Lucas Chesnel
email: lucas.chesnel@inria.fr

Sergei A Nazarov
email: srgnazarov@yahoo.co.uk

Jari Taskinen
email: jari.taskinen@helsinki.fi.

Spectrum of the Laplacian with mixed boundary conditions in a chamfered quarter of layer

Keywords: Laplacian with mixed boundary conditions, chamfered quarter of layer, Fichera layer, discrete spectrum, trapped modes Mathematics Subject Classification. 35J05, 35P15, 35Q40, 81Q10, 65N25

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Formulation of the problem

In this work, we study the spectrum of a Laplace operator with mixed boundary conditions in a chamfered quarter of layer characterized by two parameters κ 1 , κ 2 ∈ R gathered in some vector κ = (κ 1 , κ 2 ). To describe the geometry, referring to carpentry and locksmith tools, we first define the "blade"

B κ 1 := x = (x 1 , x 2 , x 3 ) ∈ R 3 | x 1 > κ 1 x 3 , x 2 ∈ R, x 3 ∈ (0, 1) (1)
(see Figure 1 left). Then we introduce the "incisor"

Ω κ := {x ∈ B κ 1 | x 2 > κ 2 x 3 } (2)
(see Figure 1 centre). Let us give names to the different components of the boundary ∂Ω κ of Ω κ . First, denote by Σ κ the union of the two "horizontal" quadrants:

Σ κ := {x ∈ ∂Ω κ | x 3 = 0 or x 3 = 1}.
Then consider the laterals sides of the incisor. Set

Γ κ 1 := ∂Ω κ ∩ B κ 1 , Γ κ 2 := ∂Ω κ \ {Γ κ 1 ∪ Σ κ }. (3)
We study the spectral problem with mixed boundary conditions

-∆ x u = λu in Ω κ u = 0 on Σ κ ∂ ν u = 0 on Γ κ 1 ∪ Γ κ 2 , ( 4
)
where ∂ ν is the outward normal derivative on ∂Ω κ . Denote by H 1 0 (Ω κ ; Σ κ ) the Sobolev space of functions of H 1 (Ω κ ) vanishing on Σ κ . Classically (see e.g. [START_REF] Ladyzhenskaya | The Boundary Value Problems of Mathematical Physics[END_REF][START_REF] Lions | Problèmes aux limites non homogènes et applications[END_REF]), the variational formulation of Problem [START_REF] Colton | Inverse acoustic and electromagnetic scattering theory[END_REF] writes

Find (λ, u) ∈ R × H 1 0 (Ω κ ; Σ κ ) \ {0} such that (∇ x u, ∇ x ψ) Ω κ = λ (u, ψ) Ω κ ∀ψ ∈ H 1 0 (Ω κ ; Σ κ ), (5)
where for a domain Ξ, (•, •) Ξ stands for the inner product of the Lebesgue spaces L 2 (Ξ) or (L 2 (Ξ)) 3 according to the case. Using the homogeneous Dirichlet condition on Σ κ for the functions in H 1 0 (Ω κ ; Σ κ ), one can prove that there holds the Friedrichs inequality

∥u; L 2 (Ω κ )∥ 2 ≤ c κ ∥∇ x u; L 2 (Ω κ )∥ 2 ∀u ∈ H 1 0 (Ω κ ; Σ κ ).
Classically (see e.g [3, §10.1], [START_REF] Reed | Methods of modern mathematical physics. IV. Analysis of operators[END_REF]Ch. VIII.6]), the variational problem [START_REF] Dauge | Dirichlet spectrum of the Fichera layer[END_REF] gives rise to the unbounded operator A κ of L 2 (Ω κ ) such that

A κ : D(A κ ) → L 2 (Ω κ ) u → A κ u = -∆u, with D(A κ ) := {u ∈ H 1 0 (Ω κ ; Σ κ ) | ∆u ∈ L 2 (Ω κ ) and ∂ ν u = 0 on Γ κ 1 ∪ Γ κ 2 }.
The operator A κ is positive definite and selfadjoint. Since Ω κ is unbounded, the embedding H 1 0 (Ω κ ; Σ κ ) ⊂ L 2 (Ω κ ) is not compact and A κ has a non-empty essential component σ ess (A κ ) ( [START_REF] Sh | Spectral theory of selfadjoint operators in Hilbert space[END_REF]Thm. 10.1.5]). Let us mention that the case

κ 1 = κ 2 = 1
plays a particular role. Indeed in this situation, if u is an eigenfunction associated with an eigenvalue of A κ , by extending u via even reflections with respect to the faces Γ κ 1 , Γ κ 2 , one gets an eigenvalue of the Dirichlet Laplacian in the Fichera layer 1 right (see [START_REF] Fichera | Asymptotic behaviour of the electric field and density of the electric charge in the neighbourhood of singular points of a conducting surface[END_REF] for the original article that gave rise to the name). This latter problem has been studied in [START_REF] Dauge | Dirichlet spectrum of the Fichera layer[END_REF][START_REF] Bakharev | Existence of the discrete spectrum in the Fichera layers and crosses of arbitrary dimension[END_REF]. More precisely, in [START_REF] Dauge | Dirichlet spectrum of the Fichera layer[END_REF] the authors give a characterization of the essential spectrum of the Dirichlet Laplacian and show that the discrete spectrum has at most a finite number of eigenvalues. The existence of discrete spectrum is proved in [START_REF] Bakharev | Existence of the discrete spectrum in the Fichera layers and crosses of arbitrary dimension[END_REF]Thm. 2].

F = j=1,2,3 x | x j ∈ (0, 1), x k > 0, k ̸ = j represented in Figure
The goal of this paper is to get similar information for the operator A κ with respect to the parameter κ. Observe that by exchanging the axes or modifying their orientations, there is no loss of generality to restrict our study to the cases

κ 1 ≥ 0, |κ 2 | ≤ κ 1 . ( 6
)
Note that we consider mixed boundary conditions in (3) because we have in mind to study the spectrum of the Dirichlet Laplacian in geometries similar to the Fichera layer of Figure 1. Indeed, playing with symmetries, we can reduce the analysis of certain of these problems to the one of (3) (see Figure 13 for an example). On the other hand, let us mention that the spectrum of the Dirichlet Laplacian in Ω κ has a rather simple structure with only essential spectrum and no discrete spectrum. This is a consequence of Proposition 8.1 in Appendix.

This note is organized as follows. In Section 2, we describe the essential spectrum of A κ (Theorem 2.1). Then in Section 3, we state the results for the discrete spectrum of A κ (the main outcome of the present work is Theorem 3.2). The next four sections contain the proof of the different items of Theorem 3.2. In Section 8, we illustrate the theory with some numericals results. Finally we establish the above mentioned result related to the Dirichlet Laplacian in Ω κ in the Appendix. Introducing the angle α 1 ∈ [0, π/2) such that κ 1 = tan α 1 , the blade (1) can also be defined as

Essential spectrum

ξ 1 ξ 2 ξ 1 = ξ 2 tan α 1 e 1 e 3 e 2 α 1
B tan α 1 = {(x 1 , x 2 , x 3 ) ∈ R 3 | (x 1 , x 3 ) ∈ Π α 1 }
where Π α 1 stands for the 2D pointed strip

Π α 1 := ξ = (ξ 1 , ξ 2 ) ∈ R 2 | ξ 1 > ξ 2 tan α 1 , ξ 2 ∈ (0, 1) (7)
(see Figure 2). To describe σ ess (A κ ), we need information on the spectrum of the auxiliary planar problem

-∆ ξ v = µv in Π α 1 v = 0 on ς α 1 ∂ ν v = 0 on γ α 1 (8)
where ς α 1 := {ξ ∈ ∂Π α 1 | ξ 2 = 0 or ξ 2 = 1} denotes the horizontal part of ∂Π α 1 and γ α 1 := ∂Π α 1 \ς σ stands for the oblique part of ∂Π α 1 .

