Rodolfo Araya
email: rodolfo.araya@udec.cl

Cristóbal Bertoglio
email: c.a.bertoglio@rug.nl

Cristian Cárcamo
email: c.e.carcamo@rug.nl

David Nolte

Sergio Uribe
email: suribe@uc.cl

Error analysis of pressure reconstruction from discrete velocities

Magnetic Resonance Imaging allows to measure the three-dimensional velocity field in blood flows. Therefore, several methods have been proposed to reconstruct the pressure field from such measurements using the incompressible Navier-Stokes equations. However, those measurements are obtained at limited spatial resolution given by the voxel dimensions in the image. Therefore, the velocity entering to the right-hand-side corresponds to a piecewise linear interpolation of the exact velocity.

In this work we propose a strategy for convergence analysis of state-of-the-art pressure reconstruction methods. The methods analyzed are the the so called Pressure Poisson Estimator (PPE) and Stokes Estimator (STE). In the theoretical error analysis, we show that many terms of different convergence order appear. However, numerical results, show in academic examples that only the PPE may profit of increasing the polynomial order, and that the STE presents a higher accuracy than the PPE. Additionally, we compare the pressure estimation methods on real MRI data, assessing the impact of different spatial resolutions and polynomial degree on each of the methods. Here, the results are aligned with the academic test cases in terms of sensitivity to polynomial order and that the STE shows to be potentially more accurate when compared to reference pressure measurements.

Introduction

The intra-arterial spatial distribution of the blood pressure can be measured by means of catheterization [START_REF] Baumgartner | ESC Guidelines for the management of grown-up congenital heart disease (new version 2010) The Task Force on the Management of Grown-up Congenital Heart Disease of the European Society of Cardiology (ESC)[END_REF]. This technique consists in inserting a catheter equipped with a pressure transducer into the vasculature of the patient and manoeuvring it, under local anaesthesia and guided by fluoroscopy, to the location of interest. Although it is the 'clinical gold standard' for pressure quantification, the invasive nature of the method is associated with a risk of complications [START_REF] Wyman | Current complications of diagnostic and therapeutic cardiac catheterization[END_REF][START_REF] Vitiello | Complications associated with pediatric cardiac catheterization[END_REF][START_REF] Omran | Silent and apparent cerebral embolism after retrograde catheterisation of the aortic valve in valvular stenosis: A prospective, randomised study[END_REF]. Given that there are recommendations to avoid its use [START_REF] Vahanian | Guidelines on the management of valvular heart disease (version2012)[END_REF], to compute the pressure difference from measured flow fields is strongly preferred. Time-resolved 3D velocity encoded magnetic resonance imaging, or 4D flow MRI, offers measuring the complete 3D velocity field within a region of interest [START_REF] Markl | 4D flow MRI[END_REF][START_REF] Soulat | 4d flow with mri[END_REF]. The measured velocities can then be inserted in the linear momentum balance of the incompressible Navier-Stokes equations (NSE) and the velocity terms laid in the right-hand-side while the pressure holds as an unknown, i.e.,for a given measurement of the velocity u, the pressure gradient ∇p is found by solving:

∇p = -f u in Ω (1)
with Ω ⊂ R d and f u := (u • ∇)u -ν∆u., where the proper function spaces will be defined for each of the methods throughout the article. An example of pressure map estimation from real 4D Flow MRI data is shown in Figure 1. Figure 1: Pressure map estimation from in an experimental phantom of a thoraxic aorta [START_REF] Urbina | Realistic aortic phantom to study hemodynamics using MRI and cardiac catheterization in normal and aortic coarctation conditions[END_REF], adapted and reprinted from [START_REF] Nolte | Validation of 4d flow based relative pressure maps in aortic flows[END_REF]: Left: The 4D Flow MRI velocity measurements. Right: Relative pressure map computed from velocity data. Shown: cuts through roughly the center of the vessels.

In practice, and as it can be appreciated in Figure 1-Left, those measurements are obtained at limited spatial resolution -given by the voxel size in the image -and therefore the velocity entering to the righthand-side corresponds to an interpolated version of the exact velocity. Therefore, there is not a unique numerical approach to compute the reconstructed pressures. A review and preliminary numerical comparison of methods can be found in [START_REF] Bertoglio | Relative pressure estimation from velocity measurements in blood flows: State-of-art and new approaches[END_REF]. Among those methods, only a few can compute pressure fields and not just averaged pressure differences between two locations. The first one is the so-called Pressure Poisson Estimator (PPE) [START_REF] Ebbers | Estimation of relative cardiovascular pressure using time-resolved three-dimensional phase contrast mri[END_REF][START_REF] Smith | A finite-element approach to the direct computation of relative cardiovascular pressure from time-resolved mr velocity data[END_REF] and it consists of applying the divergence to the NSE obtaining a pressure Poisson equation, similarly as it is used in projection methods [START_REF] Guermond | An overview of projection methods for incompressible flows[END_REF]. However, the original PPE method cannot include the viscous contribution to the pressure gradient at the level of accuracy of the measured data. Therefore, recently in [START_REF] Pacheco | A continuous finite element framework for the pressure poisson equation allowing non-newtonian and compressible flow behavior[END_REF] the PPE method was modified by adding a boundary term with the viscous contribution.

Another more modern method corresponds to the Stokes Estimator (STE) was reported in [START_REF] Svihlová | Determination of pressure data from velocity data with a view toward its application in cardiovascular mechanics. part 1. theoretical considerations[END_REF]. The STE consists in adding to the NSE the Laplacian of an artificial incompressible velocity field with null trace leading to a linear Stokes problem for both pressure and artificial velocity fields. Such artificial velocity is supposed to be zero for perfect velocity measurements. The STE has shown more accurate results than the PPE in numerically simulated data [START_REF] Svihlová | Determination of pressure data from velocity data with a view toward its application in cardiovascular mechanics. part 1. theoretical considerations[END_REF][START_REF] Bertoglio | Relative pressure estimation from velocity measurements in blood flows: State-of-art and new approaches[END_REF] and in real phantom and patient data [START_REF] Nolte | Validation of 4d flow based relative pressure maps in aortic flows[END_REF]. However, the STE method is considerably more expensive computationally than the PPE.

To the best of the authors' knowledge, neither a mathematical convergence analysis of both PPE and STE methods or a comparison among discretization schemes for each of the methods has been reported. Therefore, the purpose of this work is to propose a strategy for performing a priori error analysis and applied it to the PPE and STE methods. The strategy is based on the splitting of the solution in two components and adding their contributions to the overall error. Moreover, for both methods we studied different discretization strategies in order to verify the theoretical analysis and give insights on the costeffectiveness of each approach. In order to assess the impact of discretizations on each of the methods, calculations of pressure fields based on experimental MRI data are also included. The remainder of this work is organized as follows. In Section 2 we present and analyze the PPE method in the standard and modified variants using Continuos Galerkin approaches. Section 3 introduces the STE and analyzes the classical Taylor-Hood and a tailored PSPG discretization. Then, in Section 4 we show numerical results using three known analytical solutions for the NSE, confirming the a priori error analysis. In Section 5 the methods are assessed under different spatial resolutions and polynomial degree on experimental MRI data. Finally, in Section 6 we draw some interpretation of the results and give recommendations for the use of these methods.

The Poisson Pressure Estimator

     -∆q = ∇ • f u , in Ω -∂q ∂n = f u • n, on ∂Ω Ω q dx = 0, (2)
with n the outward normal vector of ∂Ω. We will make use of the function spaces

H := [H 1 (Ω)] d , V := [H 1 0 (Ω)] d and
H j := v ∈ H : ∆v ∈ [L 2 (Ω)] d if j = 1 v ∈ H : ∇ × (∇ × u) ∈ [L 2 (Ω)] d if j = 2 Q j := r ∈ H 1 (Ω) : Ω rdx = 0 if j = 1 r ∈ H 1 (Ω) : n • curl r ∈ L 2 (∂Ω) and Ω r dx = 0 if j = 2
Assuming u ∈ H j , the weak formulation of the ( 2) is given by: Find q ∈ Q j such that

A(q, r) = F j u (r), ∀ r ∈ Q j , (3)
where A(q, r) = (∇q, ∇r) Ω and

F j u (r) = -((u • ∇)u, ∇r) Ω + δ 1j (ν∆u, ∇r) Ω + δ 2j n × ∇r, ν∇ × u ∂Ω , (4)
where δ ij is the Kronecker delta. We refer to Standard-PPE if j = 1 and Modified-PPE if j = 2 [START_REF] Pacheco | A continuous finite element framework for the pressure poisson equation allowing non-newtonian and compressible flow behavior[END_REF].

The uniqueness of the solution of Problem (3) follows from the Lax-Milgram lemma [START_REF] Ern | Theory and Practice of Finite Elements[END_REF]. Indeed, the coercivity of the left-side is straightforward. The continuity of F j u is obtained thanks to the identity

-(∇r, ν∇ × (∇ × u)) Ω = n × ∇r, ν∇ × u ∂Ω , (5)
and then

|F j u (r)| ≤ ( (u • ∇)u 0,Ω + δ 1j ν ∆u 0,Ω + δ 2j ν ∇ × (∇ × u) 0,Ω )|r| 1,Ω 2.

Continuous Galerkin discretizations

The finite element spaces for the pressure approximation and velocity interpolation are:

Q jh := {q h ∈ Q j : q h | K ∈ P k (K) ∀K ∈ T h } , H jh := {v h ∈ H j : v h | K ∈ P 1 (K) ∀K ∈ T h } .
We will also consider the Lagrange interpolation operator J h :

Q j ∩ H k+1 (Ω) → Q jh and L h : H j ∩ [H 2 (Ω)] d → H jh such that: |q -J h q| m,Ω ≤ a k h k+1-m |q| k+1,Ω , ∀ q ∈ H k+1 (Ω), 0 ≤ m ≤ k + 1. |v -L h v| m,Ω ≤ a k h 2-m |v| 2,Ω , ∀ v ∈ H 2 (Ω), 0 ≤ m ≤ 2. ( 6
)
Thus, the Galerkin scheme associated with the continuous variational formulation (3) reads as follows: Find

q h ∈ Q jh such that A(q h , r h ) := F j u h (r h ) ∀ r h ∈ Q jh , (7)
with

F j u h (r h ) = -((L h u • ∇)L h u, ∇r h ) Ω + δ 2j F ∈E h ∩∂Ω n × ∇r h , ν∇ × L h u F , (8)
According to discrete Lax-Milgram Theorem (Generalized Lax-Milgram for j = 2), Problem (7) has a unique solution q h ∈ Q jh .

Remark 1. Note that from the definitions (4) and (8) we can assure that the problem (7) is not a Galerkin scheme of the continuous problem (3). Indeed, the scheme is not consistent.

The strategy to prove convergence is to use the known Strang's lemma for conformal and non-consistent cases.

In order to prove the convergence theorems, let us state the next result.

Lemma 1. Let us assume that u ∈ H 2 (Ω). Then,

(u • ∇)u -(L h u • ∇)L h u 0,Ω ≤ C(a 1 + a 2 C I )h|u| 2,Ω u 2,Ω . (9)
with a 1 , a 2 the error interpolation constants, C is an injection constant and C I an inverse inequality constant.

Proof. Using properties of interpolation given by [START_REF] Ebbers | Estimation of relative cardiovascular pressure using time-resolved three-dimensional phase contrast mri[END_REF] and Young inequality we obtain

(u • ∇)u -(L h u • ∇)L h u 0,Ω ≤ (u • ∇)(u -L h u) 0,Ω + ((u -L h u) • ∇)L h u 0,Ω ≤|u -L h u| 1,Ω u ∞,Ω + ∇L h u ∞,Ω u -L h u 0,Ω ≤a 1 h|u| 2,Ω C u 2,Ω + C I h -1 L h u ∞,Ω a 2 h 2 |u| 2,Ω ≤ Ca 1 h|u| 2,Ω u 2,Ω + a 2 C I h u ∞,Ω |u| 2,Ω ≤ Ca 1 h|u| 2,Ω u 2,Ω + a 2 C C I h u 2,Ω |u| 2,Ω = C(a 1 + a 2 C I )h|u| 2,Ω u 2,Ω (10)
Lemma 2. Assume that u ∈ H j ∩ H 2 (Ω) d if j = 1 and u ∈ H j with u ∂Ω ∈ [H 2 (∂Ω)] d if j = 2. Then, sup r h ∈Q jh r h =0 |F j u (r h ) -F k u h (r h )| |r h | 1,Ω ≤ C 1 h|u| 2,Ω u 2,Ω + δ 1,j ∆u 0,Ω + δ 2,j C 2 νh 1/2 |u| 2,∂Ω Proof. sup r h ∈Q jh r h =0 |F j u (r h ) -F k u h (r h )| |r h | 1,Ω ≤ (u • ∇)u -(I h u • ∇)I h u 0,Ω + δ 1,j ∆u 0,Ω + δ 2,j sup r h ∈Q jh r h =0 F ∈E h ∩∂Ω | n × ∇r h , ν ∇ × (u -I h u) ∂Ω | |r h | 1,Ω .
For the first term in above inequality we use Lemma 1 and for the third term we have

F ∈E h ∩∂Ω | n × ∇r h , ν ∇ × (u -I h u) F | |r h | 1,Ω ≤ F ∈E h ∩∂Ω ν ∇r h 0,F ∇ × (u -I h u) 0,F |r h | 1,Ω (11)
Now, thanks to [5, Lemma 1.46], we have

∇r h 0,F ≤ C tr h -1/2 K |r h | 1,K , (12)
where

C tr = (k + 1)(k + 2) 2 1/2
(see [START_REF] Warburton | On the constants in hp-finite element trace inverse inequalities[END_REF]Theorem 3]). Besides, from [20, Lemma 10.8] we get

∇ × (u -I h u) 0,F ≤ C b |u -I h u| 1,F ≤ C b a 2,I h K |u| 2,F . (13)
and then, from ( 11), ( 12) and ( 13) we arrive to

F ∈E h ∩∂Ω | n × ∇r h , ν ∇ × (u -I h u) F | |r h | 1,Ω ≤ C tr C b a 2,I ν h 1/2 |u| 2,∂Ω , being C 2 = C tr C b a 2,I
, which allows us arrive to the desired result, where

C 1 = C(a 1 + a 2 C I ).
Finally, the next theorem holds.

Theorem 1 (Main Result I). Let q ∈ Q j ∩ H k+1 (Ω) and q h ∈ Q jh solutions of (3) and (7), respectively.

