Density of 4-edge paths in graphs with fixed edge density

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by  Dániel T. Nagy

Density of 4-edge paths in graphs with fixed edge density
Dániel T. Nagy∗
January 7, 2016
6
Abstract
1
0
We investigate the number of 4-edge paths in graphs with a fixed number of vertices and edges.
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An asymptotically sharp upper bound is given to this quantity. The extremal construction is the
n
a quasi-star or the quasi-clique graph, depending on the edge density. An easy lower bound is also
J
proved. This answer resembles the classic theorem of Ahlswede and Katona about the maximal

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numberof2-edgepaths,andarecenttheoremofKenyon,Radin,RenandSadunaboutk-edgestars.

]
O
C 1 Introduction
.
h
t
a The aim of this paper is to asymptotically determine the maximal and minimal number of 4-edge paths
m
in graphs with fixed number of vertices and edges.
[

ThefirstresultofthiskindisduetoAhlswedeandKatona[1], whodescribedthegraphswithafixed
1
v number of vertices and edges containing the maximal number of 2-edge paths. To state this result, we
1
need some simple definitions.
1
2
1 Definition The quasi-clique Ce is a graph with n vertices and e edges, defined as follows. Take the
0 n
. unique representation
1 (cid:18) (cid:19)
a
0 e= +b 0≤b<a,
6 b
1
connect the first a vertices to each other, and connect the a+1-th vertex to the first b vertices
:
v
i
X Definition Thequasi-starSe isagraphwithnverticesandeedges,definedasfollows. Taketheunique
n
r representation
a
(cid:18) (cid:19) (cid:18) (cid:19)
n p
−e= +q, 0≤q <p,
2 2
connect the first n−p−1 vertices with every vertex, and connect the n−p-th vertex with the first n−q
vertices.
(n)−e
It is easy to see that Se is isomorphic to the complement of S 2 .
n n
Notation Thenumberof2-edgepathsinCe andSe isdenotedbyC(n,e)andS(n,e)respectively,while
n n
the number of k-edge stars is denoted by C (n,e) and S (n,e) respectively.
k k
∗EötvösLorándUniversity,Budapest. [email protected]
1
Theorem 1.1. (Ahlswede and Katona, 1978, [1]) Let G be a simple graph with n vertices and e
edges. Then the number of 2-edge paths in G is at most max(C(n,e), S(n,e)).
Furthermore,

S(n,e) if 0≤e≤ 1(cid:0)n(cid:1)− n,
max(C(n,e), S(n,e))= 2 2 2
C(n,e) if 1(cid:0)n(cid:1)+ n ≤e≤(cid:0)n(cid:1).
2 2 2 2
Roughlyspeaking,thistheoremstatesthatiftheedgedensityifsmallerthan 1,thenthequasi-staris
2
theextremalexample, whileforhigheredgedensitiesthequasi-cliquebecomesextremal. (Thetransition
between the two cases happens in a nontrivial way.)
Recently, Kenyon, Radin, Ren and Sadun proved a similar result for k-edge stars, using the notion of
graphons. Translating the result back to language of graphs, we get the following theorem:
Theorem 1.2. (Kenyon, Radin, Ren and Sadun, 2014, [7]) Let G bea simplegraphwith n vertices
and e edges, and let 2≤k ≤30. Then the number of k-edge stars in G is at most
max(Ck(n,e), Sk(n,e))(1+O(e−12)).
The theorem is conjectured to hold for all values of k. (The only thing left to prove this, is a
complicated extremal value problem.) Similarly to the case of the 2-edge path, C (n,e)<S (n,e) if the
k k
edge density is small, and S (n,e)<C (n,e) if it is greater. The point of transition depends on k.
k k
Now let us discuss three theorems with just one fixed parameter: the number edges. (So n is not
fixed.) We will start with a general theorem of Alon.
Theorem 1.3. (Alon, 1981, [2])LetN(G,H)denotethenumberofsubgraphsofGthatareisomorphic
to H. Assume that H is a single graph that has a spanning subgraph which is the vertex-disjoint union
of edges and cycles. Then
N(G,H)≤(1+O(e−21))N(Cne,H).
It means that for these graphs H, the asymptotically extremal example is always the quasi-clique.
Notethatthistheoremcanbeappliedinthecaseoffixednande,sincetheextremalexampleprovidedby
itisthequasi-clique. (Nomatterhowmanyverticeswearegiven,wejusthavetoconstructaquasi-clique
of e edges.)
