Petals and books: The largest Laplacian spectral gap from 1

We prove that, for any connected graph on N≥3 $N\ge 3$ vertices, the spectral gap from the value 1 with respect to the normalized Laplacian is at most 1/2. Moreover, we show that equality is achieved if and only if the graph is either a petal graph (for N $N$ odd) or a book graph (for N $N$ even). This implies that 12,32 $\left(\frac{1}{2},\frac{3}{2}\right)$ is a maximal gap interval for the normalized Laplacian on connected graphs. This is closely related to the Alon–Boppana bound on regular graphs and a recent result by Kollár and Sarnak on cubic graphs. Our result also provides a sharp bound for the convergence rate of some eigenvalues of the Laplacian on neighborhood graphs.


Introduction
A spectral gap is the maximal difference between two eigenvalues for linear operators in some class.Here, we consider Laplace operators, more precisely, the normalized Laplacian of a finite graph.Such inequalities were first studied for the Laplace operator ∆ = − n i=1 ∂ 2 ∂(x i ) 2 on a connected smooth domain Ω ⊂ R n with Dirichlet conditions, that is, ∆u = λu in Ω (1) where a (smooth) solution u that does not vanish identically is called an eigenfunction, and the corresponding value λ an eigenvalue.We choose the sign in the definition of ∆ to make it a non-negative operator.Since Ω is connected, the smallest eigenvalue λ 1 is positive, and the famous Faber-Krahn inequality [10,15,16] says that among all domains with the same 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 two eigenvalues, λ 2 λ 1 is 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 then the smallest eigenvalue is λ 1 = 0, with the constants being eigenfunctions, and Weinberger [19] proved that the second smallest eigenvalue λ 2 now is always less than or equal to that of the ball, again with equality only for the ball.While the proofs of such results can be difficult, there is a general pattern here, that the extremal cases occur only for a very particular class of domains, balls in this case.Similarly, for eigenvalue problems for the Laplace-Beltrami operator in Riemannian geometry, often the extremal case is realized by spheres (for the eigenvalue problem in Riemannian geometry, see for instance the references given in [4,11]).There are also discrete versions of those Laplacians, and naturally, their spectra have also been investigated.For the algebraic graph Laplacian, a systematic analysis of spectral gaps is presented in [14], and these authors have identified many beautiful classes of graphs with a particular structure of their spectra and spectral gaps.Here, we consider another discrete Laplacian, the normalized Laplace operator of a connected, finite, simple graph Γ = (V, E) on N ≥ 3 vertices.For a vertex v ∈ V , we denote by deg v its degree, that is, the number of its neighbors, i.e., the other vertices w ∼ v connected to v by an edge.Then the Laplacian for a function f : that is, we subtract from the value of f at v the average of the values at its neighbors.This operator generates random walks and diffusion processes on graphs, and it was first systematically studied in [6].Since its spectrum is that of an (N × N )-matrix with 1s in the diagonal, the eigenvalues below in terms of properties of the graph (see [6]), and the largest value among all graphs with N vertices is realized by the complete graph K N where λ 2 = N N −1 ; for all other graphs λ 2 ≤ 1 [6, Lemma 1.7].The largest eigenvalue λ N is always less than or equal to 2, with equality if and only if Γ is bipartite.And the gap 2 − λ N quantifies how different Γ is from being bipartite [3].In fact, the smallest possible value λ N = N N −1 is again realized only for K N .For all other graphs, λ N ≥ N +1 N −1 [8], and again, the extremal graphs, where equality is realized, can be characterized [12].
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 gaps in the middle of the spectrum.In fact, besides 0 and 2, also the eigenvalue 1 plays a special role.It arises, in particular, from vertex duplications [2].The extreme case is given by complete bipartite graphs K n,m with n + m = N .They have the eigenvalue 1 with multiplicity N − 2. In fact, if 1 occurs with this multiplicity, then by (5), 2 also has to be an eigenvalue, and the graph is bipartite.But like the eigenvalue 2, the eigenvalue 1 need not be present in a graph.Therefore, we can ask about the maximal spectral gap at 1, that is, we can ask what the maximal value of 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 can identify the class of graphs for which the maximal possible value ε = 1 2 is realized.In fact, in those cases, except for the triangle K 3 , which only has 3 2 , both values 1  2 and 3 2 are eigenvalues.Our results have fit into a larger picture.They have connections with expander graphs and random walks on graphs, including Alon-Boppana's theorem on Ramanujan graphs, Kollár-Sarnak's theorem on the maximal gap interval for cubic graphs [14], as well as Bauer-Jost's Laplacian on neighborhood graphs [3].We shall explain these relations in Section 2.

