An Exposition on Wigner's Semicircular Law

TL;DR

This paper refines the semicircular law for Hermitian random matrices using the moment method, accommodating broader conditions including infinite variance cases, and provides detailed expositions suitable for newcomers.

Contribution

It introduces a strengthened semicircular law under weaker assumptions and extends results to cases with row sums converging to normal distribution, including infinite variance scenarios.

Findings

Strengthened semicircular law under minimal assumptions

Extension to matrices with infinite variance entries

Detailed exposition suitable for beginners

Abstract

We revisit the moment method to obtain a slightly strengthened version of the usual semicircular law. Our version assumes only that the upper triangular entries of Hermitian random matrices are independent, have mean zero and variances close to $1/n$ in a certain sense, and satisfy a Lindeberg-type condition. As an application, we derive another semicircular law for the case when the sum of a row converges in distribution to the standard normal distribution, including the case where all matrix entries may have infinite variance. The appendix, making up the majority of the paper, provides for those new to the subject, a rigorous exposition of most details involved, including also a proof of a semicircular law that uses the Stieltjes transform method.

Equations

$${\lambda }_1(A)\ge \cdots \ge {\lambda }_n(A).$$

$${\mu }_A:=\frac{1}{n}\sum _{i=1}^n{\delta }_{{\lambda }_i(A)}$$

$${\mu }_{\mathrm{sc}}(dx)=\frac{1}{2\pi }\sqrt{(4-x^2{)}_{+}}dx$$

$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\biggl{|}\sum_{j=1}^{n}\Bigl{(}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}-\frac{1}{n}\Bigr{)}\biggr{|}=0,$$

$$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}\biggl{(}\sum_{j=1}^{n}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}-C\biggr{)}_{+}=0\qquad\text{for some finite $C\geq 0$,}$$

$$\lim_{n\to\infty}\frac{1}{n}\sum_{i,j=1}^{n}\operatorname{\mathbf{E}}\bigl{[}\,|w_{ij}^{(n)}|^{2}\mathrel{;}|w_{ij}^{(n)}|>\epsilon\,\bigr{]}=0\qquad\text{for every $\epsilon>0$,}$$

$$\lim_{n\to\infty}\frac{1}{n}\sum_{i,j=1}^{n}\Bigl{|}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}-\frac{1}{n}\Bigr{|}=0.$$

$$\sum_{j=1}^{n}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}\leq C\qquad\text{for all $n=1,2,\ldots$ and $i=1,\ldots,n$,}$$

$$\lim_{n\to\infty}\operatorname{\mathbf{E}}\biggl{[}\int_{\mathbf{R}}f\,d\mu_{W_{n}}\biggr{]}=\int_{\mathbf{R}}f\,d\mu_{\mathrm{sc}}$$

$$\underset{n\to \infty }{\lim }{\int }_{\mathbf{R}}{f}_{p,q}d{\mu }_{W_n}={\int }_{\mathbf{R}}{f}_{p,q}d{\mu }_{\mathrm{sc}}\text{a.s.}$$

$$\frac{1}{n}\sum_{i,j=1}^{n}\operatorname{\mathbf{E}}\bigl{[}\,|w_{ij}^{(n)}|^{2}\mathrel{;}|w_{ij}^{(n)}|>1/k\,\bigr{]}\leq 1/k$$

$$\lim_{n\to\infty}\frac{1}{n}\sum_{i,j=1}^{n}\operatorname{\mathbf{E}}\bigl{[}\,|w_{ij}^{(n)}|^{2}\mathrel{;}|w_{ij}^{(n)}|>\eta_{n}\,\bigr{]}=0.$$

$$\frac{1}{n}\operatorname{\mathbf{E}}\bigl{[}\operatorname{tr}(W_{n}-W_{n}^{\prime})^{2}\bigr{]}=\frac{1}{n}\sum_{i,j=1}^{n}\operatorname{\mathbf{E}}\bigl{[}\,|w_{ij}^{(n)}|^{2}\mathrel{;}|w_{ij}^{(n)}|>\eta_{n}\,\bigr{]}\to 0\qquad\text{as $n\to\infty$},$$

$$v_{i,j}^{(n)}:=w_{ij}^{(n)}\operatorname{\mathbf{1}}(|w_{ij}^{(n)}|\leq\eta_{n})-\operatorname{\mathbf{E}}\bigl{[}\,w_{ij}^{(n)}\mathrel{;}|w_{ij}^{(n)}|\leq\eta_{n}\,\bigr{]},$$

$$\frac{1}{n}\sum_{i,j=1}^{n}\bigl{|}\operatorname{\mathbf{E}}\bigl{[}\,w_{ij}^{(n)}\mathrel{;}|w_{ij}^{(n)}|\leq\eta_{n}\,\bigr{]}\bigr{|}^{2}=\frac{1}{n}\sum_{i,j=1}^{n}\bigl{|}\operatorname{\mathbf{E}}\bigl{[}\,w_{ij}^{(n)}\mathrel{;}|w_{ij}^{(n)}|>\eta_{n}\,\bigr{]}\bigr{|}^{2}\\ \leq\frac{1}{n}\sum_{i,j=1}^{n}\operatorname{\mathbf{E}}\bigl{[}\,|w_{ij}^{(n)}|^{2}\mathrel{;}|w_{ij}^{(n)}|>\eta_{n}\,\bigr{]}\to 0\qquad\text{as $n\to\infty$,}$$

$$\begin{split}\sum_{i,j=1}^{n}&\bigl{|}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}-\operatorname{\mathbf{Var}}\bigl{[}v_{ij}^{(n)}\bigr{]}\bigr{|}\\ &=\sum_{i,j=1}^{n}\Bigl{(}\operatorname{\mathbf{E}}\bigl{[}\,|w_{ij}^{(n)}|^{2}\mathrel{;}|w_{ij}^{(n)}|>\eta_{n}\,\bigr{]}+\bigl{|}\operatorname{\mathbf{E}}\bigl{[}\,w_{ij}^{(n)}\mathrel{;}|w_{ij}^{(n)}|\leq\eta_{n}\,\bigr{]}\bigr{|}^{2}\Bigr{)}\\ &\leq 2\sum_{i,j=1}^{n}\operatorname{\mathbf{E}}\bigl{[}\,|w_{ij}^{(n)}|^{2}\mathrel{;}|w_{ij}^{(n)}|>\eta_{n}\,\bigr{]}.\end{split}$$

$$\sum_{j=1}^{n}\Bigl{[}\bigl{(}c_{1j}^{(n)}\bigr{)}^{2}\operatorname{\mathbf{Var}}\bigl{[}w_{1j}^{(n)}\bigr{]}\Bigr{]}=C,$$

$$\sum_{i,j=1}^{n}\Bigl{[}\bigl{(}1-\bigl{(}c_{ij}^{(n)}\bigr{)}^{2}\bigr{)}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}\Bigr{]}\leq 2\biggl{(}\sum_{j=1}^{n}\operatorname{\mathbf{Var}}\bigl{[}w_{1j}^{(n)}\bigr{]}-C\biggr{)}_{+}.$$

$$\sum_{j=1}^{k-1}\Bigl{[}\bigl{(}c_{kj}\bigr{)}^{2}\operatorname{\mathbf{Var}}\bigl{[}w_{kj}^{(n)}\bigr{]}\Bigr{]}+\sum_{j=k}^{n}\operatorname{\mathbf{Var}}\bigl{[}w_{kj}^{(n)}\bigr{]}\leq C,$$

$$\sum_{j=1}^{n}\Bigl{[}\bigl{(}c_{kj}^{(n)}\bigr{)}^{2}\operatorname{\mathbf{Var}}\bigl{[}w_{kj}^{(n)}\bigr{]}\Bigr{]}=C,$$

$$\sum_{i,j=1}^{n}\Bigl{[}\bigl{(}1-\bigl{(}c_{ij}^{(n)}\bigr{)}^{2}\bigr{)}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}\Bigr{]}\leq 2\sum_{i=1}^{k}\biggl{(}\sum_{j=1}^{n}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}-C\biggr{)}_{+}.$$

$$\sum_{j=1}^{n}\Bigl{[}\bigl{(}c_{ij}^{(n)}\bigr{)}^{2}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}\Bigr{]}\leq C$$

$$\sum_{i,j=1}^{n}\Bigl{[}\bigl{(}1-\bigl{(}c_{ij}^{(n)}\bigr{)}^{2}\bigr{)}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}\Bigr{]}\leq 2\sum_{i=1}^{n}\biggl{(}\sum_{j=1}^{n}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}-C\biggr{)}_{+}.$$

$$\frac{1}{n}\operatorname{\mathbf{E}}\bigl{[}\operatorname{tr}(W_{n}-\widetilde{W}_{n})^{2}\bigr{]}=\frac{1}{n}\sum_{i,j=1}^{n}\Bigl{[}\bigl{(}1-c_{ij}^{(n)}\bigr{)}^{2}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}\Bigr{]}\\ \leq\frac{2}{n}\sum_{i=1}^{n}\biggl{(}\sum_{j=1}^{n}\operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}-C\biggr{)}_{+}\to 0\qquad\text{as $n\to\infty$}$$

$$\biggl{|}\sum_{k=1}^{\infty}\frac{1}{k!}\int_{\mathbf{R}}x^{k}\,\mu_{\mathrm{sc}}(dx)r^{k}\biggr{|}\leq\sum_{k=1}^{\infty}\frac{|2r|^{k}}{k!}<\infty$$

$$\underset{n\to \infty }{\lim }{\int }_{\mathbf{R}}{x}^{k}E{\mu }_{W_n}(dx)={\int }_{\mathbf{R}}{x}^k{\mu }_{\mathrm{sc}}(dx)\text{for all k=1,2,….}$$

$${\int }_{\mathbf{R}}{x}^{k}E{\mu }_{W_n}(dx)={\int }_{\mathbf{R}}{g}_{k,n}dE{\mu }_{W_n}=E{\int }_{\mathbf{R}}{g}_{k,n}d{\mu }_{W_n}=\frac{1}{n}E\mathrm{tr}{W}_n^k.$$

$${\int }_{\mathbf{R}}{x}^{2m}{\mu }_{\mathrm{sc}}(dx)=\frac{1}{m+1}\left(\genfrac{}{}{0ex}{}{2m}{m}\right).$$

$$\begin{array}{rl}{\int }_{\mathbf{R}}{x}^{2m}{\mu }_{\mathrm{sc}}(dx)& =\frac{1}{2\pi }{\int }_{-2}^2{x}^{2m}\sqrt{4-x^2}dx=\frac{2}{\pi }{\int }_{-\pi }^0{2}^{2m}{\cos }^{2m}\theta {\sin }^2\theta d\theta \\ & =\frac{2^{2m+1}}{\pi }\left[{\int }_{-\pi }^0{\cos }^{2m}\theta d\theta -{\int }_{-\pi }^0{\cos }^{2m+2}\theta d\theta \right].\end{array}$$

$${\int }_{-\pi }^0{\cos }^{2l}\theta d\theta =\frac{1}{2^{2l+1}}{\int }_{-\pi }^{\pi }(e^{i\theta }+e^{-i\theta }{)}^{2l}d\theta =\frac{\pi }{2^{2l}}(\genfrac{}{}{0ex}{}{2l}{l})$$