Alsonotethatthistheoremprovidesupperboundsforallgraphswithaperfectmatching,(forexample
all paths with an odd number of edges) and Hamiltonian graphs (for example complete graphs). In the
case of the triangle graph K , the asymptotically best lower bound was proved by Razborov [9].
3
Theproblemoffindingthemaximalnumberof4-edgepathsingraphswitheedges(andanunlimited
number of vertices) was solved by Bollobás and Sarkar.
Theorem 1.4. (Bollobás and Sarkar, 2003, [5]) The number of 4-edge paths among graphs with
e edges is maximized by the graph that is obtained by taking the complete bipartite graph K , and
2,(cid:100)e/2(cid:101)
deleting an edge if e is odd.
2
BollobásandSarkaralsoprovedasymptoticresultsfor2k-edgepaths. [4]Theextremalexampleinthis
caseisthecompletebipartitegraphwithk verticesinoneside. For2k+1-edgepaths,theasymptotically
extremal example is the quasi-clique. It follows from Theorem 1.3, and is also proved in [4].
Alon had a conjecture for star-forests (vertex-disjoint union of stars), which was partially verified by
Füredi.
Conjecture (Alon, 1986, [3])LetH beastar-forest. Foranye>0,thegraphmaximizingthenumber
of subgraphs isomorphic to H among graphs with e edges is a star-forest.
Theorem 1.5. (Füredi, 1992, [6]) Let H be star-forest consisting of components with a ,a ,...a
1 2 t
edges. Assume that a > log (t+1) holds for all 1 ≤ i ≤ t. Let e be sufficiently large. Then the graph
i 2
maximizing the number of subgraphs isomorphic to H among those with e edges is a star-forest with t
components.
Consideringtheaboveresults,investigatingthenumberofthe4-edgepathsseemstobethe"natural"
choiceinthecaseoffixed(n,e). Inthispaper, anasymptoticupperboundwillbegiventothisquantity.
Similarly to the case of k-edge stars, the asymptotically extremal graphs are the quasi-stars and the
quasi-cliques. We will also prove an easy asymptotic lower bound.
2 Proof of the main result
Theorem 2.1. Let G be a simple graph with n vertices and e edges. Let c= 2e. (Then 0≤c≤1.) Let
n2
N denote the number of 4-edge-paths. Then
1 1 √ √
c4n5(1−O(n−1))≤N ≤ max((1− 1−c)2((c+1) 1−c+c), c25)n5.
2 2
Proof. LetN(cid:48)denotethenumberofthesequences{v ,v ,v ,v ,v }wherev are(notnecessarilydifferent
0 1 2 3 4 i
vertices) of G and v v ∈E(G) for i=1,...4. Here, we count every 4-edge path twice (there are two
i−1 i
directions). We also count some walks of length 4 with repeated vertices. However, the number of such
walks is only O(n4). Therefore 2N ≤N(cid:48) ≤2N +O(n4), so it suffices to prove
N(cid:48) √ √
c4 ≤ ≤max((1− 1−c)2((c+1) 1−c+c), c52).
n5
Let us note that N(cid:48) is often referred to as the homomorphism density of the 4-edge path P4 in G,
n5
and denoted by t(P4,G). (See [8] for an overview in the topic of graph homomorphisms.)
First, we prove the lower bound, which is much easier. If we want to select a 4-edge walk, we
can start by choosing v , then v and v (we have to pick them from N(v )), and finally v and v
2 1 3 2 0 4
(deg(v ) and deg(v ) possibilities). So we can write N(cid:48) as below, and estimate it by using twice that
1 3
m (cid:32) m (cid:33)2
(cid:88) 1 (cid:88)
x2 ≥ x holds for all real numbers.
i n i
i=1 i=1
 2   2
(cid:88) (cid:88) 1 (cid:88) (cid:88)
N(cid:48) =  deg(vi) ≥ n  deg(vi) =
v2∈V(G) vi∈N(v2) v2∈V(G) vi∈N(v2)
3
 2   22
n1  (cid:88) deg(vi)2 ≥ n1 n1  (cid:88) deg(vi)  = n1 (cid:18)n1 (cid:0)cn2(cid:1)2(cid:19)2 =c4n5.
vi∈V(G) vi∈V(G)
Now we move on to the proof of the upper bound. Let codeg(v,w) denote the number of common
neighbours of the vertices v and w. Note that
(cid:88)
N(cid:48) = deg(v )deg(v )codeg(v ,v ),
1 3 1 3
v1,v3∈V(G)
sinceafterfixingv andv ,wehavedeg(v )candidatesforv ,deg(v )candidatesforv ,andcodeg(v ,v )
1 3 1 0 3 4 1 3
candidates for v . Obviously, codeg(v ,v )≤min(deg(v ),deg(v )), therefore
2 1 3 1 3
(cid:88)
N(cid:48) ≤ deg(v )deg(v )min(deg(v ),deg(v )).
