Polar Topologies on Sequence Spaces in Non-Archimedean Analysis

Proyecciones Journal of Mathematics Vol. 31, No 2, pp. 103-123, June 2012. Universidad Cat´olica del Norte Antofagasta - Chile

Polar topologies on sequence spaces in non-archimedean analysis

R. AMEZIANE HASSANI A. EL AMRANI UNIVERSITE´ SIDI MOHAMED BEN ABDELLAH, MOROCCO and M. BABAHMED UNIVERSITE´ MOULAY ISMAIL, MOROCCO Received : May 2011. Accepted : January 2012

Abstract The purpose of the present paper is to develop a theory of a duality in sequence spaces over a non-archimedean vector space. We introduce polar topologies in such spaces, and we give basic results characterizing compact, C-compact, complete and AK complete subsets related to these topologies. −

Key words : Locally K-convex topologies, non archimedean sequence spaces, Schauder basis, separated duality.

MSC2010 : 11F85 - 46A03 - 46A20 - 46A22 - 46A35 - 46A45 - 464A50. 104 R.AmezianeHassani,A.ElamraniandM.Babahmed

1. Introduction

The duality λ, λα , where λ is a scalar sequence space, was studied by Köthe and Toeplitzh i [7] and it has been reformulated by Köthe [6] using the theory of locally convex spaces. After, the duality λ, λβ has been studied by Chillingworth [2], Matthews [8], T. KomuraD and Y.E Komura [4]. In this work, we are interested to a duality in non-archimedean sequence spaces. We consider a separated duality X, Y of vector spaces over a non-archimedean valued field K (n.a); in [1]h Amezianei and Babahmed gave a fundamental properties of this duality. Afterwards we take E (X) and E (Y ) two vector-valued sequence spaces over X and Y respectively such that E (Y ) E (X)β that are endwed with the separated duality E (X) ,E(Y ) by⊂ the canonic bilinear form (p.108). We introduce the no- h i X X tion of polar topoogies over E (X) ; and by the linear maps πj and δj which we define in this paper; we study the polar topologies compatible with the duality E (X) ,E(Y ) using the basic duality X, Y . Finally we characterize C hcompact, AKi complete and completeh subsetsi of E (X) relatively at these− topologies. This− study was useful in the study that we made in [3]. Throughout this paper, K is a non-archimedean (n.a) non trivially valued complete fieldwithvaluation . , X and Y are two n.a topological vector spaces over K (or K vector spaces)| | that are in separated duality X, Y . The duality theory for locally K convex spaces can be found moreh exten-i sively in [1], [9] , [11] and [12]. −

2. Preliminary

A nonempty subset A of a K vector space X is called K convex if λx + µy + γz A whenever x, y, z− A, λ, µ, γ K, λ 1, −µ 1, γ 1 and λ + µ∈+ γ =1.Ais said to∈ be absolutely∈ K| convex| ≤ | if|λx≤ + µy| | ≤ A whenever x, y A, λ, µ K, λ 1, µ 1. For a− nonempty set A X∈its ∈ ∈ | | ≤ | | ≤ ⊂ K convex hull c (A) and absolutely K convex hull c0 (A) are respectively the− smallest K convex and absolutely−K convex set that contains A. If − − A is a ﬁnite set x1,...,xn we sometimes write c0 (x1,...,xn) instead of { } c0 (A) . An absolutely K convex subset of a locally K convex space X is called K closed if for every− x X the set λ : λ −K, λx A is closed in K−. If the valuation on K∈ is discrete every{| | absolutely∈ ∈K }convex set A is| K| closed. If K has a dense valuation an absolutely K convex− set A is − Polar topologies on sequence spaces in non-archimedean analysis 105

