!xn+1,n≥0,(2.4)ψ(x)=ln⁡x-12x-112x2+O(1x3),x⟶+∞,(2.5)(-1)n+1ψ(n)(x)=(n-1)!xn+n!2xn+1+O(1xn+2),n≥1,x⟶+∞.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 Lemma2.2 is contained in [4, Lemma  1.3].Lemma 2.2.
Iff(x) is completely monotonic on some interval (a,b), then so is f(x)-f(x+c) on (a,b)∩(a-c,b-c) for any c>0. Consequently, if f(x) is logarithmically completely monotonic on some interval (a,b), then so is f(x)/f(x+c) on (a,b)∩(a-c,b-c) for any c>0.Lemma 2.3 (see [15, Lemma  2.1]).
Iff′(x) is completely monotonic on an interval, then exp(-f(x)) is also completely monotonic on the same interval.Lemma 2.4.
Letai and bi(i=1,…,n) be real numbers such that 0<a1≤⋯≤an, 0<b1≤⋯≤bn, and ∑i=1kai≤∑i=1kbi for k=1,…,n. If the function f(x) is decreasing and convex on (0,+∞), then
(2.6)∑i=1nf(bi)≤∑i=1nf(ai).
If ∑i=1nai=∑i=1nbi, then one only needs f(x) 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).
Letf(x) be a convex function on [a,b], then
(2.7)f(a+b2)≤1b-a∫abf(x)dx≤f(a)+f(b)2.Lemma 2.6.
For real numbers0<s,q<1 and any integer n≥1,
(2.8)(qs-q(1-s)(1-q))n≥qns-qn(1-s)(1-qn).Proof.
We can recast the above inequality as(2.9)E(-1,s-1;q,1)≥E(-n,n(s-1);q,1),
where for positive numbers x,y and real numbers r,s with rs(r-s)(x-y)≠0, we define
(2.10)E(r,s;x,y)=(rs⋅ys-xsyr-xr)1/(s-r).
The assertion of the lemma now follows from the fact that E(r,s;x,y) is increasing in both x and y and in both r and s (see, e.g., Theorem  1 in [17] for a proof of this).Lemma 2.7.
Letm>n≥1 be two integers, then for any fixed constant 0<c<1, the function
(2.11)a(t;m,n,c)=tm-n+tn-c(1+tm)
has exactly one root when t≥1.Proof.
We have(2.12)a′(t;m,n,c)=tm-1((m-n)t-n+ntn-m-cm).
The function t↦(m-n)t-n+ntn-m-cm is clearly decreasing when t≥1. By considering the cases t=1 and t→+∞ we conclude that a′(t;m,n,c) has exactly one root when t≥1. It follows from this and Cauchy's mean value theorem that a(t;m,n,c) has at most two roots when t≥1. This combined with the observation that a(1;m,n,c)>0 and limt→+∞a(t;m,n,c)<0 yields the desired conclusion.Lemma 2.8.
Fort≥s≥0, one has
(2.13)s1-e-s⋅t-s1-e-(t-s)≥t1-e-t.Proof.
We write(2.14)f(x)=x1-e-x
and we observe that
(2.15)(f′(t)f(t))′=e-t(t2-(et/2-e-t/2)2)t2(1-e-t)2≤0,
where the last inequality follows from
(2.16)et/2-e-t/2≥t,t≥0.
The above inequality follows from the fact that the derivative of sinh(t)-t is cosh(t)-1>0 for t>0 and that sinh(0)-0=0. We now deduce that
(2.17)f′(t-s)f(t-s)-f′(t)f(t)≥0,
for t≥s≥0. This implies that the function t↦lnf(t-s)-lnf(t) is increasing for t>s. Thus we get
(2.18)ln⁡f(s)+ln⁡f(t-s)-ln⁡f(t)≥lim⁡t→s+(ln⁡f(s)+ln⁡f(t-s)-ln⁡f(t))=0,
which is the desired result.


## 3. Main Results

Theorem 3.1.
Letai and bi(i=1,…,n) be real numbers such that 0≤a1≤⋯≤an, 0≤b1≤⋯≤bn, and ∑i=1kai≤∑i=1kbi for k=1,…,n. If f′′(x) is completely monotonic on (0,+∞), then
(3.1)exp⁡(∑i=1n(f(x+ai)-f(x+bi)))
is logarithmically completely monotonic on (0,+∞).Proof.
