Download THE CIRCULAR LAW PROOF OF THE REPLACEMENT PRINCIPLE by ZHIWEI TANG

THE CIRCULAR LAW
PROOF OF THE REPLACEMENT PRINCIPLE
by
ZHIWEI TANG
A thesis submitted to the Department of Mathematics and Statistics
in conformity with the requirements for
the degree of Master of Science
Queen’s University
Kingston, Ontario, Canada
July, 2009
c Zhiwei Tang, 2009
Copyright °
Abstract
It was conjectured in the early 1950s that the empirical spectral distribution
(ESD) of an n × n matrix whose entries are independent and identically
distributed with mean zero and variance one, normalized by a factor of
√1 ,
n
converges to the uniform distribution over the unit disk on the complex
plane, which is called the circular law. The goal of this thesis is to prove
the so called Replacement Principle introduced by Tao and Vu which is a
crucial step in their recent proof of the circular law in full generality. It gives
a general criterion for the difference of the ESDs of two normalised random
matrices
√1 An , √1 Bn
n
n
to converge to 0.
i
Acknowledgements
I would like to sincerely thank my supervisors Professor James Mingo and
Professor Roland Speicher for their enormous help and valuable suggestions
in the construction of this thesis.
ii
Contents
Abstract
i
Acknowledgements
ii
1 Introduction
1
2 Notations and Definitions
5
3 The replacement principle
9
4 Proof of the replacement principle
13
5 Conclusion
34
Bibliography
35
iii
Chapter 1
Introduction
In 1955, Wigner first observed that the empirical spectral distribution of an
n × n Hermitian matrix with upper diagonal entries being independent and
identically distributed Gaussian random variables converges to the semicircle law (see Figure 1). Wigner’s result was later extended to many other
forms according to the distributions of the entries, the most general one only
requires mean zero and variance one.
Theorem 1.1 (Wigner’s semicircle law)
Let An be the n × n Her-
mitian random matrix whose upper diagonal entries are independent and
identically distributed complex random variables with mean zero and variance one. Then the empirical spectral distribution of
semi-circle distribution.
1
√1 An
n
converges to the
0.3
0.2
0.1
-1
0
1
2
Figure 1.1: The distribution of eigenvalues of a randomly generated 2000 by
2000 Hermitian matrix with entries chosen from a standard normal distribution is illustrated above.
Circular Law Conjecture: Suppose we have an n × n matrix An , where
each entry aij of An is an independently identically distributed copy of
a single random variable with mean 0 and variance 1. Numerical evidence
reveals that the eigenvalues of such a matrix appear to distribute themselves
uniformly on the unit disk in the limit n → ∞. (see Figure 2)
Conjecture 1.2 (Circular Law) Let An be the n×n random matrix whose
entries are independently identically distributed complex random variables
with mean zero and variance one. Then the empirical spectral distribution
of
√1 An
n
converges in probability to the uniform distribution on the unit disk.
There are significant difficulties in giving a rigorous proof to this conjecture since the main techniques used to handle Hermitian matrices cannot
be applied to the non-Hermitian model. However the conjecture has been
worked on intensively for many decades. The circular law was verified for
the complex Gaussian distribution in [8] and the real Gaussian distribution
2
in [4] by using Ginibre’s formula for the joint density function of the eigenvalues of An . In 1984, Girko [5] suggested a strategy to prove the general
case. In 1997 Bai succeeded in making this rigorous for continuous distributions with bounded sixth moment in [1] and this hypothesis was lowered
to (2 + η)-th moment for any η > 0 in [2]. The removal of the hypothesis of continuous distribution required some new ideas. Important partial
results were obtained by Götze-Tikhomirov ([6], [7]), Pan-Zhou [9] and TaoVu [11]. All these results are established under additional assumptions on
the distribution of the entries. Tao-Vu finally proved the conjecture in full
generality in [12]. The most important step in their proof is the so called
replacement principle. Namely it shows that the difference of the empirical
spectral distribution of two normalised random matrices
√1 An , √1 Bn
n
n
verges to 0 if they satisfy certain conditions. It is proven in [12] that if
is a Gaussian random matrix and
√1 Bn
n
con-
√1 An
n
satisfies the assumptions of conjec-
ture 1.2, then these conditions always hold, therefore we know the empirical
spectral distribution for
for the
√1 An
n
√1 Bn
n
converges to the same limiting measure as
which is the uniform distribution on the unit disk. The proof
of the replacement principle in [12] is quite sketchy, with many details left
to the reader. In this thesis we will give a complete proof of the replacement
principle with all the necessary steps and calculations.
3
-1.0
1.0
1.0
0.5
0.5
0.5
-0.5
1.0
-1.0
1.0
0.5
0.5
-0.5
1.0 -1.0
0.5
-0.5
-0.5
-0.5
-0.5
-1.0
-1.0
-1.0
Figure 1.2: Eigenvalue plots of three randomly generated 1000 by 1000
matrices. Each entry of the left one is an iid Gaussian normal random
2
variable, with probability density function √12π e−x /2 . Each entry of the
middle one is an iid uniform random variable. Each entry of the right one
1
is an iid Cauchy random variable, with probability density function π(1+x
2) .
The third example shows that one can not expect the circular law to hold
in case where the second moment does not exist.
4
1.0
Chapter 2
Notations and Definitions
Empirical spectral distributions
This thesis is focused on the convergence
in probability of empirical spectral distributions of random matrices. Let us
first fix our definitions of convergence in probability and weak boundedness
[3].
Definition 2.1 (Convergence in probability) The sequence {Xn } of random
variables is said to converge in probability to X if for every ² > 0 we have
lim P (|Xn − X| > ²) = 0.
n→∞
Definition 2.2 (Weak boundedness). Let {Xn } be a sequence of random
variables on probability space (Ω, P ), we say {Xn } is weakly bounded if we
5
have
lim lim inf P (|Xn | ≤ C) = 1
C→∞ n→∞
Remark 2.3. In order to make Definition 2.2 more convenient to use, one
can rewrite Definition 2.2 as the following equivalent statement:
Proposition 2.4 (Weak boundedness). Let {Xn } be a sequence of random
variables on probability space (Ω, P ). {Xn } is weakly bounded if and only
if given ² > 0, ∃n0 , C0 , such that P (|Xn | ≤ C0 ) > 1 − ² for n ≥ n0 .
