C om mun.Fac.Sci.U niv.A nk.Series A 1 Volum e 66, N umb er 2, Pages 130–140 (2017) D O I: 10.1501/C om mua1_ 0000000807 ISSN 1303–5991
http://com munications.science.ankara.edu.tr/index.php?series= A 1
CONVEXITY PROPERTIES AND INEQUALITIES CONCERNING
THE (p; k)-GAMMA FUNCTION
KWARA NANTOMAH
Abstract. In this paper, some convexity properties and some inequalities for the (p; k)-analogue of the Gamma function, p;k(x)are given. In particular, a (p; k)-analogue of the celebrated Bohr-Mollerup theorem is given. Further-more, a (p; k)-analogue of the Riemann zeta function, p;k(x)is introduced and some associated inequalities are derived. The established results provide the (p; k)-generalizations of some known results concerning the classical Gamma function.
1. Introduction
In a recent paper [10], the authors introduced a (p; k)-analogue of the Gamma function de…ned for p 2 N, k > 0 and x 2 R+ as
p;k(x) = Z p 0 tx 1 1 tk pk p dt (1.1) = (p + 1)!k p+1(pk)x k 1 x(x + k)(x + 2k) : : : (x + pk) (1.2) satisfying the basic properties
p;k(x + k) = pkx x + pk + k p;k(x); (1.3) p;k(ak) = p + 1 p k a 1 p(a); a 2 R+ p;k(k) = 1:
Received by the editors: August 16, 2016; Accepted: January 20, 2017.
2010 Mathematics Subject Classi…cation. Primary: 33B15, Secondary: 33E50, 26A51. Key words and phrases. (p; k)-Gamma function, convex functions, Bohr-Mollerup theorem, (p; k)-Riemann zeta function, inequality.
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The (p; k)-analogue of the Digamma function is de…ned for x > 0 as p;k(x) = d dxln p;k(x) = 1 kln(pk) p X n=0 1 nk + x (1.4) = 1 kln(pk) Z 1 0 1 e k(p+1)t 1 e kt e xtdt:
Also, the (p; k)-analogue of the Polygamma functions are de…ned as
(m) p;k(x) = dm dxm p;k(x) = p X n=0 ( 1)m+1m! (nk + x)m+1 (1.5) = ( 1)m+1 Z 1 0 1 e k(p+1)t 1 e kt t me xtdt where m 2 N, and (0)p;k(x) p;k(x).
The functions p;k(x) and p;k(x) satisfy the following commutative diagrams.
The (p; k)-analogue of the classical Beta function is de…ned as Bp;k(x; y) = p;k
(x) p;k(y) p;k(x + y)
; x > 0; y > 0: (1.6) The purpose of this paper is to establish some convexity properties and some in-equalities involving the function, p;k(x). In doing so, a (p; k)-analogue of the
Bohr-Mollerup theorem is proved. Also, a (p; k)-analogue of the Riemann zeta function, p;k(x) is introduced and some associated inequalities relating p;k(x)
and p;k(x) are derived. We present our …ndings in the following sections. 2. Convexity Properties Involving the (p; k)-Gamma function Let us begin by recalling the following basic de…nitions and concepts. De…nition 1. A function f : (a; b) ! R is said to be convex if
f ( x + y) f (x) + f (y) (2.1)
for all x; y 2 (a; b), where ; > 0 such that + = 1.
Lemma 1. Let f : (a; b) ! R be a twice di¤erentiable function. Then f is said to be convex if and only if f00(x) 0 for every x 2 (a; b).
Remark 1. A function f is said to be concave if f is convex, or equivalently, if the inequality (2.1) is reversed.
De…nition 2. A function f : (a; b) ! R+is said to be logarithmically convex if the
inequality
log f ( x + y) log f (x) + log f (y) or equivalently
f ( x + y) (f (x)) (f (y)) holds for all x; y 2 (a; b) and ; > 0 such that + = 1. Theorem 1. The function, p;k(x) is logarithmically convex.
Proof. Let x; y > 0 and ; > 0 such that + = 1. Then, by the integral representation (1.1) and by the Hölder’s inequality for integrals, we obtain
p;k( x + y) = Z p 0 t x+ y 1 1 t k pk p dt = Z p 0 t (x 1)t (y 1) 1 t k pk p( + ) dt = Z p 0 t (x 1) 1 t k pk p t (y 1) 1 t k pk p dt Z p 0 tx 1 1 t k pk p dt Z p 0 ty 1 1 t k pk p dt = ( p;k(x)) ( p;k(y)) as required.
