PROOF OF A CONJECTURE OF V. NIKIFOROV
TAMÁS TERPAI
Abstract. Using analytical tools, we prove that for any simple graph G on
n vertices and its complement G the inequality µ(G) + µ(G) ≤ 43 n − 1 holds,
where µ(G) and µ(G) denote the greatest eigenvalue of adjacency matrix of
the graphs G and G respectively.
1. Introduction and statement of results
Consider a simple graph G = (V, E) on the vertex set V = v1 , . . . , vn . Its
adjacency matrix AdjG is defined by
(
1, if (vi , vj ) ∈ E,
(AdjG )ij =
0
otherwise.
The eigenvalues of this matrix are important invariants of the underlying graph.
For example, define
µ(G) = max{λ : λ is an eigenvalue of AdjG } =
= max{|λ| : λ is an eigenvalue of AdjG },
the maximal eigenvalue of AdjG . It can also be obtained as the ℓ2 -norm of AdjG :
µ(G) = kAdjG k2 = max{kAdjG vk2 : kvk2 = 1}.
This value (together with the information of whether the graph is bipartite or not)
describes the asymptotic behaviour of the number of closed walks of a given length
as this length tends to infinity and is related to the mixing properties of the random
walk on G.
It is easy to see that if one considers a simple graph G and its complement G,
then AdjG + AdjG is a matrix with norm n − 1, so µ(G) + µ(G) ≥ n − 1 holds for all
choices of G (and this inequality is sharp e.g. for G the empty graph). In her
√ thesis,
Nosal [5] shows that if G and G are complements, then µ(G) + µ(G) ≤ 2n − 1.
Unlike the trivial lower bound, this value is not optimal: Nikiforov [4] improved it
to
√
µ(G) + µ(G) ≤ ( 2 − 10−7 )n
and conjectured an upper bound of 43 n + O(1), which is attained by the clique
√
of size roughly 32 n. Csı́kvári [2] enhanced the known upper bound to 1+2 3 n and
showed that in order to get a bound of the form cn + O(1), an asymptotic result
would suffice:
2000 Mathematics Subject Classification. 05C35.
Key words and phrases. Maximal eigenvalue, Nordhaus-Gaddum type problem.
1
2
T. TERPAI
Theorem 1 ([2, Theorem 1.2]). If for any simple graph G on n vertices and its
complement G the inequality
holds, then so does
µ(G) + µ(G) ≤ cn + o(n)
µ(G) + µ(G) ≤ cn − 1.
In this paper, we prove exactly such an upper bound:
Theorem 2. For any simple graph G on n vertices and its complement G
4
µ(G) + µ(G) ≤ n + o(n)
3
As a corollary, Theorems 1 and 2 yield the conjecture of Nikiforov.
2. Proof
2.1. Translation to analysis. In order to obtain the desired asymptotic bound,
we will consider graphs as approximations of graphons, dense graph limit objects
in the sense of Lovász and Szegedy [3]. Let K ⊆ [0, 1] × [0, 1] be a measurable set.
Define the linear operator AK : L 2 [0, 1] → L 2 [0, 1] as follows:
Z
f (t)dt for any f ∈ L 2 [0, 1].
(1)
AK f (x) =
(x,t)∈K
The definition makes sense pointwise, since on the probability measure space [0, 1]
we have kf k1 ≤ kf k2 < ∞. The resulting function AK f lies in L 2 [0, 1], because
the quadratic form
Z
Z
AK (f )(x)ḡ(x)dy
f (y)ḡ(x)dxdy =
QK (f, g) =
(x,y)∈K
[0,1]
(which we will denote by f Kg for brevity) is bounded by kf k1 kgk1 ≤ kf k2 kgk2
