TAO’S SPECTRAL PROOF OF THE SZEMERÉDI REGULARITY
LEMMA
S. CIOABA AND R.R. MARTIN
Abstract. On December 3, 2012, following the Third Abel conference, in honor of Endre
Szemerédi, Terence Tao posted on his blog a proof of the spectral version of Szemerédi’s
regularity lemma. This, in turn, proves the original version.
1. Introduction
Tao attributes this proof to Frieze and Kannan [1].
One thing to observe is that this is a statement on matrices and graphs are just a consequence.
2. Basic matrix version
Lemma 1 (Szemerédi’s regularity lemma, matrix version). Let T be a self-adjoint n × n
matrix such that tr(T 2 ) ≤ n2 . Let V be the set of n indices and let > 0. Then there exists
an M ≤ M () and
• a decomposition of T into three matrices, T = T1 + T2 + T3 , each of which is selfadjoint,
• a partition V = V0 ∪ V1 ∪ ·· · ∪ VM , and
}
• a set of pairs Σ ⊂ {0,...,M
(which contains all pairs with 0),
2
such that
• for all i, j ∈ {0, 1, . . . , M }, there exists dij such that for all a ∈ Vi and b ∈ Vj , we
have |(T1 )ab − dij | < ,
P P
• for all i, j ∈ {0, 1, . . . , M } − Σ, we have
|(T2 )ab |2 < |Vi ||Vj |,
a∈Vi b∈Vj
• for all i, j ∈ {0, 1, . . . , M } − Σ, we have n · σ(T3 ) < |Vi ||Vj |, (where σ(T3 ) is T3 ’s
largest
and
P singular value)
2
•
|Vi ||Vj | ≤ n .
(i,j)∈Σ
Proof. Enumerate V = {1, . . . , n}. Since T is self-adjoint, it has an eigenvalue decomposition
n
X
T =
λi ui u∗i ,
i=1
n
for some orthonormal basis u1 , . . . , un of C (where the vectors are column vectors) and real
eigenvalues λ1 , . . . , λn ∈ R. We arrange them in decreasing order of magnitude
|λ1 | ≥ · · · ≥ |λn |.
1
In a self-adjoint matrix,
of T 2 is the sum of the squares of the eigenvalues
Pn the trace
2
2
of T and so tr(T ) =
j=1 |λj | . So we can bound the eigenvalues by observing that
P
i
iλ2i ≤ j=1 λ2j ≤ n2 and so, for all i ∈ {1, . . . , n},
n
(1)
|λi | ≤ √ .
i
We will be given a function F : N → N that we will specify later. This function does
depend on (we suppress this in the notation) and it satisfies the inequality F (i) > i for all
integers i. We want to find an integer J for which
X
(2)
|λj |2 ≤ 3 n2 .
J≤j<F (J)
To do this, we consider the partition of {1, . . . , n} into intervals [F (k−1) (1), F (k) (1) − 1] from
k = 1, . . . , 1/3 where F (k) represents the k th composition of F with itself. Note that either
we find a J = F (k−1) (1) for which (2) is satisfied or the sum of |λj |2 for all j in some interval
is greater than 3 n2 . Since there are 1/3 intervals, this would contradict the n2 bound for
tr(T 2 ).
Thus, we have a partition of T into three matrices:
T = T1 + T2 + T3 ,
where T1 is the “structured” component
(3)
T1 :=
X
λi ui u∗i ,
i<J
and T2 is the “error” component
(4)
X
T2 :=
λi ui u∗i ,
J≤i<F (J)
and T3 is the “pseudorandom” component
(5)
T3 :=
X
λi ui u∗i .
i≥F (J)
We will partition the vertex set so that T1 is approximately constant on most clusters. The
number of such clusters will be OJ, (1). For each j < J we define a partition into clusters on
which entry ui (a complex number) varies by J n−1/2 . There is also an exceptional cluster of
size J n which comes from the vertices for which the entry of ui is large in magnitude. That
q
is, either its real or its imaginary part is larger than J n−1/2 in absolute value.