The continuous spectrum of Problem (8) coincides with the ray [π 2 , +∞). When α 1 = 0 (straight end), working with the decomposition in Fourier series in the vertical direction, one can prove that the discrete spectrum is empty. On the other hand, for all α 1 ∈ (0, π/2), it has been shown in [START_REF] Kamotskii | On eigenfunctions localized in a neighborhood of the lateral surface of a thin domain[END_REF] that there is at least one eigenvalue below the continuous spectrum (see also [START_REF] Nazarov | Discrete spectrum of cranked, branching, and periodic waveguides[END_REF] for more general shapes). Notice that by extending Π α 1 by reflection with respect to γ α 1 , we obtain a broken strip that we can also call a V-shaped domain. This allows us to exploit all the results from [START_REF] Exner | On existence of a bound state in an L-shaped waveguide[END_REF][START_REF] Nazarov | Trapped modes in angular joints of 2D waveguides[END_REF] (see also [START_REF] Dauge | Plane waveguides with corners in the small angle limit[END_REF] as well as the amendments in [START_REF] Nazarov | Asymptotics of eigenvalues of the Dirichlet problem in a skewed T -shaped waveguide[END_REF]) to get information on µ α 1 1 , the smallest eigenvalue of (8). In particular, the function α 1 → µ α 1 1 is smooth and strictly decreasing on (0, π/2). Additionally, we have lim

α 1 →0 + µ α 1 1 = π 2 , lim α 1 →(π/2) - µ α 1 1 = π 2 4 ( 9
)
(see Figure 8 for a numerical approximation of α 1 → µ α 1 1 ). By adapting the approach proposed in [5, §3.1], one establishes the next assertion. The only point to be commented here is that there holds

λ κ † := µ arctan κ 1 1 ≤ µ arctan |κ 2 | 1 (10

Discrete spectrum

For the discrete spectrum σ d (A κ ), our main results are as follows:

Theorem 3.1. For κ 1 = κ 2 = 0 (straight edges), σ d (A κ ) is empty. Theorem 3.2. Assume that κ 1 > 0. 1) σ d (A κ ) is non-empty for κ 2 ∈ [-κ 1 , 0).
2) There exists h(κ 1 ) > 0 such that:

i) σ d (A κ ) is empty for κ 2 ∈ [0, h(κ 1 )]; ii) σ d (A κ ) is non-empty for κ 2 ∈ (h(κ 1 ), κ 1 ]. 3) For κ 2 ∈ [-κ 1 , 0) ∪ (h(κ 1 ), κ 1 ], denote by λ κ 1 the first (smallest) eigenvalue of σ d (A κ ). The function κ 2 → λ κ
1 is strictly increasing on [-κ 1 , 0) and strictly decreasing on (h(κ 1 ), κ 1 ].

4)

For κ 2 ∈ (-κ 1 ; κ 1 ), σ d (A κ ) contains at most a finite number of eigenvalues.

The items 1)-3) of Theorem 3.2 are illustrated by Figure 3. Note in particular that we have the following mechanism for positive κ 2 : diminishing κ 2 from the value κ 1 makes the eigenvalue λ κ 1 to reach the threshold

λ κ † = µ arctan κ 1 1 at a certain κ 2 = h(κ 1 ) ∈ (0, κ 1 )
. The rest of the present note is dedicated to the proof of the above statements.

-κ 2 κ 2 κ 1 λ κ † = µ arctan κ 1 1 λ κ 1 h(κ 1 )
Figure 3: Picture of the behaviour of λ κ 1 , the smallest eigenvalue of σ d (A κ ), for a given κ 1 > 0 and

κ 2 ∈ [-κ 1 , 0) ∪ (h(κ 1 ), κ 1 ].

Discrete spectrum for negative κ 2

In this section, we prove the item 1) of Theorem 3.2 and so we consider the case κ 2 < 0. A direct application of the minimum principle, see e. discrete spectrum of A κ contains an eigenvalue λ κ

1 if one can find a trial function ψ ∈ H 1 0 (Ω κ ; Σ κ ) such that ∥∇ x ψ; L 2 (Ω κ )∥ 2 < λ κ † ∥ψ; L 2 (Ω κ )∥ 2 . ( 11
)
Let us construct a function satisfying [START_REF] Kato | Perturbation theory for linear operators[END_REF]. To proceed, first divide the incisor Ω κ into the two domains

Ω κ -:= {x ∈ Ω κ | x 2 < 0}, Ω κ + := {x ∈ Ω κ | x 2 > 0} = Ω (0,κ 2 ) (12)
(see Figure 4). Then for ε > 0 small, define ψ ε such that

ψ ε (x) = v(x 1 , x 3 ) in Ω κ - e -εx 2 v(x 1 , x 3 ) in Ω κ + ( 13
)
where v is an eigenfunction of the 2D problem ( 8) associated with µ α 1 1 , the smallest eigenvalue, and α 1 = arctan κ 1 . To set ideas, we choose v such that ∥v; L 2 (Π α 1 )∥ = 1. Note that ψ ε satisfies the homogeneous Dirichlet condition on Σ κ and decays exponentially at infinity. Using [START_REF] Kamotskii | On eigenfunctions localized in a neighborhood of the lateral surface of a thin domain[END_REF], we obtain

∥∇ x ψ ε ; L 2 (Ω κ + )∥ 2 -λ κ † ∥ψ ε ; L 2 (Ω κ + )∥ 2 = ∥∇ ξ v; L 2 (Π α 1 )∥ 2 + (ε 2 -µ α 1 1 )∥v; L 2 (Π α 1 )∥ 2 ¢ ∞ 0 e -2εx 2 dx 2 = ε 2 . ( 14
)
As for the integral over the prism Ω κ -with triangular cross-sections and the bevelled end, we integrate by parts and take into account the boundary conditions of (4), which yields

∥∇ x ψ ε ; L 2 (Ω κ -)∥ 2 -λ κ † ∥ψ ε ; L 2 (Ω κ -)∥ 2 = - ¢ Ω κ - v(x 1 , x 3 ) ∆ x v(x 1 , x 3 ) + µ α 1 1 v(x 1 , x 3 ) dx + ¢ Γ κ 1 v(x 1 , x 3 )∂ ν v(x 1 , x 3 ) ds =: I Ω κ -+ I Γ κ 1 . ( 15
)
Owing to [START_REF] Fichera | Asymptotic behaviour of the electric field and density of the electric charge in the neighbourhood of singular points of a conducting surface[END_REF], there holds I Ω κ -= 0. Now we focus our attention on the term I Γ κ 1 . Let (e 1 , e 2 , e 3 ) denote the canonical basis of R 3 . Set α 2 := arctan κ 2 ∈ (-π/2, 0) and define the new orthonormal basis (ẽ 1 , ẽ2 , ẽ3 ) with

ẽ1 = e 1 ; ẽ2 = cos(α 2 )e 2 -sin(α 2 )e 3 ; ẽ3 = sin(α 2 )e 2 + cos(α 2 )e 3 . ( 16
)
Observe that the component Γ κ 1 of the boundary of the incisor Ω κ is included in the plane (O, ẽ1 , ẽ3 ). Let (x 1 , x2 , x3 ) denote the coordinates in the basis [START_REF] Nazarov | Asymptotics of eigenvalues of the Dirichlet problem in a skewed T -shaped waveguide[END_REF]. We have

I Γ κ 1 = - 1 2 ¢ Γ κ 1 ∂(v 2 ) ∂ x2 dx 1 dx 3 . ( 17
)
Using that v is independent of x 2 , we obtain

0 = ∂(v 2 ) ∂x 2 = cos α 2 ∂(v 2 ) ∂ x2 + sin α 2 ∂(v 2 ) ∂ x3 . ( 18
)
Combining ( 18) and ( 17), we find

I Γ κ 1 = tan α 2 2 ¢ Γ κ 1 ∂(v 2 ) ∂ x3 dx 1 dx 3 = tan α 2 2 ¢ ∂Γ κ 1 v 2 ν • ẽ3 dℓ = tan α 2 2 ¢ L κ v 2 ν • ẽ3 dℓ, ( 19
)
where

L κ := {x ∈ R 3 | x j = κ j x 3 , j = 1, 2, x 3 ∈ (0, 1)
} and where ν stands for the outward unit normal vector to ∂Γ κ 1 (in the plane (O, ẽ1 , ẽ3 )). Using that α 2 ∈ (-π/2, 0), ν • ẽ3 > 0 on L κ (see Figure 5) and v ̸ ≡ 0 on L κ , we deduce that I Γ κ 1 < 0. Note also that the quantity I Γ κ 1 is independent of ε. Gathering ( 14) and ( 15), we infer that the inequality (11) holds for ε > 0 small enough. This is enough to guarantee that σ d (A κ ) is non-empty for negative κ 2 .

