Spectral Radii of Truncated Circular Unitary Matrices

Introduction

The early study of large random matrices was stimulated by analysis of high-dimensional data. One example is Wishart’s (1928) investigation on large covariance matrices whose statistical properties are mainly determined by eigenvalues and eigenvectors from the point view of a principal components analysis. Since then, the random matrix theory has been developed very rapidly and found many applications in areas such as heavy-nuclei atoms (Wigner, 1955), number theory (Mezzadri and Snaith, 2005), quantum mechanics (Mehta, 2005), condensed matter physics (Forrester, 2010), wireless communications (Couillet and Debbah, 2011).

The study of random matrices has greatly been motivated by Tracy and Widom’s (1994, 1996) work. They show that the largest eigenvalues of the three Hermitian matrices (Gaussian orthogonal ensemble, Gaussian unitary ensemble and Gaussian symplectic ensemble) converge to some special distributions that are now known as the Tracy-Widom laws. Subsequently, the Tracy-Widom laws have found their applications in the study of problems such as the longest increasing subsequence (Baik et al., 1999), combinatorics, growth processes, random tilings and the determinantal point processes (see, e.g., Tracy and Widom (2002), Johansson (2007) and references therein) and the largest eigenvalues in the high-dimensional statistics (see, e.g., Johnstone (2001, 2008) and Jiang (2009)). Some recent research focuses on the universality of the largest eigenvalues of matrices with non-Gaussian entries; see, for example, Tao and Vu (2011), Erdős et al. (2012) and the references therein.

A very recent paper by Jiang and Qi (2017) studies the largest radii of three rotation-invariant and non-Hermitian random matrices: the spherical ensemble, the truncation of circular unitary ensemble and the product of independent complex Ginibre ensembles. It is proved in the paper that the spectral radii converge to the Gumbel distribution and some new distributions.

where $C$ is a normalizing constant. See, e.g., Zyczkowski and Sommers (2000).

Assume $p=p_{n}$ depends on $n$ and set $c=\lim_{n\to\infty}\frac{p_{n}}{n}$. Życzkowski and Sommers (2000) show that the empirical distribution of $z_{i}$’s converges to the distribution with density proportional to $\frac{1}{(1-|z|^{2})^{2}}$ for $|z|\leq c$ if $c\in(0,1)$. Dong et al. (2012) prove that the empirical distribution goes to the circular law and the arc law as $c=0$ and $c=1$, respectively. See also Diaconis and Evans (2001) and Jiang (2009, 2010) and references therein for more results.

Jiang and Qi (2017) have proved that the spectral radius $\max_{1\leq j\leq p}|z_{j}|$ for the truncated circular unitary ensemble converges to the Gumbel distribution when the dimension of the truncated truncated circular unitary matrix is of the same order as the dimension of the original circular unitary matrix, see Theorem 1 in section 2.

In this paper we consider heavily truncated and lightly truncated circular unitary matrices and investigate the limiting distribution of the spectral radii for those truncated circular unitary matrices. Our results complement that in Jiang and Qi (2017).

The rest of the paper is organized as follows. The main results in this paper are given in section 2 and their proofs are provided in section 3.

Main Results

Consider the $p_{n}\times p_{n}$ submatrix $\mathbf{A}$, truncated from a $n\times n$ circular unitary matrix $\mathbf{U}$ in section 1. Denote the $p_{n}$ eigenvalues as $z_{1},\cdots,z_{p_{n}}$ with the joint density function given by (1.1).

For completeness, we first quote a theorem in Jiang and Qi (2017) on the limiting distribution of the spectral radii $\max_{1\leq j\leq p_{n}}|z_{j}|$ before we give our results in the paper.