Main result
Throughout the paper we fix a connected, finite, simple graph Γ = (V, E) on N ≥ 3 vertices.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 let N (v) := {w ∈ V : w ∼ v} denote the neighborhood of v.We let denote the eigenvalues of the Laplacian in (4), and we let be the spectral gap from 1. Clearly, ε ≥ 0 and this inequality is sharp since, for instance, any graph with duplicate vertices (i.e., vertices that have the same neighbours [17]) has 1 as an eigenvalue.As anticipated in the introduction, here we prove that ε ≤ 1 2 and equality is achieved 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 graph is the graph on N = 2m + 1 vertices such that (Figure 1): Figure 1: The petal graph Petal graphs are also known as Dutch windmill graphs or friendship graphs.They appear in the famous Friendship Theorem from Erdös, Rényi and Sós [9], which states that the only finite graphs with the property that every two vertices have exactly one neighbor in common are precisely the petal graphs.In fact, a proof of this result can proceed via spectral methods.The friendship assumption determines the square of the adjacency matrix, and hence the spectrum of that matrix, and one then proceeds by showing that this spectrum implies that the graph in question has to be petal graph.This is another example, and one relevant for the present paper of how the structure of a graph can be determined from its spectrum.As shown in [6], for the petal graph on N = 2m + 1 vertices, the eigenvalues are 0, 1  2 (with multiplicity m−1) and 3 2 (with multiplicity m+1).Therefore, ε = 1 2 in this case.In fact, the eigenfunctions are easily constructed.Putting f (w i ) = −f (v i ) and f (x) = 0 produces m linearly independent eigenfunctions for the eigenvalue 3  2 .Letting f (v i ) = f (w i ) = 1 for all i and f (x) = −2 produces another eigenfunction for that eigenvalue.With f (v i ) = f (w i ) for all i, i f (v i ) = 0 = f (x) we get m − 1 linearly independent eigenfunctions for the eigenvalue 1  2 .Again, this is a good example of how the structure of a graph and properties of its spectrum are tightly related.Whenever we have two neighboring vertices v, w that have all their other neighbors in common, the function with f (v) = 1, f (w) = −1, f (z) = 0 for all other vertices is an eigenfunction, and the eigenvalue depends on the number of those common neighbors.When, as here, there is only one common neighbor for such a pair, the eigenvalue is 3/2.This eigenvalue will also occur for the book graph to be introduced in a moment, which for an even number of pages is a two-fold cover of the petal graph.But the book graph will be bipartite, and hence have the eigenvalue 2, which the petal graph, not being bipartite, cannot possess.
Definition 2 (Book graph, N ≥ 4 even).Given m ≥ 1, the m-book graph is the graph on N = 2m + 2 vertices such that (Figure 2): Figure 2: The book graph Remark 1.For the book graph on N = 2m + 2 vertices, 0 and 2 are eigenvalues with multiplicity 1, since Γ is connected and bipartite.Moreover, one can check that λ = 1 ± 1 2 are eigenvalues with multiplicity m each.In fact, the corresponding eigenfunctions can be constructed as follows: , we obtain one more eigenfunction for 1 2 and similarly by letting g(v i ) = g(w i ) = −1 and g(x) = g(y) = 2, we obtain one more eigenfunction for 3  2 .
Our main result is that these, and only these, examples have the largest possible spectral gap ε = 1 2 at 1.