Absence of eigenvalues for small positive κ 2

The goal of this section is to prove an intermediate result to establish the item 2) of Theorem 3.2. Therefore we assume that κ 2 ≥ 0. In that situation, the integral I Γ κ 1 in ( 19) is positive because α 2 ∈ [0, π/2) and our argument of the previous section does not work for showing the existence of discrete spectrum. Of course, this does not yet guarantee that σ d (A κ ) is empty. Actually we will see in Section 6 that σ d (A κ ) is non-empty for certain κ with κ 2 > 0. For the moment, combining the calculations of Section 4 with the approach of [START_REF] Nazarov | Variational and asymptotic methods for finding eigenvalues below the continuous spectrum threshold[END_REF], we show the following result. Proposition 5.1. For all κ 1 > 0, there exists δ(κ 1 ) > 0 such that σ d (A κ ) is empty for

κ 2 ∈ [0, δ(κ 1 )). ( 20
)
Proof.

e 2 e 3 e 1 Ω κ 1- Ω κ 1+ Figure 6: Domains Ω κ 1-and Ω κ 1+ .
Fix κ 2 ∈ [0, min(1, κ 1 )) and divide Ω κ into the two domains

Ω κ 1-:= {x ∈ Ω κ | x 2 < 1} and Ω κ 1+ := {x ∈ Ω κ | x 2 > 1} (21)
(see Figure 6). Since Ω κ

1+ = {(x 1 , x 2 , x 3 ) ∈ R 3 | (x 1 , x 3 ) ∈ Π arctan κ 1 , x 2 > 1}, there holds ∥∇ x ψ; L 2 (Ω κ 1+ )∥ 2 ≥ λ κ † ∥ψ; L 2 (Ω κ 1+ )∥ 2 (22)
for all ψ ∈ H 1 0 (Ω κ ; Σ κ ). Below we show that there is some ϱ > 0 such that for κ 2 small enough, there holds, for all ψ ∈ H 1 0 (Ω κ ; Σ κ ),

∥∇ x ψ; L 2 (Ω κ 1-)∥ 2 ≥ (λ κ † + κ 1 ϱ)∥ψ; L 2 (Ω κ 1-)∥ 2 . ( 23
)
Combining ( 22) and ( 23) with the minimum principle yields the result of the proposition.

Remark 5.2. Note that estimate (22) implies that σ d (A κ ) is empty for all κ = (κ 1 , 0) with κ 1 ≥ 0. This proves in particular the result of Theorem 3.1.

The remaining part of the proof is dedicated to establishing (23). Consider the mixed boundaryvalue problem

-∆ x w = τ w in Ω κ 1- w = 0 on {x ∈ ∂Ω κ 1-| x 3 = 0 or x 3 = 1} ∂ ν w = 0 on {x ∈ ∂Ω κ 1-| x 3 ∈ (0, 1)}. ( 24
)
As pictured in Figure 6 left, the domain Ω κ 1-in ( 21) is a semi-infinite prism with a trapezoidal cross-section and a skewed end. When κ 2 = 0, the trapezoid is simply the unit square and the continuous spectrum of the problem (24) coincides with the ray [π 2 , +∞). In that situation, the problem (24) admits an eigenvalue at µ α 1 1 ∈ (0, π 2 ) with α 1 = arctan κ 1 (see the text above ( 9) for the definition of that quantity), a corresponding eigenfunction being w such that

w(x) = v(x 1 x 3 ),
where v is an eigenfunction of (8) associated with µ α 1 1 . Now let us consider the situation κ 2 > 0 small. Then the map

Ω κ 1-∋ x → x 1 , x 2 -κ 2 x 3 1 -κ 2 x 3 , x 3 ∈ Ω (κ 1 ,0) 1- (25)
is a diffeomorphism whose Jacobian matrix is close to the identity and whose Hessian matrix is small. Using these properties, we deduce that the discrete spectrum of the problem ( 24) is still non-empty for κ 2 small enough. This comes from the fact that the cut-off point of the essential spectrum satisfies the estimate

|τ κ 2 † -π 2 | ≤ C † κ 2
and the first (smallest) eigenvalue of the discrete spectrum, which is simple, admits the expansion

τ κ 2 1 = µ α 1 1 + κ 2 τ ′ 1 + τ κ 2 1 (26) with |τ κ 2 1 | ≤ C κ 2 2 .
Here C > 0 is a constant independent of κ 2 . These properties can be justified using classical results of the perturbation theory for linear operators, see e.g. [START_REF] Kato | Perturbation theory for linear operators[END_REF]Ch. 7], [3, Ch. 10], [18, Ch. XII]. From the minimum principle, to establish (23), we see that it suffices to show that

τ ′ 1 = dτ κ 2 1 dκ 2 κ 2 =0 > 0. ( 27
)
Let w κ 2 1 be an eigenfunction of Problem (24) associated with τ κ 2 1 . Together with (29), consider the asymptotic ansatz

w κ 2 1 (x) = v(x 1 , x 3 ) + κ 2 w ′ 1 (x) + wκ 2 1 (x) (28)
where wκ 2 1 is a small remainder. Insert (29), (28) into (24) and collect the terms of order κ 2 . We obtain

-∆ x w ′ 1 -µ α 1 1 w ′ 1 = τ ′ 1 v in Ω (κ 1 ,0) 1- w ′ 1 = 0 on {x ∈ ∂Ω κ 1-| x 3 = 0 or x 3 = 1}. ( 29
)
As for the Neumann boundary condition of (24), using in particular that on Γ κ 1 ,

∂ ν • = (1 + κ 2 2 ) -1/2 - ∂• ∂x 2 + κ 2 ∂• ∂x 3 ,
at order κ 2 , we find

- ∂w ′ 1 ∂x 2 (x 1 , 0, x 3 ) = - ∂v ∂x 3 (x 1 , x 3 ), ∂w ′ 1 ∂x 2 (x 1 , 1, x 3 ) = 0, (x 1 , x 3 ) ∈ Π α 1 . ( 30
)
Since the smallest eigenvalue µ α 1 1 is simple, there exists only one compatibility condition to satisfy to ensure that the problem (29)-(30) has a non trivial solution. It can be written as 6 Existence of eigenvalues for κ 2 close to κ 1 > 0

τ ′ 1 = τ ′ 1 ∥v; L 2 (Π α 1 )∥ 2 = - ¢ Ω (κ 1 ,0) 1- v(∆ x w ′ 1 + µ α 1 1 w ′ 1 ) dx = ¢ Γ (κ 1 ,0) 1 v(x 1 , x 3 ) ∂w ′ 1 ∂x 2 (x 1 , 0, x 3 ) ds = ¢ Π α 1 v(ξ 1 , ξ 2 ) ∂v ∂ξ 2 (ξ 1 , ξ 2 ) dξ 1 dξ 2 = 1 2 cos α 1 ¢ L (κ 1 ,0) v 2 dℓ > 0 where L (κ 1 ,0) := {(ξ 1 , ξ 2 ) ∈ R 2 | ξ 1 ξ 2 tan α 1 , ξ 2 ∈ (0,
We start this section by proving that the discrete spectrum σ d (A κ ) of the operator A κ can also be non-empty for certain positive κ 2 . This happens for example in the case κ 1 = κ 2 , which we now assume. We adapt the proof of [1, Thm. 2] and exhibit a function φ ∈ H 1 0 (Ω κ ; Σ κ ) satisfying [START_REF] Kato | Perturbation theory for linear operators[END_REF]. First, note that for κ 1 = κ 2 , the domain Ω κ is symmetric with respect to the "bisector" cross-section

Υ κ := {x ∈ Ω κ | x 1 = x 2 }.
Let us divide Ω κ into the two congruent domains

Ω κ ∧ := {x ∈ Ω κ | x 1 > x 2 } and Ω κ < := {x ∈ Ω κ | x 2 > x 1 }. (31)
Accordingly, we set

ψ ε (x) = e -εx 1 v(x 2 , x 3 ) in Ω κ ∧ e -εx 2 v(x 1 , x 3 ) in Ω κ < ( 32
)
where v is as in [START_REF] Lions | Problèmes aux limites non homogènes et applications[END_REF]. Since ψ ε is continuous on Υ κ and decays exponentially at infinity, it belongs to H 1 0 (Ω κ ; Σ κ ). Moreover, we have