The main results of the paper are the following theorems:

Under condition (2.4), $\frac{2n^{1+1/k}}{((k+1)!)^{1/k}}(\max_{1\leq j\leq p_{n}}|z_{j}|-1)$ converges weakly to the reversed Weibull distribution $W_{k}(x)$ defined as

We notice that the limiting distribution of the spectral radii depends on the dimension of truncated matrices. Our results in Theorems 2, 3 and 4 indicate that the limiting distribution of the spectral radii of the truncated circular unitary matrices is Gumbel distribution $\Lambda$ if the parameter $k_{n}=n-p_{n}$, the number of truncated columns and rows diverges. When the truncation is very light, that is, $k_{n}=n-p_{n}=k\geq 1$ is a fixed integer, the limiting distribution of the spectral radii of the truncated matrices is the reversed Weibull distribution $W_{k}$.

It is obvious that the case when $k_{n}=n-p_{n}$ is of order between $\log n$ and $(\log n)^{3}$ has not been covered in Theorems 1 to 4. We conjecture that $\max_{1\leq j\leq p_{n}}|z_{j}|$, after properly normalized, converges in distribution to the Gumbel distribution in this case.

Proofs

For random variables $\{X_{n};\,n\geq 1\}$ and constants $\{a_{n};\,n\geq 1\}$, we write $X_{n}=O_{p}(a_{n})$ if $\lim_{x\to+\infty}\limsup_{n\to\infty}P(|\frac{X_{n}}{a_{n}}|\geq x)=0$. It is well known that $\frac{X_{n}}{a_{n}b_{n}}\to 0$ in probability as $n\to\infty$ if $X_{n}=O_{p}(a_{n})$ and $\{b_{n};\,n\geq 1\}$ is a sequence of constants with $\lim_{n\to\infty}b_{n}=\infty$.

Let $U_{i}$, $i\geq 1$ be a sequence of i.i.d. random variables uniformly distributed over $(0,1)$, and $U_{1:n}\leq U_{2:n}\leq\cdots\leq U_{n:n}$ be the order statistics of $U_{1},U_{2},\cdots,U_{n}$ for each $n\geq 1$. Then from page 14 on the book by Balakrishnan and Cohen (1991), we know that the cumulative distribution function of $U_{i:n}$ is given by

for each $1\leq i\leq n$, and the probability density function (pdf) of $U_{i:n}$ is given by

This is the so-called Beta distribution, denoted by Beta($i,n-i+1$).

From (3.2), $U_{p_{n}-j+1:n-j}$ has a Beta($p_{n}-j+1,k_{n}$) distribution with pdf given by

For each $n\geq 2$, let $\{Y_{nj};\,1\leq j\leq p_{n}\}$ be independent random variables such that $Y_{nj}$ and $(U_{p_{n}-j+1:n-j})^{1/2}$ have the same distribution. Jiang and Qi (2017) have shown that $\max_{1\leq j\leq p_{n}}|z_{j}|$ and $\max_{1\leq j\leq p_{n}}Y_{nj}$ have the same distribution, that is

for $x\in(0,1)$. See the proof of Theorem 2 in Jiang and Qi (2017).

We will present some useful lemmas before we prove our main results.

Suppose $\{l_{n};\,n\geq 1\}$ is sequence of positive integers. Let $z_{nj}\in[0,1)$ be real numbers for $1\leq j\leq l_{n}$ such that $\max_{1\leq j\leq l_{n}}z_{nj}\to 0$ as $n\to\infty$. Then $\displaystyle\lim_{n\to\infty}\prod^{l_{n}}_{j=1}(1-z_{nj})\in(0,1)$ exists if and only if the limit $\displaystyle\lim_{n\to\infty}\sum^{l_{n}}_{j=1}z_{nj}=:z\in(0,\infty)$ exists and the relationship of the two limits is given by

which implies $\log(1-z_{nj})=-z_{nj}+O(z_{nj}^{2})$ uniformly over $1\leq j\leq l_{n}$ since $\max_{1\leq j\leq l_{n}}z_{nj}\to 0$ as $n\to\infty$. Therefore,

