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Some inequalities involving the polygamma functions
Journal of Inequalities and Applications volume 2019, Article number: 54 (2019)
Abstract
Let \(\psi _{n}(x)=(1)^{n1}\psi ^{(n)}(x)\), where \(\psi ^{(n)}(x)\) are the polygamma functions. We determine necessary and sufficient conditions for the monotonicity and convexity of the function
for α and \(\beta \in \mathbb{R}\), where \(\psi (x)\) is the psi function. Consequently, this yields not only some new inequalities for the polygamma functions, but also new starshaped and superadditive functions involving them. In addition, we improve a wellknown meanvalue inequality for the polygamma functions.
Introduction
We recall that the logarithmic derivative of \(\varGamma (x)\) is called the psi or digamma function denoted by
for \(x>0 \), that the derivatives \(\psi '(x)\) and \(\psi ''(x)\) are, respectively, called the trigamma and tetragamma functions, and that the derivatives
for \(n\in \mathbb{N} \) are called the polygamma functions.
A world of the most fundamental properties involving the gamma, digamma, and polygamma functions can be found in some books [1,2,3,4]. It is known that these functions are all key parts of special functions. Moreover, they also play a vital role in other areas like analysis, physics, inequality theory, and statistics. Due to their significance, they attract many scholars to explore some their useful properties. In particular, some properties such as monotonicity, convexity, and complete monotonicity yield numerous inequalities related to these functions; see, for example, [5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28].
The following recurrence and asymptotic formulas are often encountered in the literature:
For convenience, we define \(\psi _{n}(x)=(1)^{n1}\psi ^{(n)}(x)\) for \(n\in \mathbb{N}\). It follows from (1.2) that \(\psi _{n}(x)\) is strictly completely monotonic on \((0,\infty )\) (see [29, 30]). Moreover, from (1.3) and (1.5) it follows that
which easily yields \(\lim_{x \rightarrow 0^{+}}\psi _{n}(x)=\infty \) and \(\lim_{x \rightarrow \infty }\psi _{n}(x)=0\).
It was shown in [31, Thm. 1.9] that \(\psi _{n}(x)\) is logarithmically convex on \((0,\infty )\). This and Jensen’s theorem yield the following functional inequality:
for all \(x_{m}>0\) and weights \(p_{m}\) (\(m=1,2,\ldots ,n\)), that is, \(p_{m}>0\) with \(\sum_{m=1}^{n} p_{m}=1\). Further, Alzer [32, Thm. 1] provided a refinement of (1.8) with the inequality
for all \(x_{m}>0\) and weights \(p_{m}\) (\(m=1,2,\ldots ,n\)) if and only if \(\alpha \leq k\). Then, inspired by these results, we are devoted to look for \(\alpha _{1}\), \(\alpha _{2}\), \(\beta _{1}\), and \(\beta _{2} \in \mathbb{R}\) such that, for all \(n, k\in \mathbb{N}\), the double inequality
is valid for all \(x_{m}>0\) and weights \(p_{m}\) (\(m=1,2,\ldots ,n\)). Here and in what follows,
It was proven in [32, Thm. 2] that
for all \(x_{m}>0\) and weights \(p_{m}>0\) (\(m=1,2,\ldots ,n\)) if and only if \(s\geq 0\) and \(r\geq sk\), or \(s<0\) and \(r\geq s(k+1)\).
The power means \(M_{n}^{[t]}\) are defined as (see [33])
for parameter \(t\in \mathbb{R}\) and all \(x_{m}>0\) and weights \(p_{m}\) (\(m=1,2,\ldots ,n\)).
In view of (1.10), it is natural to raise a new question: under which conditions on r and s, the reversed inequality of (1.10) is still valid for all \(x_{m}>0\) and weights \(p_{m}\) (\(m=1,2,\ldots ,n\))?
The main aim of this paper is to settle the two questions posed thereinbefore. Our main results are the following.
