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# Advanced Nonlinear Studies

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Volume 16, Issue 1

# Estimates for Eigenvalues of Poly-Harmonic Operators

Guoxin Wei
/ Lingzhong Zeng
Published Online: 2016-02-01 | DOI: https://doi.org/10.1515/ans-2015-5020

## Abstract

In this paper, we study the eigenvalues of the poly-Laplacian with arbitrary order on a bounded domain in an n-dimensional Euclidean space and we obtain a lower bound which generalizes the results due to Cheng and Wei [5] and gives an improvement of the results due to Cheng, Qi and Wei [3].

MSC 2010: 35P15

## 1 Introduction

Let Ω be a bounded domain with piecewise smooth boundary $\partial \mathrm{\Omega }$ in an n-dimensional Euclidean space ${ℝ}^{n}$. Let ${\lambda }_{i}$ be the i-th eigenvalue of the Dirichlet eigenvalue problem of the poly-Laplacian with arbitrary order

$\left\{\begin{array}{cccc}& {\left(-\mathrm{\Delta }\right)}^{l}u=\lambda u\hfill & & \hfill \text{in}\mathrm{\Omega },\\ & u=\frac{\partial u}{\partial \nu }=\mathrm{\cdots }=\frac{{\partial }^{l-1}u}{\partial {\nu }^{l-1}}=0\hfill & & \hfill \text{on}\partial \mathrm{\Omega },\end{array}$(1.1)

where l is a positive integer, Δ is the Laplacian in ${ℝ}^{n}$, and ν denotes the outward unit normal vector field of the boundary $\partial \mathrm{\Omega }$. It is well known that the spectrum of this eigenvalue problem is real and discrete, that is,

$0<{\lambda }_{1}\le {\lambda }_{2}\le {\lambda }_{3}\le \mathrm{\cdots }\to +\mathrm{\infty },$

where each ${\lambda }_{i}$ has finite multiplicity and is repeated according to its multiplicity. Let $V\left(\mathrm{\Omega }\right)$ denote the volume of Ω and let ${B}_{n}$ denote the volume of the unit ball in ${ℝ}^{n}$.

When $l=1$, the eigenvalue problem (1.1) is called a fixed membrane problem. In this case, one has Weyl’s asymptotic formula

${\lambda }_{k}\sim \frac{4{\pi }^{2}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2}{n}}}{k}^{\frac{2}{n}},k\to +\mathrm{\infty }.$

From the above asymptotic formula, one can derive the formula

$\frac{1}{k}\sum _{i=1}^{k}{\lambda }_{i}\sim \frac{n}{n+2}\frac{4{\pi }^{2}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2}{n}}}{k}^{\frac{2}{n}},k\to +\mathrm{\infty }.$(1.2)

Pólya [11] proved that

${\lambda }_{k}\ge \frac{4{\pi }^{2}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2}{n}}}{k}^{\frac{2}{n}},k=1,2,\mathrm{\dots },$

if Ω is a tiling domain in ${ℝ}^{n}$ . Furthermore, he proposed the following conjecture.

#### (Pólya)

If Ω is a bounded domain in ${ℝ}^{n}$, then the k-th eigenvalue ${\lambda }_{k}$ of the fixed membrane problem satisfies

${\lambda }_{k}\ge \frac{4{\pi }^{2}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2}{n}}}{k}^{\frac{2}{n}},k=1,2,\mathrm{\dots }.$

Berezin [2] and Lieb [8] gave a partial solution for the conjecture of Pólya. In particular, Li and Yau [7] proved that

$\frac{1}{k}\sum _{i=1}^{k}{\lambda }_{i}\ge \frac{n}{n+2}\frac{4{\pi }^{2}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2}{n}}}{k}^{\frac{2}{n}},k=1,2,\mathrm{\dots }.$(1.3)

The formula (1.2) shows that the result of Li and Yau is sharp in the sense of average. From (1.3) one can infer that

${\lambda }_{k}\ge \frac{n}{n+2}\frac{4{\pi }^{2}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2}{n}}}{k}^{\frac{2}{n}},k=1,2,\mathrm{\dots },$

which gives a partial solution for the conjecture of Pólya with a factor $\frac{n}{n+2}$. Recently, Melas [9] has improved (1.3) to the estimate

$\frac{1}{k}\sum _{i=1}^{k}{\lambda }_{i}\ge \frac{n}{n+2}\frac{4{\pi }^{2}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2}{n}}}{k}^{\frac{2}{n}}+\frac{1}{24\left(n+2\right)}\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)},k=1,2,\mathrm{\dots },$

where

$I\left(\mathrm{\Omega }\right)=\underset{a\in {ℝ}^{n}}{\mathrm{min}}{\int }_{\mathrm{\Omega }}{|x-a|}^{2}𝑑x$

is called the moment of inertia of Ω.

When $l=2$, the eigenvalue problem (1.1) is called a clamped plate problem. For the eigenvalues of the clamped plate problem, Agmon [1] and Pleijel [10] obtained

${\lambda }_{k}\sim \frac{16{\pi }^{4}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{4}{n}}}{k}^{\frac{4}{n}},k\to +\mathrm{\infty }.$(1.4)

From (1.4) one can obtain

$\frac{1}{k}\sum _{i=1}^{k}{\lambda }_{i}\sim \frac{n}{n+4}\frac{16{\pi }^{4}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{4}{n}}}{k}^{\frac{4}{n}},k\to +\mathrm{\infty }.$(1.5)

Furthermore, Levine and Protter [6] proved that the eigenvalues of the clamped plate problem satisfy the inequality

$\frac{1}{k}\sum _{i=1}^{k}{\lambda }_{i}\ge \frac{n}{n+4}\frac{16{\pi }^{4}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{4}{n}}}{k}^{\frac{4}{n}}.$(1.6)

The formula (1.5) shows that the coefficient of ${k}^{\frac{4}{n}}$ is the best possible constant. By adding to its right-hand side two terms of lower order in k, Cheng and Wei [4] obtained the estimate

$\frac{1}{k}\sum _{i=1}^{k}{\lambda }_{i}\ge \frac{n}{n+4}\frac{16{\pi }^{4}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{4}{n}}}{k}^{\frac{4}{n}}+\left(\frac{n+2}{12n\left(n+4\right)}-\frac{1}{1152{n}^{2}\left(n+4\right)}\right)\frac{4{\pi }^{2}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2}{n}}}\frac{n}{n+2}\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)}{k}^{\frac{2}{n}}$$+\left(\frac{1}{576n\left(n+4\right)}-\frac{1}{27648{n}^{2}\left(n+2\right)\left(n+4\right)}\right){\left(\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)}\right)}^{2},$(1.7)

which is an improvement of (1.6). Very recently, Cheng and Wei [5] have improved the estimate (1.7) to the estimate

$\frac{1}{k}\sum _{i=1}^{k}{\lambda }_{i}\ge \frac{n}{n+4}\frac{16{\pi }^{4}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{4}{n}}}{k}^{\frac{4}{n}}+\frac{n+2}{12n\left(n+4\right)}\frac{4{\pi }^{2}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2}{n}}}\frac{n}{n+2}\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)}{k}^{\frac{2}{n}}+\frac{{\left(n+2\right)}^{2}}{1152n{\left(n+4\right)}^{2}}{\left(\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)}\right)}^{2}.$

When l is arbitrary, Levine and Protter [6] proved the estimate

$\frac{1}{k}\sum _{i=1}^{k}{\lambda }_{i}\ge \frac{n}{n+2l}\frac{{\pi }^{2l}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2l}{n}}}{k}^{\frac{2l}{n}},k=1,2,\mathrm{\dots },$(1.8)

which implies that

${\lambda }_{k}\ge \frac{n}{n+2l}\frac{{\pi }^{2l}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2l}{n}}}{k}^{\frac{2l}{n}},k=1,2,\mathrm{\dots }.$

By adding l terms of lower order of ${k}^{\frac{2l}{n}}$ to its right-hand side, Cheng, Qi and Wei [3] obtained a sharper result than (1.8), that is,

$\frac{1}{k}\sum _{i=1}^{k}{\lambda }_{i}\ge \frac{n}{n+2l}\frac{{\left(2\pi \right)}^{2l}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2l}{n}}}{k}^{\frac{2l}{n}}+\frac{n}{\left(n+2l\right)}\sum _{p=1}^{l}\frac{l+1-p}{{\left(24\right)}^{p}n\mathrm{\cdots }\left(n+2p-2\right)}\frac{{\left(2\pi \right)}^{2\left(l-p\right)}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2\left(l-p\right)}{n}}}{\left(\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)}\right)}^{p}{k}^{\frac{2\left(l-p\right)}{n}}.$(1.9)

In this paper, we study the eigenvalues of the Dirichlet eigenvalue problem (1.1) of a Laplacian with arbitrary order and prove the following result.