The lemma can be easily concluded from the above expression. $\blacksquare$

Let $\{l_{n}\}$ be a sequence of positive integers such that $l_{n}\to\infty$ and for each $n$, $\{z_{nj},~{}1\leq j\leq l_{n}\}$ are non-negative numbers such that $z_{nj}$ is non-increasing in $j$ with $z_{n1}>0$. Then for any sequence of positive integers $\{r_{n}\}$satisfying that $r_{n}<l_{n}$ for all large $n$ and $r_{n}/l_{n}\to 1$ as $n\to\infty$, we have

as $n\to\infty$. In fact, from the monotonicity of $z_{nj}$, we have $z_{nj}\leq\frac{1}{r_{n}}\sum^{r_{n}}_{j=1}z_{nj}$ for $r_{n}+1\leq j\leq l_{n}$. Hence

where $t_{n}\in(0,1)$ will be specified later in the proof of each theorem. From (3.4),

Obviously, we have for $1\leq j\leq p_{n}$.

Assume that $1\leq p_{n}<n$ and $p_{n}\to\infty$ as $n\to\infty$. Let $\{r_{n}\}$ satisfy the condition in Lemma 3.2 with $l_{n}=p_{n}$. Assume $\alpha_{n}>0$ and $\beta_{n}$ are real numbers such that $\lim_{n\to\infty}P(Y_{n1}^{2}>\beta_{n}+\alpha_{n}x)=0$ for any $x\in R$. If $(\max_{1\leq j\leq r_{n}}Y_{nj}^{2}-\beta_{n})/\alpha_{n}$ converges in distribution to a cdf $G$, then $(\max_{1\leq j\leq p_{n}}Y_{nj}^{2}-\beta_{n})/\alpha_{n}$ converges in distribution to the same distribution $G$.

Proof. Note that $(\max_{1\leq j\leq r_{n}}Y_{nj}^{2}-\beta_{n})/\alpha_{n}$ converges in distribution to the cdf $G$ if and only if

for every continuity point $x$ of $G$ with $G(x)\in(0,1)$. We need to prove the above expression is still true when $r_{n}$ is replaced by $p_{n}$. Now fix $x$, a continuity point of $G$ with $G(x)\in(0,1)$. Set $t_{n}=t_{n}(x)=\beta_{n}+\alpha_{n}x$ and define $z_{nj}$ as in (3.7). Note that (3.8) holds, $z_{n1}\to 0$ as $n\to\infty$,

By using Lemma 3.1 and (3.10) we have $\sum^{r_{n}}_{j=1}z_{nj}\to z=-\log G(x)$ which, together with Lemma 3.2, implies $\sum^{p_{n}}_{j=1}z_{nj}\to z=-\log G(x)$. Once again we have from Lemma 3.1 that

This completes the proof of the lemma. $\blacksquare$

Let $Z_{n}$ be nonnegative random variables such that $(Z_{n}^{2}-\beta_{n})/\alpha_{n}$ converges weakly to a cdf $G(x)$, where $\alpha_{n}>0$ and $\beta_{n}>0$ are constants satisfying that $\lim_{n\to\infty}\alpha_{n}/\beta_{n}=0$. Then

Proof. Set $W_{n}=(Z_{n}^{2}-\beta_{n})/\alpha_{n}$. We have $Z_{n}^{2}=\beta_{n}+\alpha_{n}W_{n}=\beta_{n}(1+\frac{\alpha_{n}}{\beta_{n}}W_{n})$. Then by Taylor’s expansion

where for $i=1,2$, $l_{i}(t)$ is a polynomial in $t$ of degree $\leq 3i$, depending on $r$ and $k$, and all of its coefficients are of order O(\big{(}\frac{r}{(r-k)k}\big{)}^{i/2}).