Proof: (=⇒) Let ² > 0 be given, ∃C0 , such that for C > C0 ,
lim inf P (|Xn | ≤ C) > 1 − ²
n→∞
thus ∃n0 , such that
P (|Xn | ≤ C) > 1 − ² (for n ≥ n0 )
by the definition of lim inf.
(⇐=) Let ² > 0 be given. By assumption ∃n0 , C0 , such that for n ≥ n0 , we
have P (|Xn | ≤ C0 ) > 1 − ². Therefore
lim inf P (|Xn | ≤ C0 ) ≥ 1 − ²
n→∞
Clearly P (|Xn | ≤ C) ≥ P (|Xn | ≤ C0 ) for all C ≥ C0 , hence lim inf n→∞ P (|Xn | ≤
C) ≥ 1 − ², i.e.
lim lim inf P (|Xn | ≤ C) = 1
C→∞ n→∞
6
Notation 2.5 Let Mn (C) denote the set of n × n complex matrices. For
A ∈ Mn (C) and s, t ∈ R, we let
µA (s, t) :=
1
|{1 ≤ i ≤ n, Reλi ≤ s, Imλi ≤ t}|
n
be the empirical spectral distribution of its eigenvalues λi ∈ C, i = 1, · · · , n,
counted by algebraic multiplicity (we use |B| to denote the cardinality of a
finite set B). Thus for any complex valued function f
Z
n
1X
f dµA =
f (λi ).
n
i=1
Clearly, µA is a discrete probability measure on C.
Notation 2.6 (Convergence in probability for µA (s, t)) Let us give the space
of probability measures on C the usual vague topology. Hence a sequence of
R
R
measures µn converges to µ vaguely if C f dµn converges to C f dµ for every
continuous and compactly supported function f : C → R.
Remark 2.7 According to Definition 2.1, we see that the random probability
measures µ √1
n
An
converge to µ∞ in probability if for every continuous and
compactly supported function f : C → R, the sequence of random variables
Z
Z
C
f (z)dµ √1
n
An
7
−
C
f (z)dµ∞
converges to zero in probability, i. e.
Z
lim P (|
n→∞
C
Z
f (z)dµ √1
n
An
−
C
f (z)dµ∞ | > ²) = 0
for every ² > 0.
We will use the following lemma, whose proof is given in the appendix of
[12].
Lemma 2.8 Let A = (aij )1≤i,j≤n ∈ Mn (C) have eigenvalues λ1 , · · · , λn ∈ C
repeated according to multiplicity. Then
n
X
|λj |2 ≤ ||A||22 =
j=1
n X
n
X
i=1 j=1
8
|aij |2 .
Chapter 3
The replacement principle
Now we are going to give the formal statement of the replacement principle.
Theorem 3.1 (Replacement principle). Suppose for each n that An , Bn
are sequences of n × n random matrices. Assume that
(i) the sequence of random variables
n1
o
1
2
2
||A
||
+
||B
||
n 2
n 2
n2
n2
is weakly bounded; and
(ii) for almost all complex numbers z, the sequence of random variables
n1
o
1
1
1
log | det( √ An − zI)| −
log | det( √ Bn − zI)|
n
n
n
n
converges to zero in probability.
Then the sequence of random measures {µ √1
n
9
An
− µ √1
n
Bn }
converges to zero
in probability, i.e. given ² > 0, we have for every continuous function f with
compact support
Z
lim P (|
n→∞
C
Z
f (z)dµ √1
n
An (z)
−
C
f (z)dµ √1
n
Bn (z)|
> ²) = 0.
Note that we do not need to require independence among the entries of An
and Bn . The proof of this theorem relies on the Stieltjes transform technique
that analyses the random probability measure µ √1
n
determinants
1
n
An
in terms of the log-
log | det( √1n An −zI)|, combined with tools from real analysis
such as the dominated convergence theorem. The details are given in the
next chapter.
From the weak boundedness of Theorem 3.1 (i) we see that for every ² > 0
there exists C² > 0, such that for n large enough,
P(
1
||An ||22 ≤ C² ) > 1 − ².
n2
Thus by Lemma 2.8, with probability greater than 1 − ², we have for the
eigenvalues λ1 , · · · , λn of An the bound
n
X
1
|λj |2 ≤ C²
n2
(3.1)
j=1
or, equivalently, that
Z
C
|z|2 dµ √1
n
An (z)
10
≤ C² .
Similarly, we have
Z
C
|z|2 dµ √1
n
Bn (z)
≤ C²
with probability greater than 1 − ². For any R > 0, we have
Z
2
R µ √1 An {z ∈ C : |z| ≥ R} ≤
n
Z
≤
|z|2 dµ √1 An (z) ≤ C² .
|z|≥R
|z|2 dµ √1
n
An (z)
n
C
Thus we have the bounds
µ √1
An {z
∈ C : |z| ≥ R} ≤ C² /R2
µ √1
Bn {z
∈ C : |z| ≥ R} ≤ C² /R2
n
and
n
for all R > 0 with probability greater than 1 − ².
Let Ω be the underlying probability space. We have that, as an immediate
consequence, for fixed ² ≥ 0 one can find an R ≥ 1 large enough such that
P (Ω1 ) > 1 − ², where Ω1 ⊆ Ω,
Ω1 = {ω|µ √1
n
An (ω) (z
∈ C : |z| ≥ R)+µ √1
n
Bn (ω) (z
∈ C : |z| ≥ R) ≤ ²}. (3.2)
In the following chapter we will establish the proof of the replacement principle. We will pass from the empirical spectral distribution to the characteristic function and show the equivalence between them in the sense of
11
convergence in probability. Then we can significantly simplify our problem
by using Girko’s identity which gives the relationship between the characteristic function and the Stieltjes-like transform.