Full text

1. 1 Introduction

2. 2 Preliminary reductions

3. 2.1 Convergence in expectation is enough

4. 2.2 Truncation

5. 2.3 Centralization

6. 2.4 Rescaling

7. 2.5 Reduction to moment convergence

8. 3 Trees and products of variances

9. 4 Computation of moments

10. 5 Gaussian convergence

11. A Probability measures on $\mathbf{R}$

12. A.1 Weak convergence

13. A.2 Expected probability measures

14. B Spectra of Hermitian matrices

15. B.1 Basic facts

16. B.2 Perturbations by small Frobenius norms

17. B.3 Perturbations by small ranks

18. B.4 Random Hermitian matrices

19. C Concentration of measure

20. D Reduction to unit variance case

21. D.1 Extension of the underlying probability space

22. D.2 Repeating what we already know

23. D.3 Replacing and rescaling

24. E The Stieltjes transform method

25. E.1 Stieltjes transform

26. E.2 Predecessor comparison

27. E.3 Derivation of a recurrence relation

28. E.3.1 From $w_{i}^{\ast}S_{W^{(i)}}(z)w_{i}$ to $\frac{1}{n}\operatorname{tr}S_{W^{(i)}}(z)$

29. E.3.2 From $\frac{1}{n}\operatorname{tr}S_{W^{(i)}}(z)$ to $\frac{1}{n}\operatorname{\mathbf{E}}\operatorname{tr}S_{W^{(i)}}(z)$

30. E.3.3 From $\frac{1}{n}\operatorname{\mathbf{E}}\operatorname{tr}S_{W^{(i)}}(z)$ to $\frac{1}{n}\operatorname{\mathbf{E}}\operatorname{tr}S_{W_{n}}(z)$

31. E.3.4 The result

32. E.4 Convergence of the Stieltjes transform

33. E.5 Computation of the limiting distribution

1 Introduction

If $A$ is an $n\times n$ Hermitian matrix, then $A$ is diagonalizable and all eigenvalues of $A$ are real. We denote the eigenvalues of $A$, counted with multiplicities, as

[TABLE]

(For details, see Subsection B.1.) We define the spectral distribution of $A$ as the Borel probability measure

[TABLE]

on $\mathbf{R}$. If $X$ is a random $n\times n$ Hermitian matrix, then $\mu_{X}$ is a random Borel probability measure on $\mathbf{R}$. (For details, see Subsection B.4.)

Definition 1.1** (The semicircle distribution).**

The Borel probability measure $\mu_{\mathrm{sc}}$ on $\mathbf{R}$ given by

[TABLE]

is called the semicircle distribution. Here, $x_{+}:=x\vee 0=\max\{x,0\}$.

Since the seminal work [Wig55] by Wigner, there have been many theorems that assume $W_{1},W_{2},\ldots$ to be random $1\times 1$, $2\times 2,\ldots$ Hermitian matrices satisfying certain conditions, and show that $\mu_{W_{n}}$ converges in some sense to $\mu_{\mathrm{sc}}$. Let us call such theorems semicircular laws.

In the main part of this paper, we study semicircular laws assuming joint independence of the upper triangular entries (we include the diagonal in both the upper and the lower triangles). We first prove a semicircular law (Theorem 1.2) with rather weak assumptions. In particular, we don’t require the entries to be identically distributed, and we allow the entries to deviate from unit variance. It is notable that other than the mostly standard reduction steps, our proof is just a simple application of the moment method.

After proving the main theorem, we apply the theorem to obtain another semicircular law (Theorem 5.2) which more or less assumes that the sum of a row converges in distribution to the standard normal distribution. This theorem allows the entries to have infinite variances.

The appendices make up about two thirds of this paper. There we provide a self-contained and rigorous account of the details (including the measure-theoretic ones) involved in the main part of the paper. In the main part of the paper, we refer to the appendix whenever we need a fact given there. After that, for completeness we provide a proof of a semicircular law (little weaker than the one proved in the main part, but still stronger than the laws appear in many textbooks) which uses the Stieltjes transform method.

We assumed no prior knowledge more advanced than one-semester courses in probability theory and combinatorics. A total newcomer to the field might want to read the Appendices A–C first, then read the main part, and then go to Appendix D–E.

Now we state our main theorem.

Theorem 1.2** (A general semicircular law).**

For each $n\in\mathbf{N}$, let $W_{n}=(w_{ij}^{(n)})_{i,j=1}^{n}$ be a random $n\times n$ Hermitian matrix (see Definition B.13) whose upper triangular entries are jointly independent, have mean zero, and have finite variances. We assume that $W_{1},W_{2},\ldots$ are defined on the same probability space. If

[TABLE]

[TABLE]

and

[TABLE]

then $\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$ as $n\to\infty$ a.s.

Remark 1.3*.*

One sufficient condition for (1.1) and (1.2) to hold is

[TABLE]

Note that we can take $C=1$ to show (1.2). An even more special case is when we have \operatorname{\mathbf{Var}}\bigl{[}w_{ij}^{(n)}\bigr{]}=1/n for all $n=1,2,\ldots$ and $i,j=1,\ldots,n$. This case is Theorem 2.9 in [BS10]. If there is some finite $C\geq 0$ such that

[TABLE]

then (1.2) holds for the same $C$. This case is more or less equivalent to Corollary 1 in [GNT15], which is proved first for matrices with Gaussian entries, and then generalized to arbitrary matrices by proving an analogue of the Lindeberg universality principle for random matrices. In this paper, we will prove Theorem 1.2 directly by the moment method without appealing to the universality principle.

Remark 1.4*.*

Theorem 1.2 assumes no dependence between $W_{1}$, $W_{2}$, …, yet it asserts an a.s. convergence. This is in contrast to some versions of the semicircular law where only convergence in probability is asserted (e.g. [AGZ10, Theorem 2.1.1]), or $\sqrt{n}W_{n}$ is assumed to be the top left $n\times n$ minor of a fixed infinite random Hermitian matrix (e.g. [Tao12, Theorem 2.4.2]). If $\mu_{1},\mu_{2},\ldots$ are Borel probability distributions on a separable metric space $S$, and $c\in S$, then the following two statements are equivalent:

1. (i)

$X_{n}\to c$ a.s. whenever $X_{1},X_{2},\ldots$ are random elements of $S$ defined on a common probability space such that each $X_{n}$ has distribution $\mu_{n}$; 2. (ii)

\sum_{n=1}^{\infty}\mu_{n}\bigl{(}\{\,x\in S\mid d(x,c)>\epsilon\,\}\bigr{)}<\infty for all $\epsilon>0$.

This can be shown using the Borel-Cantelli lemmas [Bil12, Theorem 4.3 and 4.4]. This type of strong convergence is possible in Theorem 1.2 because of a strong concentration of measure result we will use.

The rest of the paper is organized as follows. In Section 2, we will first reduce Theorem 1.2 to a form with stronger assumptions. Then we will see that the reduced semicircular law follows from some moment computations. In Section 3, we will develop a tool needed for the moment computation, and in Section 4, we will perform the actual moment computation. In Section 5, we will derive the aforementioned semicircular law which assumes Gaussian convergence of the sum of a row.

2 Preliminary reductions

Assume that $W_{n}$ satisfies the conditions of Theorem 1.2.

2.1 Convergence in expectation is enough

If we have $\operatorname{\mathbf{E}}\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$ (for the meaning of $\operatorname{\mathbf{E}}\mu_{W_{n}}$, see Theorem A.5), then

[TABLE]

for all continuous and bounded $f\colon\mathbf{R}\to\mathbf{R}$. By the concentration inequality Theorem C.3 for spectral measures and the Borel-Cantelli lemma, we have

[TABLE]

for all $p,q\in\mathbf{Q}$ with $p<q$, where $f_{p,q}\colon\mathbf{R}\to\mathbf{R}$ is $1$ on $(-\infty,p\,]$, [math] on $[\,q,\infty)$, and linear on $[\,p,q\,]$. This implies that $\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$ a.s. by Theorem A.3. Therefore, it is enough to show $\operatorname{\mathbf{E}}\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$.

2.2 Truncation

Since (1.3) holds, we have positive integers $n_{1}<n_{2}<\ldots$ such that

[TABLE]

for all $n\geq n_{k}$ for each $k\in\mathbf{N}$. If we let $\eta_{n}=1$ for all $n\in\{1,\ldots,n_{1}-1\}$, and $\eta_{n}=1/k$ for all $n\in\{n_{k},\ldots,n_{k+1}-1\}$ for each $k\in\mathbf{N}$, then $\eta_{n}\to 0$ and

[TABLE]

Let W_{n}^{\prime}:=\bigl{(}w_{ij}^{(n)}\operatorname{\mathbf{1}}(|w_{ij}^{(n)}|\leq\eta_{n})\bigr{)}_{i,j=1}^{n}. Since

[TABLE]

it is enough to show $\operatorname{\mathbf{E}}\mu_{W^{\prime}_{n}}\Rightarrow\mu_{\mathrm{sc}}$ by Corollary B.15 and Theorem A.3.

2.3 Centralization

For each $n=1,2,\ldots$ and $i,j=1,\ldots,n$, let

[TABLE]

and let $W_{n}^{\prime}-\operatorname{\mathbf{E}}W_{n}^{\prime}:=(v_{ij}^{(n)})_{i,j=1}^{n}$. Since

[TABLE]

it is enough to show $\operatorname{\mathbf{E}}\mu_{W^{\prime}_{n}-\operatorname{\mathbf{E}}W^{\prime}_{n}}\Rightarrow\mu_{\mathrm{sc}}$ by Corollary B.15 and Theorem A.3.

We claim that $W^{\prime}_{n}-\operatorname{\mathbf{E}}W^{\prime}_{n}$ satisfies all conditions $W_{n}$ is supposed to satisfy in Theorem 1.2. The fact that (1.1) and (1.2) still hold even if we replace $W_{n}$ by $W^{\prime}_{n}-\operatorname{\mathbf{E}}W^{\prime}_{n}$ follows from the following:

[TABLE]

The condition (1.3) for $W^{\prime}_{n}-\operatorname{\mathbf{E}}W^{\prime}_{n}$ easily follows from the bound $|w_{ij}^{(n)}|\leq\eta_{n}$. Since $|v_{i,j}^{(n)}|\leq 2\eta_{n}$ for all $n=1,2,\ldots$ and $i,j=1,\ldots,n$, by doubling $\eta_{1}$, $\eta_{2}$, …we have $\eta_{n}\to 0$ and $|v_{i,j}^{(n)}|\leq\eta_{n}$. Thus, from now on, we can assume that $|w_{ij}^{(n)}|\leq\eta_{n}$ for some $\eta_{1},\eta_{2},\ldots>0$ satisfying $\eta_{n}\to 0$.

2.4 Rescaling

Fix $n\in\mathbf{N}$. We will choose a number $0\leq c_{ij}^{(n)}\leq 1$ for each $i,j=1,\ldots,n$ so that $c_{ij}^{(n)}=c_{ji}^{(n)}$ always hold. Start by letting $c_{ij}^{(n)}=1$ for all $i,j=1,\ldots,n$. We start with the first row and the first column. If \sum_{j=1}^{n}\operatorname{\mathbf{Var}}\bigl{[}w_{1j}^{(n)}\bigr{]}\leq C, then do nothing. Otherwise, lower $c_{11}^{(n)},\ldots,c_{1n}^{(n)}$ (not below [math]) so that

[TABLE]

and let $c_{j1}^{(n)}:=c_{1j}^{(n)}$ for all $j=1,\ldots,n$. We note that at this point we have

[TABLE]

Assume that $k\in\{2,\ldots,n\}$, and that we’ve examined up to $(k-1)$-th row. If

[TABLE]

then do nothing. Otherwise, lower $c_{k1}^{(n)},\ldots,c_{kn}^{(n)}$ (not below [math]) so that

[TABLE]

and let $c_{jk}^{(n)}=c_{kj}^{(n)}$ for all $j=1,\ldots,n$. At this point we have

[TABLE]

This can be shown by an induction on $k$. After completing the whole process, we are left with numbers $0\leq c_{ij}^{(n)}\leq 1$ such that

[TABLE]

for all $i=1,\ldots,n$, and

[TABLE]

Let \widehat{W}_{n}=\bigl{(}c_{ij}^{(n)}w_{ij}^{(n)}\bigr{)}_{i,j=1}^{n}. Since $(1-c)^{2}\leq 1-c^{2}$ holds for any $0\leq c\leq 1$, we have

[TABLE]

by (2.2). Thus, by Corollary B.15 and Theorem A.3, it is enough to show $\operatorname{\mathbf{E}}\mu_{\widehat{W}_{n}}\Rightarrow\mu_{\mathrm{sc}}$.