To see this, simply place a vertex into the exceptional cluster if the
qcorresponding entry of
n−1/2 . Since kui k2 =
q
1, this means there can be at most J n such entries. Partition the square of length 2 J n−1/2
ui has the absolute value of either its real or imaginary part at least
J
3/2
centered at the origin of the complex plane into subsquares of side length J 3/2 n−1/2 . There
q
2
2 3/2
are 2 J n−1/2 / J 3/2 n−1/2 = 4J 4 /4 such subsquares. Partition the vertices according
to where its corresponding entry of ui lies.
2
Take the union of all vertices in the exceptional clusters and the corresponding exceptional
cluster is of size at most (J − 1) · n
< n. For the rest of the vertices, we take the common
J
refinement of the partitions defined by each ui , i < J. This defines a partition of the vertex
set V = V0 + V1 + · · · + VM in which V0 is the exceptional set. For i = 1, . . . , M , the entries
over Vi of each of u1 , . . . , uJ−1 have magnitude at most
r
√
J −1/2
2·2
n
,
(6)
and differ in magnitude by at most
√ 3/2 −1/2
2 3/2 n
.
J
(7)
Moreover,
M≤
(8)
4J 4
4
J
.
Now, let i, j ∈ {1, . . . , M }. We will show that the values of T1 over the block Vi × Vj differ
by at most 8. To see this, let a, c ∈ Vi and b, d ∈ Vj . Then
X
(T1 )ab − (T1 )cd = λi ui (a)ui (b) − λi ui (c)ui (d)
i<J
X
≤
|λi ||ui (a)ui (b) − ui (c)ui (b) + ui (c)ui (b) − ui (c)ui (d)|
i<J
≤
X
n|ui (b)||ui (a) − ui (c)| + n|ui (c)||ui (b) − ui (d)|
i<J
!
r
r
√
√
J −1/2 √ 3/2 −1/2
J −1/2 √ 3/2 −1/2
n
· 2 3/2 n
+n·2 2
n
· 2 3/2 n
≤J n·2 2
J
J
!
r
√
J −1/2 √ 3/2 −1/2
n
· 2 3/2 n
= 2J n · 2 2
J
< 8.
As a result, we can conclude that, if dij is the mean of the entries in the block Vi × Vj ,
then by the triangle inequality,
|(T1 )ab − dij | < 16.
P
P
Next we consider T2 and observe that tr(T22 ) = J≤j≤F (J) λ2i < 3 n2 . So, a,b∈V |(T2 )ab |2 <
3 n2 . Define Σ1 so that for every (i, j) 6∈ Σ1 ,
XX
(10)
|(T2 )ab |2 < |Vi ||Vj |.
(9)
a∈Vi b∈Vj
Thus,
2
X
(i,j)∈Σ1
|Vi ||Vj | ≤
X XX
(i,j)∈Σ1 a∈Vi b∈Vj
3
|(T2 )ab |2 ≤ 3 n2 .
consequently,
X
|Vi ||Vj | ≤ n2 .
(i,j)∈Σ1
p
Finally, we turn our attention
to
T
3 . The maximum eigenvalue of T3 is |λF (J) | ≤ n/ F (J).
p
We want to establish that n2 / F (J) ≤ |Vi ||Vj | for (i, j) 6∈ Σ. Because |Vi |, |Vj | ≥ n/M , it
is sufficient to show that F (J) ≥ M 4 /6 because that would verify that
n2
3 n2
p
≤
≤ |Vi ||Vj |.
M2
F (J)
J
By (8), M ≤ (4J 4 /4 ) . So the function that suffices is
4x
1 4x4
.