for any g ∈ L 2 [0, 1]. The integration kernel is a bounded function in a compact
region, in particular it is in L 2 ([0, 1]2 ), hence the defined operator AK is HilbertSchmidt (cf. [1, Prop. 12.1.1]). In particular, such an operator AK is compact, has
a discrete spectrum consisting entirely of eigenvalues with the possible exception of
the 0, and its operator norm coincides with the spectral norm, that is,
kAK k2 = max{|λ| : λ ∈ C is an eigenvalue of AK }.
If K is symmetric with respect to the diagonal (this corresponds to our choice of
non-oriented graphs), then the defining equation (1) shows that AK is self-adjoint.
This implies that all of its eigenvalues are real. Also, if f is an eigenfunction
corresponding to the eigenvalue ±kAK k2 , then |f | has to be an eigenfunction corresponding to the eigenvalue kAK k2 :
kAK k2 |f |(x) = |kAK k2 f | (x) = |AK f | (x) =
Z
Z
|f (t)| dt = AK |f |(x)
f (t)dt ≤
=
(x,t)∈K
(x,t)∈K
(note that the inequality cannot be strict on a set of positive measure, because otherwise |f |K|f | would exceed kAK k2 kf k22 ). From now on, whenever an eigenfunction
PROOF OF A CONJECTURE OF V. NIKIFOROV
3
f of an AK corresponding to the eigenvalue kAK k2 is chosen, we implicitly assume
that f is nonnegative.
Given a graph G on n vertices, which is simple apart from possible single loop
edges, we can define a set K = KG corresponding to this graph:
[
j−1 j
i−1 i
×
.
,
,
KG =
n n
n n
1≤i,j≤n
(vi ,vj ) is an edge in G
Lemma 1. µ(G) = nkAKG k2 .
Proof of Lemma 1. Take an ℓ2 -normalized eigenvector (a1 , . . . , an ) of the adjacency
matrix AdjG of G corresponding to µ(G). Then the function
√
j−1 j
f (x) = naj for x ∈
,
n n
is an eigenfunction of AKG corresponding to an eigenvalue of µ(G)
n , hence kAKG k2 ≥
µ(G)
n . On the other hand, assume that f is the eigenfunction of AKG with the
maximal eigenvalue:
AKG f
f=
.
kAKG
The right-hand side of this equation
function, constant (up to a set of
is a jstep
measure 0) on each of the intervals j−1
,
due
to the definition of AKG . Theren
n
fore so is f , there
are
values
a
,
.
.
.
,
a
such
that
f (x) = aj for almost all x in the
1
n
j−1 j
interval n , n . For such an f , the condition of being an eigenfunction of AKG
corresponding to the eigenvalue λ is equivalent to (a1 , . . . , an ) being an eigenfunction (with ℓ2 norm n) of AdjG corresponding to the eigenvalue nλ. Hence we have
the opposite inequality µ(G) ≥ nkAKG k2 as well.
We utilize this analytic viewpoint to show that there exists a measurable set K ⊆
[0, 1]2 such that for K and its complement K̄ = [0, 1]2 \ K the sum kAK k2 + kAK̄ k2
achieves its supremum on the set of all measurable subsets of [0, 1]2 . This lets us
evaluate this supremum:
Theorem 3. For any measurable symmetric set K ⊆ [0, 1]2 , the norm kAK k2 is
at most 43 .
Combining Lemma 1 and Theorem 3 immediately gives Theorem 2; the rest of
the paper is dedicated to the proof of Theorem 3.
2.2. Existence of optimum. Denote by k · k the cut norm on L 1 [0, 1]2 :