Let Ω be a bounded domain in an n-dimensional Euclidean space ${ℝ}^{n}$. Assume that $l\mathrm{\ge }\mathrm{2}$ and ${\lambda }_{i}$ is the i-th eigenvalue of the eigenvalue problem (1.1). Then, the eigenvalues satisfy

$\frac{1}{k}\sum _{j=1}^{k}{\lambda }_{j}\ge \frac{n}{n+2l}\frac{{\left(2\pi \right)}^{2l}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2l}{n}}}{k}^{\frac{2l}{n}}+\frac{l}{24\left(n+2l\right)}\frac{{\left(2\pi \right)}^{2\left(l-1\right)}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2\left(l-1\right)}{n}}}\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)}{k}^{\frac{2\left(l-1\right)}{n}}$$+\frac{l{\left(n+2\left(l-1\right)\right)}^{2}}{2304n{\left(n+2l\right)}^{2}}\frac{{\left(2\pi \right)}^{2\left(l-2\right)}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2\left(l-2\right)}{n}}}{\left(\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)}\right)}^{2}{k}^{\frac{2\left(l-2\right)}{n}}.$(1.10)

When $l=2$, Theorem 1.2 reduces to the result of Cheng and Wei [5].

When $l\ge 2$, we give an important improvement of the result in (1.9) due to Cheng, Qi and Wei [3], since the inequality (1.10) is sharper than (1.9). We will give a proof of this fact in Section 2.

## 2 Proofs of Theorem 1.2 and Remark 1.4

In this section, we give the proof of Theorem 1.2. At first, we need the following key lemma which will play an important role in the proof of Theorem 1.2. Its proof will be given in Appendix A.

Let $b\mathrm{\ge }\mathrm{2}$ be a positive real number and $\mu \mathrm{>}\mathrm{0}$. If $\psi \mathrm{:}\mathrm{\left[}\mathrm{0}\mathrm{,}\mathrm{+}\mathrm{\infty }\mathrm{\right)}\mathrm{\to }\mathrm{\left[}\mathrm{0}\mathrm{,}\mathrm{+}\mathrm{\infty }\mathrm{\right)}$ is a decreasing function such that

$-\mu \le {\psi }^{\prime }\left(s\right)\le 0$

and

$A:={\int }_{0}^{\mathrm{\infty }}{s}^{b-1}\psi \left(s\right)𝑑s>0,$

then, for any positive integer $l\mathrm{\ge }\mathrm{2}$, we have

${\int }_{0}^{\mathrm{\infty }}{s}^{b+2l-1}\psi \left(s\right)𝑑s\ge \frac{1}{b+2l}{\left(bA\right)}^{\frac{b+2l}{b}}\psi {\left(0\right)}^{-\frac{2l}{b}}+\frac{l}{6b\left(b+2l\right){\mu }^{2}}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}\psi {\left(0\right)}^{\frac{2b-2l+2}{b}}$$+\frac{l{\left(b+2\left(l-1\right)\right)}^{2}}{144{b}^{2}{\left(b+2l\right)}^{2}{\mu }^{4}}{\left(bA\right)}^{\frac{b+2l-4}{b}}\psi {\left(0\right)}^{\frac{4b-2l+4}{b}}.$

#### Proof of Theorem 1.2.

In this proof, we will use the same notation as in [3]. Let ${\stackrel{^}{\phi }}_{j}\left(z\right)$ be the Fourier transform of the trial function ${\phi }_{j}\left(x\right)$ defined as

${\phi }_{j}\left(x\right)=\left\{\begin{array}{cc}{u}_{j}\left(x\right),\hfill & x\in \mathrm{\Omega },\hfill \\ 0,\hfill & x\in {ℝ}^{n}\setminus \mathrm{\Omega },\hfill \end{array}$

where ${u}_{j}\left(x\right)$ is an orthonormal eigenfunction corresponding to the eigenvalue ${\lambda }_{j}$, $f\left(z\right):={\sum }_{j=1}^{k}{|{\stackrel{^}{\phi }}_{j}\left(z\right)|}^{2}$, and ${f}^{\ast }$ is the symmetric decreasing rearrangement of f. We assume that $A=\frac{k}{n{B}_{n}}$, $b=n$, and $\psi \left(s\right)=\varphi \left(s\right)$, where $\varphi :\left[0,+\mathrm{\infty }\right)\to \left[0,{\left(2\pi \right)}^{-n}V\left(\mathrm{\Omega }\right)\right]$ is a non-increasing function of $|x|$ and $\varphi \left(x\right)$ is defined by $\varphi \left(|x|\right):={f}^{\ast }\left(x\right)$. Then, from Lemma 2.1 we can obtain that

$\sum _{j=1}^{k}{\lambda }_{j}\ge n{B}_{n}{\int }_{0}^{\mathrm{\infty }}{s}^{n+2l-1}\varphi \left(s\right)𝑑s$$\ge \frac{n{B}_{n}{\left(\frac{k}{{B}_{n}}\right)}^{\frac{n+2l}{n}}}{n+2l}\varphi {\left(0\right)}^{-\frac{2l}{n}}+\frac{l{B}_{n}{\left(\frac{k}{{B}_{n}}\right)}^{\frac{n+2\left(l-1\right)}{n}}}{6\left(n+2l\right){\mu }^{2}}\varphi {\left(0\right)}^{\frac{2n-2l+2}{n}}+\frac{l{\left(n+2\left(l-1\right)\right)}^{2}{B}_{n}{\left(\frac{k}{{B}_{n}}\right)}^{\frac{n+2l-4}{n}}}{144n{\left(n+2l\right)}^{2}{\mu }^{4}}\varphi {\left(0\right)}^{\frac{4n-2l+4}{n}}.$(2.1)

Now, we define a function $\xi \left(t\right)$ as

$\xi \left(t\right)=\frac{n{B}_{n}}{n+2l}{\left(\frac{k}{{B}_{n}}\right)}^{\frac{n+2l}{n}}{t}^{-\frac{2l}{n}}+\frac{l{B}_{n}}{6\left(n+2l\right){\mu }^{2}}{\left(\frac{k}{{B}_{n}}\right)}^{\frac{n+2\left(l-1\right)}{n}}{t}^{\frac{2n-2l+2}{n}}+\frac{l{\left(n+2\left(l-1\right)\right)}^{2}{B}_{n}}{144n{\left(n+2l\right)}^{2}{\mu }^{4}}{\left(\frac{k}{{B}_{n}}\right)}^{\frac{n+2l-4}{n}}{t}^{\frac{4n-2l+4}{n}}.$(2.2)

Here, we assume that $l\le n+1$. The other cases (i.e., $n+1, $l\ge 2\left(n+1\right)$) can be discussed by using a similar method. After differentiating (2.2) with respect to the variable t, we derive