Theorem 1.For any connected graph Γ on N ≥ 3 vertices, Moreover, equality is achieved if and only if Γ is either a petal graph (for N odd) or a book graph (for N even).
As a consequence of Theorem 1, we can infer that both petal graphs and book graphs are uniquely characterized by their normalized Laplacian spectra.This relates to [7], where it has been proved that, among connected graphs, the petal graphs are uniquely determined by the eigenvalues of the adjacency matrix.
We prove Theorem 1 in Section 3.For completeness, we also state the following result on the value max i̸ =1 |λ i − 1|.
Proposition 2. For any connected graph Γ on N ≥ 2 vertices, Moreover, the lower bound is an equality if and only if Γ is the complete graph; the upper bound is an equality if and only if Γ is bipartite.
We also prove the following lemma, which will be needed in the proof of Theorem 1 and which is an interesting result itself, since it allows us to characterize ε for any graph.Lemma 3.For any graph Γ, Proof.We observe that the values (1 − λ 1 ) 2 , . . ., (1 − λ N ) 2 are exactly the eigenvalues of the matrix M := (Id −D ) 2 whose entries are where Id is the N ×N identity matrix and D = diag(deg 1, • • • , deg N ) is the diagonal matrix consisting of degrees.In particular, ε 2 is the smallest eigenvalue of M .Therefore, by the Courant-Fischer-Weyl min-max principle, it can be written as Now, observe that the numerator can be rewritten as It follows that This lemma will play an important role for controlling ε, because we can derive inequalities on vertex degrees in case ε ≥ 1  2 , by choosing suitable local functions f .Remark 2. A gap interval with respect to a family G of simple graphs is an open interval such that there are infinitely many graphs in G whose Laplacian spectrum does not intersect the interval.Clearly, Theorem 1 trivially implies that, for G = {connected graphs}, ( 1 2 , 3 2 ) is a maximal gap interval.This can be compared to a result by Kollár and Sarnak [14], which states that, for G = {connected regular graphs of degree 3}, ( 23 , 4 3 ) is a maximal gap interval.(The original result, Theorem 3 in [14], is formulated in terms of the adjacency matrix, but since these graphs are regular, the statement can be equivalently reformulated in terms of ∆.)We also refer to the recent work of the application of gap intervals to microwave coplanar waveguide resonators by Kollár et al. [13].This line of research has an origin in the Alon-Boppana theorem [18,5], which implies that (0, 1 3 ) is a maximal gap interval for the Laplacian on cubic graphs, and this gap is achieved by Ramanujan graphs.
Combining Theorems 1 and 16, we have the following This is rather interesting, and can be compared to Alon-Boppana's work stating that for any d-regular Ramanujan graph, See Proposition 2 for another comparison.Furthermore, our results provide a sharp bound for the convergence rate of some eigenvalues of the Laplacian on neighborhood graphs [3].
The neighborhood graph Γ [ℓ] of order ℓ of a graph Γ = (V, E) is a weighted graph whose edge weight w ij equals the probability that a random walker starting at i reaches j in ℓ steps.The neighborhood graphs {Γ [ℓ] } ∞ ℓ=1 encode properties of random walks on Γ, asymptotic ones if ℓ → ∞.We thereby gain a new source of geometric intuition for obtaining eigenvalue estimates.The graph Laplacian ∆ [ℓ] on Γ [ℓ] satisfies ∆ [ℓ]  [3]).By Theorem 1, we have: Theorem 5.For every connected graph Γ with at least three vertices, there is some eigenvalue λ [ℓ] of ∆ [ℓ] with When ℓ is even, the largest eigenvalue of and both bounds are sharp.