 Z

kW k = sup W (x, y)dxdy : S, T ⊆ [0, 1]


S×T
and let δ be the cut semidistance on the space of all measurable subsets of [0, 1]2 ,
that is,
δ (U, V ) = inf kχU − χV ◦(ϕ,ϕ) k : ϕ is a measure-preserving bijection of [0, 1] .
4
T. TERPAI
Denote by X the factor space of the space of measurable subsets of [0, 1]2 by the
equivalence relation U Ṽ ↔ δ (U, V ) = 0. On X , the function δ is a distance and
[3, Theorem 5.1] proves that this distance makes X compact.
Measure-preserving reparametrization of [0, 1] does not change the norm of an
operator AK , so we can consider the function K 7→ kAK k2 on the space X . This
function is continuous with respect to δ since (see [3]) the cut norm of AK is
within absolute constant factors to the L 1 → L ∞ norm of AK , and this latter is
continuous with respect to the L 2 norm (considered as the L 2 → L ∞ norm) due
to the finiteness of the measure on [0, 1]2 :
kAK f k22 ≤ kAK f k∞ ≤ kAK kL 1 →L ∞ kf k1 ≤ kAK kL 1 →L ∞ kf k2 .
Therefore the function K 7→ kAK k2 achieves its supremum on X , proving our claim
that there exists an optimal set K.
From now on, we will use the notations µ = µ(K) := kAK k2 and µ̄ = µ̄(K) :=
µ(K̄) = kAK̄ k2 whenever the choice K ∈ K is clear from the context. A nonnegative
normalized eigenfunction of AK (respectively, AK̄ ) corresponding to µ (respectively,
µ̄) will be denoted by f (respectively, f¯).
Our further investigation is aimed at placing restrictions on any K that achieves
the maximum of µ + µ̄.
2.3. Monotonicity of f and f¯. Assume that K maximizes the sum µ + µ̄, and
choose nonnegative eigenfunctions f and f¯ accordingly.
Lemma 2. If K is optimal with respect to µ + µ̄, then {(x, y) ∈ [0, 1]2 |f (x)f (y) >
f¯(x)f¯(y)} ⊆ K ⊆ {(x, y) ∈ [0, 1]2 |f (x)f (y) ≥ f¯(x)f¯(y)}. Also, the set {(x, y) ∈
[0, 1]2 |f (x)f (y) = f¯(x)f¯(y)} has measure 0.
Proof of Lemma 2. Change K to K̃ = {(x, y) ∈ [0, 1]2 : f (x)f (y) > f¯(x)f¯(y)}.
Then we have
Z
¯
¯
¯
(2) µ(K̃) + µ̄(K̃) ≥ f K̃f + f K̃ f = f (x)f (y)χK̃ + f¯(x)f¯(y)χK̃¯ dxdy ≥
Z
≥ f (x)f (y)χK + f¯(x)f¯(y)χK̄ dxdy = f Kf + f¯K̄ f¯ = µ(K) + µ̄(K),
(3)
contradicting the optimality of K unless the second inequality is satisfied with
equality, that is, K contains K̃. Repeating the argument with K̃ = {(x, y) ∈
[0, 1]2 : f (x)f (y) ≥ f¯(x)f¯(y)} gives the other containment.
In order for the first inequality of inequality (2) to hold, the function f has to
¯ If
be an eigenfunction of K̃ and the function f¯ has to be an eigenfunction of K̃.
the set {(x, y) ∈ [0, 1]2 : f (x)f (y) = f¯(x)f¯(y)} had positive measure, we could
modify K̃ in a small off-diagonal subrectangle of [0, 1]2 and its mirror image to
invalidate the eigenfunction condition AK̃ f = µ(K̃)f (note that the unaffected
parts of [0, 1] enforce the equality µ(K̃) = µ(K)). This proves our second claim by
contradiction.
We can now obtain an analogue of the restriction to threshold graphs of the
Kelmans transformation in [2]. Define the function R : [0, 1] → [0, +∞] as follows:
f (x)
R(x) = ¯ ,
f (x)
PROOF OF A CONJECTURE OF V. NIKIFOROV
5