K closed if and only if from x E, λx A for all λ K, λ 1itfollows that− x A. Intersections of K ∈closed sets∈ are K closed.∈ For| | ≺ an absolutely K convex∈ set A the K closed− hull of A is the− smallest subset of X that − − is K closed and contains A, it is denoted by Kc (A) . If K is discrete we − have Kc (A)=A and if K is dense, Kc (A)= λA : λ Kandλ 1 ([1] p. 220) . ∩ { ∈ | |Â } A topological vector space X over K is called locally K convex space if X has a base of zero consisting of locally K convex sets.− Let (X, τ)alocallyK convex space, τ is− define by a family of n.a. semi-norms τ continuous− over X, and if K is discrete, we can suppose − that Np = p(x)/x X K for every p ([9]) ; where ( )isafamily of n.a semi-norms{ which∈ } ⊂ defi| ne| the topology∈ Pτ. P If p is a (n.a) semi-norm over X, Bp (0, 1) is the set x X : p (x) 1 . { ∈ ≤ } A sequence (ei) is a Schauder basis for X if every x X can be written i ∈ ∞ uniquely as x = λixi where the coefficient functionals fj : x λj are i=1 7−→ continuous. X Let X a K vector space and M a subset of X, a K convex filter over M, is a filter −over M having a basis consisting of −K convex subsets of M; this basisF is called K convex basisB of K convex filter− . The order of all filters on− M induces an order− on all K convexF filters on M. A maximal element of the ordered set of K convex−filter on M is called maximal K convex filter of M. − − Let (xi)i I anetonM; for all i I, put Fi = xj/j i . (Fi)i I ∈ ∈ { ≥ } ∈ is a filter over M called filter associated to a net (xi)i I . Conversely, if ∈ =(Fi)i I is a filter over M, for all i I let xi Fi ;overI we define F ∈ ∈ ∈ the following order: i j Fj Fi . (xi)i I is a net in M called a net associated to a filter ≤. ⇔ ⊂ ∈ F Proposition 1. Let X alocallyK convex space, M a subset of X and − =(Fi)i I a maximal K convex filter over M. F 1. ∈converges or not− having any clusterpoint . F 2. Let (xi)i I a net associated to a ; if (xi)i I converges to x0, ∈ F ∈ F converges to x0.

Proof. 1. Let x0 a cluster point of and (Uj)j J a K convex F ∈ − neighbourhood base of x0, 0 = Fi Uj/i Iandj J is a K convex F { ∩ ∈ ∈ } − ﬁlter which converges to x0 and it is coarsest than , then = 0 . F F F 2. x0 is a clusterpoint of (xi)i I , then it is a clusterpoint of , and so ∈ F converges to x0. F 106 R.AmezianeHassani,A.ElamraniandM.Babahmed

Proposition 2. Let X, Y two K vector spaces, f : X Y alinearmap − −→ and =(Fi)i I a maximal K convex ﬁlter over X that having us a K convexF basis;∈ f( ) is a K convex− basis of a maximal K convexB ﬁlter over− Y. B − −

A subset A of a locally K convex space X is compactoid if for each − neighbourhood U of zero there exist x1,...,xn X such that A U + ∈ ⊂ c0 (x1,...,xn) . An absolutely K convex subset A of X is said to be C compact if every convex filter on A has− a clusterpoint on A. − K is C compact if and only if K is spherically complete. − Proposition 3. Let M be a subset of X. The following are equivalent: (i). M is C compact; (ii). Every maximal− K convex filter over M converges; (iii). Any family of closed− and K convex subsets of M whose intersection is empty contains a finite subfamily− whose intersection is empty.

Let a basis of a filter on a subset M of X; the smallest K convex filter containingB , is called KF convex filter generated by and is− denoted by B − B c( ). We show that c ( )= F M/there exists B : c(B) F , F B F B { ⊂ ∈ B ⊂ } and c( )isK convex basis of c( ), that is to say c( )= (c( )). B − F B F B F B If (xi)i I is a net in X;(xi)i I converges to x0 if and only if the filter ∈ ∈ K convex associated with (xi)i I converges to x0. − ∈ Proposition 4. Let X, Y two K vector spaces, f : X Y alinear map, M a subset of X and abaseof− filter on M. Then f−→( ) is a base of B B filter on f(M), and we have c(f( )) = f( c( )). F B F B

(ω (X) ,τω (X)) = the linear space of all sequences in X endowed with the product topology τω (X) which is generated by the family of n.a semi- norms (pn)n IN, p ( ) ,pn (x)=p (xn) for all x =(xn)n ω (X)and all p ( ) ,∈ if X∈ isP a locally K-convex space and ( )isafamilyof∈ n.a semi-norms∈ P which deﬁne his topology; this space is notedP ω (K)(orω, for short) in case when X = K. A sequence space over X is a subspace of ω (X) . We deﬁne the following sequence spaces over X c0 (X)= (xk) ω (X):(xk) converges to zero { k ∈ k } c (X)= (xk) ω (X):(xk) converges in X , { k ∈ k } ϕ (X)= (xk) ω (X): there exists k0 IN : xk =0for all k k0 , { k ∈ ∈ ≥ } m (X)= (xk) ω (X):(xk) is bounded in X . { k ∈ k } Polar topologies on sequence spaces in non-archimedean analysis 107