1 3 1 3
v1,v3∈V(G)
Definition Let G be a simple graph with n vertices labeled w ,w ,...w . A : [0,1)2 → [0,1] is the
1 2 n G
function that is 1 on all rectangles [i−1, i)×[j−1, j) where w w ∈E(G), and 0 elsewhere.
n n n n i j
Definition Let A := [0,1)2 → [0,1) be an integrable function satisfying A(x,y) = A(y,x) for all
0≤x,y <1. Then for all 0≤x<1 let
(cid:90) 1
(cid:96)(x)= A(x,y)dy
0
and let
(cid:90) 1(cid:90) 1
S(A)= (cid:96)(x)(cid:96)(y)min((cid:96)(x),(cid:96)(y))dxdy.
0 0
Note that A satisfies (cid:82)1(cid:82)1A (x,y)dxdy = c and A (x,y) = A (y,x). If x ∈ [i−1, i), then
G 0 0 G G G n n
(cid:96)(x)= deg(wi), so
n
1 (cid:88) N(cid:48)
S(A )= deg(w )deg(w )min(deg(w ),deg(w ))≥ .
G n5 i j i j n5
1≤i,j≤n
Definition Let 0 ≤ c ≤ 1. Then let A (c) : [0,1)2 → [0,1] be the function satisfying A (x,y) = 1 if
1 1
√
min(x,y)<1− 1−c and A (x,y)=0 otherwise. Let A (c):[0,1)2 →[0,1] be the function satisfying
1 2
√
(cid:82)1(cid:82)1
A (x,y)=1ifmax(x,y)< candA (x,y)=0otherwise. (Itiseasytoseethat A (x,y)dxdy =c
2 2 0 0 i
holds for i=1,2.) See Figure 1.
Theorem 2.1 will be an easy consequence of the following theorem.
Theorem 2.2. Let 0 ≤ c ≤ 1 and K ∈ N+ fixed numbers. Assume that A := [0,1)2 → [0,1) is a
(cid:82)1(cid:82)1
function satisfying A(x,y)dxdy = c and A(x,y) = A(y,x) for all 0 ≤ x,y < 1, and that there are
0 0
some numbers 0 = q < q < q < ··· < q = 1 such that A is constant on [q ,q )×[q ,q ) for all
0 1 2 K i−1 i j−1 j
1≤i,j ≤K. Then
√ √
S(A)≤max(S(A1(c)),S(A2(c)))=max((1− 1−c)2((c+1) 1−c+c), c52).
4
Figure 1: The functions A (c) and A (c).
1 2
Proof. We will use the following notations. I =[q ,q ), t =q −q =|I |, (cid:96) =(cid:96)(x) for any x∈I ,
i i−1 i i i i−1 i i i
A isthevalueofAintherectangleI ×I . Wewillrefertothesetsoftheform[0,1)×I andI ×[0,1)
i,j i j i i
as rows and columns respectively.
The function S is continuous on a compact set defined by the conditions, so its maximum is attained
for some A. Let A be a function maximizing S, and let
(cid:90) 1(cid:90) 1(cid:90) 1(cid:90) 1
(cid:88)
T(A)= |A(x ,y )−A(x ,y )|dx dy dx dy = t t t t |A −A |.
1 1 2 2 1 1 2 2 a1 b1 a2 b2 a1,b1 a2,b2
0 0 0 0 1≤a1,a2,b1,b2≤K
Byasimilarcompactnessargument,theminimumofT isalsoattainedforsomeA(amongthosethat
maximize S). Such an A can not have four rectangles I ×I ,I ×I ,I ×I and I ×I satisfying
i1 j1 i1 j2 i2 j1 i2 j2
A <A and A >A .
i1,j1 i2,j1 i1,j2 i2,j2
Forsomeε>0,replacethevaluesA ,A ,A andA byA + ε ,A − ε ,A −
i1,j1 i2,j1 i1,j2 i2,j2 i1,j1 ti1tj1 i2,j1 ti2tj1 i1,j2
ε and A + ε respectively. By choosing a small enough ε, the value of A remains greater in
ti1tj2 i2,j2 ti2tj2
I ×I andI ×I thaninI ×I andI ×I respectively. Notethatsuchachangedoesnotchange
i2 j1 i1 j2 i1 j1 i2 j2
the values (cid:96)(x), therefore not changing S(A). (To see that, take a line that intersects two of the four
rectangles where the value of A changes. It increases in one of them, while decreasing in the other one.
