# Several Turán-Type Inequalities for the Generalized Mittag-Leffler Function

Authors: Xiang Kai Dou; Li Yin; Xiu-Li Lin
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, several Turán-type inequalities for the more generalized Mittag-Leffler function are proved. In addition, we also gave affirmative answers to two open problems posed by Mehrez and Sitnik.


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## Body

## 1. Introduction

The Mittag-Leffler function was first defined by Mittag-Leffler in 1903 [1]. In this paper, he defined the function by
(1)Eαz=∑n=0∞znΓαn+1,z,α∈ℂ,Reα>0,where Γ· is a classical gamma function. In 1905, Wiman [2] generalized Eαz as
(2)Eα,βz=∑n=0∞znΓαn+β,z,α,β∈ℂ,Reα>0,Reβ>0.In 1971, Prabhakar [3] introduced the function
(3)Eα,βγz=∑n=0∞γnΓαn+βznn!,z,α,β,γ∈ℂ,Reα>0,Reβ>0.Later, Dorrego and Cerutti [4] introduced the k-Mittag-Leffler function
(4)Ek,α,βγz=∑n=0∞γn,kΓkαn+βznn!,z,α,β,γ∈ℂ,k∈ℝ,Reα>0,Reβ>0.where γn,k is the k-Pochhammer symbol defined by γn,k=Γkγ+nk/Γkγ. Recently, a generalization of the k-Mittag-Leffler function
(5)Ek,α,βγ,τz=∑n=0∞γnτ,kΓkαn+βznn!,z,α,β,γ,τ∈ℂ,k∈ℝ,Reα>0,Reβ>0.was introduced and studied in [5].The Mittag-Leffler function plays an important role in various branches of applied mathematics and engineering sciences, such as chemistry, biology, statistics, thermodynamics, mechanics, quantum physics, informatics, and signal processing. In 1930, the most known result in this field is an explicit formula for the resolvent of Riemann-Liouville fractional integral proved by E. Hille and J. Tamarkin. Based on these important formulas, many results are based still for solving fractional integral and differential equations. More properties and numerous applications of the Mittag-Leffler function to fractional calculus are collected, for instance, in References [6, 7]. In particular, we also refer to References [3, 8–10]. On the recent introduction of the Mittag-Leffler function and its generalizations, the reader may see [11, 12]. There are further related generalizations of the Mittag-Leffler function.Recently, Mehrez and Sitnik ([13, 14]) obtained some Turán-type inequalities for the Mittag-Leffler function by considering monotonicity for the special ratio of sections for series of the Mittag-Leffler function. In the course of their research, they used a new method. We call it the Mehrez-Sitnik method (the reader can refer to [15–17]). And then they applied this method to the Fox-Wright function and got a lot of interesting new results.Turán-type inequalities which initiated a new field of research on inequalities for special functions were proved by Paul Turán, it states(6)Pnx2−Pn+1xPn−1x≥0,where −1<x<1, n∈N, and Pn· stands for the classical Legendre polynomial.In this paper, we mainly consider a more general generalization(7)Ek,α,β,δγ,τz=∑n=0∞γnτ,kΓkαn+βznδn,z,α,β,γ∈ℂ,τ∈0,1∪ℕ,k∈ℝ,where δ≠0,−1,−2,⋯ and Reα>0,Reβ>0. It is clear that Ek,α,β,1γ,τz=Ek,α,βγ,τz and Ek,α,β,1γ,1z=Ek,α,βγz. In the following, we mainly prove the monotonicity of ratios for sections of series of generalized Mittag-Leffler functions; the result is also closely connected with Turán-type inequalities.