Define $V_{p_{n}-j+1:n-j}$ as in (3.25). Assume that $k_{n}=n-p_{n}\to\infty$ and $k_{n}/n\to 0$ as $n\to\infty$. Then for any $\delta_{n}>0$ such that $\delta_{n}\to\infty$ and $\delta_{n}=o(k_{n}^{1/6})$

uniformly over $0\leq x\leq\delta_{n}$, $1\leq j\leq p_{n}-k_{n}$ as $n\to\infty$.

Proof. Set $\beta_{nj}(x)=\frac{p_{n}-j}{n-j}+\frac{((p_{n}-j)k_{n})^{1/2}}{(n-j)^{3/2}}x=\frac{p_{n}-j}{n-j}(1+\frac{k_{n}^{1/2}}{(n-j)^{1/2}(p_{n}-j)^{1/2}}x)$. Then $1-\beta_{nj}(x)=\frac{n-p_{n}}{n-j}-\frac{((p_{n}-j)k_{n})^{1/2}}{(n-j)^{3/2}}x=\frac{k_{n}}{n-j}(1-\frac{(p_{n}-j)^{1/2}}{(n-j)^{1/2}k_{n}^{1/2}}x)$, and the density function of $V_{p_{n}-j+1:n-j}$ is given by

To estimate $h_{j}(x)$, we need Stirling’s formula:

and Taylor’s expansion: $1-t=\exp(\log(1-t))=\exp(-t-\frac{1}{2}t^{2}+O(t^{3}))$ as $t\to 0$. By applying Stirling’s formula to $k_{n}!$, $(p_{n}-j)!$ and $(n-j)!$, the product in (3.16) is equal to $\frac{1+o(1)}{\sqrt{2\pi}}(1+o(1))$ for $|x|=o(k_{n}^{1/6})$ uniformly over $1\leq j\leq p_{n}-k_{n}$ as $n\to\infty$. By applying Taylor’s expansion to $(1+\frac{k_{n}^{1/2}}{(n-j)^{1/2}(p_{n}-j)^{1/2}}x)^{p_{n}-j}$ and $(1-\frac{(p_{n}-j)^{1/2}}{(n-j)^{1/2}k_{n}^{1/2}}x)^{k_{n}}$, the product in (3.16) is equal to

for $|x|=o(k_{n}^{1/6})$ uniformly over $1\leq j\leq p_{n}-k_{n}$ as $n\to\infty$. Therefore, we have

uniformly over $1\leq j\leq p_{n}-k_{n}$ as $n\to\infty$. Next we will give an estimate of the upper bound of $h_{j}(x)$ for large $x$. Note that $\beta_{nj}(x)<1$ if and only if $x<\frac{k_{n}^{1/2}(p_{n}-j)^{1/2}}{(n-j)^{1/2}}=O(k_{n}^{1/2})$ uniformly over $1\leq j\leq p_{n}-k_{n}$ and thus $\frac{k_{n}^{1/2}x}{(n-j)^{1/2}(p_{n}-j)^{1/2}}\leq O(\frac{k_{n}^{1/2}}{n^{1/2}})\to 0$ uniformly over $0<x<\frac{k_{n}^{1/2}(p_{n}-j)^{1/2}}{(n-j)^{1/2}}$, $1\leq j\leq p_{n}-k_{n}$ as $n\to\infty$. Now by applying Taylor’s expansion to $(1+\frac{k_{n}^{1/2}}{(n-j)^{1/2}(p_{n}-j)^{1/2}}x)^{p_{n}-j}$ and inequality

to $(1-\frac{(p_{n}-j)^{1/2}}{(n-j)^{1/2}k_{n}^{1/2}}x)^{k_{n}}$ we get

uniformly over $1\leq j\leq p_{n}-k_{n}$ as $n\to\infty$.