Theorem 1.1
Let α, \(\beta \in \mathbb{R}\) and \(n\in \mathbb{N}\). Define the function
Then \(F(x;\alpha ,\beta )\) is strictly increasing and concave for \(\beta \leq 0\) if and only if \(\alpha \geq n\) and strictly decreasing and convex for \(\beta \geq \frac{1}{2}\) if and only if \(\alpha \leq n\).
Applying the asymptotic formulas (1.4) and (1.5) to the function \(F(x;n,\beta )\), we get \(\lim_{x\rightarrow \infty }F(x;n, \beta )=0\). This in combination with Theorem 1.1 yields the following bounds for \(\psi _{n}(x)\).
Corollary 1
Let \(n\in \mathbb{N}\), \(\alpha \leq 0\), and \(\beta \geq \frac{1}{2}\). Then we have the double inequality
for \(x>\alpha \).
As a matter of fact, the double inequality (1.11) was first proved by Batir [34, Thm. 2.1]. In addition, using the concavity and convexity of \(F(x;n,\beta )\) and applying Jensen’s inequality give the following double inequality.
Theorem 1.2
Let \(k\geq 1\) and \(n\geq 2\) be integers, and let \(\alpha _{1}\), \(\alpha _{2}\in \mathbb{R}\), \(\beta _{1}\geq \frac{1}{2}\), and \(\beta _{2}\leq 0\). Then we have the double inequality
for all \(x_{m}>\beta _{2}\) and \(p_{m}>0\) (\(m=1,2,\ldots ,n\)) with \(\sum_{m=1}^{n} p_{m}=1\) if and only if \(\alpha _{1}\leq k\) and \(\alpha _{2}\geq k\).
Theorem 1.3
Let \(k\geq 1\) and \(n\geq 2\) be integers, and let r, \(s\in \mathbb{R}\). Then
for all \(x_{m}>0\) and \(p_{m}>0\) (\(m=1,2,\ldots ,n\)) with \(\sum_{m=1} ^{n} p_{m}=1\) if and only if \(s\geq 1\) and \(r\leq s(k+1)\), or \(s<1\) and \(r\leq skM(s)\), where
Remark 1
In brief, Theorem 1.3 can be restated as follows: inequality (1.13) is valid for all \(x_{m}>0\) and weights \(p_{m}\) (\(m=1,2, \ldots , n\)) if and only if \(r\leq skM(s)\).
In the last section, we give some sharp inequalities and starshaped functions, which are consequences of Theorem 1.1.
Some lemmas
In this section, we establish monotonicity properties of some functions closely related to the function \(e^{x}\).
According to Theorem 2.2 in [35], we obtain the following:
Lemma 1
([35, Theorem 2.2])
Let \(z\in \mathbb{R}\). Then the function
is strictly increasing on \((0,\infty )\) if and only if \(z\in (\infty ,0)\cup (\frac{1}{2},1)\) and strictly decreasing on the same interval if and only if \(z\in (0,\frac{1}{2})\cup (1,\infty )\).
Lemma 2
Let a, \(x_{1}\), and \(x_{2}\in \mathbb{R}\) with \(x_{2}>x_{1}>0\).

(1)
Let
$$ H_{a}(x)=\frac{e^{a x_{1} x}e^{(a1)x_{1} x}}{e^{a x_{2} x}e^{(a1)x _{2} x}}. $$Then \(H_{a}(x)\) is strictly increasing on \((0,\infty )\) if and only if \(a\leq 0\) and strictly decreasing on the same interval if and only if \(a\geq \frac{1}{2}\).

(2)
The function
$$ h(x)=\frac{1e^{x_{1} (1x)}}{1e^{x_{2} (1x)}}\frac{1e^{x_{1} x}}{1e ^{x_{2} x}}\frac{e^{a x_{1} x}}{e^{ax_{2} x}} $$is strictly decreasing on \((0,\infty )\) for \(a\geq \frac{1}{2}\).