12
Chapter 4
Proof of the replacement
principle
We begin with a probabilistic version of the dominated convergence theorem.
Lemma 4.1 (Dominated convergence). Let (X, ν) be a finite measure space.
For each integer n ≥ 1, let fn (x, ω) : X × Ω → R be a sequence of random
functions which are jointly measurable with respect to X and Ω. Assume
that
(i) (Uniform integrability) There exists δ > 0 such that the sequence of
R
random variables { X |fn (x)|1+δ dν(x)} is weakly bounded.
(ii) (Pointwise convergence in probability) For ν-almost every x ∈ X, the
sequence of random variables fn (x) converges to zero in probability.
R
Then the sequence of random variables { X fn (x)dν(x)} converges to zero
in probability.
13
Proof: We can normalise ν to be a probability measure. It suffices to show
that given ² > 0, δ > 0, ∃N , such that:
¯
³¯ Z
´
¯
¯
P ¯
fn (x)dν(x)¯ < ² > 1 − δ (for n ≥ N)
X
Fix ² > 0. By our first assumption, ∃n0 , C² , such that for n ≥ n0 , we have
³Z
P
X
´
|fn (x)|1+δ dν(x) ≤ C² > 1 − ²
For every M > 0 and ω ∈ Ω, we have:
Z
M
1+δ
ν({x | |fn (x, ω)| ≥ M }) ≤
Z
≤
X
{x| |fn (x,ω)|≥M }
|fn (x, ω)|1+δ dν(x)
|fn (x, ω)|1+δ dν(x)
i.e.
Z
ν{x| |fn (x, ω)| ≥ M } ≤
{x| |fn (x,ω)|≥M }
|fn (x, ω)|1+δ dν(x)/(M 1+δ )
Now Hölder’s inequality tells us:
¯Z
¯ Z
¯
¯
fn (x, ω)I(|fn (x, ω)| ≥ M )dν(x)¯ ≤
|fn (x, ω)I(|fn (x, ω)| ≥ M )|dν(x)
¯
X
X
´1/(1+δ) ³ Z
³Z
´ δ
1+δ
1+δ
1+δ
|I(|fn (x, ω)| ≥ M )| δ dν(x)
|fn (x, ω)| dν(x)
≤
X
X
´1/(1+δ)
³Z
δ
(ν{x| |fn (x, ω)| ≥ M }) 1+δ
|fn (x, ω)|1+δ dν(x)
≤
X
14
1
≤ δ
M
Z
X
|fn (x, ω)|1+δ dν(x)
where I(E) denotes the indicator of an event E.
By our second hypothesis, we have limn→∞ P (|fn (x, ω)| ≥ ²) = 0 except for
a subset of X of ν-measure 0. The dominated convergence theorem implies
that
Z
0=
lim P (|fn (x, ω)| ≥ ²)dν(x)
X n→∞
Z
= lim
n→∞ X
Z
= lim
n→∞ X
P (|fn (x, ω)| ≥ ²)dν(x)
Z
[ I{ω̃| |fn (x, ω̃)| ≥ ²}dP (ω̃)]dν(x)
(4.1)
Ω
By Fubini’s theorem and Markov’s inequality, we conclude that
Z Z
[ I{x| |fn (x, ω̃)| ≥ ²} dν(x)] dP (ω)
n→∞ Ω X
³Z
´
= lim E
I{x| |fn (x, ω̃)| ≥ ²}dν(x)
n→∞
XZ
² ´
² ³
P
I{x| |fn (x, ω̃)| ≥ ²}dν(x) ≥
≥ lim
n→∞ M
M
X
(4.1) = lim
which tells us that, for every M , limn→∞ P (ν({x | |fn (x)| ≥ ²}) ≥
²
M)
= 0.
Choose M > (C² /²)1/δ . There exists n1 , such that P (ν({x | |fn (x)| ≥ ²}) ≥
²
M)
< ² for n ≥ n1 . Now fix n ≥ max{n0 , n1 }.
Now we let Ω1 = {ω |
R
X
|fn (x, ω)|1+δ dν(x) ≤ C² }. Then P (Ω1 ) > 1 − ².
15
For ω ∈ Ω1 we have
¯Z
¯
¯
¯
fn (x, ω)I(|fn (x, ω)| ≥ M )dν(x)¯ ≤ C² /M δ < ²
¯
X
Next let Ω2 = {ω | ν({x | |fn (x, ω)| ≥ ²}) <
²
M )}.
We have P (Ω2 ) > 1 − ².
Then for ω ∈ Ω2 , we have
¯Z
¯
¯
{x| |fn (x,ω)|≤M }
¯
¯
fn (x, ω)dν(x)¯
¯Z
¯
≤¯
{x| |fn (x,ω)|<²}
¯ ¯Z
¯ ¯
fn (x, ω)dν(x)¯ + ¯
{x| ²≤|fn (x,ω)|≤M }
Z
≤²+M
X
¯
¯
fn (x, ω)dν(x)¯
I{x| |fn (x, ω)| ≥ ²}dν(x)
≤ 2²
Now let Ω3 = Ω1 ∩ Ω2 , clearly P (Ω3 ) > 1 − 2².
¯R
¯
¯
¯
For the ω ∈ Ω3 and n as above, we have ¯ X fn (x, ω) dν(x)¯ < 3². Thus for
every n ≥ max{n0 , n1 } we have found an Ω3 with P (Ω3 ) > 1 − 2² such that
¯R
¯
¯
¯
for ω ∈ Ω3 we have ¯ X fn (x, ω) dν(x)¯ < 3². Hence for n ≥ max{n0 , n1 }
¯
³¯ R
´
¯
¯
we have P ¯ X fn (x) dν(x)¯ < 3² > 1 − 2². Therefore we have completed
the proof of Lemma 3.1.