Assume that $0\leq x\leq\delta_{n}$. From (3.19) we have

uniformly over $0\leq x\leq\delta_{n}$, $1\leq j\leq p_{n}-k_{n}$ as $n\to\infty$. Therefore, to complete the proof of the lemma, it suffices to show that

uniformly over $0\leq x\leq\delta_{n}$, $1\leq j\leq p_{n}-k_{n}$ as $n\to\infty$.

uniformly if $y\to\infty$. Since $(\delta_{n}k_{n}^{1/6})^{1/2}=o(k_{n}^{1/6})$, by using (3.18) we have

proving (3.20). This completes the proof of the lemma. $\blacksquare$

2 Proofs of the Theorems

where $A_{n}=c_{n}+\frac{1}{2}(1-c_{n}^{2})^{1/2}(n-1)^{-1/2}a_{n}$, $B_{n}=\frac{1}{2}(1-c_{n}^{2})^{1/2}(n-1)^{-1/2}b_{n}$,

with $a(x)=(\log x)^{1/2}-(\log x)^{-1/2}\log(\sqrt{2\pi}\log x)$ and $b(x)=(\log x)^{-1/2}$ for $x>3$.

Since $Y_{nj}^{2}$ and $U_{p_{n}-j+1:n-j}$ are identically distributed, we have

Part 1. First we show (3.21) under condition (2.1). We will prove that

Let $j_{n}=[p_{n}^{5/8}]$, the integer part of $p_{n}^{5/8}$. For $1\leq j\leq j_{n}$, define

Then we see that uniformly over $1\leq j\leq j_{n}$,

uniformly for all $1\leq j\leq j_{n}$ as $n\to\infty$.

In Lemma 3.6, take $r=n-j$ and $k=n-p_{n}$ to have

uniformly over $1\leq j\leq j_{n}$ as $n\to\infty$, where

and where, for $i=1,2$, $l_{i}(t)$ is a polynomial in $t$ of degree $\leq 3i$, depending on $n$, and all of its coefficients are of order $O((1/p_{n})^{i/2})$ uniformly over $1\leq j\leq j_{n}$ as $n\to\infty$. Now, by taking $B=(u_{nj},\infty)$ we obtain

uniformly for $1\leq j\leq j_{n}$ as $n\to\infty$. From L’Hospital’s rule, we have that for any $r\geq 0$

Since $\min_{1\leq j\leq j_{n}}u_{nj}\to\infty$ as $n\to\infty$ by (3.24), it follows from (3.27) that

holds uniformly over $1\leq j\leq j_{n}$. Furthermore, since the coefficients of $l_{i}(t)$ are uniformly bounded by $O((1/p_{n})^{i/2})$ for $i=1,2$, we have

In Lemma 3.5, by taking $x_{n}=nc_{n}^{2}/(1-c_{n}^{2})$ and and define $c_{n,j}$ such that $u_{nj}=(j-1)c_{n,j}+a(x_{n})+b(x_{n})x$ for $1\leq j\leq j_{n}$ with $c_{n,1}=x_{n}^{-1/2}$. It follows from (3.24) that $c_{n,j}=x_{n}^{-1/2}(1+o(1))$ uniformly over $1\leq j\leq j_{n}$ as $n\to\infty$, which implies $\lim_{n\to\infty}\max_{1\leq j\leq j_{n}}|c_{n,j}x_{n}^{1/2}-1|=0$. Then we get

We will show that the second term and the third term on the line below (3.2) converge to zero as $n\to\infty$. By noting that $\frac{(1-c_{n}^{2})^{1/2}}{n^{1/2}c_{n}}\sim p_{n}^{-1/2}$ we have

uniformly over $1\leq j\leq j_{n}$. Thus, it follows from (3.13) that

as $n\to\infty$. Therefore, by combining (3.2), (3.26) and (3.29) we get

for all large $n$. Then we estimate $z_{nj_{n}}$ by using (3.26) with $j=j_{n}$ and (3.30)

as $n\to\infty$. From (3.8) we have $\sum^{p_{n}}_{j=j_{n}+1}z_{nj}\leq p_{n}z_{nj_{n}}=O(p_{n}^{-1/2})\to 0$ as $n\to\infty$, which together with (3.31) yields