Remark 2. Since every logarithmically convex function is also convex [13, p. 66], it follows that the function p;k(x) is convex.
Remark 3. Theorem 1 was proved in [10] by using a di¤ erent procedure. In the present work, we provide a much simpler alternative proof by using the Hölder’s inequality for integrals.
The next theorem is the (p; k)-analogue of the celebrated Bohr-Mollerup theorem. Theorem 2. Let f (x) be a positive function on (0; 1). Suppose that
(a) f (k) = 1,
(b) f (x + k) =x+pk+kpkx f (x), (c) ln f (x) is convex. Then, f (x) = p;k(x).
Proof. De…ne by e (x) = f (x)
p;k(x) for x > 0, p 2 N and k > 0. Then by (a) we
obtain
e (k)= f (k)
p;k(k)
= 1 implying that (k) = 0. Also by (b), we obtain
e (x+k)= f (x + k)
p;k(x + k)
= f (x)
p;k(x)
= e (x) which implies (x + k) = (x). Thus (x) is periodic with period k.
Next we want to show that (x) = ln f (x) ln p;k(x) is a constant. That is 0(x) = 0 , lim
h!0
(x + h) (x)
h = 0:
By (c) and Theorem 1, the functions ln f (x) and ln p;k(x) are convex. This implies
ln f (x) and ln p;k(x) are continuous. Then for " > 0, there exist 1; 2 > 0 such
that jln f(x + h) ln f (x)j < jhj" 2 whenever jhj < 1 and jln p;k(x + h) ln p;k(x)j < jhj" 2 whenever jhj < 2: Let = minf 1; 2g. Then for jhj < , we have
(x + h) (x) h = ln f (x + h) ln p;k(x + h) ln f (x) + ln p;k(x) h ln f (x + h) ln f (x) h + ln p;k(x + h) ln p;k(x) h < " 2+ " 2 = "
proving that 0(x) = 0. Since (x) is a constant and (k) = 0, then (x) = 0 for every x. Hence e0= f (x)
p;k(x). Therefore f (x) = p;k(x).
Theorem 3. The function, Bp;k(x; y) as de…ned by (1.6) is logarithmically convex
on (0; 1) (0; 1).
Proof. For x; y > 0, let Bp;k(x; y) be de…ned as in (1.6). Then
ln Bp;k(x; y) = ln p;k(x) + ln p;k(y) ln p;k(x + y):
Without loss of generality, let y be …xed. Then,
(ln Bp;k(x; y))00= 0p;k(x) 0p;k(x + y) > 0
since 0p;k(x) is decreasing for x > 0. This completes the proof. Remark 4. Theorem 3 is a (p; k)-analogue of Theorem 6 of [1].
Corollary 1. Let p 2 N and k > 0. Then the inequality
0
p;k(x) 0p;k(y) 0p;k(x) + 0p;k(y) 0p;k(x + y) (2.2)
is valid for x; y > 0.
Proof. This follows from the logarithmic convexity of Bp;k(x; y). Let
(x; y) = ln Bp;k(x; y) = ln p;k(x) + ln p;k(y) ln p;k(x + y):
Since (x; y) is convex on (0; 1) (0; 1), then its discriminant, is positive semide…nite. That is,
@2 @x2 > 0; = @2 @x2 @2 @y2 @2 @x@y @2 @y@x 0 implying that 0 p;k(x) 0p;k(x + y) 0p;k(y) 0p;k(x + y) 0p;k(x + y) 2 0: Thus, 0 p;k(x) 0p;k(y) 0p;k(x) + 0p;k(y) 0p;k(x + y) 0
which completes the proof.
Theorem 4. Let x; y > 0 and ; > 0 such that + = 1. Then
p;k( x + y) p;k(x) + p;k(y): (2.3)
Proof. It su¢ ces to show that p;k(x) is convex on (0; 1). By (1.5) we obtain
00 p;k(x) = p X n=0 2 (nk + x)3 > 0:
Then (2.3) follows from De…nition 1.
Theorem 5. Let p 2 N, k > 0 and a > 0. Then the function Q(x) = ax
p;k(x) is
convex on (0; 1).
