NEW GENERALIZATIONS OF SOME INEQUALITIES FOR k-SPECIAL ...

LE MATEMATICHE Vol. LXX (2015) – Fasc. I, pp. 103–113 doi: 10.4418/2015.70.1.8

NEW GENERALIZATIONS OF SOME INEQUALITIES FOR k-SPECIAL AND q, k-SPECIAL FUNCTIONS SABRINA TAF - BOCHRA NEFZI - LATIFA RIAHI

In this work, we establish some new inequalities involving some kspecial and q, k-special functions, by using the technique of A. McD. Mercer [11].

1.

Introduction

Let L be a positive linear functional defined on a subspace C∗ (I) ⊂ C(I), where I is the interval (0, a) with a > 0 or (0, +∞). Let f and g be two functions continuous on I which are strictly increasing and strictly positive on I. In [11], A. Mcd. Mercer, posed the following result: Supposing that f , g ∈ C∗ (I) such that f (x) → 0, g(x) → 0 as x → 0+ and gf is strictly increasing, we define the function φ by φ =g

L( f ) , L(g)

and let F be a function defined on the ranges of f and g such that the compositions F( f ) and F(g) each belong to C∗ (I). a) If F is convex then L[F( f )] ≥ L[F(φ )]. (1) Entrato in redazione: 25 marzo 2014 AMS 2010 Subject Classification: 33B15, 33D05. Keywords: k-Gamma function, q, k-Gamma function, k-Beta function, q, k-Beta function, k-Zeta function.

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SABRINA TAF - BOCHRA NEFZI - LATIFA RIAHI

b) If F is concave then L[F( f )] ≤ L[F(φ )].

(2)

The objective of this paper is to use the technique of A. McD. Mercer [11] to develop some new inequalities for k-Gamma, k-Beta and k-Zeta functions then q, k-Gamma and q, k-Beta functions which are a generalization of some inequalities studied in [11, 13]. Note that for α ∈ R, the function F(t) = t α is convex if α < 0 or α > 1 and concave if 0 < α < 1. Then, for f and g satisfying the conditions (1) and (2), we have: L( f α ) > L(φ α ) if α < 0 or α > 1 and L( f α ) < L(φ α ) if 0 < α < 1. Substituting for φ this reads: [L(g)]α [L( f )]α > (resp. 1 (resp. 0 < α < 1). In particular, if we take f (x) = xβ and g(x) = xδ with β > δ > 0, we obtain the following useful inequality: [L(xδ )]α [L(xβ )]α ≷ , L(xαδ ) L(xαβ )

(3)

where ≷ correspond to the case (α < 0 or α > 1) and (0 < α < 1) respectively. In recent years, many authors have studied Gamma and Beta functions. For more information see [1], [2], [3], [5], [6], [10], [13].

2.

Basic Results

Throughout this paper, we will fix q ∈ (0, 1). For the convenience of the reader, we provide in this section a summary of the mathematical notions and definitions used in this paper (see [7], [9] , [12]). We write for a ∈ C, [a]q =

1 − qa . 1−q

The q-Jackson integrals from 0 to a and from 0 to ∞ are defined by (see[8]) Z a 0

∞

f (x)dq x = (1 − q)a ∑ f (aqn )qn , n=0

SOME INEQUALITIES FOR k-SPECIAL AND q, k-SPECIAL FUNCTIONS

Z ∞ 0

105

∞

f (x)dq x = (1 − q)

f (qn )qn ,

∑

n=−∞

provided the sums converge absolutely. For k > 0, the Γk function is defined by (see[4], [10]) x

n!kn (nk) k −1 Γk (x) = lim , x ∈ C \ kZ− , n→∞ (x)n,k where (x)n,k = x(x + k)(x + 2k) . . . (x + (n − 1)k). The above definition is a generalization of the definition of Γ(x) function. For x ∈ C with ℜ(x) > 0, the function Γk (x) is given by the integral (see [4]) Z ∞

Γk (x) =

tk

t x−1 e− k dt,

0

and satisfies the following properties: (see [5], [10]) 1. Γk (x + k) = xΓk (x) 2. (x)n,k =

Γk (x + nk) Γk (x)

3. Γk (k) = 1 Definition 2.1. Let x, y, s,t ∈ R and n ∈ N, we note by n−1

1. (x + y)nq,k := ∏ (x + q jk y) j=0 ∞

jk 2. (1 + x)∞ q,k := ∏ (1 + q x) j=0

3. (1 + x)tq,k :=

(1 + x)∞ q,k (1 + qt x)∞ q,k

.

s ks t We have (1 + x)s+t q,k = (1 + x)q,k (1 + q x)q,k .