Proof
(1) Differentiating \(H_{a}(x)\) with respect to x leads to
which implies
where
It is not difficult to obtain that
and
Hence the function \(H_{a}(x)\) is strictly increasing on \((0,\infty )\) for \(a\leq 0\) and strictly decreasing on the same interval for \(a\geq \frac{1}{2}\). Now, we assume that the function \(H_{a}(x)\) is strictly increasing on \((0,\infty )\). Firstly, we need to notice that
Then, for \(x>0\), we have:
Letting x tend to ∞, we obtain \(a\leq 0\).
If \(H_{a}(x)\) is strictly decreasing on \((0,\infty )\), then we have \(a\geq \frac{1}{2}\) as x tends to 0 in the reversed inequality of (2.4).
(2) We write
From part (1) of the lemma we conclude that, for \(a\geq \frac{1}{2}\), \(h(x)\) is the product of two positive decreasing functions on \((0,\infty )\). Therefore it is easy to get the desired result.
This completes the proof of Lemma 2. □
Lemma 3
Let α, \(\beta \in \mathbb{R}\) and m, \(n\in \mathbb{N}\). The function
is strictly increasing on \((\max (\alpha ,\beta ),\infty )\) for \(\alpha \beta \leq 0\) and strictly decreasing on \((\max (\alpha , \beta ),\infty )\) for \(\alpha \beta \geq \frac{1}{2}\).
Proof
Let \(k=m+n\), \(\theta =\alpha \beta \), and \(y=x+\beta \). Then we have
If we determine conditions for θ such that the function \(L(y)\) is strictly increasing and strictly decreasing on \((\max ( \theta ,0),\infty )\), then the desired result follows.
For convenience, we replace y by x in \(L(y)\). Using the wellknown formula (1.2) and applying the convolution theorem for the Laplace transform, we get
where
Differentiating \(L(x)\), we get
so that using the convolution theorem for the Laplace transform again in (2.6) yields
where
We will prove that
and
Making the substitution \(s=\frac{t}{2}(1+y)\), we obtain
For \(\delta >0\) and \(y\in (0,1)\), we define
Combining (2.10) with (2.11), we see that
Hence, to prove (2.8) and (2.9), it suffices to show that
and
From (2.11) a direct computation immediately gives
Substituting \(s=\delta (1+y)x\) into the first integral and \(s=\delta (1y)x\) into the second, we obtain
where
It is sufficient to show that
and
which imply that (2.13) and (2.14) are valid.
Firstly, we assume that \(\theta =0\). Clearly, (2.18) is equivalent to
where
Differentiating \(\phi (x)\) with respect to x gives
Using Lemma 1, we get \(\phi '(x)>0\) for \(x\in (0,\frac{1}{2})\). Since \(\phi (0)=0\), we conclude that \(\phi (x)>0\) for \(x\in (0, \frac{1}{2})\). In combination with the fact that \(\phi (x)\) is symmetric about \(x=\frac{1}{2}\) on \((0,1)\), this establishes (2.20).
We suppose that \(\theta <0\). It is obvious that
From (2.20) we see that
Hence, taking into consideration (2.21) and (2.22), we complete the proof of (2.18).
Next, we consider the case \(\theta \geq \frac{1}{2}\). Then (2.19) can be equivalently written as
where
From part (2) of Lemma 2 it is easy to prove (2.23). Therefore the proof of (2.19) is complete.
The proof of Lemma 3 is complete. □
Using Lemma 3 and limit relations (1.6) and (1.7), we have the following:
Corollary 2
Let m, \(n\in \mathbb{N}\), \(\alpha \geq 0\) and \(\beta \geq \frac{1}{2}\). Then, for all \(x>0\), we have:
where \((m+n2)_{n1}=\frac{(m+n2)!}{(n1)! (m1 )!}\). All bounds are optimal.
Remark 2
Setting \(m=n=1\) in Corollary 2, we get the wellknown inequality
with the optimal lower bound. This inequality was proved in many papers by different methods (see [8, 12, 15, 16, 34]) and used to establish many significant results related to the gamma and polygamma functions in [12, 13].
Proofs of the main results
We give a proof of Theorem 1.1.