${\xi }^{\prime }\left(t\right)=\frac{{B}_{n}}{n+2l}{\left(\frac{k}{{B}_{n}}\right)}^{\frac{n+2l}{n}}{t}^{-\frac{2l}{n}-1}\left[-2l+\frac{l\left(2n-2l+2\right)}{6n{\mu }^{2}}{\left(\frac{k}{{B}_{n}}\right)}^{-\frac{2}{n}}{t}^{\frac{2n+2}{n}}$$+\frac{l\left(4n-2l+4\right){\left(n+2\left(l-1\right)\right)}^{2}}{144{n}^{2}\left(n+2l\right){\mu }^{4}}{\left(\frac{k}{{B}_{n}}\right)}^{-\frac{4}{n}}{t}^{\frac{4n+4}{n}}\right].$(2.3)

Putting

$\zeta \left(t\right)={\xi }^{\prime }\left(t\right)\frac{n+2l}{{B}_{n}}{\left(\frac{k}{{B}_{n}}\right)}^{-\frac{n+2l}{n}}{t}^{\frac{2l}{n}+1}$

and noticing that $\mu \ge {\left(2\pi \right)}^{-n}{B}_{n}^{-\frac{1}{n}}V{\left(\mathrm{\Omega }\right)}^{\frac{n+1}{n}}$, from (2.3) we have

$\zeta \left(t\right)=-2l+\frac{l\left(2n-2l+2\right)}{6n{\mu }^{2}}{\left(\frac{k}{{B}_{n}}\right)}^{-\frac{2}{n}}{t}^{\frac{2n+2}{n}}+\frac{l\left(4n-2l+4\right){\left(n+2\left(l-1\right)\right)}^{2}}{144{n}^{2}\left(n+2l\right){\mu }^{4}}{\left(\frac{k}{{B}_{n}}\right)}^{-\frac{4}{n}}{t}^{\frac{4n+4}{n}}$$\le -2l+\frac{l\left(2n-2l+2\right)}{6n{\left(2\pi \right)}^{-2n}{B}_{n}^{-\frac{2}{n}}V{\left(\mathrm{\Omega }\right)}^{\frac{2\left(n+1\right)}{n}}}{\left(\frac{k}{{B}_{n}}\right)}^{-\frac{2}{n}}{t}^{\frac{2n+2}{n}}+\frac{l\left(4n-2l+4\right){\left(n+2\left(l-1\right)\right)}^{2}}{144{n}^{2}\left(n+2l\right){\left(2\pi \right)}^{-4n}{B}_{n}^{-\frac{4}{n}}V{\left(\mathrm{\Omega }\right)}^{\frac{4\left(n+1\right)}{n}}}{\left(\frac{k}{{B}_{n}}\right)}^{-\frac{4}{n}}{t}^{\frac{4n+4}{n}}.$(2.4)

Since the right-hand side of (2.4) is an increasing function of t, if the right-hand side of (2.4) is not larger than $0$ at $t={\left(2\pi \right)}^{-n}V\left(\mathrm{\Omega }\right)$, that is,

$\zeta \left(t\right)\le -2l+\frac{l\left(2n-2l+2\right)}{6n}{k}^{-\frac{2}{n}}\frac{{B}_{n}^{\frac{4}{n}}}{{\left(2\pi \right)}^{2}}+\frac{l\left(4n-2l+4\right){\left(n+2\left(l-1\right)\right)}^{2}}{144{n}^{2}\left(n+2l\right)}{k}^{-\frac{4}{n}}\frac{{B}_{n}^{\frac{8}{n}}}{{\left(2\pi \right)}^{4}}\le 0,$(2.5)

we can claim from (2.5) that ${\xi }^{\prime }\left(t\right)\le 0$ on $\left(0,{\left(2\pi \right)}^{-n}V\left(\mathrm{\Omega }\right)\right]$. If ${\xi }^{\prime }\left(t\right)\le 0$, then $\xi \left(t\right)$ is a decreasing function on $\left(0,{\left(2\pi \right)}^{-n}V\left(\mathrm{\Omega }\right)\right]$. In fact, by a direct calculation, we can obtain

$\zeta \left(t\right)\le -2l+\frac{l\left(2n-2l+2\right)}{6n}+\frac{l\left(4n-2l+4\right){\left(n+2\left(l-1\right)\right)}^{2}}{144{n}^{2}\left(n+2l\right)}\le 0,$

since

$\frac{{B}_{n}^{\frac{4}{n}}}{{\left(2\pi \right)}^{2}}<1.$

On the other hand, since $0<\varphi \left(0\right)\le {\left(2\pi \right)}^{-n}V\left(\mathrm{\Omega }\right)$ and the right-hand side of (2.1) is $\xi \left(\varphi \left(0\right)\right)$, which is a decreasing function of $\varphi \left(0\right)$ on $\left(0,{\left(2\pi \right)}^{-n}V\left(\mathrm{\Omega }\right)\right]$, we can replace $\varphi \left(0\right)$ by ${\left(2\pi \right)}^{-n}V\left(\mathrm{\Omega }\right)$ in (2.1), which gives the inequality

$\frac{1}{k}\sum _{j=1}^{k}{\lambda }_{j}\ge \frac{n}{n+2l}\frac{{\left(2\pi \right)}^{2l}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2l}{n}}}{k}^{\frac{2l}{n}}+\frac{l}{24\left(n+2l\right)}\frac{{\left(2\pi \right)}^{2\left(l-1\right)}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2\left(l-1\right)}{n}}}\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)}{k}^{\frac{2\left(l-1\right)}{n}}$$+\frac{l{\left(n+2\left(l-1\right)\right)}^{2}}{2304n{\left(n+2l\right)}^{2}}\frac{{\left(2\pi \right)}^{2\left(l-2\right)}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2\left(l-2\right)}{n}}}{\left(\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)}\right)}^{2}{k}^{\frac{2\left(l-2\right)}{n}}.$

This completes the proof of the theorem. ∎

Next, we will prove that the inequality (1.10) is sharper than (1.9).

#### Proof of Remark 1.4.

Under the same assumptions as in Lemma 2.1, letting $b=n$ and $A=\frac{k}{n{B}_{n}}$, from

$\mu \ge {\left(2\pi \right)}^{-n}{B}_{n}^{-\frac{1}{n}}V{\left(\mathrm{\Omega }\right)}^{\frac{n+1}{n}}$

we obtain that

$\frac{{\left(bA\right)}^{-\frac{2}{b}}\psi {\left(0\right)}^{2+\frac{2}{b}}}{{\mu }^{2}}\le \frac{{\left(2\pi \right)}^{-2n-2}V{\left(\mathrm{\Omega }\right)}^{2+\frac{2}{n}}}{{\left(2\pi \right)}^{-2n}{B}_{n}^{-\frac{2}{n}}V{\left(\mathrm{\Omega }\right)}^{\frac{2\left(n+1\right)}{n}}}{\left(\frac{k}{{B}_{n}}\right)}^{-\frac{2}{n}}=\frac{{\left(2\pi \right)}^{-2}}{{\left({B}_{n}\right)}^{-\frac{4}{n}}}{k}^{-\frac{2}{n}}<\frac{{\left(2\pi \right)}^{-2}}{{\left({B}_{n}\right)}^{-\frac{4}{n}}}<1$

and then we have

$\frac{1}{b+2l}\sum _{p=2}^{l}\frac{\left(l+1-p\right)}{{6}^{p}b\mathrm{\cdots }\left(b+2p-2\right){\mu }^{2p}}{\left(bA\right)}^{\frac{b+2\left(l-p\right)}{b}}\psi {\left(0\right)}^{\frac{2pb-2\left(l-p\right)}{b}}$$<\frac{1}{b+2l}\sum _{p=2}^{l}\frac{\left(l+1-p\right)}{{6}^{p}b\mathrm{\cdots }\left(b+2p-2\right){\mu }^{4}}{\left(bA\right)}^{\frac{b+2l-4}{b}}\psi {\left(0\right)}^{\frac{4b-2l+4}{b}}$$<\frac{l-1}{36\left(b+2l\right)b\left(b+2\right){\mu }^{4}}\sum _{p=0}^{\mathrm{\infty }}\frac{1}{{6}^{p}{\left(b+2\right)}^{p}}{\left(bA\right)}^{\frac{b+2l-4}{b}}\psi {\left(0\right)}^{\frac{4b-2l+4}{b}}$$=\frac{l-1}{6b\left(b+2l\right)\left(6\left(b+2\right)-1\right){\mu }^{4}}{\left(bA\right)}^{\frac{b+2l-4}{b}}\psi {\left(0\right)}^{\frac{4b-2l+4}{b}}.$(2.6)

