# A supplement to the convergence rate in a theorem of Heyde

## Abstract

Let $\left\{X,{X}_{n},n\ge 1\right\}$ be a sequence of i.i.d. random variables with zero mean, set ${S}_{n}={\sum }_{k=1}^{n}{X}_{k}$, $E{X}^{2}={\sigma }^{2}>0$, and $\lambda \left(ϵ\right)={\sum }_{n=1}^{\mathrm{\infty }}P\left(|{S}_{n}|\ge nϵ\right)$. In this paper, the authors discuss the rate of approximation of ${\sigma }^{2}$ by ${ϵ}^{2}\lambda \left(ϵ\right)$ under suitable conditions, improve the results of Klesov (Theory Probab. Math. Stat. 49:83-87, 1994), and extend the work He and Xie (Acta Math. Appl. Sin. 2012, doi:10.1007/s10255-012-0138-6).

MSC:60F15, 60G50.

## 1 Introduction and main results

Let $\left\{X,{X}_{n},n\ge 1\right\}$ be a sequence of i.i.d. random variables, set ${S}_{n}={\sum }_{k=1}^{n}{X}_{k}$, and $\lambda \left(ϵ\right)={\sum }_{n=1}^{\mathrm{\infty }}P\left(|{S}_{n}|\ge nϵ\right)$. Heyde  proved that

$\underset{ϵ\to 0}{lim}{ϵ}^{2}\lambda \left(ϵ\right)={\sigma }^{2},$

whenever $E{X}^{2}={\sigma }^{2}<\mathrm{\infty }$ and $EX=0$.

There are various extensions of this result: Chen , Gut and Spǎtara , Lanzinger and Stadtmüller . Liu and Lin  introduced a new kind of complete moment convergence; Klesov  studied the rate of approximation of ${\sigma }^{2}$ by ${ϵ}^{2}\lambda \left(ϵ\right)$ and proved the following Theorem A.

Theorem A Let$\left\{X,{X}_{n},n\ge 1\right\}$be a sequence of i.i.d. random variables with zero mean, if$E{X}^{2}={\sigma }^{2}>0$, and$E{|X|}^{3}<\mathrm{\infty }$, then

${ϵ}^{2}\lambda \left(ϵ\right)-{\sigma }^{2}=o\left({ϵ}^{1/2}\right),\phantom{\rule{1em}{0ex}}\mathit{\text{as}}ϵ\to 0.$

Recently, He and Xie  obtained Theorem B which improved Theorem A. Gut and Steinebach  extended the results of Klesov .

Theorem B Let$\left\{X,{X}_{n},n\ge 1\right\}$be a sequence of i.i.d. random variables, and$0<\delta \le 1$, if

$EX=0,\phantom{\rule{2em}{0ex}}E{X}^{2}={\sigma }^{2}>0\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}E{|X|}^{2+\delta }<\mathrm{\infty },$

then

${ϵ}^{2}\lambda \left(ϵ\right)-{\sigma }^{2}=\left\{\begin{array}{cc}O\left(ϵ\right),\hfill & \delta =1,\hfill \\ o\left({ϵ}^{\delta }\right),\hfill & 0<\delta <1.\hfill \end{array}$

Let G be the set of functions $g\left(x\right)$ that are defined for all real x and satisfy the following conditions: (a) $g\left(x\right)$ is nonnegative, even, nondecreasing in the interval $x>0$, and $g\left(x\right)\ne 0$ for $x\ne 0$; (b) $\frac{x}{g\left(x\right)}$ is nondecreasing in the interval $x>0$.