We now take the standard step of passing from the ESDs µ √1
n
the characteristic functions m √1
n
An ,
m √1
n
Bn
Z
n
An (u, v)
=
C
eiuRe(z)+ivIm(z) dµ √1
n
16
µ √1
n
Bn
to
: R2 → C, which are defined by
the formulae:
m √1
An ,
An (z)
Z
m √1
n
Bn (u, v)
thus the functions m √1
n
An ,
=
C
m √1
n
eiuRe(z)+ivIm(z) dµ √1
n
Bn
Bn (z)
are continuous and are bounded uni-
formly in magnitude by 1.
Lemma 4.2 The following statements are equivalent:
(i) The sequence of random probability measures µ √1
n
An
− µ √1
n
Bn
converges
to zero in probability.
(ii) For almost every u, v, the sequence of random characteristic functions
m √1
n
An (u, v)
− m √1
n
Bn (u, v)
converges to zero in probability.
P roof. We first show that (i) implies (ii). Fix u, v, and let ² > 0 be arbitrary.
We can find an R depending on C² and ², n0 such that
µ √1
n
An ({z
∈ C : |z| ≥ R}) + µ √1
n
Bn ({z
∈ C : |z| ≥ R}) ≤ ² (n ≥ n0 )
In particular, we have
Z
eiuRe(z)+ivIm(z) [dµ √1 Bn − dµ √1 An ]
m √1 Bn (u, v) − m √1 An (u, v) =
n
n
n
n
C
Z
=
eiuRe(z)+ivIm(z) [dµ √1 Bn − dµ √1 An ]
n
n
|z|<R
Z
+
eiuRe(z)+ivIm(z) [dµ √1 Bn − dµ √1 An ]
n
n
|z|≥R
Z
= ψ(z/R)eiuRe(z)+ivIm(z) [dµ √1 Bn − dµ √1 An ] + ²
n
17
n
where ψ is any smooth compactly supported function that equals one on
the unit ball. But since by hypothesis µ √1
n
An
− µ √1
n
Bn
converges to zero
in probability, so the above integral converges to zero in probability. The
claim then follows.
Now we prove (ii) implies (i). Since continuous compactly supported functions are the uniform limit of smooth compactly supported functions, it
R
R
suffices to show that C f dµ √1 An − C f dµ √1 Bn converges to zero in proban
n
bility for every smooth compactly supported function f : C → C.
Now fix a smooth compactly supported function f : C → C. Let fˆ denote
the Fourier transform of f . Then using Parseval’s identity [10] , we can
write
Z
Z
C
f dµ √1
n
An −
C
Z Z
f dµ √1
n
Bn
=
R
R
fˆ(u, v)(m √1
n
An (u, v)−m √1n Bn (u, v))dudv
Note that the Fourier transform of a smooth compactly supported function is
smooth and rapidly decreasing. In particular, the measure dν = fˆ(u, v)dudv
is finite. By the hypothesis and the fact that the function m √1
n
m √1
n
Bn (u, v)
An (u, v)
−
is bounded and so clearly obeys the moment condition required
in Lemma 4.1, the claim now follows.
Definition 4.3 Now we are going to use the Stieltjes-like transform g √1
n
An (w)
C → R, defined almost everywhere by the formula that
Z
g √1
n
An (w) (z) := 2Re
C
1
n
z−w
2 X Re(z − √n λj )
.
dµ
=
1
|z − w|2 √n An (w) n
|z − √1n λj |2
j=1
18
(4.2)
:
Lemma 4.4 (Girko’s identity). For every non-zero u, v we have
u2 + v 2
m √1 An (u, v) =
n
4πiu
Z Z
( g √1
R
n
R
An (s
+ it)eius+ivt dt)ds,
where the inner integral is absolutely integrable for almost every s, and the
outer integral is absolutely convergent.
P roof. Since
n
1 X i(uRe( √1n λj )+vIm( √1n λj ))
e
m √1 An (u, v) =
n
n
j=1
it suffices from (4.2) to show that
i(uRe(w)+vIm(w))
e
u2 + v 2
=
2πiu
Z Z
Re(s + it − w) ius+ivt
(
e
dt)ds
2
R R |s + it − w|
for each complex number w. Note that
R
|s|
itv
R s2 +t2 e dt
= πe−|sv| , so this gives
us:
Z
R
Re(s + it − w) ius+ivt
e
dt = πsgn(s − Re(w))e−|v||s−Re(w)| eius eivIm(w)
|s + it − w|2
for every s 6= Re(w), and the claim follows by the Laplace transform
Z
∞
sin(su)e−|v|s ds =
0
u2
u
+ v2
.
Similarly we can define g √1 Bn in the same way. To conclude the proof of
n
19
Theorem 3.1, it thus suffices to show
Z Z
( (g √1
R
n
R
An (s
+ it) − g √1
n
Bn (s
+ it))eius+ivt dt)ds
converges to zero in probability under the assumptions of Theorem 3.1.
Now let ψ : R → R+ be a smooth cutoff function which equals 1 on [−1, 1],
|ψ(x)| ≤ 1 for all x and is supported on [−2, 2].
Lemma 4.5 (Truncation in s,t). Fix u 6= 0, v 6= 0. Let w ∈ C.
(i) The integral
Z
Z
|
R
R
Re(s + it − w) ius+ivt
e
dt|(1 − ψ(s/R2 ))ds
|s + it − w|2
(4.3)
is bounded as a function of w, and for |w| ≤ R, is less than any given ² > 0,
if R is large enough.
(ii)
Z
Z
|
R
R
Re(s + it − w) ius+ivt
e
(1 − ψ(t/R2 ))dt|ψ(s/R2 )ds
|s + it − w|2
(4.4)
is bounded, and for |w| ≤ R, is less than any given ² > 0, if R is large
enough.