and let ϕ : [0, 1] → [0, 1] be the following measure-preserving function:
ϕ(x) = k{t ∈ [0, 1] : either R(t) > R(x) or R(t) = R(x) and t < x}k.
While ϕ is not bijective, it is 1-to-1 up to a set of measure 0, and R ◦ ϕ−1 (where
defined) is monotone decreasing. Applying a bijection ψ that differs from ϕ−1 only
on a set of measure 0 to functions f and f¯ and applying (ψ, ψ) to K we still have
to obtain a maximal value of µ + µ̄; indeed, ψ is measure-preserving. For these
(x)
is almost everywhere monotone decreasing, so for
modified functions the ratio ff¯(x)
almost all x ∈ [0, 1] and almost all y ∈ [0, 1] such that x ≥ y we have the inequality
f (y)
f (x)
≤ ¯ .
f¯(x)
f (y)
In particular, if for some t ∈ [0, 1] we have
f (x)f (t) ≥ f¯(x)f¯(t),
so that (x, t) ∈ K by Lemma 2, then we also have
f (y)f (t) ≥
f (x)f¯(y)f (t)
≥ f¯(y)f¯(t)
f¯(x)
and consequently (y, t) ∈ K. That is, the set K satisfies (up to a set of measure 0)
the property
(x, y) ∈ K, x ≥ x′ ≥ 0, y ≥ y ′ ≥ 0 ⇒ (x′ , y ′ ) ∈ K,
which we call monotonicity. The complement K̄ satisifies a similar property
(x, y) ∈ K̄, x ≤ x′ ≤ 1, y ≤ y ′ ≤ 1 ⇒ (x′ , y ′ ) ∈ K̄.
These two properties correspond to each other by reflection of [0, 1], so we will
assume the first one in the proofs and omit the symmetric statements for K̄.
Note that monotonicity of K implies that f is monotone decreasing: if x ≤ y,
then (x, t) ∈ K implies (y, t) ∈ K and we have
Z
Z
f (t)dt = AK f (y) = µf (y).
f (t)dt ≥
µf (x) = AK f (x) =
(x,t)∈K
(y,t)∈K
The same argument shows that f¯ has to be monotone increasing.
2.4. Stability under small deformations. We call f and f¯ optimal (with respect
to µ + µ̄) if the choice of K defined by Lemma 2 makes f an eigenfunction of AK ,
the function f˜ is an eigenfunction of AK̄ and µ(K) + µ(K̄) is the maximal possible.
The partial results above, for example, show that optimal f and f¯ can be chosen
to nonnegative and monotone.
We can get further information about optimal f and f¯ by checking that (infinitesimally) rescaling their domain [0, 1] can not increase the sum µ + µ̄.
Lemma 3. If f and f¯ are optimal with respect to µ+µ̄, then µf 2 (x)+µ̄f¯2 (x) = µ+µ̄
for almost all x ∈ [0, 1].
6
T. TERPAI
Proof of Lemma 3. Take an orientation-preserving homeomorphism φ : [0, 1] →
[0, 1] and consider the reparametrizations fnew (x) = f (φ(x))φ′ (x) and f¯new (x) =
f¯(φ(x))φ′ (x). Then
Z
Z
Z
f (u)2 φ′ (φ−1 (u))du,
f (φ(x))2 φ′ (x)2 dx =
fnew (x)2 dx =
u=φ(x)
[0,1]
[0,1]
[0,1]
and similarly for kf¯new k22 . Let Knew be the preimage of K under φ, then
Z
f (φ(x))f (φ(y))φ′ (x)φ′ (y)χ{(φ(x),φ(y))∈K}dxdy =
fnew Knew fnew =
[0,1]2
Z
=
u=φ(x)
v=φ(y) [0,1]2
f (u)f (v)χ{(u,v)∈K} dudv = f Kf = µ
and in the same way f¯new K̄new f¯new = f¯K̄ f¯ = µ̄. We will consider a family of
homeomorphisms φ, which stretch one interval I by 1 + ε + o(ε), shrink another
interval J of the same length as I by 1 − ε + o(ε) and only shift the components of
the complement of these intervals. Then we have
Z
Z
Z
Z
f (x)2 dx + f (x)2 (ε + o(ε))dx − f (x)2 (ε + o(ε))dx =
fnew (x)2 dx =
I
[0,1]
[0,1]
J