Proof. Recall that p;k(x) is logarithmically convex. Thus, p;k( x + y) ( p;k(x)) ( p;k(y))
for x; y > 0 and ; > 0 such that + = 1. Then,
Q( x + y) = a x+ y p;k( x + y) a x+ y( p;k(x)) ( p;k(y)) : (2.4)
Also recall from the Young’s inequality that
u v u + v (2.5)
for u; v > 0 and ; > 0 such that + = 1. Let u = ax
p;k(x) and v = ay p;k(y).
Then (2.5) becomes
a x+ y( p;k(x)) ( p;k(y)) ax p;k(x) + ay p;k(y) = Q(x) + Q(y): (2.6)
Combining (2.4) and (2.6) yields Q( x + y) Q(x) + Q(y) which concludes the proof.
Theorem 6. Let p 2 N and k > 0. Then the functions A(x) = x p;k(x) is strictly
convex on (0; 1).
Proof. Direct computations yield
A00(x) = 2 0p;k(x) x 00p;k(x) which by (1.5) implies A00(x) = 2 p X n=0 1 (nk + x)2 2 p X n=0 x (nk + x)3 = 2 p X n=0 nk (nk + x)3 > 0:
Thus, A(x) is convex.
Remark 5. Corollary 1 and Theorems 4, 5 and 6 provide generalizations for some results proved in [14] and [6].
De…nition 3 ([12],[15]). Let f : I (0; 1) ! (0; 1) be a continuous function. Then f is said to be geometrically (or multiplicatively) convex on I if any of the following conditions is satis…ed.
f (px1x2) p f (x1)f (x2); (2.7) or more generally f n Y i=1 x i i ! n Y i=1 [f (xi)] i; n 2 (2.8)
where x1; x2; : : : ; xn 2 I and 1; 2; : : : ; n > 0 with Pni=1 i = 1. If
inequali-ties (2.7) and (2.8) are reversed, then f is said to be geometrically (or multiplica-tively) concave on I.
Lemma 2 ([12]). Let f : I (0; 1) ! (0; 1) be a di¤erentiable function. Then f is a geometrically convex function if and only if the functionxff (x)0(x) is nondecreasing. Lemma 3 ([12]). Let f : I (0; 1) ! (0; 1) be a di¤erentiable function. Then f is a geometrically convex function if and only if the function f (x)f (y) xy
yf 0 (y) f (y)
holds for any x; y 2 I.
Theorem 7. Let f (x) = ex p;k(x) for p 2 N and k 1. Then f is geometrically
convex and the inequality ey ex x y y[1+ p;k(y)] p;k(x) p;k(y) ey ex x y x[1+ p;k(x)] (2.9) is valid for x > 0 and y > 0..
Proof. We proceed as follows.
ln f (x) = x + ln p;k(x) implying
f0(x)
Then, xf0(x) f (x) 0 = 1 + p;k(x) + x 0p;k(x) = 1 + 1 kln(pk) p X n=0 1 nk + x+ p X n=0 x (nk + x)2 = 1 + 1 kln(pk) + p X n=1 x (nk + x)2 1 nk + x = 1 + 1 kln(pk) p X n=1 nk (nk + x)2 , h(x): Then h0(x) = 2Pp
n=0(nk+x)nk 3 > 0 implying that h is increasing. Moreover,
h(0) = 1 + 1 kln(pk) p X n=1 1 nk = 1 + 1 kln k + 1 k ln p p X n=1 1 n ! > 1 + 1 kln k 1 k > 0
since ln p Ppn=1n1 > 1 (See eqn. (6) of [2]). Then for x > 0, we have h(x) > h(0) > 0. Thus xff (x)0(x) is nondecreasing. Therefore, by Lemmas 2 and 3, f is geometrically convex and as a result, f (x)f (y) xy
yf 0 (y) f (y) . Consequently, we obtain ex p;k(x) ey p;k(y) x y y[1+ p;k(y)] (2.10) and ey p;k(y) ex p;k(x) y x x[1+ p;k(x)] : (2.11)
Now combining (2.10) and (2.11) yields the result (2.9) as required.
Remark 6. In particular, by replacing x and y respectively by x + k and x +k2, inequality (2.9) takes the form:
1 p ek x + k x + k 2 !(x+k 2)[1+ p;k(x+k2)] p;k(x + k) p;k(x + k2) 1 p ek x + k x + k 2 !(x+k)[1+ p;k(x+k)] : (2.12)
Remark 7. Theorem 7 gives a (p; k)-analogue of the previous results: [3, Theorem 1], [15, Theorem 1.2, Corollary 1.5] and [6, Theorem 3.5]. In particular, by letting k = 1, we recover the result of [6].
