arXiv:2110.08751v1 [math.CO] 17 Oct 2021

Petals and Books:

The largest Laplacian spectral gap from 1

Jurgen Jost1,2, Raffaella Mulas∗1, and Dong Zhang1

1Max Planck Institute for Mathematics in the Sciences, Leipzig,Germany

2Santa Fe Institute for the Sciences of Complexity, Santa Fe, NewMexico, USA

Abstract

We prove that, for any graph on N ≥ 3 vertices, the spectral gapfrom the value 1 with respect to the normalized Laplacian is at most1/2. Moreover, we show that inequality is achieved if and only if thegraph is either a petal graph (for N odd) or a book graph (for N even).

Keywords: Spectral graph theory; Normalized Laplacian; Spectralgaps; Eigenvalue 1

1 Introduction

A spectral gap is the maximal difference between two eigenvalues for linearoperators in some class. Here, we consider Laplace operators, more pre-cisely, the normalized Laplacian of a finite graph. Such inequalities werefirst studied for the Laplace operator ∆ = −

∑ni=1

∂2

∂(xi)2on a connected

smooth domain Ω ⊂ Rn with Dirichlet conditions, that is,

∆u = λu in Ω (1)

u = 0 on ∂Ω (2)

where a (smooth) solution u that does not vanish identically is called aneigenfunction, and the corresponding value λ an eigenvalue. We choose thesign in the definition of ∆ to make it a non-negative operator. Since Ωis connected, the smallest eigenvalue λ1 is positive, and the famous Faber-Krahn inequality [7, 11, 12] says that among all domains with the same

∗Email address: [email protected]

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volume, the smallest possible value is realized by a ball of that volume.By a result of Ashbaugh and Benguria [1], the ratio between the first twoeigenvalues, λ2

λ1is largest for the ball. In either case, equality is assumed

precisely for the ball. When we replace the Dirichlet boundary condition(2) by the Neumann boundary condition

∂u

∂n= 0 on ∂Ω, (3)

then the smallest eigenvalue is λ1 = 0, with the constants being eigen-functions, and Weinberger [13] proved that the second eigenvalue λ2 nowis always less or equal than that of the ball, again with equality only forthe ball. While the proofs of such results can be difficult, there is a gen-eral pattern here, that the extremal cases occur only for a very particularclass of domains, balls in this case. Similarly, for eigenvalue problems for theLaplace-Beltrami operator in Riemannian geometry, often the extremal caseis realized by spheres (for the eigenvalue problem in Riemannian geometry,see for instance the references given in [4, 8]).

There are also discrete versions of those Laplacians, and naturally, theirspectra have also been investigated. For the algebraic graph Laplacian, asystematic analysis of spectral gaps is presented in [10], and these authorshave identified many beautiful classes of graphs with a particular structureof their spectra and spectral gaps. Here, we consider another discrete Lapla-cian, the normalized Laplace operator of a connected, finite, simple graphΓ = (V,E) on N ≥ 3 nodes. For a vertex v ∈ V , We denote by deg v itsdegree, that is, the number of its neighbors, i.e., the other vertices w ∼ vconnected to v by an edge. Then the Laplacian for a function f : V → R is

∆f(v) = f(v)− 1

deg v

∑w∼v

f(w), (4)

that is, we subtract from the value of f at v the average of the values at itsneighbors. This operator generates random walks and diffusion processeson graphs, and it was first systematically studied in [5]. Since its spectrumis that of an (N ×N)-matrix with 1s in the diagonal, the eigenvalues λ1 ≤λ2 ≤ · · · ≤ λN satisfy

N∑i=1

λi = N. (5)

The first eigenvalue now is λ1 = 0, but since Γ is connected, the secondeigenvalue λ2 is positive. There are several inequalities controlling λ2 frombelow in terms of properties of the graph (see [5]), and the largest valueamong all graphs with N vertices is realized by the complete graph KN

where λ2 = NN−1 ; for all other graphs λ2 ≤ 1. The largest eigenvalue λN is

always ≤ 2, with equality if and only if Γ is bipartite. And the gap 2 − λ2

2

quantifies how different Γ is from being bipartite [3]. In fact, the smallestpossible value λN = N

N−1 is again realized only for KN . For all other graphs,

λN ≥ N+1N−1 [6], and again, the extremal graphs, where equality is realized,

can be characterized [9].Thus, the situation for the spectral gaps at the ends of the spectrum,

that is, at 0 and near 2 has been clarified. But we may also ask about gapsin the middle of the spectrum. In fact, besides 0 and 2, also the eigenvalue 1plays a special role. It arises, in particular, from vertex duplications [2]. Theextreme case is given by complete bipartite graphs Kn,m with n + m = N .They have the eigenvalue 1 with multiplicity N −2. In fact, if 1 occurs withthis multiplicity, then by (5), 2 also has to be an eigenvalue, and the graphis bipartite. But like the eigenvalue 2, the eigenvalue 1 need not be presentin a graph. Therefore, we can ask about the maximal spectral gap at 1, thatis, we can ask what the maximal value of

ε := mini|1− λi|

could be. In this paper, we show that for any graph with N ≥ 3 vertices,ε ≤ 1

2 (for the graph with two vertices, the eigenvalues are 0 and 2, therefore,in this particular case, the gap is 1), and as the title already reveals, we canidentify the class of graphs for which the maximal possible value ε = 1

2 isrealized. In fact, in those cases, except for the triangle K3, which only has32 , both values 1

2 and 32 are eigenvalues.

2 Main result

Throughout the paper we fix a connected, finite, simple graph Γ = (V,E)on N ≥ 3 nodes. We let d denote the smallest vertex degree, we let C(V )denote the space of functions f : V → R and, given a vertex v, we letN (v) := w ∈ V : w ∼ v denote the neighborhood of v. We let

λ1 = 0 < λ2 ≤ . . . ≤ λN

denote the eigenvalues of the Laplacian in (4), and we let

ε := mini|1− λi|

be the spectral gap from 1. Clearly, ε ≥ 0 and this inequality is sharp since,for instance, any graph with duplicate vertices has 1 as an eigenvalue. Asanticipated in the introduction, here we prove that ε ≤ 1

2 and equality isachieved if and only if Γ belongs to one of the following two optimal classes.

Definition 1 (Petal graph, N ≥ 3 odd). Given m ≥ 1, the m–petal graphis the graph on N = 2m+ 1 nodes such that (Figure 1):

3

Figure 1: The petal graph

• V = x, v1, . . . , vm, w1, . . . , wm;

• E = (x, vi)mi=1 ∪ (x,wi)mi=1 ∪ (vi, wi)mi=1.

As shown in [5], for the petal graph on N = 2m+ 1 vertices, the eigen-values are 0, 1

2 (with multiplicity m − 1) and 32 (with multiplicity m + 1).

Therefore, ε = 12 in this case.

In fact, the eigenfunctions are easily constructed. Putting f(wi) = −f(vi)and f(x) = 0 produces m linearly independent eigenfunctions for the eigen-value 3

2 . Letting f(vi) = f(wi) = 1 for all i and f(x) = −2 producesanother eigenfunction for that eigenvalue. With f(vi) = f(wi) for all i,∑

i f(vi) = 0 = f(x) we get m − 1 linearly independent eigenfunctions forthe eigenvalue 1

2 .