## 2. Definition of thek-Gamma Function and Lemmas

In 2007, Díaz and Pariguan [18] defined the k-analogue of the gamma function for k>0 and x>0 as
(8)Γkx=∫0∞tx−1e−tk/kdt=limn→∞n!knnkx/k−1xx+k⋯x+n−1k,where limk→1Γkx=Γx. Similarly, we may define the k-analogue of the digamma and polygamma functions as
(9)ψkx=ddxlnΓkx,ψkmx=dmdxmψkx.It is well known that thek-analogues of the digamma and polygamma functions satisfy the following recursive formula and series identities (see [18]):
(10)Γkx+k=xΓkx,x>0,ψkx=lnk−γk−1x+∑n=1∞xnknk+x,(11)ψkmx=−1m+1m!∑n=0∞1nk+xm+1=−1m+1∫0∞11−e−kttme−xtdt.For more properties of these functions, the reader may see Reference [19].Lemma 1 (see [16]).
Letan and bnn=0,1,2,⋯ be real numbers such that bn>0 and an/bnn≥0 is increasing (decreasing); then, a0+a1+⋯+an/b0+b1+⋯+bn is increasing (decreasing).Lemma 2 (see [20]).
Letan and bnn=0,1,2,⋯ be real numbers and let the power series Ax=∑n=0∞anxn and Bx=∑n=0∞bnxn be convergent if ∣x∣<r. If bn>0n=0,1,2,⋯ and the sequence an/bnn≥0 is (strictly) increasing (decreasing), then the function Ax/Bx is also (strictly) increasing (decreasing) on 0,r.