Remark 8. Results of type (2.9) and (2.12) can also be found in [9]. 3. Inequalities involving the (p; k)-Riemann zeta function De…nition 4. For p 2 N, k > 0 and x > 0, let p;k(x) be the (p; k)-analogue of the Riemann zeta function, (x). Then p;k(x) is de…ned as
p;k(x) = 1 p;k(x) Z p 0 tx k 1 + pktk p 1 dt; x > k: (3.1)
The functions p;k(x) satis…es the commutative diagram:
where p(x) and k(x) respectively denote the p and k analogues of the Riemann zeta function. See [5] and [4] for instance.
Lemma 4 ([7]). Let f and g be two nonnegative functions of a real variable, and m, n be real numbers such that the integrals in (3.2) exist. Then
Z b a g(t) (f (t))mdt Z b a g(t) (f (t))n dt Z b a g(t) (f (t))m+n2 dt !2 : (3.2)
Theorem 8. Let p 2 N, k > 0 and x > 0. Then the inequality x + pk + k x + pk + 2k: p;k(x) p;k(x + k) x x + k: p;k(x + k) p;k(x + 2k) ; x > k (3.3) holds. Proof. Let g(t) = 1 1+tk pk p 1, f (t) = t, m = x k, n = x + k, a = 0 and b = p. Then (3.2) implies Z p 0 tx k 1 + tpkk p 1 dt Z p 0 tx+k 1 + tpkk p 1 dt 0 @ Z p 0 tx 1 + pktk p 1 dt 1 A 2
which by relation (3.1) gives
p;k(x) p;k(x) p;k(x + 2k) p;k(x + 2k) p;k(x + k) p;k(x + k) 2
Then by the functional equation (1.3), inequality (3.4) can be rearranged to obtain the desired result (3.3).
Remark 9. (1)
(i) By letting p ! 1 in (3.3), we obtain the result of Theorem 3.1 of [4]. (ii) By setting k = 1 in (3.3), we obtain the result of Theorem 6 of [5]. (iii) By letting p ! 1 and k = 1 in (3.3), we obtain the result of Theorem 2.2
of [7].
Theorem 9. Let p 2 N and k > 0. Then for x > k, y > k, 1 + 1 = 1 such that x +y > k, the inequality p;k x +y ( p;k(x)) 1 ( p;k(y)) 1 p;k(x) 1 p;k(y) 1 p;k x +y (3.5) holds.
Proof. We employ the Hölder’s inequality: Z b a f (t)g(t) dt Z b a (f (t)) dt !1 Z b a (g(t)) dt !1 (3.6) where > 1, 1 + 1 = 1. Let f (t) = t x k 1+tk pk p 1 1, g(t) = t y k 1+tk pk p 1 1, a = 0
and b = p. Then (3.6) implies Z p 0 tx+y k 1 + pktk p 1 dt 0 @ Z p 0 tx k 1 + pktk p 1 dt 1 A 1 0 @ Z p 0 ty k 1 + pktk p 1 dt 1 A 1 : By relation (3.1) we obtain p;k x + y p;k x+ y p;k(x) p;k(x) 1 p;k(y) p;k(y) 1
which when rearranged gives (3.5) as required.
Remark 10. (1)
(i) By letting p ! 1 in (3.5), we obtain the result of Theorem 3.3 of [4]. (ii) By letting p ! 1 and k = 1 in (3.5), we obtain the result of Theorem 7 of
[5].
(iii) In particular, let k = 1 in (3.5). Then by replacing x and y respectively by x 1 and y + 1, we obtain p x 1+y+1 ( p(x 1)) 1 ( p(y + 1)) 1 p(x 1) 1 p(y + 1) 1 p x 1+y+1
which corresponds to Theorem 2.7 of [8].
Lemma 5([11]). Let f : (0; 1) ! (0; 1) be a di¤erentiable, logarithmically convex function. Then the function
g(x) = (f (x))
f ( x) ; 1
is decreasing on its domain.
Lemma 6. Let p 2 N, k > 0 and 1. Then the inequality [ p;k(y + k)] p;k( y + k) [ p;k(x + k)] p;k( x + k) 1 (3.7) holds for 0 x y.