∥∇ x ψ; L 2 (Ω κ ∧ )∥ 2 -µ α 1 1 ∥ψ; L 2 (Ω κ ∧ )∥ 2 = - ¢ Ω κ ∧ e -2εx 1 v(x 2 , x 3 ) ∆ x v(x 2 , x 3 ) + (µ α 1 1 + ε 2 )v(x 2 , x 3 ) dx + ¢ Υ κ e -εx 1 v(x 2 , x 3 )∂ ν e -εx 1 v(x 2 , x 3 ) ds =: I ε Ω κ + I ε Υ κ . ( 33
)
Using that v solves (8), we get

I ε Ω κ = O(ε). Now consider the integral I ε Υ κ . Define the new orthonormal basis (ê 1 , ê2 , ê3 ) with ê1 = √ 2 2 e 1 + √ 2 2 e 2 ; ê2 = - √ 2 2 e 1 + √ 2 2 e 2 ; ê3 = e 3 . ( 34
)
Remark that Υ κ is included in the plane (O, ê1 , ê3 ). Let (x 1 , x2 , x3 ) denote the coordinates in the basis (34). We have

I ε Υ κ = ¢ Υ κ e -ε √ 2(x 1 -x 2 )/2 v ∂ ∂ x2 e -ε √ 2(x 1 -x 2 )/2 v dx 1 dx 3 = ¢ Υ κ ε √ 2 2 e -ε √ 2(x 1 -x 2 ) v 2 dx 1 dx 3 + ¢ Υ κ 1 2 e -ε √ 2(x 1 -x 2 ) ∂(v 2 ) ∂ x2 dx 1 dx 3 . ( 35
)
Exploiting the exponential decay of v(ξ) as ξ 1 → +∞, (see Section 2), one finds that the first integral of the right hand side above is O(ε). For the second one, using that v is independent of x 1 , we can write

0 = ∂(v 2 ) ∂x 1 = √ 2 2 ∂(v 2 ) ∂ x1 - √ 2 2 ∂(v 2 ) ∂ x2 .
Remarking also that x1 = 0 on Υ κ , this gives

¢ Υ κ 1 2 e -ε √ 2(x 1 -x 2 ) ∂(v 2 ) ∂ x2 dx 1 dx 3 = ¢ Υ κ 1 2 e -ε √ 2 x1 ∂(v 2 ) x1 dx 1 dx 3 = ¢ Υ κ ε √ 2 2 e -ε √ 2 x1 v 2 dx 1 dx 3 + ¢ L κ 1 2 e -ε √ 2 x1 v 2 ν • ê1 dℓ (36)
where

L κ := {x ∈ R 3 | x j = κ j x 3 , j = 1, 2, x 3 ∈ (0, 1)
} and where ν stands for the outward unit normal vector to Υ κ (in the plane (O, ê1 , ê3 )). Using that κ 1 = κ 2 > 0, we find ν • ê1 < 0 on L κ . Since there holds v ̸ ≡ 0 on L κ , gathering ( 35) and ( 36), we deduce that we have I ε Υ κ < 0 for ε small enough. From (33), we deduce

∥∇ x ψ; L 2 (Ω κ ∧ )∥ 2 -µ α 1 1 ∥ψ; L 2 (Ω κ ∧ )∥ 2 < 0
for ε small enough. Then by symmetry, we obtain

∥∇ x ψ; L 2 (Ω κ )∥ 2 -µ α 1 1 ∥ψ; L 2 (Ω κ )∥ 2 = 2∥∇ x ψ; L 2 (Ω κ ∧ )∥ 2 -2µ α 1 1 ∥ψ; L 2 (Ω κ ∧ )∥ 2 < 0.
We conclude that the inequality ( 11) is satisfied by the function (32) which proves the following statement.

Theorem 6.1. For κ 1 = κ 2 > 0, the discrete spectrum σ d (A κ ) of the operator A κ is not empty.

Since the eigenvalues of the discrete spectrum are stable with respect to small perturbations of the operator, Theorem 6.1 and diffeomorphisms similar to (25) imply that σ d (A κ ) is not empty for κ 2 in a neighbourhood of κ 1 . With Proposition 5.1, this allows us to introduce h(κ 1 ) ∈ (0, κ 1 ) as the infimum of the numbers δ such that σ d (A κ ) is non-empty for all κ 2 ∈ (δ, κ 1 ].

On the other hand, we have the following monotonicity result: Proposition 6.2. Consider some κ = (κ 1 , κ 2 ) with κ 1 > 0 and κ 2 ∈ (0, κ 1 ] such that A κ has a non-empty discrete spectrum. Let λ κ 1 denote the first (smallest) eigenvalue of σ d (A κ ). For ε > 0 small, set κ ε := (κ 1 , κ 2 + ε) and denote by λ κ ε 1 the first eigenvalue of σ d (A κ ε ). Then, we have

λ κ ε 1 < λ κ 1 . ( 37
)
Proof. Using again the minimum principle, we can write

λ κ ε 1 = min ψ ε ∈H 1 0 (Ω κ ε ;Σ κ ε )\{0} ∇ x ψ ε ; L 2 (Ω κ ε ) 2 ψ ε ; L 2 (Ω κ ε ) 2 . ( 38
)
Now define the function ψ ε such that

ψ ε (x) = u κ x 1 , κ 2 x 2 κ 2 + ε , x 3 ,
where u κ ∈ H 1 0 (Ω κ ; Σ κ ) is an eigenfunction associated with the first eigenvalue of σ d (A κ ). Clearly ψ ε is a non zero element of H 1 0 (Ω κ ε ). Besides, we find

ψ ε ; L 2 (Ω κ ε ) 2 = κ 2 + ε κ 2 u κ ; L 2 (Ω κ ) 2 and ∇ x ψ ε ; L 2 (Ω κ ε ) 2 = κ 2 κ 2 + ε ∂u κ ∂x 2 ; L 2 (Ω κ ) 2 + κ 2 + ε κ 2 j=1,3 ∂u κ ∂x j ; L 2 (Ω κ ) 2 .
According to (38), these identities imply

λ κ ε 1 ≤ ∥u κ ; L 2 (Ω κ )∥ -2 κ 2 2 (κ 2 + ε) 2 ∂u κ ∂x 2 ; L 2 (Ω κ ) 2 + j=1,3 ∂u κ ∂x j ; L 2 (Ω κ ) 2 ≤ ∥∇ x u κ ; L 2 (Ω κ )∥ 2 ∥u κ ; L 2 (Ω κ )∥ 2 = λ κ 1 .
The strict inequality in (37) follows from the fact the derivative ∂u κ /∂x 2 cannot be null in the whole domain Ω κ . This completes the proof of the proposition.

According to relation (37), the function κ 2 → λ κ 1 is strictly decreasing on (h(κ 1 ), κ 1 ). Besides, Proposition 6.2 ensures that σ d (A κ ) cannot be non-empty for some h(κ 1 ) ∈ (0, h(κ 1 )) otherwise σ d (A κ ) would be non-empty for all κ 2 ∈ ( h(κ 1 ), κ 1 ] which contradicts the definition of h(κ 1 ). This completes the proof of the item 2) of Theorem 3.2. Remark 6.3. For κ 2 ∈ [-κ 1 , 0), we have seen in Section 4 that σ d (A κ ) is non-empty. Let λ κ 1 denote the smallest eigenvalue of σ d (A κ ). By adapting the proof of Proposition 6.2, one establishes that the map κ 2 → λ κ 1 is strictly increasing on [-κ 1 , 0). Together with Proposition 6.2, this shows the item 3) of Theorem 3.2.