Over m (X)wedeﬁne the sequence of n.a semi-norms (p)p ( ) by: ∈ P p (x)=sup p (xk) for all x =(xk)k m (X) . k ∈ Let τ (X) be the topology on m (X)deﬁned with the sequence of n.a ∞ semi-norms (p)p ( ) . ∈ P 3. Polar topologies

Let X and Y two K vector spaces placed in separating duality X, Y . If A is a subset of X, we− denote by A = y Y/ x, y 1 for allh x i A ◦ { ∈ |h i| ≤ ∈ } the polar of A and A◦◦ = x X/ x, y 1 for all y A◦ the bipolar of A. { ∈ |h i| ≤ ∈ } A◦ is absolutely K convex and σ(Y,X) bounded. − − σ(Y,X) For each absolutely K convex subset A of Y , Kc A = A ([1] , − ◦◦ corollary 4.3, p. 233). A subset A of Y is said to be ³X closed´ if for every y Y A,thereexitsx X such that x, y 1and− x, A 1. Intersections∈ \ of X closed∈ sets are X closed.|h Fori| Â a subset|hA ofi|Y≤the − − X closed hull Xc (A)ofA is the smallest X closed subset of Y that − − contains A. For each subset A of Y , Xc (A)=A◦◦([1] , proposition 2.5, p. 224). Using these two results and by [1], theorem 4.2, p. 233 we have: for all absolutely K convex subset A of Y , A is X closed, if and only if, A is K closed and σ−(Y,X) closed. − −Let be a family of−σ (Y,X) bounded subsets of Y such that (a) A is directed by inclusion,− (b) YA = A, A [∈A (c) there exists λ0 K, λ0 > 1 such that λ0A , for all A . ∈ | | ∈ A ∈ A A topology τ on X is called polar topology of convergence, if τ has a fundamental system of zero neighbourhood (AF.S.N− ) consisting of − A◦/A . { A vector∈ A} topology τ on X is called polar topology if there exists a family of σ (Y,X) bounded subsets of Y which has the properties (a) , (b)and (Ac) , such that−τ is a polar topology of convergence. it is deﬁned by the A− family of n.a. semi-norms (PA)A , where PA(x)=sup x, y /y A . If is the family of all subsets∈A of Y that are: {|h i| ∈ } 1.A Absolutely K convex, weakly bounded and weakly C compacts, − − we have the C compact topology τc (X, Y )=τc, 2. Absolutely− convex and σ (Y,X) compact, we have the Mackey − topology τm (X, Y )=τm, 108 R.AmezianeHassani,A.ElamraniandM.Babahmed

3. σ (Y,X) bounded and X closed,wehavetheX closed topology − − − τe (X, Y )=τe. 4. σ(Y,X) bounded, we have the strong topology τb(X, Y ). − AlocallyK convex topology τ on X is called compatible with the duality X, Y or− (X, Y ) compatible if Y is isomorphic to the topological dual of Xh providedi with− the topology τ. The weak topology σ(X, Y )isthe coarsest topology among all topologies (X, Y ) compatible, and the upper bound topology of all topologies (X, Y ) compatible− topology is the finest among all the topologies (X, Y ) compatible.− − We say that X is semi-reflexive if X is isomorphic to the strong topological dual of Y and if τ is a locally K convex topology on X we say that − X is τ reflexive if X is semi-reflexive and τ = τb(X, X0 ). − For further information about polar topology of convergence and general properties of locally K convex spaces we referA to− [1], [11] and [12]. − If A ω (X) , the β dual of A is the subspace of ω (Y ) which is define β ⊂ − by A = (yn) ω (Y ):lim xn,yn =0for all(xn) A .Ais called n ∈ n h i n ∈ perfect if Anββ = A. If A is perfect then ϕ (X) A. For all Ao ω (X), Aβ is perfect. We define Bβ if B ω (Y )onthesameway.⊂ ⊂ ⊂ A subset D of ω (X) is said to be solid if for every x =(xk) D and k ∈ α =(αk)k ω such that αk 1 for all k,wehaveαx =(αkxk)k D. The solid hull∈ S (D)ofD|isthesmallestsolidsetofsequencecontaining| ≤ ∈ D. A topology on E (X) , with respect the duality E (X) ,E(X)β , will be called solid if the elements of the determining familyD of weakly boundedE subsets of E (X)β are solids sets. Let E (X)andE (Y ) be two sequence spaces on X and Y respectively such that E (Y ) E (X)β,wedefine on the pair (E (X) ,E(Y )) the fol- ⊂ ∞ lowing duality (xn)n , (yn)n = xn,yn for all (xn)n E (X)andall h i n=1 h i ∈ X (yn) E (Y ) . n ∈ If ϕ(X) E(X)andϕ(Y ) E(Y ), the duality E(X),E(Y ) is sepa- rate. ⊂ ⊂ h i In the sequel E(X),E(Y ) denotes a duality of this type. h i S(E(Y )) [S(E(X))]β and S(E(X)),S(F (Y )) is a separating duality extending the⊂ separating dualityh E(X),F(Y ) , therefore,i we can assume that E(X)andF (Y )aresolid. h i For all j 1, we consider the following linear mappings: ≥ Polar topologies on sequence spaces in non-archimedean analysis 109