This results in a 0 net change in the integral of A over that line, since if one of the rectangles intersect
the line in a segment λ times as long as the other one, then its area is λ times greater, so the change in
the value of A is λ times smaller.)
NowweshowthatthevalueT(A)decreasesduringthistransformation. T(A)isthesumofdifferences
betweenthevaluesA ,weightedwiththeareasoftheserectangles. AssumethatthevalueofAisgreater
i,j
inr thaninr fortworectanglesr andr . IfwedecreasethevalueofAinarectangler with ε ,
1 2 1 2 1 Area(r1)
and increase it in r with ε for a small enough ε, then T(A) decreases. To see that, note that
2 Area(r2)
(cid:12) (cid:18) (cid:19)(cid:12)
(cid:12) ε ε (cid:12)
Area(r1)Area(r2)|Ar1 −Ar2|>Area(r1)Area(r2)(cid:12)(cid:12)Ar1 − Area(r ) − Ar2 + Area(r ) (cid:12)(cid:12)
1 2
and for any rectangle r (cid:54)∈{r ,r }
3 1 2
Area(r )Area(r )|A −A |+Area(r )Area(r )|A −A |≥
1 3 r1 r3 2 3 r2 r3
(cid:12) (cid:12) (cid:12) (cid:12)
(cid:12) ε (cid:12) (cid:12) ε (cid:12)
Area(r1)Area(r3)(cid:12)(cid:12)Ar1 − Area(r ) −Ar3(cid:12)(cid:12)+Area(r2)Area(r3)(cid:12)(cid:12)Ar2 + Area(r ) −Ar3(cid:12)(cid:12).
1 2
5
Applyingthisto(r ,r )=(I ×I ,I ×I )and(r ,r )=(I ×I ,I ×I )thedesiredresultfollows.
1 2 i2 j1 i1 j1 1 2 i1 j2 i2 j2
The symmetry of A can be ruined by this transformation, but replacing A(x,y) by A(x,y)+A(y,x) for
2
all 0 ≤ x,y ≤ 1 fixes this while not increasing T(A) and not changing S(A). (The fact that T(A) does
not increase can be verified by the above calculation dealing with the decrease of A in a high-valued
rectangle and the its increase in a lower valued one.)
Rearrange the intervals I such that (cid:96) ≥ (cid:96) ≥ ··· ≥ (cid:96) . The property we just proved for the
i 1 2 K
rectangles implies that for any four rectangles of the form I ×I ,I ×I ,I ×I and I ×I , we
i1 j1 i1 j2 i2 j1 i2 j2
have
A <A ⇒A ≤A .
i1,j1 i2,j1 i1,j2 i2,j2
Now we prove that A is decreasing in both variables. (Since A(x,y) = A(y,x), it is enough to show
that for one variable.) Assume to the contrary that for some i <i and j we have A <A . Then
1 2 i1,j i2,j
for all 1≤p≤K we have A ≤A . It results in (cid:96) <(cid:96) , a contradiction.
i1,p i2,p i1 i2
This decreasing property implies that if (cid:96) =(cid:96) then A is identical in I ×[0,1) and I ×[0,1) so
i i+1 i i+1
we can merge all such intervals and assume that (cid:96) >(cid:96) >···>(cid:96) for some k ≤K.