Definition 2 (Book graph, N ≥ 4 even). Given m ≥ 1, the m–book graphis the graph on N = 2m+ 2 nodes such that (Figure 2):

• V = x, y, v1, . . . , vm, w1, . . . , wm;

• E = (x, vi)mi=1 ∪ (y, wi)mi=1 ∪ (vi, wi)mi=1.

Figure 2: The book graph

4

Remark 1. For the book graph on N = 2m + 2 nodes, 0 and 2 are eigen-values with multiplicity 1, since Γ is connected and bipartite. Moreover, onecan check that λ = 1± 1

2 are eigenvalues with multiplicity m each. In fact,the corresponding eigenfunctions can be constructed as follows:

1. By letting∑

i f(vi) = 0, f(x) = f(y) = 0, f(wi) = ∓f(vi), we obtainm− 1 linearly independent eigenfunctions;

2. By letting f(vi) = −f(wi) = 1 and f(x) = −f(y) = 2, we obtain onemore eigenfunction for 1

2 and similarly by letting g(vi) = g(wi) = −1and g(x) = g(y) = 2, we obtain one more eigenfunction for 3

2 .

Hence, also in this case, ε = 12 .

Our main result is that these, and only these, examples have the largestpossible spectral gap ε = 1

2 at 1.

Theorem 1. For any connected graph Γ on N ≥ 3 nodes,

ε ≤ 1

2.

Moreover, equality is achieved if and only if Γ is either a petal graph (for Nodd) or a book graph (for N even).

We prove Theorem 1 in Section 3. For completeness, we also state thefollowing result on the value maxi 6=1 |λi − 1|.

Proposition 2. For any connected graph Γ on N ≥ 2 nodes,

1

N − 1≤ max

i 6=1|λi − 1| ≤ 1.

Moreover, the lower bound is an equality if and only if Γ is the completegraph; the upper bound is an equality if and only if Γ is bipartite.

Proof. By (5), using the fact that λ1 = 0, it follows that∑N

i=2(λi − 1) = 1.Thus,

1

N − 1≤ 1

N − 1·

(N∑i=2

|λi − 1|

)≤ max

i 6=1|λi − 1|

and equality holds if and only if λ2 = . . . = λN = NN−1 , that is, if and only

if Γ is the complete graph.The other claim follows from the fact that

0 < λ2 ≤ . . . ≤ λN ≤ 2

and λN = 2 if and only if Γ is bipartite.

5

We also prove the following lemma, which will be needed in the proofof Theorem 1 and which is an interesting result itself, since it allows us tocharacterize ε for any graph.

Lemma 3. For any graph Γ,

ε2 = minf∈C(V )\0

∑w∈V

1degw

( ∑v∈N (w)

f(v)

)2

∑w∈V

degw · f(w)2.

Proof. We observe that the values (1 − λ1)2, . . . , (1 − λn)2 are exactly theeigenvalues of the matrix M := (Id−∆)2 whose entries are

Muv =∑

w∈N (u)∩N (v)

1

degw√

deg u · deg v, for u, v ∈ V.

In particular, ε2 is the smallest eigenvalue of M . Therefore, by the Courant–Fischer–Weyl min-max principle, it can be written as


∑u,v∈V

∑w∈N (u)∩N (v)

1degw

√deg u·deg vf(u) · f(v)∑

w∈Vf(w)2

.

Now, observe that the numerator can be rewritten as∑u,v∈V

∑w∈N (u)∩N (v)

f(u) · f(v)

degw√

deg u · deg v=∑w∈V

∑u,v∈N (w)

f(u) · f(v)

degw√

deg u · deg v

=∑w∈V

1

degw

∑v∈N (w)

f(v)√deg v

2

.

It follows that


∑w∈V

1degw

( ∑v∈N (w)

f(v)√deg v

)2

∑w∈V

f(w)2

= minf∈C(V )\0

∑w∈V

1degw

( ∑v∈N (w)

f(v)

)2

∑w∈V

degw · f(w)2.

6

This lemma will play an important role for controlling ε, because we canderive inequalities on vertex degrees in case ε ≥ 1

2 , by choosing suitable localfunctions f .

Remark 2. A gap interval with respect to a family G of simple graphs is anopen interval such that there are infinitely many graphs in G whose Laplacianspectrum does not intersect the interval. Clearly, Theorem 1 trivially impliesthat, for G = connected graphs, (12 ,

32) is a maximal gap interval. This

can be compared to a result by Kollar and Sarnak [10], which states that,for G = connected regular graphs of degree 3, (23 ,

43) is a maximal gap

interval. (The original result, Theorem 3 in [10], is formulated in terms ofthe adjacency matrix, but since these graphs are regular, the statement canbe equivalently reformulated in terms of ∆.)

3 Proof of the main result

This section is dedicated to the proof of Theorem 1, that we split intotwo main parts. In particular, in Section 3.1 we prove that ε ≤ 1

2 for anyconnected graph Γ on N ≥ 3 nodes, while in Section 3.2 we prove that ε = 1

2if and only if Γ is either a petal graph or a book graph.

3.1 Upper bound

In this section we prove the first claim of Theorem 1, namely

Theorem 4. For any connected graph Γ on N ≥ 3 nodes,

ε ≤ 1

2.

Before proving it by contradiction, we show several properties that aconnected graph on N ≥ 3 nodes with ε > 1

2 would have. We begin byobserving that, by Lemma 3, such a graph should satisfy

∑w∈V

1

degw

∑v∈N (w)

f(v)

2

>1

4

∑w∈V

degw · f(w)2, ∀f ∈ C(V ) \ 0. (6)

Lemma 5. Let Γ be a connected graph on N ≥ 3 nodes such that ε > 12 .

Then, for any v ∈ V , there exists w ∈ N (v) such that degw ≤ 3.

Proof. Let f be such that f(v) := 1 and f(u) := 0 for all u 6= v. Then, (6)implies ∑

w∈N (v)

1

degw>

1

4deg v.

7

Now, if degw ≥ 4 for all w ∈ N (v), then∑w∈N (v)

1

degw≤

∑w∈N (v)

1

4=

1

4deg v,

which is a contradiction.

Given two vertices u and v, we denote by N (u)4N (v) the symmetricdifference of N (u) and N (v), i.e., the set of vertices that are neighbors ofeither u or v.

Lemma 6. Let Γ be a connected graph on N ≥ 3 nodes such that ε > 12 . If

N (u) ∩N (v) 6= ∅, then N (u)4N (v) 6= ∅ and∑w∈N (u)4N (v)

(1

degw− 1

4

)>

1

2. (7)

Proof. Let f be such that f(v) := 1, f(u) := −1 and f(w) := 0 for allw ∈ V \ u, v. Then, by (6),∑

w∈N (u)4N (v)

1

degw>

1

4(deg v + deg u),

which also implies N (u)4N (v) 6= ∅. Moreover, since

deg v + deg u = N (u)4N (v) + 2|N (u) ∩N (v)|,

the above inequality can be rewritten as∑w∈N (u)4N (v)

(1

degw− 1

4

)>

1

4· 2|N (u) ∩N (v)|.