Proof
Differentiation yields
and therefore
From Lemma 3 we conclude that \(F_{\alpha ,\beta }'(x)>0\) for \(\beta \geq \frac{1}{2}\) and \(F_{\alpha ,\beta }'(x)<0\) for \(\beta \leq 0\). In view of Corollary 2, it follows that \(F_{\alpha ,\beta }(x)<0\) for \(\beta \geq \frac{1}{2}\) and \(\alpha \leq n\) and \(F_{\alpha ,\beta }(x)>0\) for \(\beta \leq 0\) and \(\alpha \geq n\). By relations (3.1) and (3.2) it is easy to see that \(F'(x;\alpha ,\beta )<0\) and \(F''(x;\alpha ,\beta )>0\) for \(\beta \geq \frac{1}{2}\) and \(\alpha \leq n\) and that \(F'(x;\alpha , \beta )>0\) and \(F''(x;\alpha ,\beta )<0\) for \(\beta \leq 0\) and \(\alpha \geq n\).
Next, we assume that \(F(x;\alpha ,\beta )\) is strictly increasing and concave for \(\beta \leq 0\). From (3.1) we obtain \(F_{\alpha , \beta }(x)>0\), so that Lemma 3 and \(\lim_{x\rightarrow \infty }F_{\alpha ,\beta }(x)=\alpha n\) imply \(\alpha \geq n\).
Besides, from (3.2) it follows that
Letting x tend to ∞ and utilizing (1.7), we get \(\alpha \geq n\).
By similar arguments the necessary condition of the case where \(F(x;\alpha ,\beta )\) is strictly decreasing and convex for \(\beta \geq \frac{1}{2}\) can be proved.
The proof of Theorem 1.1 is complete. □
Proof
of Theorem 1.2 Using the concavity and convexity in Theorem 1.1 and Jensen’s inequality, we can easily prove the sufficiency of the theorem. Assume that the lefthand side of (1.12) is valid and \(x_{m}\) are not all equal. Setting \(x_{2}=\cdots =x_{n}=y\), from (1.12) we get
Letting \(x_{1}\) tend to y and applying L’Hospital’s rule, we obtain
so that (1.7) and (3.5) lead to \(\alpha _{1}\leq k\) as y tends to ∞. If the righthand side of (1.12) is valid, then by letting y tend to ∞ in the reversed inequality of (3.5), we have \(\alpha _{2}\geq k\).
The proof of Theorem 1.2 is complete. □
Next we give a proof of Theorem 1.3.
Proof of Theorem 1.3
Firstly, we need to show that \(M(s)\) for \(s<1\) is well defined. It follows from (1.6) and (1.7) that \(\lim_{x\rightarrow \infty }f_{k+1}(x)f_{k}(x)+sf _{k}(x)=0\) and \(\lim_{x\rightarrow 0}f_{k+1}(x)f_{k}(x)+{sf_{k}(x)=s}\). According to [32, Lemma 2], it is obvious that \(0\leq M(s)<1\) for \(s<0\) and \(s\leq M(s)<1\) for \(0\leq s<1\).
Since the power mean is increasing with respect to its order (see [33, p. 159]) and \(\psi _{k}(x)\) is decreasing, it suffices to show that (1.13) holds for \(s\geq 1\) and \(r=s(k+1)\) and for \(s<1\) and \(r=skM(s)\).
In fact, it is still necessary to make use of the techniques of the proof of [32, Thm. 2], and we cannot present it in detail. In view of the proof of [32, Thm. 2], we will prove that \(\varPhi (x)\) is increasing on \((0,\infty )\) for \(s\geq 1\) and \(r=s(k+1)\) and for \(0\neq s<1\) and \(r=skM(s)\), where
Furthermore, we get
where
Case 1. \(s\geq 1\) and \(r=s(k+1)\). From [32, Lemma 2] we conclude that \(\varPsi (x)\) is strictly increasing on \((0,\infty )\), so that \(\varPsi (x)>\lim_{x\rightarrow 0^{+}}\varPsi (x)=0\). Hence \(\varPhi (x)\) is increasing on \((0,\infty )\).