Z
Z
= 1 + ε  f (x)2 dx − f (x)2 dx + o(ε).
I
Rescaling fnew
by
J
and f¯new back to L 2 norm 1 causes fnew Knew fnew to be multiplied
kfnew k−2
2


Z
Z
= 1 − ε  f (x)2 dx − f (x)2 dx + o(ε),
I
so f Kf changes by
J


Z
Z
εµ  f (x)2 dx − f (x)2 dx + o(ε).
J
I
In the same way, f¯K̄ f¯ changes by


Z
Z
εµ̄  f¯(x)2 dx − f¯(x)2 dx + o(ε),
J
I
so the total change of f Kf + f¯K̄ f¯ is

Z
Z
ε
µf (x)2 + µ̄f¯(x)2 dx −
J
I

µf (x)2 + µ̄f¯(x)2 dx + o(ε).
If µf (x) + µ̄f¯(x)2 is not an almost everywhere constant function, then we can
choose intervals I and J such that ε has a nonzero coefficient in the formula above,
and then choosing ε sufficiently small and of the right sign causes a first-order
increase in the sum f Kf +f¯K̄ f¯ and correspondingly an increase in µ+µ̄. This would
2
PROOF OF A CONJECTURE OF V. NIKIFOROV
7
be a contradiction with the optimality of f Kf + f¯K̄ f¯ = µ + µ̄, so µf (x)2 + µ̄f¯(x)2
is constant almost everywhere as claimed.
Lemma 4. If f and f¯ are optimal with respect to µ + µ̄ and K is a set on which
this maximum is achieved, then f is (up to a function with support of measure 0)
a step function with at most 4 steps.
Proof of Lemma 4. We can assume that K is compact: the boundary of a monotone
set in [0, 1]2 has measure 0 (it has to have linear measure at most 2). Denote by
0 ≤ xK ≤ 1 the unique number for which the boundary ∂K intersects the diagonal
in the point (xK , xK ); let xK = 0 if K = ∅ and xK = 1 if K̄ = ∅.
Assume that x0 is a point in (0, xK ) (the other case x0 ∈ (xK , 1) is symmetric,
with the roles of K and K̄ reversed) where µf (x0 )2 + µ̄f¯(x0 )2 = µ + µ̄ is satisfied
and assume that there is an interval neighbourhood U = (x1 , x2 ) of x0 with the
property that the L 2 average of f on U is f (x0 ) and U × U is disjoint from ∂K
(that is, U ⊆ (0, xK )). Change f on U to constant f (x0 ), change f¯ on U to constant
¯
f¯(x0 ). Then the norms
R kf k2 and kf k2 stay the same, and we will show that the
sum f Kf + f¯K̄ f¯ = max{f (x)f (y), f¯(x)f¯(y)} increases unless f is constant on
U.
The integrand max{f (x)f (y), f¯(x)f¯(y)} stayed the same where f and f¯ did not
change, so we can focus on the cross-shaped area C = U × [0, 1] ∪ [0, 1] × U . Before,
the integral evaluated to
Z
Sold = max{f (x)f (y), f¯(x)f¯(y)}dxdy =
C
=2
Z
max{f (x)f (y), f¯(x)f¯(y)}dxdy −
Z
max{f (x)f (y), f¯(x)f¯(y)}dxdy =
U×U
U×[0,1]


Z
Z
Z
Z


= 2  f (x) f (y)χK (x, y)dydx + f¯(x) f¯(y)χK̄ (x, y)dydx −
U
U
[0,1]

−
Z
U
[0,1]
2
f (x)dx =

 
2
Z
Z
Z
= 2  f (x)AK (f )(x)dx + f¯(x)AK̄ (f¯)(x)dx −  f (x)dx =
U
U
U
 
2
Z
Z
Z
= 2 µ f 2 (x)dx + µ̄ f¯2 (x)dx −  f (x)dx =

U
U
U
2
 
Z
Z
Z
= 2 µ f 2 (x0 )dx + µ̄ f¯2 (x0 )dx −  f (x)dx =

U
U
U

= 2λ(U )(µ + µ̄) − 
Z
U
2
f (x)dx
8
T. TERPAI
After the change, the integral on C evaluated to
Z
max{fnew (x)f (x0 ), f¯new (x)f¯(x0 )}dxdy−
Snew = 2
U×[0,1]
−
Z
max{f (x0 )f (x0 ), f¯(x0 )f¯(x0 )}dxdy =
U×U


= 2 λ(U )f (x0 )
Z
fnew (x)χK (x, x0 )dx+
[0,1]
+λ(U )f¯(x0 )
Z


f¯new (x)χK̄ (x, x0 )dx − λ(U )2 f (x0 )2 =
[0,1]


= 2λ(U ) f (x0 )
Z
[0,1]
+f¯(x0 )
Z
Z
f (x)χK (x, x0 )dx − f (x0 ) f (x)dx + f (x0 ) f (x0 )dx+
Z
[0,1]

U
U


f¯(x)χK̄ (x, x0 )dx − λ(U )2 f (x0 )2 =
Z
= 2λ(U ) f (x0 )AK (f )(x0 ) − f (x0 ) f (x)dx + λ(U )f (x0 )2 +
U
+f¯(x0 )AK̄ (f¯)(x0 ) − λ(U )2 f (x0 )2 =
Z
= 2λ(U )(µ + µ̄) − 2λ(U )f (x0 ) f (x)dx + λ(U )2 f (x0 )2
U
So the sum f Kf + f¯K̄ f¯ changed by
Snew − Sold