By a direct calculation, we derive

$l\left(6\left(b+2\right)-1\right){\left(b+2\left(l-1\right)\right)}^{2}>24b\left(b+2l\right)\left(l-1\right)>0.$

In fact,

$l\left(6\left(b+2\right)-1\right){\left(b+2\left(l-1\right)\right)}^{2}-24b\left(b+2l\right)\left(l-1\right)=4b\left(l-1\right)\left(6b\left(l-1\right)-l\right)+l\left(6b+11\right)\left({b}^{2}+4{\left(l-1\right)}^{2}\right)$$>24{b}^{2}{\left(l-1\right)}^{2}+4bl\left(l-1\right)\left(6\left(l-1\right)-1\right)>0,$

that is,

$\frac{24b\left(b+2l\right)\left(l-1\right)}{l\left(6\left(b+2\right)-1\right){\left(b+2\left(l-1\right)\right)}^{2}}<1.$(2.7)

Therefore, from (2.6) and (2.7) we get

$\frac{1}{b+2l}\sum _{p=2}^{l}\frac{\left(l+1-p\right)}{{6}^{p}b\mathrm{\cdots }\left(b+2p-2\right){\mu }^{2p}}{\left(bA\right)}^{\frac{b+2\left(l-p\right)}{b}}\psi {\left(0\right)}^{\frac{2pb-2\left(l-p\right)}{b}}<\frac{l-1}{6b\left(b+2l\right)\left(6\left(b+2\right)-1\right){\mu }^{4}}{\left(bA\right)}^{\frac{b+2l-4}{b}}\psi {\left(0\right)}^{\frac{4b-2l+4}{b}}$$=\frac{24b\left(b+2l\right)\left(l-1\right)}{l\left(6\left(b+2\right)-1\right){\left(b+2\left(l-1\right)\right)}^{2}}$$×\frac{l{\left(b+2\left(l-1\right)\right)}^{2}}{144{b}^{2}{\left(b+2l\right)}^{2}{\mu }^{4}}{\left(bA\right)}^{\frac{b+2l-4}{b}}\psi {\left(0\right)}^{\frac{4b-2l+4}{b}}$$<\frac{l{\left(b+2\left(l-1\right)\right)}^{2}}{144{b}^{2}{\left(b+2l\right)}^{2}{\mu }^{4}}{\left(bA\right)}^{\frac{b+2l-4}{b}}\psi {\left(0\right)}^{\frac{4b-2l+4}{b}}.$(2.8)

Taking

$b=n,A=\frac{k}{n{B}_{n}},\psi \left(0\right)={\left(2\pi \right)}^{-n}V\left(\mathrm{\Omega }\right),\mu =2{\left(2\pi \right)}^{-n}\sqrt{V\left(\mathrm{\Omega }\right)I\left(\mathrm{\Omega }\right)}$(2.9)

and substituting (2.9) into (2.8), we have

$\frac{n}{n+2l}\frac{{\left(2\pi \right)}^{2l}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2l}{n}}}{k}^{\frac{2l}{n}}+\frac{l}{24\left(n+2l\right)}\frac{{\left(2\pi \right)}^{2\left(l-1\right)}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2\left(l-1\right)}{n}}}\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)}{k}^{\frac{2\left(l-1\right)}{n}}+\frac{l{\left(n+2\left(l-1\right)\right)}^{2}}{2304n{\left(n+2l\right)}^{2}}\frac{{\left(2\pi \right)}^{2\left(l-2\right)}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2\left(l-2\right)}{n}}}{\left(\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)}\right)}^{2}{k}^{\frac{2\left(l-2\right)}{n}}$$>\frac{n}{n+2l}\frac{{\left(2\pi \right)}^{2l}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2l}{n}}}{k}^{\frac{2l}{n}}+\frac{n}{\left(n+2l\right)}\sum _{p=1}^{l}\frac{l+1-p}{{\left(24\right)}^{p}n\mathrm{\cdots }\left(n+2p-2\right)}\frac{{\left(2\pi \right)}^{2\left(l-p\right)}}{{\left({B}_{n}V\left(\mathrm{\Omega }\right)\right)}^{\frac{2\left(l-p\right)}{n}}}{\left(\frac{V\left(\mathrm{\Omega }\right)}{I\left(\mathrm{\Omega }\right)}\right)}^{p}{k}^{\frac{2\left(l-p\right)}{n}}.$

This completes the proof of the remark. ∎

## A Proof of Lemma 2.1

In this section, we give the proof of Lemma 2.1.

#### Proof of Lemma 2.1.

Letting

$\varrho \left(t\right)=\frac{\psi \left(\frac{\psi \left(0\right)}{\mu }t\right)}{\psi \left(0\right)},$

we have $\varrho \left(0\right)=1$ and $-1\le {\varrho }^{\prime }\left(t\right)\le 0$. Without loss of generality, we can assume that

$\psi \left(0\right)=1\mathit{ }\text{and}\mathit{ }\mu =1.$

Define

${D}_{l}:={\int }_{0}^{\mathrm{\infty }}{s}^{b+2l-1}\psi \left(s\right)𝑑s.$

One can assume that ${D}_{l}<\mathrm{\infty }$, otherwise there is nothing to prove. Since ${D}_{l}<\mathrm{\infty }$, we can conclude that

$\underset{s\to \mathrm{\infty }}{lim}{s}^{b+2l-1}\psi \left(s\right)=0.$

Putting $h\left(s\right)=-{\psi }^{\prime }\left(s\right)$ for $s\ge 0$, we get

$0\le h\left(s\right)\le 1\mathit{ }\text{and}\mathit{ }{\int }_{0}^{\mathrm{\infty }}h\left(s\right)𝑑s=\psi \left(0\right)=1.$

By making use of integration by parts, one has

${\int }_{0}^{\mathrm{\infty }}{s}^{b}h\left(s\right)𝑑s=b{\int }_{0}^{\mathrm{\infty }}{s}^{b-1}\psi \left(s\right)𝑑s=bA$

and

${\int }_{0}^{\mathrm{\infty }}{s}^{b+2l}h\left(s\right)𝑑s\le \left(b+2l\right){D}_{l},$

since $\psi \left(s\right)>0$. By the same assertion as in [9], one can infer that there exists an $ϵ\ge 0$ such that

${\int }_{ϵ}^{ϵ+1}{s}^{b}𝑑s={\int }_{0}^{\mathrm{\infty }}{s}^{b}h\left(s\right)𝑑s=bA$(A.1)

and

${\int }_{ϵ}^{ϵ+1}{s}^{b+2l}𝑑s\le {\int }_{0}^{\mathrm{\infty }}{s}^{b+2l}h\left(s\right)𝑑s\le \left(b+2l\right){D}_{l}.$(A.2)

Letting

$\mathrm{\Theta }\left(s\right)=b{s}^{b+2l}-\left(b+2l\right){\tau }^{2l}{s}^{b}+2l{\tau }^{b+2l}-2l{\tau }^{b+2\left(l-1\right)}{\left(s-\tau \right)}^{2},$

we can prove that $\mathrm{\Theta }\left(s\right)\ge 0$. By integrating the function $\mathrm{\Theta }\left(s\right)$ from ϵ to $ϵ+1$, from (A.1) and (A.2) we deduce, for any $\tau >0$, that

$b\left(b+2l\right){D}_{l}-\left(b+2l\right){\tau }^{2l}bA+2l{\tau }^{b+2l}\ge \frac{l}{6}{\tau }^{b+2\left(l-1\right)}.$(A.3)

Defining

$f\left(\tau \right):=\left(b+2l\right){\tau }^{2l}bA-2l{\tau }^{b+2l}+\frac{l}{6}{\tau }^{b+2\left(l-1\right)},$

for any $\tau >0$, we can obtain from (A.3) that

${D}_{l}={\int }_{0}^{\mathrm{\infty }}{s}^{b+2l-1}\psi \left(s\right)𝑑s\ge \frac{f\left(\tau \right)}{b\left(b+2l\right)}.$

Then, taking

$\tau ={\left(bA\right)}^{\frac{1}{b}}{\left(1+\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{\frac{1}{b}},$

we have

$f\left(\tau \right)={\left(bA\right)}^{\frac{b+2l}{b}}\left(b-\frac{l\left(b+2\left(l-1\right)\right)}{6\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right){\left(1+\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{\frac{2l}{b}}$$+\frac{l}{6}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}{\left(1+\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{\frac{b+2\left(l-1\right)}{b}}.$

Next, we consider the following four cases.