In addition, we assume that u ∈ H j ∩ [H 2 (Ω)] d and u ∈ H j with u ∂Ω ∈ [H 2 (∂Ω)] d , for j = 1 and j = 2 respectively. Then,

|q -q h | 1,Ω ≤ a k h k |q| k+1,Ω + C 1 h|u| 2,Ω u 2,Ω + δ 1,j ∆u 0,Ω + δ 2,j C 2 νh 1/2 |u| 2,∂Ω with k ≥ 1.
Proof. Thanks to the Strang's Lemma (see [START_REF] Ern | Theory and Practice of Finite Elements[END_REF]Lemma 2.27]) we have that

|q -q h | 1,Ω ≤ sup r h ∈Q jh r h =0 |F j u (r h ) -F j u h (r h )| |r h | 1,Ω + 2 inf r h ∈Q jh |q -r h | 1,Ω
The bound for the first term in the right-hand side follows directly from Lemma 2. For the second term, we will consider the interpolation operator, and then inf

r h ∈Q jh |q -r h | 1,Ω ≤|q -J h q| 1,Ω ≤ a k h k |q| k+1,Ω Corollary 1.
Let the hypothesis of Theorem 1 hold with Ω is convex polygonal domain. Then,

q -q h 0,Ω ≤ C r a k h k+1 |q| k+1,Ω + C 1 C p h|u| 2,Ω u 2,Ω + δ 1,j C p ∆u 0,Ω + δ 2,j C 2 C p νh 1/2 |u| 2,∂Ω
with C r the regularity constant and k ≥ 1.

Proof.

The proof starts taking qh ∈ Q jh such that satisfy Equation (3). It follows from the triangle inequality that q -q h 0,Ω ≤ q -qh 0,Ω + qh -q h 0,Ω

The bound estimation for q -qh 0,Ω is obtained solving the problem

-∆r = q -qh ,
with the Dirichlet conditon zero and applying the known Aubin-Nitsche Lemma [7, Lemma 2.31] and interpolation properties leading to

q -qh 0,Ω ≤ C r h|q -qh | 1,Ω ≤ C r a k h k+1 |q| k+1,Ω . (15)
For the second term of the right-hand side in ( 14) we proceed as follows:

|q h -q h | 1,Ω ≤ sup r h ∈Q jh r h =0 A(q h -q h , r h ) |r h | 1,Ω ≤ sup r h ∈Q jh r h =0 F j u (r h ) -F j u h (r h ) |r h | 1,Ω .
From Lemma 2 and Poincaré inequality, we get

qh -q h 0,Ω ≤ C p |q h -q h | 1,Ω ≤ C 1 C p h|u| 2,Ω u 2,Ω + δ 1,j C p ∆u 0,Ω + δ 2,j C 2 C p νh 1/2 |u| 2,∂Ω . (16)
Hence, the result is a direct consequence of the estimates ( 15) and ( 16).

The Stokes Estimator

The continuous problem

The STE consists then in adding the Laplacian of an incompressible auxiliary velocity w ∈ V and to the left-hand side of (1):

-∆w + ∇q = -f u in Ω ∇ • w = 0 in Ω. ( 17
)
Let us define the space P = L 2 0 (Ω). Hence, we can define the weak problem of ( 17) as:

Find (w, q) ∈ V × P such that B((w, q), (v, r)) = G u (v, r) ∀(v, r) ∈ V × P, (18)
where

B((w, q), (v, r)) := (∇w, ∇v) Ω -(q, ∇•v) Ω +(r, ∇•w) Ω and G u (v, r) := -((u•∇)u, v) Ω -ν(∇u, ∇v) Ω . ( 19
)
For the analysis, we will use the following norm (v, r) V×P := |v| 1,Ω + r 0,Ω , and if F is an linear functional operator we use the norm

F (V×P ) := sup (v,r)∈V×P (v,r) =0 |F (v, r)| (v, r) V×P (20)
Note that the problem ( 18) is well posed thanks to the V-ellipticity of the bilinear form (∇w, ∇v) Ω , and (q, ∇ • w) Ω satisfy an inf-sup condition (cf. [7, Prop. 2.36]).

Lemma 3. There exists a positive constant C B such that

B ≤ C B .
Proof. Using triangular inequality, Cauchy-Schwarz inequality, together to the inequalities 1

≤ √ d and √ a 2 + b 2 ≤ a + b, with a, b ≥ 0, we get |B((w, q), (v, r))| ≤ |w| 1,Ω |v| 1,Ω + √ d q 0,Ω |v| 1,Ω + √ d r 0,Ω |w| 1,Ω ≤ √ d|w| 1,Ω |v| 1,Ω + √ d q 0,Ω |v| 1,Ω + √ d r 0,Ω |w| 1,Ω ≤ √ d(|w| 2 1,Ω + |w| 2 1,Ω + q 0,Ω ) 1/2 (|v| 2 1,Ω + |v| 2 1,Ω + r 0,Ω ) 1/2 ≤ C B (|w| 2 1,Ω + q 2 0,Ω ) 1/2 (|v| 2 1,Ω + r 2 0,Ω ) 1/2 ≤ C B (w, q) V×P (v, r) V×P ,
and the result is obtained directly, with C B = 2 √ d.

Discrete spaces

Let us denote by {T h } a regular family partition of Ω composed of triangular elements K of diameter h K . We will denote by h the mesh size, where h = max {h K : K ∈ T h }. Now, for each h let W h and P h be finite-dimensional spaces such that:

W h := v h ∈ [H 1 (Ω)] d : v h | K ∈ [P l (K)] d ∀K ∈ T h , P h := {q h ∈ P : q h | K ∈ P k (K) ∀K ∈ T h } , H h := w h ∈ H : w h | K ∈ [P 1 (K)] d ∀K ∈ T h .
For our error analysis we will need to make use of some known results.

Theorem 2. For all w ∈ [P l (K)] d there holds,

∇w 0,K ≤ √ κ l |∂K| |K| w 0,K . ( 21
)
For d = 2, it holds κ 1 = 6 and κ 2 = 45 2 . Proof. See [15, Theorem 2]
Corollary 2. For d = 2 and l = 1 there holds

∇w 0,K ≤ C I h -1 K w 0,K , (22)
where

C I = 6 √ κ 1 sin 2 (θ)
, being θ the minimun angle of the element K.

Proof. The proof is a direct consequence of the minimum angle condition and Theorem 2.

Taylor-Hood discretization

For the discrete STE we set V h = W h ∩ V where inf-sup stable pairs of finite elements requires the use of different spaces for velocity and pressure and for this reason we take Taylor-Hood, where l = k + 1. Otherwise, it is not possible to use conforming spaces of lowest order for the discrete velocity. Furthermore, we will consider the property of interpolation operator

I h : V ∩ [H k+1 (Ω)] d → V h : |w -I h w| m,Ω ≤ a k h k+1-m |w| k+1,Ω , ∀w ∈ [H k+1 (Ω)] d , 0 ≤ m ≤ k + 1, (23)
Thereby, the discrete version of the problem (18) reads as follows:

Find (w h , q h ) ∈ V h × P h such that B((w h , q h ), (v h , r h )) = G u h (v h , r h ) ∀(v h , r h ) ∈ V h × P h , (24)
where the bilinear form B is like in the continuous case, and

G u h (v h , r h ) := -((L h u • ∇)L h u, v h ) Ω -ν(∇L h u, ∇v h ) Ω , (25)
with

L h : [H 2 (Ω)] d -→ H h a Lagrange interpolant.
Lemma 4. There exists a constant β 1 , independent of h, such that sup

(v h ,r h )∈V h ×P h (v h ,r h ) =0 B((w h , q h ), (v h , r h )) (v h , r h ) V×P ≥ β 1 (w h , q h ) V×P , ∀ (w h , q h ) ∈ V h × P h .
Proof. See equation [8, (1.39)] and [8, Corollary 4.1] For the next result, we consider the pair ( wh , qh )

∈ V h × P h , such that B(( wh , qh ), (v h , r h )) = G u (v h , r h ) ∀(v h , r h ) ∈ V h × P h , (26)
where

G u (v h , r h ) = -((u • ∇)u, v h ) 0,Ω -ν(∇u, ∇v h
) Ω with the continuous velocity u. Let us recall the following convergence result.

Lemma 5. Let (w, q) and ( wh , qh ) solutions of (18) and (26) respectively. Assume that (w,

q) ∈ [H 1 0 (Ω) ∩ H k+1 (Ω)] d × [L 2 0 (Ω) ∩ H k (Ω)], with k ≥ 1.
Then, there exists C > 0 independent of h such that

(w -wh , q -qh ) V×P ≤ C 1 C 2 h k |w| k+1,Ω + |q| k,Ω , with C 2 = 1 + C B β 1
and β 1 being the constant given in Lemma 4.

Proof. See [START_REF] Ern | Theory and Practice of Finite Elements[END_REF]Lemma 2.44] In order to show the convergence of q h (Main Result I, see later Theorem 3), we set the following Lemma.

Lemma 6. Let (w h , q h ), ( wh , qh ) ∈ V h × P h , solutions of (24) and (26) respectively, and β 1 the constant given in Lemma 4. Then,

( wh -w h , qh -q h ) V×P ≤ β -1 1 G u -G u h (V×P ) with (G u -G u h )(v h , r h ) := -((u • ∇)u -(L h u • ∇)L h u, v h ) Ω -ν(∇u -∇L h u, ∇v h ) Ω .
Proof. By Lemma 4 together with the Cauchy -Schwarz inequality, we arrive to the inequality

β 1 ( wh -w h , qh -q h ) V×P ≤ sup (v h ,r h )∈V h ×P h (v h ,r h ) =0 B(( wh -w h , qh -q h ), (v h , r h )) (v h , r h ) V×P = sup (v h ,r h )∈V h ×P h (v h ,r h ) =0 G u (v h , r h ) -G u h (v h , r h ) (v h , r h ) V×P = sup (v h ,r h )∈V h ×P h (v h ,r h ) =0 (G u -G u h )(v h , r h ) (v h , r h ) V×P ≤ G u -G u h (V×P ) .
Lemma 7. Let G u and G u h be as in [START_REF] Soulat | 4d flow with mri[END_REF] and (25) respectively and denote by (V × P ) the dual space of the product space V × P . In addition, we assume that

u ∈ [H 2 (Ω)] d . Then G u -G u h (V×P ) ≤ h|u| 2,Ω C p C(a 1 + a 2 C I ) u 2,Ω + ν a 1 .
Proof.

G u (v h , r h ) -G u h (v h , r h ) 0,Ω ≤ (u • ∇)u -(L h u • ∇)L h u) 0,Ω v h 0,Ω + ν ∇u -∇L h u 0,Ω ∇v h 0,Ω .
For the first term of the right-hand side, we use Lemma 1.

For the second term of the right-hand-side, we have from the interpolation bounds:

∇(u -L h u) 0,Ω ≤ a 1 h|u| 2,Ω .
Finally, using the above inequalities and Poincaré inequality we get

(G u -G u h )(v h , r h ) 0,Ω ≤ h|u| 2,Ω C p C (a 1 + a 2 C I ) u 2,Ω + ν a 1 |v h | 1,Ω ≤ h|u| 2,Ω C p C (a 1 + a 2 C I ) u 2,Ω + ν a 1 (v h , r h ) V×P ,
where C p is the Poincaré constant. Thereby, we arrive straight to the result of the lemma.¸ Finally, we can derive the first main convergence result.

Theorem 3 (Main Result II). Assume that (w, q) ∈ [H 1 0 (Ω) ∩ H k+1 (Ω)] d × [L 2 0 (Ω) ∩ H k (Ω)] and u ∈ [H 2 (Ω)] d . Then, |w -w h | 1,Ω + q -q h 0,Ω ≤C 1 C 2 h k |w| k+1,Ω + |q| k,Ω + β -1 1 h|u| 2,Ω ρ C p C (a 1 + a 2 C I ) u 2,Ω + µ a 1 .
Proof. The proof follows from Lemmas 5, 6, and 7.

Stabilized PSPG discretization

Let us consider again the Stokes problem given as in [START_REF] Rhebergen | A hybridizable discontinuous galerkin method for the navier-stokes equations with pointwise divergence-free velocity field[END_REF] and its respective variational formulation [START_REF] Smith | A finite-element approach to the direct computation of relative cardiovascular pressure from time-resolved mr velocity data[END_REF]. We will now analyze the PSPG Stabilization [START_REF] Hugues | A new finite element formulation for computational fluis dynamics: V. Circumventing the Babuska-Brezzi condition: A stable Petrov-Galerkin formulation of the Stokes problem accommodating equal-order interpolations[END_REF] with the end of comparing the error of convergence between the pressure obtained with both schemes. We want to use spaces of finite element of order k for the velocity and the pressure, i.e., k = l, by means of the following stabilized formulation.

B s ((w h , q h )(v h , r h )) = G s u h (v h , r h ) (27)
where

B s ((w h , q h )(v h , r h )) := B((w h , q h )(v h , r h )) + K∈T h δ h 2 K (∇q h , ∇r h ) K G s u h (v h , r h ) := G u h (v h , r h ) + K∈T h δ h 2 K (-f u h , ∇r h ) K with B((•, •), (•, •)) and G u h (•,
•) defined as in [START_REF] Vitiello | Complications associated with pediatric cardiac catheterization[END_REF], and

f u h := (L h u • ∇)L h u. ( 28
)
Remark 2. Note that the term ∆w h is not included in the stabilization. This is possible to do while keeping strong consistency since w = 0. Our choice allows also to avoid conditional well-posedness of the discrete solution as in standard PSPG stabilized formulations.

Let us define the mesh-dependent norm on the product space

V × P (v, r) 2 h := B s ((v, r), (v, r)) = ∇v 2 0,Ω + K∈T h δh 2 K ∇r 2 0,K . ( 29
) Remark 3. It is possible to prove that (v h , r h ) h (v h , r h ) V×P for all (v h , r h ) ∈ V h × P h .
Indeed, applying the inequality [START_REF] Svihlová | Determination of pressure data from velocity data with a view toward its application in cardiovascular mechanics. part 1. theoretical considerations[END_REF] and the previous assumptions we get

(v h , r h ) 2 h := ∇v h 2 0,Ω + K∈T h δh 2 K ∇r h 2 0,K ≤ ∇v h 2 0,Ω + δC 2 I r h 2 0,Ω ≤ max 1, δC 2 I ( ∇v h 2 0,Ω + r h 2 0,Ω ),
and then,

(v h , r h ) h ≤ C eq (v h , r h ) V×P , (30)
where

C eq = max 1, δC 2 I 1/2 Lemma 8. B s ≤ C B s = max C B , √ δC eq C I (31)
Proof. Using the inequalities ( 22) and (30), Theorem 3 and Cauchy-Schwarz inequality we obtain that

|B s ((w h , q h ), (v h , r h ))| ≤ B (w h , q h ) V×P (v h , r h ) V×P + K∈T h δh 2 K ∇q h 0,K ∇r h 0,K ≤ C B (w h , q h ) V×P (v h , r h ) V×P + K∈T h δh 2 K ∇q h 2 0,K 1/2 K∈T h δh 2 K ∇r h 2 0,K 1/2 ≤ C B (w h , q h ) V×P (v h , r h ) V×P + K∈T h δC 2 I q h 2 0,K 1/2 (v h , r h ) h ≤ C B s (w h , q h ) V×P (v h , r h ) V×P ,
and then the result follows.