2
Z
Z
= λ(U )2 f (x0 )2 − 2λ(U )f (x0 ) f (x)dx +  f (x)dx =
U

= λ(U )f (x0 ) −
Z
U
U
2
f (x)dx ≥ 0.
Equality is satisfied when f (x0 ) is also the L 1 average of f on U , which is equivalent
to f being constant on U .
Therefore for all points x0 ∈ (0, xK ) for which the equality of Lemma 3 is satisfied
there are no intervals U for which x0 ∈ U ⊂ (0, xK ) and L 2 -average of f is f (x0 ).
Hence for all points x satisfying the equality of Lemma 3 f is constant on either
(0, x] or [x, xK ). Since the equality of Lemma 3 is satisfied on a dense set of points,
f restricted to (0, xK ) is a step function with at most two steps, the same is true
for f¯ on (xK , 1). Applying Lemma 3 gives us that if f is constant on an interval
I ⊆ [0, 1], then so is f¯ and vice versa, therefore the maximal intervals on which f and
f¯ are constant coincide and the statement of the lemma follows immediately.
PROOF OF A CONJECTURE OF V. NIKIFOROV
0
I
I−
0
ε−
I+
J
9
1
ε+
I−
ε−
K
I
ε+
I+
J
K
1
Let n + 1 denote the number of the maximal intervals on which f and f¯ are
constant.
Lemma 5. There are no functions f and f¯ that are optimal with respect to µ + µ̄
and have n = 3.
Proof of Lemma 5. The constant intervals of f (and f¯) partition [0, 1] into n+1 = 4
intervals, let these be I− , I, I+ and J in increasing order of their midpoints. These
intervals give a partition of [0, 1]2 into (n + 1)2 rectangles on which the function
(x, y) 7→ max{f (x)f (y), f¯(x)f¯(y)} is constant, so we may assume that K is a union
of some of these rectangles. We will attempt the same procedure as in Lemma 4
on a “middle” square of the partition. The intersection point of the boundary ∂K
with the diagonal is the vertex of two “diagonal” squares of this partition, I × I
and I+ × I+ . One of squares is in K, the other in K̄; one intersects ∂K in a single
point, the other in two line segments. Switching K and K̄ if necessary, we can
assume that it is I × I that is in K and meets ∂K in a single point.
Set furthermore a− = f |I− , a = f |I , a+ = f |I+ and ā+ = f¯|I+ . Recall that
a− > a > a+ ≤ ā+ since f has to be monotone decreasing and the square I+ × I+
is in K̄, so a2+ ≤ ā2+ . In fact, ā+ > a+ ; assume otherwise, then the expression
f Kf + f¯K̄ f¯ does not change if K is modified in any way in the square I+ × I+ . In
particular, we can change K to a set K̃ that contains a nontrivial subset of I+ × I+
and thus prevent f and f¯ from being eigenfunctions of AK̃ and AK̃¯ respectively. But
¯ f¯
then changing f and f¯ could increase the sum µ + µ̄ = f Kf + f¯K̄ f¯ = f K̃f + f¯K̃
– a contradiction.
Our construction will shrink I− by ε− and shrink I+ by ε+ ; I will expand accordingly to keep the remaining fourth interval fixed. K is adjusted correspondingly.
The values of f and f¯ on the constant intervals will stay the same. As long as
ε− a2− + ε+ a2+ = (ε− + ε+ )a2 , the functions f and f¯ stay at L 2 norm 1. This holds
if for some ε we have ε− = ε(a2 − a2+ ) and ε+ = ε(a2− − a2 ).
10
T. TERPAI
The value of f Kf + f¯K̄ f¯ is a quadratic function of ε, with a vanishing linear part
as we will see immediately. To extract the quadratic part, denote by Knew,x the set
{(x, y) ∈ [0, 1]2 |fnew (x)f (y) > f¯new (x)f¯(y)}, that is, we apply the reparametrization only to the x coordinate. Similarly, we get Knew,y from K by reparametrizing