20
Proof: (i) Since
Z
R
Re(s + it − w) ius+ivt
e
dt = πsgn(s − Re(w))e−|v||s−Re(w)| eius eivIm(w)
|s + it − w|2
for every s 6= Re(w), then we have
Z
Re(s + it − w) ius+ivt
e
dt| = πe−|v||s−Re(w)| .
|s + it − w|2
|
R
Now it suffices to show
Z
πe−|v||s−Re(w)| (1 − ψ(s/R2 ))ds
(4.5)
R
is bounded. When |Re(w)| < R2
Z
Z
(4.5) =
πe
−|v||s−Re(w)|
R
³Z
R
∞
−|v|(s−Re(w))
≤π
e
Z
πe−|v||s−Re(w)| ψ(s/R2 )ds
−
−
Re(w)
ds +
Re(w)
R2
Z
e−|v|(Re(w)−s) ds
´
−∞
πe−|v||s−Re(w)| ds
−R2
³Z
R2
= 2π/|v| − π
Z
−|v|(s−Re(w))
e
Rew)
ds +
e−|v|(Re(w)−s) ds
−R2
Re(w)
2π −|v|(R2 −Re(w))
2
(e
+ e−|v|(R +Re(w)) )
|v|
4π
≤
.
|v|
=
When Re(w) ≥ R2
Z
−R2
(4.5) =
πe−|v||s−Re(w)| (1 − ψ(s/R2 ))ds
−∞
21
´
Z
+
∞
πe−|v||s−Re(w)| (1 − ψ(s/R2 ))ds
R2
Z −R2
Z
−|v|(Re(w)−s)
≤ π(
e
ds +
−∞
Z ∞
+
e−|v|(s−Re(w)) )ds
Re(w)
e−|v|(Re(w)−s) ds
R2
Re(w)
4π
≤
.
|v|
Similarly we have (4.5) ≤
4π
|v|
for Re(w) ≤ −R2 . Therefore we get an explicit
upper bound for (4.3), namely
4π
|v| .
If |w| ≤ R, then |Re(w)| ≤ R, and thus one can easily see that for fixed w,
(4.3) goes to 0 if R is large enough.
(ii)First we are trying to show that there are positive constants A,B,C such
that:
Z
|
R
is bounded by
A
R4
Re(w − (s + it)) ius+ivt
e
(1 − ψ(t/R2 ))dt|
|w − (s + it)|2
+
B
R2 |x−s|
+
C
,
|x−s|2
(4.6)
where w = x + yi. Since the eius term
has magnitude 1 and can be discarded, to prove (4.6) is bounded, it suffices
to show
Z
I=|
R
is bounded by
A
R4
x−s
eivt (1 − ψ(t/R2 ))dt|
(x − s)2 + (y − t)2
B
C
+ R2 |x−s|
+ |x−s|
2 . Using integration by parts, integrating
the eivt term and differentiating the others, we have:
¯1
x−s
¯
I = ¯ eivt
(1 − ψ(t/R2 )) |∞
−∞
iv
(x − s)2 + (y − t)2
22
Z
(s − x)(2t − 2y)
1 ivt
e (
(1 − ψ(t/R2 ))
((x − s)2 + (y − t)2 )2
R iv
¯
x−s
1
¯
0
2
+
(−
)ψ
(t/R
))dt
¯
2
2
2
(x − s) + (y − t)
R
−
Note that the first term appears in I vanishes, so
Z
1 ivt (s − x)(2t − 2y)
e
(1 − ψ(t/R2 ))dt|
((x − s)2 + (y − t)2 )2
R iv
Z
1 ivt
x−s
1
+|
e
(− 2 )ψ 0 (t/R2 )dt|
2
2
(x − s) + (y − t)
R
R iv
I≤|
Now we integrate the integrals above by parts again separately:
¯Z
¯
I1 = ¯
¯
1 ivt (s − x)(2t − 2y)
¯
2
e
(1
−
ψ(t/R
))dt
¯
2
2
2
((x − s) + (y − t) )
R iv
¯
1
(s − x)(2t − 2y)
¯
= ¯ − 2 eivt
(1 − ψ(t/R2 )) |∞
−∞
v
((x − s)2 + (y − t)2 )2
Z
1 ivt
(s − x)2 − 3(t − y)2
+
e
2(s
−
x)
(1 − ψ(t/R2 ))dt
2
((x − s)2 + (y − t)2 )3
R v
Z
¯
1 ivt (s − x)(2t − 2y)
1
¯
0
2
+
e
(−
)ψ
(t/R
))dt
¯
2
2 + (y − t)2 )2
2
v
((x
−
s)
R
R
Again, the first term in the RHS disappears.
Z
I1 ≤
Since
1
(s − x)2 − 3(t − y)2
(2(s
−
x)
(1 − ψ(t/R2 ))|dt
2
2 + (y − t)2 )3
v
((x
−
s)
R
Z
1
(s − x)(2t − 2y)
1
+ | 2
( 2 )ψ 0 (t/R2 ))|dt.
2
2
2
R v ((x − s) + (y − t) ) R
|
√
(s − x)2 − 3(t − y)2
3 3
|2(s − x)
|dt =
((x − s)2 + (y − t)2 )3
4|s − x|2
R
Z
23
and by the fact that
Z
|
0
∞
(a2
2ax
1
dx| =
2
2
+x )
|a|
we obtain:
√
1
(s − x)2 − 3(t − y)2
3 3
2
| 2 (2(s − x)
(1 − ψ(t/R ))|dt ≤ 2
((x − s)2 + (y − t)2 )3
4v |s − x|2
R v
Z
Z
|
R
1
(s − x)(2t − 2y)
2
1
1
1
( 2 )ψ 0 (t/R2 ))|dt ≤ 2
A1
2
2
2
2
v ((x − s) + (y − t) ) R
v |x − s| R2
Let A1 > 0 be such that |ψ 0 (t/R2 )| ≤ A1 . Similarly, we have
Z
1 ivt
x−s
1
e
(− 2 )ψ 0 (t/R2 )dt|
2
2
(x − s) + (y − t)
R
R iv
1
x−s
1
= | 2 eivt
( )ψ 0 (t/R2 ) |∞
−∞
v
(x − s)2 + (y − t)2 R2
Z
1 ivt
(x − s)(2t − 2y)
1
+
e (
(− 2 )ψ 0 (t/R2 )
2
2
2
2
((x − s) + (y − t) )
R
R v
x−s
1
+
(− )ψ 00 (t/R2 ))dt|
(x − s)2 + (y − t)2 R4
1
2
1
π 1
≤ 2
A1 + 2 4 A2
2
v |x − s| R
v R
I2 = |
Again let A2 > 0 be such that |ψ 00 (t/R2 )| ≤ A2 . Hence we finally get:
√
4A1
1
πA2 1
3 3
+ 2 2
+ 2 4
I≤ 2
4v |s − x|2
v R |x − s|
v R
which is exactly what we want.