In the next lemmas we will consider the pair ( wh , qh ) ∈ V h × P h which are solution of the equation

B s ((w h , q h )(v h , r h )) = G s u (v h , r h ), ∀ (v h , r h ) ∈ V h × P h (32)
where

G s u (v h , r h ) := G u (v h , r h ) + K∈T h δ h 2 K (f u , ∇r h ) K .
We highlight that the solvability of the problem (32) has been guaranteed in [START_REF] Hugues | A new finite element formulation for computational fluis dynamics: V. Circumventing the Babuska-Brezzi condition: A stable Petrov-Galerkin formulation of the Stokes problem accommodating equal-order interpolations[END_REF]. Lemma 9. Let (w, q) and ( wh , qh ) solutions of (18) and (32) respectively. Assume that (w,

q) ∈ [H 1 0 (Ω) ∩ H k+1 (Ω)] d × [L 2 0 (Ω) ∩ H k (Ω)] . Then, there is C > 0 independent of h such that |w -wh | 1,Ω + q -qh 0,Ω ≤ C 1 C 3 h k |w| k+1,Ω + |q| k,Ω with C 3 = 1 + B s .
Proof.

We note that (w, q) and ( wh , qh ) satisfy the orthogonality property

B s ((w -wh , q -qh ), (v h , r h )) = 0 ∀(v h , r h ) ∈ V h × P h .
Indeed, thanks to the consistency of bilinear form B we get

B s ((w -wh , q -qh ), (v h , r h )) = B((w -wh , q -qh ), (v h , r h )) + K∈T h δh 2 K (∇q, ∇r h ) K - K∈T h δh 2 K (∇q h , ∇r h ) K = K∈T h δh 2 K (f u , ∇r h ) K - K∈T h δh 2 K (f u , ∇r h ) K = 0.
By the triangle inequality we can get, (w -wh , q -qh )

V×P = |w -I h w + I h w -w h | 1,Ω + q -J h q + J h q -q h 0,Ω ≤ (w -I h w, q -J h q) V×P + ( wh -I h w, qh -J h q) V×P . (33)
For the second term of the right-hand side, we must consider the result earned in [START_REF] Hugues | A new finite element formulation for computational fluis dynamics: V. Circumventing the Babuska-Brezzi condition: A stable Petrov-Galerkin formulation of the Stokes problem accommodating equal-order interpolations[END_REF] from where we get ( wh -I h w, qh -J h q) V×P ≤ sup

(v h ,r h )∈V h ×P h (v h ,r h ) =0 B s (( wh -I h w, qh -J h q), (v h , r h )) (v h , r h ) h = sup (v h ,r h )∈V h ×P h (v h ,r h ) =0 B s ((w -I h w, q -J h q), (v h , r h )) (v h , r h ) h ≤ B s (w -I h w, q -J h q) V×P ,
and so, from this inequality and (33) we obtain (w -wh , q -qh ) V×P ≤ 1 + B s (w -I h w, q -J h q) V×P and thereby we arrive to

|w -wh | 1,Ω + q -qh 0,Ω ≤ C 1 C 3 h k |w| k+1,Ω + |q| k,Ω , .
Lemma 10. Let ( wh , qh ) and (w h , q h ) be solutions of (32) and (27), respectively. Additionally, we assume that u ∈ [H 2 (Ω)] d . Then, the following bound is satisfied:

( wh -w h , qh -q h ) h ≤ G u -G u h (V×P ) + √ δh 2 T (a 1 , a 2 , u, h)|u| 2,Ω + √ δνh ∆u 0,Ω , Proof.
Let e w h := whw h and e q h := qh -q h . Then, thanks to the stability of B s given in (29) we have (e w h , e q h ) h = B s ((e w h , e q h )(e w h , e q h )) (e w h , e q h ) h ≤ sup

(v h ,r h )∈V h ×P h (v h ,r h ) =0 B s ((e w h , e q h )(v h , q h )) (v h , q h ) h = sup (v h ,r h )∈V h ×P h (v h ,r h ) =0 G s u (v h , r h ) -G s u h (v h , r h ) (v h , q h ) h = sup (v h ,r h )∈V h ×P h (v h ,r h ) =0 G u (v h , r h ) -G u h (v h , r h ) - K∈T h δh 2 K (f u -f u h , ∇r h ) K (v h , q h ) h
We take the term within sum, making use of the Cauchy-Schwarz inequality and proceeding similarly as in [START_REF] Hugues | A new finite element formulation for computational fluis dynamics: V. Circumventing the Babuska-Brezzi condition: A stable Petrov-Galerkin formulation of the Stokes problem accommodating equal-order interpolations[END_REF], we obtain

- K∈T h δh 2 K (f u -f u h , ∇r h ) K = K∈T h δh 2 K ((u • ∇)u -(L h u • ∇)L h u, ∇r h ) K -ν K∈T h δh 2 K (∆u, ∇r h ) K ≤ K∈T h δh 2 K (u • ∇)u -(L h u • ∇)L h u 0,Ω ∇r h 0,K + ν K∈T h δh 2 K ∆u 0,K ∇r h 0,K ≤ K∈T h δh 2 K (u • ∇)u -(L h u • ∇)L h u 2 0,K 1/2 K∈T h δh 2 K ∇r h 2 0,K 1/2 + ν K∈T h δh 2 K ∆u 2 0,K 1/2 K∈T h δh 2 K ∇r h 2 0,K 1/2 ≤ √ δh 2 C(a 1 + a 2 C I )|u| 2,Ω u 2,Ω (v h , r h ) h + ν √ δ K∈T h h 2 K ∆u 2 0,K 1/2 (v h , r h ) h ≤ √ δh 2 C(a 1 + a 2 C I )|u| 2,Ω u 2,Ω (v h , r h ) h + ν √ δh ∆u 0,Ω (v h , r h ) h .
As a main result of this section, by employing the approximation properties and a priori estimates, we obtain the next result.

Theorem 4 (Main Result III). Assume that the hypothesis of Theorem 3 hold. Then,

|w -w h | 1,Ω + q -q h 0,Ω ≤C 1 C 3 h k |w| k+1,Ω + |q| k,Ω + ν √ δh ∆u 0,Ω + h|u| 2,Ω C p C (a 1 + a 2 C I ) u 2,Ω + ν a 1 + √ δhC p C (a 1 + a 2 C I ) u 2,Ω .
Proof. The proof follows from combining the results of Lemmas 7, 9 and 10. 4. Numerical results for the convergence analysis

In this section we present some numerical examples to illustrate the theoretical results previously described. The legends in the plots follow the notation:

• e 1 (q): Pressure error in L 2 -norm with P 1

• e 2 (q): Pressure error in L 2 -norm with P 2 , with e i (q) := q -q h 0,Ω q 0,Ω .

In addition, for Modified-PPE and Standard-PPE we will use the legend PPEvisc and PPE respectively. For the STE computed using Taylor Hood spaces and PSPS we will use the legend STE (TH) and STE (PSPG) respectively.

Every numerical routine has been sorted out using the open-source finite element libraries FEniCS [START_REF] Alnaes | The FEniCS Project Version 1.5[END_REF].

Example 1. For the first example, we consider the exact solution of the two dimensional Kovasznay flow u(x, y) = 1 -e λx cos(2πy)

λ 2π e λx sin(2πy)

, p(x, y) = 1 2 e λx -(e 3λ -e -λ ),
where

Ω = - 1 2 , 3 2
× 0, 2 and the parameter λ is given by λ = 1 2ν -1 4ν 2 + 4π 2 . For this ilustration we have taken the Reynold number as in [START_REF] Rhebergen | A hybridizable discontinuous galerkin method for the navier-stokes equations with pointwise divergence-free velocity field[END_REF] which is given by Re = 1 ν .

The convergence results for Example are shown in Figure 2 and examples of pressure and velocity fields in Figures 3456. First, it can be appreciated the lack of convergence of the PPE, while adding the viscous terms recovers it. Also, the STE appears to be more accurate than the PPE(visc) and it seems not to profit from the increase of polynomial order. Moreover, the STE-PSPG appears to deliver more accurate results than the STE-TH. Finally, it is worth to say that the sensitivity of all methods with respect to the polynomial order decreases when increasing the Reynolds number.

Example 2. Next we turn to the testing the scheme, where the computational domain is the rectangle Ω = [0, 1] 2 and we consider the exact solution of the Navier-Stokes equation given by u(x, y) = ( ν 4 e x sin(νy), 1 4 e x cos(νy)) and p(x, y) = -

ν 2 e 2x + ν 4 (e 2 -1) (34)
The convergence results for Example are shown in Figure 7 and the examples of pressure and velocity fields in Figures 891011. Here, the same remarks given about the results in Example 4 apply, except that for higher Reynolds number the STE methods appear to keep the sensitivity (though worsening) when increasing the polynomial order. Experimental MRI data was used to assess the impact of discretization in the pressure estimation methods in realistic data and flow regimes. The setup consisted of a 3D printed, MR compatible phantom of the thoracic aorta with 60 % of obstruction in order to produce a typical obstruction. A blood mimicking fluid was pumped into the phantom obtaining physiological velocities. The phantom was equipped with a catheterization unit to measure invasively and simultaneously the pressure gradient across the obstruction. 4D Flow MRI was acquired with an isotropic voxel size of 0.9 mm and 25 time instants along the emulated cardiac cycle. We refer to [START_REF] Urbina | Realistic aortic phantom to study hemodynamics using MRI and cardiac catheterization in normal and aortic coarctation conditions[END_REF][START_REF] Montalba | Variability of 4D flow parameters when subjected to changes in MRI acquisition parameters using a realistic thoracic aortic phantom[END_REF][START_REF] Nolte | Validation of 4d flow based relative pressure maps in aortic flows[END_REF] for the technical details of the experiment. The 4D Flow data is shown in Figure 1.

Two tetrahedral meshes for the pressure computations were constructed. The first one was created using the original 0.9 mm resolution where the nodes of the mesh correspond to the voxels center. The second mesh has 2 mm resolution created using linear interpolation on the first mesh. Pressure maps were computed from all 4D Flow data sets with the PPE, PPEvisc and STE methods. Due to the pulsatile nature of the experiment, the term

-(∂ τ u, ∇r) Ω and -(∂ τ u, r) Ω ,
with ∂ τ the backward finite difference operator between two measured time instants, were added to the right-hand-side of the PPE and STE methods, respectively. This implies that the convergence analysis does not fully apply to the this experimental setup, however, the goal is merely to give an idea on how the discretization setup and methods compare in a scenario of practical relevance. The pressure differences, to be compared with the corresponding catheter values, were defined as differences of the pressure averages over two spheres with a radius of 4 mm at locations proximally (ascending aorta) and distally to the obstruction.

For the PPE and PPEvisc continuous Galerkin finite elements with k = 1, 2, 3 were considered.

For the STE, both Taylor-Hood (TH) and PSPG cases were computed, the latter with stabilization parameter δ = 0.01 as in the previous section for the convergence analysis. In the 2 mm element size mesh, k = 1, 2 was tested for both TH and PSPG. In the 0.9 mm element size mesh, only k = 1 was used for TH (due to the very high computational cost of higher order) and k = 1, 2 was used for PSPG. These methods were implemented using the FEM library FEniCS [START_REF] Alnaes | The FEniCS Project Version 1.5[END_REF].

Figure 12 shows the results of the pressure estimation, where the catheter pressure values show that the 4D Flow based pressure estimation deliver reasonable values. However, note that the catheter measurements cannot be considered as ground truth, since the precision of the pressure measurements can be considered within a few mmHg [START_REF] Nolte | Validation of 4d flow based relative pressure maps in aortic flows[END_REF][START_REF] Brüning | Uncertainty quantification for non-invasive assessment of pressure drop across a coarctation of the aorta using CFD[END_REF][START_REF] De Vecchi | Catheter-induced errors in pressure measurements in vessels: an in-vitro and numerical study[END_REF]. It can be noted that:

• PPE and PPEvisc deliver visually the same results, which may occur due to the fact that viscous effects are negligible in this type of (patho-)physiological flows.

• PPE and PPEvisc allow for a larger pressure gradient when increasing k in the coarse mesh.

• PPE and PPEvisc are less sensitive to k for the finest mesh.

• STE methods allow to recover larger pressure differences than PPE methods.

• STE-PSPG delivers equal or better results than STE-TH for k = 1, what is consistent with the convergence results of the numerical tests in the previous section.

• STE-PSPG seems not to profit from increasing polynomial order, what is consistent with convergence results for high Reynolds numbers.

• If one may take the catheter measurements as a ground truth, STE-PSPG would deliver the most accurate results.

Conclusions

In this article we have analyzed theoretically and numerically some strategies used to recover pressure fields from discrete velocities using the incompressible Navier-Stokes equations. We analyze two methods, the STE and PPE. While the STE is implemented using the classical Taylor-Hood finite element spaces and pressure-stabilizing Petrov-Galerkin (PSPG), the PPE is implemented with the traditional continuous Galerkin method. For the PPE, two versions have been studied, the standard one without (the viscous term) and a modified one including the viscous term. The error analysis shows that all methods, except the standard PPE, converge to the exact solution when decreasing the element size of the image mesh h.With respect to the convergence rate, terms of several orders appear in the error analysis. Numerical results show that for the PPEvisc linear order dominates in the presented test cases. For the STE, convergence in the numerical examples vary depending on the testcase and the polynomial order. Numerical results in academic test cases show that with increasing Reynolds number, the results appear to be less sensitive to an increase in polynomial order, in particular for the STE while the PPEvisc shows some improvements. In many of the cases, the error also appears to decrease faster with h for the STE than the PPEvisc. Among both STE discretizations, PSPG appears to be equally or sometimes more accurate than Taylor-Hood approximations. The computations with real MRI data are aligned with these observations. Therefore, it appears that STE-PSPG can be the method of choice with the best accuracy and reasonable computational cost.