only the y coordinate. Clearly, the expressions fnew Knew,x f and f Knew,y fnew are
linear in ε, hence δ = fnew Knew fnew − fnew Knew,x f − f Knew,y fnew + f Kf has
the same ε-quadratic part as fnew Knew fnew . The same can be done with f¯ and
K̄, yielding δ̄ = f¯new K̄new f¯new − f¯new K̄new,x f¯ − f¯K̄new,y f¯new + f¯K̄ f¯. The sum
fnew Knew,x f + f¯new K̄new,x f¯ is constant µ + µ̄ since the integral calculating this
sum evaluates to µ + µ̄ on each vertical segment regardless of our reparametrization
by Lemma 3; in the same way f Knew,y fnew + f¯K̄new,y f¯new is also constant. Writing out the definitions, we get that the integrals in the expression for δ + δ̄ vanish
everywhere in [0, 1]2 except on the four little squares at the vertices of I × I and
correspondingly
δ + δ̄ = ε2− (a2 − 2aa− + a2− ) + 2ε− ε+ (a2 − aa− − aa+ + a− a+ )+
2
+ ε2+ (a2 − 2aa+ + ā2+ ) = (ε− (a− − a) − ε+ (a − a+ )) + ε2+ (ā2+ − a2+ ) =
2
= ε2 (a− − a)(a2 − a2+ ) − (a − a+ )(a2− − a2 ) + (a2− − a2 )2 (ā2+ − a2+ )
is a positive multiple of ε2 since a > a+ and ā+ > a+ . Therefore averaging out
f and f¯ around I as in Lemma 4 works here as well, and n = 3 can not give an
optimal eigenvalue sum.
2.5. Brute force. Finally, we perform the calculations for the cases n = 0, 1 and
2. Since the weighted square sum µf 2 + µ̄f¯2 is constant, we can pick increasing
angles
αi ∈ [0, π/2] such
q
q that on the ith interval (i runs from 0 to n) we have f =
µ+µ̄
¯
cos αi and f = µ+µ̄ sin αi . The condition for f and f¯ to be eigenfunctions
µ
µ̄
of AK and AK̄ respectively is that their jumps are 1/µ (resp. 1/µ̄) times their
integral on the corresponding constant intervals. Namely, let li denote the length
of the ith interval. Then n − i is the index corresponding to the jump from αi to
αi+1 and we have
r
r
µ + µ̄
µ + µ̄
(cos αi − cos αi+1 ) = ln−i
cos αn−i
µ
µ
µ
and symmetrically
r
r
µ + µ̄
µ + µ̄
µ̄
(sin αi+1 − sin αi ) = ln−i
sin αn−i .
µ̄
µ̄
Dividing these yields
α
−α
α
+α
µ cos αi − cos αi+1
µ 2 sin i+12 i sin i+12 i
αi+1 + αi
µ
cos αn−i
=
=
,
= tan
sin αn−i
µ̄ sin αi+1 − sin αi
µ̄ 2 sin αi+12−αi cos αi+12+αi
µ̄
2
or in an equivalent form
tan αn−i tan
αi + αi+1
µ̄
= .
2
µ
PROOF OF A CONJECTURE OF V. NIKIFOROV
11
Denote the ratio µ̄µ by λ for brevity. Calculating li from the equations above and
setting the sum of li to 1 gives that
n−1
(4)
1
cos αn X cos αi − cos αi+1
+
= .
cos α0
cos
α
µ
n−i
i=0
Also, the function f has to satisfy the condition kf k2 = 1, which is equivalent to
(5)
cos αn cos α0 +
n−1
X
i=0
(cos αi − cos αi+1 ) cos αn−i =
1
.
µ + µ̄
For n = 0 we have f and f¯ constant, K is either the whole square or its complement, µ + µ̄ = 1.
For n = 1 switch f and f¯ if needed to achieve α0 = 0, and denote α1 by α. Then
α
tan α tan = λ
2
2 sin α2 cos α2 sin α2
1
1
λ=
−1
=
−1=
cos α
1 − 2 sin2 α2
1 − 2 sin2 α2 cos α2
1
cos α =
λ+1
and we can check the other two conditions:
1
1 − cos α
λ2 + λ + 1
= cos α +
=
µ
cos α
λ+1
1
2λ + 1
= cos α + (1 − cos α) cos α =
µ + µ̄
(λ + 1)2
By definition of λ we have 1/µ = (λ + 1)/(µ + µ̄). Substituting the values obtained
above, we get that λ2 + λ + 1 = 2λ + 1, so λ2 = λ. The case when λ = 0, which
is equivalent to µ̄ = 0, has been dealt with in the case n = 0. Hence the only
possibility is λ = 1 and this is exactly the case of K being a square with side length
(λ+1)2
4
2
3 , having µ + µ̄ = 2λ+1 = 3 .