Since N (u) ∩N (v) 6= ∅, we have that |N (u) ∩N (v)| ≥ 1, implying that∑w∈N (u)4N (v)

(1

degw− 1

4

)>

1

2.


1. If deg u = degw = 1, then there is no vertex v with u ∼ v ∼ w.

2. There are no vertices u ∼ v ∼ w with

1 ≤ deg u ≤ 2, 1 ≤ deg v ≤ 2 and 1 ≤ degw ≤ 2.

8

Proof. The first statement is a direct consequence of Lemma 6.We now prove the second statement. Assume the contrary, that is, thereexist three vertices u ∼ v ∼ w with deg u,deg v,degw ⊆ 1, 2. We onlyneed to check the following two cases.

Case 1: deg u = 1, deg v = 2 and degw = 2.

Since N (u) ∩ N (w) 6= ∅, by (7), the vertex w1 ∈ N (w) \ v hasdegree 1. Hence, Γ is a path of length 3 (Figure 3). But this is acontradiction, since ε = 1

2 for the path of length 3, which is also the1-book graph.

• • • •u v w w1

Figure 3: A graph arising in the proof of Lemma 7


Since N (u) ∩ N (w) 6= ∅, by (7), the vertices w1 ∈ N (w) \ v andu1 ∈ N (u) \ v have degree 1. Hence, Γ is a path of length 4. Butthis is also a contradiction, since ε < 1

2 in this case.


Then, there are no vertices u ∼ v ∼ w with

2 ≤ deg u ≤ 3, 2 ≤ deg v ≤ 3 and 2 ≤ degw ≤ 3.

Proof. Assume the contrary, that is, there exist three vertices u ∼ v ∼ wwith deg u,deg v,degw ⊆ 2, 3. We need to check five cases.


By applying (7) three times, we have that:

• There exists w1 ∈ N (u)4N (w) with degw1 = 1, and withoutloss of generality, we may assume that w1 ∼ w;

• There exists v1 ∈ N (v)4N (w1) = N (v)\w with deg v1 = 1;

• There exists u1 ∈ N (u)4N (v1) = N (u) \ v with deg u1 = 1.

Hence, Γ is the graph in Figure 4.

By letting f(u1) := f(w1) := −1, f(v) := 1 and f(v1) := f(u) :=f(w) := 0, from (6) we derive a contradiction.

9

• • • ••

•

u v w w1u1

v1



If there is only one vertex x in N (u)4N (w), then by (7), it musthave degree 1. Also, by construction, it is clear that x ∼ w. More-over, by (7), the only vertex y ∈ N (w) \ v, x has degree 1. Butthis is a contradiction, since we must also have y ∼ u.

Otherwise, there are three vertices in N (u)4N (w) and at least twoof them have degree 2. In this case, we reduce to Case 1.


If deg x ≥ 2 for all x ∈ N (u)4N (w), then by (7), there are at leasttwo vertices x and y in N (u)4N (w) with degree 2, and thereforewe reduce to Case 1.

There are two sub-cases left:

a) If there exists u1 ∼ u with deg u1 = 1, then by (7) thereexists v1 ∼ v with deg v1 = 1 and there exists w1 ∼ w withdegw1 = 1. Therefore, Γ has the same local structure as thegraph in Figure 5. By letting f(u1) := f(w1) := −1, f(v) := 1and f |V \v,u1,w1 := 0, from (6) we get 1 > 1

4(1 + 3 + 1), whichis a contradiction.

• • • ••

•

u v w w1u1

v1 ?


b) If there exists w1 ∼ w with degw1 = 1, then there exists v1 ∼ vwith deg v1 = 1 and there exists u1 ∼ u with deg u1 = 1.Hence, as in the previous sub-case, Γ has the same local struc-ture as the graph in Figure 5, which brings to a contradiction.


If deg x ≥ 2 for all x ∈ N (u)4N (w), then by (7) there are at leasttwo vertices x and y in N (u)4N (w) of degree 2, and thus we reduceto Case 1.

10

Otherwise, there exists u1 ∼ u with deg u1 = 1. Applying Lemma6 to the vertex pair u1 and v, we derive a contradiction.


Similar to the above cases, by repeatedly applying (7) we can seethat there exist v1 ∼ v, u1 ∼ u and w1 ∼ w with deg v1 = deg u1 =degw1 = 1. Therefore, Γ has the same local structure as the graphin Figure 6. Similarly to Case 3, also in this case we derive acontradiction.

• • • ••

•

u v w w1u1

v1 ??


We now fix the following notations. Given a vertex v, we let

N2,3(v) := w ∼ v : 2 ≤ degw ≤ 3,N≥4(v) := w ∼ v : degw ≥ 4,N2(v) := w ∼ v : degw = 2.


Then, for each v ∈ V , |N2,3(v)| ≤ 1.

Proof. Fix v ∈ V and suppose the contrary, that is, |N2,3(v)| ≥ 2.

Case 1: N (v) = N≥4(v) ∪N2,3(v).

In this case, for any two vertices vi, vj ∈ N2,3(v),

minx∈N (vi)4N (vj)

deg x = 1,

because otherwise, we can reduce to Lemma 8. Therefore, exceptfor at most one vertex in N2,3(v), any other vertex in N2,3(v) isadjacent to a vertex of degree 1. Now, let f(x) := −1 if deg x = 1and x is adjacent to some vertex in N2,3(v), let f(v) := 1 and let

11

f(y) := 0 otherwise. Then, by (6) we obtain

1

2+

1

4|N≥4(v)| ≥

∑w∈V

1

degw

∑v∈N (w)

f(v)

2

>1

4

∑w∈V

degw · f(w)2

≥ 1

4(deg v + |N2,3(v)| − 1)

≥ 1

4(|N≥4(v)|+ 3),


Case 2: There exists w ∈ N (v) with degw = 1. Then, by Lemma 7,

N (v) \ w = N≥4(v) ∪N2,3(v).

Similarly to the previous case, this implies that

1 +1

4|N≥4(v)| ≥

∑w∈V

1

degw

∑v∈N (w)

f(v)

2

>1

4

∑w∈V

degw · f(w)2

≥ 1

4(deg v + |N2,3(v)|)

≥ 1

4(|N≥4(v)|+ 4),



If w ∼ v and degw = 1, then deg v ≤ 2.

Proof. Assume the contrary, i.e., w ∼ v, degw = 1 and deg v ≥ 3. Then,similarly to the proof of Lemma 9, we have that for any u ∼ v, there exists

12

u1 ∼ u with deg u1 = 1. Therefore,

1 ≥∑w∈V

1

degw

∑v∈N (w)

f(v)

2

>1

4

∑w∈V

degw · f(w)2

≥ 1

4(deg v + |N (v)| − 1)

≥ 1

4(2 deg v − 1) > 1,



Then, there exist three vertices u ∼ v ∼ w with deg u ≤ 3 and degw ≤ 3.

Proof. By Lemma 5, there exist u1 ∼ u2 with deg u1 ≤ 3 and deg u2 ≤ 3. Ifv ∼ u2, then by (7) there exists w ∈ N (u1)4N (v) with degw ≤ 3.If w ∈ N (u1) \ N (v), then we have w ∼ u1 ∼ u2 with degw ≤ 3 anddeg u2 ≤ 3; while if w ∈ N (v) \ N (u1), then we have w ∼ v ∼ u2 withdegw ≤ 3 and deg u2 ≤ 3.