πX : E(X) X δX : X E(X) j −→ j −→ (xn) xj a δj(a) −→ −→ where δj (a)isthesequencewitha in the j-th place and 0’s elsewhere. Y Y We deﬁne also πj and δj . n [n] ith Let x =(xk) ω(X), for all n 1 x = δj(xj)iscalledthen ∈ ≥ j=1 section of x. X X X Y Y X Y X We have: πj oδj = idX ,πj oδj = idY , (πj )∗/Y = δj and (δj )∗/F (Y )= Y πj where u∗ is the algebraic adjoint of the linear map u.

Proposition 5. Let A be a subset of E(X) if A is solid, A◦ is solid and we have: A =[A ϕ(X)]◦. ◦ ∩ Deﬁnition 1. Let A a subset of ω(X). X X a. Is said that A is δj saturated if for all (xn) A, δj (xj) A. −X X ∈ ∈ b. It is said that A is δ saturated if A is δj saturated for all j 1. X− − X ≥ c. It is said that A is π saturated if: xj π (A) for all j 1 − ∈ j ≥ ⇒ (xn) A. ∈ If A is solid, A is δX saturated. ϕ(X)isδX saturated− and not πX saturated. − − If p is a n.a. semi-norm on X, (xn) ω(X)/ sup p(xn) 1 is ½ ∈ n ≤ ¾ πX saturated. −The following results are demonstrated in a direct:

Proposition 6. Let A be a subset of E(X). 1. If A is πX saturated, S(A) is πX saturated. X − − X 2. If A is δ saturated, S(A) and c0 (A) are δ saturated, and A◦ is δY saturated and− πY saturated. − −3. πX (A) ◦ πY (−A ) for all j 1. j ⊂ j ◦ ≥ 4. Ifh A is δiX saturated, πX(A) ◦ = πY (A ). j − j j ◦ 5. If A is δX saturated, h i − X Y A◦ = πj πj (A◦) = (yk) F (Y )/ sup xk,yk 1 for all (xk) A . ∈ k |h i| ≤ ∈ h i ½ ¾ 6. S(A) S(A ); and if A is δX saturated, A = S(A) = S(A ). ◦ ⊂ ◦ − ◦ ◦ ◦ 110 R.AmezianeHassani,A.ElamraniandM.Babahmed

X X 7. If A is δ saturated and F (Y ) closed, πj (A) is Y closed for all j 1. − − − ≥ X X 8. If A is π saturated and πj (A) is Y closed for all j 1,Ais F (Y ) closed. − − ≥ − Corollary 1. Let A be a subset of E(X) δX saturated and πX saturated. − X − For A is F (Y ) closed, it is necessary and enough that πj (A) be Y closed for all j 1. − − ≥ Proposition 7. Let A be an absolutely K convex subset of E(X). X − X 1. If A is K closed and δj saturated, πj (A) is K closed. −X − X − 2. If A is π saturated and πj (A) is K closed for all j 1,Ais K closed. − − ≥ −

Proposition 8. Let τ be a topology on E(X) and τj the topology im- X age reciprocal of τ by the linear map δj on X. If τ admits as S.F.N of Y ◦ 0 A /A , then π (A) /A is a F.S.N. of 0 for τj. { ◦ ∈ A} j ∈ A nh i o Proof. ([1] ,proposition2.9) .