The altered matrix $\widehat{W}_{n}$ has an advantage over $W_{n}$ that (2.1) holds. Also, the modulus of each entry of $\widehat{W}_{n}$ is still bounded by $\eta_{n}$. We claim that $\widehat{W}_{n}$ also satisfies all conditions $W_{n}$ is assumed to satisfy in Theorem 1.2. First, each entry of $\widehat{W}_{n}$ obviously has mean zero. Also, since each entry of $\widehat{W}_{n}$ has modulus less than or equal to the corresponding entry of $W_{n}$, the condition (1.3) is satisfied by $\widehat{W}_{n}$. The condition (1.2) for $\widehat{W}_{n}$ obviously holds as we have an even stronger property (2.1). Finally, (1.1) for $\widehat{W}_{n}$ follows from (2.2) and the fact that (1.2) is satisfied by $W_{n}$. This proves our claim, and so from now on, we can also assume that (1.4) is true.

2.5 Reduction to moment convergence

On top of the assumptions of Theorem 1.2, we now also have the following.

1. (i)

There are $\eta_{1},\eta_{2},\ldots>0$ with $\lim_{n\to\infty}\eta_{n}=0$ such that $|w_{ij}^{(n)}|\leq\eta_{n}$ for all $n=1,2,\ldots$ and $i,j=1,\ldots,n$. 2. (ii)

There is some finite $C\geq 0$ such that (1.4) holds.

Since $|w_{ij}^{(n)}|\leq\eta_{n}$, every eigenvalue of $W_{n}$ has absolute value at most $n\eta_{n}$. So, $\operatorname{\mathbf{E}}\mu_{W_{n}}$ is supported on $[-n\eta_{n},n\eta_{n}]$, and in particular $\operatorname{\mathbf{E}}\mu_{W_{n}}$ has moments of all orders. As

[TABLE]

for any $r\in\mathbf{R}$ by the ratio test, $\mu_{\mathrm{sc}}$ is determined by its moments by [Bil12, Theorem 30.1]. Thus, by the moment convergence theorem [Bil12, Theorem 30.2], it is enough to show

[TABLE]

For each $k=1,2,\ldots$, since there are continuous bounded $g_{k,n}:\mathbf{R}\to\mathbf{R}$ with $g_{k,n}(x)=x^{k}$ for all $x\in[-n\eta_{n},n\eta_{n}]$, we have

[TABLE]

On the other hand, we can directly compute the moments of $\mu_{\mathrm{sc}}$ as follows.

Lemma 2.1**.**

For any $m=1,2,\ldots$, we have

[TABLE]

Proof.

A trigonometric substitution $x=2\cos\theta$ gives

[TABLE]

As

[TABLE]

for any $l=1,2,\ldots$, we have

[TABLE]

∎

Note that $\int_{\mathbf{R}}x^{k}\,\mu_{\mathrm{sc}}(dx)=0$ whenever $k\in\mathbf{N}$ is odd. Thus, it is enough to show that

[TABLE]

and that

[TABLE]

These will be proved in Section 4 by using the content of Section 3.

3 Trees and products of variances

Our graphs will be undirected. We allow graphs to have loops, but don’t allow them to have multiple edges. Let $G$ be a finite graph. For any $n\in\mathbf{N}$, denote by $I(G,n)$ the collection of all injections from $V(G)$ into $\{1,\ldots,n\}$. Given any $F\in I(G,n)$ and $e\in E(G)$ with ends $u,v$, we let

[TABLE]

It is well-defined since each $W_{n}$ is Hermitian. Then we let

[TABLE]

Here $P$ stands for “product.” Also, the notation $\rho_{e,F}^{(n)}$ will no longer appear.

Lemma 3.1**.**

If $T$ is a finite tree with $m$ edges, $u\in V(T)$, $n\in\mathbf{N}$, and $i\in\{1,\ldots,n\}$, then

[TABLE]

Proof.

If $m=0$, then (3.1) obviously holds. (We define the product of zero terms as $1$.) To proceed by induction, assume that (3.1) holds for $m$, and let $T$ be a tree with $m+1$ edges. Choose any leaf $w$ of $T$ different from $u$, and let $x$ be the only vertex of $T$ adjacent to $w$. Since

[TABLE]

for all $j\in\{1,\ldots,n\}$, we have

[TABLE]

by the induction hypothesis. ∎

Lemma 3.2**.**

For any finite tree $T$,

[TABLE]

Proof.

Let $m:=E(T)$. If $m=0$, then (3.2) obviously holds. To proceed by induction, assume that the result holds for trees with $m$ edges, and let $T$ be a tree with $m+1$ edges. Let $u\in V(T)$ be a leaf of $T$, and $w$ be the only vertex of $T$ adjacent to $u$ in $T$. Note that

[TABLE]

by the induction hypothesis. By Lemma 3.1, we have

[TABLE]

Combining (3.3), (3.4), and the fact that

[TABLE]

we can conclude that (3.2) holds. ∎

4 Computation of moments

Fix a $k\in\mathbf{N}$. Let us call any $n$-tuple $(i_{0},\ldots,i_{k})$ with $i_{0}=i_{k}$ a closed walk of length $k$. If $\mathbf{i}=(i_{0},\ldots,i_{k})$ is a closed walk, we let $G(\mathbf{i})$ be the graph (possibly having loops but having no multiple edges) with the vertex set $V(\mathbf{i}):=\{i_{0},\ldots,i_{k}\}$ and the edge set

[TABLE]

Two closed walks $\mathbf{i}=(i_{0},\ldots,i_{k})$ and $\mathbf{j}=(j_{0},\ldots,j_{k})$ are said to be isomorphic if for any $s,t=0,\ldots,k$ we have $i_{s}=i_{t}$ if and only if $j_{s}=j_{t}$. If $t\in\mathbf{N}$, then a canonical closed walk of length $k$ on $t$ vertices is a closed walk $\mathbf{c}=(c_{0},\ldots,c_{k})$ with $V(\mathbf{c})=\{1,\ldots,t\}$ such that

1. (i)

$c_{0}=c_{k}=1$ and 2. (ii)

$c_{t}\leq\max\{c_{0},\ldots,c_{t-1}\}+1$ for each $t=1,\ldots,k$.

Let $\Gamma(k,t)$ denote the set of such walks. It is straightforward to show that any closed walk is isomorphic to exactly one canonical closed walk. For any $\mathbf{c}\in\Gamma(k,t)$, let $L(n,\mathbf{c})$ denote the set of all closed walks $(i_{0},\ldots,i_{k})$ with $i_{0},\ldots,i_{k}\in\{1,\ldots,n\}$ which are isomorphic to $\mathbf{c}$.

Note that

[TABLE]

Here the upper bound of $t$ is (rather arbitrarily) set to $k+1$ since $\Gamma(k,t)$ is empty for any $t>k+1$. We will compute

[TABLE]

for each $t\in\mathbf{N}$ and $\mathbf{c}\in\Gamma(k,t)$.

Lemma 4.1**.**

Let $t\in\mathbf{N}$ and $\mathbf{c}=(c_{0},\ldots,c_{k})\in\Gamma(k,t)$. If $\mathbf{c}$ walks on some edge $\{i,j\}$ exactly once, i.e. $\{c_{s-1},c_{s}\}=\{i,j\}$ for exactly one $s\in\{1,\ldots,k\}$, then

[TABLE]

for any $n\in\mathbf{N}$ and $(i_{0},\ldots,i_{k})\in L(n,\mathbf{c})$.

Proof.

Since the upper triangular entries of $W_{n}$ are jointly independent, $\prod_{s=1}^{k}w_{i_{s-1}i_{s}}^{(n)}$ can be broken into $w_{ij}^{(n)}$ or $w_{ji}^{(n)}$, and a random variable independent from $w_{ij}$. Since $\operatorname{\mathbf{E}}w_{ij}^{(n)}=0$, the desired conclusion follows. ∎

Lemma 4.2**.**

Let $t\in\mathbf{N}$ and $\mathbf{c}=(c_{0},\ldots,c_{k})\in\Gamma(k,t)$. Assume that $\mathbf{c}$ doesn’t walk on any edge exactly once, i.e. for each $s=1,\ldots,k$ there is a $r\in\{1,\ldots,k\}$ with $r\neq s$ such that $\{c_{s-1},c_{s}\}=\{c_{r-1},c_{r}\}$. Then we have $t\leq k/2+1$, and the following hold.

1. (i)

If $t<k/2+1$, then

[TABLE] 2. (ii)

If $t=k/2+1$, then

[TABLE]

Proof.

As each edge of $G(\mathbf{c})$ is walked on at least twice by $\mathbf{c}$, the graph $G(\mathbf{c})$ has at most $k/2$ edges. Since $G(\mathbf{c})$ is a connected graph with $t$ vertices, we have $t\leq k/2+1$, and $G(\mathbf{c})$ has a spanning tree $S$ with $t-1$ edges. If $\mathbf{i}=(i_{0},\ldots,i_{k})\in L(n,\mathbf{c})$, then there is an injection $F_{\mathbf{i}}\colon\{1,\ldots,t\}\to\{1,\ldots,n\}$ with $i_{s}=F_{\mathbf{i}}(c_{s})$ for all $s=0,\ldots,k$.

(i) Assume $t<k/2+1$. Using the bound $|w_{ij}^{(n)}|\leq\eta_{n}$, and the fact that $\mathbf{c}$ walks on any edge of $G(\mathbf{c})$ at least twice, we can derive

[TABLE]

Note that $\lim_{n\to\infty}\eta_{n}^{k-2(t-1)}=0$ since $t<k/2+1$. Since the map $L(n,\mathbf{c})\to I(S,n)$ given by $\mathbf{i}\mapsto F_{\mathbf{i}}$ is a bijection, we have

[TABLE]

by Lemma 3.2.

(ii) Assume $t=k/2+1$. Since $S$ has $k/2$ edges and each edge of $S$ is walked on twice by $\mathbf{c}$, we see that $S=G(\mathbf{c})$. As each edge of $G(\mathbf{c})$ is traversed once in each direction, i.e. for each $s=1,\ldots,k$ there is an $r\in\{1,\ldots,k\}$ with $r\neq s$ such that $c_{s-1}=c_{r}$ and $c_{s}=c_{r-1}$, we have

[TABLE]

by Lemma 3.2. ∎

Proof of (2.3) and

(2.4).

Lemma 4.1 and 4.2 tell us that we have (4.3) if and only if $\mathbf{c}$ doesn’t walk on any edge exactly once and $t=k/2+1$. Otherwise, we have (4.2). If $k$ is odd, then $k/2+1$ is not an integer, and so we cannot have $t=k/2+1$. So, for any odd $k\in\mathbf{N}$, we have

[TABLE]

by (4.1).