Then, there exist three vertices u ∼ v ∼ w such that deg u = 1, deg v = 2and degw ∈ 2, 3.

Proof. Fix u ∼ v ∼ w with deg u ≤ 3 and degw ≤ 3 as in Lemma 11. ByLemma 7, at most one between u and w has degree 1, and by Lemma 9, atmost one between u and w has degree 2 or 3. Therefore, we may assume thatdeg u = 1 and degw ∈ 2, 3. By Lemma 10, this implies that deg v = 2.

We are now finally able to prove Theorem 4.

Proof of Theorem 4. Assume by contradiction that there exists a connectedgraph Γ on N ≥ 3 nodes such that ε > 1

2 . Then, by Lemma 12, there existu ∼ v ∼ w such that deg u = 1, deg v = 2 and degw ∈ 2, 3. By (7), thereexists a vertex w1 ∈ N (w) \ v satisfying degw1 = 1, and by Lemma 10,degw = 2. Therefore Γ is the path of length 3 in Figure 3. But as we knowthis is a contradiction, since ε = 1

2 for the path of length 3, which is alsothe 1-book graph.

13

3.2 Optimal cases

In the previous section we proved that ε ≤ 12 for any connected graph Γ on

N ≥ 3 nodes. Hence, in order to prove Theorem 1, it is left to prove thatε = 1

2 if and only if Γ is either a petal graph or a book graph. We dedicatethis section to the proof of this claim, that we further split into three parts:

1. In Section 3.2.1 we prove that, if the smallest vertex degree d is ≥ 3,then ε < 1

2 . In fact, we prove an even stronger result, since we showthat, in this case,

ε ≤√d− 1

d<

1

2.

2. In Section 3.2.2 we show that, if d = 2, then ε = 12 if and only if Γ is

either a petal graph or a book graph.

3. In Section 3.2.3 we show that, if d = 1 and Γ is not the 1–book graph,then ε < 1

2 .

Before, we observe that, by Lemma 3 and Theorem 4, if a connected graphΓ on N ≥ 3 nodes is such that ε = 1

2 , then

∑w∈V

1

degw

∑v∈N (w)

f(v)

2

≥ 1

4

∑w∈V

degw · f(w)2, ∀f ∈ C(V ) \ 0. (8)

Moreover, with the same proof as Lemma 6, one can prove the following

Lemma 13. Let Γ be a connected graph on N ≥ 3 nodes such that ε = 12 .

If N (u) ∩N (v) 6= ∅, then N (u)4N (v) 6= ∅ and∑w∈N (u)4N (v)

(1

degw− 1

4

)≥ 1

2. (9)

3.2.1 The case d ≥ 3

In this section we prove the following

Theorem 14. For any connected graph Γ on N ≥ 3 nodes with smallestdegree d ≥ 3,

ε ≤√d− 1

d<

1

2.

Before that theorem, we shall prove several preliminary results.

14

Lemma 15. Let Γ be a connected graph on N ≥ 3 nodes with smallest

degree d ≥ 3 and ε >√d−1d . Then,

∑w∈V

1

degw

∑v∈N (w)

f(v)

2

>d− 1

d2

∑w∈V

degw · f(w)2, ∀f ∈ C(V ) \ 0.

(10)Moreover, if N (u) ∩N (v) 6= ∅, then N (u)4N (v) 6= ∅ and∑

w∈N (u)4N (v)

(1

degw− d− 1

d2

)> 2 · d− 1

d2. (11)

Proof. The first claim follows directly from Lemma 3, while the second claimcan be proved as Lemma 6.


degree d ≥ 3 and ε >√d−1d . If N (u) ∩N (v) 6= ∅, then

|w ∈ N (u)4N (v) : degw ∈ d, d+ 1| ≥ 2d− 1.

Proof. Assume the contrary, then∑w∈N (u)4N (v)

(1

degw− d− 1

d2

)≤

∑w∈N (u)4N (v)degw∈d,d+1

(1

degw− d− 1

d2

)

≤ (2d− 2)

(1

d− d− 1

d2

)= 2 · d− 1

d2,

which contradicts (11).


degree d ≥ 3 and ε >√d−1d . If N (u) ∩N (v) 6= ∅ and

|w ∈ N (u)4N (v) : degw ∈ d, d+ 1| ≤ 2d,

then |w ∈ N (u)4N (v) : degw = d| ≥ 2d− 3.

Proof. If not, then∑w∈N (u)4N (v)

(1

degw− d− 1

d2

)

≤∑

w∈N (u)4N (v)degw=d

(1

degw− d− 1

d2

)+

∑w∈N (u)4N (v)degw=d+1

(1

degw− d− 1

d2

)

< (2d− 4)

(1

d− d− 1

d2

)+ 2d

(1

d+ 1− d− 1

d2

)=

2d− 4

d2+

2

d(d+ 1)<

2d− 4

d2+

2

d2= 2 · d− 1

d2,

which contradicts (11).

15

Lemma 18. Let Γ be a connected graph on N ≥ 3 nodes with smallest degree

d ≥ 3 and ε >√d−1d . Then, there exist u and v with N (u) ∩ N (v) 6= ∅ and

deg u,deg v ∈ d, d+ 1.

Proof. Fix three vertices u′ ∼ w′ ∼ v′. By Lemma 16, we either have

|w ∈ N (u′) \ N (v′) : degw ∈ d, d+ 1| ≥ d

or|w ∈ N (v′) \ N (u′) : degw ∈ d, d+ 1| ≥ d.

Without loss of generality, we may assume that there are two vertices uand v in N (u′) \ N (v′) with deg u,deg v ∈ d, d + 1. This implies thatN (u) ∩N (v) 6= ∅, hence u and v satisfy the claim.

Lemma 19. Let Γ be a connected graph on N ≥ 3 nodes with smallest degree

d ≥ 3 and ε >√d−1d . Then, there exist u and v with N (u) ∩ N (v) 6= ∅ and

deg u = deg v = d.

Proof. Fix u′ and v′ that satisfy Lemma 18. Then,

|N (u′)4N (v′)| ≤ deg u′ + deg v′ − 2 ≤ 2d.

By Lemma 17, this implies that

|w ∈ N (u′)4N (v′) : degw = d| ≥ 2d− 3 ≥ 3.

Hence, without loss of generality, we can assume that there are two verticesu and v in N (u′)\N (v′) with deg u = deg v = d. This proves the claim.

We can now prove Theorem 14.

Proof of Theorem 14. We first observe that the second inequality followsfrom the fact that the sequence√

d− 1

d

d≥2

=

1

2,

√2

3,

√3

4,2

5, . . .

is strictly decreasing. Hence, it is left to show that ε ≤√d−1d . Assume, by

contradiction, that ε >√d−1d . By Lemma 19, there exist u and v such that

N (u)∩N (v) 6= ∅ and deg u = deg v = d. This implies that |N (u)4N (v)| ≤2d− 2, but together with Lemma 16, this brings to a contradiction.