Case 1. $b\mathrm{\ge }\mathrm{2}\mathit{}l$. For $t>0$, from Taylor’s formula we have

${\left(1+t\right)}^{\frac{2l}{b}}\ge 1+\frac{2l}{b}t+\frac{2l\left(2l-b\right)}{2{b}^{2}}{t}^{2}+\frac{2l\left(2l-b\right)\left(2l-2b\right)}{6{b}^{3}}{t}^{3}+\frac{2l\left(2l-b\right)\left(2l-2b\right)\left(2l-3b\right)}{24{b}^{4}}{t}^{4}$

and

${\left(1+t\right)}^{\frac{b+2\left(l-1\right)}{b}}\ge 1+\frac{2\left(l-1\right)+b}{b}t+\frac{\left(2\left(l-1\right)+b\right)\left(l-1\right)}{{b}^{2}}{t}^{2}+\frac{\left(2\left(l-1\right)-b\right)\left(l-1\right)\left(2\left(l-1\right)+b\right)}{3{b}^{3}}{t}^{3}.$

Putting

$t=\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}},$

we have from

${\left(bA\right)}^{\frac{2}{b}}\ge \frac{1}{{\left(b+1\right)}^{\frac{2}{b}}}\ge \frac{1}{3}>\frac{1}{4}$

that $t<\frac{1}{3}$ and $b-2lt>\frac{4l}{3}>0$ (see also [5]). Then, we obtain

$\left(b-\frac{l\left(b+2\left(l-1\right)\right)}{6\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right){\left(1+\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{\frac{2l}{b}}$$=\left(b-2lt\right){\left(1+t\right)}^{\frac{2l}{b}}$$\ge \left(b-2lt\right)\left(1+\frac{2l}{b}t+\frac{2l\left(2l-b\right)}{2{b}^{2}}{t}^{2}+\frac{2l\left(2l-b\right)\left(2l-2b\right)}{6{b}^{3}}{t}^{3}+\frac{2l\left(2l-b\right)\left(2l-2b\right)\left(2l-3b\right)}{24{b}^{4}}{t}^{4}\right)$$\ge b-\frac{l\left(2l+b\right)}{b}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{2}-\frac{\left(2l-b\right)\left(8{l}^{2}+4lb\right)}{6{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{3}$$-\frac{\left(2l-b\right)\left(2l-2b\right)\left(12{l}^{2}+6lb\right)}{24{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{4}$

and also

${\left(1+\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{\frac{b+2\left(l-1\right)}{b}}={\left(1+t\right)}^{\frac{b+2\left(l-1\right)}{b}}$$\ge 1+\frac{2\left(l-1\right)+b}{b}\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)$$+\frac{\left(2\left(l-1\right)+b\right)\left(l-1\right)}{{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{2}$$+\frac{\left(2\left(l-1\right)-b\right)\left(l-1\right)\left(2\left(l-1\right)+b\right)}{3{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{3}.$

Therefore, we have

$f\left(\tau \right)=\left(b+2l\right){\tau }^{2l}bA-2l{\tau }^{b+2l}+\frac{l}{6}{\tau }^{b+2\left(l-1\right)}$$\ge {\left(bA\right)}^{\frac{b+2l}{b}}\left[b-\frac{l\left(2l+b\right)}{b}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{2}-\frac{\left(2l-b\right)\left(8{l}^{2}+4lb\right)}{6{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{3}$$-\frac{\left(2l-b\right)\left(2l-2b\right)\left(12{l}^{2}+6lb\right)}{24{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{4}\right]$$+\frac{l}{6}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}\left[1+\frac{2\left(l-1\right)+b}{b}\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)$$+\frac{\left(2\left(l-1\right)+b\right)\left(l-1\right)}{{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{2}$$+\frac{\left(2\left(l-1\right)-b\right)\left(l-1\right)\left(2\left(l-1\right)+b\right)}{3{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{3}\right]$$=b{\left(bA\right)}^{\frac{b+2l}{b}}+\frac{l}{6}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}+\frac{l{\left(b+2\left(l-1\right)\right)}^{2}}{144b\left(b+2l\right)}{\left(bA\right)}^{\frac{b+2l-4}{b}}+{\eta }_{1},$

where

${\eta }_{1}=\frac{2l\left(l+b-3\right)\left(b+2l\right)}{3{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-6}{b}}$$+\frac{l\left(b+2\left(l-1\right)\right)\left(4\left(l-1\right)\left(2\left(l-1\right)-b\right)-3\left(2l-b\right)\left(l-b\right)\right)}{72{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-8}{b}}.$

Since

${\left(bA\right)}^{\frac{2}{b}}\ge \frac{1}{{\left(b+1\right)}^{\frac{2}{b}}}\ge \frac{1}{3}>\frac{1}{4}$

and $b\ge 2l$, we have

${\eta }_{1}\ge \frac{2l\left(l+b-3\right)\left(b+2l\right)}{12{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-8}{b}}$$+\frac{l\left(b+2\left(l-1\right)\right)\left(4\left(l-1\right)\left(2\left(l-1\right)-b\right)-3\left(2l-b\right)\left(l-b\right)\right)}{72{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-8}{b}}$$=\frac{l\left(9{b}^{3}+\left(35l-26\right){b}^{2}+\left(36{l}^{2}-90l\right)b+\left(4{l}^{3}-36{l}^{2}+48l-16\right)\right)}{72{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-8}{b}}$$\ge \frac{l\left(72{l}^{3}+\left(70{l}^{2}-52l\right)b+\left(36{l}^{2}-90l\right)b-36{l}^{2}\right)}{72{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-8}{b}}$$\ge \frac{l\left(\left(72{l}^{3}-36{l}^{2}\right)+\left(140l-52l\right)b+\left(72l-90l\right)b\right)}{72{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-8}{b}}$$\ge 0,$

which implies

$f\left(\tau \right)\ge b{\left(bA\right)}^{\frac{b+2l}{b}}+\frac{l}{6}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}+\frac{l{\left(b+2\left(l-1\right)\right)}^{2}}{144b\left(b+2l\right)}{\left(bA\right)}^{\frac{b+2l-4}{b}}.$

Case 2. $\mathrm{2}\mathit{}l\mathrm{-}\mathrm{2}\mathrm{\le }b\mathrm{<}\mathrm{2}\mathit{}l\mathrm{.}$ By using Taylor’s formula, for $t>0$, we obtain the inequalities

${\left(1+t\right)}^{\frac{2l}{b}}\ge 1+\frac{2l}{b}t+\frac{2l\left(2l-b\right)}{2{b}^{2}}{t}^{2}+\frac{2l\left(2l-b\right)\left(2l-2b\right)}{6{b}^{3}}{t}^{3}$

and

${\left(1+t\right)}^{\frac{b+2\left(l-1\right)}{b}}\ge 1+\frac{2\left(l-1\right)+b}{b}t+\frac{\left(2\left(l-1\right)+b\right)\left(l-1\right)}{{b}^{2}}{t}^{2}+\frac{\left(2\left(l-1\right)+b\right)\left(l-1\right)\left(2\left(l-1\right)-b\right)}{3{b}^{3}}{t}^{3}.$