Assume that $k$ is even. Let $U$ be the set of all $\mathbf{c}\in\Gamma(k,k/2+1)$ which traverses each edge of $G(\mathbf{c})$ twice. Then by Lemma 4.2 (ii) and (4.1), we have

[TABLE]

A Dyck path of length $k$ is a finite sequence $(x_{0},\ldots,x_{k})$ satisfying the following:

1. (i)

$x_{0}=x_{k}=0$; 2. (ii)

$x_{s}\geq 0$ for all $s=0,\ldots,k$; 3. (iii)

$|x_{s}-x_{s-1}|=1$ for all $s=1,\ldots,k$.

Given an $\mathbf{c}=(c_{0},\ldots,c_{k})\in U$, let $D(\mathbf{c}):=(x_{0},\ldots,x_{k})$ where $x_{s}$ is the distance between $1$ ($=c_{0}$) and $c_{s}$ in $G(\mathbf{c})$. Then it is clear that $D(\mathbf{c})$ is indeed a Dyck path, and it is not difficult to see that $D$ is a bijection from $U$ to the set of all Dyck paths of length $k$. It is well-known that there are exactly $\frac{1}{k/2+1}\binom{k}{k/2}$ Dyck paths of length $k$; see [vLW01, Example 14.8]. Thus, we indeed have

[TABLE]

This finishes the proof of the semicircular law Theorem 1.2. ∎

5 Gaussian convergence

The paper [Jun18] considers real symmetric random matrices $W_{1},W_{2},\ldots$ with size $1\times 1$, $2\times 2$, …whose upper triangular entries are i.i.d. In that paper, it is shown that if the sum of a row of $W_{n}$ converges in distribution to the standard normal distribution $N(0,1)$ as $n\to\infty$, then $\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$ as $n\to\infty$ a.s. We prove this fact generalized to random matrices with non-i.i.d. entries in this section. By doing so, we will demonstrate how one can apply Theorem 1.2 to obtain a semicircular law for random matrices whose entries might have infinite variances.

The type of convergence described in the following fact will appear many times in this section.

Proposition 5.1** (Uniform convergence of triangular arrays).**

Let $S$ be a topological space, $m_{1},m_{2},\ldots\in\mathbf{N}$, and $(s_{ni})_{i=1}^{m_{n}}$ be a finite sequence in $S$ for each $n\in\mathbf{N}$. For any $s\in S$, the following two conditions are equivalent:

1. (i)

$s_{ni_{n}}\to s$* as $n\to\infty$ for any choice of $i_{n}\in\{1,\ldots,m_{n}\}$ for each $n\in\mathbf{N}$;* 2. (ii)

for any neighborhood $N$ of $s$, there exists some $n_{0}\in\mathbf{N}$ such that $s_{ni}\in N$ for all $n\geq n_{0}$ and $i=1,\ldots,m_{n}$.

Proof.

We omit the straightforward proof. ∎

The following is the main result of this section.

Theorem 5.2** (Gaussian convergence semicircular law).**

For each $n\in\mathbf{N}$, let $W_{n}=(w_{ij}^{(n)})_{i,j=1}^{n}$ be a random $n\times n$ real symmetric matrix whose upper triangular entries are jointly independent and have symmetric distributions. We assume that $W_{1},W_{2},\ldots$ are defined on the same probability space. Assume that $(W_{n})_{n\in\mathbf{N}}$ is a null array in the sense that $w_{i_{n}j_{n}}^{(n)}\Rightarrow 0$ as $n\to\infty$ for any choice of $i_{n},j_{n}\in\{1,\ldots,n\}$ for each $n\in\mathbf{N}$. If also $\sum_{j=1}^{n}w_{i_{n}j}^{(n)}\Rightarrow N(0,1)$ for any choice of $i_{n}\in\{1,\ldots,n\}$ for each $n\in\mathbf{N}$, then $\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$ as $n\to\infty$ a.s.

The following two facts will be used in the proof of Theorem 5.2.

Theorem 5.3** (Gaussian convergence).**

For each $n\in\mathbf{N}$, let $X_{n1},\ldots,X_{nn}$ be jointly independent real-valued random variables. Assume that $X_{ni_{n}}\Rightarrow 0$ as $n\to\infty$ regardless of how we choose $i_{n}\in\{1,\ldots,n\}$ for each $n\in\mathbf{N}$. Then $\sum_{i=1}^{n}X_{ni}\Rightarrow N(0,1)$ as $n\to\infty$ if and only if the following conditions hold:

1. (i)

$\lim_{n\to\infty}\sum_{i=1}^{n}\operatorname{\mathbf{P}}(|X_{ni}|>\epsilon)=0$* for all $\epsilon>0$;* 2. (ii)

$\lim_{n\to\infty}\sum_{i=1}^{n}\operatorname{\mathbf{E}}[\,X_{ni}\mathrel{;}|X_{ni}|\leq 1\,]=0$; 3. (iii)

$\lim_{n\to\infty}\sum_{i=1}^{n}\operatorname{\mathbf{Var}}[X_{ni}\operatorname{\mathbf{1}}(|X_{ni}|\leq 1)]=1$.

Proof.

See [Kal02, Theorem 5.15]. ∎

Theorem 5.4** (Bernstein’s inequality).**

Suppose that $X_{1},\ldots,X_{n}$ are independent real-valued random variables, each with mean [math], and each bounded by $1$. If $S=X_{1}+\cdots+X_{n}$, then

[TABLE]

Proof.

The proof of [Bil99, M20] with a slight change works. ∎

Proof of Theorem 5.2.

By Theorem 5.3, we have

[TABLE]

for any choice of $i_{n}\in\{1,\ldots,n\}$ for each $n\in\mathbf{N}$, for any $\epsilon>0$. Then Proposition 5.1 implies

[TABLE]

Let $v_{ij}^{(n)}:=w_{ij}^{(n)}\operatorname{\mathbf{1}}(|w_{ij}^{(n)}|\leq 1)$ and $W^{\prime}_{n}=(v_{ij}^{(n)})_{i,j=1}^{n}$. Since

[TABLE]

by bounding $\sum_{1\leq i\leq j\leq n}\operatorname{\mathbf{1}}(|w_{ij}^{(n)}|>1)$ from above we would be able to apply Theorem B.11. For any given $\epsilon>0$, we have some $n_{0}\in\mathbf{N}$ such that

[TABLE]

by (5.1). Since $\operatorname{\mathbf{1}}(|w_{ij}^{(n)}|>1)$, $1\leq i\leq j\leq n$, are jointly independent, Bernstein’s inequality (Theorem 5.4) implies

[TABLE]

As $\sum_{n=1}^{\infty}\exp(-\epsilon n/4)<\infty$ for each $\epsilon>0$, we have

[TABLE]

and therefore

[TABLE]

by (5.2). By Theorem B.11, it now suffices to show $\mu_{W^{\prime}_{n}}\Rightarrow\mu_{\mathrm{sc}}$ as $n\to\infty$ a.s.

We claim that $W^{\prime}_{n}$ satisfies all conditions of Theorem 1.2. Since each $w_{ij}^{(n)}$ is symmetric, each entry of $W^{\prime}_{n}$ has mean zero. By Theorem 5.3, we have

[TABLE]

for any choice of $i_{n}\in\{1,\ldots,n\}$ for each $n\in\mathbf{N}$. So, by using Proposition 5.1, we can see that the conditions 1.1 and 1.2 with $w_{ij}^{(n)}$ replaced by $v_{ij}^{(n)}$ hold. Finally, the condition (1.3) with $w_{ij}^{(n)}$ replaced by $v_{ij}^{(n)}$ follows from

[TABLE]

and (5.1). ∎

Appendix A Probability measures on $\mathbf{R}$

A.1 Weak convergence

Definition A.1** (The space $\Pr(\mathbf{R})$).**

Let $\Pr(\mathbf{R})$ denote the set of all Borel probability measures on $\mathbf{R}$. We equip $\Pr(\mathbf{R})$ with the smallest topology that makes $\mu\mapsto\int_{\mathbf{R}}f\,d\mu$ continuous for all continuous bounded $f:\mathbf{R}\to\mathbf{R}$. Then we equip $\Pr(\mathbf{R})$ with the Borel $\sigma$-algebra.

Note that if $\mu,\mu_{1},\mu_{2},\ldots\in\Pr(\mathbf{R})$, then we have $\mu_{n}\Rightarrow\mu$ if and only if $\mu_{n}\to\mu$ under the topology of $\Pr(\mathbf{R})$.

Definition A.2** (Lévy metric).**

If $F$ and $G$ are distribution functions, then the Lévy distance between $F$ and $G$ is defined by

[TABLE]

It is not difficult to show that $L$ is indeed a metric on $\Pr(\mathbf{R})$. For any given $\mu\in\Pr(\mathbf{R})$, let $F_{\mu}$ denote the distribution function of $\mu$.

Theorem A.3** (Characterizations of weak convergence).**

If $\mu$, $\mu_{1}$, $\mu_{2}$, $\ldots\in\Pr(\mathbf{R})$, then the following are equivalent:

1. (i)

$\mu_{n}\Rightarrow\mu$; 2. (ii)

$\int_{\mathbf{R}}f_{p,q}\,d\mu_{n}\to\int_{\mathbf{R}}f_{p,q}\,d\mu$* for all $p,q\in\mathbf{Q}$ with $p<q$, where $f_{p,q}\colon\mathbf{R}\to\mathbf{R}$ is the function which has value $1$ on $(-\infty,p\,]$, has value [math] on $[\,q,\infty)$, and is linear on $[\,p,q\,]$;* 3. (iii)

$L(F_{\mu_{n}},F_{\mu})\to 0$.

Proof.

(i) implies (ii): Directly follows from the definition of convergence in distribution.

(ii) implies (i): Assume that $\int_{\mathbf{R}}f_{p,q}\,d\mu_{n}\to\int_{\mathbf{R}}f_{p,q}\,d\mu$ for all $p,q\in\mathbf{Q}$ with $p<q$, and let $F,F_{1},F_{2},\ldots$ be the distribution functions of $\mu,\mu_{1},\mu_{2},\ldots$. Let $x$ be any continuity point of $\mu$, and let $\epsilon>0$ be given. Since $F$ is right continuous, we have $F(x+\delta)\leq F(x)+\epsilon$ for some $\delta>0$. If we choose any $p,q\in\mathbf{Q}$ with $x<p<q<x+\delta$, then

[TABLE]

As $F$ is also left continuous at $x$, a similar reasoning yields $F(x)-\epsilon\leq\liminf_{n\to\infty}F_{n}(x)$. Since $\epsilon>0$ is arbitrary, we have $F_{n}(x)\to F(x)$.

(i) implies (iii): Let $\epsilon>0$ be given. Choose continuity points $x_{0},\ldots,x_{k}\in\mathbf{R}$ of $F_{\mu}$ such that $x_{0}<\cdots<x_{k}$, $F_{\mu}(x_{0})\leq\epsilon$, $F_{\mu}(x_{k})\geq 1-\epsilon$, and

[TABLE]

Let $N\in\mathbf{N}$ be such that $n\geq N$ implies

[TABLE]

Let $x\in\mathbf{R}$ be arbitrarily given. If $x>x_{k}$, then

[TABLE]

for any $n\in\mathbf{N}$. If $x\in(x_{i-1},x_{i}]$ where $i\in\{1,\ldots,n\}$, then

[TABLE]

for any $n\geq N$. If $x\leq x_{0}$, then

[TABLE]

for any $n\geq N$. Similarly we can show that $F_{\mu}(x-2\epsilon)-2\epsilon\leq F_{\mu_{n}}(x)$ for all $x\in\mathbf{R}$ and $n\geq N$. So, $L(F_{\mu_{n}},F_{\mu})\leq 2\epsilon$ for all $n\geq N$. As $\epsilon>0$ was arbitrary, we have $L(F_{\mu_{n}},F_{\mu})\to 0$.