Note that one can get an explicit upper bound for (4.6) which stems from
24
the classical Poisson integral identity
R∞
a
−∞ a2 +(x−x0 )2 dx
= πsgn(a). The
eius+ivt (1 − ψ(t/R2 )) term has magnitude at most 1 and thus can be discarded.
Now given M > 0. In order to get an upper bound for (4.4), we integrate
(4.4) as follows:
Z
Z
(4.4) =
|
Z
|s−Re(λj )|>M
Z
R
Re(λj − (s + it)) ius+ivt
e
(1 − ψ(t/R2 ))dt|ψ(s/R2 )ds
|λj − (s + it)|2
Re(λj − (s + it)) ius+ivt
e
(1 − ψ(t/R2 ))dt|ψ(s/R2 )ds
2
|s−Re(λj )|≤M
R |λj − (s + it)|
Z
1
1
≤K
( 2+
)2 ψ(s/R2 )ds + 2M π
|s − Re(λj )|
|s−Re(λj )|>M R
+
|
(for some positive constant K)
Z
1
1
≤ 2K
+
ds + 2M π
T
4
(s − Re(λj ))2
[−2R2 ,2R2 ] {|s−Re(λj )|≤M } R
1
2M
1
2
4
− 4 −
+
) + 2M π
= 2K( 2 −
2
2
R
2R + Re(λj )
R
2R − Re(λj ) M
As one can see (4.4) is bounded as R → ∞.
Finally, if |w| ≤ R, then |x| ≤ R and |y| ≤ R; also note that the integrand
of (4.4) vanishes unless |s| ≤ 2R2 and |t| ≥ R2 . So |s − x| ≤ 2R2 and
|t − y| ≥ R2 − R > R2 /2 for R sufficiently large. Consequently we have
1
(x−s)2 +(y−t)2
≤
4
.
R4
Now given M1 > 0, M2 > 0, we integrate I1 and I2 by
parts separately exactly as above to obtain
Z
I1 ≤
1
(s − x)2 − 3(t − y)2
2(s
−
x)
(1 − ψ(t/R2 ))|dt
2
2 + (y − t)2 )3
v
((x
−
s)
R
Z
1
(s − x)(2t − 2y)
1
+ | 2
( 2 )ψ 0 (t/R2 )|dt
2
2
2
R v ((x − s) + (y − t) ) R
|
25
Z
1
(s − x)2 − 3(t − y)2
1 − ψ(t/R2 )
| 2 2(s − x)
|dt
((x − s)2 + (y − t)2 )2 (s − x)2 + (t − y)2
R v
Z
1
1
(s − x)(2t − 2y)
ψ 0 (t/R2 )
(
|dt
+ | 2
)
2
2 3/2 R2 ((x − s)2 + (t − y)2 )1/2
R v ((x − s) + (y − t) )
π 4
3π 4
4 A1 1
≤ 2 4+ 2 4+ 2 2 2
v R
v R
v R R
M1
≤ 4
R
=
Similarly, we have
Z
(x − s)(2t − 2y)
1 0
1
|
ψ (t/R2 )|dt
I2 ≤ 2
v R ((x − s)2 + (t − y)2 )2 R2
Z
1
x−s
1
+ 2
|
( 4 )ψ 00 (t/R2 )|dt
2
2
v R (x − s) + (t − y) R
π A2
4 A1
≤ 2 4+ 2 4
v R
v R
M2
≤ 4
R
We conclude that I ≤ I1 + I2 ≤
M
R4
for some M > 0. Clearly (4.4) ≤
4R2 M
R4
since |s| ≤ 2R2 , the second part of claim (ii) now follows and this completes
the proof of Lemma 4.5.
Lemma 4.6 For some R > 1 large enough, we have
Z Z
( g √1
R
and
R
Z Z
( g √1
R
R
n
n
+ it)eius+ivt dt)(1 − ψ(s/R2 ))ds
(4.7)
+ it)eius+ivt (1 − ψ(t/R2 ))dt)ψ(s/R2 )ds
(4.8)
An (s
An (s
both converge to zero in probability.
26
Proof: Given ²1 > 0, ²2 > 0. From (3.2), (4.2), Lemma 4.4 and the triangle
inequality we obtain
Z Z
| ( g √1
R
n
R
+ it)eius+ivt dt)(1 − ψ(s/R2 ))ds|
An (s
Z Z
n
2X
Re(w − (s + it)) ius+ivt
=
| (
e
dt)(1 − ψ(s/R2 ))ds|
2
n
|w
−
(s
+
it)|
R R
j=1
Z
Z
n
Re(w − (s + it)) ius+ivt
2X
|
e
dt|(1 − ψ(s/R2 ))ds
≤
2
n
|w
−
(s
+
it)|
R
j=1 R
Z Z
n
X
Re(w − (s + it)) ius+ivt
2
|
=
e
dt|(1 − ψ(s/R2 ))ds
2
n
|w
−
(s
+
it)|
R
R
2
j=1,|w|≤R
n
X
2
+
n
Z
|
j=1,|w|>R2
≤ 2²1 +
Z
R
R
Re(w − (s + it)) ius+ivt
e
dt|(1 − ψ(s/R2 ))ds
|w − (s + it)|2
8π
²2
|v|
Similarly we have:
|
Z Z
( g √1
R
R
n
An (s
+ it)eius+ivt (1 − ψ(t/R2 ))dtψ(s/R2 )ds|
Z Z
n
Re(w − (s + it)) ius+ivt
2X
=
| (
e
(1 − ψ(t/R2 ))dt)ψ(s/R2 )ds|
2
n
|w
−
(s
+
it)|
R R
j=1
Z
Z
n
2X
Re(w − (s + it)) ius+ivt
≤
|
e
(1 − ψ(t/R2 ))dt|ψ(s/R2 )ds
n
|w − (s + it)|2
R
R
j=1
Z Z
n
2 X
Re(w − (s + it)) ius+ivt
=
|
e
(1 − ψ(t/R2 ))dt|ψ(s/R2 )ds
2
n
|w
−
(s
+
it)|
R
R
j=1,|w|≤R
n
X
2
+
n
Z
j=1,|w|>R R
Z
|
R
Re(w − (s + it)) ius+ivt
e
(1 − ψ(t/R2 ))dt|ψ(s/R2 )ds
|w − (s + it)|2
≤ 2²1 + K²2
27
for some K > 0. Now we have finished the proof of Lemma 4.6.