We recall the two q, k-analogues of the exponential functions (see [5]) ∞ x Eq,k =

∑q

xn = (1 + (1 − qk )x)∞ q,k [n]qk !

kn(n−1) 2

n=0

and

∞

exq,k =

xn

1

∑ [n]q ! = (1 − (1 − qk )x)∞

n=0

k

.

q,k

These q, k-exponential functions satisfy the following relations:

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SABRINA TAF - BOCHRA NEFZI - LATIFA RIAHI k

x = Eq x Dqk Eq,k q,k

Dqk exq,k = exq,k ,

−x x −x Eq,k eq,k = exq,k Eq,k = 1.

and

The q, k-Gamma function is defined by [5] Γq,k (x) =

(1 − qk )∞ q,k x

k −1 (1 − qx )∞ q,k (1 − q)

, x > 0.

When k = 1 it reduces to the known q-Gamma function Γq . It satisfies the following functional equation: Γq,k (x + k) = [x]q Γq,k (x),

Γq,k (k) = 1

and it has the following integral representation (see[5]) Γq,k (x) =

Z ( [k]q ) 1k (1−qk )

k k

t

0

x−1

− q[k]tq

Eq,k

dqt,

x > 0.

The k-Beta function is defined by (see [4]) Z ∞

Bk (t, s) =

xt−1 (1 + xk )−

t+s k

t, s > 0.

dx,

0

By using the following change of variable xk = u and u = Bk (t, s) =

1 k

Z 1

s

t

u k −1 (1 − u) k −1 du,

y , we obtain 1−y

s,t, k > 0.

0

It is well-known that Bk (s,t) =

Γk (s)Γk (t) . Γk (s + t)

The q, k-Beta function is defined by (see [5]) 1

Bq,k (t, s) =

−t [k]q k

Z [k]qk

xt−1 (1 − qk

0

xk ks −1 ) dq x, [k]q q,k x

By using the following change of variable u =

1 k

s > 0,t > 0.

, the last equation becomes

[k]q Z 1

Bq,k (t, s) =

0

s

−1

k ut−1 (1 − qk uk )q,k dq u,

s > 0,t > 0.

It verifies the relation Bq,k (t, s) =

Γq,k (t)Γq,k (s) , Γq,k(t+s)

s,t > 0.


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The function Bq,k satisfies the following formulas for s,t > 0 t

1. Bq,k (t, ∞) = (1 − q) k Γq,k (t) 2. Bq,k (t + k, s) =

[t]q [s]q Bq,k (t, s + k)

3. Bq,k (t, s + k) = Bq,k (t, s) − qs Bq,k (t + k, s) 4. Bq,k (t, s + k) = 5. Bq,k (t, k) =

[s]q [s+t]q Bq,k (t, s)

1 [t]q .

Definition 2.2. We define the function ζk as (see [10]) ζk (s) =

Z ∞ s−k t

1 Γk (s)

0

et − 1

dt, s > k.

Note that when k = 1 we obtain the known Riemann Zeta function ζ (s).

3.

Main results

We start with the following theorem: Theorem 3.1. Let f be the function defined by (2n)

f (x) =

[Γk

(2n)

Γk

(k + x)]α

(k + αx)

then for all α > 1 (resp.0 < α < 1) f is decreasing (resp. increasing) on (0, ∞). Proof. The k-Gamma function is infinitely differentiable on (0, ∞) and we have (n)

Γk (x) =

Z ∞

tk

t x−1 [log(t)]n e− k dt,

n ∈ N.

0

We consider the subspace C∗ (I) obtained from C(I) by requiring its members to satisfy: (i) ω(x) = O(xθ ) (for any θ > −k) as x → 0, (ii) ω(x) = O(xϕ ) (for any finite ϕ) as x → +∞. Then, for ω ∈ C∗ (I) we define Z ∞

L(ω) = 0

tk

ω(t)t k−1 (log(t))2n e− k dt.