1 2 k
Note that S(A) can be expressed as
k k
(cid:88)(cid:88)
S(A)= t t (cid:96) (cid:96) min((cid:96) ,(cid:96) ). (1)
i j i j i j
i=1j=1
Consider an A that meets the theorem’s requirements, maximizes S(A) and is decreasing in both
variables. We state that there can not be two rectangles in the same row (or column) where the value
of A is neither 0 nor 1. Assume that for some 1 ≤ a < b ≤ k and 1 ≤ p ≤ k we have 0 < A < 1 and
a,p
0<A <1. Pick some (cid:15)∈R and change A and A to A + ε while changing A and A to
b,p a,p p,a a,p tatp b,p p,b
A − ε . This transformation changes only two (cid:96) values: (cid:96) becomes (cid:96) + ε and (cid:96) becomes (cid:96) − ε.
b,p tbtp a a ta b b tb
If |ε| is smallenough then 0<A + ε ,A − ε <1 and theorder ofthe (cid:96) values is preserved. Now
a,p tatp b,p tbtp
we show that S(A) is a strictly convex function of ε in a neighborhood of 0. Consider the k2 terms in
the expression (1). The terms including other terms than a and b are obviously convex functions of ε,
since ε appears at a power of at most 2 in them, and it has a positive coefficient when it has power 2.
So the terms of S(A) corresponding to pairs of indices other than (a,a),(a,b),(b,a) and (b,b) are convex
functions of ε. All we have to prove is that the sum of the terms corresponding to these four pairs is
strictly convex at ε=0.
(cid:18) ε (cid:19)3 (cid:18) ε(cid:19)3 (cid:18) ε (cid:19)(cid:18) ε(cid:19)2
t2 (cid:96) + +t2 (cid:96) − +2t t (cid:96) + (cid:96) − .
a a t b b t a b a t b t
a b a b
Differentiating twice with respect to ε and substituting ε=0, we get
4t (cid:96)
6(cid:96) −2(cid:96) + a a.
a b t
b
It is positive, because a < b implies (cid:96) > (cid:96) . Therefore S(A) is a strictly convex function of ε in a
a b
neighborhood of 0, so can not have a maximum at 0. This proves that the A under investigation has at
most one rectangle in every row and column with a value different from 0 or 1, as depicted in Figure 2.
Now we will prove that actually there are no such rectangles at all. Since A is decreasing in both
variables, each row (and column) starts with some 1-valued rectangles, then it might include a single
6
Figure 2: Example of a function considered at this point in the proof
rectangle with value between 0 and 1, then it contains 0-valued rectangles. (Of course, a row or column
notnecessarilycontainsallthreetypesofrectangles.) IfAhasasinglerectangleofsizet×twithnonzero
value, then A’s value is tc2 there. (So tc2 ≤1.) Then S(A)=t2·(t· tc2)3 = ct3 ≤c52. If there are multiple
nonzero-valued rectangles, then A =1.
1,1
Assume that 0 < λ = A = A < 1 for some j. (Then j ∈ {k−1,k}). We will show that it is
1,j j,1
possible to modify A to increase S(A), so this case is not possible. (From now on we will modify the
lengths of the intervals too, not just the value of A in the rectangles.) Divide the interval I into two
j
intervals I and I of length t =λ·t and t =(1−λ)·t respectively. Then divide I ×I into two
j(cid:48) j(cid:48)(cid:48) j(cid:48) j j(cid:48)(cid:48) j 1 j
rectangles of size t ×t and t ×t , and set A to be 1 and 0 respectively in them. Modify A in I ×I
1 j(cid:48) 1 j(cid:48)(cid:48) j 1
similarly to keep A symmetric.
After this modification, we will get (cid:96) = (cid:96)j and (cid:96) = 0. This means that the terms with j(cid:48)(cid:48) can be
j(cid:48) λ j(cid:48)(cid:48)
ignored in (1). The only terms to change in (1) are t becoming t = λ·t and (cid:96) becoming (cid:96) = (cid:96)j.
j j(cid:48) j j j(cid:48) λ
Since the power of (cid:96) is not smaller than the power t in any term of (1), and greater than it in t2·(cid:96)3,
j j j j
the value of S(A) increases by this modification. So we can assume that A takes only 0 and 1 values in
I ×[0,1) and [0,1)×I . To show that noninteger values are not possible in the other places, we need a
1 1
technical lemma.
Notethatthevariablest and(cid:96) appearinginthefollowinglemmashouldbeconsideredrealnumbers
i i
with no connection to any function A:[0,1)2 →[0,1], but the same notations are used, since the lemma
will be applied in such settings.