Finiteness of the discrete spectrum

Finally, we establish the item 4) of Theorem 3.2 and so assume that κ 2 ∈ (-κ 1 , κ 1 ). Set again α 1 = arctan κ 1 , α 2 = arctan κ 2 . Since |α 2 | < α 1 , similarly to [START_REF] Kamotskii | On eigenfunctions localized in a neighborhood of the lateral surface of a thin domain[END_REF], we have

λ κ † = µ α 1 1 < µ |α 2 | 1 = µ α 2 1 . ( 39
)
We remind the reader that µ α j 1 stands for the smallest eigenvalue of the 2D problem [START_REF] Fichera | Asymptotic behaviour of the electric field and density of the electric charge in the neighbourhood of singular points of a conducting surface[END_REF] set in the pointed strip Π α j appearing in [START_REF] Exner | On existence of a bound state in an L-shaped waveguide[END_REF]. Observe that Π α 2 can be obtained from Π -α 2 by a symmetry with respect to the line ξ 2 = 1/2 and a translation, which ensures that µ -α 2 1 = µ α 2 1 and so µ For R > 0, define the truncated pointed strip 7 left) and consider the problem

|α 2 | 1 = µ α 2 1 . ξ 1 ξ 2 ξ 1 = ξ 2 tan α 2 α 2 R e 2 e 3 e 1 Ω κ III Ω κ II Ω κ I R e 2 e 1 e 3
Π α 2 (R) := {(ξ 1 , ξ 2 ) ∈ Π α 2 | ξ 1 < R} (see Figure
-∆ ξ v = µv in Π α 2 (R) v = 0 on {ξ ∈ ∂Π α 2 (R) | ξ 2 = 0 or ξ 2 = 1} ∂ ν v = 0 on {ξ ∈ ∂Π α 2 (R) | ξ 2 ̸ = 0 and ξ 2 ̸ = 1}. ( 40
)
Denote by µ α2 1 (R) the smallest eigenvalue of (40). Since µ α 2 1 (R) converges to µ α 2 1 as R → +∞, according to (39), we can fix R > |κ 2 | such that

µ α 2 1 (R) > λ κ † . ( 41
)
Then let us divide Ω κ into the three domains

Ω κ I := {x ∈ Ω κ | x 1 > κ 1 and x 2 < R}, Ω κ II := {x ∈ Ω κ | x 2 > R}, Ω κ III := {x ∈ Ω κ | x 1 < κ 1 and x 2 < R}
(see the representation of Figure 7 right). Using (41), we obtain

∥∇ x ψ; L 2 (Ω κ I )∥ 2 ≥ λ κ † ∥ψ; L 2 (Ω κ I )∥ 2 ∀ψ ∈ H 1 0 (Ω κ ; Σ κ ). ( 42
)
On the other hand, from (22), we get

∥∇ x ψ; L 2 (Ω κ II )∥ 2 ≥ λ κ † ∥ψ; L 2 (Ω κ II )∥ 2 ∀ψ ∈ H 1 0 (Ω κ ; Σ κ ). ( 43
)
Besides, since Ω κ III is bounded, the max-min principle ([3, Thm. 10.2.2]) guarantees that there is

n ∈ N := {0, 1, 2, . . . } such that λ κ † ≤ max E⊂En inf ψ∈E\{0} ¢ Ω κ III |∇ψ| 2 dx ¢ Ω κ III ψ 2 dx , ( 44
)
where E n denotes the set of subspaces of H 44), we deduce that there holds

1 0 (Ω κ III ; Σ 0 ∩ ∂Ω κ ) := {φ ∈ H 1 (Ω κ III ) | φ = 0 on Σ 0 ∩ ∂Ω κ } of codimension n. Gathering (42)-(
λ κ † ≤ max E⊂ Ẽn inf ψ∈E\{0} ¢ Ω κ |∇ψ| 2 dx ¢ Ω κ ψ 2 dx
, where this times Ẽn stands for the set of subspaces of H1 0 (Ω κ ; Σ 0 ) of codimension n. From the max-min principle, this proves that σ d (A κ ) contains at most n (depending on κ) eigenvalues. Remark 7.1. Our simple proof above does not work to show that σ d (A κ ) is discrete when κ 2 = ±κ 1 . However we do not expect particular phenomenon and think the result also holds in this case. It is proved in [START_REF] Dauge | Dirichlet spectrum of the Fichera layer[END_REF]Thm. 1.2] when κ 2 = κ 1 = 1.

Numerics and discussion

In this section, we illustrate some of the results above. In Figure 8, we represent an approximation of the first eigenvalue of the 2D problem (8) set in the pointed strip Π α 1 with respect to α 1 ∈ (0; 9π/20). We use a rather crude method which consists in truncating the domain at ξ 2 = 12 (see the picture of Figure 7 left) and imposing homogeneous Dirichlet boundary condition on the artificial boundary. Then we compute the spectrum in this bounded geometry by using a classical P2 finite element method. To proceed, we use the library Freefem++ [START_REF] Hecht | New development in freefem++[END_REF] and display the results with Matlab 1 and Paraview 2 . The values we get are coherent with the ones recalled in [START_REF] Hecht | New development in freefem++[END_REF].

In Figure 9-11, we fix κ 1 = 1 (equivalently α 1 = π/4) and compute the first eigenvalue of σ d (A κ ) for different κ 2 . For κ 1 = 1, the bound of the essential spectrum of A κ is λ κ † ≈ 0.929π 2 (see Figure 8 as well as [START_REF] Dauge | Dirichlet spectrum of the Fichera layer[END_REF]). For each of the three κ 2 , in agreement with Theorem 3.2, we find an eigenvalue below the essential spectrum. Actually, in each situation our numerical experiments seem to indicate that there is only one eigenvalue in the discrete spectrum, which is a result that we have not proved. Interestingly, for κ 2 = -0.1 (Figure 10), the eigenfunction is not particularly localized at the intersection of the obliques sides. We emphasize that here we simply compute the spectrum of the Laplace operator with mixed boundary conditions in the bounded domain {x ∈ Ω κ |x 1 < 6 and x 2 < 6}. At x 1 = 6 and x 2 = 6, we impose homogeneous Neumann boundary condition. Admittedly, this is a very naive approximation, especially in the case of Figure 10 where the eigenfunction has a poor decay at infinity. In the latter situation, the corresponding eigenvalue is very close to λ κ † and we do not give its value because our approximation probably lacks precision.

In Figure 12, we represent eigenfunctions associated with two different eigenvalues of σ d (A κ ) for κ = (3, -3). For the 2D problem [START_REF] Fichera | Asymptotic behaviour of the electric field and density of the electric charge in the neighbourhood of singular points of a conducting surface[END_REF] in the pointed strip, the cardinal of the discrete spectrum can be made as large as desired by considering sufficiently sharp angles. We imagine that a similar phenomenon occurs in our geometry Ω κ . However to prove such a result is an open problem. At least the numerics of Figure 12 suggest that we can have more than one eigenvalue in σ d (A κ ).

As mentioned in the introduction, the fact that the discrete spectrum of A κ for κ = (1, 1) is not empty ensures that the Dirichlet Laplacian in the so-called Fichera layer F of Figure 1 right admits an eigenvalue below the essential spectrum. This can be proved by playing with symmetries and reconstructing F from three versions of Ω (1,1) . Now, gluing six domains Ω (1,-1) , we can create the cubical structure pictured in Figure 13. Then, from Theorem 3.2 which guarantees that σ d (A κ ) contains at least one eigenvalue, we deduce that the Dirichlet Laplacian in this geometry has at least one eigenvalue.

Figure 1 :

 1 Figure 1: Domains B κ 1 (left) and Ω κ (centre). Fichera layer F (right).

Figure 2 :

 2 Figure 2: Domain Π α 1 corresponding with a cut of the blade B κ 1 in the plane x 2 = 0.

Figure 4 :

 4 Figure 4: Domains Ω κ -and Ω κ + .

Figure 5 :

 5 Figure 5: Domain Γ κ 1 in the plane (O, ẽ1 , ẽ3 ).

Figure 7 :

 7 Figure 7: Left: truncated pointed strip Π α 2 (R). Right: bottom view of the decomposition of Ω κ .

Figure 8 :

 8 Figure 8: Curve α 1 → µ α 1 1 for α 1 ∈ (0; 9π/20). According to Theorem 2.1, this gives the bound λ κ † of the essential spectrum of A κ . The two pictures correspond to eigenfunctions associated with µ α 1 1 for α 1 = π/4 and α 1 = 9π/20.

Figure 9 :

 9 Figure 9: Two views of an eigenfunction associated with the first eigenvalue of σ d (A κ ) for κ = (1, -1). We find λ κ 1 ≈ 0.81π 2 .

Figure 10 :

 10 Figure 10: Eigenfunction associated with the first eigenvalue of σ d (A κ ) for κ = (1, -0.1).

Figure 11 :

 11 Figure 11: Eigenfunction associated with the first eigenvalue of σ d (A κ ) for κ = (1, 1). We find λ κ 1 ≈ 0.90π 2 .

Figure 12 :

 12 Figure 12: Eigenfunctions associated with two different eigenvalues of σ d (A κ ) for κ = (3, -3).

Figure 13 :

 13 Figure 13: Cubical structure obtained by gluing six domains Ω κ with κ = (1, -1).

  ) because |κ 2 | ≤ κ 1 implies arctan |κ 2 | ≤ arctan κ 1 and because α 1 → µ α 11 is decreasing. Thus the lower bound of σ ess (A κ ) is characterized by the sharpest edge of Ω κ .