Now from Lemma 4.6 (and their counterparts for g √1
n
Bn )
and the triangle
inequality, we thus see that to prove that
Z Z
( (g √1
R
n
R
An (s
+ it) − g √1
n
Bn (s
+ it))eius+ivt dt)ds
(4.9)
converges to zero in probability, it suffices to show that
Z Z
R
R
(g √1
n
An (s
+ it) − g √1
n
Bn (s
+ it))eius+ivt ψ(t/R2 )ψ(s/R2 )dtds (4.10)
converges to zero in probability for every fixed R ≥ 1 large enough. Indeed,
assume (4.10) converges to zero in probability. Observe that
Z Z
|
(g √1
RZ RZ
=|
R
R
n
An (s
(g √1
n
+ it) − g √1
An (s
n
Bn (s
+ it) − g √1
n
+ it))eius+ivt |
Bn (s
+ it))eius+ivt (1 − ψ(s/R2 ) + ψ(s/R2 )
−ψ(t/R2 )ψ(s/R2 ) + ψ(t/R2 )ψ(s/R2 ))dtds|
Z Z
≤ | ( g √1 An (s + it) − g √1 Bn (s + it))eius+ivt dt)(1 − ψ(s/R2 ))ds|
n
RZ RZ n
+| ( g √1 An (s + it) − g √1 Bn (s + it))eius+ivt (1 − ψ(t/R2 ))dt)ψ(s/R2 )ds|
n
n
ZR Z R
+|
(g √1 An (s + it) − g √1 Bn (s + it))eius+ivt ψ(t/R2 )ψ(s/R2 )dtds|
R
R
n
n
Note that the integrands here are now jointly absolutely integrable in t, s,
and so we may now freely interchange the order of integration. Now from
the assumption and Lemma 4.6, one can see (4.9) converges to zero in prob-
28
ability.
Observe that g √1
An (w) (z)
n
A (w) (z)
n n
g √1
is a locally integrable function on C, and that
∂
2
1
log | det( √ An (w) − zI)|
∂Re(z) n
n
=
(4.11)
Fix R. Using (4.11) and integration by parts in the s variable, we have
Z Z
(4.10) =
R
2
=
n
R
(g √1
Z Z
n
An (s
+ it) − g √1
n
Bn (s
+ it))eius+ivt ψ(t/R2 )ψ(s/R2 )dtds
eius+ivt ψ(t/R2 )ψ(s/R2 )
R
R
1
1
∂
× (log | det( √ An − zI)| − log | det( √ Bn − zI)|)dt
∂s
n
n
Z
2
=
eius+ivt ψ(t/R2 )ψ(s/R2 )
n R
1
1
×(log | det( √ An − zI)| − log | det( √ Bn − zI)|) |∞
−∞ dt
n
n
Z Z
2
1
1
−
(log | det( √ An − zI)| − log | det( √ Bn − zI)|)
n R R
n
n
∂ ius+ivt
× e
ψ(t/R2 )ψ(s/R2 )dsdt
∂s
Therefore we can rewrite (4.10) in the form of
Z Z
R
R
fn (s, t)φu,v,R (s, t)dsdt,
29
where
fn (s, t) :=
1
1
2
2
log | det( √ An − zI)| −
log | det( √ Bn − zI)|
n
n
n
n
and
φu,v,R (s, t) := −
∂ ius+ivt
(e
ψ(t/R2 )ψ(s/R2 )).
∂s
Note: For the values of t and s where | det( √1n An − zI)| or | det( √1n Bn − zI)|
may equal zero the integration by parts is not justified, but this is not going
to be a problem since there are only finitely many values of such t and s,
with measure 0.
Thus to complete the proof of theorem 3.1 it will suffice to show that
Z Z
R
R
|fn (s, t)||φu,v,R (s, t)|dsdt
converges to zero in probability. We shall apply Lemma 4.1 to prove this.
Note that φu,v,R (s, t)dsdt = dν is a finite measure. It remains to check the
uniform integrability of fn , we will take δ = 1 in this case.
Lemma 4.7 Given any positive numbers c1 , c2 , there exists some M > 0
such that
Z
(log |z − (s + it)|)2 dsdt ≤ M (1 + |z|2 )
|s|≤c1 ,|t|≤c2
for all z ∈ C.
30
(4.12)
Proof: Fix z ∈ C, say z = x + yi . Put D = {(s, t)||z − (s + it)| ≤ 1},
R = [−c1 , c1 ] × [−c2 , c2 ]. We can rewrite the integral in (4.12) by setting
s0 = s − x, t0 = t − y. Thus it suffices to show that for all c1 , c2 > 0, there
exists M > 0 such that
Z
(log
p
s02 + t02 )2 ds0 dt0 ≤ M (1 + |z|2 )
R0
where R0 = [−c1 −x, c1 −x]×[−c2 −y, c2 −y]. We also put D0 = {(s0 , t0 )|s02 +
T
t02 ≤ 1}. Then for the region D0 R0 , where the function has a singularity,
we can estimate the integral by the local integrability of log2 in the plane,
i.e.