## 3. Main Results

Our results read as follows.Theorem 3.
Letn∈ℕ, α>0, β>0, γ>0, k>0, τ∈0,1∪ℕ, and δ∈0,1. We define the function Ek,α,β,δγ,τ,nz on 0,∞ by
(12)Ek,α,β,δγ,τ,nz=Ek,α,β,δγ,τz−∑m=0nγmτ,kzmΓkαm+βδm=∑m=n+1∞γmτ,kzmΓkαm+βδm.
Then, the Turán-type inequality(13)Ek,α,β,δγ,τ,nz·Ek,α,β,δγ,τ,n+2z≤Ek,α,β,δγ,τ,n+1z2,holds true.Proof.
Using the formulas(14)Ek,α,β,δγ,τ,nz=Ek,α,β,δγ,τ,n+1z+γn+1τ,k·zn+1δn+1·Γkαn+1+β,(15)Ek,α,β,δγ,τ,n+2z=Ek,α,β,δγ,τ,n+1z−γn+2τ,k·zn+2δn+2·Γkαn+2+β,we have
(16)Ek,α,β,δγ,τ,nz·Ek,α,β,δγ,τ,n+2z−Ek,α,β,δγ,τ,n+1z2=Ek,α,β,δγ,τ,n+1z·γn+1τ,k·zn+1δn+1·Γkαn+1+β−γn+2τ,k·zn+2δn+2·Γkαn+2+β−γn+1τ,k·γn+2τ,k·z2n+3δn+1·δn+2·Γkαn+1+β·Γkαn+2+β=γn+1τ,kzn+1Ek,α,β,δγ,τ,n+2zδn+1·Γkαn+1+β−γn+2τ,kzn+2Ek,α,β,δγ,τ,n+1zδn+2·Γkαn+2+β=γn+1τ,kδn+1·Γkαn+1+β∑m=n+3∞γmτ,kzm+n+1Γkαm+βδm−γn+2τ,kδn+2·Γkαn+2+β∑m=n+2∞γmτ,kzm+n+2Γkαm+βδm=∑m=n+3∞γn+1τ,kγmτ,kδn+1δmΓkαn+1+βΓkαm+β−γn+2τ,kγm−1τ,kδn+2δm−1Γkαn+2+βΓkαm−1+β=∑m=n+3∞Amγ,τk,α,β,δδn+2δmΓkαn+2+βΓkαm−1+βΓkαn+1+βΓkαm+β,where
(17)Amγ,τk,α,β,δ=δ+n+1γn+1τ,kγmτ,kΓkαn+2+βΓkαm−1+β−δ+m−1γn+2τ,kγm−1τ,kΓkαn+1+βΓkαm+β.
On the other hand, we have(18)Amγ,τk,α,β,δ=δ+n+1γn+1τ,kγmτ,kΓkαn+2+βΓkαm−1+β−δ+m−1γn+2τ,kγm−1τ,kδ+n+1γn+1τ,kγmτ,kΓkαn+1+βΓkαm+β=δ+n+1γn+1τ,kγmτ,kΓkαn+2+βΓkαm−1+β−δ+m−1δ+n+1·Γkγ+n+2τkΓkγ+m−1τkΓkγ+n+1τkΓkγ+mτkΓkαn+1+βΓkαm+β≤δ+n+1γn+1τ,kγmτ,kΓkαn+2+βΓkαm−1+β−Γkαn+1+βΓkαm+β.
Taking into account the inequality [21],
(19)Γkαm+2+βΓkαm+1+β≤Γkαm+βΓkαm−1+β,which holds for all α>0, β>0, and m≥3, and clearly, we have Amγ,τk,α,β,δ≤0. This in turn implies that inequality (13) holds.Corollary 4.
Letn∈ℕ, α>0, β>0, γ>0, k>0, and q∈0,1∪ℕ. Then,
(20)E1,α,β,1γ,q,nz·E1,α,β,1γ,q,n+2z≤E1,α,β,1γ,q,n+1z2.Proof.
By takingδ=k=1 and q=τ in Theorem 3, we easily obtain the above Turán-type inequality. The proof is complete.Remark 5.
It is worth noting that Mehrez and Sitnik posed an open problem 1 in [13]: “find the generalization of the inequality in the following inequality
(21)Eα,βγ,q,nz·Eα,βγ,q,n+2z≤Eα,βγ,q,n+1z2,where α>0, β>0, γ>0, q∈0,1∪ℕ, and z>0.”
Clearly, Corollary4 gives an affirmative answer to problem 1 in [13].Theorem 6.
Letn∈ℕ, α>0, β>0, γ>0, k>0, τ∈0,1∪ℕ, and δ∈0,1. We define the function Hk,α,β,δγ,τ,nz on 0,∞ by
(22)Hk,α,β,δγ,τ,nz=Ek,α,β,δγ,τ,nz·Ek,α,β,δγ,τ,n+2zEk,α,β,δγ,τ,n+1z2,z>0.
Then, the functionz⟶Hk,α,β,δγ,τ,nz is increasing on 0,∞. So the Turán-type inequality
(23)δ+nγ+n+1τkδ+n+1γ+nτk·Γkαn+1+β2Γkαn+βΓkαn+2+βEk,α,β,δγ,τ,n+1z2≤Ek,α,β,δγ,τ,nz·Ek,α,β,δγ,τ,n+2z,is valid for n∈N, α>0, β>0, γ>0, τ∈ℂ, and δ∈0,1. The constant on the left-hand side of inequality is sharp.Proof.
By using the Cauchy product, we have(24)Hk,α,β,δγ,τ,nz=Ek,α,β,δγ,τ,nz·Ek,α,β,δγ,τ,n+2zEk,α,β,δγ,τ,n+1z2.=∑i=0∞∑j=0iγn+jτ,k·γn+2+i−jτ,k·z2n+2+i/Γkαn+j+β·Γkαn+2+i−j+β·δn+j·δn+2+i−j∑i=0∞∑j=0iγn+1+jτ,k·γn+1+i−jτ,k·z2n+2+i/Γkαn+1+j+β·Γkαn+1+i−j+β·δn+1+j·δn+1+i−j.
We define the sequenceVjk,α,β,δ,γ by