Figure 3 :Figure 4 :Figure 5 :Figure 6 :Figure 8 :Figure 9 :Figure 10 :Figure 11 :

 3456891011 Figure 3: P 1 -interpolated reference velocity and pressure fields (top) and reconstructed pressure fields with order k = 1 (bottom) for ν = 1 in Example 4 (Kovaznay flow).

Figure 12 :

 12 Figure 12: Pressure difference in the obstruction over time error computed from 4D Flow and compared with the catheter values.

Acknowledgements

The authors would like to thank to Douglas Pachecho, Institute of Computational Mathematics, TU Graz, Austria, for the discussions about the PPE and to Abner Poza, Dept. of Mathematics and Applied Physics, UCSC, Chile, for the suggestions received. R.A. was partially supported by ANID-Chile through the projects Centro de Modelamiento Matemático (AFB170001) of the PIA Program: Concurso Apoyo a Centros Científicos y Tecnológicos de Excelencia con Financiamiento Basal, and FONDECYT project 1211649. C.B. received funding from the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation programme (grant agreement No852544 -CardioZoom). C.C. was partially supported by ANID-Chile through scholarship PCHA/Doctorado Nacional. S.U. acknowledges funding from the Millennium Nucleus for Cardiovascular Magnetic Resonance and FONDECYT grant 1181057.


L Cadilhac
email: lcadilhac@impan.pl

F Sukochev
email: f.sukochev@unsw.edu.au

D Zanin
email: d.zanin@unsw.edu.au

LORENTZ-SHIMOGAKI-ARAZY-CWIKEL THEOREM REVISITED

We present a new approach to Lorentz-Shimogaki and Arazy-Cwikel Theorems which covers all range of p, q ∈ (0, ∞] for function spaces and sequence spaces.

Introduction

Descriptions of interpolation spaces for couples of L p -spaces, 1 ≤ p ≤ ∞ were extensively researched at the end of the 70's and in the 80's, providing satisfying answers to most problems which were considered relevant at the time.

However, new questions arising from noncommutative analysis recently highlighted some gaps in our knowledge of this subject, especially for the case of p < 1. In this paper, we revisit some important results of the literature ( [START_REF] Lorentz | Interpolation theorems for the pairs of spaces (L p , L ∞ ) and (L 1 , L q )[END_REF], [START_REF] Arazy | A new characterization of the interpolation spaces between L p and L q[END_REF], [START_REF] Sparr | Interpolation of weighted Lp-spaces[END_REF]), generalising them and thus filling some of the holes that were revealed in the theory. In particular, we answer a question asked in [START_REF] Levitina | Cwikel estimates revisited[END_REF] by Levitina and the last two authors and already partially studied in [START_REF] Cwikel | An alternative characterization of normed interpolation spaces between ℓ 1 and ℓ q[END_REF] regarding the interpolation theory of sequence spaces (see Theorem 1.2). Besides this new result, this paper introduces a general approach which covers the range of all 0 ≤ p ≤ ∞ and is self-contained. It puts an emphasis on the use of the space L 0 of all finitely supported measurable functions. As far as the authors know this space rarely appears in interpolation theory (however, see [START_REF] Hudzik | An interpolation theorem in symmetric function fspaces[END_REF], [START_REF] Astashkin | Interpolation of operators in quasinormed groups of measurable functions[END_REF] and [START_REF] Huang | Interpolation between L 0 (M, τ ) and L∞(M, τ )[END_REF]). We provide evidence that L 0 is a suitable "left endpoint" on the interpolation scale of L p -spaces, despite its possessing an atypical structure (it is not even an F -space).

Recall that a function space E is an interpolation space for the couple (L p , L q ) if any linear operator T bounded on L p and L q is also bounded on E (see Definition 1.6). This notion provides a way of transfering inequalities well-known in L p -spaces to more exotic ones. To both understand the range of applicability of this technique and be able to check whether it applies to a given function space E, we are interested in simple descriptions of interpolation spaces for the couple (L p , L q ). This problem has a long history starting with seminal Calderón-Mityagin theorem [START_REF] Mityagin | An interpolation theorem for modular spaces[END_REF] and [START_REF] Calderón | Spaces between L 1 and L ∞ and the theorem of Marcinkiewicz[END_REF] on the couple (L 1 , L ∞ ) and followed by Lorentz and Shimogaki's [START_REF] Lorentz | Interpolation theorems for the pairs of spaces (L p , L ∞ ) and (L 1 , L q )[END_REF] results on the couples (L 1 , L q ) and (L p , L ∞ ) with 1 ≤ p, q ≤ ∞, which are stated in terms of various submajorizations, see also [START_REF] Kalton | Interpolation of Banach spaces[END_REF]Theorem 7.2].

We consider two orders: head majorization and tail majorization. Head majorization coincides with the usual notion of (sub)majorization and already appears in Calderón's work. It is defined on L 1 + L ∞ by: g ≺≺ hd f ⇔ ∀t > 0, t 0 µ(s, g)ds ≤ t 0 µ(s, f )ds. If moreover f, g ∈ L 1 and f 1 = g 1 , then we write g ≺ hd f . Above and in the remainder of the text µ(g) : t → µ(t, g) denotes the right-continuous decreasing rearangement of g. Note that µ(g) is well-defined if and only if g belongs to L 0 +L ∞ .

Tail majorization is defined on L 0 + L 1 by:

g ≺≺ tl f ⇔ ∀t > 0, ∞ t µ(s, g)ds ≤ ∞ t µ(s, f )ds.
If moreover f, g ∈ L 1 and f 1 = g 1 , then we write g ≺ tl f . Remark that g ≺ tl f if and only if f ≺ hd g.

It is well-known that the order ≺≺ hd is strongly linked to the interpolation theory of the couples (L p , L ∞ ). We show that similarly, the order ≺≺ tl is linked to the couples (L 0 , L q ). Combining these two tools, we recover in a self-contained manner characterisations of interpolation spaces for couples of arbitrary L p -spaces, 0 < p ≤ ∞ earlier obtained in [START_REF] Cadilhac | Interpolation, majorization and noncommutative khintchine inequalities[END_REF]. Note that tail majorization coincides with the weak supermajorization of [START_REF] Dodds | On p-convexity and q-concavity in noncommutative symmetric spaces[END_REF].

Let X be the linear space of all measurable functions. If not precised otherwise, the underlying measure space we are working on is (0, ∞) equipped with the Lebesgue measure m. We obtain the following: Theorem 1.1. Let E ⊂ X be a quasi-Banach function space (a priori, not necessarily symmetric). Let p, q ∈ (0, ∞) such that p < q. Then: (a) E is an interpolation space for the couple (L p , L ∞ ) if and only if there exists c p,E > 0 such that for any f ∈ E and g ∈ L p + L ∞ ,

|g| p ≺≺ hd |f | p ⇒ g ∈ E and g E ≤ c p,E f E ;
(b) E is an interpolation space for the couple (L 0 , L q ) if and only if there exists c q,E > 0 such that for any f ∈ E and g ∈ L 0 + L q , |g| q ≺≺ tl |f | q ⇒ g ∈ E and g E ≤ c q,E f E ;

(c) E is an interpolation space for the couples (L 0 , L q ) and (L p , L ∞ ) if and only if it is an interpolation space for the couple (L p , L q ). This extends results of Lorentz-Shimogaki and Arazy-Cwikel to the quasi-Banach setting and contributes to the two first questions asked by Arazy in [9, p.232] in the particular case of L p -spaces, 0 < p < ∞. As mentioned before, our approach places L 0 as a left endpoint on the interpolation scale of L p -spaces, in sharp contrast to earlier results which focused mostly on Banach spaces and had L 1 playing this part. An advantage of our approach is that it naturally encompasses every symmetric quasi-Banach space since they are all interpolation spaces for the couple (L 0 , L ∞ ) (see [START_REF] Hudzik | An interpolation theorem in symmetric function fspaces[END_REF], [START_REF] Astashkin | Interpolation of operators in quasinormed groups of measurable functions[END_REF]). On the contrary, there exist some symmetric Banach spaces which are not interpolation spaces for the couple (L 1 , L ∞ ) (see [START_REF] Sedaev | Fully symmetric function spaces without an equivalent Fatou norm[END_REF]). This led to some difficulties which were customarily be circumvented with the help of various technical conditions such as the Fatou property (as appears for example in [START_REF] Astashkin | Interpolation between L 1 and Lp, 1 < p < ∞[END_REF]).

Our strategy in this paper is totally different from the techniques used in [START_REF] Mityagin | An interpolation theorem for modular spaces[END_REF][START_REF] Calderón | Spaces between L 1 and L ∞ and the theorem of Marcinkiewicz[END_REF][START_REF] Lorentz | Interpolation theorems for the pairs of spaces (L p , L ∞ ) and (L 1 , L q )[END_REF][START_REF] Sparr | Interpolation of weighted Lp-spaces[END_REF][START_REF] Cwikel | Monotonicity properties of interpolation spaces[END_REF][START_REF] Hudzik | An interpolation theorem in symmetric function fspaces[END_REF][START_REF] Arazy | A new characterization of the interpolation spaces between L p and L q[END_REF][START_REF] Cwikel | Interpolation Spaces and Allied Topics in Analysis[END_REF][START_REF] Astashkin | Interpolation of operators in quasinormed groups of measurable functions[END_REF][START_REF] Cwikel | An alternative characterization of normed interpolation spaces between ℓ 1 and ℓ q[END_REF][START_REF] Kalton | Interpolation of Banach spaces[END_REF][START_REF] Astashkin | Interpolation between L 1 and Lp, 1 < p < ∞[END_REF][START_REF] Cadilhac | Interpolation, majorization and noncommutative khintchine inequalities[END_REF][START_REF] Sergey | A version of Calderón-Mityagin theorem for the class of rearrangement invariant groups[END_REF] and is based on partition lemmas, which were originally developed in a deep paper due to Braverman and Mekler [START_REF] Braverman | The Hardy-Littlewood property for symmetric spaces[END_REF], which lies outside of the realm of interpolation theory. The approach of Braverman and Mekler was subsequently revised and redeveloped in [START_REF] Sukochev | Orbits in symmetric spaces[END_REF] and precisely this revision consitutes the core of our approach in this paper.

We restate partition lemmas based on [START_REF] Sukochev | Orbits in symmetric spaces[END_REF]Proposition 19] in Section 3. These lemmas allow us to restrict head and tail majorizations to very simple situations and reduce the problem to functions taking at most two values. Then, we deduce interpolation results from those structural lemmas.

Note that this scheme of proof is quite direct and in particular, does not involve at any point duality related arguments which are applicable only to Banach spaces ( [START_REF] Lorentz | Interpolation theorems for the pairs of spaces (L p , L ∞ ) and (L 1 , L q )[END_REF]) or more generally to L-convex quasi-Banach spaces ( [START_REF] Kalton | Convexity conditions for nonlocally convex lattices[END_REF], [START_REF] Popa | Rearrangement invariant p-spaces on [0, 1], 0 < p < 1, containing a complemented subspace isomorphic to l 2[END_REF]).

In Section 5, we pursue the same type of investigation, but in the setting of sequence spaces. The non-diffuse aspect of the underlying measure generates substantial technical difficulties. In particular, we require a new partition lemma which is not as efficient as those in Section 3 (compare Lemmas 5.2 and 3.8). This deficiency has been first pointed out to the authors by Cwikel. However, we are still able to resolve the conjecture of [START_REF] Levitina | Cwikel estimates revisited[END_REF] (in the affirmative) by combining Lemma 5.1 with a Boyd-type argument which we borrow from Montgomery-Smith [START_REF] Stephen | The Hardy operator and Boyd indices[END_REF]. In particular, we substantially strengthen the results in [START_REF] Cwikel | An alternative characterization of normed interpolation spaces between ℓ 1 and ℓ q[END_REF]. Here is the precise statement that we obtain: Theorem 1.2. Let E ⊂ ℓ ∞ be a quasi-Banach sequence space and q ≥ 1. The following conditions are equivalent: (a) there exists p < q such that E is an interpolation space for the couple (ℓ p , ℓ q ); (b) there exists c > 0 such that for any

u ∈ E and v ∈ ℓ ∞ |v| q ≺≺ tl |u| q ⇒ v ∈ E and v E ≤ c u E ; (c) for any u ∈ E and v ∈ ℓ ∞ , |v| q ≺≺ tl |u| q ⇒ v ∈ E.
We freely use results of Cwikel [START_REF] Cwikel | Monotonicity properties of interpolation spaces[END_REF] and the first author [START_REF] Cadilhac | Interpolation, majorization and noncommutative khintchine inequalities[END_REF] to avoid repeating too many similar arguments.

Preliminaries

Interpolation spaces. The reader is referred to [START_REF] Bergh | Interpolation spaces. An introduction[END_REF] for more details on interpolation theory and [START_REF] Kreȋn | Interpolation of linear operators[END_REF] for an introduction to symmetric spaces. In the remainder of this section, p and q will denote two nonnegative reals such that p ≤ q.

Let (Ω, m) be any measure space (in particular the following definitions apply to N equipped with the counting measure i.e sequence spaces). As previously mentioned, L 0 (Ω) ⊂ X (Ω) denotes the set of functions whose supports have finite measures, it is naturally equipped with the group norm

f 0 = m(supp f ), f ∈ L 0 (Ω).
The "norm" of a linear operator T : L 0 (Ω) → L 0 (Ω), is defined as follows:

T L0→L0 = sup f ∈L0 m(supp(T f )) m(supp(f ))
.

Definition 1.3. A linear space E ⊂ X (Ω) becomes a quasi-Banach function space when equipped with a complete quasi-norm . E such that

• if f ∈ E and g ∈ X (Ω) are such that |g| ≤ |f | , then g ∈ E and g E ≤ f E . Definition 1.4. A quasi-Banach function space E ⊂ X (Ω) is called symmetric if • if f ∈ E and g ∈ X (Ω) are such that µ(f ) = µ(g) then g ∈ E and g E = f E ;
Definition 1.5 (Bounded operator on a couple of quasi-Banach function spaces).