	Theorem 2.1. The essential spectrum σ ess (A κ ) of the operator A κ coincides with the ray [λ κ † , +∞) where λ κ † is defined in (10).
	Remark 2.2.

Matlab, http://www.mathworks.com/.

Paraview, http://www.paraview.org/.

Appendix

Here we show that the Dirichlet Laplacian in Ω κ has no isolated eigenvalue nor eigenvalues embedded in the essential spectrum.

for some λ ∈ C. Then there holds u ≡ 0 in Ω κ .

Proof. The result is clear when λ ∈ C\(0, +∞). Let us apply the Rellich trick [START_REF] Rellich | Über das asymptotische Verhalten der Lösungen von ∆u + λu = 0 in unendlichen Gebieten[END_REF] to deal with the case λ > 0. If u solves (45), the function

Therefore we get

Using this in (47) gives ∂u/∂ν = 0 on Γ κ 2 . Since we also have u = 0 on Γ κ 2 and ∆ x u + λu = 0 in Ω κ , the theorem of unique continuation (see e.g. [START_REF] Bers | Partial differential equations[END_REF], [START_REF] Colton | Inverse acoustic and electromagnetic scattering theory[END_REF]Thm. 8.6]) guarantees that u ≡ 0 in Ω κ .


Frédéric Bayart

F R Éd

Éric Bayart

ε-HYPERCYCLIC OPERATORS THAT ARE NOT δ-HYPERCYCLIC FOR δ < ε

Keywords: Mathematics Subject Classification. 47A16, 47B37 hypercyclic operators, operator weighted shifts

ϵ-hypercyclic operators that are not δ-hypercyclic for δ < ϵ

Introduction

Let X be a separable infinite dimensional Banach space. During the last decades the properties of the orbits of operators acting on X have been widely studied. In particular, the notion of hypercyclic operators, namely operators with a dense orbit, has drawn the attention of many mathematicians (see for instance [START_REF] Bayart | Dynamics of linear operators[END_REF]). It seems natural in this context to investigate operators having orbits with a property slightly weaker than denseness. Does this imply that the operator admits a dense orbit? For instance, N. Feldman has shown in [START_REF] Feldman | Perturbations of hypercyclic vectors[END_REF] that if there is an orbit of T ∈ L(X) which meets every ball of radius d > 0, then T is hypercyclic. The following definition concerning operators admitting an orbit which intersects every cone of aperture ε has been introduced in [START_REF] Badea | Epsilon-hypercyclic operators[END_REF]. Definition 1.1. Let ε ∈ (0, 1). A vector x ∈ X is called an ε-hypercyclic vector for T ∈ L(X) provided for every non-zero vector y ∈ X, there exists an integer n ∈ N such that ∥T n x-y∥ ≤ ε∥y∥. The operator T is called ε-hypercyclic if it admits an ε-hypercyclic vector.

In [START_REF] Badea | Epsilon-hypercyclic operators[END_REF], the authors have shown that for every ε ∈ (0, 1), there exists an ε-hypercyclic operator on ℓ 1 (N) which is not hypercyclic. This was refined in [START_REF] Bayart | Epsilon-hypercyclic operators on a Hilbert space[END_REF] and [START_REF] Tapia-García | An epsilon-hypercyclicity criterion and its application on classical Banach spaces[END_REF] where similar examples are given on ℓ 2 (N) and on more general spaces. Moreover it is pointed out in [START_REF] Tapia-García | An epsilon-hypercyclicity criterion and its application on classical Banach spaces[END_REF]Remark 4.7] that the ε-hypercyclic operator which is considered in that paper is not even δ-hypercyclic for some δ ∈ (0, ε). This leaves open the following natural question: let X be a Banach space, let 0 < δ < ε < 1. Can we distinguish the class of δ-hypercyclic operators and that of ε-hypercyclic operators acting on X? We give a positive answer for a large class of Banach spaces. To state our result we recall some terminology. Let (e n ) n≥0 be a basis of X (namely every x ∈ X writes uniquely n≥0 x n e n ) and let C ≥ 1. We say that (e n ) n≥0 is C-unconditional if for any N ≥ 0, for any finite sequences of scalars (a n ) n=0,...,N and (b n ) n=0,...,N such that

|b n | ≤ |a n | for all n = 0, . . . , N , then N n=0 b n e n ≤ C N n=0
a n e n .

Let us fix now X and Y two Banach spaces and suppose that (f n ) is a 1-unconditional basis of Y . We denote by Y X the vector space

Y X := (x n ) ∈ X N : +∞ n=0 ∥x n ∥ X f n ∈ Y
and we endow it by

∥(x n )∥ = +∞ n=0 ∥x n ∥ X f n Y .
It is standard that Y X is a Banach space. Our main theorem now reads.

Theorem 1.2. Let X be an infinite dimensional separable Banach space with a 1-unconditional basis, let Y be an infinite dimensional separable Banach space with a normalized 1-unconditional basis such that the associated backward shift operator is continuous. For all ε ∈ (0, 1), there exists an operator on Z = Y X which is not δ-hypercyclic for all δ ∈ (0, ε) and which is δ-hypercyclic for all δ ∈ (ε, 1).

Observe that if X is either c 0 (N) or ℓ p (N), p ∈ [1, +∞), then X is isometric to X X by using the canonical basis of X. Therefore, it satisfies the assumptions of the previous theorem. Recall also that if T is ε-hypercyclic for all ε ∈ (0, 1), then it is hypercyclic (see [START_REF] Badea | Epsilon-hypercyclic operators[END_REF]Theorem 1.3]). We will need a way to prove that an operator is ε-hypercyclic. We state here a variant of the ε-hypercyclicity criterion given in [START_REF] Tapia-García | An epsilon-hypercyclicity criterion and its application on classical Banach spaces[END_REF]Theorem 1.2]. Its proof is completely similar. Theorem 1.3. Let X be an infinite dimensional separable Banach space, let T ∈ L(X) and let ε ∈ (0, 1). Assume that there exist a dense subset D of X, a sequence (u(k)) dense in X such that that, for all k ≥ 0, u(k) = u(l) for infinitely many integers l, a sequence (v(k)) of vectors in X and an increasing sequence of positive integers (n k ) such that

• lim k→+∞ ∥T n k x∥ = 0 for all x ∈ D; • lim k→+∞ ∥v(k)∥ = 0; • for all k ≥ 0, ∥T n k v(k) -u(k)∥ ≤ ε∥u(k)∥.
Then T is δ-hypercyclic for all δ > ε.

The remaining part of the paper is devoted to the proof of Theorem 1.2.

Proofs

2.1.

A geometric lemma in dimension 2. The construction ultimately relies on the following fact regarding normed spaces of dimension 2. It deals with the distance of some fix vector to lines depending on a parameter. Lemma 2.1. Let F be a normed space of dimension 2, let (u, v) be a normalized basis of F , let (u * , v * ) be the dual basis and assume that ∥u * ∥ = ∥v * ∥ = 1. For all ε ∈ (0, 1), there exists

ω ∈ [ε, ε(1 -ε) -1 ] such that min y∈C ∥(y -1)u + yωv∥ = ε. Proof. When ω = ε, min y∈C ∥(y -1)u + yωv∥ ≤ ∥ωv∥ ≤ ε. When ω = ε(1 -ε) -1 , for all y ∈ C, ∥(y -1)u + yωv∥ ≥ max(|y -1|, |y|ω). Now, if |y| ≥ 1/(1 + ω), |y|ω ≥ ω/(1 + ω) ≥ ε and if |y| ≤ 1/(1 + ω), |y -1| ≥ 1 - 1 1 + ω = ε.
Therefore, min y∈C ∥(y -1)u + yωv∥ ≥ ε. The result follows by continuity of ω → min y∈C ∥(y -1)u + yωv∥. □ Remark 2.2. If ∥au + bv∥ = (|a| p + |b| p ) 1/p for some p ∈ (1, +∞), then it is easy to prove that the value of ω is given by ω

(1 + ω p p-1 ) p-1 p = ε
and that the minimum is attained at

y = 1 1 + ω p p-1 . When p = 1, ω = ε and y = 1. When p = ∞, ω = ε
1-ε and y = 1 1+ω = 1 -ε. This corresponds to the extremal cases of Lemma 2.1.