Proposition 9. For all j 1,πX is (σ(E(X),F(Y )),σ(X, Y )) continuous ≥ j − and δX is (σ(X, Y ),σ(E(X),F(Y ))) continuous. j −

X X Proof. (πj )∗(Y ) F (Y )and(δj )∗(F (Y )) Y, and the result follows from ([9] ,p.128)⊂. ⊂

Proposition 10. 1. πX (A) ◦ =(δY ) 1(A ) for all A E(X). j j − ◦ ⊂ 2. δX(B) ◦ =(πYh) 1(B i) for all B X. j j − ◦ ⊂ 3. πh X (A) i B δY (B ) A for all A E(X) and for all B X. j ⊂ ⇒ j ◦ ⊂ ◦ ⊂ ⊂ 4. δX (B) A πY (A ) B for all A E(X) and for all B X. j ⊂ ⇒ j ◦ ⊂ ◦ ⊂ ⊂ 5. (πX ) 1(D )= δY (D) ◦ for all D Y. j − ◦ j ⊂ 6. (δX ) 1(C )= hπY (C) i◦ for all C F (Y ). j − ◦ j ⊂ 7. (πX ) (D) C h πX(Ci ) D for all D Y and for all C E(Y ). j ∗ ⊂ ⇒ j ◦ ⊂ ◦ ⊂ ⊂ 8. (δX) (C) D δX (D ) C for all D Y and for all C E(Y ). j ∗ ⊂ ⇒ j ◦ ⊂ ◦ ⊂ ⊂ Proof. ([1] ,proposition2.8) . Polar topologies on sequence spaces in non-archimedean analysis 111

A polar topology of convergence on E(X) is said solid, if all A is solid. Thus, any polar,A− solid topology admits a F.S.N from 0 consisting∈ A of solid subsets . If τ is the polar topology of convergence on E(X) such that A is δY saturated for all A ,τAcoincides− with the polar topology of S( )−convergence (proposition∈ A 6), and then τ is a polar and solid topology . A −

Proposition 11. Let τ be a polar topology of convergence over E(X) A− X and τj the topology image reciprocal of τ by the linear map δj on X. Y 1. τj is the polar topology of πj ( ) convergence. X A − Y Y 2. π is (τ,τj) continuous if and only if δ oπ (A) for all A . j − j j ∈ A ∈ A Proof. ([1] ,proposition3.8).

Proposition 12. If τ is the weak topology (resp. Mackey, resp. C compact, resp. − E(X) closed; resp. strong) of E(X) for all j 1,τj is the weak topology (resp.− Mackey, resp. C compact, resp. X closed;≥ resp. strong) on X − − Proof. ([1] ,proposition3.9).

Proposition 13. Let τ a polar topology of convergence on E(X), for all j 1, we have: A− ≥ X 1. δj is (τj,τ) continuous; − X 2. If τ is solid, πj is (τ,τj) continuous; X − X 3. If π is (τ,τj) continuous, δ is (τj,τ) closed. j − j − Y Proof. 1. τj is a polar topology of πj ( ) convergence, and we have: A − δX πY (A) ◦ A for all A . j j ⊂ ◦ ∈ A ³h2. If τ isi solid,´ we have : πX (A ) πY (A) ◦ for all A . j ◦ ⊂ j ∈ A h i Y ◦ 3. Let M aclosedin(X, τj), there exists A such that π (A) ∈ A j ⊂ M , therefore A δX (M )= δX (M) ◦ . h i ◦ ◦ ⊂ j ◦ j h i Let τ be a locally K convex topology on E(X) such that E(X)be τ polar; if τ is (E(X) F (−Y )) compatible, τ is a polar topology of − convergence, where is constituted− of σ(F (Y ),E(X)) bounded and A− A − 112 R.AmezianeHassani,A.ElamraniandM.Babahmed

E(X) closed subsets of F (Y ), ([1 ], theorem 4.3). For all j 1,τj is the − Y ≥ polar topology of π ( ) convergence on X and X is τj polar if all A j A − − ∈ A is δY saturated, πX (A)isσ(Y,X) bounded and X closed (Proposition − j − − 6), and then τj is (X, Y ) compatible. − If K is spherically complete, we have the following theorem:

Theorem 1. Suppose that K be spherically complete, and let τ alo- cally K convex topology on E(X); if τ is (E(X),F(Y )) compatible, τj is (X, Y−) compatible, for all j 1. − − ≥ Proof. τ is a polar topology of convergence, where consists of absolutely K convex , σ(E(Y ),E(X))A bounded and σ(E(Y )A,E(X)) C compact subsets of F (Y )([1],theorem− 4.4) . For all j 1,πY −is − ≥ j (σ(F (Y ),E(X)),σ(Y,X)) continuous, then πY (A) is absolutely K convex, − j − σ(Y,X) bounded and σ(Y,X) C compact for all A and then τj is (X, Y ) −compatible. − − ∈ A − Theorem 2. Let τ a solid and polar topology on E(X); if E(X) is τ barreled, − X is τj barreled for all j 1. − ≥ X X Proof. Let B a τj barrel in X; δ is (τj,τ) closed, then δ (B)is − j − j a τ barrel into E(X)andthen(δX ) 1(δX(B)) is a neighborhood of 0 in − j − j (X, τj)thenB is a neighborhood of 0 for τj.

Remark 1. Instead of assuming that τ is solid, we can assume only that X π be (τ,τj) continuous for all j 1. j − ≥ X A subset A of E(X)saidtobeδ stable if for all x =(xk) E(X)such X− ∈ that there exists j 1satisfyingδ (xj) A, then x A. ≥ j ∈ ∈ Let A E(X) such that A δX(a)/a Xandj 1 = φ, A is δX ⊂ ∩ j ∈ ≥ stable. n o

Deﬁnition 2. Let τ a vector topology on E(X); we say that E(X) is δXτ barreled if every τ barrel δX stable, is a neighborhood of 0. − − − If E(X)isτ barreled, it is δX τ barreled. − − Theorem 3. Let τ a polar and solid topology on E(X); if there exists j X ≥ 1 such that X is τj barreled, E(X) is δ τ barreled − − Polar topologies on sequence spaces in non-archimedean analysis 113

X X Proof. Let B a τ barrel δ stable in E(X); δj is (τj,τ) continuous, X 1 − − X 1 − so (δ ) (B)isaτj barrel, and then (δ ) (B) is a neighborhood of 0 in j − − j − X 1 X 1 (X, τj) and hence (πj )− (δj )− (B) is a neighborhood of 0 in (E(X),τ).B is δX stable, then (πX ) h1 (δX ) 1(Bi) B and then B is a neighborhood − j − j − ⊂ of 0 in (E(X),τ). h i

Theorem 4. Suppose that X and Y are semi-reflexive, and let τ atopol- ogy on E(X) which is (E(X),F(Y )) compatible. If E(X) is τ reflexive, − − X is τj reflexive for every j 1. − ≥

Proof. τ = τb(E(X),E(X)0 )=τb(E(X),F(Y )); so for all j 1 τj = ≥ τb(X, Y ) (Proposition 12). Y is semi-reﬂexive, then τj is (X, Y ) compatible − ([ 1], proposition 5.9) and then τj = τb(X, (X, τj)0 ).

Corollary 2. If K is spherically complete and τ is a topology on E(X) which is (E(X),F(Y )) compatible and solid such that E(X) is τ barreled, − − then X is τj reﬂexive for any j 1. ≥

Proof. For all j 1,τj is (X, Y ) compatible (theorem 1) and X is ≥ − τj barreled for all j 1, then X is τj reﬂexive ([1] ,theorem5.2) . − ≥ −

4. Compactness and C compactness − X Let τ a polar topology on E(X) such that π be (τ,τj) continuous for all j − j 1. If M is a compact subset of (E(X),τ); πX (M)isacompactsubset ≥ j of (X, τj) for all j 1. ≥ In order to study the converse, we introduce the notion of TK convergent net. −

i Deﬁnition 3. Anet(x )i I in E(X) is called TK convergent if for all i ∈ − j 1, (xj)i I is convergent in (X, τj). ≥ ∈ Theorem 5. Let M a subset of E(X); M is relatively compact in (E(X),τ) if and only if: X (i.) π (M) is relatively compact in (X, τj) for all j 1; j ≥ (ii.) All TK convergent net in M, converges in (E(X),τ). − 114 R.AmezianeHassani,A.ElamraniandM.Babahmed