Case 2. \(0\neq s<1\) and \(r=skM(s)\). Then \(\varPsi (x)\geq 0\) is equivalent to
and therefore \(\varPhi (x)\) is increasing on \((0,\infty )\).
We consider the case \(s=0\). From Theorem 2 of [32] it follows that \(r<0\). Let \(x_{1}\geq x_{2}\geq \cdots \geq x_{n}> 0\). Then we define
where
Differentiating \(F_{m}(x)\) gives
where
If \(r=M(0)\), then we conclude that
which, together with (3.8), implies \(F'_{m}(x)\leq 0\). By the techniques of the proof in [32, Thm. 2] we have \(F(x_{1}, \ldots ,x_{n})\leq 0\).
We assume that (1.13) is valid for all \(x_{m}>0\) and weights \(p_{m}\) (\(m=1,2,\ldots, n\)). We set \(x_{2}=\cdots =x_{n}=y\) and \(x_{1}=x\) in (1.13). Then, for \(s\neq 0\), we obtain
with \(f(y,y)=0\). In fact, a direct computation of the partial derivative of \(f(x,y)\) gives
In combination with \(f(y,y)=0\), this establishes
which implies that \(\varPsi (y)\geq 0\) is equivalent to
If \(s\geq 1\), then letting y tend to 0 in (3.9) yields \(r\leq s(k+1)\). For \(0\neq s<1\), from the definition of \(M(s)\) and (3.9) it follows that \(r\leq skM(s)\). Finally, we prove the case \(s=0\). In the same way, from (1.13) we obtain
Since \(f(y,y)=0\) together with
we obtain
Hence (3.10), together with the definition of \(M(s)\), leads to \(r\leq M(0)\).
The proof of Theorem 1.3 is complete. □
Remark 3
Define
where \(f_{k}(x)\) is defined in Theorem 1.3. From [32, Lemma 2] we get
which implies \(f_{k+1}(x)f_{k}(x)>0\). For \(s\geq 0\), it is easy to see that
Since \(\lim_{x\rightarrow \infty }f_{k+1}(x)f_{k}(x)+sf_{k}(x)=0\), we conclude that \(m(s)=0\). For \(s<0\), from [32, Lemma 2] and (1.6) it follows that
which, together with \(\lim_{x\rightarrow 0}f_{k+1}(x)f_{k}(x)+sf_{k}(x)=s\), implies \(m(s)=s\). As a consequence, the reversed inequality of (1.13) is valid for all \(x_{m}>0\) and weights \(p_{m}\) (\(m=1,2,\ldots, n\)) if and only if \(r\geq skm(s)\).
Inequalities
In this section, we use Theorem 1.1 to yield some new inequalities for the polygamma functions.
We recall that a function f is said to be superadditive on an interval I if
If −f is superadditive, then f is called subadditive on I (see [36]). A function f is said to be starshaped on \((0,\infty )\) if
is valid for all \(x>0\) and \(a\in (0,1)\). There is a close relation between the above two notions that starshaped functions are superadditive (see [37,38,39]). Particularly, Trimble et al. [39] show that a function \(f(x)\) is starshaped on \((0,\infty )\) if and only if \(f(x)/x\) is nondecreasing on \((0,\infty )\).
Theorem 4.1
For \(n\in \mathbb{N}\) and \(\beta \in \mathbb{R}\), let \(G(x)=e^{n \psi (x+\beta )}\psi _{n}(x)\).

(1)
For \(\beta \leq 0\), we have the inequality
$$ \frac{1}{(n1)!}< \frac{G(x+y)}{G(x)G(y)}< +\infty $$(4.1)for \(x,y>\beta \).

(2)
For \(\beta \geq \frac{1}{2}\), we have the inequality
$$ 0< \frac{G(x+y)}{G(x)G(y)}< \frac{1}{(n1)!} $$(4.2)for \(x,y>0\). All bounds are optimal.