Lemma 2.3. Let t ,t ,...,t be positive reals and let (cid:96) > (cid:96) > ··· > (cid:96) ≥ 0. Assume that there is
1 2 k 1 2 k
a neighborhood H of t such that for some α ∈ {β −1,β +1} and x ∈ H we can replace the numbers
β
t ,t ,(cid:96) by t (x)=t +t −x, t (x)=x and (cid:96) (x)=(cid:96) +tβ((cid:96) −(cid:96) ) respectively, without changing the
α β β α α β β β α x β α
nonnegativity of the variables and preserving the order of the (cid:96)’s. The other variables are left unchanged:
t (x)=t , if i(cid:54)∈{α,β} and (cid:96) (x)=(cid:96) , if i(cid:54)=β. (Roughly speaking, this transformation preserves the sum
i i i i
k
(cid:88)
c= t (x)(cid:96) (x), while changing only three of the values: two neighboring t’s and the (cid:96) corresponding to
i i
i=1
one of them.) Then the function
k k
(cid:88)(cid:88)
S(x)= t (x)t (x)(cid:96) (x)(cid:96) (x)min((cid:96) (x),(cid:96) (x)) (2)
i j i j i j
i=1j=1
is strictly convex at x=t . Therefore it has no maximum there.
β
7
Proof. Consider the formula (2) and select all the terms depending on x. We can ignore the terms where
one of the indices is α or β, and the other is greater than max(α,β) because
(cid:18) (cid:19)
t
(t +t −x)t (cid:96) (cid:96)2+xt (cid:96) + β((cid:96) −(cid:96) ) (cid:96)2 =t t (cid:96) (cid:96)2+t t (cid:96) (cid:96)2.
α β p α p p α x β α p α p α p β p β p
does not depend on x. The sum of the other terms depending on x can be written as 2S (x)+S (x).
1 2
Here S (x) is the sum of the terms corresponding to pairs of indices where one of the elements is α or
1
β and the other one in smaller than α and β. S (x) denotes the sum of the terms corresponding to the
2
pairs of indices (α,α),(α,β),(β,α) and (β,β).
S (x)=(cid:32)(t +t −x)·(cid:96)2 +x·(cid:18)(cid:96) + tβ((cid:96) −(cid:96) )(cid:19)2(cid:33)·min((cid:88)α,β)−1t (cid:96) ,
1 α β α α x β α p p
p=1
(cid:18) t (cid:19)3
S (x)=(t +t −x)2·(cid:96)3 +x2· (cid:96) + β((cid:96) −(cid:96) ) +
2 α β α α x β α
(cid:18) (cid:19) (cid:18) (cid:19)
t t
2(t +t −x)x(cid:96) (cid:96) + β((cid:96) −(cid:96) ) min (cid:96) ,(cid:96) + β((cid:96) −(cid:96) ) .
α β α α x β α α α x β α
We have to show that S (x) and S (x) are strictly convex at x=t , therefore S(x) does not takes its
1 2 β
maximum there. We will start with S (x). We can disregard the constant factor at the right, as it does
1
not change convexity. The left factor can be expressed as λ x2+λ x+λ +λ x−1. Since λ =(cid:96)2 >0
2 1 0 −1 2 α
and λ =t ((cid:96) −(cid:96) )2 >0, S is strictly convex.
−1 β β α 1
Now we consider S (x). First, assume that β =α+1, and therefore (cid:96) >(cid:96) . In this case we have
2 α β
(cid:18) t (cid:19)3 (cid:18) t (cid:19)2
S (x)=(t +t −x)2·(cid:96)3 +x2· (cid:96) + β((cid:96) −(cid:96) ) +2(t +t −x)x(cid:96) (cid:96) + β((cid:96) −(cid:96) ) .
2 α β α α x β α α β α α x β α
Differentiating twice by x and setting x=t we get
β
2t−1((cid:96) −(cid:96) )2(cid:0)((cid:96) +(cid:96) )t +2(cid:96) t (cid:1)>0.
β β α α β β α α
If β =α−1, and therefore (cid:96) <(cid:96) , we have
α β
(cid:18) t (cid:19)3 (cid:18) t (cid:19)
S (x)=(t +t −x)2·(cid:96)3 +x2· (cid:96) + β((cid:96) −(cid:96) ) +2(t +t −x)x(cid:96)2 (cid:96) + β((cid:96) −(cid:96) ) .
2 α β α α x β α α β α α x β α
Differentiating twice by x and setting x=t we get
β
2((cid:96) −(cid:96) )3 >0.
β α
In both cases, S (x) is strictly convex at x=t . This concludes the proof of the lemma.