X X Proof. N.C.] πj is (τ,τj) continuous for all j 1, then πj (M)is − i ≥ relatively compact in (X, τj). Let (x )i I a TK convergent net in M. For i ∈ − i all j 1letxj X such that (xj)i I converges to xj in (X, τj). (x )i I has ≥ ∈ ∈ ∈ aclusterpointz =(zn)in(E(X),τ). For all j 1,zj is a cluster point of i ≥ i (xj)i I in (X, τj); then zj = xj. (xn) is the unique cluster point of (x )i I , ∈ i ∈ therefore (x )i I converges to (xn)in(E(X),τ). ∈i S.C.] Let (x )i I anetinM, and let the family of σ(F (Y ),E(X)) bounded ∈ A − subset of F (Y ) which deﬁnes the topology τ. For any j 1,τj is the polar Y ≥ topology of πj ( ) convergence on X. A − i Let x1 aclusterpointof(x1)i I in (X, τ1). For all A and for all ∈ ∈ A iA Y ◦ i I, there exists iA >isuch that x π (A) . Consider the sub ∈ 1 ∈ 1 family (iA)A of I, it is ordered by: iA iBh A i B for all A, B . ∈A ≤ ⇔ ⊂ ∈ A (iA)A is a ﬁlter on the right family. Let A0 ; iA iA0 A0 ∈A ∈ A ≥ ⇒ ⊂ Y ◦ Y ◦ iA Y ◦ iA A π1 (A) π1 (A0) x1 x1 π1 (A0) . Therefore (x1 )A ⇒ ⊂ ⇒ − ∈ ∈A convergesh to ix1 inh (X, τ1).i h i iA Let x2 a cluster point of (x2 )A in (X, τ2). for all A , there exists ∈A ∈ A l1(iA) Y ◦ l1(iA) >iA such that x x2 π (A) . 2 − ∈ 2 h Y i ◦ Y ◦ l1(iA) Let A0 ; iA iA A A0 π (A) π (A0) x ∈ A ≥ 0 ⇒ ⊃ ⇒ 2 ⊂ 2 ⇒ 2 − Y ◦ l1(iA) h i h i x2 π2 (A0) . Therefore (x2 )A converges to x2 in (X, τ2). Let ∈ ∈A h i l1(iA) x3 a cluster point of (x3 )A in (X, τ3). For all A , there exists ∈A ∈ A l2ol1(iA) Y ◦ l2ol1(iA) l2(l1(iA)) >l1(iA) such that x3 x3 π3 (A) . (x3 )A − ∈ ∈A converges to x3 in (X, τ3). h i Inductively, for all j 3andforallA , there exists ljolj 1o....l1(iA) > − ≥ lj o...ol1(iA) ∈ A lj 1o.....ol1(iA) such that (xj+1 )A converges to xj+1 in (X, τj+1). − i l (i ) l ol (i ) l ∈o....olA (i ) Put y =(x A ,x1 A ,x2 1 A , ...., x k 1 A ,....)A . ∈A iA l1(iA) l2ol1(iA) lko....ol1(iA) For all j 1, (xj ,xj ,xj , ...., xj ,....)A converges ≥ ∈A to xj in (X, τj); therefore y is TK convergent, and hence it converges to x − i in (E(X),τ). Hence x is a cluster point of (x )i I , and then M is relatively compact. ∈

Corollary 3. Let M a subset of E(X), Miscompactin(E(X),τ) if and only if: X (i.) πj (M) is compact in (X, τj) for all j 1, (ii.) Any TK convergent net in M converges≥ to an element of M in (E(X),τ). −

To give version of theorem 5 using the ﬁlters, we need introduce the Polar topologies on sequence spaces in non-archimedean analysis 115 following deﬁnition:

Definition 4. Let M a subset of E(X) and a filter on M; we say that is TK convergent if for all j 1 the filter generatedF by πX( ) convergesF − ≥ j F in (X, τj).

Every convergent filter is TK convergent, and if is a TK convergent filter and is a filter finer than− , is TK convergent.F − F0 F F 0 − Proposition 14. Let M a subset of E(X). 1. If = (Fi)i I is a TK convergent filter on M, any net associated to is TKF convergent.∈ − F i− 2. If (x )i I is a TK convergent net, the K convex filter associated i ∈ − − to (x )i I is TK convergent. ∈ − Theorem 6. Let M a subset of E(X); M is compact in (E(X),τ) if and only if: X (i.) πj (M) is compact in (X, τj) for all j 1; (ii.) Any TK convergent filter on M converges≥ to an element of M. − Proof. N.C.] Let a TK convergent filter on M. For any j 1 X F − ≥ let xj X such that πj ( )convergestoxj in (X, τj). has at least one ∈ F F X cluster point z =(zn)inM. For all j 1,zj is a cluster point of π ( ), ≥ j F therefore zj = xj;then(xn)istheuniqueclusterpointof in M, so F F converges to (xn)in(M,τ). X S.C.] Let amaximalfilter on M; for all j 1 πj ( ) is a maximal X F ≥ F filter on πj (M), therefore it converges to xj in (X, τj), and then is TK convergent, therefore it converges to an element of M. F −