(25)Vjk,α,β,δ,γ=δn+1+j·δn+1+i−j·γn+jτ,k·γn+2+i−jτ,kδn+j·δn+2+i−j·γn+1+jτ,k·γn+1+i−jτ,k·Ujk,α,β,where the sequence Ujk,α,β is defined by
(26)Ujk,α,β=Γkαn+1+j+β·Γkαn+1+i−j+βΓkαn+j+β·Γkαn+2+i−j+β.
In [21], the authors proved that the sequence Ujk,α,β is increasing for all j=0,1,⋯. Thus,
(27)Vj+1k,α,βVjk,α,β=Ki,jk,α,βUj+1k,α,βUjk,α,β≥Ki,jk,α,β.
By using the inequality(28)δ+n+1+jδ+n+1+i−jδ+n+jδ+n+i−j≥1,we have
(29)Ki,jk,α,β=γ+n+jτkγ+n+i−jτkγ+n+1+jτkγ+n+1+i−jτkδ+n+1+jδ+n+1+i−jδ+n+jδ+n+i−j≥1.
So the sequenceVjk,α,β,δ,γ is increasing for j=0,1,⋯, and n∈N, α>0, β>0, γ>0,τ∈ℂ, and δ∈0,1. This implies that the ratios
(30)Hk,α,β,δγ,τ,nz=∑i=0∞∑j=0iγn+jτ,k·γn+2+i−jτ,k/Γkαn+j+β·Γkαn+2+i−j+β·δn+j·δn+2+i−j∑i=0∞∑j=0iγn+1+jτ,k·γn+1+i−jτ,k/Γkαn+1+j+β·Γkαn+1+i−j+β·δn+1+j·δn+1+i−j,increase by Lemma 1. So the function z⟶Hk,α,β,δγ,τ,nz is increasing on 0,∞ by Lemma 2. Finally, it is easy to see that
(31)limz→∞Hk,α,β,δγ,τ,nz=δ+nγ+n+1τkδ+n+1γ+nτk·Γkαn+1+β2Γkαn+βΓkαn+2+β.
So the constantδ+nγ+n+1τk/δ+n+1γ+nτk·Γkαn+1+β2/Γkαn+βΓkαn+2+β is the best possible for which the inequality holds for all n∈N, α>0, β>0, γ>0, τ∈ℂ, and δ∈0,1.Corollary 7.
Forn∈ℕ, α>0, β>0, γ>0, k>0, and q∈0,1∪ℕ, the function z⟶H1,α,β,1γ,q,nz is increasing on 0,∞.Proof.
By puttingδ=k=1 and q=τ, we can get the above inequality. The proof is complete.Remark 8.
In [13], Mehrez and Sitnik also posed another open problem 2: “for z∈0,∞, find the monotonicity of the function
(32)Eα,βγ,q,nz·Eα,βγ,q,n+2zEα,βγ,q,n+1z2,where n∈ℕ, α>0, β>0, γ>0, q∈0,1∪ℕ, and z>0?”
Here, Corollary7 gives an affirmative answer to problem 2 in [13].Theorem 9.
Forn∈ℕ, α>0, β>0, γ>0, k>0, τ∈0,1∪ℕ, δ∈0,1, and fixed z>0, the function f:β↦ΓkβEk,α,β,δγ,τz is strictly log-convex on 0,∞. As a result, we have the following inequality:
(33)Ek,α,β+k,δγ,τz2<β+kβEk,α,β,δγ,τzEk,α,β+2k,δγ,τz.Proof.
Simple computation yields(34)∂∂βlogΓkβΓkαk+β=ψkβ−ψkαk+β,(35)∂2∂β2logΓkβΓkαk+β=ψ′kβ−ψ′kαk+β<0,where we apply that the function ψkx is concave on ℝ. Therefore, we get that the function β↦Γkβ/Γkαk+β is strictly log-convex on 0,∞. Using the fact that the sum of the log-convex functions is also log-convex, we obtain that the function f is strictly log-convex on 0,∞.
Due to inequality (33), we easily know
(36)logfβ+β+22<logfβ+logfβ+22.
That is,(37)Ek,α,β+k,δγ,τz2<ΓkβΓkβ+2kΓkβ+k2Ek,α,β,δγ,τzEk,α,β+2k,δγ,τz.
Using the definition ofΓkx, we easily obtain
(38)ΓkβΓkβ+2kΓkβ+k2=β+kβ.
The proof is complete.Corollary 10.
Forα>0, β>0, γ>0, k>0, τ∈0,1∪ℕ, δ∈0,1, β2>β1>0, and fixed z>0, we have
(39)Ek,α,β1+k,δγ,τzEk,α,β1,δγ,τz<β2β1Ek,α,β2+k,δγ,τzEk,α,β2,δγ,τz.Proof.
Since the functionfβ is strictly log-convex, we obtain that the function
(40)fβ+kfβ=Γkβ+kEk,α,β+k,δγ,τzΓkβEk,α,β,δγ,τz,is strictly increasing on 0,∞. Taking 0<β1<β2, we have
(41)Γkβ+kEk,α,β1+k,δγ,τzΓkβEk,α,β1,δγ,τz<Γkβ+kEk,α,β2+k,δγ,τzΓkβEk,α,β2,δγ,τz.
This completes the proof.



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