It suffices to show that(3.2)-∑i=1n(f′(x+ai)-f′(x+bi))
is completely monotonic on (0,+∞) or for k≥1,
(3.3)(-1)k∑i=1nf(k)(x+ai)≥(-1)k∑i=1nf(k)(x+bi).
By Lemma 2.4, it suffices to show that (-1)kf(k)(x) is decreasing and convex on (0,+∞) or equivalently, (-1)kf(k+1)(x)≤0 and (-1)kf(k+2)(x)≥0 for k≥1. The last two inequalities hold since we assume that f′′(x) is completely monotonic on (0,+∞). This completes the proof.As a direct consequence of Theorem3.1, we now generalize a result of Alzer [2, Theorem  10].Corollary 3.2.
Letai and bi (i=1,…,n) be real numbers such that 0≤a1≤⋯≤an, 0≤b1≤⋯≤bn, and ∑i=1kai≤∑i=1kbi for k=1,…,n. Then,
(3.4)x⟼∏i=1nΓq(x+ai)Γq(x+bi)
is logarithmically completely monotonic on (0,+∞).Proof.
Apply Theorem3.1 to f(x)=lnΓq(x) and note that f′′(x)=ψq′(x) is completely monotonic on (0,+∞) and this completes the proof.Theorem 3.3.
Letf′′(x) be completely monotonic on (0,+∞), then for 0≤s≤1, the functions
(3.5)x⟼exp⁡(-(f(x+1)-f(x+s)-(1-s)f′(x+1+s2))),x⟼exp⁡(f(x+1)-f(x+s)-(1-s)2(f′(x+1)+f′(x+s)))
are logarithmically completely monotonic on (0,+∞).Proof.
We may assume0≤s<1. We will prove the first assertion and the second one can be shown similarly. It suffices to show that
(3.6)f′(x+1)-f′(x+s)-(1-s)f′′(x+1+s2)
is completely monotonic on (0,+∞) or for k≥1,
(3.7)11-s∫x+sx+1(-1)k+1f(k+1)(t)dt≥(-1)k+1f(k+1)(x+1+s2).
The last inequality holds by Lemma 2.5 and our assumption that f′′(x) is completely monotonic on (0,+∞). This completes the proof.Corollary 3.4.
For0≤s≤1, the functions
(3.8)x⟼Γq(x+s)Γq(x+1)exp⁡((1-s)ψq(x+1+s2)),x⟼Γq(x+1)Γq(x+s)exp⁡(-(1-s)2(ψq(x+1)+ψq(x+s)))
are logarithmically completely monotonic on (0,+∞).Proof.
Apply Theorem3.3 to f(x)=lnΓq(x) and note that f′′(x)=ψq′(x) is completely monotonic on (0,+∞) and this completes the proof.By applying Lemma2.5 to f(x)=-ψq(x), we obtain the following theorem.Theorem 3.5.
For positivex and 0≤s≤1,
(3.9)exp⁡((1-s)2(ψq(x+1)+ψq(x+s)))≤Γq(x+1)Γq(x+s)≤exp⁡((1-s)ψq(x+1+s2)).The upper bound in Theorem3.5 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<q<1.Theorem 3.6.
Let0<s<1 and 0<q<1. Let u(q,s) be defined by (1.7) and let the function gq(x;s,1,u(q,s)) be defined as in Theorem 1.1. Then gq(x;s,1,u(q,s)) is logarithmically completely monotonic on (0,+∞).Proof.
Define(3.10)hq(x)=-ln⁡gq(x;s,1,u(q,s))=-ln⁡Γq(x+1)+ln⁡Γq(x+s)+(1-s)ln⁡(1-qx+u(q,s)1-q).
It suffices to show that hq′(x) is completely monotonic on (0,+∞). We have
(3.11)hq′(x)=-ψq(x+1)+ψq(x+s)-(1-s)ln⁡qqx+u(q,s)1-qx+u(q,s).
Using the expression (2.1), we can rewrite hq′(x) as
(3.12)hq′(x)=-ln⁡q(∑n=1∞qnx(qn-qns)1-qn+(1-s)qx+u(q,s)1-qx+u(q,s)),0<q<1.