(iii) implies (i): Let $x\in\mathbf{R}$ be a continuity point of $F_{\mu}$, and let $\epsilon>0$ be given. Since $F_{\mu}$ is continuous at $x$, there is a $\delta\in(0,\epsilon/2)$ such that $|F_{\mu}(x)-F_{\mu}(y)|\leq\epsilon/2$ for all $|y-x|\leq\delta$. Let $N\in\mathbf{N}$ be such that $L(F_{\mu_{n}},F_{\mu})<\delta$ for all $n\geq N$. Then,

[TABLE]

for all $n\geq N$. Now observe that

[TABLE]

and

[TABLE]

for all $n\geq N$. Since $\epsilon>0$ was arbitrary, $F_{\mu_{n}}(x)\to F_{\mu}(x)$. ∎

A.2 Expected probability measures

Lemma A.4**.**

For any Borel $A\subset\mathbf{R}$, the map $e_{A}\colon\Pr(\mathbf{R})\to[0,1]$ defined by $e_{A}(\mu):=\mu(A)$ is measurable.

Proof.

For any $x\in\mathbf{R}$ and $\epsilon>0$, let $f_{x,x+\epsilon}\colon\mathbf{R}\to\mathbf{R}$ be the map which is $1$ on $(-\infty,x\,]$, [math] on $[\,x+\epsilon,\infty)$, and linear on $[x,x+\epsilon]$. Then the map $\mu\mapsto\int_{\mathbf{R}}f_{x,x+\epsilon}\,d\mu$ is continuous, and so measurable. Since $\int_{\mathbf{R}}f_{x,x+1/n}\,d\mu\to\mu((-\infty,x])$ as $n\to\infty$ by bounded convergence, the map $e_{(-\infty,x]}$ is measurable for any $x\in\mathbf{R}$. Let $\mathcal{C}$ be the collection of all Borel $A\subset\mathbf{R}$ such that $e_{A}$ is measurable. If $A_{1},A_{2},\ldots\in\mathcal{C}$ are disjoint, then

[TABLE]

is measurable, and so $\bigcup_{n=1}^{\infty}A_{n}\in\mathcal{C}$. If $A\in\mathcal{C}$, then $e_{\mathbf{R}\setminus A}=1-e_{A}$ is measurable, and so $\mathbf{R}\setminus A\in\mathcal{C}$. These show that $\mathcal{C}$ is a $\lambda$-system containing $(-\infty,x\,]$ for all $x\in\mathbf{R}$. As the rays $(-\infty,x\,]$ form a $\pi$-system that generates the Borel $\sigma$-algebra of $\mathbf{R}$, the $\pi$-$\lambda$ theorem concludes the proof. ∎

Theorem A.5**.**

Let $\mu$ be a random element of $\Pr(\mathbf{R})$. Then there exists a unique $\operatorname{\mathbf{E}}\mu\in\Pr(\mathbf{R})$ satisfying

[TABLE]

for all $x\in\mathbf{R}$. The probability measure $\operatorname{\mathbf{E}}\mu$ satisfies

[TABLE]

for all continuous and bounded $f:\mathbf{R}\to\mathbf{R}$.

Proof.

As uniqueness is easy, we only need to show the existence. Define $F:\mathbf{R}\to\mathbf{R}$ by $f(x):=\operatorname{\mathbf{E}}[F_{\mu}(x)]$. Since $F_{\mu}$ is surely nondecreasing, $f$ is nondecreasing. Since $F_{\mu}(n)\to 1$ and $F_{\mu}(-n)\to 0$ as $n\to\infty$ surely, $f(n)\to 1$ and $f(-n)\to 0$ as $n\to\infty$ by bounded convergence. If $x_{1}\geq x_{2}\geq\cdots$ and $x_{n}\to x\in\mathbf{R}$, then $F_{\mu}(x_{n})\to F_{\mu}(x)$ surely by the right continuity of distribution functions, and so $f(x_{n})\to f(x)$ by bounded convergence. This shows that $f$ is right continuous, and so the proof that $f$ is a distribution function is finished.

Let $\operatorname{\mathbf{E}}\mu$ denote the Borel probability measure on $\mathbf{R}$ with distribution $f$. For any $-\infty<a\leq b<\infty$, we have

[TABLE]

If $(a_{1},b_{1}],(a_{2},b_{2}],\ldots$ are disjoint, then

[TABLE]

by monotone convergence. Since any open subset of $\mathbf{R}$ is a countable union of disjoint open intervals, and any open interval is a countable union of disjoint bounded intervals of the form $(a,b]$, we see that

[TABLE]

holds for any open $U\subset\mathbf{R}$.

Now we show (A.1). By linearity of integral and expectation, we may assume that $f$ is nonnegative. For each $t\geq 0$, let $U_{t}:=\{\,x\in\mathbf{R}\mid f(x)>t\,\}$. We want to apply Tonelli’s theorem to the map $G:\Pr(\mathbf{R})\times[0,\infty)\to[0,1]$ given by $(\nu,t)\mapsto\nu(U_{t})$, so we first show that this map is jointly measurable. For each $n\in\mathbf{N}$, let $G_{n}:\Pr(\mathbf{R})\times[0,\infty)\to[0,1]$ by

[TABLE]

Since $\nu\mapsto\nu(U)$ is measurable for any open $U$, each $G_{n}$ is measurable. As $G_{n}$ increases to $G$, we can conclude that $G$ is measurable. Now we can use Tonelli’s theorem to conclude that

[TABLE]

∎

Appendix B Spectra of Hermitian matrices

B.1 Basic facts

Recall the following version of the spectral theorem from linear algebra.

Theorem B.1** (Spectral Theorem).**

Let $V$ be an $n$-dimensional complex inner product space. For any self-adjoint linear operator $T:V\to V$, there exists an orthonormal basis of $V$ consisting of eigenvectors of $V$.

Proof.

See [Tao12, Theorem 1.3.1] or [HK71, Theorem 9 in Section 9.5]. ∎

Also recall the following.

Proposition B.2**.**

Any eigenvalue of a self-adjoint linear operator on a complex inner product space is real.

Proof.

Let $\lambda$ be an eigenvalue of a self-adjoint linear operator $T$ on a complex inner product space $V$. If $Tv=\lambda v$ and $v\neq 0$, then

[TABLE]

and so $\bar{\lambda}=\lambda$. ∎

The following naturally follows from Theorem B.1 and Proposition B.2.

Corollary B.3**.**

For any $n\times n$ Hermitian matrix $A$, there exists an $n\times n$ diagonal matrix $D$ with real entries and an $n\times n$ unitary matrix $U$ such that $A=UDU^{\ast}$.

Definition B.4** (Ordered eigenvalues).**

If $A$ is an $n\times n$ Hermitian matrix, then we denote the eigenvalues of $A$ counted with multiplicities as

[TABLE]

Definition B.5** (Spectral distributions).**

If $A$ is an $n\times n$ Hermitian matrix, then the spectral distribution of $A$ is the Borel probability measure on $\mathbf{R}$ defined by

[TABLE]

We write $F_{A}$ as a shorthand for $F_{\mu_{A}}$.

Theorem B.6** (Courant-Fischer minimax theorem).**

Let $A$ be an $n\times n$ Hermitian matrix. For each $1\leq i\leq n$, we have

[TABLE]

and

[TABLE]

where $V$ ranges over the subspaces of $\mathbf{C}^{n}$.

Proof.

We only need to show the first equality, since the second follows by applying the first to $-A$. By the spectral theorem, we may assume that $A$ is diagonal with $\lambda_{i}(A)$ at the $(i,i)$-entry. Note that if $v=\begin{pmatrix}v_{1}&\cdots&v_{n}\end{pmatrix}^{T}$, then we have $v^{\ast}Av=\sum_{i=1}^{n}\lambda_{i}(A)|v_{i}|^{2}$. If $V$ is the subspace spanned by $e_{1},\ldots,e_{i}$, then $\inf_{v\in V:\lVert v\rVert=1}v^{\ast}Av=\lambda_{i}(A)$. To show the other direction, let $V$ be any $i$-dimensional subspace of $\mathbf{C}^{n}$. If $W$ is the subspace spanned by $e_{i},\ldots,e_{n}$, then $V\cap W\neq\{\mathbf{0}\}$ follows from

[TABLE]

If we choose any $w\in V\cap W$ with $\lVert w\rVert=1$, then

[TABLE]

∎

Theorem B.7** (Cauchy interlacing law).**

If $A_{n}$ is an $n\times n$ Hermitian matrix and $A_{n-1}$ is the top left $(n-1)\times(n-1)$ minor of $A_{n}$, then

[TABLE]

for any $1\leq i\leq n-1$.

Proof.

For any $v\in\mathbf{C}^{n-1}$, we have

[TABLE]

So, by Theorem B.6, we have

[TABLE]

By applying this result to $-A_{n}$, we have

[TABLE]

∎

B.2 Perturbations by small Frobenius norms

We will show that spectral distributions are stable under two types of perturbations. The first can be described using the following norm.

Definition B.8**.**

If $A=(a_{ij})_{i,j=1}^{n}$ is an $n\times n$ complex matrix, then the Frobenius norm of $A$ is given by

[TABLE]

Note that the Frobenius norm is just the $L^{2}$-norm on $\mathbf{C}^{n^{2}}$. If $A$ is a Hermitian matrix, then $\lVert A\rVert_{F}^{2}=\operatorname{tr}(A^{2})$. The following inequality tells us that the ordered tuple of eigenvalues is stable under perturbations with small Frobenius norms.

Theorem B.9** (Hoffman-Wielandt inequality).**

If $A$ and $B$ are $n\times n$ Hermitian matrices, then

[TABLE]

Proof.

Recall that eigenvalues and traces are similarity invariant. So, by the spectral theorem, we have

[TABLE]

Thus, it is enough to show

[TABLE]

(Recall that the right side of the desired inequality is equal to $\operatorname{tr}(A-B)^{2}$.) Again by the spectral theorem, we may assume that $A$ is diagonal with $\lambda_{i}(A)$ at its $i$-th entry, and write $B=UDU^{\ast}$ for some unitary $U$ where $D$ is the diagonal matrix with $\lambda_{i}(B)$ as its $i$-th entry. If $u_{ij}$ denotes the $(i,j)$-entry of $U$, we have

[TABLE]

It is enough to show that if $a_{1}\geq\cdots\geq a_{n}$ and $b_{1}\geq\cdots\geq b_{n}$, then the maximum of $\sum_{i,j=1}^{n}v_{ij}a_{i}b_{j}$ where $v_{ij}=v_{ji}\geq 0$ and $\sum_{j=1}^{n}v_{ij}=1$ is obtained when $(v_{ij})_{i,j=1}^{n}=I$; that is, $v_{11}=\cdots=v_{nn}=1$ and $v_{ij}=0$ whenever $i\neq j$. Let $(v_{ij})_{i,j=1}^{n}$ where $v_{ij}=v_{ji}\geq 0$ and $\sum_{j=1}^{n}v_{ij}=1$ is given. If $(v_{ij})\neq I$, we have $v_{kk}<1$ for some $k$. Since $\sum_{j=1}^{n}v_{kj}=1$, we have $v_{k\ell}>0$ for some $\ell\neq k$. Let $w_{kk}=v_{kk}+v_{k\ell}$, $w_{\ell\ell}=v_{\ell\ell}+v_{k\ell}$, $w_{k\ell}=w_{\ell k}=0$, and $w_{ij}=v_{ij}$ for all other $(i,j)$’s. Then we have $w_{ij}=w_{ji}\geq 0$ and $\sum_{j=1}^{n}w_{ij}=1$. Also,

[TABLE]

Note that $(w_{ij})$ has more $1$’s on the diagonal than $(v_{ij})$. If we repeat this procedure, we will arrive at $I$, and this shows our claim. ∎

From the Hoffman-Wielandt inequality (Theorem B.9), it follows that the spectral distribution is also stable under perturbations of small Frobenius norms.