Let ${G}_{0}$ be the set of functions $g\left(x\right)\in G$ satisfying the supplementary condition (c) ${lim}_{x\to \mathrm{\infty }}\frac{g\left({x}^{2}\right)}{xg\left(x\right)}=0$. Obviously, the function $g\left(x\right)={|x|}^{\delta }$ with $0<\delta <1$ belongs to ${G}_{0}$ and does not belong to ${G}_{0}$ if $\delta =1$. The purpose of this paper is to generalize Theorem B to the case where the condition $E{|X|}^{2+\delta }<\mathrm{\infty }$ is replaced by a more general condition $E{|X|}^{2}g\left(X\right)<\mathrm{\infty }$ in which the function g belongs to some subset of G. Denote ${T}_{g}\left(v\right)=E{X}^{2}g\left(X\right)I\left(|X|>v\right)$, ${T}_{g}\left(v\right)$ is a nonnegative nonincreasing function in the interval $v>0$, and ${lim}_{v\to \mathrm{\infty }}{T}_{g}\left(v\right)=0$ with $E{X}^{2}g\left(X\right)<\mathrm{\infty }$. Now we state our results as follows.

Theorem 1.1 Let$\left\{X,{X}_{n};n\ge 1\right\}$be a sequence of i.i.d. random variables with zero mean and$E{X}^{2}={\sigma }^{2}>0$, if$E{X}^{2}g\left(X\right)<\mathrm{\infty }$for some function$g\left(x\right)\in G$, and

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{ng\left(\sqrt{n}\right)}<\mathrm{\infty },$
(1.1)

then

${ϵ}^{2}\lambda \left(ϵ\right)-{\sigma }^{2}=O\left({ϵ}^{1/2}\right)+o\left(1\right)\left({h}_{1}\left(ϵ\right)+{f}_{1}\left(ϵ\right)\right),\phantom{\rule{1em}{0ex}}\mathit{\text{as}}ϵ\to 0,$
(1.2)

where${f}_{1}\left(ϵ\right)={\sum }_{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{1}{ng\left(\sqrt{n}\right)}$, ${h}_{1}\left(ϵ\right)={ϵ}^{2}{\sum }_{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{1}{g\left(\sqrt{n}\right)}$.

Theorem 1.2 Under the conditions of Theorem 1.1, and$g\left(x\right)\in {G}_{0}$, then

${ϵ}^{2}\lambda \left(ϵ\right)-{\sigma }^{2}=o\left(1\right)\left({h}_{1}\left(ϵ\right)+{f}_{1}\left(ϵ\right)\right),\phantom{\rule{1em}{0ex}}\mathit{\text{as}}ϵ\to 0.$
(1.3)

Throughout this paper, we suppose that C denotes a constant which only depends on some given numbers and may be different at each appearance, and that $\left[x\right]$ denotes the integer part of x.

## 2 Proofs of the main results

Before we prove the main results we state some lemmas. Lemma 2.1 is from . $\mathrm{\Phi }\left(x\right)$ is the standard normal distribution function, $\mathrm{\Phi }\left(x\right)=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{x}{e}^{-{t}^{2}/2}\phantom{\rule{0.2em}{0ex}}dt$.

Lemma 2.1 Let$\left\{X,{X}_{n},n\ge 1\right\}$be a sequence of i.i.d. standard normal distribution random variables. Then

${ϵ}^{2}\lambda \left(ϵ\right)={ϵ}^{2}\sum _{n=1}^{\mathrm{\infty }}\frac{2}{\sqrt{2\pi }}{\int }_{ϵ\sqrt{n}}^{\mathrm{\infty }}{e}^{-{t}^{2}/2}\phantom{\rule{0.2em}{0ex}}dt=1-\frac{{ϵ}^{2}}{2}+O\left({ϵ}^{3}\right),\phantom{\rule{1em}{0ex}}\mathit{\text{as}}ϵ\to 0.$
(2.1)

If$\left\{{X}_{n},n\ge 1\right\}$is a sequence of independent random variables with zero mean and finite variance, and put$E{X}_{j}^{2}={\sigma }_{j}^{2}$, ${B}_{n}={\sum }_{j=1}^{n}{\sigma }_{j}^{2}$, Bikelisobtained the following inequality: for every x, where${V}_{j}\left(x\right)=P\left({X}_{j}is the distribution function of the random variable${X}_{j}$. By applying the above inequality to the sequence of i.i.d. random variables with zero mean and variance 1, and letting$|x|=ϵ\sqrt{n}$, we have the following lemma.