Expanding (1-qx+u(q,s))-1, we may further rewrite hq′(x) as
(3.13)hq′(x)=-ln⁡q∑n=1∞qnx1-qnwq,n(s),0<q<1,
where
(3.14)wq,n(c)=qn-qns+(1-s)qnu(q,s)(1-qn).
In order for hq′(x) to be completely monotonic on (0,+∞), it suffices to show wq,n(s)≥0 for 0<s<1. This is just Lemma 2.6 and this completes the proof.Theorem3.6 implies that gq(x;s,1,u(q,s))>limx→+∞gq(x;s,1,u(q,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<q<1.Forα≥0, let(3.15)fα(x)=-ln⁡Γ(x)+(x-12)ln⁡x-x+112ψ′(x+α).
Alzer [19, Theorem  1] showed that for fixed 0<s<1, the function exp(-(fα(x+s)-fα(x+1))) is strictly completely monotonic on (0,+∞) if and only if α≥1/2 and the function exp(fα(x+s)-fα(x+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 fα′(x) is strictly completely monotonic on (0,+∞) if and only if α≥1/2 and that -fα′(x) is strictly completely monotonic on (0,+∞) if and only if α=0. We now establish the above assertions in the following.Proposition 3.7.
Forα≥0, let fα(x) be defined as in (3.15). Then fα′(x) is strictly completely monotonic on (0,+∞) if and only if α≥1/2 and -fα′(x) is strictly completely monotonic on (0,+∞) if and only if α=0.Proof.
Asψ′(x) is completely monotonic on (0,+∞), we may just focus on the case α=1/2 or 0. We have
(3.16)fα′(x)=-ψ(x)+ln⁡x-12x+112ψ′′(x+α),fα′′(x)=-ψ′(x)+1x+12x2+112ψ(3)(x+α).
Using the asymptotic expressions (2.4) and (2.5), we see that limx→+∞fα(n)(x)=0 for any integer n≥0. It is then easy to see that the “if” part of the assertions of the proposition will follow if we can show that f1/2′′(x+1)-f1/2′′(x) is strictly completely monotonic on (0,+∞) and f0′′(x)-f0′′(x+1) is completely monotonic on (0,+∞). Using (2.3), it is easy to see that
(3.17)f1/2′′(x+1)-f1/2′′(x)=x2+x+1/84x2(1+x)2(x+1/2)4=14(1+x)2(x+1/2)4(1+1x+18x2).
It is easy to see from this and the ring structure of completely monotonic functions on (0,+∞) that f1/2′′(x+1)-f1/2′′(x) is strictly completely monotonic on (0,+∞). Similarly, one shows that f0′′(x)-f0′′(x+1) is completely monotonic on (0,+∞).
To prove the “only if” part of the assertions of the proposition, we note that(3.18)fα′′(x+1)-fα′′(x)=(4α-2)x3+(6α2-1)x2+4α3x+α42x2(1+x)2(x+α)4,
from which we see easily by considering the case x→+∞ that fα′′(x) fails to be increasing if α<1/2. Similarly, by considering the case x→0+ that fα′′(x) fails to be decreasing if α>0 and this completes the proof.


## 4. Some Inequalities Involving Polygamma Functions

For integersp≥m≥n≥q≥0 and any real number c, we define (4.1)Fp,m,n,q(x;c)=(-1)m+nψ(m)(x)ψ(n)(x)-c(-1)p+qψ(p)(x)ψ(q)(x).
Here we set ψ(0)(x)=-1 for convenience.Forn≥2, a result of Alzer and Wells [20, Theorem  2.1] asserts that the function Fn+1,n,n,n-1(x;c) is completely monotonic on (0,+∞) if and only if c≤(n-1)/n and -Fn+1,n,n,n-1(x;c) is completely monotonic on (0,+∞) if and only if c≥n/(n+1).We denote(4.2)cp,m,n,q=(m-1)!(n-1)!(p-1)!(q-1)!,q≥1;cp,m,n,0=(m-1)!(n-1)!(p-1)!,dp,m,n,q=m!n!p!q!.
and note that 0<cp,m,n,q,dp,m,n,q<1 when p+q=m+n,p>m. We now extend the result of Alzer and Wells to the following.Theorem 4.1.