F (x) ≥ 6
4
Let Σ be the pairs (i, j) ∈ {0, 1, . . . , M } such that either (i, j) ∈ Σ1 , i = 0, j = 0 or
n
min(|Vi |, |Vj |) ≤ M
. Thus,
X
X
X
|Vi ||Vj | ≤
|Vi ||Vj | + 2|V0 ||V | + 2
|Vi ||V |
(i,j)∈Σ
|Vi |<n/M
(i,j)∈Σ1
n
n ≤ 5n2 .
M
Note that in this proof we use coefficients of 1q6 and 5 in (9) and (11), respectively.
We can, of course, choose /16 rather than but we chose these parameters to make the
computations somewhat more transparent.
2
(11)
≤ n2 + 2n · n + 2M
3. Spectral version
Lemma 2 (Szemerédi’s regularity lemma, spectral version). Let T be a self-adjoint n × n
matrix such that tr(T 2 ) ≤ n2 . Let V be the set of n indices and let > 0. Then there exists
a partition V = V1 ∪ · · · ∪ VM for some M ≤ M () with the property that, for all pairs
(i, j) ∈ {1, . . . , M }2 outside of an exceptional set Σ, one has
(12)
∗
|vB
(T − dij I)vA | ≤ |Vi ||Vj |
whenever supp(vA ) ⊂ Vi , kvA k22 ≤ |Vi |, supp(vB ) ⊂ Vj and kvB k22 ≤ |Vj |, for some real
number dij . Furthermore, we have
X
(13)
|Vi ||Vj | ≤ n2 .
(i,j)∈Σ
Proof. Enumerate V = {1, . . . , n}. Since T is self-adjoint, it has an eigenvalue decomposition
n
X
T =
λi ui u∗i ,
i=1
for some orthonormal basis u1 , . . . , un of Cn (where the vectors are column vectors) and real
eigenvalues λ1 , . . . , λn ∈ R. We arrange them in decreasing order of magnitude
|λ1 | ≥ · · · ≥ |λn |.
4
In a self-adjoint matrix,
of T 2 is the sum of the squares of the eigenvalues
Pn the trace
2
2
of T and so tr(T ) =
j=1 |λj | . So we can bound the eigenvalues by observing that
P
i
iλ2i ≤ j=1 λ2j ≤ n2 and so, for all i ∈ {1, . . . , n},
n
(14)
|λi | ≤ √ .
i
We will be given a function F : N → N that we will specify later. This function does
depend on (we suppress this in the notation) and it satisfies the inequality F (i) > i for all
integers i. We want to find an integer J for which
X
(15)
|λj |2 ≤ 3 n2 .
J≤j<F (J)
To do this, we consider the partition of {1, . . . , n} into intervals [F (k−1) (1), F (k) (1) − 1] from
k = 1, . . . , 1/3 where F (k) represents the k th composition of F with itself. Note that either
we find a J = F (k−1) (1) for which (15) is satisfied or the sum of |λj |2 for all j in some interval
is greater than 3 n2 . Since there are 1/3 intervals, this would contradict the n2 bound for
tr(T 2 ).
Thus, we have a partition of T into three matrices:
T = T1 + T2 + T3 ,
where T1 is the “structured” component
(16)
T1 :=
X
λi ui u∗i ,
i<J
and T2 is the “error” component
(17)
X
T2 :=
λi ui u∗i ,
J≤i<F (J)
and T3 is the “pseudorandom” component
(18)
T3 :=
X
λi ui u∗i .
i≥F (J)
We will partition the vertex set so that T1 is approximately constant on most clusters. The
number of such clusters will be OJ, (1). For each j < J we define a partition into clusters on
which entry ui (a complex number) varies by J n−1/2 . There is also an exceptional cluster of
size J n which comes from the vertices for which the entry of ui is large in magnitude. That
q
is, either its real or its imaginary part is larger than J n−1/2 in absolute value.