Let X and Y be quasi-Banach function spaces. We say that a linear operator T is bounded on (X, Y ) if T is defined from X + Y to X + Y and restricts to a bounded operator from X to X and from Y to Y . Set:

T (X,Y )→(X,Y ) = max T X→X , T Y →Y .
Les us now recall the precise abstract definition of an interpolation space (see [START_REF] Bergh | Interpolation spaces. An introduction[END_REF], [START_REF] Kreȋn | Interpolation of linear operators[END_REF]). Definition 1.6 (Interpolation space between quasi-Banach function spaces). Let X, Y and Z be quasi-Banach function spaces on Ω. We say that Z is an interpolation space for the couple (X, Y ) if X ∩ Y ⊂ Z ⊂ X + Y and any bounded operator on (X, Y ) restricts to a bounded operator on Z. Denote by Int(X, Y ) the set of interpolation spaces for the couple (X, Y ).

Theorem 1.7. Let V be a separable topological linear space and let A, B, C ⊂ V be quasi-Banach spaces. If C is an interpolation space for the couple (A, B), then there exists a constant c(A, B, C) > 0 such that for any bounded operator T on (A, B),

T C→C ≤ c(A, B, C) • max{ T A→A , T B→B }.
Proof. In [18, Lemma I.4.3], the assertion is proved for Banach spaces. The argument for quasi-Banach spaces is identical (because it relies on the closed graph theorem, which holds for F -spaces, hence, for quasi-Banach spaces).

If, in particular, 0 < p < q < ∞ and if E is an interpolation space for the couple (L p (Ω), L q (Ω)), then there exists a constant c > 0 called interpolation constant of E for (L p (Ω), L q (Ω)) such that for any bounded operator T on (L p (Ω), L q (Ω)),

T E→E ≤ c T (Lp(Ω),Lq(Ω))→(Lp(Ω),Lq(Ω)) .
Finally, let us recall the definition of the K-functional associated to a couple of quasi-Banach spaces. For any t > 0, and f ∈ X + Y , let

K t (f, X, Y ) := inf g+h=f g X + t h Y .
In addition to that, we need to define an interpolation space between L 0 and a quasi-Banach space. Definition 1.8 (Bounded operator on a couple (L 0 (Ω), Y ) for a quasi-Banach function space Y ). Let Y be a quasi-Banach function space. We say that a linear operator T is bounded on (L 0 (Ω), Y ) if T is defined from L 0 (Ω) + Y to L 0 (Ω) + Y and restricts to a bounded operator from L 0 (Ω) to L 0 (Ω) and from Y to Y. Definition 1.9 (Interpolation space for a couple (L 0 (Ω), Y ) for a quasi-Banach function space Y ). Let Y and Z be quasi-Banach function spaces on Ω. We say that Z is an interpolation space for the couple

(L 0 (Ω), Y ) if L 0 (Ω) ∩ Y ⊂ Z ⊂ L 0 (Ω) +
Y and any bounded operator on (L 0 (Ω), Y ) restricts to a bounded operator on Z. Denote by Int(L 0 (Ω), Y ) the set of interpolation spaces for the couple (L 0 (Ω), Y ).

Symmetry of interpolation spaces. In this subsection, we show that a quasi-Banach interpolation space for a couple of symmetric spaces can always be remormed into a symmetric space. Note that similar results can be found in the literature, see for example [START_REF] Kreȋn | Interpolation of linear operators[END_REF]Theorem 2.1].

As usual, we will use the term measure preserving for a map ω between measure spaces (Ω 1 , A 1 , m 1 ) and (Ω 2 , A 2 , m 2 ) verifying,

∀A ∈ A 1 , ω -1 (A) ∈ A 2 and m 2 (ω -1 (A)) = m 1 (A). Lemma 1.10. Assume that Ω is (0, 1), (0, ∞) or N. Let 0 ≤ f, g ∈ L 0 (Ω) + L ∞ (Ω)
and let ε > 0. Assume that µ(f ) = µ(g). There exists a measure preserving map

ω : supp(g) → supp(f ) such that (1 + ε)(f • ω) ≥ µ(f ). Proof. Case 1: Suppose first that µ(∞, f ) = µ(∞, g) = 0.
Define, for any n ∈ Z,

F n = t : (1 + ε) n < f (t) ≤ (1 + ε) n+1 , G n = t : (1 + ε) n < g ≤ (1 + ε) n+1 .
By assumption, m(F n ) = m(G n ) for every n ∈ Z. Let ω n : G n → F n be an arbitrary measure preserving bijection. Define measure preserving map ω : supp(g) → supp(f ) by concatenating

ω n : G n → F n , n ∈ Z. For every t ∈ G n , we have f (ω(t)) ≥ (1 + ε) n , g ≤ (1 + ε) n+1 .

Thus,

(1 + ε)f (ω(t)) ≥ g, t ∈ supp(g). This completes the proof of Step 1.

Case 2. Let δ such that (1 + δ) 2 = (1 + ε). Let a = µ(∞, f ) = µ(∞, g) > 0. Define for any n ≥ 1

F n = t : a(1+δ) n < f (t) ≤ a(1+δ) n+1 , G n = t : a(1+δ) n < g ≤ a(1+δ) n+1 ,
and

F 0 = t : (1 + δ) -1 a ≤ f (t) ≤ (1 + δ)a , G 0 = t : 0 < g ≤ (1 + δ)a .

By asumption for any

n ≥ 1, m(G n ) = m(F n ) and m(G 0 ) = m(F 0 ) = ∞. For any n ≥ 0, choose a measure preserving bijection ω n from G n to F n .
Define the measure preserving map ω : supp(g) → supp(f ) by concatenating the ω n 's. For any n ≥ 0 and any t ∈ G n ,

f (ω(t)) ≥ a(1 + δ) n-1 , g ≤ a(1 + δ) n+1 . Thus, (1 + δ) 2 f (ω(t)) = (1 + ε)f (ω(t)) ≥ g, t ∈ supp(g). Lemma 1.11. Assume that Ω is (0, 1), (0, ∞) or N. Let E, A, B ⊂ (L 0 + L ∞ )(Ω)
be quasi-Banach function spaces. Assume that A and B are symmetric and that E is an interpolation space for the couple (A, B). Then E admits an equivalent symmetric quasi-norm.

Proof. Let f ∈ E and g ∈ L 0 + L ∞ . Assume that µ(g) ≤ µ(f ). By Lemma 1.10, there exists a map ω : supp(g) → supp(f ) such that for any t ∈ supp(g),

2 |f • ω(t)| ≥ |g(t)| .
Define, for any h ∈ X (Ω),

T (h) :=    g f • ω h • ω on supp(g) 0 elsewhere.
Since ω is measure preserving, T is bounded on A and B of norm less than 2. Let c E be the interpolation constant of E for the couple (A, B) (as in Theorem 1.7).

We know that T f = g ∈ E and

(1.1)

g ≤ 2c E f . Define, for any f ∈ E f E ′ = inf µ(g)≥µ(f ) g E . By (1.1), f E ′ ≤ f E ≤ 2c E f E ′ and (E, • E ′ ) is a symmetric space.
Remark 1.12. It is not difficult to see that if the underlying measure space Ω contains both a continuous part and atoms, Lemma 1.11 is no longer true for A = L p (Ω), B = L q (Ω) and p < 1. However, one can observe that if A and B are fully symmetric (i.e. interpolation spaces between L 1 (Ω) and L ∞ (Ω)), Lemma 1.11 remains valid for any Ω. This is reminiscent of the conditions required in [25, Section 4].

2. Interpolation for the couple (L 0 , L q )

In this section, Ω = (0, ∞) (for brevity, we omit Ω in the notations). We investigate some basic properties of the interpolation couple (L 0 , L q ). First, we provide a statement analogous to Theorem 1.7 and applicable to L 0 .

Since the closed graph theorem does not apply to L 0 (it is not an F -space), our proof uses a concrete constructions that relies on the structure of the underlying measure space.

For any f ∈ X , denote by M f the multiplication operator g → f • g.

Theorem 2.1. Let E be a quasi-Banach function space and q ∈ (0, ∞]. Assume that E is an interpolation space for the couple (L 0 , L q ). Then, there exists a constant c such that for any contraction T on (L 0 , L q ), T E→E ≤ c.

Proof. Let (A n ) n≥1 be a partition of (0, ∞) such that m(A n ) = ∞ for every n ≥ 1. Let γ n : A n → A c n be a measure preserving bijective transform. Set (U n x)(t) = x(γ n (t)), t ∈ A n 0, t ∈ A c n , (V n x)(t) = x(γ -1 n (t)), t ∈ A c n 0, t ∈ A n .
Obviously, U n and V n are bounded operators on the couple (L 0 , L q ). By assumption,

U n , V n : E → E are bounded mappings. Note that V n U n = M χ A c n , n ≥ 1.
Let us argue by contradiction. For any n ≥ 1, choose an operator T n which is a contraction on (L 0 , L q ) and such that (2.1)

T n E→E ≥ 4 n • max{ U n E→E , V n E→E , 1} 2 .
It is immediate that

T n = M χA n T n M χA n + M χ A c n T n M χA n + M χA n T n M χ A c n + M χ A c n T n M χ A c n = = T 1,n + V n T 2,n + T 3,n U n + V n T 4,n U n ,
where

T 1,n = M χA n T n M χA n , T 2,n = U n T n M χA n , T 3,n = M χA n T n V n , T 4,n = U n T n V n .
By quasi-triangle inequality, we have

T n E→E ≤ C 2 E • 4 k=1 T k,n E→E • max{ U n E→E , V n E→E .1} 2 .
Let k n ∈ {1, 2, 3, 4} be such that

T kn,n E→E = max 1≤k≤4 T k,n E→E .
We, therefore, have

(2.2) T n E→E ≤ 4C 2 E T kn,n E→E • max{ U n E→E , V n E→E .1} 2 . Set S n = T kn,n . Note that S n L0→L0 ≤ 1, S n Lq→Lq ≤ 1. A combination of (2.1) and (2.2) yields S n E→E ≥ 4 n-1 C -2 E , n ≥ 1. Note that S n = M χA n S n M χA n . Set S = n≥1 S n .
Since the S n 's are in direct sum, S L0→L0 = sup n≥1 S n L0→L0 ≤ 1 and

S Lq→Lq = sup n≥1 S n Lq→Lq ≤ 1.
Moreover, E is an interpolation space for the couple (L 0 , L q ), it follows that S :

E → E is bounded. For any n ≥ 1, choose f n ∈ E such that f n E ≤ 1 and S n f n E ≥ 4 n-2 C -2 E . Recall that S n = S n M χA n .
Hence, we may assume without loss of generality that

f n is supported on A n . Thus, S(f n ) = S n (f n ) and S(f n ) E = S n (f n ) E ≥ 4 n-2 C -2
E . This contradicts the boundedness of S.

Remark 2.2. Theorem 2.1 above remains true for other underlying measure spaces:

• for sequence spaces. Indeed, in the proof of Theorem 2.1, we only use properties of the underlying measure space in the first sentence, namely when we consider a partition of (0, ∞) into countably many sets, each of them isomorphic to (0, ∞). Since a partition satisfying the same property exists for Z + , Theorem 2.1 remains true for interpolation spaces between ℓ 0 and ℓ q . • for (0, 1). The same general idea applies in this case but some modification have to be made because the maps γ n introduced in the proof cannot be assumed to be measure preserving. The details are left to the reader.

Lemma 2.3. Let E, Y ⊂ L 0 + L ∞ be quasi-Banach function spaces. Assume that
Y is symmetric and that E is an interpolation space for the couple (L 0 , Y ). Then E admits an equivalent symmetric quasi-norm.

Proof. The argument follows that in Lemma 1.11 mutatis mutandi.

The following assertion is a special case Theorem 1.1 and an important ingredient in the proof of the latter theorem.

Theorem 2.4. Let E be a quasi-Banach function space and q ∈ (0, ∞). Assume that L 0 ∩ L q ⊂ E ⊂ L 0 + L q and that for any f ∈ E and g ∈ L 0 + L q ,

|g| q ≺≺ tl |f | q ⇒ g ∈ E.
Then E is an interpolation space for the couple (L 0 , L q ). The rest of this section is occupied with the proof of Theorem 2.4. Let C E be the concavity modulus of E, that is, (the minimal) constant such that

x 1 + x 2 E ≤ C E ( x 1 E + x 2 E ), x 1 , x 2 ∈ E.
It allows to write a triangle inequality with infinitely many summands. This inequality should be understood in the usual sense: if scalar-valued series in the right hand side converges, then the series on the left hand side converges in E and the inequality holds.

Lemma 2.5. Let E be a quasi-Banach space. We have

n≥1 x n E ≤ n≥1 C n E x n E , (x n ) n≥1 ⊂ E.
Proof. The proof is standard and is, therefore, omitted.

Lemma 2.6. Let E ⊂ L 0 + L q be a symmetric quasi-Banach function space. If

f E ≤ 1, then µ(f )χ (1,∞) q ≤ c E . Proof. Assume the contrary. Choose (f n ) n≥1 ∈ E such that f n E ≤ 1 and µ(f n )χ (1,∞) q ≥ n. Set h n = k≥0 µ(k + 1, f n )χ (k,k+1) , n ≥ 1.
It is immediate that h n E ≤ 1 and h n q ≥ n.

Let C E be the concavity modulus of the space E and let m ≥ 4C E be a natural number. Set

h = n≥1 (2C E ) -n h m n .
By Lemma 2.5, we have

h E ≤ n≥1 C n E (2C E ) -n h m n E ≤ n≥1 2 -n = 1.
Since h ∈ E, it follows that h ∈ L 0 + L q . Since h = µ(h) is constant on the interval (0, 1), it follows that h ∈ L q . Thus,

h q ≥ (2C E ) -n h m n q ≥ (2C E ) -n • m n ≥ 2 n , n ≥ 1.
It follows that h / ∈ L q . This contradiction completes the proof.

Lemma 2.7. Let E be as in Theorem 2.4. If T is a contraction on (L 0 , L q ), then T : E → E.

Proof. Take f ∈ E. Fix t > 0 and let A = A 1 A 2 , where

A 1 = {|f | > µ(t, f )} and A 2 ⊂ {|f | = µ(t, f )} is such that m(A 2 ) = t -m(A 1 ).
Obviously, m(A) = t and

µ(s, f χ A c ) = µ(s + t, f ), s > 0. Now, we write T f = T (f χ A ) + T (f χ A c ). Since T : L q → L q is a contraction, it follows that T (f χ A c ) q q ≤ f χ A c q q = ∞ t µ q (s + t, f )ds = ∞ t µ q (s, f )ds, T (f χ A ) 0 ≤ f χ A 0 ≤ t. Now, ∞ t µ q (s, T f )ds = inf{ T f -h q q : h 0 ≤ t}. Taking h = T (f χ A ), we write ∞ t µ q (s, T f )ds ≤ T f -T (f χ A ) q q = T (f χ A c ) q q ≤ ∞ t µ q (s, f )ds.
Since t > 0 is arbitrary, it follows that

|T f | q ≺≺ tl |f | q .
By assumption, T f ∈ E. Since f ∈ E is arbitrary, it follows that T : E → E.