Corollary B.10**.**

If $A$ and $B$ are $n\times n$ Hermitian matrices, then

[TABLE]

(For the definition of $\lVert\cdot\rVert_{F}$, see Definition B.8.)

Proof.

For any $x\in\mathbf{R}$ and $\epsilon>0$, we will show

[TABLE]

and

[TABLE]

Let i:=\bigl{|}\{\,\ell\mid\lambda_{\ell}(A)>x\,\}\bigr{|} and j:=\bigl{|}\{\,\ell\mid\lambda_{\ell}(B)>x+\epsilon\,\}\bigr{|}. Since $\lambda_{k}(A)\leq x$ and $\lambda_{k}(B)>x+\epsilon$ for each $i<k\leq j$, we have

[TABLE]

As $j-i=n(F_{A}(x)-F_{B}(x+\epsilon))$, (B.1) follows. Now let i^{\prime}:=\bigl{|}\{\,\ell\mid\lambda_{\ell}(A)>x\,\}\bigr{|} and j^{\prime}:=\bigl{|}\{\,\ell\mid\lambda_{\ell}(B)>x-\epsilon\,\}\bigr{|}. Then,

[TABLE]

and so (B.2) follows.

Now let $\epsilon>0$ be such that $\epsilon^{3}:=\frac{1}{n}\lVert A-B\rVert_{F}^{2}$. Then, $\frac{1}{\epsilon^{2}n}\lVert A-B\rVert_{F}^{2}=\epsilon$. Since

[TABLE]

for all $x\in\mathbf{R}$, we have $L(F_{A},F_{B})\leq\epsilon$, and thus the desired claim follows. ∎

B.3 Perturbations by small ranks

The second type of perturbation is the low-rank perturbation.

Theorem B.11** (Rank inequality).**

If $A$ and $B$ are $n\times n$ Hermitian matrices, then

[TABLE]

where $\lVert f\rVert_{\infty}:=\sup_{x\in\mathbf{R}}|f(x)|$.

Note that $L(F_{A},F_{B})\leq\lVert F_{A}-F_{B}\rVert_{\infty}$.

Proof.

Let $k:=\operatorname{rank}(A-B)$. Note that replacing $A$ and $B$ with $UAU^{\ast}$ and $UBU^{\ast}$ for some unitary $U$ doesn’t change each side of the desired inequality. So, using Corollary B.3, we may assume that $A-B$ is diagonal. By swapping rows and columns, we can further assume that

[TABLE]

where $A_{22}$ is a $(n-k)\times(n-k)$ matrix. If $x\in[\lambda_{i+1}(A_{22}),\lambda_{i}(A_{22}))$, then $\lambda_{k+i+1}(A)\leq\lambda_{i+1}(A_{22})$ and $\lambda_{i}(A_{22})\leq\lambda_{i}(A)$ by the Cauchy interlacing law (Theorem B.7), and so

[TABLE]

By the same reasoning, we also have

[TABLE]

and so

[TABLE]

Even if $x<\lambda_{n-k}(A_{22})$ or $x\geq\lambda_{1}(A_{22})$, this inequality can be proved by a similar argument. Now the desired inequality follows since $x$ is arbitrary. ∎

The following is a generalization of Theorem B.11.

Corollary B.12**.**

If $A$ and $B$ are $n\times n$ Hermitian matrices and $f:\mathbf{R}\to\mathbf{R}$ satisfies $\lVert f\rVert_{TV}\leq 1$, then

[TABLE]

Proof.

Let $-\infty<t_{1}<\cdots<t_{m}<\infty$ be such that

[TABLE]

Define $g:\mathbf{R}\to\mathbf{R}$ by letting $g(t_{i})=f(t_{i})$ for each $i=1,\ldots,m$, extending linearly between $t_{i}$ and $t_{i+1}$ for each $i=1,\ldots,m-1$, and setting to constants on $(-\infty,t_{1}]$ and $[t_{m},\infty)$. Note that $g^{\prime}:\mathbf{R}\to\mathbf{R}$ exists as an integrable function, and we have

[TABLE]

for any $i=1,\ldots,m$. Also, as $\lVert f\rVert_{TV}\leq 1$, we have

[TABLE]

Observe that

[TABLE]

Similarly we have

[TABLE]

and so

[TABLE]

by Theorem B.11. ∎

B.4 Random Hermitian matrices

Definition B.13** (Random Hermitian matrices).**

Let $H_{n}$ denote the space of all $n\times n$ Hermitian matrices. We equip $H_{n}$ with the standard Euclidean metric (and so the metric topology and the Borel $\sigma$-algebra) by identifying $H_{n}$ with $C^{n(n-1)/2}\times\mathbf{R}^{n}$, which is thought to represent the lower triangle of an $n\times n$ Hermitian matrix. A random element of $H_{n}$ is called a random $n\times n$ Hermitian matrix.

From Hoffman-Wielandt inequality (Theorem B.9), it follows that the map $\lambda:H_{n}\to\mathbf{R}^{n}$ given by $\lambda(A):=(\lambda_{1}(A),\ldots,\lambda_{n}(A))$ is continuous. This fact combined with the following lemma shows that the spectral distribution of a random Hermitian matrix is measurable.

Lemma B.14**.**

The map $\mathbf{R}^{n}\to\Pr(\mathbf{R})$ given by

[TABLE]

is continuous (and so is measurable).

Proof.

For any continuous bounded $f:\mathbf{R}\to\mathbf{R}$, the map $(x_{1},\ldots,x_{n})\in\mathbf{R}^{n}\mapsto\int_{\mathbf{R}}f\,d(\frac{1}{n}\sum_{i=1}^{n}\delta_{x_{i}})=\frac{1}{n}\sum_{i=1}^{n}f(x_{i})$ is continuous. Thus the given map is continuous by the definition of the topology of weak convergence. ∎

The following is a “random version” of Corollary B.10. If $X$ is a random $n\times n$ Hermitian matrix, we let $\operatorname{\mathbf{E}}F_{X}$ be the distribution function of $\operatorname{\mathbf{E}}\mu_{X}$, i.e. $(\operatorname{\mathbf{E}}F_{X})(x)=\operatorname{\mathbf{E}}[F_{X}(x)]$.

Corollary B.15**.**

If $X$ and $Y$ are random $n\times n$ Hermitian matrices, then

[TABLE]

(For the definition of $\lVert\cdot\rVert_{F}$, see Definition B.8.)

Proof.

If \operatorname{\mathbf{E}}\bigl{[}\lVert X-Y\rVert_{F}^{2}\bigr{]}=\infty, there is nothing to prove; so we may assume \operatorname{\mathbf{E}}\bigl{[}\lVert X-Y\rVert_{F}^{2}\bigr{]}<\infty. By applying (B.1) and (B.2) to $X$ and $Y$, and taking the expectation, we have

[TABLE]

for all $x\in\mathbf{R}$ and $\epsilon>0$. As in the proof of Corollary B.10, let $\epsilon>0$ be such that \epsilon^{3}:=\frac{1}{n}\operatorname{\mathbf{E}}\bigl{[}\lVert X-Y\rVert_{F}^{2}\bigr{]}. Since

[TABLE]

for all $x\in\mathbf{R}$, we have $L(\operatorname{\mathbf{E}}F_{X},\operatorname{\mathbf{E}}F_{Y})\leq\epsilon$, and thus the desired inequality holds. ∎

Appendix C Concentration of measure

Lemma C.1** (Hoeffding’s lemma).**

Let $-\infty<a<b<\infty$. If $X$ is a $[a,b]$-valued random variable, then

[TABLE]

Proof.

We may assume $\operatorname{\mathbf{E}}[X]=0$. Note that $a\leq 0\leq b$. For any $x\in[a,b]$,

[TABLE]

and so

[TABLE]

Since $\operatorname{\mathbf{E}}[X]=0$ and $\operatorname{\mathbf{E}}[X^{2}]\leq(b-a)^{2}$, we have

[TABLE]

If $b-a<1$, then

[TABLE]

If $b-a\geq 1$, then

[TABLE]

In any case, we have \operatorname{\mathbf{E}}[e^{X}]\leq\exp\bigl{(}2(b-a)^{2}\bigr{)}. ∎

Theorem C.2** (McDiarmid’s inequality).**

Let $S_{1},\ldots,S_{n}$ be measurable spaces, and $F\colon S_{1}\times\cdots\times S_{n}\to\mathbf{R}$ be a bounded measurable function. Assume that

[TABLE]

for any $i\in\{1,\ldots,n\}$, $x_{j}\in S_{j}$ for each $j\in\{1,\ldots,n\}$, and $x^{\prime}_{i}\in S_{i}$, where $c_{i}>0$ doesn’t depend on $x_{1},\ldots,x_{n},x^{\prime}_{i}$. If $X_{1},\ldots,X_{n}$ are independent random elements of $S_{1},\ldots,S_{n}$, then

[TABLE]

for any $\lambda>0$ where $\sigma=\sqrt{c_{1}^{2}+\cdots+c_{n}^{2}}$.

Proof.

Let us first show

[TABLE]

for any $t>0$ by induction on $n$. If $n=0$, in which case $S_{1}\times\cdots\times S_{n}$ is a singleton, $F$ is essentially a constant, and $\sigma=0$, there is nothing to prove. We now proceed by induction on $n$.

Note that we may assume that each $X_{i}$ is the projection $\pi_{i}\colon S_{1}\times\cdots\times S_{n}\to S_{i}$. Let $\mu$ and $\mu_{n}$ be the distributions of $(X_{1},\ldots,X_{n-1})$ and $X_{n}$. Let $G:S_{1}\times\cdots\times S_{n-1}\to\mathbf{R}$ be defined by

[TABLE]

For any $i\in\{1,\ldots,n-1\}$, $x_{j}\in S_{j}$ for each $j\in\{1,\ldots,n-1\}$, and $x^{\prime}_{i}\in S_{i}$, we have

[TABLE]

Since

[TABLE]

the induction hypothesis implies

[TABLE]

Define $H_{t}:S_{1}\times\cdots\times S_{n-1}\to\mathbf{R}$ by

[TABLE]

Whenever $x_{i}\in S_{i}$ is fixed for each $i\in\{1,\ldots,n-1\}$, we have

[TABLE]

by Hoeffding’s lemma (Lemma C.1). Thus, (C.3) and (C.2) yield

[TABLE]

finishing the proof of (C.1).

We can now finish the proof. Observe that

[TABLE]

By some calculus one can find that $t=\lambda/4\sigma$ minimizes the right side, yielding

[TABLE]

Applying this result to $-F$, we obtain

[TABLE]

∎

The following inequality was found independently by Guntuboyina and Leeb [GL09], and Bordenave, Caputo, and Chafaï [BCC11].

Theorem C.3** (Concentration for spectral measures).**

Let $X$ be a random $n\times n$ Hermitian matrix whose rows of the lower triangle are jointly independent. If $f:\mathbf{R}\to\mathbf{R}$ satisfies $\lVert f\rVert_{TV}\leq 1$, and $t>0$, then

[TABLE]

Proof.

