Some Monotonicity Properties of Gamma and $q$-gamma Functions

We prove some properties of completely monotonic functions and apply them to obtain results on gamma and $q$-gamma functions.


Introduction
The q-gamma function is defined for positive real numbers x and q = 1 by We note here [29] the limit of Γ q (x) as q → 1 − gives back the well-known Euler's gamma function: Note also from the definition, for positive x and 0 < q < 1, we see that lim q→1 Γ q (x) = Γ(x). For historical remarks on gamma and q-gamma functions, we refer the reader to [29], [3] and [6]. There exists an extensive and rich literature on inequalities for the gamma and q-gamma functions. For the recent developments in this area, we refer the reader to the articles [25], [3], [6]- [8], [32] and the references therein. Many of these inequalities follow from the monotonicity properties of functions which are closely related to Γ (resp. Γ q ) and its logarithmic derivative ψ (resp. ψ q ). 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) k f (k) (x) ≥ 0, x ∈ (a, b), k ≥ 0. Lemma 2.4 below asserts that f (x) = e −h(x) is completely monotonic on an interval if h ′ is. Following [23], we call such functions f (x) logarithmically completely monotonic.
We note here that lim q→1 ψ q (x) = ψ(x) (see [30]) and that ψ ′ and ψ ′ q are completely monotonic functions on (0, +∞) (see [26], [8]). 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 ψ ′ q . 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 3], [25,Theorem 2.5]). Let a + 1 ≥ b > a, α = max(−a, −c) and for q > 0, define where Γ 1 (x) = Γ(x). Then g q (x; a, b, c) is logarithmically completely monotonic on (α, +∞) if c ≤ (a + b − 1)/2 and 1/g q (x; a, b, c) is logarithmically completely monotonic on (α, +∞) if c ≥ a.
It follows immediately from the above theorem that for 0 < q < 1 and 0 < s < 1, one has [25] for x > 0: Alzer later [6] determined the best values u(q, s), v(q, s) such that the inequalities: Motivated by the above results, we will show that the function g q (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.
Let P n,k (1 ≤ k ≤ n) be the set of all vectors m = (m 1 , . . . , m k ) whose components are natural numbers such that 1 ≤ m ν < m µ ≤ n for 1 ≤ ν < µ ≤ k and let P n,0 be the empty set. Recently, Grinshpan and Ismail [23] proved the following result: .
Then F n (x, q) is a logarithmically completely monotonic function of x on (0, +∞). Here we define We will show in Section 4 that the above theorem follows from a simple observation on logarithmically completely monotonic functions which also enables us to deduce some other useful inequalities.
The derivatives of ψ(x) are known as polygamma functions and in Section 5 we will prove some inequalities involving the polygamma functions.
Formulas for the volumes of geometric bodies sometimes involve the gamma function. One then expects to obtain inequalities involving such volumes via relevant inequalities involving gamma functions. In Section 6, we will have a brief discussion on inequalities for the volume of the unit ball in R n .
The set of all the completely monotonic functions on an interval can be 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 function, part of Lemma 2.3 is contained in [25, Lemma 1.3]: is also completely monotonic on the same interval.
Lemma 2.5. Let a i and b i (i = 1, . . . , n) be real numbers such that 0 < a 1 ≤ · · · ≤ a n , If the function f (x) is decreasing and convex on (0, +∞), then then one only needs f (x) to be convex for the above inequality to hold. The above lemma is similar to Lemma 2 in [3], except here we assume a i , b i 's to be positive and f (x) defined on (0, +∞). We leave the proof to the reader by pointing out that it follows from the theory of majorization, for example, see the discussions in Chap. 1, §28 − §30 of [15]. .
The function r → L r (a, b) is strictly increasing on R.