2.2.

The construction of a sequence of operators on X. As the previous constructions of ε-hypercyclic operators which are not δ-hypercyclic, our operator will be an operator weighted shift. The next part of the proof consists in defining his weights. We denote by (e n ) (resp. (f n )) the 1-unconditional basis of X (resp. Y ) which appears in the statement of Theorem 1.2. We may assume that (e n ) is normalized which implies (by 1-unconditionality) that (e * n ) is normalized too. The strategy is the following. At each step k we will define weights

A m k +1 , • • • , A m k+1 such that the products A m k +1 • • • A j , j = m k + 1, .
. . , m k+1 leave e 0 invariant, send e k onto the line defined by Lemma 2.1 and e l onto a multiple of e l for l ̸ = k. Therefore, provided e * 0 (u) is small, A m k +1 • • • A j (u) can be close to e 0 , but not too close. We proceed with the details. We set

λ = 3 ε(1 -ε) and κ = (1 + λ) + max(1 + ε(1 -ε) -1 , 2/ε).
We exhibit two sequences of integers (m k ) k≥1 and (r k ) k≥1 and a sequence of operators (A j ) j≥1 on X such that, for all k ≥ 1,

(i) A n e 0 = e 0 for all n = m k + 1, . . . , m k+1 ; (ii) A n is invertible, ∥A n ∥ ≤ κ for all n = m k + 1, . . . , m k+1 ; (iii) A m k +1 • • • A m k+1 = Id.
We initialize the construction by setting m 1 = 0. We assume that the construction has been done until step k -1 to do it at step k ≥ 1. We thus have to define m k+1 , r k and (A j ) j=m k +1,...,m k+1 . We set F k = span(e 0 , e k ) and G k = span(e l : l ̸ = 0, k) so that Let r k > 0 be a very large integer (more precise conditions on r k will be given later) and let us set

X = F k ⊕ G k . Let ω k ∈ [ε, ε(1 -ε) -1 ]
m k+1 = m k + r k + k + 1. For j = 1, . . . , r k + k + 1, we define A m k +j by • A m k +j (e 0 ) = e 0 . •                A m k +1 (e k ) = e 0 + ω k e k A m k +2 (e k ) = • • • = A m k +r k (e k ) = 2e k A m k +r k +1 (e k ) = • • • = A m k +r k +k-1 (e k ) = e k A m k +r k +k (e k ) = 1 2 r k -1 e k A m k +r k +k+1 (e k ) = -1 ω k e 0 + 1 ω k e k . • for l ̸ = 0, k, A m k +j (e l ) = λe l , j = 1, . . . , r k + k, A m k +r k +k+1 (e l ) =
k ∈ [ε, ε(1 -ε) -1 ].
To go further with the properties of (A j ) we need to compute

A m k +1 • • • A m k +j e k for j = 1, . . . , r k + k + 1. We find A m k +1 • • • A m k +j e k =              e 0 + ω k e k j = 1 2 j-1 e 0 + 2 j-1 ω k e k j = 2, . . . , r k 2 r k -1 e 0 + 2 r k -1 ω k e k j = r k + 1, . . . , r k + k -1 e 0 + ω k e k j = r k + k e k j = r k + k + 1.
We then deduce the following formula, which will be equally important:

A -1 m k +j • • • A -1 m k +1 e k =                    -1 ω k e 0 + 1 ω k e k j = 1 -1 ω k e 0 + 1 2 j-1 ω k e k j = 2, . . . , r k -1 ω k e 0 + 1 2 r k -1 ω k e k j = r k + 1, . . . , r k + k -1 -1 ω k e 0 + 1 ω k e k j = r k + k e k j = r k + k + 1.
2.3. The operator. We now glue together these maps. We formally define T on Z = Y X by T (u 0 , u 1 , . . . ) = (A 1 u 1 , A 2 u 2 , . . . ).

Let K 1 be the norm of the backward shift operator associated to (f n ). Then for u = (u 0 , u 1 , . . . ),

∥T u∥ = +∞ n=1 ∥A n u n ∥ X f n-1 Y ≤ K 1 +∞ n=1 ∥A n ∥ • ∥u n ∥ X f n Y ≤ K 1 κ∥u∥
which implies that T is well defined and maps boundedly Z into itself.

2.4. T is not δ-hypercyclic for any δ ∈ (0, ε). By contradiction, assume that T is δ-hypercyclic for some δ ∈ (0, ε) and let u = (u 0 , u 1 , . . . ) be a δ-hypercyclic vector for T .

Observe that ∥u n ∥ → 0 so that u n,0 := e * 0 (u n ) → 0. Therefore it is possible to fix K > 0 such that |K -u n,0 |ε > δK for any n ≥ 0.

We set v = (Ke 0 , 0, . . . ). Let n ≥ 1 be such that ∥v -T n u∥ ≤ δ∥v∥ and let k ≥ 1 be such that n ∈ [m k + 1, m k+1 ]. Let us write u n = u n,0 e 0 + w n with e * 0 (w n ) = 0. Then by using (i) and (iii),

∥v -T n u∥ ≥ ∥Ke 0 -A 1 • • • A n (u n )∥ ≥ ∥Ke 0 -A m k +1 • • • A n (u n,0 e 0 + w n )∥ ≥ ∥(K -u n,0 )e 0 -A m k +1 • • • A n (w n )∥. If n = m k+1 , then A m k +1 • • • A n (w n ) = w n and ∥v -T n u∥ ≥ |K -u n,0 | ≥ ε|K -u n,0 | > δ∥v∥. If n ̸ = m k+1 , then A m k +1 • • • A n (w n ) = x n e 0 + x n ω k e k + w ′ n for some w ′ n ∈ G k and some x n ∈ C. Therefore ∥v -T n u∥ ≥ (K -u n,0 ) -x n e 0 + x n ω k e k ≥ ε|K -u n,0 | > δ∥v∥
where we have used [START_REF] Badea | Epsilon-hypercyclic operators[END_REF]. In both cases, we find a contradiction.

2.5.

T is δ-hypercyclic for all δ ∈ (ε, 1). Let δ ∈ (ε, 1) and let us prove that T is δhypercyclic by applying Theorem 1.3. Let (u(k)) be a dense sequence in Z such that each u(k) may be written u(k) = (u 0 (k), . . . , u k-1 (k), 0, . . . ) with u j (k) ∈ span(e 0 , . . . , e k-1 ) and ∥u j (k)∥ ≤ k. Moreover for any k ≥ 1, we assume that there exist infinitely many integers ℓ with u(k) = u(ℓ). We want to find a sequence of vectors (v(k)) in Z and a sequence of integers (n k ) such that ∥v(k)∥ → 0 and ∥T n k v(k) -u(k)∥ ≤ ε∥u(k)∥ for all k ≥ 1. We will define v(k) = (0, . . . , 0, v 0 (k), . . . , v k-1 (k), 0, . . . ) where v 0 (k) is at the (m k + r k )-th position. Let k ≥ 1, let j ∈ {0, . . . , k -1} and let us write

u j (k) = k-1 s=0 u j,s (k)e s .
Let l be the unique integer such that m l ≤ j < m l+1 . We will search v j (k) under the form

v j (k) = k-1 s=1 s̸ =l u j,s (k) λ r k +j λ j-m l e s + xe l + ye k
where x and y will be chosen so that

∥A j+1 • • • A m k +r k +j (v j (k)) -u j (k)∥ ≤ ε∥u j (k)∥
and ∥v j (k)∥ ≤ k -2 . Upon this has been done, we can easily apply Theorem 1.3 to deduce that T is δ-hypercyclic for δ > ε. Indeed, T has a dense generalized kernel and, for all k ≥ 1,

∥v(k)∥ = k-1 j=0 ∥v j (k)∥f m k +r k +j ≤ k-1 j=0 ∥v j (k)∥ ≤ 1 k . Moreover T m k +r k (v(k)) = (A 1 • • • A m k +r k (v 0 (k)), . . . , A k • • • A k +r k +k-1 (v k-1 (k)), 0, . . . ).
Therefore,

∥u(k) -T m k +r k (v(k))∥ = k-1 j=0 ∥A j+1 • • • A m k +r k +j (v j (k)) -u j (k)∥ f j ≤ ε k-1 j=0 ∥u j (k)∥f j ≤ ε∥u(k)∥.