We can also prove from (3.26) that $z_{n1}\to 0$ as $n\to\infty$. In view of (3.22) and Lemma 3.1 we conclude (3.23), ie.,

where $\alpha_{n}=c_{n}(1-c_{n}^{2})^{1/2}(n-1)^{-1/2}b_{n}$ and $\beta_{n}=c_{n}^{2}+c_{n}(1-c_{n}^{2})^{1/2}(n-1)^{-1/2}a_{n}$. Since

as $n\to\infty$, we can apply Lemma 3.4 and get that

Recall that $A_{n}=c_{n}+\frac{1}{2}(1-c_{n}^{2})^{1/2}(n-1)^{-1/2}a_{n}=c_{n}(1+o(1))$ and $B_{n}=\frac{1}{2}(1-c_{n}^{2})^{1/2}(n-1)^{-1/2}b_{n}\sim\frac{1}{2}(n-1)^{-1/2}(\log p_{n})^{-1/2}$. Then

ie., (3.21) holds. The proof of Part 1 is completed.

Part 2. We will show (3.21) under condition (2.2). First, it follows from condition (2.2) that

as $n\to\infty$. Noting that $n\leq n^{2}/k_{n}\leq n^{2}$, we get that

is of order $(\log n)^{1/2}$ as $n\to\infty$.

Use the same notation as in Part 1. Recall that $p_{n}/n\to 1$, $k_{n}=n-p_{n}=o(n)$ and $\log n=o(k_{n}^{1/3})$ as $n\to\infty$. In order to use both Lemmas 3.5 and 3.7, we take $x_{n}=\frac{nc_{n}^{2}}{1-c_{n}^{2}}$. Define $j_{n}=[5(\log n)^{1/2}\sqrt{x_{n}}]+1$. Then $j_{n}\sim\frac{5n(\log n)^{1/2}}{\sqrt{k_{n}}}=o(n)$, which implies $1\leq j_{n}\leq p_{n}-k_{n}$ for all large $n$. Define $c_{n,j}$ for $1\leq j\leq j_{n}$ in the same way as in Part 1. Similar to the proof of (3.24) we can show that

uniformly for $1\leq j\leq j_{n}$ as $n\to\infty$. Then $u_{nj}=O((\log n)^{1/2})=o(k_{n}^{1/6})$ uniformly for $1\leq j\leq j_{n}$ as $n\to\infty$. We can also verify that all conditions in Lemma 3.5 are satisfied. Thus, from (3.14) and (3.12) we have

Note that (3.14) holds uniformly over $1\leq j\leq p_{n}-k_{n}$ and $u_{nj_{n}}\geq 4(\log n)^{1/2}$ for all large $n$. By employing (3.14) with $j=j_{n}$ and $x=4(\log n)^{1/2}$ and using equation (3.8) and Lemma 3.7 we have

Thus, we obtain that $\sum^{p_{n}}_{j=1}z_{nj}\to e^{-x}$ for any $x$. Then equation (3.23) follows from equation (3.22) and Lemma 3.1. The rest of the proof will follow from the same lines in the proof of the first part. Again Lemma 3.4 will be used. The details are omitted. $\blacksquare$

Proof of Theorem 3. Recall that $a_{n}$ is given by

By using Stirling’s formula (3.17), we have under condition (2.3) that

Since $te^{-t}$ is strictly increasing in $(0,1)$, for all large $n$ such that $y_{n}<1$ define $\varepsilon_{n}$ as the unique solution to $te^{-t}=y_{n}$ in $(0,1)$, that is, $\varepsilon_{n}e^{-\varepsilon_{n}}=y_{n}$, which implies that $\varepsilon_{n}\to 0$ and $\varepsilon_{n}\sim y_{n}$ as $n\to\infty$ and

for all large $n$. Then it follows from the the first inequality in (3.37) that

for all large $n$, which together with (3.36) implies that $a_{n}\leq k_{n}\varepsilon_{n}$ for all large $n$,and thus $a_{n}=o(k_{n})$ as $n\to\infty$. By plugging $y=a_{n}$ in (3.37) and using $a_{n}\leq k_{n}\varepsilon_{n}$ for large $n$ we conclude