If we set u = 1 − s, y = q x so that 0 < u < 1 and 0 < y < 1 and set Then we have g ′ (u) = uy u (1 − y u ) 3 (2 + ln y u )(1 − y u ) + 2y u ln y u . It is easy to check that the function h(w) = 2(1 − w) + (1 + w) ln w is concave on (0, 1). This combined with the observation that h ′ (1) = 0 implies that h ′ (w) ≥ 0 for 0 < w ≤ 1 and hence h(w) ≤ h(1) = 0 for 0 < w ≤ 1. Apply this with w = q u implies that g ′ (u) ≤ 0. We deduce from this that f ′′ (x) ≥ 0. Hence it remains to show that f ′ (1) ≥ 0. In this case we regard f ′ (1) as a function of s with 0 < s < 1 and define It suffices to show a(s) ≥ 0 and calculation yields If we set z = q s−1 so that z ≥ 1 and define b(z) = (z − 1) 2 − z(ln z) 2 , then it is easy to check that b(z) is convex for z ≥ 1. As b ′ (1) = 0, this implies that b ′ (z) ≥ 0 for z ≥ 1 so that b(z) ≥ b(1) = 0 for z ≥ 1. It follows from this that a ′ (s) ≥ 0 so that a(s) ≥ a(0) = 0 and this completes the proof.
Lemma 2.9. Let m > n ≥ 1 be two integers, then for any fixed constant 0 < c < 1, the function has exactly one root when t ≥ 1.
The function t → (m − n)t −n + nt n−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 lim t→+∞ a(t; m, n, c) < 0 yields the desired conclusion.
Proof. We write where the last inequality follows from The above inequality can be established by using the Taylor expansion of e t/2 − e −t/2 at t = 0. We now deduce that which is the desired result.
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 is completely monotonic on (0, +∞) or for k ≥ 1, The last inequality holds by Lemma 2.6 and our assumption that f ′′ (x) is completely monotonic on (0, +∞). This completes the proof.
By applying Lemma 2.6 to f (x) = −ψ q (x), we obtain The upper bound in Theorem 3.3 is due to Ismail and Muldoon [25]. Our proof here is similar to that of Corollary 3 in [31], which asserts for positive x, We further note the following integral analogue of Theorem 3.5 in [8]: It follows from this that for positive x and 0 ≤ s ≤ 1, We note that the left-hand side inequality above is contained in [20,Lemma 1] and [19,Theorem 4]. Now by Lemma 2.7, observing that Note that Alzer [8,Lemma 2.4] has shown that ψ(e x ) is strictly concave on R and it follows that also note that ψ(x) is an increasing function on (0, +∞) and (x + 1)(x + s) ≥ x + s 1/2 , we see that the inequalities in (3.2) refine the case q → 1 in Theorem 3.3 and the following result of Kershaw [28], which states that for positive x and 0 ≤ s ≤ 1, We now show the lower bound above and the corresponding one in (3.1) are not comparable in general (see [21, p. 856] for a similar discussion). In fact, by Theorem 3.7 of [8], we have This gives one inequality on letting x → 0. On the other hand, using the well-known series representation (see, for example, [25, (1.8)]): with γ = 0.57721 . . . denoting Euler's constant, we obtain for x > 1, Our next result refines the left-hand side inequality of (1.2): Theorem 3.4. Let 0 < s < 1 and 0 < q < 1. Let u(q, s) be defined by (1.3) and let the function g q (x; s, 1, u(q, s)) be defined as in Theorem 1.1. Then g q (x; s, 1, u(q, s)) is logarithmically completely monotonic on (0, +∞).

Proof.
Define It suffices to show that h ′ q (x) is completely monotonic on (0, +∞). We have where w q,n (c) = q n − q ns + (1 − s)q nu(q,s) (1 − q n ). In order for h ′ q (x) to be completely monotonic on (0, +∞), it suffices to show w q,n (s) ≥ 0 for 0 < s < 1. This is just Lemma 2.8 and this completes the proof.
The above theorem implies that g q (x; s, 1, u(q, s)) > lim x→+∞ g q (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 (1.2).

Applications of Lemma 2.3
In this section, we look at a few applications of Lemma 2.3 and we first deduce Theorem 1.2 from it. As was noted in [23], we have F n (x, q) = F n−1 (x, q)/F n−1 (x + a n , q) if F n and F n−1 are defined by the same parameters a k . It follows from Lemma 2.3 by induction that we only need to prove the case n = 1, which follows from [23, Lemma 2.1, 3.1].