Proof of the main result
This section is dedicated to the proof of Theorem 1, that we split into two main parts.In particular, in Section 3.1 we prove that ε ≤ 1 2 for any connected graph Γ on N ≥ 3 vertices, while in Section 3.2 we prove that ε = 1 2 if and only if Γ is either a petal graph or a book graph.

Upper bound
In this section we prove the first claim of Theorem 1, namely Theorem 6.For any connected graph Γ on N ≥ 3 vertices, Before proving it by contradiction, we show several properties that a connected graph on N ≥ 3 vertices with ε > 1 2 would have.We begin by observing that, by Lemma 3, such a graph should satisfy Lemma 7. Let Γ be a connected graph on N ≥ 3 vertices such that ε > 1 2 .Then, for any v ∈ V , there exists w ∈ N (v) such that deg w ≤ 3.
Proof.Let f be such that f (v) := 1 and f (u) := 0 for all u ̸ = v.Then, (6) implies which is a contradiction.
Given two vertices u and v, we denote by N (u)△N (v) the symmetric difference of N (u) and N (v), i.e., the set of vertices that are neighbors of either u or v.
Proof.Let f be such that f (v) := 1, f (u) := −1 and f (w) := 0 for all w ∈ V \ {u, v}.Then, by (6), the above inequality can be rewritten as 2. There are no distinct vertices u ∼ v ∼ w with Proof.The first statement is a direct consequence of Lemma 8. We now prove the second statement.Assume the contrary, that is, there exist three vertices u ∼ v ∼ w with {deg u, deg v, deg w} ⊆ {1, 2}.We only need to check the following two cases. Case Since N (u) ∩ N (w) ̸ = ∅, by (7), the vertex w 1 ∈ N (w) \ {v} has degree 1.Hence, Γ is a path of length 3 (Figure 3).But this is a contradiction, since ε = 1 2 for the path of length 3, which is also the 1-book graph.Since N (u) ∩ N (w) ̸ = ∅, by (7), the vertices w 1 ∈ N (w) \ {v} and u 1 ∈ N (u) \ {v} have degree 1.Hence, Γ is a path of length 4. But this is also a contradiction, since ε < 1 2 in this case.
Lemma 10.Let Γ be a connected graph on N ≥ 3 vertices such that ε > 1 2 .Then, there are no vertices u ∼ v ∼ w with Proof.Assume the contrary, that is, there exist three vertices u ∼ v ∼ w with {deg u, deg v, deg w} ⊆ {2, 3}.We need to check five cases.By applying (7) three times, we have that: • There exists w 1 ∈ N (u)△N (w) with deg w 1 = 1, and without loss of generality, we may assume that w 1 ∼ w; Hence, Γ is the graph in Figure 4.If there is only one vertex x in N (u)△N (w), then by (7), it must have degree 1.Also, by construction, it is clear that x ∼ w.Moreover, by (7), the only vertex y ∈ N (w) \ {v, x} has degree 1.But this is a contradiction, since we must also have y ∼ u.
Otherwise, there are three vertices in N (u)△N (w) and at least two of them have degree 2. In this case, we reduce to Case 1.If deg x ≥ 2 for all x ∈ N (u)△N (w), then by (7), there are at least two vertices x and y in N (u)△N (w) with degree 2, and therefore we reduce to Case 1.
There are two sub-cases left: a) If there exists u 1 ∼ u with deg u 1 = 1, then by (7) there exists v 1 ∼ v with deg v 1 = 1 and there exists w 1 ∼ w with deg w 1 = 1.Therefore, Γ has the same local structure as the graph in Figure 5.By letting f (u Similar to the above cases, by repeatedly applying (7) we can see that there exist Therefore, Γ has the same local structure as the graph in Figure 6.Similarly to Case 3, also in this case we derive a contradiction.
Figure 6: A graph arising in the proof of Lemma 10 We now fix the following notations.Given a vertex v, we let

In this case, for any two vertices
because otherwise, we can reduce to Lemma 10.Therefore, except for at most one vertex in N 2,3 (v), any other vertex in N 2,3 (v) is adjacent to a vertex of degree 1.Now, let f (x) := −1 if deg x = 1 and x is adjacent to some vertex in N 2,3 (v), let f (v) := 1 and let f (y) := 0 otherwise.Then, by (6) we obtain which is a contradiction.
Case 2: There exists w ∈ N (v) with deg w = 1.Then, by Lemma 9, Similarly to the previous case, this implies that which is a contradiction.
Proof.Assume the contrary, i.e., w ∼ v, deg w = 1 and deg v ≥ 3.Then, similarly to the proof of Lemma 11, we have that for any u ∼ v, there exists which is a contradiction.We are now finally able to prove Theorem 6.
Proof of Theorem 6. Assume by contradiction that there exists a connected graph Γ on N ≥ 3 vertices such that ε > 1 2 .Then, by Lemma 14, there exist u ∼ v ∼ w such that deg u = 1, deg v = 2 and deg w ∈ {2, 3}.By (7), there exists a vertex w 1 ∈ N (w) \ {v} satisfying deg w 1 = 1, and by Lemma 12, deg w = 2. Therefore Γ is the path of length 3 in Figure 3.But as we know this is a contradiction, since ε = 1 2 for the path of length 3, which is also the 1-book graph.