16

3.2.2 The case d = 2

Here we prove the following

Theorem 20. For any connected graph Γ on N ≥ 3 nodes with smallestdegree d = 2,

ε ≤ 1

2,

with equality if and only if Γ is either a petal graph or a book graph.

Also in this section, we first prove several preliminary results. We startby showing that, if Γ = CN is the cycle graph on N ≥ 3 vertices, then ε = 1

2if and only if either N = 3 (in which case Γ = C3 is the 1–petal graph) orN = 6 (in which case Γ = C6 is the 2–book graph).

Proposition 21. If Γ = CN is the cycle graph on N ≥ 3 vertices, thenε = 1

2 if and only if N ∈ 3, 6.

Proof. As shown in [5], for the cycle graph CN on N ≥ 3 nodes, the eigen-values are 1 − cos (2πk/N), for k = 0, . . . , N − 1. We already know thatε = 1

2 if N ∈ 3, 6 since, in this case, Γ is either the 1–petal graph or the2–book graph. Therefore, it is left to show that ε < 1

2 for N ≥ 4, N 6= 6.We consider three cases.

1. If N = 4, then 1 is an eigenvalue, therefore ε = 0.

2. IfN = 5, by letting k = 1 we can see that 1−cos (2π/5) is an eigenvaluethat has distance cos (2π/5) ≈ 0.3 from 1. Hence, ε < 1/2.

3. Let N > 6 and let

k ∈N − 2

4,N − 1

4,N

4,N + 1

4

.

Then,2k

N≤ 2

N· N + 1

4=N + 1

2N<

2

3,

since N > 3, and

2k

N≥ 2

N· N − 2

4=N − 2

2N>

1

3,

since N > 6. Hence,2k

N∈(

1

3,2

3

),

implying that ∣∣∣∣cos2πk

N

∣∣∣∣ < 1

2,

therefore ε < 1/2.

17

Proposition 22. Let Γ be a connected graph on N ≥ 3 nodes with smallestdegree d = 2 and ε = 1

2 . If three vertices are such that u ∼ v ∼ w anddeg u = deg v = degw = 2, then Γ = C3 or C6.

Proof. If u ∼ w, then clearly Γ = C3. If u 6∼ w, then (9) applied toN (u)4N (w) implies that the vertices u1 ∈ N (u)\v and w1 ∈ N (w)\vare such that deg u1 = degw1 = 2. If u1 = w1, then clearly Γ = C4.Otherwise, we can repeat the process and obtain any cycle graph CN . ByProposition 21, the claim follows.

Lemma 23. Let Γ be a connected graph on N ≥ 3 nodes with smallestdegree d = 2 and ε = 1

2 . Then, there exist two adjacent vertices u ∼ v suchthat deg u = deg v = 2.

Proof. We illustrate this proof in Figure 7.Fix u ∼ v ∼ w such that deg v = 2. Then, by (9), there exists w1 ∈N (u)4N (w) with degw1 ∈ 2, 3. Without loss of generality, we assumethat w1 ∼ w. Again by (9), there are at least two vertices in N (v)4N (w1)of degree 2. If deg u = 2, then we have done. If deg u ≥ 3, then degw1 = 3,deg u ≤ 4 and deg x = 2 for any x ∈ N (w1) \ w. Now, let x1, x2 ∈N (w1) \ w. By applying (9) to x1 and x2, we infer that there exists avertex adjacent to x1 of degree 2. Since also deg x1 = 2, this proves theclaim.

• • • • •

•

u v w w1 x1

x2

Figure 7: The vertices in the proof of Lemma 23


2 . For any vertex v of degree 2, there exists u ∼ vwith deg u = 2.

Proof. Suppose the contrary, then there are u ∼ v ∼ w with deg u,degw ≥ 3and deg v = 2.

Claim 1: For all x ∈ N (u)4N (w), deg x ≥ 3.

If not, without loss of generality we can assume that there existsx ∈ N (u) \ N (w) with deg x = 2. This implies that

N (v)4N (x) = w, y,

18

for some vertex y. By the assumption, deg y ≥ 2 and degw ≥ 3.Hence, applying (9) to N (v)4N (x) implies(

1

2− 1

4

)+

(1

3− 1

4

)≥(

1

deg y− 1

4

)+

(1

degw− 1

4

)≥ 1

2,


Claim 2: There are at least six vertices in N (u)4N (w) of degree 3, andthus deg u+ degw ≥ 8.

If there are at most five vertices in N (u)4N (w) of degree 3, thenthe other vertices in N (u)4N (w) have degree at least 4. Applying(9) to N (u)4N (w), we then obtain

5

(1

3− 1

4

)≥

∑x∈N (u)4N (w)

(1

deg x− 1

4

)≥ 1

2,


Claim 3: Assume that there exists w1 ∼ w with degw1 = 3. Then, x1, x2 ∈N (w1) \ w are such that deg x1 = deg x2 = 2.

If not, then without loss of generality we can assume that deg x1 ≥3. Applying (9) to N (v)4N (w1), this implies

5

12=

(1

3− 1

4

)+

(1

3− 1

4

)+

(1

2− 1

4

)≥(

1

deg u− 1

4

)+

(1

deg x1− 1

4

)+

(1

deg x2− 1

4

)=

∑x∈N (v)4N (w1)

(1

deg x− 1

4

)≥ 1

2,


Now, if x1 ∼ x2, then by letting f(x1) := −1, f(w1) := 1 and f := 0otherwise, (9) brings to a contradiction. Hence, x1 6∼ x2. By applying (9)again, we can infer that there exist y1 ∼ x1 and y2 ∼ x2 such that deg y1 =deg y2 = 2. If y1 = y2, then we reduce to Proposition 22. If y1 6= y2, then asimilar reasoning implies y1 6∼ y2. By applying then (9) to N (y1)4N (w1),we infer that there exists z1 ∼ y1 with deg z1 = 2, and therefore we againreduce to Proposition 22. In any case, we obtain a contradiction.

Proposition 25. Let Γ be a connected graph on N ≥ 3 nodes with smallestdegree d = 2 and ε = 1

2 . If there exist u ∼ v ∼ w such that deg u = deg v = 2and degw = 3, then Γ is the book graph on N = 8 nodes.