Lemma 2.2 Let$\left\{X,{X}_{n},n\ge 1\right\}$be a sequence of i.i.d. random variables with zero mean and$E{X}^{2}=1$. Then for any given$ϵ>0$, we have where$V\left(x\right)=P\left(Xis the distribution function of a random variable X.

Proof of Theorem 1.1 Without loss of generality, we suppose that ${\sigma }^{2}=1$, $0<ϵ<1$, and write

${ϵ}^{2}\lambda \left(ϵ\right)=I+{ϵ}^{2}\sum _{n=1}^{\mathrm{\infty }}\frac{2}{\sqrt{2\pi }}{\int }_{ϵ\sqrt{n}}^{\mathrm{\infty }}{e}^{-{t}^{2}/2}\phantom{\rule{0.2em}{0ex}}dt,$

where

$I={ϵ}^{2}\sum _{n=1}^{\mathrm{\infty }}\left(P\left(|{S}_{n}|>nϵ\right)-\frac{2}{\sqrt{2\pi }}{\int }_{ϵ\sqrt{n}}^{\mathrm{\infty }}{e}^{-{t}^{2}/2}\phantom{\rule{0.2em}{0ex}}dt\right).$

Applying Lemma 2.1, we obtain

${ϵ}^{2}\lambda \left(ϵ\right)=I+1-\frac{{ϵ}^{2}}{2}+O\left({ϵ}^{3}\right),$

then

${ϵ}^{2}\lambda \left(ϵ\right)-1=-\frac{{ϵ}^{2}}{2}+{ϵ}^{2}\sum _{n=1}^{\mathrm{\infty }}{R}_{n}+O\left({ϵ}^{3}\right),$

here ${R}_{n}=P\left(|{S}_{n}|>nϵ\right)-\frac{2}{\sqrt{2\pi }}{\int }_{ϵ\sqrt{n}}^{\mathrm{\infty }}{e}^{-{t}^{2}/2}\phantom{\rule{0.2em}{0ex}}dt$. By Lemma 2.2,

$|{R}_{n}|\le {R}_{1n}+{R}_{2n},$

where We obtain

${ϵ}^{2}\lambda \left(ϵ\right)-1={ϵ}^{2}\sum _{n=1}^{\mathrm{\infty }}{R}_{1n}+{ϵ}^{2}\sum _{n=1}^{\mathrm{\infty }}{R}_{2n}+O\left({ϵ}^{2}\right).$
(2.2)

Firstly, we estimate ${ϵ}^{2}{\sum }_{n=1}^{\mathrm{\infty }}{R}_{1n}$. Note that

${ϵ}^{2}\sum _{n=1}^{\mathrm{\infty }}{R}_{1n}={ϵ}^{2}\sum _{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}{R}_{1n}+{ϵ}^{2}\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}{R}_{1n}=:{T}_{1}+{T}_{2}.$

Applying the condition $E{X}^{2}g\left(X\right)<\mathrm{\infty }$, we have

$\underset{n\to \mathrm{\infty }}{lim}{\int }_{|u|>\sqrt{n}}{u}^{2}g\left(u\right)\phantom{\rule{0.2em}{0ex}}dV\left(u\right)=0.$

Therefore, for any $\eta >0$, there is an integer ${N}_{0}$ such that ${\int }_{|u|>\sqrt{n}}{u}^{2}g\left(u\right)\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\le \eta$, whenever $n>{N}_{0}$. Hence

$\begin{array}{rcl}{T}_{1}& \le & C{ϵ}^{2}\sum _{n=1}^{{N}_{0}}{\int }_{|u|>\sqrt{n}}{u}^{2}\phantom{\rule{0.2em}{0ex}}dV\left(u\right)+C{ϵ}^{2}\sum _{n={N}_{0}+1}^{\left[\frac{1}{{ϵ}^{2}}\right]}{\left(1+ϵ\sqrt{n}\right)}^{-2}{\int }_{|u|>\left(1+ϵ\sqrt{n}\right)\sqrt{n}}{u}^{2}\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C{ϵ}^{2}{N}_{0}+C{ϵ}^{2}\eta \sum _{n={N}_{0}+1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{1}{{\left(1+ϵ\sqrt{n}\right)}^{2}g\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}\\ \le & C{ϵ}^{2}\left({N}_{0}+\eta \sum _{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{1}{g\left(\sqrt{n}\right)}\right)\\ =& C{h}_{1}\left(ϵ\right)\left(\frac{{N}_{0}}{{\sum }_{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{1}{g\left(\sqrt{n}\right)}}+\eta \right)\\ \le & C{h}_{1}\left(ϵ\right)\left({N}_{0}ϵ+\eta \right)\\ =& o\left({h}_{1}\left(ϵ\right)\right),\end{array}$
(2.3)