Putting

$t=\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}},$

we have $b-2lt>\frac{l}{3}>0$,

$\left(b-\frac{l\left(b+2\left(l-1\right)\right)}{6\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right){\left(1+\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{\frac{2l}{b}}$$=\left(b-2lt\right){\left(1+t\right)}^{\frac{2l}{b}}$$\ge \left(b-2lt\right)\left(1+\frac{2l}{b}t+\frac{2l\left(2l-b\right)}{2{b}^{2}}{t}^{2}+\frac{2l\left(2l-b\right)\left(2l-2b\right)}{6{b}^{3}}{t}^{3}\right)$$=b-\frac{l\left(b+2l\right)}{b}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{2}-\frac{\left(2l-b\right)\left(8{l}^{2}+4lb\right)}{6{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{3}$$-\frac{4{l}^{2}\left(2l-b\right)\left(2l-2b\right)}{6{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{4},$

and

$\left(1+\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}{\right)}^{\frac{b+2\left(l-1\right)}{b}}={\left(1+t\right)}^{\frac{b+2\left(l-1\right)}{b}}$$\ge 1+\frac{2\left(l-1\right)+b}{b}\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)$$+\frac{\left(2\left(l-1\right)+b\right)\left(l-1\right)}{{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{2}$$+\frac{\left(2\left(l-1\right)+b\right)\left(l-1\right)\left(2\left(l-1\right)-b\right)}{3{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{3}.$

Furthermore, using the same method as in Case 1, we deduce that

$f\left(\tau \right)=\left(b+2l\right){\tau }^{2l}bA-2l{\tau }^{b+2l}+\frac{l}{6}{\tau }^{b+2\left(l-1\right)}$$\ge {\left(bA\right)}^{\frac{b+2l}{b}}\left[b-\frac{l\left(b+2l\right)}{b}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{2}-\frac{\left(2l-b\right)\left(8{l}^{2}+4lb\right)}{6{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{3}$$-\frac{4{l}^{2}\left(2l-b\right)\left(2l-2b\right)}{6{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{4}\right]$$+\frac{l}{6}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}\left[1+\frac{2\left(l-1\right)+b}{b}\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)$$+\frac{\left(2\left(l-1\right)+b\right)\left(l-1\right)}{{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{2}$$+\frac{\left(2\left(l-1\right)+b\right)\left(l-1\right)\left(2\left(l-1\right)-b\right)}{3{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{3}\right]$$=b{\left(bA\right)}^{\frac{b+2l}{b}}+\frac{l}{6}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}+\frac{l{\left(b+2\left(l-1\right)\right)}^{2}}{144b\left(b+2l\right)}{\left(bA\right)}^{\frac{b+2l-4}{b}}+{\eta }_{2},$

where

${\eta }_{2}=\frac{2l\left(l+b-3\right)\left(b+2l\right)}{3{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-6}{b}}$$+\frac{l\left(2\left(l-1\right)+b\right)\left(\left(l-1\right)\left(2\left(l-1\right)-b\right)\left(b+2l\right)-l\left(2l-b\right)\left(2l-2b\right)\right)}{18{b}^{3}\left(b+2l\right)}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-8}{b}}$$\ge \frac{2l\left(l+b-3\right)\left(b+2l\right)}{9{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-8}{b}}+\frac{l\left(2\left(l-1\right)+b\right)\left(l-1\right)\left(2\left(l-1\right)-b\right)}{18{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-8}{b}}$$=\left(\frac{4bl\left(l+b-3\right)\left(b+2l\right)}{18{b}^{3}}+\frac{l\left(2\left(l-1\right)+b\right)\left(l-1\right)\left(2\left(l-1\right)-b\right)}{18{b}^{3}}\right){\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-8}{b}}$$\ge \left(\frac{4bl\left(l+b-3\right)\left(b+2l\right)}{18{b}^{3}}+\frac{lb\left(b+2l\right)\left(2\left(l-1\right)-b\right)}{18{b}^{3}}\right){\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-8}{b}}$$\ge \frac{bl\left(b+2l\right)\left(6l+3b-14\right)}{18{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-8}{b}}$$\ge 0,$

since

${\left(bA\right)}^{\frac{2}{b}}\ge \frac{1}{{\left(b+1\right)}^{\frac{2}{b}}}\ge \frac{1}{3}.$

Therefore, we have

$f\left(\tau \right)\ge b{\left(bA\right)}^{\frac{b+2l}{b}}+\frac{l}{6}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}+\frac{l{\left(b+2\left(l-1\right)\right)}^{2}}{144b\left(b+2l\right)}{\left(bA\right)}^{\frac{b+2l-4}{b}}.$

Case 3. $l\mathrm{\le }b\mathrm{<}\mathrm{2}\mathit{}l\mathrm{-}\mathrm{2}$. By using Taylor’s formula, for $t>0$, we have

${\left(1+t\right)}^{\frac{2l}{b}}\ge 1+\frac{2l}{b}t+\frac{l\left(2l-b\right)}{{b}^{2}}{t}^{2}+\frac{l\left(2l-b\right)\left(2l-2b\right)}{3{b}^{3}}{t}^{3}$

and

${\left(1+t\right)}^{\frac{b+2\left(l-1\right)}{b}}\ge 1+\frac{2\left(l-1\right)+b}{b}t+\frac{\left(2\left(l-1\right)+b\right)\left(l-1\right)}{{b}^{2}}{t}^{2}.$

Putting

$t=\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}>0,$

we have $b-2lt>0$,

$\left(b-\frac{l\left(b+2\left(l-1\right)\right)}{6\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right){\left(1+\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{\frac{2l}{b}}$$=\left(b-2lt\right){\left(1+t\right)}^{\frac{2l}{b}}$$\ge \left(b-2lt\right)\left(1+\frac{2l}{b}t+\frac{l\left(2l-b\right)}{{b}^{2}}{t}^{2}+\frac{l\left(2l-b\right)\left(2l-2b\right)}{3{b}^{3}}{t}^{3}\right)$$=b-\frac{l\left(b+2l\right)}{b}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{2}-\frac{\left(2l-b\right)\left(4{l}^{2}+2lb\right)}{3{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{3}$$-\frac{4{l}^{2}\left(2l-b\right)\left(l-b\right)}{3{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{4},$

and

${\left(1+\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{\frac{b+2\left(l-1\right)}{b}}={\left(1+t\right)}^{\frac{b+2\left(l-1\right)}{b}}$$\ge 1+\frac{2\left(l-1\right)+b}{b}\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}$$+\frac{\left(2\left(l-1\right)+b\right)\left(l-1\right)}{{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{2}.$

By the same argument as in Case 2, we can deduce that

$f\left(\tau \right)=\left(b+2l\right){\tau }^{2l}bA-2l{\tau }^{b+2l}+\frac{l}{6}{\tau }^{b+2\left(l-1\right)}$$\ge {\left(bA\right)}^{\frac{b+2l}{b}}\left[b-\frac{l\left(b+2l\right)}{b}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{2}-\frac{\left(2l-b\right)\left(4{l}^{2}+2lb\right)}{3{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{3}$$-\frac{4{l}^{2}\left(2l-b\right)\left(l-b\right)}{3{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{4}\right]+\frac{l}{6}\left(bA\right){}^{\frac{b+2\left(l-1\right)}{b}}\left[1+\frac{2\left(l-1\right)+b}{b}\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}$$+\frac{\left(2\left(l-1\right)+b\right)\left(l-1\right)}{{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{2}\right]$$=b{\left(bA\right)}^{\frac{b+2l}{b}}+\frac{l}{6}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}+\frac{l{\left(b+2\left(l-1\right)\right)}^{2}}{144b\left(b+2l\right)}{\left(bA\right)}^{\frac{b+2l-4}{b}}+{\eta }_{3},$

where

${\eta }_{3}=\frac{2l\left(l+b-3\right)\left(b+2l\right)}{3{b}^{2}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{3}{\left(bA\right)}^{\frac{b+2l-6}{b}}-\frac{4{l}^{2}\left(2l-b\right)\left(l-b\right)}{3{b}^{3}}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{4}{\left(bA\right)}^{\frac{b+2l-8}{b}}\ge 0.$