19

Proof. If w ∼ u, by letting f(w) := 1, f(u) := −1, and f := 0 otherwise,(8) implies that

1

2+

1

3+

1

degw1≥ 1

4(2 + 3),

therefore the vertex w1 ∈ N (w) \ u, v has degree 2. Similarly, there existsx ∼ w1 with deg x = 2. By letting now f(w) := 2, f(u) := f(v) := −1,f(w1) := 1, f(x) := −1 and f := 0 otherwise, then (8) implies that

2 · 1

2+

1

3+

1

2+

1

2+

1

3≥ 1

4(2 + 2 + 3 · 22 + 2 + 2),

which is a contradiction. Therefore, w 6∼ u.Now, if the vertex u1 ∈ N (u) \ v has degree 2, then we reduce to Propo-sition 22. Thus, deg u1 ≥ 3. Applying (9) to the vertices u and w, we inferthat the two vertices w1, w2 ∈ N (w) \ v satisfy degw1 = degw2 = 2.Again, applying (9) to the vertex pairs (v, w1) and (v, w2), we infer that thevertices x1 ∼ w1 and x2 ∼ w2 satisfy deg x1 = deg x2 = 2.If N (x1)∩N (x2) = ∅, then the vertices y1 ∼ x1 and y2 ∼ x2 have to satisfydeg y1, deg y2 ≥ 3. By letting f(w) := 1, f(x1) := f(x2) := −1 and f := 0otherwise, (8) implies

1

3+

1

3+

1

2≥ 1

4(3 + 2 + 2),

which is a contradiction. Hence, N (x1) ∩ N (x2) 6= ∅ and, similarly, alsoN (u) ∩N (x1) 6= ∅ and N (u) ∩N (x2) 6= ∅.In particular, there exists z such that z ∼ u, z ∼ x1 and z ∼ x2. By lettingnow f(w) := 1, f(u) := f(x1) := f(x2) := −1 and f := 0 otherwise, by (8)we obtain

1

deg z· 32 ≥ 1

4(3 + 2 + 2 + 2),

which implies that deg z ≤ 4.We now claim that deg z = 3. Suppose the contrary, then deg z = 4. Letz1 ∈ N (z) \ u, x1, x2. If there exists z2 ∼ z1 of degree 2, then f(z2) :=f(u) := −1, f(z) := 1 and f := 0 otherwise together with (8) gives

1

3+ 3 · 1

2≥ 1

4(2 + 4 + 2),

which is a contradiction. Therefore, for any a ∈ N (z1), deg a ≥ 3. By(9) applied to z2 and v, we infer that there exist at least three verticesof degree 3 in N (z1), and deg z1 ≥ 4. Let b1, b2 ∈ N (z1) be such thatdeg b1 = deg b2 = 3. Then, (9) applied to b1 and b2 implies that there is atleast one vertex of degree 2 in N (b1)4N (b2). Without loss of generality, wecan assume that there exists b′ ∈ N (b1) \ N (b2) such that deg b′ = 2. ByLemma 24, there exists c′ ∼ b′ with deg c′ = 2. But this is a contradiction,

20

since we proved that all vertices adjacent to b1 must have degree 2.This proves that deg z = 3, which implies that Γ must be the book graphon N = 8 vertices.


2 . If Γ is not in the classes described in Proposition22 and Proposition 25, then there is no vertex of degree 3.

Proof. Assume the contrary, that is, there exists a vertex u with deg u = 3.We first claim that, for any v ∈ N (u), deg v ≥ 3. This is true because,otherwise, there would be a vertex v ∼ u with deg v = 2, and by Lemma24, also a vertex w ∼ v with degw = 2. Hence, Γ would be in the classdescribed in Proposition 25, but we are assuming that this is not the case.This shows that deg v ≥ 3 for any v ∈ N (u).Now, we prove that there are no vertices v and w such that u ∼ v ∼ w anddegw ∈ 2, 3. This follows from the fact that, otherwise, applying (9) tou and w together with the above remarks implies that there exists w1 ∼ wwith degw1 = 2, while degw = 3, which is a contradiction.But on the other hand, for any x, y ∈ N (u), by (9) there exists w ∈N (x)4N (y) with degw ∈ 2, 3, which contradicts the above argument.

We are now able to prove Theorem 20.

Proof of Theorem 20. We shall illustrate this proof in Figure 8 below.Observe that, if Γ is in one of the classes described in Proposition 22 andProposition 25, then the claim follows. Therefore, we can assume that thisis not the case. We can then apply Lemma 26 and infer that there is novertex of degree 3. Also, if all vertices have degree < 3, from the fact thatd = 2 it follows that all vertices have degree 2, but this is not possible, sincewe are assuming that Γ does not satisfy the condition in Proposition 22.Therefore, we can assume that there exists a vertex with degree ≥ 4.Moreover, we can also assume that there exists one vertex u with deg u ≥ 4such that N2(u) 6= ∅. This can be seen by the fact that, if not, then allvertices of degree 2 would only be adjacent to vertices of degree 2, contra-dicting the fact that Γ is connected.Now, from the fact that N2(u) 6= ∅, we can infer two facts about N (u) =N2(u) tN≥4(u) :

1. For any v ∈ N2(u), there exists a unique w ∼ v with degw = 2. Thisfollows immediately from Lemma 24 and from the fact that deg u ≥ 4.

2. If N≥4(u) 6= ∅, then for any v ∈ N≥4(u) there exists w ∼ v withdegw = 2. This can be shown by fixing v′ ∈ N2(u) and then applying(9) to N (v′)4N (v).

21

As a consequence of the first fact, we can write

N2(u) = v1, . . . , vk, vk+1, . . . , vk+2l,

where:

• For each i = 1, . . . , k, there exists a unique wi ∈ V \ N2(u) such thatvi ∼ wi and degwi = 2;

• vk+1 ∼ vk+2, . . . , vk+2l−1 ∼ vk+2l.

Now, for i = 1, . . . , k, since vi ∼ wi and deg vi = degwi = 2, we can inferthat xi ∈ N (wi) \ vi must have degree ≥ 4, because otherwise, Γ wouldsatisfy the assumption of Proposition 22, but we are assuming that this isnot the case. Therefore, in particular, w1, . . . , wk must be pairwise distinct,while x1, . . . , xk do not need to be pairwise distinct.Up to relabeling x1, . . . xk, we assume that

x′1 := x1 = . . . = xk1 ,

x′2 := xk1+1 = . . . = xk1+k2 ,

. . .

x′p+q := xk1+...+kp+q−1+1 = . . . = xk1+...+kp+q = xk,

where:

• k1, . . . , kp+q are such that k = k1 + . . .+ kp+q,

• x′1, . . . , x′p+q are pairwise distinct,

• x′1, . . . , x′p 6∈ N (u) while x′p+1, . . . , x

′p+q ∈ N (u),

• q is such that q ≤ deg u− k − 2l.

Following the above notation, we also relabel the vertices v1, . . . , vk andw1, . . . , wk as

v1, . . . , vk1 , . . . , vk1+...+kp+q ,

w1, . . . , wk1 , . . . , wk1+...+kp+q .

Now, we write

N≥4(u) \ x′p+1, . . . , x′p+q = y1, . . . , yr.

Then, as observed above, for each i = 1, . . . , r there exists y′i ∼ yi withdeg y′i = 2. Furthermore, by Lemma 24, for each i = 1, . . . , k there alsoexists zi ∼ y′i ∼ yi with deg zi = deg y′i = 2. Clearly, y′1, . . . , y

′r are pairwise

22

...

•

•

•

•

•

•

••

•

•

•

v1v2

w1

w2

vk1

vk+1

vk+2

vk+2l−1

vk+2l

...

• x′1

...

•vk1+...+kp−1+1 wk1+...+kp−1+1

...•vk1+...+kp wk1+...+kp

•

x′p• •

u

•• y1y′1

...

•• y2y′2

•• yry′r

•• •

•

· · ·•

••

•

•

•

x′p+1vk1+...+kp+1wk1+...+kp+1vk1+...+kp+1

wk1+...+kp+1· · · · · ·

· · · vk1+...+kp+q−1+1

wk1+...+kp+q−1+1

vk1+...+kp+q

wk1+...+kp+q

x′p+q

Figure 8: Illustration of the proof of Theorem 20

distinct, while z1, . . . , zr may have some overlap with y′1, . . . , y′r.