We remark here the above corollary is essentially Theorem 4 and 5 in [23], except that Theorem 4 of [16] is originally stated as the function is completely monotonic on (1/2, +∞). One can also obtain this result by modifying our approach above.
We now look at some functions involving the q-gamma function. Ismail et al. [24,Theorem 2.2] proved that for q ∈ (0, 1) the function (1 − q) x Γ q (x) is logarithmically completely monotonic on (0, +∞). It follows from this and Lemma 2.
Proof. As φ ′ (x) is completely monotonic on (0, +∞), we may just focus on the case α = 1/2 or 0. We have Using the asymptotic expressions (2.5) and (2.6), we see that lim x→+∞ f (n) α (x) = 0 for any integer n ≥ 0. It is then easy the see that the assertions of the proposition will follows if we can show that f ′′ 1/2 (x + 1) − f ′′ 1/2 (x) is strictly completely monotonic on (0, +∞) and f ′′ 0 (x) − f ′′ 0 (x + 1) is completely monotonic on (0, +∞). Using (2.4), it is easy to see that It is easy to see from this and the ring structure of completely monotonic functions on (0, +∞) that f ′′ 1/2 (x + 1) − f ′′ 1/2 (x) is strictly completely monotonic on (0, +∞). Similarly, one shows that is completely monotonic on (0, +∞) and this completes the proof. We note here that recently, Alzer and Batir [10] showed that the following function (x > 0, c ≥ 0) is completely monotonic if and only if c ≥ 1/3 and −G c (x) is completely monotonic if and only if c = 0. We point out here that using the idea in the proof of Proposition 4.1, one can give another proof of the above result of Alzer and Batir and we shall leave the details to the reader.
We denote and d p,m,n,q = m!n! p!q! .
and note that 0 < c p,m,n,q , d p,m,n,q < 1 when p + q = m + n, p > m. We now extend the result of Alzer and Wells to the following Theorem 5.1. Let p > m ≥ n > q ≥ 0 be integers satisfying m + n = p + q. The function F p,m,n,q (x; c p,m,n,q ) is completely monotonic on (0, +∞). The function −F p,m,n,q (x; d p,m,n,q ) is also completely monotonic on (0, +∞) when q > 0, .
Proof. We first prove the assertion for F p,m,n,q (x; c p,m,n,q ) with q ≥ 1 and the proof here uses the method in [13]. Using the integral representation (2.3) for (−1) n+1 ψ (n) (x) and using * for the Laplace convolution, we get It suffices to show g(t) ≥ 0 and by a change of variable s → ts we can recast it as 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 where the function a(t; m, n, c) is as defined in Lemma 2.9. Note that (1 + s)/(1 − s) ≥ 1 for 0 ≤ s < 1, hence by Lemma 2.9, there is a unique number 0 < s 0 < 1 such that a 1 + s 0 1 − s 0 ; p − q, n − q, c p,m,n,q = 0.
We further note it is shown in the proof of [13, Lemma 2.2] that the function 2 is a decreasing function on (0, 1) so that for 0 ≤ s ≤ 1, Hence Note that the integral above is (by reversing the process above on changing variables) It follows that F p,m,n,0 (x; c p,m,n,0 ) where the last inequality follows from Lemma 2.10.
It remains to show the assertion for −F p,m,n,q (x; d p,m,n,q ). In this case we use the series repre- We note the following Binet-Cauchy identity: We now apply the above identity with We note that the second factor on the right-hand side above is completely monotonic on (0, +∞) and also that 1 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 above is also completely monotonic on (0, +∞). Hence by the ring structure of completely monotonic functions on (0, +∞) again we deduce that −F p,m,n,q (x; d p,m,n,q ) is completely monotonic on (0, +∞).