Therefore, we have

$f\left(\tau \right)\ge b{\left(bA\right)}^{\frac{b+2l}{b}}+\frac{l}{6}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}+\frac{l{\left(b+2\left(l-1\right)\right)}^{2}}{144b\left(b+2l\right)}{\left(bA\right)}^{\frac{b+2l-4}{b}}.$

Case 4. $\mathrm{2}\mathrm{\le }b\mathrm{<}l$. Since $2\le b, there exists a positive integer k such that $2\le k-1\le \frac{2l}{b} and, for $t>0$, we have

${\left(1+t\right)}^{\frac{2l}{b}}\ge 1+\frac{2l}{b}t+\frac{1}{2!}\frac{2l}{b}\left(\frac{2l}{b}-1\right){t}^{2}+\frac{1}{3!}\frac{2l}{b}\left(\frac{2l}{b}-1\right)\left(\frac{2l}{b}-2\right){t}^{3}+\mathrm{\cdots }+\frac{1}{\left(k+1\right)!}\frac{2l}{b}\left(\frac{2l}{b}-1\right)\mathrm{\cdots }\left(\frac{2l}{b}-k\right){t}^{k+1}$$=1+\sum _{p=0}^{k}\left\{\frac{1}{\left(p+1\right)!}\prod _{q=0}^{p}\left(\frac{2l}{b}-q\right)\right\}{t}^{p+1}$

and

${\left(1+t\right)}^{\frac{b+2l}{b}}\le 1+\frac{b+2l}{b}t+\frac{1}{2!}\frac{b+2l}{b}\frac{2l}{b}{t}^{2}+\frac{1}{3!}\frac{b+2l}{b}\frac{2l}{b}\left(\frac{2l}{b}-1\right){t}^{3}+\mathrm{\cdots }$$+\frac{1}{\left(k+1\right)!}\frac{b+2l}{b}\frac{2l}{b}\left(\frac{2l}{b}-1\right)\mathrm{\cdots }\left(\frac{2l}{b}-\left(k-1\right)\right){t}^{k+1}$$=1+\sum _{p=0}^{k}\left\{\frac{1}{\left(p+1\right)!}\prod _{q=0}^{p}\left(\frac{2l}{b}-q+1\right)\right\}{t}^{p+1},$

and also

${\left(1+t\right)}^{\frac{b+2\left(l-1\right)}{b}}\ge 1+\frac{2\left(l-1\right)+b}{b}t+\frac{1}{2!}\frac{\left(2\left(l-1\right)+b\right)}{b}\frac{2\left(l-1\right)}{b}{t}^{2}+\frac{1}{3!}\frac{\left(2\left(l-1\right)+b\right)}{b}\frac{2\left(l-1\right)}{b}\left(\frac{2\left(l-1\right)}{b}-1\right){t}^{3}+\mathrm{\cdots }$$+\frac{1}{k!}\frac{\left(2\left(l-1\right)+b\right)}{b}\frac{2\left(l-1\right)}{b}\mathrm{\cdots }\left(\frac{2\left(l-1\right)}{b}-\left(k-2\right)\right){t}^{k}$$-|\frac{1}{\left(k+1\right)!}\frac{\left(2\left(l-1\right)+b\right)}{b}\frac{2\left(l-1\right)}{b}\mathrm{\cdots }\left(\frac{2\left(l-1\right)}{b}-\left(k-1\right)\right)|{t}^{k+1}$$=1+\sum _{p=0}^{k-1}\left\{\frac{1}{\left(p+1\right)!}\prod _{q=0}^{p}\left(\frac{2\left(l-1\right)}{b}-q+1\right)\right\}{t}^{p+1}-|\frac{1}{\left(k+1\right)!}\prod _{q=0}^{k}\left(\frac{2\left(l-1\right)}{b}-q+1\right)|{t}^{k+1}.$

Putting

$t=\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}$

and $f\left(\tau \right)={\left(bA\right)}^{\frac{b+2l}{b}}h\left(\tau \right)$, where

$h\left(\tau \right)=\left(b+2l\right){\left(1+t\right)}^{\frac{2l}{b}}-2l{\left(1+t\right)}^{\frac{b+2l}{b}}+\frac{1}{6}{\left(bA\right)}^{-\frac{2}{b}}{\left(1+t\right)}^{\frac{b+2\left(l-1\right)}{b}},$

then, for $2\le b, we have

$h\left(\tau \right)\ge \left(b+2l\right)\left\{1+\sum _{p=0}^{k}\left[\frac{1}{\left(p+1\right)!}\prod _{q=0}^{p}\left(\frac{2l}{b}-q\right)\right]{t}^{p+1}\right\}-2l\left\{1+\sum _{p=0}^{k}\left[\frac{1}{\left(p+1\right)!}\prod _{q=0}^{p}\left(\frac{2l}{b}-q+1\right)\right]{t}^{p+1}\right\}$$+\frac{l}{6}{\left(bA\right)}^{-\frac{2}{b}}\left\{1+\sum _{p=0}^{k-1}\left[\frac{1}{\left(p+1\right)!}\prod _{q=0}^{p}\left(\frac{2\left(l-1\right)}{b}-q+1\right)\right]{t}^{p+1}-|\frac{1}{\left(k+1\right)!}\prod _{q=0}^{k}\left(\frac{2\left(l-1\right)}{b}-q+1\right)|{t}^{k+1}\right\}$$=b+\frac{l}{6}{\left(bA\right)}^{-\frac{2}{b}}+\sum _{p=1}^{k}\left\{\frac{b+2l}{\left(p+1\right)!}\frac{2l}{b}\left[\prod _{q=1}^{p}\left(\frac{2l}{b}-q\right)-\prod _{q=1}^{p}\left(\frac{2l}{b}-q+1\right)\right]\right\}{t}^{p+1}$$+\sum _{p=0}^{k-1}\left\{\frac{l{\left(bA\right)}^{-\frac{2}{b}}}{6\left(p+1\right)!}\prod _{q=0}^{p}\left(\frac{2\left(l-1\right)}{b}-q+1\right)\right\}{t}^{p+1}-|\frac{l{\left(bA\right)}^{-\frac{2}{b}}}{6\left(k+1\right)!}\prod _{q=0}^{k}\left(\frac{2\left(l-1\right)}{b}-q+1\right)|{t}^{k+1}$$=b+\frac{l}{6}{\left(bA\right)}^{-\frac{2}{b}}-\sum _{p=1}^{k}\left\{\frac{p2l\left(b+2l\right)}{b\left(p+1\right)!}\prod _{q=1}^{p-1}\left(\frac{2l}{b}-q\right)\right\}{t}^{p+1}+\sum _{p=1}^{k}\left\{\frac{l{\left(bA\right)}^{-\frac{2}{b}}}{6p!}\prod _{q=0}^{p-1}\left(\frac{2\left(l-1\right)}{b}-q+1\right)\right\}{t}^{p}$$-|\frac{l{\left(bA\right)}^{-\frac{2}{b}}}{6\left(k+1\right)!}\prod _{q=0}^{k}\left(\frac{2\left(l-1\right)}{b}-q+1\right)|{t}^{k+1}$$=b+\frac{l}{6}{\left(bA\right)}^{-\frac{2}{b}}-\sum _{p=1}^{k}\left\{\frac{p}{{b}^{p}\left(p+1\right)!}\prod _{q=0}^{p}\left(2l-\left(q-1\right)b\right)\right\}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{p+1}$$+\sum _{p=1}^{k}\left\{\frac{l{\left(bA\right)}^{-\frac{2}{b}}}{6{b}^{p}p!}\prod _{q=0}^{p-1}\left(2\left(l-1\right)-\left(q-1\right)b\right)\right\}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{p}$$-|\frac{l{\left(bA\right)}^{-\frac{2}{b}}}{6{b}^{k+1}\left(k+1\right)!}\prod _{q=0}^{k}\left(2\left(l-1\right)-\left(q-1\right)b\right)|{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}{\left(bA\right)}^{-\frac{2}{b}}\right)}^{k+1}.$