We illustrate the above vertices in Figure 8 and we observe that

N (u) = N2(u) tN≥4(u)

= v1, . . . , vk, vk+1, . . . , vk+2l t x′p+1, . . . , x′p+q t y1, . . . , yr,

where the above vertices are all pairwise distinct, therefore

deg u = k + 2l + q + r.

Now, by letting

f(u) := 2,

f(vk+1) := . . . := f(vk+2l) := f(w1) := . . . := f(wk) := −1,

f(v1) := . . . := f(vk) := 1 if k ≤ 2l,

f(v1) := . . . := f(v2l) := 1,

f(v2l+t) := (−1)t for t = 1, . . . , k − 2l, if 2l < k,

f(y′1) := . . . := f(y′r) := −1,

f := 0 otherwise

and

ck,l :=

(2l − k)2, if 2l ≥ k,1, if 2l < k and k is odd,

0, if 2l < k and k is even,

23

from (8) we can infer that

ck,ldeg u

+2k + 2l

2+

p∑i=1

k2ideg x′i

+

p+q∑i=p+1

(ki − 2)2

deg x′i+ r · 1

2+

r∑i=1

1

deg yi

=∑w∈V

1

degw

∑v∈N (w)

f(v)

2

≥ 1

4

∑w∈V

degw · f(w)2

=1

4(deg u · 22 + 2(2k + 2l) + 2r)

= deg u+ k + l +r

2= 2k + 3l + q +

3

2r.

We now use the following known facts:

• deg yi ≥ 4, for each i = 1, . . . , r;

• deg x′i ≥ maxki, 4, for i = 1, . . . , p+ q;

• (ki − 2)2 ≤ k2i (since ki ≥ 1), for i = p+ 1, . . . , p+ q.

Putting everything together, we obtain that

2k + 3l + q +3

2r

≤ck,l

deg u+ k + l +

p+q∑i=1

k2ideg x′i

+ r · 1

2+

r∑i=1

1

4

≤ck,l

k + 2l + q + r+ k + l +

p+q∑i=1

k2iki

+3

4r,

which implies

k + 2l + q +3

4r ≤

ck,lk + 2l + q + r

+

p+q∑i=1

ki = k +ck,l

k + 2l + q + r.

Therefore,

2l + q +3

4r ≤

ck,lk + 2l + q + r

≤

(2l−k)2k+2l ≤ 2l if 2l ≥ k,1

deg u ≤14 if 2l < k,

which implies that r = q = 0, either l = 0 or k = 0 (but they cannot beboth zero), and deg x′i = ki ≥ 4, for i = 1, . . . , p.We then have two cases:

24

1. If k = 0, then Γ is a petal graph with 2l + 1 vertices.

2. If l = 0, then we can write

Γ = (Γ1 t . . . t Γp)/u1, . . . , up,

where for i = 1, . . . , p, Γi as a ki-book graph and ui as one of itsvertices of degree ki, and the above notation means that Γ is given bythe union of Γ1, . . . ,Γp by identifying u1, . . . , up as one vertex, whichis u.In this case, by letting f(x′1) := 2, f(wi) := (−1)i for i = 1, . . . , k1,f(v1) := . . . := f(vk) := −1 and f := 0 otherwise, by (8) we can inferthat deg u = k1 (with the same proof that we used for inferring thatdeg x′1 = k1). As a consequence, p = 1, therefore Γ is a book graph onN = 2k + 2 vertices.

This proves the claim.

3.2.3 The case d = 1

Given N ≥ 3, we let PN denote the path graph on N nodes, and we observethat P4 coincides with the 1–book graph. It is left to prove the following

Theorem 27. For any connected graph Γ on N ≥ 3 nodes with smallestdegree d = 1, if Γ 6= P4 then

ε <1

2.

Proof. Suppose the contrary, then there exists a connected graph Γ 6= P4

on N ≥ 3 nodes with ε = 12 and d = 1. We fix such Γ = (V,E) so that it

is the graph with these properties that has the smallest possible order, thatis, if Γ′ 6= P4 is a connected graph with smallest degree 1 and 3 ≤ N ′ < Nvertices, then ε(Γ′) < 1

2 . If we show that the existence of Γ brings to acontradiction, then we are done.Fix u ∈ V of degree 1, and let v ∼ u be its only neighbour. Then, by lettingf(u) := 1 and f := 0 otherwise, (8) gives 1

deg v ≥14 , i.e., deg v ≤ 4.

Now, assume first that deg v = 4. Let Γ := Γ \ u, let ε := ε(Γ) and let fbe a function on V (Γ) = V \ u that is an eigenfunction for (Id−∆(Γ))2

with eigenvalue ε2. Then,

ε2 =

1deg v−1

(∑x∈N (v)\u f(x)

)2+∑

w∈V \v,u1

degw

(∑x∈N (w) f(x)

)2(deg v − 1)f(v)2 +

∑w∈V \v,u degwf(w)2

.

(12)Moreover, since ε2 = (1− λ)2 for some eigenvalue λ of Γ, from (4) it followsthat

(1− λ)f(v) =1

deg v − 1

∑x∈N (v)\u

f(x)

,

25

therefore

ε · (deg v − 1) · |f(v)| =

∣∣∣∣∣∣∑

x∈N (v)\u

f(x)

∣∣∣∣∣∣and similarly, for all w ∈ V \ u, v,

ε · degw · |f(w)| =

∣∣∣∣∣∣∑

x∈N (w)

f(x)

∣∣∣∣∣∣ .In particular, without loss of generality, we may assume

ε · (deg v − 1) · f(v) =∑

x∈N (v)\u

f(x)

andε · degw · f(w) =

∑x∈N (w)

f(x), ∀w ∈ V \ u, v,

as the proof can be easily adapted otherwise.Now, since we are assuming that ε = ε(Γ) = 1

2 , we have that, independentlyof the value of f(u) ∈ R that we choose, (8) implies

∑w∈V

1degw

(∑x∈N (w) f(x)

)2∑w∈V degwf(w)2

≥ 1

4,

or equivalently,

f(v)2 +1

4·

∑x∈N (v)

f(x)

2

+∑

w∈V \v,u

1

degw

∑x∈N (w)

f(x)

2

≥ 1

4

∑w∈V

degwf(w)2,

which by the above observations can also be rewritten as

f(v)2 +1

4(f(u) + ε(deg v − 1)f(v))2 +

∑w∈V \v,u

1

degw(ε · degw · f(w))2

≥ 1

4

f(u)2 + 4 · f(v)2 +∑

w∈V \v,u

degwf(w)2

.

We now consider two cases.

26

Case 1. If f(v) = 0, then the above inequality becomes

(ε2 − 1

4

)·

∑w∈V \v,u

degwf(w)2

≥ 0.