Proof
Since the proof of (4.1) is similar to that of (4.2), we provide the proof of (4.1). We define
Partial differentiation yields
where \(g_{1}(x)=G'(x)/G(x)\). For \(\beta \leq 0\), due to Theorem 1.1, we conclude that \(g_{1}(x)\) is decreasing on \((\beta , \infty )\), so that \(\partial g(x,y)/\partial x<0\). By the double inequality
which can be derived from [40], we get \(\lim_{x\rightarrow \infty }G(x)=(n1)!\) for all \(\beta \in \mathbb{R}\), \(\lim_{x\rightarrow 0}G(x)=\infty \) for \(\beta >0\), and \(\lim_{x\rightarrow \beta }G(x)=0\) for \(\beta \leq 0\). Together with \(\partial g(x,y)/\partial x<0\) and Theorem 1.1, this establishes
from which inequality (4.1) easily follows. Finally, we check that the bounds of (4.1) are optimal. From
and
we get the desired result. □
Theorem 4.1 immediately leads to the upper and lower bounds for the ratio \(\psi _{n}(x)\psi _{n}(y)/ \psi _{n}(x+y)\).
Corollary 3
Let \(n\in \mathbb{N}\). We have the double inequality
for \(x, y>0\).
From Theorem 4.1 we observe that, for \(\beta \leq 0\), the function \(\xi _{1}(x)=\ln (e^{n\psi (x)}\psi _{n}(x\beta ))\ln (n1)!\) is superadditive on \((0,\infty )\), and for \(\beta \geq \frac{1}{2}\), the function \(\xi _{2}(x)=\ln (e^{n\psi (x+\beta )}\psi _{n}(x))\ln (n1)!\) is subadditive on \((0,\infty )\).
In fact, \(\xi _{1}(x)\) and \(\xi _{2}(x)\) are not only superadditive but also starshaped on \((0,\infty )\).
Theorem 4.2
The functions \(\xi _{1}(x)\) and \(\xi _{2}(x)\) are both starshaped on \((0,\infty )\).
Proof
Since a function \(f(x)\) is starshaped on \((0,\infty )\) if and only if \(f(x)/x\) is nondecreasing on \((0,\infty )\) (see [39]), it suffices to prove that \(\xi _{1}(x)/x\) and \(\xi _{2}(x)/x\) is increasing on \((0,\infty )\). Since
by Theorem 1.1 we complete the proof of the theorem. □
Remark 4
Theorems 4.1 and 4.2 provide two distinct ways to prove that \(\xi _{1}(x)\) and \(\xi _{2}(x)\) are superadditive.
Conclusions
In this paper, Theorem 1.1 gives necessary and sufficient conditions for \(\alpha \in \mathbb{R}\) such that the function \(F(x;\alpha ,\beta )\) is strictly increasing (decreasing) and concave (convex) on \((\max (0,\beta ),\infty )\) for \(\beta \in (\infty ,0]\) (\(\beta \in [1/2,\infty )\)). Consequently, Theorem 1.1 provides a alternative proof of [34, Thm. 2.1] and yields a double meanvalue inequality (Theorem 1.2). In Theorem 1.3, it is stated that there exist necessary and sufficient conditions such that the reversed inequality of (1.10) is valid, which enhances Theorem 2 of [32].
Moreover, we find several new inequalities for the polygamma functions. In Theorem 4.1, we prove two sharp inequalities for the function \(G(x)\), which determine the upper and lower bounds for the ratio \(\psi _{n}(x)\psi _{n}(y)/\psi _{n}(x+y)\) (Corollary 3). Finally, two starshaped functions are posed in Theorem 4.2.
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The authors would like to express their sincere thanks to the editors and reviewers for their remarkable comments, suggestions, and ideas that helped to improve this paper. In particular, I sincerely appreciate that Lifei Zheng, as my supervisor, helped and encouraged me during the Master.
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Liang, L., Zhao, B. & Li, A. Some inequalities involving the polygamma functions. J Inequal Appl 2019, 54 (2019). https://doi.org/10.1186/s1366001919995
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DOI: https://doi.org/10.1186/s1366001919995
Keywords
 Polygamma functions
 Inequalities
 Psi function
 Starshaped functions
 Superadditive functions