Optimal cases
In the previous section we proved that ε ≤ 1 2 for any connected graph Γ on N ≥ 3 vertices.Hence, in order to prove Theorem 1, it is left to prove that ε = 1 2 if and only if Γ is either a petal graph or a book graph.We dedicate this 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 greater than or equal to 3, then ε < 1 2 .In fact, we prove an even stronger result, since we show that, in this case, 2. In Section 3.2.2we show that, if d = 2, then ε = 1 2 if and only if Γ is either a petal graph or a book graph.
3. In Section 3.2.3we 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 6, if a connected graph Γ on N ≥ 3 vertices is such that ε = 1 2 , then Moreover, with the same proof as Lemma 8, one can prove the following 3.2.1The case d ≥ 3 In this section we prove the following Theorem 16.For any connected graph Γ on N ≥ 3 vertices with smallest degree d ≥ 3, Before that theorem, we shall prove several preliminary results.
Lemma 17.Let Γ be a connected graph on N ≥ 3 vertices with smallest degree d ≥ 3 and ε > Proof.The first claim follows directly from Lemma 3, while the second claim can be proved as Lemma 8.
Lemma 18.Let Γ be a connected graph on N ≥ 3 vertices with smallest degree d ≥ 3 and ε > Proof.Assume the contrary, then which contradicts (11).Proof.If not, then which contradicts (11).Proof.Fix u ′ and v ′ that satisfy Lemma 20.Then, Hence, without loss of generality, we can assume that there are two vertices u and v in N (u ′ ) \ N (v ′ ) with deg u = deg v = d.This proves the claim.
We can now prove Theorem 16.
Proof of Theorem 16.We first observe that the second inequality follows from the fact that the sequence

The case d = 2
Here we prove the following Theorem 22.For any connected graph Γ on N ≥ 3 vertices with smallest degree d = 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 start by showing that, if Γ = C N is the cycle graph on N ≥ 3 vertices, then ε = 1 2 if and only if either N = 3 (in which case Γ = C 3 is the 1-petal graph) or N = 6 (in which case Γ = C 6 is the 2-book graph).

Let N > 6 and let
for some vertex y.By the assumption, deg y ≥ 2 and deg w ≥ 3. Hence, applying (9) to N (v)△N (x) implies which is a contradiction.If there are at most five vertices in N (u)△N (w) of degree 3, then the other vertices in N (u)△N (w) have degree at least 4. Applying (9) to N (u)△N (w), we then obtain which is a contradiction.
(2 + 3), therefore the vertex w 1 ∈ N (w) \ {u, v} has degree 2. Similarly, there exists x ∼ w 1 with deg x = 2.By letting now f (w which is a contradiction.Therefore, w ̸ ∼ u.Now, if the vertex u 1 ∈ N (u) \ {v} has degree 2, then we reduce to Proposition 24.Thus, deg u 1 ≥ 3. Applying (9) to the vertices u and w, we infer that the two vertices w 1 , w 2 ∈ N (w) \ {v} satisfy deg Again, applying (9) to the vertex pairs (v, w 1 ) and (v, w 2 ), we infer that the vertices x 1 ∼ w 1 and x 2 ∼ w 2 satisfy deg which is a contradiction.Hence, N (x 1 ) ∩ N (x 2 ) ̸ = ∅ and, similarly, also In particular, there exists z such that z ∼ u, z ∼ x 1 and z ∼ x 2 .By letting now f (w which implies that deg z ≤ 4. We now claim that deg z = 3. Suppose the contrary, then deg z = 4. Let := 1 and f := 0 otherwise together with (8) gives which is a contradiction.Therefore, for any a ∈ N (z 1 ), deg a ≥ 3.By (9) applied to z 2 and v, we infer that there exist at least three vertices of degree 3 in N (z 1 ), and deg z 1 This proves that deg z = 3, which implies that Γ must be the book graph on N = 8 vertices.
Lemma 28.Let Γ be a connected graph on N ≥ 3 vertices with smallest degree d = 2 and ε = 1 2 .If Γ is not in the classes described in Proposition 24 and Proposition 27, 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 Lemma 26, also a vertex w ∼ v with deg w = 2. Hence, Γ would be in the class described in Proposition 27, 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 and deg w ∈ {2, 3}.This follows from the fact that, otherwise, applying (9) to u and w together with the above remarks implies that there exists w 1 ∼ w with deg w 1 = 2, while deg w = 3, which is a contradiction.But on the other hand, for any x, y ∈ N (u), by (9) there exists w ∈ N (x)△N (y) with deg w ∈ {2, 3}, which contradicts the above argument.
We are now able to prove Theorem 22.
Proof of Theorem 22.We shall illustrate this proof in Figure 8 below.Observe that, if Γ is in one of the classes described in Proposition 24 and Proposition 27, then the claim follows.Therefore, we can assume that this is not the case.We can then apply Lemma 28 and infer that there is no vertex of degree 3. Also, if all vertices have degree < 3, from the fact that d = 2 it follows that all vertices have degree 2, but this is not possible, since we are assuming that Γ does not satisfy the condition in Proposition 24.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 ≥ 4 such that N 2 (u) ̸ = ∅.This can be seen by the fact that, if not, then all vertices of degree 2 would only be adjacent to vertices of degree 2, contradicting the fact that Γ is connected.Now, from the fact that N 2 (u) ̸ = ∅, we can infer two facts about N (u) = N 2 (u) ⊔ N ≥4 (u) : 1.For any v ∈ N 2 (u), there exists a unique w ∼ v with deg w = 2.This follows immediately from Lemma 26 and from the fact that deg u ≥ 4.
2. If N ≥4 (u) ̸ = ∅, then for any v ∈ N ≥4 (u) there exists w ∼ v with deg w = 2.This can be shown by fixing v ′ ∈ N 2 (u) and then applying (9) to As a consequence of the first fact, we can write where: • For each i = 1, . . ., k, there exists a unique w i ∈ V \ N 2 (u) such that v i ∼ w i and deg w i = 2; Now, for i = 1, . . ., k, since v i ∼ w i and deg v i = deg w i = 2, we can infer that x i ∈ N (w i ) \ {v i } must have degree ≥ 4, because otherwise, Γ would satisfy the assumption of Proposition 24, but we are assuming that this is not the case.Therefore, in particular, w 1 , . . ., w k must be pairwise distinct, while x 1 , . . ., x k do not need to be pairwise distinct.Up to relabeling x 1 , . . .x k , we assume that where: • k 1 , . . ., k p+q are such that k = k 1 + . . .+ k p+q , Following the above notation, we also relabel the vertices v 1 , . . ., v k and w 1 , . . ., w k as v 1 , . . ., v k 1 , . . ., v k 1 +...+k p+q , w 1 , . . ., w k 1 , . . ., w k 1 +...+k p+q .