To see this, simply place a vertex into the exceptional cluster if the
qcorresponding entry of
n−1/2 . Since kui k2 =
q
1, this means there can be at most J n such entries. Partition the square of length 2 J n−1/2
ui has the absolute value of either its real or imaginary part at least
J
3/2
centered at the origin of the complex plane into subsquares of side length J 3/2 n−1/2 . There
q
2
2 3/2
are 2 J n−1/2 / J 3/2 n−1/2 = 4J 4 /4 such subsquares. Partition the vertices according
to where its corresponding entry of ui lies.
5
Take the union of all vertices in the exceptional clusters and the corresponding exceptional
cluster is of size at most (J − 1) · n
< n. For the rest of the vertices, we take the common
J
refinement of the partitions defined by each ui , i < J. This defines a partition of the vertex
set V = V0 + V1 + · · · + VM in which V0 is the exceptional set. For i = 1, . . . , M , the entries
over Vi of each of u1 , . . . , uJ−1 have magnitude at most
r
√
J −1/2
2·2
(19)
n
,
and differ in magnitude by at most
√ 3/2 −1/2
2 3/2 n
.
J
(20)
Moreover,
M≤
(21)
4J 4
4
J
.
Now, let i, j ∈ {1, . . . , M }. We will show that the values of T1 over the block Vi × Vj differ
by at most 8. To see this, let a, c ∈ Vi and b, d ∈ Vj . Then
X
(T1 )ab − (T1 )cd = λi ui (a)ui (b) − λi ui (c)ui (d)
i<J
X
≤
|λi ||ui (a)ui (b) − ui (c)ui (b) + ui (c)ui (b) − ui (c)ui (d)|
i<J
≤
X
n|ui (b)||ui (a) − ui (c)| + n|ui (c)||ui (b) − ui (d)|
i<J
!
r
r
√
√
J −1/2 √ 3/2 −1/2
J −1/2 √ 3/2 −1/2
n
· 2 3/2 n
+n·2 2
n
· 2 3/2 n
≤J n·2 2
J
J
!
r
√
J −1/2 √ 3/2 −1/2
n
· 2 3/2 n
= 2J n · 2 2
J
< 8.
As a result, we can conclude that, if dij is the mean of the entries in the block Vi × Vj ,
then by the triangle inequality and Cauchy-Schwarz,
XX
∗
|vB
(T1 − dij I)vA | ≤
|(T1 )ab − dij )| |vA (a)||vB (b)|
a∈Vi b∈Vj
< 8kvA k1 kvB k1
≤ 8|Vi ||Vj |.
The last step follows
p from a basic vector norm inequality which gives kvA k1 ≤
|Vi | and kvB k1 ≤ |Vj |kvB k2 ≤ |Vj |.
P
Next we consider T2 and observe that tr(T22 ) = J≤j≤F (J) λ2i ≤ 3 n2 .
6
p
|Vi |kvA k2 ≤
So,
P
a,b∈V
(22)
|(T2 )ab |2 ≤ 3 n2 . Define Σ1 so that for every (i, j) 6∈ Σ1 ,
XX
|(T2 )ab |2 ≤ |Vi ||Vj |.
a∈Vi b∈Vj
Thus,
2
X
X XX
|Vi ||Vj | ≤
|(T2 )ab |2 ≤ 3 n2 .
(i,j)∈Σ1 a∈Vi b∈Vj
(i,j)∈Σ1
consequently,
X
|Vi ||Vj | ≤ n2 .
(i,j)∈Σ1
P
P
So, for any (i, j) 6∈ Σ1 , use the fact that a∈Vi b∈Vj |(T2 )ab |2 ≤ |Vi ||Vj | we use CauchySchwarz to obtain the following bound
2
X X
2
∗
(T2 )ab vA (a) · vB (b)
|vB T2 vA | = a∈Vi b∈Vj



XX
XX
≤
|(T2 )ab |2  
|vA (a)|2 |vB (b)|2 
a∈Vi b∈Vj
a∈Vi b∈Vj
= 2 |Vi ||Vj | kvA k22 kvB k22
∗
|vB
T2 vA | ≤ |Vi ||Vj |.