where ${h}_{1}\left(ϵ\right)={ϵ}^{2}{\sum }_{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{1}{g\left(\sqrt{n}\right)}$. For ${T}_{2}$, noting that $g\left(x\right)\in G$, we have the following inequality:

$\begin{array}{rcl}{T}_{2}& \le & C{ϵ}^{2}\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{1}{n{ϵ}^{2}}{\int }_{|u|>\sqrt{n}\left(1+ϵ\sqrt{n}\right)}{u}^{2}\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{1}{ng\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}{\int }_{|u|>\sqrt{n}\left(1+ϵ\sqrt{n}\right)}{u}^{2}g\left(u\right)\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{1}{ng\left(\sqrt{n}\right)}{\int }_{|u|>\frac{1}{ϵ}}{u}^{2}g\left(u\right)\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C{T}_{g}\left(\frac{1}{ϵ}\right){f}_{1}\left(ϵ\right).\end{array}$
(2.4)

Next, we estimate the second term of (2.2). Note that

$\begin{array}{rcl}{ϵ}^{2}\sum _{n=1}^{\mathrm{\infty }}{R}_{2n}& =& C{ϵ}^{2}\sum _{n=1}^{\mathrm{\infty }}{n}^{-1/2}{\left(1+ϵ\sqrt{n}\right)}^{-3}{\int }_{|u|\le {\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}}{|u|}^{3}\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ +C{ϵ}^{2}\sum _{n=1}^{\mathrm{\infty }}{n}^{-1/2}{\left(1+ϵ\sqrt{n}\right)}^{-3}{\int }_{{\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}<|u|<\sqrt{n}\left(1+ϵ\sqrt{n}\right)}{|u|}^{3}\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ =:& {J}_{1}+{J}_{2}.\end{array}$

For ${J}_{1}$, we can write

$\begin{array}{rcl}{J}_{1}& =& C{ϵ}^{2}\left(\sum _{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}+\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\right){n}^{-1/2}{\left(1+ϵ\sqrt{n}\right)}^{-3}{\int }_{|u|\le {\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}}{|u|}^{3}\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ =:& {J}_{11}+{J}_{12}.\end{array}$

Noting that $\frac{x}{g\left(x\right)}$ is nondecreasing in the interval $x>0$, we have

$\begin{array}{rcl}{J}_{11}& =& C{ϵ}^{2}\sum _{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{1}{\sqrt{n}{\left(1+ϵ\sqrt{n}\right)}^{3}}{\int }_{|u|\le {\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}}{|u|}^{3}\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C{ϵ}^{2}\sum _{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{1}{{n}^{1/4}{\left(1+ϵ\sqrt{n}\right)}^{5/2}g\left({\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}\right)}{\int }_{|u|\le {\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}}{u}^{2}g\left(u\right)\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C{ϵ}^{2}\sum _{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{1}{{n}^{1/4}g\left({n}^{1/4}\right)}\\ =& C{h}_{2}\left(ϵ\right),\end{array}$
(2.5)

where ${h}_{2}\left(ϵ\right)={ϵ}^{2}{\sum }_{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{1}{{n}^{1/4}g\left({n}^{1/4}\right)}$.