Now, by letting
We now use the following known facts: • deg y i ≥ 4, for each i = 1, . . ., r; , for i = p + 1, . . ., p + q.Putting everything together, we obtain that which implies that r = q = 0, either l = 0 or k = 0 (but they cannot be both zero), and deg x ′ i = k i ≥ 4, for i = 1, . . ., p.We then have two cases: 1.If k = 0, then Γ is a petal graph with 2l + 1 vertices.

The case d = 1
Given N ≥ 3, we let P N denote the path graph on N vertices, and we observe that P 4 coincides with the 1-book graph.It is left to prove the following Theorem 29.For any connected graph Γ on N ≥ 3 vertices with smallest degree d = 1, if Γ ̸ = P 4 then ε < 1 2 .
Proof.Suppose the contrary, then there exists a connected graph Γ ̸ = P 4 on N ≥ 3 vertices with ε = 1 2 and d = 1.We fix such Γ = (V, E) so that it is the graph with these properties that has the smallest possible order, that is, if Γ ′ ̸ = P 4 is a connected graph with smallest degree 1 and 3 ≤ N ′ < N vertices, then ε(Γ ′ ) < 1  2 .If we show that the existence of Γ brings to a contradiction, then we are done.Fix u ∈ V of degree 1, and let v ∼ u be its only neighbour.Then, by letting f (u) := 1 and f := 0 otherwise, (8) gives Moreover, since ε2 = (1 − λ) 2 for some eigenvalue λ of Γ, from (4) it follows that and similarly, for all w ∈ V \ {u, v}, In particular, without loss of generality, we may assume ε as the proof can be easily adapted otherwise.Now, since we are assuming that ε = ε(Γ) = 1 2 , we have that, independently of the value of f (u) ∈ R that we choose, (8) implies or equivalently, which by the above observations can also be rewritten as We now consider two cases.Since ε2 ≤ 1 4 by Theorem 6 and f (v) = 0 implies f | V \{v,u} ̸ = 0, we deduce that ε2 = 1  4 .But this brings to a contradiction.In fact, by construction, it is clear that Γ is a connected graph on N − 1 ≥ 4 vertices, since v has degree 3 in Γ.Hence, if Γ has vertices of degree 1, then by the assumption on the graphs with less than N vertices, Γ must be P 4 , but this is not possible because of the degree of v. Similarly, if Γ does not have vertices of degree 1, then by Theorem 22, 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 8 vertices.But in this case, one can directly check that ε = ε(Γ) < 1  2 , which is a contradiction.In fact, the case in which ε is replaced by −ε is similar.
Case 2: ε2 > 1 4 and deg v = 2.In this case, Γ = P 2 and Γ = P 4 , which is a contradiction since we are assuming that Γ ̸ = P 4 .Case 3: ε2 > 1 4 and deg v = 3.In this case, since we cannot have Γ = P 2 (as shown in the previous case), Γ must have exactly two connected components, each of them is either P 2 or a single vertex.But in any of these cases, one can show that ε = ε(Γ) < 1  2 , which is a contradiction.This proves the claim.