Our next lemma demonstrates the boundedness of the mapping T : E → E established in Lemma 2.7.

Lemma 2.8. Let E be as in Theorem 2.4. If T is a contraction on (L 0 , L q ), then the mapping T : E → E is bounded.

Proof. Using Lemma 2.3, we may assume without loss of generality that E is symmetric.

Let us show that the graph of T :

E → E is closed. Assume the contrary. Choose a sequence (f n ) n≥1 ⊂ E such that f n → f ∈ E and T f n → g = T f in E. Replacing (f n ) n≥1 with (f n -f ) n≥1 if necessary, we may assume without loss of generality that f = 0. That is, f n → 0 in E and T f n → g = 0 in E.
Passing to a subsequence, if needed, we may assume without loss of generality that

f n E χ (0, 1 n ) E → 0, n → ∞.
Clearly,

f n E = µ(f n ) E ≥ µ(f n )χ (0, 1 n ) E ≥ µ( 1 n , f n ) χ (0, 1 n ) E . It follows that µ( 1 n , f n ) → 0 as n → ∞. Set f 1n = f n χ {|fn|>µ( 1 n ,fn)} , n ≥ 1, f 2n = f n χ {µ(1,fn)<|fn|≤µ( 1 n ,fn)} , n ≥ 1, f 3n = f n χ {|fn|≤µ(1,fn)} , n ≥ 1.
Since T : L 0 → L 0 is a contraction, it follows that

T (f 1n ) 0 ≤ f 1n 0 ≤ 1 n , n ≥ 1.
In particular, T (f 1n ) → 0 in measure. Since T : L q → L q is a contraction, it follows that

T (f 2n ) q ≤ f 2n q .
Since f 2n is supported on a set of measure 1, it follows that

f 2n q ≤ f 2n ∞ = µ( 1 n , f n ) → 0, n → ∞.
In particular, T (f 2n ) → 0 in measure. Since T : L q → L q is a contraction, it follows that

T (f 3n ) q ≤ f 3n q ≤ min{µ(f n ), µ(1, f n )} q .
By Lemma 2.6, we have

min{µ(f n ), µ(1, f n )} q ≤ c q,E f n E → 0, n → ∞.
In particular, T (f 3n ) → 0 in measure.

Combining the last 3 paragraphs, we conclude that T f n → 0 in measure. However, T f n → g in E and, therefore, in measure. It follows that g = 0, which contradicts our choice of the sequence (f n ) n≥1 . This contradiction shows that the graph of T : E → E is closed. By the closed graph theorem, T : E → E is bounded.

Proof of Theorem 2.4. Let T be a bounded operator on (L 0 , L q ). Fix n ∈ N such that

T L0→L0 ≤ n, T Lq→Lq ≤ n.
Let σ 1 n be a dilation operator defined by the usual formula (σ

1 n f )(t) = f (nt), n ≥ 0. Set S = 1 n σ 1 n • T so that T = nσ n • S. Obviously, S L0→L0 ≤ 1, S Lq→Lq ≤ 1.
By Lemma 2.8, S : E → E is a bounded mapping. Since E is a symmetric quasi-Banach function space, it follows that σ n : E → E is bounded; this implies T : E → E.

Theorem 2.4 applies in particular to L p -spaces, p ≤ q. We decided to add more precise statement and to provide a direct proof of the latter.

Corollary 2.9. Let p, q ∈ (0, ∞) such that p < q. Then, L p is an interpolation space for the couple (L 0 , L q ). More precisely, if T is a contraction on (L 0 , L q ), then T is a contraction on L p .

Proof. Let us first consider characteristic functions. Let E be a set with finite measure. Since T is a contraction on L 0 , the measure of the support of T (χ E ) is less than m(E). So by Hölder's inequality, setting r = (p -1 -q -1 ) -1 , we have:

T (χ E ) p ≤ T (χ E ) q • m(E) 1/r ≤ χ E q • m(E) 1 r = χ E p .
First, consider the case p ≤ 1. Let f ∈ L p be a step function, i.e.

f = i∈N a i χ Ei ,
where a i ∈ C and the sets E i are disjoint sets with finite measure. By the ptriangular inequality:

T f p p ≤ i∈N |a i | p T (χ Ei ) p p ≤ i∈N |a i | p χ Ei p p = f p p .
Since T : L p → L p is bounded by Theorem 2.4 and since step functions are dense in L p , it follows that T : L p → L p is a contraction (for p ≤ 1). Now consider the case p > 1. Since p < q, it follows that q > 1. By the preceding paragraph, T : L 1 → L 1 is a contraction. By complex interpolation, T : L p → L p is also contraction.

3. Construction of contractions on (L 0 ,L q ) and (L p ,L ∞ ) Let p, q ∈ (0, ∞). In this section, we are interested in the following question. Given functions f and g in L 0 +L q (resp. L p +L ∞ ), does there exist a bi-contraction T on (L 0 ,L q ) (resp. (L p ,L ∞ )) such that T (f ) = g? We show that such an operator exists provided that |g| q ≺≺ tl |f | q (resp. |g| p ≺≺ hd |f | p . This directly implies a necessary condition for a symmetric space to be an interpolation space for the couple (L 0 , L q ) (resp. (L p , L ∞ )) which will be exploited in the next section.

Our method of proof is very direct. We construct the bi-contraction T as direct sums of very simple operators. This is made possible by three partition lemmas that enable us to understand the orders ≺≺ tl and ≺≺ hd as direct sums of simple situations.

3.1. Partition lemmas. We state our first lemma without proof since it essentially repeats that of Proposition 19 in [START_REF] Sukochev | Orbits in symmetric spaces[END_REF].

Lemma 3.1. Let f, g ∈ L 1 be positive decreasing step functions. Assume that g ≺ hd f. There exists a partition {I k , J k } k≥0 of (0, ∞) such that (i) for every k ≥ 0, I k and J k are disjoint intervals of finite length;

(ii) (I k ∪ J k ) ∩ (I l ∪ J l ) = ∅ for k = l; (iii) f and g are constant on I k and on J k ; (iv) g| I k ∪J k ≺ hd f | I k ∪J k for every k ≥ 0;
Remark 3.2. Note that in [START_REF] Sukochev | Orbits in symmetric spaces[END_REF], Proposition 19 is proved for couples of functions f, g of the form

f = i∈Z f i 1 [ai,ai+1) , g = i∈Z g i 1 [ai,ai+1] ,
with (a i ) an increasing sequence in (0, ∞) and f i , g i ∈ C for any i ∈ N. However, the proof applies with very little modification to couples of decreasing positive step functions as in Lemma 3.1 above.

Scholium 3.3. Let f, g ∈ X be positive decreasing functions. Let ∆ ⊂ (0, ∞) be an arbitrary measurable set.

(i) if f, g ∈ L 1 + L ∞ are such that [0,t]∩∆ g ≤ [0,t]∩∆ f, t > 0, then gχ ∆ ≺≺ hd f χ ∆ . (ii) if f, g ∈ L 0 + L 1 are such that (t,∞)∩∆ g ≤ (t,∞)∩∆ f, t > 0, then gχ ∆ ≺≺ tl f χ ∆ .
Our second partition lemma shows that the order ≺≺ hd can be reduced for our purpose to a direct sum of ≺ hd and ≤. Lemma 3.4. Let f, g ∈ L 1 + L ∞ be such that f = µ(f ), g = µ(g) and g ≺≺ hd f. There exists a collection {∆ k } k≥0 of pairwise disjoint sets such that

(i) g| ∆ k ≺ hd f | ∆ k for every k ≥ 0; (ii) g ≤ f on the complement of k≥0 ∆ k ;
Proof. Consider the set {g > f }. Since g -f is right-continuous, it follows that, for every t ∈ {g > f }, there exists an ǫ > 0 (which depends on t) such that [t, t + ǫ) ⊂ {g > f }. Hence, connected components of the set {g > f } are intervals (closed or not) not reduced to points. Let us enumerate these intervals as (a k , b k ), k ≥ 0 (it does not really matter for us if boundary points of these intervals belong to {g > f }). We have

t 0 (f -g) + - t 0 (f -g) -= t 0 (f -g) ≥ 0.
Let

H(t) = inf u : u 0 (f -g) + = t 0 (f -g) -.
Obviously, H is a monotone function, H(t) ≤ t for all t > 0 and

H(t) 0 (f -g) + = t 0 (f -g) -. Set ∆ k = (a k , b k ) ∪ (H(a k ), H(b k )) ∩ {f ≥ g} .
Note that

a k 0 (f -g) + = b k 0 (f -g) + ≥ b k 0 (f -g) - and, therefore, H(b k ) ≤ a k . We claim that ∆ k ∩ ∆ l = ∅ for k = l. Indeed, let a k < b k ≤ a l < b l . We have H(a k ) ≤ H(b k ) ≤ H(a l ) ≤ H(b l ). Thus, (H(a k ), H(b k )) ∩ (H(a l ), H(b l )) = ∅. We now have ∆ k ∩ ∆ l = (∆ k ∩ {f < g}) ∩ (∆ l ∩ {f < g}) (∆ k ∩ {f ≥ g}) ∩ (∆ l ∩ {f ≥ g}) .
Obviously,

(∆ k ∩ {f < g}) ∩ (∆ l ∩ {f < g}) = (a k , b k ) ∩ (a l , b l ) = ∅, (∆ k ∩ {f ≥ g}) ∩ (∆ l ∩ {f ≥ g}) = (H(a k ), H(b k )) ∩ (H(a l ), H(b l )) ∩ {f ≥ g} = ∅.
This proves the claim. We now claim that

[0,t]∩∆ k (f -g) ≥ 0. If t ≤ a k , then, taking into account that H(b k ) ≤ a k , we infer that [0, t] ∩ ∆ k ⊂ {f ≥ g} and the claim follows immediately. If t ∈ (a k , b k ), then [0,t]∩∆ k (f -g) = (H(a k ),H(b k )) (f -g) + - t a k (f -g) -≥ ≥ (H(a k ),H(b k )) (f -g) + - b k a k (f -g) -= 0.
This proves the claim. It follows from the claim and Scholium 3.

3 that gχ ∆ k ≺≺ hd f χ ∆ k . Since ∆ k g = ∆ k f, the first assertion follows. By construction, (a k , b k ) ⊂ ∆ k . Thus, {g > f } = k≥0 (a k , b k ) ⊂ k≥0 ∆ k .
The second assertion is now obvious.

Finally, the third partition lemma deals with describing the order ≺≺ tl in terms of ≺ tl and ≤. Lemma 3.5. Let f, g ∈ L 0 + L 1 be such that f = µ(f ), g = µ(g) and g ≺≺ tl f. There exists a collection {∆ k } k≥0 of pairwise disjoint sets such that

(i) f | ∆ k ≺ hd g| ∆ k for every k ≥ 0; (ii) g ≤ f on the complement of k≥0 ∆ k ;
Proof. Consider the set {g > f }. Similarly to the previous proof, connected components of the set {g > f } are intervals (closed or not) not reduced to points. Let us enumerate these intervals as (a k , b k ), k ≥ 0. We have

∞ t (f -g) + - ∞ t (f -g) -= ∞ t (f -g) ≥ 0.
Let

H(t) = sup u : ∞ u (f -g) + = ∞ t (f -g) -.
Obviously, H is a monotone function, H(t) ≥ t for all t > 0 and

∞ H(t) (f -g) + = ∞ t (f -g) -. Set ∆ k = (a k , b k ) ∪ (H(a k ), H(b k )) ∩ {g ≤ f } . Note that ∞ b k (f -g) + = ∞ a k (f -g) + ≥ ∞ a k (f -g) - and, therefore, H(a k ) ≥ b k . We claim that ∆ k ∩ ∆ l = ∅ for k = l. Indeed, let a k < b k ≤ a l < b l . We have H(a k ) ≤ H(b k ) ≤ H(a l ) ≤ H(b l ). Thus, (H(a k ), H(b k )) ∩ (H(a l ), H(b l )) = ∅. We now have ∆ k ∩ ∆ l = (∆ k ∩ {f < g}) ∩ (∆ l ∩ {f < g}) (∆ k ∩ {f ≥ g}) ∩ (∆ l ∩ {f ≥ g}) .
Obviously,

(∆ k ∩ {f < g}) ∩ (∆ l ∩ {f < g}) = (a k , b k ) ∩ (a l , b l ) = ∅, (∆ k ∩ {f ≥ g}) ∩ (∆ l ∩ {f ≥ g}) = (H(a k ), H(b k )) ∩ (H(a l ), H(b l )) ∩ {f ≥ g} = ∅.
This proves the claim. We now claim that

(t,∞)∩∆ k (f -g) ≥ 0. If t ≥ b k , then, taking into account that H(a k ) ≥ b k , we infer that (t, ∞) ∩ ∆ k ⊂ {f ≥ g} and the claim follows immediately. If t ∈ (a k , b k ), then (t,∞)∩∆ k (f -g) = (H(a k ),H(b k )) (f -g) + - b k t (f -g) -≥ ≥ (H(a k ),H(b k )) (f -g) + - b k a k (f -g) -= 0.
This proves the claim. It follows from the claim and Scholium 3.

3 that gχ ∆ k ≺≺ tl f χ ∆ k . Since ∆ k g = ∆ k f,
it follows that gχ ∆ k ≺ tl f χ ∆ k , which immediately implies the first assertion. By construction, (a k , b k ) ⊂ ∆ k . Thus,

{g > f } = k≥0 (a k , b k ) ⊂ k≥0 ∆ k .
The second assertion is now obvious.

Construction of operators.

We repeat the same structure as in the previous subsection, proving four lemmas, each one dealing with a certain order : ≺ hd , ≺ tl , ≺≺ hd and finally ≺≺ tl .

Lemma 3.6. Let p ∈ (0, ∞). Let f, g ∈ L p (0, ∞), assume that |g| p ≺ hd |f | p , f = µ(f ) and g = µ(g).
There exists a linear operator T : X (0, ∞) → X (0, ∞) such that g = T (f ) and

T Lp→Lp ≤ 2 • 3 1 p , T L∞→L∞ ≤ 2 • 2 1 p .
Proof.