Similarly, we can obtain

$\begin{array}{rcl}{J}_{12}& =& C{ϵ}^{2}\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{1}{\sqrt{n}{\left(1+ϵ\sqrt{n}\right)}^{3}}{\int }_{|u|\le {\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}}{|u|}^{3}\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C{ϵ}^{2}\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{1}{{n}^{1/4}{\left(1+ϵ\sqrt{n}\right)}^{5/2}g\left({\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}\right)}{\int }_{|u|\le {\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}}{u}^{2}g\left(u\right)\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C{ϵ}^{2}\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{1}{{ϵ}^{5/2}{n}^{3/2}g\left({n}^{1/4}\right)}\\ =& C\frac{1}{\sqrt{ϵ}}{f}_{2}\left(ϵ\right),\end{array}$
(2.6)

where ${f}_{2}\left(ϵ\right)={\sum }_{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{1}{{n}^{3/2}g\left({n}^{1/4}\right)}$.

For ${J}_{2}$, we write

$\begin{array}{rcl}{J}_{2}& =& C{ϵ}^{2}\left(\sum _{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}+\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\right){n}^{-1/2}{\left(1+ϵ\sqrt{n}\right)}^{-3}{\int }_{{\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}<|u|<\sqrt{n}\left(1+ϵ\sqrt{n}\right)}{|u|}^{3}\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ =:& {J}_{21}+{J}_{22}.\end{array}$

Using the properties of $g\left(x\right)$ by simple calculation, it follows that

$\begin{array}{rcl}{J}_{21}& =& C{ϵ}^{2}\sum _{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}{n}^{-1/2}{\left(1+ϵ\sqrt{n}\right)}^{-3}{\int }_{{\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}<|u|<\sqrt{n}\left(1+ϵ\sqrt{n}\right)}{|u|}^{3}\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C{ϵ}^{2}\left(\sum _{n=1}^{{N}_{0}}+\sum _{n={N}_{0}+1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\right)\frac{1}{{\left(1+ϵ\sqrt{n}\right)}^{2}g\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}\\ ×{\int }_{{\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}<|u|<\sqrt{n}\left(1+ϵ\sqrt{n}\right)}{u}^{2}g\left(u\right)\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C{ϵ}^{2}\left(\sum _{n=1}^{{N}_{0}}+\sum _{n={N}_{0}+1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\right)\frac{1}{g\left(\sqrt{n}\right)}{\int }_{|u|>{n}^{1/4}}{u}^{2}g\left(u\right)\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C{ϵ}^{2}\left({N}_{0}+\eta \sum _{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{1}{g\left(\sqrt{n}\right)}\right)\\ =& o\left({h}_{1}\left(ϵ\right)\right),\end{array}$
(2.7)

and

$\begin{array}{rcl}{J}_{22}& \le & C{ϵ}^{2}\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}{n}^{-\frac{1}{2}}{\left(1+ϵ\sqrt{n}\right)}^{-3}{\int }_{{\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}<|u|<\sqrt{n}\left(1+ϵ\sqrt{n}\right)}{|u|}^{3}\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{1}{ng\left(\sqrt{n}\right)}{\int }_{{\left(\sqrt{n}\left(1+ϵ\sqrt{n}\right)\right)}^{1/2}<|u|<\sqrt{n}\left(1+ϵ\sqrt{n}\right)}{u}^{2}g\left(u\right)\phantom{\rule{0.2em}{0ex}}dV\left(u\right)\\ \le & C{T}_{g}\left(\frac{1}{\sqrt{ϵ}}\right)\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{1}{ng\left(\sqrt{n}\right)}\\ \le & C{T}_{g}\left(\frac{1}{\sqrt{ϵ}}\right){f}_{1}\left(ϵ\right).\end{array}$
(2.8)

From (2.2) to (2.8), we conclude that

${ϵ}^{2}\lambda \left(ϵ\right)-1\le C\frac{1}{\sqrt{ϵ}}{f}_{2}\left(ϵ\right)+C{T}_{g}\left(\frac{1}{\sqrt{ϵ}}\right){f}_{1}\left(ϵ\right)+o\left(1\right){h}_{1}\left(ϵ\right)+C{h}_{2}\left(ϵ\right).$
(2.9)