Let $S_{i}:=\mathbf{C}^{i}$ and $X_{i}:=(X_{i1},\ldots,X_{ii})$ for each $i=1,\ldots,n$. Given $(x_{1},\ldots,x_{n})\in S_{1}\times\cdots\times S_{n}$, let $H(x_{1},\ldots,x_{n})$ be the Hermitian matrix whose $i$th row of the lower triangle is $x_{i}$ for each $i=1,\ldots,n$. Let $(x_{1},\ldots,x_{n})\in S_{1}\times\cdots\times S_{n}$ and $x_{i}^{\prime}\in S_{i}$. If we change a row or a column of a matrix, then the change in rank is at most $1$. Since $H(x_{1},\ldots,x_{i-1},x_{i}^{\prime},x_{i+1},\ldots,x_{n})$ can be obtained from $H(x_{1},\ldots,x_{n})$ by changing a row and then changing a column, the rank of

[TABLE]

is at most $2$. Thus, Corollary B.12 tells us that

[TABLE]

Let $X_{i}$ be the $i$th row of the lower triangle of $X$. Then, $X_{1},\ldots,X_{n}$ are independent, and $X=H(X_{1},\ldots,X_{n})$. By applying Theorem C.2 to $F:S_{1}\times\cdots\times S_{n}\to\mathbf{R}$ given by $F(x_{1},\ldots,x_{n}):=\int_{\mathbf{R}}f\,\mu_{H(x_{1},\ldots,x_{n})}$ and $X_{1},\ldots,X_{n}$, we obtain

[TABLE]

for any $\lambda>0$. Our desired result follows by letting $\lambda=t\sqrt{n}/2$. ∎

Appendix D Reduction to unit variance case

The Stieltjes transform method, which is the topic of the next section, is able to prove a semicircular law (Theorem E.1) which assumes that every entry of $W_{n}$ has variance excatly $1/n$. However, it seems not so easy to reduce Theorem 1.2 itself to the case the Stieltjes transform can handle. This section provides an alternative semicircular law, which is somewhat weaker than 1.2, that can still be reduced to what the Stieltjes transform can handle. If you’re satisfied by the reduced version (Theorem E.1), feel free to skip to Section E. Otherwise, the following is the alternative semicirular law. It was pointed out in the Remark following Theorem 1.2 as a special case of Theorem 1.2.

Theorem D.1** (A semicircular law).**

For each $n\in\mathbf{N}$, let $W_{n}=(w_{ij}^{(n)})_{i,j=1}^{n}$ be a random $n\times n$ Hermitian matrix whose upper triangular entries are jointly independent, have mean zero, and have finite variances. We assume that $W_{1},W_{2},\ldots$ are defined on the same probability space. If

[TABLE]

and

[TABLE]

then $\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$ as $n\to\infty$ a.s.

D.1 Extension of the underlying probability space

Let $(\Omega,\mathcal{B},\operatorname{\mathbf{P}})$ be the probability space on which $W_{1},W_{2},\ldots$ are defined. If $(\Omega^{\prime},\mathcal{B}^{\prime},\operatorname{\mathbf{P}}^{\prime})$ is another probability space, and $T\colon\Omega^{\prime}\to\Omega$ is a measurable map such that $\operatorname{\mathbf{P}}(A)=\operatorname{\mathbf{P}}^{\prime}(T^{-1}(A))$ for all $A\in\mathcal{B}$, then the random matrices $W_{n}\circ T$ satisfy all conditions of Theorem D.1. Assume that we proved $\mu_{W_{n}\circ T}\Rightarrow\mu_{\mathrm{sc}}$ $\operatorname{\mathbf{P}}^{\prime}$-a.s. Since

[TABLE]

we will have $\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$ $\operatorname{\mathbf{P}}$-a.s. if we can show that

[TABLE]

For any $p,q\in\mathbf{Q}$ with $p<q$, let $f_{p,q}$ be defined as in Theorem A.3. Since $\int_{\mathbf{R}}f_{p,q}\,d\mu_{X}$ is a real-valued random variable for any random Hermitian matrix $X$, the event

[TABLE]

is measurable. So, (D.2) follows from Theorem A.3, and thus $\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$ $\operatorname{\mathbf{P}}$-a.s. follows. This shows that we can think that $W_{n}\circ T$’s and $\Omega^{\prime}$ are the given random matrices and the underlying space. By considering $\Omega^{\prime}=\Omega\times\{0,1\}^{\mathbf{N}}$, we may assume that we have i.i.d. random variables $\xi_{ij}^{(n)}$’s, where $n\in\mathbf{N}$ and $1\leq i,j\leq n$, independent from $W_{1},W_{2},\ldots$, which satisfy

[TABLE]

D.2 Repeating what we already know

The first three steps of Section 2 (that is, until centralization) works for our case with a slight change. Applying those steps, we can now assume the following, and need to prove $\operatorname{\mathbf{E}}\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$.

1. (i)

The upper triangular entries of $W_{n}$ are jointly independent and have mean zero. 2. (ii)

We have (D.1). 3. (iii)

There are $\eta_{1},\eta_{2},\ldots>0$ such that $|w_{ij}^{(n)}|\leq\eta_{n}$ and $\eta_{n}\to 0$ as $n\to\infty$.

D.3 Replacing and rescaling

Let

[TABLE]

and define $W^{\prime}=(v_{ij}^{(n)})_{i,j=1}^{n}$ by

[TABLE]

Note that

[TABLE]

as $n\to\infty$. Since $(a-b)^{2}\leq|a-b|(a+b)=|a^{2}-b^{2}|$ for any $a,b\geq 0$, we also have

[TABLE]

Combining previous two displays, we obtain

[TABLE]

and so it is enough to show $\operatorname{\mathbf{E}}\mu_{W^{\prime}_{n}}\Rightarrow\mu_{\mathrm{sc}}$ by Corollary B.15 and Theorem A.3.

If $i,j\notin E^{(n)}$, then \operatorname{\mathbf{Var}}\bigl{[}v_{ij}^{(n)}\bigr{]}\geq\tfrac{1}{2n}, and so

[TABLE]

As $|v_{ij}^{(n)}|\leq 1/\sqrt{n}$ for any $(i,j)\in E^{(n)}$, by letting $\eta^{\prime}_{n}:=\eta_{n}\vee(1/\sqrt{n})$ we have $\eta^{\prime}_{n}\to 0$ and $|v_{ij}^{(n)}|\leq\eta^{\prime}_{n}$. Since the upper triangular entries of $W^{\prime}_{n}$ are jointly independent random variables with mean zero and variance $1/n$, we can now assume that the upper triangular entries of $W_{n}$ have variance $1/n$, and there are $\eta_{1},\eta_{2},\ldots>0$ such that $\eta_{n}\to 0$ and $|w_{ij}^{(n)}|\leq\eta_{n}$.

Appendix E The Stieltjes transform method

Theorem E.1** (A unit-variance semicircular law).**

For each $n\in\mathbf{N}$, let $W_{n}=(w_{ij}^{(n)})_{i,j=1}^{n}$ be a random $n\times n$ Hermitian matrix whose upper triangular entries are jointly independent random variables with mean zero and variance $1/n$. We assume that $W_{1},W_{2},\ldots$ are defined on the same probability space. If there are $\eta_{1},\eta_{2},\ldots>0$ with $|w_{ij}^{(n)}|\leq\eta_{n}$ and $\eta_{n}\to 0$ as $n\to\infty$, then $\operatorname{\mathbf{E}}\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$ as $n\to\infty$ a.s.

For the readers who skipped Section D: note that the first step in Section 2 lets us upgrade the conclusion of Theorem E.1 to $\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$ as $n\to\infty$ a.s.

E.1 Stieltjes transform

Let $\mathbf{C}_{+}:=\{z\in\mathbf{C}\mid\Im z>0\}$. Weak convergence of probability measures on $\mathbf{R}$ can be coded in terms of Stieltjes transforms.

Definition E.2** (Stieltjes transform).**

Let $\mu$ be a positive, finite Borel measure on $\mathbf{R}$. The Stieltjes transform $s_{\mu}:\mathbf{C}_{+}\to\mathbf{C}$ of $\mu$ is given by

[TABLE]

Note that $|1/(x-z)|\leq\Im z$ for all $x\in\mathbf{R}$. So $x\mapsto 1/(x-z)$ is a continuous and bounded function, and it also follows that $|s_{\mu}(z)|\leq\Im z$.

Theorem E.3** (Stieltjes inversion formula).**

If $\mu$ is a positive, finite Borel measure on $\mathbf{R}$, then the following hold:

1. (i)

for any $b>0$, $a\in\mathbf{R}\mapsto\frac{1}{\pi}\Im s_{\mu}(a+ib)$ is a nonnegative function with $\int_{\mathbf{R}}\frac{1}{\pi}\Im s_{\mu}(a+ib)\,da=\mu(\mathbf{R})$; 2. (ii)

$\frac{1}{\pi}\Im s_{\mu}(a+ib)\,da\Rightarrow\mu$* as $b\downarrow 0$.* 3. (iii)

$b\Im s_{\mu}(ib)\to\mu(\mathbf{R})$* as $b\to\infty$;*

Proof.

If $\mu=0$, then there is nothing to prove. By renormalization, we may, and will, assume $\mu(\mathbf{R})=1$. Note that

[TABLE]

Let $X$ be a real-valued random variable with distribution $\mu$, and $C$ be a standard Cauchy random variable (i.e. the law of $C$ has density $\frac{1}{\pi(x^{2}+1)}$) independent of $X$. Then $X+bC\to X$ as $b\downarrow 0$ a.s., and thus in distribution. Since the right side of (E.1) is the density of the law of $X+bC$, both (i) and (ii) are proved. As

[TABLE]

by dominated convergence, (iii) is also proved. ∎

Theorem E.4** (Stieltjes continuity theorem).**

If $\mu,\mu_{1},\mu_{2},\ldots$ are Borel probability measures on $\mathbf{R}$, then $\mu_{n}\Rightarrow\mu$ if and only if $s_{\mu_{n}}(z)\to s_{\mu}(z)$ for all $z\in\mathbf{C}_{+}$.

Proof.

The “only if” direction follows immediately from the definition of weak convergence. To show the “if” direction, assume that $s_{\mu_{n}}(z)\to s_{\mu}(z)$ for all $z\in\mathbf{C}_{+}$. Whenever $n_{1}<n_{2}<\cdots$ and $\mu_{n_{k}}\to\nu$ vaguely for some finite $\nu$, we have $s_{\mu_{n_{k}}}(z)\to s_{\nu}(z)$ for all $z\in\mathbf{C}_{+}$ since $x\mapsto 1/(x-z)$ vanishes at infinity. This implies $s_{\nu}(z)=s_{\mu}(z)$ for all $z\in\mathbf{C}_{+}$, and thus $\nu=\mu$ by Theorem E.3. As any subsequence of $(\mu_{n})_{n\in\mathbf{N}}$ has a vaguely convergent further subsequence, it follows that any subsequence of $(\mu_{n})_{n\in\mathbf{N}}$ has a further subsequence converging vaguely to $\mu$. This shows $\mu_{n}\to\mu$ vaguely, and so $\mu_{n}\Rightarrow\mu$. ∎

E.2 Predecessor comparison

In the remainder of this section, we will show that $s_{\operatorname{\mathbf{E}}\mu_{W_{n}}}(z)\to s_{\mu_{\mathrm{sc}}}(z)$ for all $z\in\mathbf{C}_{+}$. To do so, we will first express $s_{\operatorname{\mathbf{E}}\mu_{W_{n}}}(z)$ in terms of the resolvent $(W_{n}-zI)^{-1}$. By the spectral theorem, for any Hermitian matrix $A$ the matrix $A-zI$ is invertible for any $z\in\mathbf{C}_{+}$. Let $S_{A}(z):=(A-zI)^{-1}$ for any Hermitian $A$ and $z\in\mathbf{C}_{+}$. Using the spectral theorem, we can also see that

[TABLE]

Thus, we have

[TABLE]

and so it is enough to show that

[TABLE]

for all $z\in\mathbf{C}_{+}$.