2 β
Nowwecancontinuetheproofofthetheorem. AssumethatAisnotentirely0-1valued. LetI ×[0,1)
p
be the first column containing a rectangle with a value different from 0 and 1. We already proved that
p≥2. LetI ×I theuniquerectangleinthep-thcolumnwith0<A <1. SinceA =A ,weknow
p q p,q p,q q,p
that p ≤ q. We will show that A admits the type of transformation described in Lemma 2.3, therefore
doesnotmaximizeS(A). Wewilldescribetransformationsineachcasethatchangeonlytwoneighboring
t values and the (cid:96) corresponding to one of them.
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Casea: First,assumethatp<q <kandA =1. Thentherows[0,1)×I and[0,1)×I differ
p−1,q+1 q q+1
onlyinthep-thcolumn. WecanmovethepointseparatingtheintervalsI andI , keepingA =0
q q+1 p,q+1
while adjusting A such that the integral of A over the whole square remains unchanged. We apply
p,q
these changes to the other side of the main diagonal to keep A symmetric. During this transformation
the only (cid:96) value to change is (cid:96) . So Lemma 2.3 can be applied with α=q+1,β =q.
q
Figure 3: Illustration for Case a (left) and Case b (right).
Case b: Now assume that p<q =k or p<q <k and A =0. Then the columns I ×[0,1)
p−1,q+1 p−1
and I ×[0,1) differ only in the q-th row. We can move the point separating the intervals I and
p p−1
I , keeping A = 1 while adjusting A such that the integral of A over the whole square stays the
p p−1,q p,q
same. We apply these changes to the other side of the main diagonal to keep A symmetric. During this
transformation the only (cid:96) value to change is (cid:96) . So Lemma 2.3 can be applied with α=p−1,β =p.
p
Figure 4: Illustration for Case c (left) and Case d (right).
Case c: Assume that p = q < k and A = 1. Then the rows [0,1)×I and [0,1)×I differ
p−1,p+1 p p+1
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onlyinthep-thcolumn. WecanmovethepointseparatingtheintervalsI andI , keepingA =0
p p+1 p,p+1
while adjusting A such that the integral of A over the whole square stays the same. (We apply these
p,p
changestothetheintervalsdefiningtherowsandcolumnssimultaneouslytokeepAsymmetric.) During
this transformation the only (cid:96) value to change is (cid:96) . So Lemma 2.3 can be applied with α=p+1,β =p.
p
Case d: Now assume that p=q =k or p=q <k and A =0. Then the columns I ×[0,1)
p−1,p+1 p−1
and I ×[0,1) differ only in the p-th row. We can move the point separating the intervals I and I ,
p p−1 p
keepingA =1whileadjustingA suchthattheintegralofAoverthewholesquarestaysthesame.
p−1,p p,p
(We apply these changes to the the intervals defining the rows and columns simultaneously to keep A
symmetric.) During this transformation the only (cid:96) value to change is (cid:96) . So Lemma 2.3 can be applied
p
with α=p−1,β =p.
Withthis,wehavecoveredallthepossibilities. Fromnowon,wecanassumethatA:[0,1)2 →{0,1}.
Since A is 0-1 valued and decreasing in both variables, there exists some k(cid:48) ∈ {k−1,k} such that k
of the (cid:96) values is positive and A = 1 if and only if i+j ≤ k(cid:48)+1. Now we will show that A can not
i,j
maximize S(A) if k(cid:48) ≥4.
Assume that k(cid:48) ≥ 4. It is possible to move the point separating the intervals I and I , while
k(cid:48)−1 k(cid:48)
keeping (cid:96) unchanged and adjusting (cid:96) to keep the integral over the whole square unchanged. When
k(cid:48)−1 k(cid:48)
thesechangesareappliedtotheothersideofthemaindiagonaltopreservethesymmetry, wesethatthe
point separating I and I moves, (cid:96) remains unchanged, but (cid:96) changes. During this transformation,
1 2 1 2
the following values change (see Figure 5):
t →x,
k(cid:48)
t →t +t −x,
k(cid:48)−1 k(cid:48)−1 k(cid:48)
(cid:96)k(cid:48) →(cid:96)k(cid:48)−1+ txk(cid:48)((cid:96)k(cid:48) −(cid:96)k(cid:48)−1),
(cid:96) →(cid:96) −x,
2 1
t → t2((cid:96)1−(cid:96)2),
2 x
t →t +t − t2((cid:96)1−(cid:96)2).
1 1 2 x
Figure 5: The case k(cid:48) ≥4
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