Z
Z
D
T
Z
(log |z − (s + it)|)2 dsdt =
R
Z
(log r)2 rdrdθ ≤
≤
D0
0
2π
Z
D
1
T
0
(log
p
s02 + t02 )2 ds0 dt0
R0
(log r)2 rdrdθ
0
= π/2
For the area R0 \D0 , we can estimate the integral by estimating log
√
s02 + t02
over R0 \D0 . For |z| ≤ max{2||(c1 , c2 )||, 1 + ||(c1 , c2 )||}, then the function
(log |z − (s + it)|)2 is a continuous function on the bounded region R0 \D0 .
R
So we can find some M 0 > 0, such that: R0 \D0 (log |z − (s + it)|)2 dsdt ≤ M 0 .
R
Thus in this case we can bound |s|≤c1 ,|t|≤c2 (log |z−(s+it)|)2 dsdt by M 0 +π/2.
For |z| > max{2||(c1 , c2 )||, 1 + ||(c1 , c2 )||}, we have
1 ≤ |z| − ||(c1 , c2 )|| ≤ ||(x, y) − (c1 , c2 )|| ≤
31
p
s02 + t02
and
p
s02 + t02 ≤ |z| + ||(s, t)|| ≤ |z| + ||(c1 , c2 )|| ≤ 2|z|
√
This implies that log s02 + t02 ≤ log(2|z|) ≤ 2|z|. Thus we obtain a very
√
R
crude estimation of R0 (log s02 + t02 )2 ds0 dt0 by 16c1 c2 |z|2 . Now we find
M = max{M 0 + π/2, 16c1 c2 } such that
Z
(log |z − (s + it)|)2 dsdt ≤ M (1 + |z|2 )
|s|≤c1 ,|t|≤c2
This completes the proof of Lemma 4.7.
From linear algebra, we have the identity
n
1
1
1
1X
log | det( √ An − zI)| =
log | √ λj − (s + it)|
n
n
n
n
(4.13)
j=1
and similarly for Bn . From the boundedness and compact support of φu,v,R
and Lemma 4.7, we observe that
Z Z
R
1
1
(log | √ λ − (s + it)|)2 |φu,v,R (s, t)|dsdt ≤ M (1 + |λ|2 )
n
n
R
for all λ ∈ C and some M > 0.
Now given ² > 0, there exist C1 , C2 and M such that
Z Z
R
R
|fn (s, t)|2 |φu,v,R (s, t)|dsdt
(triangle inequality)
32
(4.14)
Z Z
≤
4
1
(| log | det( √ An − zI)|
2
n
R R n
1
+| log | det( √ Bn − zI)|)2 |φu,v,R (s, t)|dsdt
n
(by (4.13))
Z Z
n
4 X
1 (A )
≤
(
| log | √ λj n − (s + it)||
2
n
R R n
j=1
+
n
X
1 (B )
| log | √ λj n − (s + it)||)2 |φu,v,R (s, t)|dsdt
n
j=1
(Cauchy − Schwartz inequality)
Z Z
n
1 (A )
8X
≤
((log | √ λj n − (s + it)|)2
n
R R n
j=1
1 (B )
+(| log | √ λj n − (s + it)|)2 )|φu,v,R (s, t)|dsdt
n
(by (4.14))
n
8M X
1 (A )
1 (B )
≤
(1 + |λj n |2 + 1 + |λj n |2 )
n
n
n
j=1
= 16M +
n
X
8M
j=1
n2
(An ) 2
|λj
| +
n
X
8M
j=1
n2
(Bn ) 2
|λj
|
By (3.1) we can estimate the last term by 8M (C1 + C2 + 2) with probability
greater than 1 − ², i.e. fn has uniform integrability with δ = 1. By the
hypothesis of Theorem 3.1 fn (s, t) converges to zero in probability for almost
every s, t. The claim now follows directly from Lemma 4.1. The proof of
Theorem 3.1 is now complete.
33
Chapter 5
Conclusion
Now we have shown that the difference of the empirical spectral distribution
of two normalised random matrices
√1 An , √1 Bn
n
n
converges to 0 if they
satisfy conditions (i) and (ii) in Theorem 3.1. The remainder of the proof of
Conjecture 1.2 is given in [12]. The main burden is to show that condition
(ii) of Theorem 3.1 is satisfied. We shall not discuss this part of the proof
in this thesis.
34
Bibliography
[1] Z. D. Bai, Circular law, Ann. Probab. 25, 494-529, 1997
[2] Z. D. Bai and J. Silverstein, Spectral analysis of large dimensional random matrices, Mathematics Monograph Series 2, Science Press, Beijing,
2006.
[3] K. L. Chung, A Course in Probability Theory (Third Edition), 68-90,
2001.
[4] A. Edelman, Probability that a random real Gaussian matrix has k real
eigenvalues, related distributions, and the Circular Law, Journal of Multivariate Analysis, 203-232, 1997.
[5] V. L. Girko, Circular law, Theory Probab. Appl. , 694-706, 1984.
[6] F. Götze and A.N. Tikhomirov, On the circular law, preprint, 2007
[7] F. Götze and A.N. Tikhomirov, The Circular Law for random matrices,
preprint, 2007
[8] M.L. Mehta, Random Matrices and the Statistical Theory of Energy
Levels, Academic Press, New York, NY, 1967.
35
[9] G. Pan and W. Zhou, Circular Law, extreme singular values and potential theory, preprint, 2007
[10] W. Rudin, Functional Analysis (Second Edition), 186-189, 1991.
[11] T. Tao and V. Vu, Random matrices: The Circular Law, Communication in Contemporary Mathematics 10, 261-307, 2008.
[12] T. Tao, V. Vu, M. Krishnapur, Random matrices: Universality of ESDs
and the circular law, arXiv:0807.4898v5, 2009.
36