Since

$\frac{1}{\sqrt{ϵ}}{f}_{2}\left(ϵ\right)\le \frac{C}{\sqrt{ϵ}}\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{1}{{n}^{3/2}}\le C\sqrt{ϵ},$

and

${h}_{2}\left(ϵ\right)={ϵ}^{2}\sum _{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{1}{\sqrt{n}g\left(\sqrt{n}\right)}\le C{ϵ}^{2}\sum _{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{1}{\sqrt{n}}\le C\sqrt{ϵ},$

by (2.9), we have

${ϵ}^{2}\lambda \left(ϵ\right)-1=O\left({ϵ}^{1/2}\right)+o\left(1\right)\left({f}_{1}\left(ϵ\right)+{h}_{1}\left(ϵ\right)\right).$

This completes the proof of Theorem 1.1. □

Proof of Theorem 1.2 By the conditions $g\left(x\right)\in {G}_{0}$, and ${lim}_{x\to \mathrm{\infty }}\frac{g\left({x}^{2}\right)}{xg\left(x\right)}=0$, for any $\eta >0$, there is an integer ${N}_{1}$ such that $\frac{g\left(\sqrt{n}\right)}{\sqrt{n}g\left(\sqrt{n}\right)}\le \eta$, whenever $n>{N}_{1}$. We have

$\begin{array}{rcl}{h}_{2}\left(ϵ\right)& \le & {ϵ}^{2}\sum _{n=1}^{{N}_{1}}\frac{1}{\sqrt{n}g\left(\sqrt{n}\right)}+{ϵ}^{2}\sum _{n={N}_{1}}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{\eta }{g\left(\sqrt{n}\right)}\\ \le & C{ϵ}^{2}{N}_{1}+{ϵ}^{2}\sum _{n={N}_{1}+1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{\eta }{g\left(\sqrt{n}\right)}\\ \le & C{ϵ}^{2}{N}_{1}+{ϵ}^{2}\sum _{n=1}^{\left[\frac{1}{{ϵ}^{2}}\right]}\frac{\eta }{g\left(\sqrt{n}\right)}\\ =& o\left(1\right){h}_{1}\left(ϵ\right),\end{array}$
(2.10)

and

$\begin{array}{rcl}\frac{1}{\sqrt{ϵ}}{f}_{2}\left(ϵ\right)& \le & \frac{1}{\sqrt{ϵ}}\sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{\eta }{{n}^{5/4}g\left(\sqrt{n}\right)}\\ \le & \sum _{n=\left[\frac{1}{{ϵ}^{2}}\right]+1}^{\mathrm{\infty }}\frac{\eta }{ng\left(\sqrt{n}\right)}=o\left(1\right){f}_{1}\left(ϵ\right).\end{array}$
(2.11)

By (2.9)-(2.11), note that ${T}_{g}\left(\frac{1}{\sqrt{ϵ}}\right)=o\left(1\right)$, as $ϵ\to 0$, we have

This completes the proof of Theorem 1.2. □

Remark 2.1 If $g\left(x\right)={|x|}^{\delta }$, $0<\delta <1$, then ${f}_{1}\left(ϵ\right)=O\left({ϵ}^{\delta }\right)$, ${h}_{1}\left(ϵ\right)=O\left({ϵ}^{\delta }\right)$. By Theorem 1.2, we get

Remark 2.2 If $g\left(x\right)=|x|$, $\delta =1$, then $\frac{1}{\sqrt{ϵ}}{f}_{2}\left(ϵ\right)=O\left(ϵ\right)$, ${f}_{1}\left(ϵ\right)=O\left(ϵ\right)$, ${h}_{1}\left(ϵ\right)=O\left(ϵ\right)$, ${h}_{2}\left(ϵ\right)=O\left(ϵ\right)$. By (2.9), we get

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## Acknowledgements

The authors are very grateful to the referees and editors for their valuable comments and some helpful suggestions that improved the clarity and readability of the paper.

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He, J., Xie, T. A supplement to the convergence rate in a theorem of Heyde. J Inequal Appl 2012, 195 (2012). https://doi.org/10.1186/1029-242X-2012-195 