Letp>m≥n>q≥0 be integers satisfying m+n=p+q. The function Fp,m,n,q(x;cp,m,n,q) is completely monotonic on (0,+∞). The function -Fp,m,n,q(x;dp,m,n,q) is also completely monotonic on (0,+∞) when q>0.Proof.
We first prove the assertion forFp,m,n,q(x;cp,m,n,q) with q≥1. The proof here uses the method in [20]. Using the integral representation (2.2) for (-1)n+1ψ(n)(x) and using * for the Laplace convolution, we get
(4.3)Fp,m,n,q(x;cp,m,n,q)=∫0∞e-xtg(t)dt,
where
(4.4)g(t)=tm1-e-t*tn1-e-t-cp,m,n,qtp1-e-t*tq1-e-t=∫0t((t-s)msn-cp,m,n,q(t-s)psq)h(t-s)h(s)ds,
with
(4.5)h(s)=11-e-s.
It suffices to show that g(t)≥0. By change of variables s→ts we can recast it as
(4.6)g(t)=tm+n+1∫01((1-s)msn-cp,m,n,q(1-s)psq)h(t(1-s))h(ts)ds.
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→(1-s)/2 for the first one and s→(1+s)/2 for the second one. We now combine them to get
(4.7)g(t)=(t2)m+n+1∫01a(1+s1-s;p-q,n-q,cp,m,n,q)(1-s2)q(1-s)p-q×h(t(1-s)2)h(t(1+s)2)ds,
where the function a(t;m,n,c) is defined as in Lemma 2.7. Note that (1+s)/(1-s)≥1 for 0≤s<1, hence by Lemma 2.7, there is a unique number 0<s0<1 such that
(4.8)a(1+s01-s0;p-q,n-q,cp,m,n,q)=0.
We further note it is shown in the proof of [20, Lemma  2.2] that the function
(4.9)s⟼(1-s2)h(t(1-s)2)h(t(1+s)2)
is a decreasing function on (0,1) so that for 0≤s≤1,
(4.10)a(1+s1-s;p-q,n-q,cp,m,n,q)(1-s2)q(1-s)p-qh(t(1-s)2)h(t(1+s)2)≥a(1+s1-s;p-q,n-q,cp,m,n,q)(1-s2)q-1(1-s)p-q(1-s02)×h(t(1-s0)2)h(t(1+s0)2).
Hence
(4.11)g(t)≥(t2)m+n+1(1-s02)h(t(1-s0)2)h(t(1+s0)2)⋅∫01a(1+s1-s;p-q,n-q,cp,m,n,q)(1-s2)q-1(1-s)p-qds.
Note that the integral above is (by reversing the process above on changing variables)
(4.12)2m+n-1∫01((1-s)m-1sn-1-cp,m,n,q(1-s)p-1sq-1)ds=0,
where the last step follows from the well-known beta function identity
(4.13)B(x,y)=∫01tx-1(1-t)y-1dt=Γ(x)Γ(y)Γ(x+y),x,y>0,
and the well-known fact Γ(n)=(n-1)! for n≥1.
Now we prove the assertion forFp,m,n,0(x;cp,m,n,0). In this case p=m+n and we note that
(4.14)cm+n,m,n,0=B(m,n)=∫01sm-1(1-s)n-1ds,
and we use this to write
(4.15)cm+n,m,n,0tm+n1-e-t=∫0tsm-1(t-s)n-1t1-e-tds.
It follows that
(4.16)Fp,m,n,0(x;cp,m,n,0)=∫0∞e-xt(tm1-e-t*tn1-e-t-cm+n,m,n,0tm+n1-e-t)dt=∫0∞e-xt(∫0tsm-1(t-s)n-1(s1-e-s⋅t-s1-e-(t-s)-t1-e-t)ds)dt≥0,
where the last inequality follows from Lemma 2.8.
It remains to show the assertion for-Fp,m,n,q(x;dp,m,n,q). In this case we use the series representation in (2.2) for (-1)n+1ψ(n)(x) to get
(4.17)-Fp,m,n,q(x;dp,m,n,q)=m!n!((∑i=0∞1(x+i)p+1)(∑j=0∞1(x+j)q+1)-(∑i=0∞1(x+i)m+1)(∑j=0∞1(x+j)n+1)).