Step 1: First, let us assume that f and g are step functions Apply Lemma 3.1 to the functions f p and g p and let I k and J k be as in Lemma 3.1. Without loss of generality, the interval I k is located to the left of the interval J k .

For every k ≥ 0, define the mapping S k : X (I k ∪ J k ) → X (I k ∪ J k ) as below. The construction of this mapping will depend on whether

f p | J k ≤ 1 2 g p | J k or f p | J k > 1 2 g p | J k . If f p | J k ≤ 1 2 g p | J k , then g p | J k • m(J k ) ≤ g p | I k • m(I k ) + g p | J k • m(J k ) = = f p | I k • m(I k ) + f p | J k • m(J k ) ≤ f p | I k • m(I k ) + 1 2 g p | J k • m(J k ).
Therefore,

g p | J k • m(J k ) ≤ 2f p | I k • m(I k ).
Let l k be a linear bijection from J k to I k . We set

S k x = g| I k f | I k • xχ I k + g| J k f | I k • (x • l k )χ J k .
Clearly, S k is a contraction in the uniform norm. Let x ∈ L p . We have

S k x p p ≤ g p | I k f p | I k • xχ I k p p + g p | J k f p | I k • (x • l k )χ J k p p ≤ ≤ g p | I k f p | I k • x p p + g p | J k f p | I k • m(J k ) m(I k ) • x p p ≤ 3 x p p .
Also, we have

S k x ∞ ≤ x ∞ . If f p | J k > 1 2 g p | J k , then we set S k = M gf -1 . Clearly, S k x ∞ ≤ 2 1 p x ∞ and S k x p ≤ 2 1 p x p .
We define S : X → X by:

S = k≥0 S k .
Remark that for any r ∈ [0, ∞], S r→r = sup k≥0 S k Lr→Lr . Hence,

S Lp→Lp ≤ 3 1 p , S L∞→L∞ ≤ 2 1 p .
It remains only to note that Sf = g.

Step 2: Now, only assume that f and g are positive and non-increasing. Define for any n ∈ Z,

a n = sup t ∈ (0, ∞) : f (t) ≥ 2 n 2 and b n = sup t ∈ (0, ∞) : g(t) ≥ 2 n 2 .
Let A be the σ-algebra generated by the intervals (a n , a n+1 ) and (b n , b n+1 ). Define

f p 0 = E[f p | A] and g p 0 = E[g p | A].
Note that f 0 and g 0 are step functions such that

g p 0 ≺ hd f p 0 , 2 -1 2 f ≤ f 0 ≤ 2 1 2 f and 2 -1 2 g ≤ g 0 ≤ 2 1 2 g.
Apply Step 1 to f 0 and g 0 to obtain an operator S and set

T = M f f -1 0 • S • M g0g -1 .
Clearly, T f = g and

T Lp→Lp ≤ 2 S Lp→Lp ≤ 2 • 2 1 p and T L∞→L∞ ≤ 2 S L∞→L∞ ≤ 2 2 • 2 1 p .
Lemma 3.7. Let f, g ∈ L q (0, ∞) be positive non-increasing functions such that g q ≺ tl f q . Let d > 1. There exists a linear operator T : X (0, ∞) → X (0, ∞) such that g = T (f ) and

T L0→L0 ≤ 4, T Lq→Lq ≤ 2 • 3 1 q .
Proof. Following step 2 of Lemma 3.6, we are reduced to dealing with step functions. Apply Lemma 3.1 to the functions g q and f q and let (I k ) k≥1 and (J k ) k≥1 be as in Lemma 3.1. Without loss of generality, the intervals I k is located to the left of the intervals J k .

Let k ≥ 1. Define the mappings S k :

X (I k ∪ J k ) → X (I k ∪ J k ) as below. Note that since f q | I k ∪J k ≺ hd g q | I k ∪J k , g| I k ≥ f | I k ≥ f | J k ≥ g| J k .
The construction of S k will depend on whether f χ

I k q ≤ f χ J k q or f χ I k q > f χ J k q . If f χ I k q ≤ f χ J k q , then m(I k ) ≤ m(J k ) and
g q 0 | I k • m(I k ) ≤ 2f q | J k • m(J k ). Let l k : I k → J k be a linear bijection. We set S k x = g| I k f | J k (x • l k )χ I k + g f xχ J k . Let x ∈ L q . S k x 0 ≤ xχ J k 0 + (x • l k )χ I k 0 ≤ 1 + m(I k ) m(J k ) x 0 .
Thus, S k L0→L0 ≤ 2. We have

S k x q q = g q | I k f q | J k (x • l k )χ I k q q + g q | J k f q | J k xχ J k q q ≤ ≤ g q | I k f q | J k • m(I k ) m(J k ) • x q q + g q | J k f q | J k x q q ≤ 3 x q q . If f χ I k q > f χ J k q , then g q | I k • m(I k ) ≤ 2f q | I k • m(I k ) and, therefore, g 0 ≤ 2 1 q f 0 . We set S k = M gf -1 . Obviously, S k L0→L0 ≤ 1 and S k Lq→Lq ≤ 2 1 q . We set S = k≥0 S k .
Since S : X (0, ∞) → X (0, ∞) is defined as a direct sum:

S L0→L0 = sup k≥1 S k L0→L0 ≤ 2, S Lq→Lq = sup k≥1 S k Lq→Lq ≤ 3 1 q .
It remains only to note that Sf = g.

Lemma 3.8. Let f, g ∈ (L p + L ∞ )(0, ∞) be such that |g| p ≺≺ hd |f | p , f = µ(f ) and g = µ(g).
There exists a linear operator T : X → X such that g = T (f ) and

T Lp→Lp ≤ 2 • 3 1 p , T L∞→L∞ ≤ 2 • 2 1 p .
Proof. Let (∆ k ) k≥0 be as in Lemma 3.4 and let ∆ ∞ be the complement of k≥0 ∆ k . By Lemma 3.6, there exists

T k : X (∆ k ) → X (∆ k ) such that T k (f ) = g on ∆ k and such that T k Lp(X k )→Lp(X k ) ≤ 2 • 9 1 p , T k L∞(X k )→L∞(X k ) ≤ 2 • 4 1 p . Set T ∞ = M g f on X (∆ ∞ ). We now set T = T ∞ k≥0 T k .
Obviously, T f = g on (0, ∞) and

T Lp→Lp ≤ 2 • 9 1 p , T L∞→L∞ ≤ 2 • 4 1 p .
Lemma 3.9. Let f, g ∈ (L 0 + L q )(0, ∞) be such that |g| q ≺≺ tl |f | q , f = µ(f ) and g = µ(g). There exists a linear operator T : X → X such that g = T (f ) and

T L0→L0 ≤ 4, T Lq→Lq ≤ 2 • 3 1 q .
Proof. Without loss of generality, g = µ(g) and f = µ(f ). Let (∆ k ) k≥0 be as in Lemma 3.5 and let ∆ ∞ be the complement of k≥0 ∆ k . By Lemma 3.7, there exists

T k : X (∆ k ) → X (∆ k ) such that T k (f ) = g on ∆ k and such that T k L0(X k )→L0(X k ) ≤ 8, T k Lq(X k )→Lq(X k ) ≤ 2 • 9 1 q .
Set T ∞ = M g f on X (∆ ∞ ). We now set

T = T ∞ k≥0 T k .
Obviously, T f = g on (0, ∞) and

T L0→L0 ≤ 4, T Lq→Lq ≤ 2 • 3 1 q .
4. Interpolation spaces for the couple (L p , L q )

In this section, we obtain characterizations of interpolation spaces for the couple (L p , L q ), in terms of the majorization notions studied earlier. The necessity of the condition we consider is a direct consequence of the constructions explained in the previous section. The fact that it is sufficient is shown by linking majorization to the K-functional. Theorem 4.1. Let 0 ≤ p < q ≤ ∞. Let E be a quasi-Banach function space such that E ∈ Int(L p , L q ). There exist c p,E and c q,E in R >0 such that:

(i) suppose p = 0: for any f ∈ E and g ∈ L p + L ∞ if |g| p ≺≺ hd |f | p , then g ∈ E and g E ≤ c p,E f E ; (ii) suppose q = ∞: for any f ∈ E and g ∈ L 0 + L q if |g| q ≺≺ tl |f | q , then g ∈ E and g E ≤ c q,E f E .
Proof. By Lemma 1.11 (for p > 0) or Lemma 2.3 (for p = 0), we may assume without loss of generality that E is a symmetric function space. Assume that p = 0. Let f ∈ E and let g ∈ L p + L ∞ be such that |g| p ≺≺ hd |f | p . Since E is symmetric, we may assume without loss of generality that f = µ(f ) and g = µ(g). By Lemma 3.8, there exists an operator T such that T (f ) = g and

T (Lp,L∞)→(Lp,L∞) ≤ 2 • 3 1 p .
Recall that L q is an interpolation space for the couple (L p , L ∞ ) (one can take, for example, real or complex interpolation method). Let c p,q be the interpolation constant of L q for the couple (L p , L ∞ ). We have

T Lq→Lq ≤ c p,q • 2 • 3 1 p .
Let c E be the interpolation constant of E for (L p , L q ) (see Theorem 1.7). Then,

T E→E ≤ c E • max{1, c p,q } • 2 • 3 1 p . Thus, g E ≤ T E→E f E ≤ c E • max{1, c p,q } • 2 • 3 1 p f E .
This proves the first assertion. The proof of the second one follows mutatis mutandi using Lemma 2.9 instead of complex interpolation and (for p = 0) Theorem 2.1 instead of Theorem 1.7. Lemma 4.2. Assume that 0 < p < q < ∞. Let f, g ∈ L p + L q such that f = µ(f ) and g = µ(g). Suppose that at every t > 0, one of the following inequalities holds

t 0 g p ds ≤ t 0 f p ds or ∞ t g q ds ≤ ∞ t f q ds.
Then, there exist g 1 , g 2 ∈ (L p + L q ) + satisfying: g = g 1 + g 2 , g p 1 ≺≺ hd f p and g q 2 ≺≺ tl f q . Proof. Set:

A = t > 0 : t 0 g(s) p ds ≤ t 0 f (s) p ds , B = t > 0 : ∞ t g(s) q ds ≤ ∞ t f (s) q ds . Let u + (t) = inf{s ∈ A : s ≥ t}, u -(t) = sup{s ∈ A : s ≤ t}, v + (t) = inf{s ∈ B : s ≥ t}, v -(t) = sup{s ∈ B : s ≤ t}. Note that, for t / ∈ A, f p χ (u-(t),u+(t)) ≺ hd g p χ (u-(t),u+(t)) and, therefore, g(u + (t) -0) ≤ f (u + (t) -0). Set h 1 (t) = g(u + (t) -0), t > 0. By definition, u + (t) ≥ t for all t > 0. Since g is decreasing, it follows that h 1 ≤ g. Set h 2 = gχ B . Since u + (t) = t for t ∈ A, it follows that h 1 = g on A. Thus, h 1 + h 2 ≥ gχ A + gχ B ≥ g.
We claim that Since h 1 = µ(h 1 ), the claim follows.

We claim that

∞ t µ(s, h 2 ) q ds ≤ ∞ t f (s) q ds.
For t ∈ B, we have

∞ t h 2 (s) q ds ≤ ∞ t g(s) q ds ≤ ∞ t f (s) q ds.
For t / ∈ B, we have

∞ t h 2 (s) q ds = ∞ v+(t) h 2 (s) q ds ≤ ≤ ∞ v+(t) g(s) q ds = ∞ v+(t) f (s) q ds ≤ ∞ t f (s) q ds.
In either case,

∞ t µ(s, h 2 ) q ds ≤ ∞ t h 2 (s) q ds ≤ ∞ t f (s) q ds.
This proves the claim. Setting

g 1 = h 1 h 1 + h 2 g, g 2 = h 2 h 1 + h 2 g,
we complete the proof.

Theorem 4.3. Let 0 ≤ p < q ≤ ∞ with either p = 0 or q = ∞. Let E be a quasi-Banach function space. Assume that there exist c p,E and c q,E in R >0 such that:

(i) if p = 0: for any f ∈ E and g ∈ L 0 + L p , if |g| p ≺≺ hd |f | p , then g ∈ E and g E ≤ c p,E f E ; (ii) if q = ∞: for any f ∈ E and g ∈ L 0 + L q , if |g| q ≺≺ tl |f | q , then g ∈ E and g E ≤ c q,E f E ;
Then E belongs to Int(L p , L q ).

Proof. Assume that p = 0. Let us show that the first condition implies E ⊂ L p + L ∞ . Indeed, assume the contrary and choose f ∈ E such that µ(f )χ (0,1) / ∈ L p . Let

f n = min{µ( 1 n , f ), µ(f )χ (0,1) }, n ≥ 1.
Obviously, f n E ≤ µ(f )χ (0,1) E ≤ f E . On the other hand, f n p p χ (0,1) ≺≺ hd f p n . By the first condition on E, we have f n p χ (0,1) E ≤ c p,E f E . However f n p ↑ µ(f )χ (0,1) p = ∞. This contradiction shows that our initial assumption was incorrect. Thus, E ⊂ L p + L ∞ .

A similar argument shows that the second condition implies E ⊂ L 0 + L q . Thus, a combination of both conditions implies E ⊂ L p + L q .

Let T be a contraction on (L p , L q ) and f ∈ E. To conclude the proof, it suffices to show that T f belongs to E. First, note that

K(t, T f, L p , L q ) ≤ K(t, f, L p , L q ).
Assume that p > 0 and q < ∞. Let α -1 = 1 p -1 q . By Holmstedt formula for the K-functional (see [START_REF] Holmstedt | Interpolation of quasi-normed spaces[END_REF]), there exists a constant c p,q > 0 such that for any t ∈ R >0 :

t α 0 µ(s, T f ) p ds 1 p + t ∞ t α µ(s, T f ) q ds 1 q ≤ ≤ c p,q t α 0 µ(s, f ) p ds 1 p + t ∞ t α µ(s, f ) q ds 1 q .
Hence, for any given t > 0, we either have

t α 0 µ(s, T f ) p ds ≤ t α 0 µ(s, c p,q f ) p ds or ∞ t α µ(s, T f ) q ds ≤ ∞ t α
µ(s, c p,q f ) q ds. By Lemma 4.2, one can write (4.1) µ(T f ) = g 1 + g 2 , g p 1 ≺≺ hd (c p,q µ(f )) p , g 2 ≺≺ tl (c p,q µ(f )) q . By assumption, we have

g 1 E ≤ c p,E f E , g 2 E ≤ c q,E f E .
By triangle inequality, we have

T f E ≤ c p,q,E f E .
Assume now that p > 0 and q = ∞. This case is simpler since by the Holmstedt formula (see [START_REF] Holmstedt | Interpolation of quasi-normed spaces[END_REF]), there exists c p ∈ R >0 such that for any t ∈ R >0 :

t p 0 µ(s, T f ) p ds 1 p ≤ c p t p 0 µ(s, f ) p ds 1 p .
This means that |T f | p ≺≺ hd |c p f | p so by assumption (1), T f belongs to E and

T f E ≤ c p c p,E f E .
The case of p = 0 and q < ∞ is given by Theorem 2.4. Theorem 1.1 claimed in the introduction compiles some results of this section. Finally, using (a), (b), Theorem 4.1 and 4.3, for 0 < p < q < ∞, one obtains (c).