Figure 3 :
Figure 3: A graph arising in the proof of Lemma 9

Figure 4 :
Figure 4: A graph arising in the proof of Lemma 10

Figure 5 :
Figure 5: A graph arising in the proof of Lemma 10

Lemma 13 .
Let Γ be a connected graph on N ≥ 3 vertices such that ε > 1 2 .Then, there exist three vertices u ∼ v ∼ w with deg u ≤ 3 and deg w ≤ 3. Proof.By Lemma 7, there exist u 1 ∼ u 2 with deg u 1 ≤ 3 and deg u 2 ≤ 3.If v ∼ u 2 , then by (7) there exists w ∈ N (u 1 )△N (v) with deg w ≤ 3.If w ∈ N (u 1 ) \ N (v), then we have w ∼ u 1 ∼ u 2 with deg w ≤ 3 and deg u 2 ≤ 3; while if w ∈ N (v) \ N (u 1 ), then we have w ∼ v ∼ u 2 with deg w ≤ 3 and deg u 2 ≤ 3. Lemma 14.Let Γ be a connected graph on N ≥ 3 vertices such that ε > 1 2 .Then, there exist three vertices u ∼ v ∼ w such that deg u = 1, deg v = 2 and deg w ∈ {2, 3}.Proof.Fix u ∼ v ∼ w with deg u ≤ 3 and deg w ≤ 3 as in Lemma 13.By Lemma 9, at most one between u and w has degree 1, and by Lemma 11, at most one between u and w has degree 2 or 3. Therefore, we may assume that deg u = 1 and deg w ∈ {2, 3}.By Lemma 12, this implies that deg v = 2.

5 , 1 d
. . . is strictly decreasing.Hence, it is left to show that ε ≤ .By Lemma 21, there exist u and v such thatN (u) ∩ N (v) = ∅ and deg u = deg v = d.This implies that |N (u)△N (v)| ≤ 2d − 2,but together with Lemma 18, this brings to a contradiction.

Proposition 24 .
Let Γ be a connected graph on N ≥ 3 vertices with smallest degree d = 2 and ε = 1 2 .If three vertices are such that u ∼ v ∼ w and deg u = deg v = deg w = 2, then Γ = C 3 or C 6 .Proof.If u ∼ w, then clearly Γ = C 3 .If u ̸ ∼ w, then (9) applied to N (u)△N (w) implies that the vertices u 1 ∈ N (u) \ {v} and w 1 ∈ N (w) \ {v} are such that deg u 1 = deg w 1 = 2.If u 1 = w 1 , then clearly Γ = C 4 .Otherwise, we can repeat the process and obtain any cycle graph C N .By Proposition 23, the claim follows.Lemma 25.Let Γ be a connected graph on N ≥ 3 vertices with smallest degree d = 2 and ε = 1 2 .Then, there exist two adjacent vertices u ∼ v such that deg u = deg v = 2.

Figure 7 :
Figure 7: The vertices in the proof of Lemma 25

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

≥ 4 .
Let b 1 , b 2 ∈ N (z 1 ) be such that deg b 1 = deg b 2 = 3.Then, (9) applied to b 1 and b 2 implies that there is at least one vertex of degree 2 in N (b 1 )△N (b 2 ).Without loss of generality, we can assume that there exists b ′ ∈ N (b 1 ) \ N (b 2 ) such that deg b ′ = 2.By Lemma 26, there exists c ′ ∼ b ′ with deg c ′ = 2.But this is a contradiction, since we proved that all vertices adjacent to b 1 must have degree 2.

Figure 8 :
Figure 8: Illustration of the proof of Theorem 22

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