To understand the limiting behavior of $\frac{1}{n}\operatorname{\mathbf{E}}\operatorname{tr}S_{W_{n}}(z)$, we relate it with the $(n-1)\times(n-1)$ minors of $W_{n}$ using the Schur complement formula, which will be presented below. For each $i\in\{1,\ldots,n\}$, let $W^{(i)}$ be the $(n-1)\times(n-1)$ matrix obtained by removing the $i$-th row and column from $W_{n}$. Also, let $w_{i}$ denote the $i$-th column of $W_{n}$ with $w_{ii}$ removed. (So, $w_{i}$ is an $(n-1)$-dimensional column vector.) Let us denote the $(i,j)$-entry of a matrix $A$ by $A(i,j)$. Recall that if $A$ is an invertible matrix, then

[TABLE]

where $C_{ji}(A)$ is the $(i,j)$-cofactor of $A$. So we have

[TABLE]

where $I_{n-1}$ is the $(n-1)\times(n-1)$ identity matrix.

Proposition E.5** (Schur complement formula).**

Consider a matrix

[TABLE]

with complex entries, where $A$ and $D$ are square matrices and $A$ is invertible. Then we have

[TABLE]

Proof.

Note that

[TABLE]

Since

[TABLE]

and

[TABLE]

we have

[TABLE]

by the multiplicativity of determinant. ∎

By Proposition E.5, we have

[TABLE]

and so

[TABLE]

Summing over $i=1,\ldots,n$ and taking the expectation, we obtain

[TABLE]

(The fact that the expectation on the right side is well-defined follows from Lemma E.6 below.) We will show that the right side of (E.2) gets close to

[TABLE]

as $n$ grows, and obtain a recursive relation involving the limit of $\frac{1}{n}\operatorname{\mathbf{E}}\operatorname{tr}S_{W_{n}}(z)$.

E.3 Derivation of a recurrence relation

The following fact will be used repeatedly. In particular, it will guarantee that many denominators we face in the computation below are nonzero.

Lemma E.6**.**

If $A$ is an $n\times n$ Hermitian matrix and $z\in\mathbf{C}_{+}$, then the following hold:

1. (i)

$\operatorname{tr}S_{A}(z)\in\mathbf{C}_{+}$; 2. (ii)

\operatorname{tr}\bigl{(}(A-zI)(A-\bar{z}I)\bigr{)}^{-1}\leq n/(\Im z)^{2}; 3. (iii)

$u^{\ast}S_{A}(z)u\in\mathbf{C}_{+}$* for any $u\in\mathbf{C}^{n}$.*

Proof.

Let $A=UDU^{\ast}$ where $U$ is unitary and $D$ is real diagonal with diagonal entries $d_{1},\ldots,d_{n}$. (i) Since trace is similarity invariant, we have

[TABLE]

(ii) Also,

[TABLE]

(iii) Finally, if we let $U^{\ast}u=(u_{1},\ldots,u_{n})$, then

[TABLE]

∎

In the rest of this subsection, we fix $z\in\mathbf{C}_{+}$, and transform

[TABLE]

step-by-step to obtain

[TABLE]

(asymptotically) in the end.

E.3.1 From $w_{i}^{\ast}S_{W^{(i)}}(z)w_{i}$ to $\frac{1}{n}\operatorname{tr}S_{W^{(i)}}(z)$

Instead of numbering the rows and columns of $W^{(i)}$ using $1,\ldots,n-1$, let us use $1,\ldots,\hat{i},\ldots,n$ as if $W^{(i)}$ still lies in $W_{n}$. For $j,k\neq i$, let $b_{jk}$ denote the $(j,k)$-entry of $S_{W^{(i)}}(z)$. Since

[TABLE]

we have

[TABLE]

by Lemma E.6 (ii). The fact that $b_{jk}$’s are in $L^{2}$ will guarantee that all terms in the computation below are well-defined and finite. Using the fact that $W^{(i)}$ and $w_{i}$ are independent, and each entry of $W_{n}$ is of mean zero and variance $1/n$, we have

[TABLE]

Note that the last term in the last line must be real. Since

[TABLE]

by the Cauchy-Schwarz inequality, we have

[TABLE]

by (E.3). It follows that

[TABLE]

(We are fixing $z\in\mathbf{C}_{+}$.) Therefore, by E.6 (i) and (iii), we have

[TABLE]

as $n\to\infty$.

E.3.2 From $\frac{1}{n}\operatorname{tr}S_{W^{(i)}}(z)$ to $\frac{1}{n}\operatorname{\mathbf{E}}\operatorname{tr}S_{W^{(i)}}(z)$

Since the maps $x\in\mathbf{R}\mapsto\Re(1/(x-z))$ and $x\in\mathbf{R}\mapsto\Im(1/(x-z))$ have bounded variations, Theorem C.3 implies that there are $c,C>0$ (depending on $z$, but we are fixing $z\in\mathbf{C}_{+}$) such that

[TABLE]

for any $n\in\mathbf{N}$, $t>0$, and a random $n\times n$ Hermitian matrix $X$. So, we have

[TABLE]

for any $n\in\mathbf{N}$ and $i=1,\ldots,n$. It follows that

[TABLE]

by dominated convergence. Therefore,

[TABLE]

as $n\to\infty$.

E.3.3 From $\frac{1}{n}\operatorname{\mathbf{E}}\operatorname{tr}S_{W^{(i)}}(z)$ to

$\frac{1}{n}\operatorname{\mathbf{E}}\operatorname{tr}S_{W_{n}}(z)$

Let $\overline{W}^{(i)}$ be the $n\times n$ matrix obtained by replacing all the entries in the $i$-th row and column of $W$ by [math]. Since $\overline{W}^{(i)}$ has the same (multi)set of eigenvalues as $W^{(i)}$ except that it has one more zero eigenvalue, we have

[TABLE]

By the Hoffman-Wielandt inequality (Theorem B.9), we have

[TABLE]

Combining the results of the previous two displays, we obtain

[TABLE]

for all $n\in\mathbf{N}$ and $i=1,\ldots,n$, and so

[TABLE]

as $n\to\infty$. Therefore,

[TABLE]

as $n\to\infty$.

E.3.4 The result

Combining (E.2) and the final results of the previous three subsubsections, we obtain

[TABLE]

for any $z\in\mathbf{C}_{+}$.

E.4 Convergence of the Stieltjes transform

Let $s_{n}(z):=\frac{1}{n}\operatorname{\mathbf{E}}\operatorname{tr}S_{W_{n}}(z)$. Fix $z\in\mathbf{C}_{+}$ for now, and let us write $s_{n}=s_{n}(z)$. If there are $n_{1}<n_{2}<\cdots$ such that $|s_{n_{k}}|\to\infty$ as $k\to\infty$, then we would have

[TABLE]

which contradicts (E.4). Thus $\{s_{n}\mid n\in\mathbf{N}\}$ is bounded, and therefore any subsequence of $(s_{n})_{n\in\mathbf{N}}$ has a convergent subsequence. If we show that any convergent subsequence of $(s_{n})_{n\in\mathbf{N}}$ should converge to a number independent of the subsequence we choose, then we will have the convergence of $(s_{n})_{n\in\mathbf{N}}$ to that number.

Assume $s_{n_{k}}\to s\in\mathbf{C}_{+}\cup\mathbf{R}$ as $k\to\infty$. Since the left side of (E.4) converges to \bigl{|}s+1/(z+s)\bigr{|} along $n_{1}<n_{2}<\cdots$, we have

[TABLE]

Solving the quadratic equation, we obtain

[TABLE]

We need to decide which branch of $\sqrt{z^{2}-4}$ we use. For simplicity, we will define $\sqrt{z^{2}-4}$ only for $z\in\mathbf{C}_{+}\cup\mathbf{R}$, and it will suffice. Choose the branch of $\sqrt{z-2}$ and $\sqrt{z+2}$ defined on $\mathbf{C}_{+}\cup\mathbf{R}$ which are continuous and have nonnegative imaginary part for all $z\in\mathbf{C}_{+}\cup\mathbf{R}$. Then let $\sqrt{z^{2}-4}:=\sqrt{z-2}\sqrt{z+2}$. This will make $\sqrt{z^{2}-4}$ continuous and have nonnegative imaginary part on $\mathbf{C}_{+}\cup\mathbf{R}$.

Since $s_{n_{k}}=\frac{1}{n_{k}}\operatorname{\mathbf{E}}\operatorname{tr}S_{W_{n_{k}}}(z)\in\mathbf{C}_{+}$ for all $k\in\mathbf{N}$, we have $\Im s\geq 0$. On the other hand, $(-z-\sqrt{z^{2}-4})/2$ has a negative imaginary part. Thus we have

[TABLE]

Since this is true for any subsequence $(s_{n_{k}})_{k\in\mathbf{N}}$ of $(s_{n})_{n\in\mathbf{N}}$ converging to $s$, we have

[TABLE]

E.5 Computation of the limiting distribution

Lemma E.7**.**

If $\mu$ is a positive, finite Borel measure on $\mathbf{R}$ satisfying $s_{\mu}(z)=(-z+\sqrt{z^{2}-4})/2$ for all $z\in\mathbf{C}_{+}$, then $\mu(dx)=\sqrt{(4-x^{2})_{+}}\,dx$.

Proof.

As

[TABLE]

we have $\mu(\mathbf{R})=1$ by Theorem E.3 (iii). Since $z\mapsto(-z+\sqrt{z^{2}-4})/2$ is continuous on $\mathbf{C}_{+}\cup\mathbf{R}$, we have

[TABLE]

for each fixed $a\in\mathbf{R}$. Since $\frac{1}{\pi}\Im s_{\mu}(a+ib)\,da$ is a probability density (by Theorem E.3 (i)) converging pointwise to the probability density $\frac{1}{2\pi}\sqrt{(4-a^{2})_{+}}$ as $b\downarrow 0$, we have $\frac{1}{\pi}\Im s_{\mu}(a+ib)\,da\Rightarrow\frac{1}{2\pi}\sqrt{(4-a^{2})_{+}}\,da$ as $b\downarrow 0$ by Scheffé’s theorem ([Bil12, Theorem 16.12]). Now $\mu(dx)=\frac{1}{2\pi}\sqrt{(4-x^{2})_{+}}\,dx$ follows from Theorem E.3 (ii). ∎

The proof of Lemma E.7 shows how one can figure out what the limiting spectral distribution should be in the first place. Now we finish our proof of the semicircular law. Since $\operatorname{\mathbf{E}}\mu_{W_{1}},\operatorname{\mathbf{E}}\mu_{W_{2}},\ldots$ are probability measures, there are integers $n_{1}<n_{2}<\cdots$ such that $\operatorname{\mathbf{E}}\mu_{W_{n_{k}}}\to\mu$ vaguely as $k\to\infty$ for some positive, finite measure $\mu$. Since $s_{\operatorname{\mathbf{E}}\mu_{W_{n_{k}}}}(z)\to s_{\mu}(z)$ as $k\to\infty$ for all $z\in\mathbf{C}_{+}$, (E.5) implies $s_{\mu}(z)=(-z+\sqrt{z^{2}-4})/2$ for all $z\in\mathbf{C}_{+}$. So, we have $\mu=\mu_{\mathrm{sc}}$ by Lemma E.7. Now $\operatorname{\mathbf{E}}\mu_{W_{n}}\Rightarrow\mu_{\mathrm{sc}}$ follows from (E.5) and Theorem E.4. Interestingly, we were able to avoid an actual computation of $s_{\mu_{\mathrm{sc}}}$, in which we might have used something like the residue theorem or the Cauchy integral formula.