We note the following Binet-Cauchy identity:
(4.18)(∑i=0naici)(∑i=0nbidi)-(∑i=0naidi)(∑i=0nbici)=∑0≤i<j≤n(aibj-ajbi)(cidj-cjdi).
We now apply the above identity with
(4.19)ai=1(x+i)m+1,bi=1(x+i)q+1,ci=1(x+i)p-m,di=1
to get
(4.20)-Fp,m,n,q(x;dp,m,n,q)=m!n!∑0≤i<j(1(x+i)m+1(x+j)q+1-1(x+j)m+1(x+i)q+1)(1(x+i)p-m-1(x+j)p-m).
We note that the second factor on the right-hand side above is completely monotonic on (0,+∞) and also that
(4.21)1(x+i)m+1(x+j)q+1-1(x+j)m+1(x+i)q+1=1(x+i)q+11(x+j)q+1(1(x+i)m-q-1(x+j)m-q).
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 -Fp,m,n,q(x;dp,m,n,q) is completely monotonic on (0,+∞).We note here that whenq=0, the function -Fm+n,m,n,0(x;dm+n,m,n,0) is not completely monotonic on (0,+∞) in general, as we observe for example that it follows from (2.5) that as x→+∞, -F2,1,1,0(x;d2,1,1,0)<0.Corresponding tom=n=1,p=2,q=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):(4.22)(ψ′(x))2+ψ′′(x)≥0,x>0.
We note that inequality (4.22) follows from the limiting case c→0 of the following inequalities for any 0<c<1 and x>0:(4.23)1c(ψ(x+c)-ψ(x))2>ψ′(x)-ψ′(x+c)>(ψ(x+c)-ψ(x))2,
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 Δ0,c;1(x) or Δ0,c;1/c(x) 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 aq-analogue to (4.23).Theorem 4.2.
Let0<q<1 and 0<c<1 be fixed. Then for any x>0,
(4.24)1-q1-qc(ψq(x+c)-ψq(x))2>qx(ψq′(x)-ψq′(x+c))>(ψq(x+c)-ψq(x))2.
The above inequalities reverse when c>1.Proof.
We first prove the left-hand side inequality of (4.24). For this, we define
(4.25)f(x)=qx(ψq′(x+c)-ψq′(x))+1-q1-qc(ψq(x+c)-ψq(x))2.
Applying (2.1), we obtain
(4.26)f(x)=(ln⁡q)2∑n=1∞nq(n+1)x(qnc-1)1-qn+1-q1-qc(ln⁡q∑n=1∞qnx(qnc-1)1-qn)2=(1-q)(ln⁡q)21-qc∑n=2∞qnxgn(c;q),
where
(4.27)gn(c;q)=∑k=1n-1((1-qkc)(1-q(n-k)c)(1-qk)(1-qn-k)-(1-qc)(1-q(n-1)c)(1-q)(1-qn-1)).
It suffices to show that gn(c;q)>0 for 0<c<1 and gn(c;q)<0 for c>1 when n≥2. For this, we let y=qc so that 0<y<1 and it suffices to show the function (1-yk)(1-yn-k)/((1-y)(1-yn-1)) is increasing for 0<y<1 and 1≤k≤n-1. On taking the logarithmic derivative of the above function, we see that it suffices to show that h(k;y)+h(n-k;y)≤h(1;y)+h(n-1;y), where
(4.28)h(z;y)=z1-yz.
We now regard h(z;y) as a function of z and note that
(4.29)h′′(z;y)=(ln⁡y)yz(1-yz)3u(yz),u(t)=2-2t+ln⁡t+tln⁡t.
It is easy to see that u′(t)>0 for 0<t<1 so that u(t)<u(1)=0 for 0<t<1. It follows that h′′(z;y)>0. We then deduce from this and Lemma 2.4 that h(k;y)+h(n-k;y)≤h(1;y)+h(n-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 (1-yk)(1-yn-k)/(1-yn-1) is decreasing for 0<y<1 and 1≤k≤n-1. This follows from the observation that both functions (1-yk)/(1-yn-1) and 1-yn-k are decreasing and this completes the proof.



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*Source: 375715-2010-12-30.xml*