Since ε2 ≤ 14 by Theorem 4 and f(v) = 0 implies f |V \v,u 6= 0, we

deduce that ε2 = 14 . But this brings to a contradiction. In fact, by

construction, it is clear that Γ is a connected graph on N − 1 ≥ 4nodes, since v has degree 3 in Γ. Hence, if Γ has vertices of degree1, then by the assumption on the graphs with less than N vertices,Γ must be P4, but this is not possible because of the degree of v.Similarly, if Γ does not have vertices of degree 1, then by Theorem20, Γ is either a petal graph or a book graph, and in particular,since v has degree 3 in Γ, then Γ must be the book graph on 8vertices. But in this case, one can directly check that ε = ε(Γ) < 1

2 ,which is a contradiction.

Case 2. If f(v) 6= 0, then the above inequality becomes

3

2·εf(u)f(v)+

9

4·ε2f(v)2+

(ε2 − 1

4

)·

∑w∈V \v,u

degwf(w)2

≥ 0.

If ε 6= 0, we can always derive a contradiction from the above in-equality by choosing an appropriate value of f(u) ∈ R. Therefore,ε = 0, implying that the inequality holds if and only if f |V \v,u = 0.But in this case, (12) is not satisfied, which is a contradiction.

Hence, deg v = 4 always brings to a contradiction, implying that deg v ∈2, 3, since we already know that deg v ≤ 4.

Let now Γ := Γ \ u, v, let ε := ε(Γ) and let f be a function on V (Γ) =

V \u, v that is an eigenfunction for (Id−∆(Γ))2 with eigenvalue ε2. Then,

ε2 =

∑w∈N (v)\u

1degw−1

(∑x∈N (w)\v f(x)

)2+∑

w∈V \N (v)1

degw

(∑x∈N (w) f(x)

)2∑

w∈N (v)\u(degw − 1)f(w)2 +∑

w∈V \N (v) degwf(w)2

and, similarly to the first part of the proof, without loss of generality wecan assume that∑

x∈N (w)\v f(x) = ε(degw − 1)f(w), ∀w ∈ N (v) \ u,∑x∈N (w) f(x) = εdegwf(w), ∀w ∈ V \ N (v).

In fact, the case in which ε is replaced by −ε is similar.

27

Since ε = ε(Γ) = 12 , (8) holds for any value of f(u) and f(v). Therefore,

f(v)2 +1

deg v

f(u) +∑

w∈N (v)\u

f(w)

2

+∑

w∈V \N (v)

1

degw(εdegwf(w))2

+∑

w∈N (v)\u

1

degw(f(v) + ε(degw − 1)f(w))2

≥ 1

4

f(u)2 + deg vf(v)2 +∑

w∈V \v,u

degwf(w)2

.

This is equivalent to

f(v)2 +1

deg v

f(u) +∑

w∈N (v)\u

f(w)

2

+

(ε2 − 1

4

)·

∑w∈V \N (v)

degwf(w)2

+

∑w∈N (v)\u

1

degw(f(v) + ε(degw − 1)f(w))2

≥ 1

4

f(u)2 + deg vf(v)2 +∑

w∈N (v)\u

degwf(w)2

,

which can be also rewritten as1 +∑

w∈N (v)\u

1

degw− deg v

4

f(v)2 +

(1

deg v− 1

4

)f(u)2

+2

deg v· f(u)

∑w∈N (v)\u

f(w)

+ f(v)

∑w∈N (v)\u

2

degw· ε(degw − 1)f(w)

+

1

deg v

∑w∈N (v)\u

f(w)

2

≥∑

w∈N (v)\u

(1

4· degwf(w)2 − 1

degw(ε(degw − 1)f(w))2

)

+

(1

4− ε2

) ∑w∈V \N (v)

degwf(w)2.

Using the fact that deg v ∈ 2, 3, we can finally complete the proof byconsidering the following three cases.

Case 1: ε2 ≤ 14 .

28

In this case, by letting

f(u) := −∑

w∈N (v)\u

f(w)

and f(v) := 0, the above inequality implies that

0 ≥∑

w∈N (v)\u

(1

4· degwf(w)2 − 1


)(13)

+

(1

4− ε2

)·

∑w∈V \N (v)

degwf(w)2

.

Since, for w ∈ N (v) \ u,1

4· degwf(w)2 − 1


≥ f(w)2

4·(

degw − (degw − 1)2

degw

)=

f(w)2

4· 2 degw − 1

degw≥ 0,

the inequality (13) must be an equality, implying that f |N (v)\u = 0

and ε2 = 14 . Thus, we have two cases:

(a) If Γ is connected, then it is either a book graph or a petalgraph.

(b) If Γ is not connected, then it has two connected componentsand at least one of them is either a book graph or a petalgraph.

In both cases it is possible to find an eigenfunction f for (Id−∆(Γ))2

with eigenvalue ε2, such that f |N (v)\u 6= 0, which contradicts theabove implication.

Case 2: ε2 > 14 and deg v = 2.

In this case, Γ = P2 and Γ = P4, which is a contradiction since weare assuming that Γ 6= P4.

Case 3: ε2 > 14 and deg v = 3.

In this case, since we cannot have Γ = P2 (as shown in the previouscase), Γ must have exactly two connected components, each of themis either P2 or a single vertex. But in any of these cases, one canshow that ε = ε(Γ) < 1

2 , which is a contradiction.

This proves the claim.

29

Acknowledgments

Raffaella Mulas was supported by the Max Planck Society’s Minerva Grant.

References

[1] M.S. Ashbaugh and R.D. Benguria. A sharp bound for the ratio of thefirst two eigenvalues of Dirichlet Laplacians and extensions. Annals ofMathematics, pages 601–628, 1992.

[2] A. Banerjee and J. Jost. On the spectrum of the normalized graphLaplacian. Linear algebra and its applications, 428(11-12):3015–3022,2008.

[3] F. Bauer and J. Jost. Bipartite and neighborhood graphs and thespectrum of the normalized graph Laplacian. Comm. Anal. Geom.,21:787 – 845, 2013.

[4] I. Chavel. Eigenvalues in Riemannian geometry. Academic Press, 1984.

[5] F. Chung. Spectral graph theory. American Mathematical Soc., 1997.

[6] K. Das and S. Sun. Extremal graph on normalized Laplacian spectralradius and energy. Electron. J. Linear Algebra, 29(1):237–253, 2016.

[7] G. Faber. Beweis, dass unter allen homogenen Membranen von gleicherFlache und gleicher Spannung die kreisformige den tiefsten Grundtongibt. Verlag d. Bayer. Akad. d. Wiss., 1923.

[8] J. Jost. Riemannian geometry and geometric analysis. Springer, 2017.

[9] J. Jost, R. Mulas, and F. Munch. Spectral gap of the largest eigenvalueof the normalized graph Laplacian. Communications in Mathematicsand Statistics, 2021.

[10] A.J. Kollar and P. Sarnak. Gap sets for the spectra of cubic graphs.arXiv:2005.05379.

[11] E. Krahn. Uber eine von Rayleigh formulierte Minimaleigenschaft desKreises. Mathematische Annalen, 94(1):97–100, 1925.

[12] E. Krahn. Uber Minimaleigenschaften der Kugel in drei und mehr Di-mensionen. Mattiesen, 1926.

[13] H.F. Weinberger. An isoperimetric inequality for the N-dimensionalfree membrane problem. Journal of Rational Mechanics and Analysis,5(4):633–636, 1956.

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Documents

arXiv:2110.08751v1 [math.CO] 17 Oct 2021