So let us compute

A j+1 • • • A m k +r k +j (v j (k)) =: z j (k). z j (k) = A -1 j • • • A -1 1 A 1 • • • A m k +r k +j (v j (k)) = A -1 j • • • A -1 m l +1 A m k +1 • • • A m k +r k +j (v j (k)) = A -1 j • • • A -1 m l +1    k-1 s=1 s̸ =l λ j-m l u j,s (k)e s + λ r k +j xe l + 2 r k -1 ye 0 + 2 r k -1 yω k e k    .
The easiest case is when j = m l . In that case,

z j (k) = 2 r k -1 ye 0 + k-1 s=1 s̸ =l u j,s (k)e s + λ r k +j xe l + 2 r k -1 yω k e k .
We simply choose x = 1 λ r k +j u j,l (k) and y = y k 2 r k -1 u j,0 (k) so that by (1)

∥z j (k) -u j (k)∥ = |u j,0 (k)| • ∥(y k -1)e 0 + y k ω k e k ∥ ≤ ε|u j,0 (k)| ≤ ε∥u j (k)∥ whereas ∥v j (k)∥ ≤ k-1 s=1 ∥u j (k)∥ λ r k + |y k | • ∥u j (k)∥ 2 r k -1 ≤ k -2
provided r k is sufficiently large.

Let us now turn to j > m l . In that case, there exists 0 ≤ t j ≤ j -m l such that

z j (k) = k-1 s=1 s̸ =l u j,s (k)e s + λ r k +j ω l 2 t j xe l + 2 r k -1 y - λ r k +j x ω l e 0 + 2 r k -1 ω k y λ j-m l e k .
If we set to simplify the notations

λ j = λ j-m l µ j = 2 t j x ′ = λ r k +j ω l 2 t j x y ′ = 2 r k -1 λ j-m l y a = u j,0 (k) b = u j,l (k) then z j (k) -u j (k) = λ j y ′ -µ j x ′ -a e 0 + (x ′ -b)e l + y ′ ω k e k .
We are now ready to choose x ′ and y ′ , namely x and y. We indeed set

x ′ = b and y ′ = a + µ j b λ j so that

z j (k) -u j (k) = a + µ j b λ j ω k e k .
Therefore, by 1-unconditionality of (e k ), since t j ≤ j -m l , λ ≥ 2 and

ω k ≤ (1 -ε) -1 , ∥z j (k) -u j (k)∥ ≤ 1 + µ j λ j ω k ∥u j (k)∥ ≤ 1 λ + 2 λ ω k ∥u j (k)∥ ≤ 3 λ(1 -ε) ∥u j (k)∥ ≤ ε∥u j (k)∥
by our choice of λ. Finally observe that

∥v j (k)∥ ≤ k-1 s=1 s̸ =l ∥u j (k)∥ λ r k + ω l 2 t j λ r k ∥u j (k)∥ + λ j-m l 2 r k -1 × 3 λ ∥u j (k)∥ ≤ k k-1 s=1 1 λ r k + ω l 2 m k λ r k + 3λ m k -1 2 r k -1 ≤ k -2
provided r k has been chosen large enough.

2.6. Concluding remarks and questions. Observe that in Theorem 1.2, we assume a priori that (f n ) is normalized which was not the case for (e n ). Indeed, if we normalize (f n ), this could destroy the continuity of the associated backward shift operator. Following [START_REF] Tapia-García | An epsilon-hypercyclicity criterion and its application on classical Banach spaces[END_REF], we can slightly enlarge the scale of spaces where it is possible to produce such an example. Indeed, observe that, by adjusting (m k ) and (r k ) during the construction (we may ask that r k is bigger than any prescribed value), it is possible to ensure that T satisfies the ε-hypercyclicity criterion with the sequence (n k ) chosen as a subsequence of a prescribed sequence (p k ). Arguing like in [START_REF] Tapia-García | An epsilon-hypercyclicity criterion and its application on classical Banach spaces[END_REF]Theorem 4.10], we get therefore the following corollary.

Corollary 2.3. Let Z be a separable Banach space. Assume that Z admits an infinite dimensional complemented subspace Y X, where X is an infinite dimensional separable Banach space with a 1-unconditional basis and Y is an infinite dimensional separable Banach space with a normalized 1-unconditional basis such that the associated backward shift operator is continuous. Then for all ε ∈ (0, 1), there exists an operator on Z which is not δ-hypercyclic for all δ ∈ (0, ε) and which is δ-hypercyclic for all δ ∈ (ε, 1).

Writing V = Y X and Z = V ⊕ W , the main step is to define T = T 1 ⊕ T 2 where T 2 is a hypercyclic operator satisfying the hypercyclicity criterion and T 1 is the operator defined above. In particular, if Z contains a complemented copy of c 0 (N) or of ℓ p (N), p ∈ [1, +∞), then it satisfies the assumptions of Corollary 2.3.

To conclude, we observe that we can give an additional property of our operator when it is defined on ℓ 1 .

Theorem 2.4. Assume that X = Y = ℓ 1 . Then T is not ε-hypercyclic. Arguing as above, we find ℜe(x n ) ≥ 0 so that εa ≥ |ℜe(u n,0 ) + ℜe(x n ) -a| + ℜe(x n )ε.

We thus find εa ≥ ℜe(u n,0 ) + ℜe(x n ) -a + ℜe(x n )ε εa ≥ a -ℜe(x n ) -ℜe(u n,0 ) + ℜe(x n )ε which in turn yields ℜe(x n ) + ℜe(u n,0 ) 1 + ε ≤ a ≤ ℜe(x n ) + ℜe(u n,0 ) 1 -ε .

In particular, ℜe(u n,0 ) must be positive and I n is contained in an interval of length cℜe(u n,0 ) for some c > 0. But since we are working on ℓ 1 , n |ℜe(u n,0 )| < +∞, which contradicts that I = n I n has infinite length. □ Corollary 2.5. Let ε ∈ (0, 1). There exists an operator T on ℓ 1 which is δ-hypercyclic operator for all δ ∈ (ε, 1) and which is not ε-hypercyclic.

This leads to the following natural question:

Question 2.6. Let ε ∈ (0, 1). Does there exist an operator which is δ-hypercyclic if and only if δ ∈ [ε, 1)?

  be given by Lemma 2.1 for F = F k and let y k ∈ C minimizing y → ∥(y -1)e 0 + yω k e k ∥. Since (e n ) is a 1-unconditional basis of X, we deduce from the definition of ω k and y k that (1) min y∈C, w∈G k ∥(y -1)e 0 + yω k e k + w∥ = ∥(y k -1)e 0 + y k ω k e k ∥ = ε.

1 λ

 1 r k +k e l . The invertibility of each A n comes from the invertibility of its restriction to F k and to G k . Furthermore we prove ∥A n ∥ ≤ κ. For n = m k + 1, . . . , m k+1 , for a, b ∈ C and w ∈ G k , A n (ae 0 + be k + w) ≤ |a| + |b| max(1 + ω k , 2, 2/ω k ) + λ∥w∥ ≤ κ∥ae 0 + be k + w∥ where we have taken into account that ω

Proof.

  By contradiction assume that u is an ε-hypercyclic vector for T . Let us introduce M = max |u n,0 |, v = (e 0 , 0, . . . ), I = 2M 1-ε , +∞ and for n ∈ N,I n = I ∩ {a ∈ R : ∥T n u -av∥ ≤ εa} so that I = n I n .We first observe that if n = m k for some k, then I n is empty. Indeed, for these values of n, sinceA 1 • • • A n u n = u n , we get ∥T n u -av∥ ≥ ∥u n -ae 0 ∥ ≥ |u n,0 -a| = ε∥v∥.Let N = {n ∈ N :I n ̸ = ∅}. For n ∈ N , let a ∈ I n . Since n ̸ = m k , we know that A 1 • • • A n (u n ) = (u n,0 + x n )e 0 +x n εe k + w n for some x n ∈ C and some w n ∈ ℓ 1 with e * 0 (w n ) = e * k (w n ) = 0. Therefore εa ≥ ∥T n u -av∥ ≥ |u n,0 + x n -a| + |x n |ε.

Laboratoire de Mathématiques Blaise Pascal UMR 6620 CNRS, Université Clermont Auvergne, Campus universitaire Cézeaux, 3 place Vasarely, 63178 Aubière Cedex, France. Email address: frederic.bayart@uca.fr