The last step follows from kvA k22 ≤ |Vi | and kvB k22 ≤ |Vj |.
Finally, we turn our attention to T3 . Let vA and p
vB be vectors such that kvA k22 , kvB k22 ≤ n.
Since the maximum eigenvalue of T3 is |λF (J) | ≤ n/ F (J), we have, first by Cauchy-Schwarz,
p
∗
|vB
T3 vA | ≤ |λF (J) |kvA k2 kvB k2 ≤ n2 / F (J).
Let Σ be the pairs (i, j) ∈ {0, 1, . . . , M } such that either (i, j) ∈ Σ1 , i = 0, j = 0 or
n
min(|Vi |, |Vj |) ≤ M
. Thus,
X
X
X
|Vi ||Vj | ≤
|Vi ||Vj | + 2|V0 ||V | + 2
|Vi ||V |
(i,j)∈Σ
(23)
|Vi |<n/M
(i,j)∈Σ1
n
≤ n2 + 2n · n + 2M n ≤ 5n2 .
M
If (i, j) 6∈ Σ, then for all A ⊆ Vi and B ⊆ Vj ,
∗
∗
∗
∗
|vB
(T − dij )vA | ≤ |vB
(T1 − dij )vA | + |vB
T2 vA | + |vB
T3 vA |
p
≤ 8|Vi ||Vj | + |Vi ||Vj | + n2 / F (J).
p
Finally, we want to establish that n2 / F (J) ≤ |Vi ||Vj | for (i, j) 6∈ Σ. This would
establish that
(24)
∗
|vB
(T − dij )vA | ≤ 10|Vi ||Vj |.
7
Because |Vi |, |Vj | ≥ n/M , it is sufficient to show that F (J) ≥ M 4 /6 because that would
verify that
3 n2
n2
p
≤
≤ |Vi ||Vj |.
M2
F (J)
J
By (21), M ≤ (4J 4 /4 ) . So the function that suffices is
4x
1 4x4
F (x) ≥ 6
.
4
Note that in this proof we use coefficients of 10 and 5 in (24) and (23), respectively.
We can, of course, choose /10 rather than but we chose these parameters to make the
computations somewhat more transparent.
2
4. Graph version
Lemma 3 (Szemerédi’s regularity lemma, spectral version). Let G = (V, E) be a graph on n
vertices and let > 0. Then there exists a partition V = V1 ∪ · · · ∪ VM for some M ≤ M ()
with the property that, for all pairs (i, j) ∈ {1, . . . , M }2 outside of an exceptional set Σ, one
has
(25)
|E(A, B) − dij |A||B|| |Vi ||Vj |
whenever A ⊂ Vi , B ⊂ Vj , for some real number dij , where
E(A, B) := |{(a, b) ∈ A × B : {a, b} ∈ E}|
is the number of edges between A and B. Furthermore, we have
X
(26)
|Vi ||Vj | |V |2 .
(i,j)∈Σ
Proof. Here we do the proof directly, even though this is a direct consequence of the matrix
version above.
Enumerate V = {1, . . . , n}. Let T be the incidence matrix of G and note that since T is
self-adjoint, it has an eigenvalue decomposition
n
X
T =
λi ui u∗i ,
i=1
for some orthonormal basis u1 , . . . , un of Cn (where the vectors are column vectors) and real
eigenvalues λ1 , . . . , λn ∈ R. We arrange them in decreasing order of magnitude
|λ1 | ≥ · · · ≥ |λn |.