Remark 4.4. In the spirit of Corollary 2.4, we could have used a non-quantitative condition to deal with the case of q = ∞ in Theorem 4.3. Let E be a quasi-Banach function space and p, q ∈ (0, ∞). This means that the two following conditions are equivalent:

(i) for any f ∈ E, g ∈ L p + L ∞ , |g| p ≺≺ hd |f | p ⇒ g ∈ E.
(ii) there exists c > 0 such that for any f ∈ E, g ∈ L p + L ∞ ,

|g| p ≺≺ hd |f | p ⇒ g ∈ E and g E ≤ c f E .
Similarly, the two following conditions are equivalent:

(i) for any f ∈ E, g ∈ L 0 + L q , |g| q ≺≺ tl |f | q ⇒ g ∈ E.
(ii) there exists c > 0 such that for any f ∈ E, g ∈ L 0 + L q , |g| q ≺≺ tl |f | q ⇒ g ∈ E and g E ≤ c f E .

Interrpolation spaces for couples of ℓ p -spaces

In this section, we show that our approach to the Lorentz-Shimogaki and Arazy-Cwikel theorems also applies to sequence spaces. We follow a structure similar to the previous sections, proving partition lemmas, then constructing bounded operators on couples (ℓ p , ℓ q ) with suitable properties to finally draw conclusions on the interpolation spaces of the couple (ℓ p , ℓ q ). Additional arguments involving Boyd indices will be required to prove Theorem 1.2.

We identify sequences with bounded functions on (0, ∞) which are almost constant on intervals of the form (

k, k + 1), k ∈ Z + by i : ℓ ∞ → L ∞ (u k ) k∈Z + → ∞ k=0 u k 1 (k,k+1) .
5.1. An interpolation theorem for the couple (ℓ p , ℓ q ). We start with a partition lemma playing, for sequence spaces, the role of Lemma 3.5.

Lemma 5.1. Let a = (a n ) n∈Z + , b = (b n ) n∈Z + be two positive decreasing sequences such that b ≺≺ tl a. There exists a sequence (∆ n ) n∈Z + of subsets of Z + such that:

(i) for every k ∈ Z + , we have |{n ∈ Z + : k ∈ ∆ n }| ≤ 3. (ii) k∈∆n a k ≥ b n for any n ∈ Z + .
Proof. Define I = {n ∈ Z + : b n > a n }. For n / ∈ I, set ∆ n = {n}. For any n ∈ Z + , define:

i n = sup i : ∞ k=i a k ≥ ∞ k=n b k ,
and for n ∈ I, ∆ n = {i n , . . . , i n+1 } .

From the definition of i n , we have

k≥in a k ≥ k≥n b k .
From the definition of i n+1 , we have

k>in+1 a k < k≥n+1 b k .
Taking the difference of these inequalities, we infer that

k∈∆n a k ≥ b n .
This proves the second condition. Note that since b ≺≺ tl a, i n ≥ n for any n

∈ Z + . Hence, if n ∈ I, b n > a n ≥ a in+1 . Furthermore, by definition of i n+1 , k>in+1 a k < ∞ k=n+1 b k so ∞ k=in+1 a k < ∞ k=n b k .
Hence, by definition of i n , i n+1 > i n for n ∈ I.

Let us now check the first condition. Suppose there exist distinct numbers

n 1 , n 2 , n 3 ∈ I such that k ∈ ∆ n1 , ∆ n2 , ∆ n3 . Without loss of generality, n 1 < n 2 < n 3 . Since k ∈ ∆ n1 , it follows that k ≤ i n1+1 ≤ i n2 . Since k ∈ ∆ n3 , it follows that k ≥ i n3 ≥ i n2+1 . Hence, i n2+1 ≤ k ≤ i n2 and, therefore, i n2+1 = i n2 . Since n 2 ∈ I, it follows i n2+1 > i n2 . This contradiction shows that |{n ∈ I : k ∈ ∆ n }| ≤ 2. By definition, k also belongs to at most one set ∆ n , n / ∈ I. Consequently, |{n ∈ Z + : k ∈ ∆ n }| ≤ 3.
From the partition lemma, we deduce an operator lemma similar to Lemma 3.8. It extends Proposition 2 in [START_REF] Sergey | A version of Calderón-Mityagin theorem for the class of rearrangement invariant groups[END_REF], which is established there for the special case p = 1 by completely different method. Proof. We can assume that both sequences are non-negative and decreasing. Apply

p p = n∈Z + |ϕ n (x)| p = n∈Z + ϕ n (x1 ∆n) p ≤ n∈Z + k∈∆n |x k | p = k∈Z + |{n : k ∈ ∆ n }| |x k | p ≤ 3 x p p . (x)
The second estimate is clear, using once again the fact that an integer k belongs to at most three ∆ n 's.

The following remarks were communicated to the authors by Cwikel and Nilsson.

Remark 5.3. A bounded linear operator on ℓ p , p ≤ 1 extends automatically to a bounded linear operator on ℓ 1 .

Proof. Indeed, let (e n ) n∈Z + be the canonical basis of ℓ ∞ . Let T be a contraction on ℓ p , p < 1. Then by Hölder's inequality T (e n ) 1 ≤ T (e n ) p ≤ T p→p . By the triangle inequality, for any finite sequence a = (a n ) n∈Z + , We conclude this subsection with a new interpolation theorem.

T (a n ) 1 ≤ n∈Z + |a n | T (e n ) 1 ≤ a 1 T p→p .
Theorem 5.6. Let p < q ∈ (0, ∞] such that q ≥ 1. Let E be a quasi-Banach sequence space. Then E belongs to Int(ℓ p , ℓ q ) if and only if there exists c p,E and c q,E in R >0 such that: (i) for any u ∈ E and v ∈ ℓ ∞ , if |v| p ≺≺ hd |u| p , then v ∈ E and v E ≤ c p,E u E ;

(ii) if q < ∞ : for any u ∈ E and v ∈ ℓ ∞ , if |v| q ≺≺ tl |u| q , then v ∈ E and v E ≤ c q,E u E .

Proof. The proof of the "only if" implication is identical to the proof of Theorem 4.1 using Lemmas 5.2 and 5.5 instead of Lemmas 3.7 and 3.6. The "if" implication is given by [7, Theorem 4.7].

Upper Boyd index.

Let us now recall the definition of the upper Boyd index, in the case of sequence spaces. For any n ∈ N define the dilation operator:

D n : ℓ ∞ → ℓ ∞ (u k ) k∈Z + → u ⌊ k n ⌋ k∈Z + .
Let E be a symmetric function space. Define the Boyd index associated to E by:

β E = lim k→∞ log D k E→E log k .
Note that since E is a quasi-Banach space, β E < ∞.

In this next proposition, we relate the upper Boyd index to an interpolation property. We follow [START_REF] Stephen | The Hardy operator and Boyd indices[END_REF]Theorem 2]. 1 p E ≤ c p,E u E , 0 ≤ u ∈ E. Proof. Let E p be the p-concavification of E, that is,

E p = {f : |f | 1 p ∈ E}, f Ep = |f | 1 p p E .
Obviously, E p is a quasi-Banach space. Apply Aoki-Rolewicz theorem to the space E p and fix q = q p,E > 0 such that For every u ∈ E, we have (V (u p ))

1 p qp E = V (u p ) q Ep = ∞ n=0 1 2 n (D 2 n u) p q Ep ≤ ≤ C p,E ∞ n=0 1 2 n (D 2 n u) p q Ep = C p,E ∞ n=0 2 -nq D 2 n u qp E .
Let r ∈ (p, β -1 E ). By the definition of β E , there exists c p,E > 0 such that D n E→E ≤ c p,E n 1 r for any n ∈ N. Therefore, (V (u p ))

1 p qp E ≤ C p,E • c q p,E • ∞ n=0 2 -nq 2 nqp r u qp E = C p,E • c q p,E • 2 q 2 q -2 qp r • u qp E .
Lemma 5.9. If x = µ(x), then Cx ≤ 3V x for every x ∈ ℓ ∞ .

Proof. Let k ≥ 0. Since x is decreasing, it follows that

(Cx)(2 k -1) = 1 2 k x(0) + k-1 i=0 2 i+1 -1 j=2 i x(j) ≤ 1 2 k x(0) + k-1 i=0 2 i x(2 i ) .
On the other hand, we have

(V x)(2 k+1 -1) = n≥0 2 -n x 2 k+1 -1 2 n = = k n=0 2 -n x(2 k+1-n -1) + ∞ n=k+1 2 -n x(0) = = 1 2 k x(0) + k i=0 2 i x(2 i+1 -1) .
Again using the fact that x is decreasing, we obtain We are now ready to deliver a complete resolution of the conjecture stated by Levitina and two of the authors in [START_REF] Levitina | Cwikel estimates revisited[END_REF].

1

 1

Proof of Theorem 1 . 1 .

 11 The assertion (a) is obtained by combining Theorem 4.1 and Theorem 4.3 with q = ∞.The assertion (b) is derived similarly from Theorem 4.1 and Theorem 4.3 by applying them with p = 0.

Lemma 5 . 2 . 1 p

 521 Let p ≥ 1. Let a, b ∈ ℓ p such that |b| p ≺≺ tl |a| p .Then there exists an operator T : ℓ p → ℓ p such that:(i) T (a) = b.(ii) T p→p ≤ 3 and T 0→0 ≤ 3.

  Lemma 5.1 to |a| p and |b| p . For every n ∈ Z + , choose a linear form ϕ n on ℓ p of norm less than 1, supported on ∆ n and such that ϕ n (a) = ϕ n (a1 ∆n ) = b n . Define T : x ∈ ℓ p → (ϕ n (x)) n∈Z + . By construction, T (a) = b. Let us check the norm estimates. Let x ∈ ℓ p , T

Hence, T extends to a contraction on ℓ 1 . 5 . 4 . 1 N 1 Lemma 5 . 5 .

 1541155 Remark The condition p ≥ 1 in Lemma 5.2 is necessary. Proof. Let us show that Lemma 5.2 cannot be true for p < 1. Assume by contradiction that there exists c > 0 such that for any finite sequences a and b in ℓ p such that |b| p ≺≺ tl |a| p there exists T with T p→p ≤ c and T (a) = b. By Remark 5.3 above, we also have T 1→1 ≤ c. In particular, b 1 ≤ c a 1 . By considering b = e 1 and a = /p N i=1 e i for N large enough, one obtains a contradiction. We will not prove a sequence version of Lemma 3.7 to avoid the repetition of too many similar arguments. Fortunately, the expected result already appears in the literature, see [10, Theorem 3]: Let p > 0. Let a, b ∈ ℓ ∞ such that |b| p ≺≺ hd |a| p . Then there exists an operator T : ℓ p → ℓ p such that: (i) T (a) = b. (ii) T p→p ≤ 8 1/p and T ∞→∞ ≤ 2 1/p .

Proposition 5 . 7 .(Lemma 5 . 8 .

 5758 Let E be a quasi-Banach symmetric sequence space. Let p < 1 βE . There exists a constant C such that for any u ∈ E and v ∈ ℓ ∞ , satisfying|v| p ≺≺ hd |u| p , we have v ∈ E and v E ≤ C u E . Define the map V : ℓ ∞ → ℓ ∞ by setting V u = ∞ n=0 2 -n D 2 n u,and the map C : ℓ ∞ → ℓ ∞ by If p < 1 βE , then (V (u p ))

2 i≤ 3 1 p

 21 x(2 i+1 -1).Combining the three previous inequalities, we have just shown that for any k ≥ 0,(Cx)(2 k -1) ≤ 3(V x)(2 k+1 -1). Now let n ≥ 0 and choose k such that n ∈ [2 k -1, 2 k+1 -1]. Since Cx and V x are decreasing, we have:(Cx)(n) ≤ (Cx)(2 k -1) ≤ 3(V x)(2 k+1 -1) ≤ 3(V x)(n).Proof of Proposition 5.7. Without loss of generality, u = µ(u) and v = µ(v). Since v p ≺≺ hd u p , it follows that|v| p ≤ C(|v| p ) ≤ C(|u| p ) ≤ 3V (|u| p ),where we used Lemma 5.9 to obtain the last inequality. By Lemma 5.8, we have v E (V (|u| p ))1 p E ≤ c p,E u p .

  For t /∈ A, we have h 1 (s) = g(u + (t)) for all s ∈ (u -(t), u + (t)). Thus,

	t	h 1 (s) p ds =	u-(t)	h 1 (s) p ds +	t	h 1 (s) p ds ≤
	0				0		u-(t)
		≤	u-(t)	g(s) p ds +	t	g(u + (t)) p ds.
			0				u-(t)
	Since
	u-(t)			u-(t)
		g(s) p ds =			f (s) p ds, g(u + (t)) ≤ f (u + (t)),
	0				0
	it follows that
	t		u-(t)			t	t
	h 1 (s) p ds ≤		f (s) p ds +	f (u + (t)) p ds ≤	f (s) p ds.
	0		0				u-(t)	0
			t	µ(s, h 1 ) p ds ≤	t	f (s) p ds.
			0				0
	Indeed, for t ∈ A, we have
		t	h 1 (s) p ds ≤	t	g(s) p ds ≤	t	f (s) p ds.
		0			0	0

Proof of Theorem 1.2. Let E be a quasi-Banach sequence space. Let q ≥ 1. Recall that Theorem 1.2 states that the two following conditions are equivalent: (a) there exists p < q such that E is an interpolation space for the couple (ℓ p , ℓ q ); (b) there exists c > 0 such that for any u ∈ E and |v| q ≺≺ tl |u| q , then v ∈ E and v E ≤ c u E .

Note that by