Furthermore, we have

$f\left(\tau \right)\ge b{\left(bA\right)}^{\frac{b+2l}{b}}+\frac{l}{6}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}+\frac{l{\left(b+2\left(l-1\right)\right)}^{2}}{144b\left(b+2l\right)}{\left(bA\right)}^{\frac{b+2l-4}{b}}$$-\sum _{p=2}^{k}\left\{\frac{p}{{b}^{p}\left(p+1\right)!}\prod _{q=0}^{p}\left(2l-\left(q-1\right)b\right)\right\}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{p+1}{\left(bA\right)}^{\frac{b+2l-2p-2}{b}}$$+\sum _{p=2}^{k}\left\{\frac{l}{6{b}^{p}p!}\prod _{q=0}^{p-1}\left(2\left(l-1\right)-\left(q-1\right)b\right)\right\}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{p}{\left(bA\right)}^{\frac{b+2l-2p-2}{b}}$$-|\frac{l}{6{b}^{k+1}\left(k+1\right)!}\prod _{q=0}^{k}\left(2\left(l-1\right)-\left(q-1\right)b\right)|{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{k+1}{\left(bA\right)}^{\frac{b+2l-2k-4}{b}}$$=b{\left(bA\right)}^{\frac{b+2l}{b}}+\frac{l}{6}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}+\frac{l{\left(b+2\left(l-1\right)\right)}^{2}}{144b\left(b+2l\right)}{\left(bA\right)}^{\frac{b+2l-4}{b}}+{\eta }_{4},$

where

${\eta }_{4}=\sum _{p=2}^{k}\left\{\frac{\left(b+2\left(l-1\right)\right)2l}{12{b}^{p}p!}\left[\prod _{q=1}^{p-1}\left(2\left(l-1\right)-\left(q-1\right)b\right)-\frac{p}{p+1}\prod _{q=1}^{p-1}\left(2l-qb\right)\right]\right\}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{p}{\left(bA\right)}^{\frac{b+2l-2p-2}{b}}$$-|\frac{l}{6{b}^{k+1}\left(k+1\right)!}\prod _{q=0}^{k}\left(2\left(l-1\right)-\left(q-1\right)b\right)|{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{k+1}{\left(bA\right)}^{\frac{b+2l-2k-4}{b}}.$

From $k-2\le \frac{2\left(l-1\right)}{b} we have

$\frac{k-2-i}{k+1-i}\le \frac{\frac{2\left(l-1\right)}{b}-i}{k+1-i}<\frac{k-i}{k+1-i}.$(A.4)

Then, from (A.4) we have

$|\frac{\frac{2\left(l-1\right)}{b}-i}{k+1-i}|\le 1,i=0,1,2,\mathrm{\dots },k-1.$

Note that

$2\left(l-1\right)-\left(q-1\right)b\ge 2l-qb\ge 0,p=2,3,\mathrm{\dots },k,$

and

$\prod _{q=1}^{p-1}\left(2\left(l-1\right)-\left(q-1\right)b\right)-\frac{p}{p+1}\prod _{q=1}^{p-1}\left(2l-qb\right)\ge \prod _{q=1}^{p-1}\left(2\left(l-1\right)-\left(q-1\right)b\right)-\prod _{q=1}^{p-1}\left(2l-qb\right)\ge 0.$

Therefore, we obtain

$\sum _{p=2}^{k}\left\{\frac{\left(b+2\left(l-1\right)\right)2l}{12{b}^{p}p!}\left[\prod _{q=1}^{p-1}\left(2\left(l-1\right)-\left(q-1\right)b\right)-\frac{p}{p+1}\prod _{q=1}^{p-1}\left(2l-qb\right)\right]\right\}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{p}{\left(bA\right)}^{\frac{b+2l-2p-2}{b}}$$\ge \frac{\left(b+2\left(l-1\right)\right)2l}{24{b}^{2}}\left(2\left(l-1\right)-\frac{2\left(2l-b\right)}{3}\right){\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{2}{\left(bA\right)}^{\frac{b+2l-6}{b}}.$

From

${\left(bA\right)}^{\frac{2}{b}}\ge \frac{1}{{\left(b+1\right)}^{\frac{2}{b}}}\ge \frac{1}{3}$

we have

${\eta }_{4}\ge \frac{\left(b+2\left(l-1\right)\right)2l}{24{b}^{2}}\left(2\left(l-1\right)-\frac{2\left(2l-b\right)}{3}\right){\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{2}{\left(bA\right)}^{\frac{b+2l-6}{b}}$$-|\frac{l}{6{b}^{k+1}\left(k+1\right)!}\prod _{q=0}^{k}\left(2\left(l-1\right)-\left(q-1\right)b\right)|{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{k+1}{\left(bA\right)}^{\frac{b+2l-2k-4}{b}}$$=\frac{l\left(b+2\left(l-1\right)\right)}{12{b}^{2}}\left\{\left(2\left(l-1\right)-\frac{2\left(2l-b\right)}{3}\right)$$-2b|\frac{1}{{b}^{k}\left(k+1\right)!}\prod _{q=1}^{k}\left(2\left(l-1\right)-\left(q-1\right)b\right)|{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{k-1}{\left(bA\right)}^{\frac{-2k+2}{b}}\right\}\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right){}^{2}\left(bA\right){}^{\frac{b+2l-6}{b}}$$\ge \frac{l\left(b+2\left(l-1\right)\right)}{12{b}^{2}}\left\{\frac{2l+2b-6}{3}-2b|\prod _{q=0}^{k-1}\left(\frac{\frac{2\left(l-1\right)}{b}-q}{k+1-q}\right)|{\left(\frac{1}{4}\right)}^{k-1}\right\}{\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{2}{\left(bA\right)}^{\frac{b+2l-6}{b}}$$\ge \frac{l\left(b+2\left(l-1\right)\right)}{12{b}^{2}}\left(\frac{2l+2b-6}{3}-\frac{b}{8}\right){\left(\frac{b+2\left(l-1\right)}{12\left(b+2l\right)}\right)}^{2}{\left(bA\right)}^{\frac{b+2l-6}{b}}$$\ge 0,$

which implies that

$f\left(\tau \right)\ge b{\left(bA\right)}^{\frac{b+2l}{b}}+\frac{l}{6}{\left(bA\right)}^{\frac{b+2\left(l-1\right)}{b}}+\frac{l{\left(b+2\left(l-1\right)\right)}^{2}}{144b\left(b+2l\right)}{\left(bA\right)}^{\frac{b+2l-4}{b}}.$

This completes the proof of the lemma. ∎

The authors wish to express their gratitude to Professor Qing-Ming Cheng for his continuous encouragement and his enthusiastic help. The authors would also like to thank the referees for their useful comments and suggestions.

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## About the article

Revised: 2014-11-11

Accepted: 2014-11-16

Published Online: 2016-02-01

Published in Print: 2016-02-01

Funding Source: National Natural Science Foundation of China

Award identifier / Grant number: 11371150

Award identifier / Grant number: 11401268

The first author acknowledges the support of the NSFC (Grant No. 11371150) and of the project Pearl River New Star of Guangzhou (Grant No. 2012J2200028). The second author acknowledges the support of the NSFC (Grant No. 11401268).

Citation Information: Advanced Nonlinear Studies, Volume 16, Issue 1, Pages 31–44, ISSN (Online) 2169-0375, ISSN (Print) 1536-1365,

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