In a self-adjoint matrix,
the trace of T 2 is the sum of the squares of the eigenvalues of
P
n
T and so tr(T 2 ) = j=1 |λj |2 . In addition, it is the sum of the degrees of degrees of the
graph, hence tr(T 2 ) = 2|E(G)| ≤ n2 . So we can bound the eigenvalues by observing that
P
iλ2i ≤ ij=1 λ2j ≤ n2 and so, for all i ∈ {1, . . . , n},
n
(27)
|λi | ≤ √ .
i
8
We will be given a function F : N → N that we will specify later. This function does
depend on (we suppress this in the notation) and it satisfies the inequality F (i) > i for all
integers i. We want to find an integer J for which
X
(28)
|λj |2 ≤ 3 n2 .
J≤j<F (J)
To do this, we consider the partition of {1, . . . , n} into intervals [F (k−1) (1), F (k) (1) − 1] from
k = 1, . . . , 1/3 where F (k) represents the k th composition of F with itself. Note that either
we find a J = F (k−1) (1) for which (28) is satisfied or the sum of |λj |2 for all j in some interval
is greater than 3 n2 . Since there are 1/3 intervals, this would contradict the n2 bound for
tr(T 2 ).
Thus, we have a partition of T into three matrices:
T = T1 + T2 + T3 ,
where T1 is the “structured” component
(29)
T1 :=
X
λi ui u∗i ,
i<J
and T2 is the “error” component
(30)
X
T2 :=
λi ui u∗i ,
J≤i<F (J)
and T3 is the “pseudorandom” component
(31)
T3 :=
X
λi ui u∗i .
i≥F (J)
We will partition the vertex set so that T1 is approximately constant on most clusters. The
number of such clusters will be OJ, (1). For each j < J we define a partition into clusters on
which entry ui (a complex number) varies by J n−1/2 . There is also an exceptional cluster of
size J n which comes from the vertices for which the entry of ui is large in magnitude. That
q
is, either its real or its imaginary part is larger than J n−1/2 in absolute value.
To see this, simply place a vertex into the exceptional cluster if the
qcorresponding entry of
n−1/2 . Since kui k2 =
q
1, this means there can be at most J n such entries. Partition the square of length 2 J n−1/2
ui has the absolute value of either its real or imaginary part at least
J
3/2
centered at the origin of the complex plane into subsquares of side length J 3/2 n−1/2 . There
q
2 3/2
2
are 2 J n−1/2 / J 3/2 n−1/2 = 4J 4 /4 such subsquares. Partition the vertices according
to where its corresponding entry of ui lies.
Take the union of all vertices in the exceptional clusters and the corresponding exceptional
cluster is of size at most (J − 1) · n
< n. For the rest of the vertices, we take the common
J
refinement of the partitions defined by each ui , i < J. This defines a partition of the vertex
set V = V0 + V1 + · · · + VM in which V0 is the exceptional set. For i = 1, . . . , M , the entries
9
over Vi of each of u1 , . . . , uJ−1 have magnitude at most
r
√
J −1/2
n
.
(32)
2·2
and differ in magnitude by at most
√ 3/2 −1/2
2 3/2 n
.
J
(33)
Moreover,
M≤
(34)
4J 4
4
J
.
Now, let i, j ∈ {1, . . . , M }. We will show that the values of T1 over the block Vi × Vj differ
by at most 8. To see this, let a, c ∈ Vi and b, d ∈ Vj . Then
X
λi ui (a)ui (b) − λi ui (c)ui (d)
(T1 )ab − (T1 )cd = i<J
X
|λi ||ui (a)ui (b) − ui (c)ui (b) + ui (c)ui (b) − ui (c)ui (d)|
≤
i<J
≤
X
n|ui (b)||ui (a) − ui (c)| + n|ui (c)||ui (b) − ui (d)|
i<J
!
r
r
√
√
J −1/2 √ 3/2 −1/2
J −1/2 √ 3/2 −1/2
≤J n·2 2
n
· 2 3/2 n
+n·2 2
n
· 2 3/2 n
J
J
!
r
√
J −1/2 √ 3/2 −1/2
= 2J n · 2 2
n
· 2 3/2 n
J
< 8.