Define $z_{nj}$ as in (3.7) with $t_{n}=t_{n}(x)=1-\frac{a_{n}}{n}(1-\frac{x}{k_{n}})$ for any fixed $x$. Then $1-t_{n}=\frac{a_{n}}{n}(1-\frac{x}{k_{n}})=o(\frac{k_{n}}{n})=o(\frac{\log n}{n})$ as $n\to\infty$, where we have used the fact that $k_{n}=n-p_{n}\to\infty$ and $k_{n}=o(\log n)$ from (2.3). This implies $n(1-t_{n})^{2}\to 0$ as $n\to\infty$.

It is easy to verify the following expression

where $0\leq d(t)\leq t^{2}$ for $0\leq t\leq 1/2$. Then

Furthermore, by using Stirling’s formula (3.17), we get

uniformly over $1\leq j\leq j_{n}$, where $j_{n}:=p_{n}-k_{n}^{3}$. Then from (3.9) we obtain that

Note that $n(1-t_{n})=o(k_{n})$. Then it follows from (3.39) and (3.37) that

uniformly over $1\leq j\leq j_{n}$, and thus

The second integral above is dominated by the first one since $t^{k_{n}}e^{-t}$ is increasing over $(0,k_{n})$ and $k_{n}^{3}(1-t_{n})/(n(1-t_{n}))=k_{n}^{3}/n\to 0$ as $n\to\infty$. Therefore, in view of (3.37) and (3.38) we have

as $n\to\infty$. Therefore, it follows from Lemma 3.2 that $\sum^{p_{n}}_{j=1}z_{nj}\to e^{-x}$ as $n\to\infty$. It is easy to conclude that $z_{n1}\to 0$ as $n\to\infty$ from (3.39), (3.37) and the above estimates. Accordingly, by taking $\beta_{n}=1-\frac{a_{n}}{n}$ and $\alpha_{n}=\frac{a_{n}}{nk_{n}}$ with $G(x)=\Lambda(x)$ in Lemmas 3.3 and 3.4 we conclude that

as $n\to\infty$. This completes the proof of the theorem. $\blacksquare$

Fix $x<0$. Let $t_{n}=t_{n}(x)=1+\frac{((k+1)!)^{1/k}}{n^{1+1/k}}x$. Then $t_{n}\in(0,1)$ for all large $n$. Since $n(1-t_{n})\to 0$ as $n\to\infty$, we have

Therefore, we have from (3.7) and (3.9) that

uniformly over $1\leq j\leq p_{n}=n-k$. Since

we have $\max_{1\leq j\leq p_{n}}z_{nj}=O(1/n)\to 0$ as $n\to\infty$. To complete the proof of (3.40), by using (3.5) we need to show that

Let $\{j_{n}\}$ be a sequence of integers such that $j_{n}\to\infty$ and $j_{n}/n\to 0$ as $n\to\infty$. Then

uniformly over $1\leq j\leq n-j_{n}$, which implies that

as $n\to\infty$. This proves (3.41) and thus we obtain (3.40).

Finally, the theorem follows from Lemma 3.4 with $\alpha_{n}=\frac{((k+1)!)^{1/k}}{n^{1+1/k}}$ and $\beta_{n}=1$. This completes the proof. $\blacksquare$

Acknowledgements We would like to thank an anonymous referee for his/her careful reading of the original version of the paper and pointing out some imperfections in the proofs. Gui’s work was partially supported by the program for the Fundamental Research Funds for the Central Universities (2014RC042).

References