Proof. Note that the function f (x) = p;k(x + k) is di¤erentiable and
logarith-mically convex. Then by Lemma 5, G(x) = [ p;k(x+k)]
p;k( x+k) is decreasing and for
0 x y, we have G(y) G(x) G(0) yielding the result. Theorem 10. Let p 2 N, k > 0 and 1. Then the inequality
p;k(y + k) p;k(y + k) p;k( y + k) p;k( y + k) p;k(x + k) p;k( x + k) (3.8) is satis…ed for 0 < x y.
Proof. Let H be de…ned x > 0 by
H(x) = p;k(x + k) p;k(x + k) = Z p 0 tx 1 + tk pk p 1 dt: (3.9)
Then for x; y > 0 and a; b > 0 such that a + b = 1, we have H(ax + by) = Z p 0 tax+by 1 +pktk p 1 dt = Z p 0 tax 1 + tk pk p 1 a tby 1 +pktk p 1 bdt 0 @Z p 0 tx 1 + tk pk p 1 dt 1 A a0 @Z p 0 ty 1 + tk pk p 1 dt 1 A b = (H(x))a(H(y))b:
Therefore, H(x) is logarithmically convex. Then by Lemma 5, the function T (x) = p;k(x + k) p;k(x + k)
is decreasing. Hence for 0 < x y, we have p;k(y + k) p;k(y + k) p;k( y + k) p;k( y + k) p;k(x + k) p;k(x + k) p;k( x + k) p;k( x + k) : Then by the right hand side of (3.7), we obtain
p;k(y + k) p;k(y + k) p;k( y + k) p;k( y + k)
p;k(x + k) p;k( x + k)
concluding the proof.
References
[1] S. S. Dragomir, R. P. Agarwal and N. S. Barnett, Inequalities for Beta and Gamma Functions via Some Classical and New Integral Inequalities, Journal of Inequalities and Applications, 5 (2000), 103-165.
[2] B. N. Guo and F. Qi, Sharp bounds for harmonic numbers, Applied Mathematics and Com-putation, 218(3)(2011), 991-995.
[3] J. D. Keµcki´c and P. M. Vasi´c, Some inequalities for the Gamma function, Publ. Inst. Math. Beograd N. S., 11(1971), 107-114.
[4] C. G. Kokologiannaki and V. Krasniqi, Some properties of the k-Gamma function, Le Matem-atiche, LXVIII (2013), Fasc. I, 13-22.
[5] V. Krasniqi, T. Mansour and A. Sh. Shabani, Some Monotonicity Properties and Inequalities for and Functions, Mathematical Communications, 15(2)(2010), 365-376.
[6] V. Krasniqi and A. S. Shabani, Convexity Properties and Inequalities for a Generalized Gamma Function, Applied Mathematics E-Notes, 10(2010), 27-35.
[7] A. Laforgia and P. Natalini, Turan type inequalities for some special functions, J. Ineq. Pure Appl. Math., 7(1)(2006), Art. 22, 1-5.
[8] F. Merovci, Turan type inequalities for p-Polygamma functions, Le Matematiche, LXVIII (2013), Fasc. II, 99-106.
[9] K. Nantomah, Some inequalities bounding certain ratios of the (p; k)-Gamma function, New Trends in Mathematical Sciences, 4(4)(2016), 329-336.
[10] K. Nantomah. E. Prempeh and S. B. Twum, On a (p; k)-analogue of the Gamma function and some associated Inequalities, Moroccan Journal of Pure and Applied Analysis, 2(2)(2016), 79-90.
[11] E. Neuman, Inequalities involving a logarithmically convex function and their applications to special functions, J. Inequal. Pure Appl. Math., 7(1)(2006) Art. 16.
[12] C. P. Nilculescu, Convexity according to the geometric mean, Mathematical Inequalities and Applications, 3(2)(2000), 155-167.
[13] C. Niculescu and L. E. Persson, Convex Functions and their Applications, A Contemporary Approach, CMS Books in Mathematics, Vol. 23, Springer-Verlag, New York, 2006.
[14] J. Sandor, Selected Chapters of Geometry, Analysis and Number Theory, RGMIA Mono-graphs, Victoria University, 2005.
[15] Xiao-Ming Zhang, Tie-Quan Xu and Ling-bo Situ, Geometric convexity of a function in-volving Gamma function and application to inequality theory, J. Inequal. Pure Appl. Math., 8(1)(2007), Art. 17.
Current address : Kwara Nantomah: Department of Mathematics, Faculty of Mathematical Sciences, University for Development Studies, Navrongo Campus, P. O. Box 24, Navrongo, UE/R, Ghana.