As a result, we can conclude that, if dij is the mean of the entries in the block Vi × Vj ,
then by the triangle inequality and Cauchy-Schwarz,
XX
|1∗B (T1 − dij I)1A | ≤
|(T1 )ab − dij |
a∈A b∈B
< 8|A||B|
≤ 8|Vi ||Vj |.
P
Next we consider T2 and observe that tr(T22 ) = J≤j≤F (J) λ2i ≤ 3 n2 .
P
So, a,b∈V |(T2 )ab |2 ≤ 3 n2 . Define Σ1 so that for every (i, j) 6∈ Σ1 ,
XX
(35)
|(T2 )ab |2 ≤ |Vi ||Vj |.
a∈Vi b∈Vj
Thus,
2
X
(i,j)∈Σ1
|Vi ||Vj | ≤
X XX
(i,j)∈Σ1 a∈Vi b∈Vj
10
|(T2 )ab |2 ≤ 3 n2 .
consequently,
X
|Vi ||Vj | ≤ n2 .
(i,j)∈Σ1
P
P
So, for any (i, j) 6∈ Σ1 , use the fact that a∈Vi b∈Vj |(T2 )ab |2 ≤ |Vi ||Vj | we use CauchySchwarz to obtain the following bound
2
X X
2
∗
|1B T2 1A | = (T2 )ab a∈A b∈B
!
XX
≤
|(T2 )ab |2 |A||B|
=
|1∗B T2 1A |
a∈A b∈B
2 |Vi ||Vj | |A||B|
≤ |Vi ||Vj |.
Finally,
we turn our attention to T3 . Since the maximum eigenvalue of T3 is |λF (J) | ≤
p
n/ F (J), we have, first by Cauchy-Schwarz,
p
|1∗B T3 1A | ≤ |λF (J) ||A||B| ≤ n2 / F (J).
Let Σ be the pairs (i, j) ∈ {0, 1, . . . , M } such that either (i, j) ∈ Σ1 , i = 0, j = 0 or
n
min(|Vi |, |Vj |) ≤ M
. Thus,
X
X
X
|Vi ||Vj | ≤
|Vi ||Vj | + 2|V0 ||V | + 2
|Vi ||V |
(i,j)∈Σ
|Vi |<n/M
(i,j)∈Σ1
n
(36)
≤ n2 + 2n · n + 2M n ≤ 5n2 .
M
If (i, j) 6∈ Σ, then for all A ⊆ Vi and B ⊆ Vj ,
|1∗B (T − dij )1A | ≤ |1∗B (T1 − dij 1A | + |1∗B T2 1A | + |1∗B T3 1A |
p
≤ 8|Vi ||Vj | + |Vi ||Vj | + n2 / F (J).
p
Finally, we want to establish that n2 / F (J) ≤ |Vi ||Vj | for (i, j) 6∈ Σ. This would
establish that
|1∗B (T − dij )1A | ≤ 10|Vi ||Vj |.
(37)
Because |Vi |, |Vj | ≥ n/M , it is sufficient to show that F (J) ≥ M 4 /6 because that would
verify that
3 n2
n2
p
≤
≤ |Vi ||Vj |.
M2
F (J)
J
By (34), M ≤ (4J 4 /4 ) . So the function that suffices is
4x
1 4x4
F (x) ≥ 6
.
4
Note that in this proof we use coefficients of 10 and 5 in (37) and (36), respectively.
We can, of course, choose /10 rather than but we chose these parameters to make the
11
2
computations somewhat more transparent.
References
[1] Frieze, Alan; Kannan, Ravi The regularity lemma and approximation schemes for dense problems. 37th
Annual Symposium on Foundations of Computer Science (Burlington, VT, 1996), 1220, IEEE Comput.
Soc. Press, Los Alamitos, CA, 1996.
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