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The operator L is well-defined on C∗ (I) and it is a positive linear functional on C∗ (I). Using the inequality (3), we obtain for β > δ > 0 (2n)

[Γk

(2n)

Γk

(k + δ )]α

(k + αδ )

(2n)

≷

[Γk

(2n)

Γk

(k + β )]α

.

(k + αβ )

Theorem 3.1 is thus proved. Corollary 3.2. For all x ∈ [0, k], we have: (2n)

[Γk

(2n)

(2k)]α

≤

(2n) Γk (k + αk)

and

(2n)

(2n) [Γk (k)]α−1

≤

[Γk

(2n)

Γk

[Γk

(k + x)]α

(2n)

≤ [Γk

(2n) Γk (k + αx)

(k + x)]α

(2n)

≤

(k + αx)

[Γk

(2n)

Γk

(k)]α−1 ,

(2k)]α

,

α ≥ 1,

0 < α ≤ 1.

(k + αk)

Corollary 3.3. For all x ∈ [0, k], we have [Γk (k + x)]α kα ≤ ≤ 1, Γk (αk + k) Γk (k + αx)

α ≥ 1,

[Γk (k + x)]α kα ≤ , Γk (k + αx) Γk (αk + k)

0 < α ≤ 1.

and 1≤

Theorem 3.4. Let f be the function defined by (2n)

f (x) =

[Γq,k (k + x)]α (2n)

Γq,k (k + αk) then for all α > 1 (resp. 0 < α < 1) f is decreasing (resp. increasing) on (0, ∞). Proof. Since that Γq,k is an infinitely differentiable function on (0, +∞), we have (n) Γq,k (x)

=

Z ( [k]q ) 1k (1−qk )

k k

t

x−1

0 [k]

1

n

− q[k]tq

(log(t)) Eq,k

dqt,

x > 0, n ∈ N.

We consider I = (0, ( (1−qqk ) ) k ) and the subspace C∗ (I) obtained from C(I) by requiring its members to satisfy: (i) ω(x) = O(xθ ) (for any θ > −k) as x → 0,

SOME INEQUALITIES FOR k-SPECIAL AND q, k-SPECIAL FUNCTIONS [k]

109

1

(ii) ω(x) = O(1) as x → ( (1−qqk ) ) k . Then we have, L(ω) =

Z ( [k]q ) 1k (1−qk )

k k

ω(t)t

k−1

0

2n

− q[k]tq

(log(t)) Eq,k

dqt.

The operator L is a positive linear functional on C∗ (I). By applying the inequality (3), we obtain for β > δ > 0 (2n)

[Γq,k (k + δ )]α (2n)

Γq,k (k + αδ )

(2n)

≷

[Γq,k (k + β )]α (2n)

.

Γq,k (k + αβ )

Theorem 3.4 is thus proved. In particular, we have the following results Corollary 3.5. For all x ∈ [0, k], we have: (2n)

(2n)

[Γq,k (2k)]α (2n) Γq,k (k + αk)

and

≤

[Γq,k (k + x)]α (2n) Γq,k (k + αx)

≤

[Γq,k (k + x)]α (2n)

α ≥ 1,

(2n)

(2n)

(2n) [Γq,k (k)]α−1

(2n)

≤ [Γq,k (k)]α−1 ,

≤

Γq,k (k + αx)

[Γq,k (2k)]α (2n)

,

0 < α ≤ 1.

Γq,k (k + αk)

Corollary 3.6. For all x ∈ [0, k], we have [Γq,k (k + x)]α kα ≤ ≤ 1, Γq,k (αk + k) Γq,k (k + αx) and 1≤

[Γq,k (k + x)]α kα ≤ , Γq,k (k + αx) Γq,k (αk + k)

α ≥ 1,

0 < α ≤ 1.

Remark 3.7. Applying Theorem 3.1 and Theorem 3.4 for k = 1, we obtain the Theorem 2.1 and Theorem 3.2 in [13]. Theorem 3.8. For s > 0, let f be the function defined by f (x) =

[Bk (k(x + 1), s)]α Bk (k(αx + 1), s)

then for all α > 1 (resp. 0 < α < 1) f is decreasing (resp. increasing) on (0, ∞).