We point out here corresponding to m = n = 1, p = 2, q = 0 in Theorem 5.1, it was shown in the proof of [8, (4.39)] and in [14, Lemma 1.1] the following special case (in fact with strict inequality) We now give another proof of the above inequality by establishing the following Proposition 5.1. Let 0 < c < 1 be fixed. Then for any x > 0, The above inequalities reverse when c > 1.
Proof. We first prove the left-hand side inequality of (5.2). For this, we define Applying the relations (2.4), we obtain We now apply (2.3) to rewrite the last expression above as We now define for t > 0, Similarly g ′ (t) ≤ 0 when c > 1. From now on we will assume 0 < c < 1 and the case c > 1 can be discussed similarly. We conclude that g(t) > g(0) = 0 for t > 0 when 0 < c < 1 and it follows that f (x + 1) − f (x) < 0 in this case. As lim x→+∞ f (x) = 0, we conclude that f (x) > 0 which completes the proof of the left-hand side inequality of (5.2). The proof of the right-hand side inequality of (5.2) follows similarly except it is easier and we shall leave it to the reader.
We note here the left-hand side inequality of (5.2) was established in the proof of [17, Theorem 1.1] and we rewrite it as for x > 0, 0 < c < 1. From this we get back (5.1) via taking the limit c → 0. We note also that the right-hand side inequality was proven in [9, Lemma 7] and our proof here is simpler for both inequalities of (5.2). We now prove a q-analogue to Proposition 5.1: Theorem 5.2. Let 0 < q < 1 and 0 < c < 1 be fixed. Then for any x > 0, The above inequalities reverse when c > 1.
Proof. We first prove the left-hand side inequality of (5.3). For this, we define Applying (2.2), we obtain It suffices to show that g c (x; q) > 0 for 0 < c < 1 and g c x; q) < 0 for c > 1 when n ≥ 3. For this, we let y = q c so that 0 < y < 1 and it suffices to show the function (1−y k )(1−y n−k )/((1−y)(1−y n−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 h(z; y) = z 1 − y z . We now regard h(z; y) as a function of z and note that 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 2.5 that h(k; y)+h(n−k; y) ≤ h(1; y)+h(n−1; y) holds and the completes the proof for the left-hand side inequality of (5.3). For the right-hand side inequality of (5.3), one proceeds similarly to the above argument to see that it suffices to show the function (1 − y k )(1 − y n−k )/(1 − y n−1 ) is decreasing for 0 < y < 1 and 1 ≤ k ≤ n − 1. This follows from the observation that both functions (1 − y k )/(1 − y n−1 ) and 1 − y n−k are decreasing and this completes the proof.
Similar to our discussion above, we have Corollary 5.1. For 0 < q < 1 and x > 0, A result of Alzer [7, Lemma 2] (see also [8, Lemma 2.1]) asserts that the function is strictly decreasing from [0, +∞) onto (n, n + 1] for any integer n ≥ 1. This result implies the second assertion in Lemma 2.2 and that the function x → x n+1 (−1) n+1 ψ (n) (x) is strictly increasing for x > 0. In a similar fashion, we have Proposition 5.2. Let a ≥ 1/2 be fixed, the for any integer n ≥ 1, the function r n,a (x) = xψ (n+1) (x + a) ψ (n) (x + a) is a decreasing function for x ≥ 0. with c 1 , d 1 constants and c 1 as previously defined. One checks easily that (t/(1 − e −t )) ′′ > 0 for t > 0 and this gives another proof of Alzer's result mentioned above.
Proof. Differentiation gives It follows from (2.2) that for n ≥ 1, 0 < q < 1, It follows from this that where we define the empty sum to be 0. As the function q → (1 − q k )/(1 − q m ) is a decreasing function for 0 < q < 1 with 1 ≤ k ≤ m. We conclude that for 0 < q < 1, where the last inequality follows easily by induction on k and this completes the proof. 6. Inequalities for the Volume of the Unit Ball in R n In this section, we apply some of our results in the previous sections to study inequalities for the volume Ω n of the unit ball in R n : Ω n = π n/2 Γ(1 + n/2) .
There exists many inequalities involving Ω n , we refer the reader to [4] and the references therein.