For n = 2 we again may assume that α0 = 0, and the equations defining α1 and
α2 are the following:
α1
α1 + α2
λ = tan
tan α2 = tan
tan α1 .
2
2
The condition ensuring that λ, α1 and α2 correspond to a solution comes from
expressing (4) in terms of λ and (5):
1
1 − cos α1
cos α1 − cos α2
+
=
= cos α2 +
µ
cos α1
α2
= (λ + 1) (cos α2 + (cos α1 − cos α2 ) cos α1 + (1 − cos α1 ) cos α2 ) ,
or in an equivalent form
1
cos α1 cos α2
cos α2 + 1 +
−
−
= (λ + 1)(2 cos α2 + cos2 α1 − 2 cos α1 cos α2 ).
cos α2 cos α2 cos α1
First we reformulate these in terms of the variables s = tan α21 and t = tan α22 ,
which are allowed to run from 0 to tan π4 = 1. We have
cos α1 = 2 cos2
α1
1 − s2
2
−
1
=
−1=
2
1 + s2
1 + s2
12
T. TERPAI
and similarly
cos α2 =
1 − t2
,
1 + t2
so the restrictions have the form
2ts
2s(t + s)
λ=
=
1 − t2
(1 − s2 )(1 − ts)
1 + t2
(1 − s2 )(1 + t2 ) (1 + s2 )(1 − t2 )
1 − t2
+1+
−
−
=
2
2
1+t
1−t
(1 + s2 )(1 − t2 ) (1 − s2 )(1 + t2 )
(6)
(1 − s2 )2
(1 − s2 )(1 − t2 )
1 − t2
+
−
2
= (λ + 1) 2
1 + t2
(1 + s2 )2
(1 + s2 )(1 + t2 )
The condition that s and t define a single value for λ is hence
2s(t + s)
2ts
=
1 − t2
(1 − s2 )(1 − ts)
t(1 − s2 )(1 − ts) = (t + s)(1 − t2 )
(7)
t − ts2 − t2 s + t2 s3 = t + s − t3 − t2 s
t3 + t2 s3 − ts2 − s = 0
Denote by ∆ the difference of the two sides of (6). Multiplying ∆ by (1 − t2 )(1 +
2ts
t2 )(1 − s2 )(1 + s2 )2 and substituting λ = 1−t
2 yields the following formula:
(1 − t2 )(1 + t2 )(1 − s2 )(1 + s2 )2 ∆ = −2s 4t4 s + 3t3 s6 + 3t3 s4 − 7t3 s2 + t3 +
+ 4t2 s5 − 4t2 s3 − 8t2 s − 5ts6 + 3ts4 + ts2 + t + 4s3 .
˜ = −(2s)−1 (1 − t2 )(1 + t2 )(1 − s2 )(1 + s2 )2 ∆. We claim that ∆
˜ and
Define ∆
consequently ∆ does not vanish under the condition (7) within the restrictions
0 < s < 1, 0 < t < 1 except at the point corresponding to the value λ = 1, that
˜ and
is, s = tan π/10, t = tan π/5. To prove this claim, calculate the resultant of ∆
(7) considered as polynomials in t:
˜ t3 + t2 s3 − ts2 − s = 75s25 − 250s23 − 440s21 + 1330s19 + 1299s17−
res ∆,
− 2532s15 − 2008s13 + 2116s11 + 1505s9 − 706s7 − 432s5 + 42s3 + s =
3 2
2
= s s4 − 1
s +1
5s4 − 10s2 + 1 15s4 − 50s2 − 1 .
q
π
,
The only real root of this expression between 0 and 1 is s0 = 1 − √25 = tan 10
p
√
π
out of the three values of t satisfying (7) only t0 = 5 − 2 5 = tan 5 lies between
0 and 1, and the corresponding λ is
q
√ √5−2
√
2
(5 − 2 5) √
2( 5 − 2)
2ts
5
√
= √
= 1.
λ0 =
=
1 − t2
2 5−4
2 5−4
So the angles that define f and f¯ are α0 = 0, α1 = π5 and α2 = 2π
5 . Substituting
these values into (5), we get that in this particular case
−1
2 √
2π
π
4
2π
2 π
= ( 5 + 1) < .
− 2 cos
cos + cos
µ + µ̄ = 2 cos
5
5
5
5
5
3
In all the cases, we got µ + µ̄ < 43 , finishing the proof of Theorem 3.
PROOF OF A CONJECTURE OF V. NIKIFOROV
13
References
[1] J.P. Aubin: Applied Functional Analysis, John Wiley and Sons, New York (1979)
[2] P. Csikvári: On a conjecture of V. Nikiforov, Discrete Mathematics, 309/13 (2009), 4522–4526
[3] L. Lovász, B. Szegedy: Szemerédi’s Lemma for the analyst, Geom. Func. Anal. 17 (2007),
252–270
[4] V. Nikiforov, Eigenvalue problems of Nordhaus-Gaddum type, Discrete Mathematics, 307/6
(2007), 774–780
[5] E. Nosal, Eigenvalues of graphs, Master’s Thesis, University of Calgary (1970)
Alfréd Rényi Institute of Mathematics, Reáltanoda u. 13-15, 1053 Budapest, Hungary
E-mail address: [email protected]