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Proof. The k-Beta function is defined by 1 Bk (t, s) = k

Z 1

t

s

x k −1 (1 − x) k −1 dx.

0

We consider the interval I = (0, 1) and the subspace C∗ (I) obtained from C(I) by requiring its members to satisfy: (i) ω(x) = O(xθ ) (for θ > −1 ) as x → 0 (ii) ω(x) = O(1) as x → 1. For ω ∈ C∗ (I) , we define Z 1

L(ω) =

s

ω(x)(1 − x) k −1 dx.

0

L is a positive linear functional on C∗ (I). Applying the inequality (3), we obtain for β > δ > 0 [Bk (k(δ + 1), s)]α [Bk (k(β + 1), s)]α ≷ . Bk (k(αδ + 1), s) Bk (k(αβ + 1), s) Theorem 3.8 is thus proved. Corollary 3.9. For x ∈ [0, k] and s > 0, we have [Bk (k(x + 1), s)]α k2(α−1) (αk2 + s) [Bk (k2 , s)]α ≤ ≤ [Bk (k, s)]α−1 α(k2 + s)α Bk (αk2 , s) Bk (k(αx + 1), s)

α ≥ 1.

Theorem 3.10. For s > 0, let f be the function defined by f (x) =

[Bq,k (k + x, s)]α Bq,k (k + αx, s)

then for all α > 1 (resp. 0 < α < 1) f is decreasing (resp. increasing) on (0, ∞). Proof. The q, k-Beta function is defined by Z 1

Bq,k (t, s) =

0

s

−1

k xt−1 (1 − qk xk )q,k dq x.

We consider the interval I = (0, 1) and the subspace C∗ (I) obtained from C(I) by requiring its members to satisfy: (i) ω(x) = O(xθ ) (for any θ > −k) as x → 0 (ii) ω(x) = O(1) as x → 1 Then we put, Z 1

L(ω) = 0

s

−1

k dq x. ω(x)xk−1 (1 − qk xk )q,k


111

L is defined on C∗ (I) and it is a positive linear functional on C∗ (I). Applying the inequality (3), we obtain for β > δ > 0 [Bq,k (k + δ , s)]α [Bq,k (k + β , s)]α ≷ Bq,k (k + αδ , s) Bq,k (k + αβ , s) Theorem 3.10 is thus proved. Corollary 3.11. For x ∈ [0, k] and s > 0, we have [k]αq [αk + s]q Bαq,k (k, s) [Bq,k (k + x, s)]α ≤ ≤ [Bq,k (k, s)]α−1 [k + s]αq [αk]q Bq,k (αk, s) Bq,k (k + αx, s)

α ≥ 1.

Theorem 3.12. Let f be the function defined by f (x) =

[ζk (x + k + 1)Γk (x + k + 1)]α ζk (αx + k + 1)Γk (αx + k + 1)

then for all α > 1 (resp. 0 < α < 1) f is decreasing (resp. increasing) on (0, ∞). Proof. From the definition of Zeta function, we can write Z ∞

ζk (s)Γk (s) =

0

xs−k

1 dx, ex − 1

s > k.

We consider the subspace C∗ (I) obtained from C(I) by requiring its members to satisfy: (i) ω(x) = O(xθ ) (for any θ > −1) as x → 0 (ii) ω(x) = O(xϕ ) (for any finite ϕ) as x → +∞. Then we have, Z ∞ x dx. L(ω) = ω(x) x e −1 0 The linear functional L is well-defined on C∗ (I) and it is positive. Applying the inequality (3), we obtain β > δ > 0 [ζk (δ + k + 1)Γk (δ + k + 1)]α [ζk (β + k + 1)Γk (β + k + 1)]α ≷ ζk (αδ + k + 1)Γk (αδ + k + 1) ζk (αβ + k + 1)Γk (αβ + k + 1) Theorem 3.12 is thus proved. Remark 3.13. Applying Theorem 3.12 for k = 1, we obtain the inequality for ζ function proved in [11].

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SABRINA TAF Department of Mathematics Faculty SEI, UMAB University of Mostaganem, Algeria e-mail: [email protected] BOCHRA NEFZI Department of Mathematics Faculty of Science of Tunis, Tunisia e-mail: [email protected] LATIFA RIAHI Department of Mathematics Faculty of Science of Tunis, Tunisia e-mail: [email protected]