Deﬁnition 5. Let M a subset of E(X),wesaythatM is an AK—complete [n] subset of (E(X),τ) if every x =(xn) element of E(X) such that (x ) is a Cauchy sequence in (M,τ); x M and (x[n]) converges to x in (E(X),τ). ∈ We say that M is relatively AK complete if its closure M in (E(X),τ) is AK complete. − If M− is complete, it is AK complete. Any closed subset of a set AK− complete is AK complete. − − In the following result, we characterize the subsets solid and relatively compact of (E(X),τ). 116 R.AmezianeHassani,A.ElamraniandM.Babahmed

Theorem 7. Let M a solid subset of E(X),Mis relatively compact in (E(X),τ) if and only if: X (i.) πj (M) is relatively compact in (X, τj) for all j 1, i ≥ (ii.) x[i] →∞ x uniformly on M in (E(X),τ), (iii.) M is−→ relatively AK complete in (E(X),τ). − Proof. N.C.] If M is relatively compact, M is relatively complete, andthenitisrelativelyAK complete. − i Suppose we did not (ii.) there exists A a sequence ( x)i in M and ∈ A i [ji] i a strictly increasing sequence of integers (ji)i such that x x/A◦ for i [j ] i − ∈ all i 1. The sequence ( x i x)i is TK convergent to 0, so it converges to 0≥ in (E(X),τ)whichisabsurd.− − α α S.C.] Let ( x)α D anetinM such that for all j 1( xj)α D converges ∈ ≥ i ∈ α [i] [i] X α to xj in (X, τj). Let A for all i 1 x x = δn ( xn xn) A◦ ∈ A ≥ − n=1 − ∈ X for α suﬃciently large. So for all i 1 αx[i] α x[i] in (E(X),τ)in particular x[i] M for all i 1. Using≥ this convergence−→ and (ii), we can choose α as x[i]∈ x[j] =(x[i] α≥x[i])+(αx[i] αx)+(αx αx[j])+(αx[j] x[j]) − − [−i] − − ∈ A◦ for i, j suﬃciently great. Therefore (x ) is a Cauchy net in M and then i + x[i] → ∞ x in (E(X),τ). From this convergence and (ii), we can choose i −→ α α α [i] α [i] [i] [i] such that x x =( x x )+( x x )+(x x) A◦ for α Large α− − − − ∈ enough, so ( x)α D converges to x in (E(X),τ) and hence M is relatively compact (theorem∈ 5).

Corollary 4. Let M a solid subset of E(X); M is compact in (E(X),τ) if and only if: X (i.) πj (M) is compact in (X, τj) for all j 1, i ≥ (ii.) x[i] →∞ x uniformly on M in (E(X),τ) (iii.) M is−→AK complete in (E(X),τ). − Corollary 5. The envelope solid of a relatively compact subset of (E(X),τ) is not necessarily relatively compact.

[i] Proof. Let x =(xn) E(X) such that (x )i does not converge to [i] ∈ x in (E(X),τ)so(z )i does not converge to z uniformly on S(x)andthen S(x) is not relatively compact.

i Proposition 15. 1. Let (x )i I anetinE(X); if is a K convex ﬁlter i X ∈ F − associated with (x )i I ,πj ( ) is a K convex ﬁlter associated with a net i ∈ F − (xj)i I for all j 1. ∈ ≥ Polar topologies on sequence spaces in non-archimedean analysis 117

i 2. Let a K convex ﬁlter on E(X); if (x )i I is a net associated to i F − X ∈ , (xj)i I is a net associated to πj ( ) for all j 1. F ∈ F ≥ Theorem 8. Let M a K convex subset of E(X); M is C compact in (E(X),τ) if and only if: − − X (i.) πj (M) is C compact in (X, τj) for all j 1, (ii.) Any K convex− and TK convergent ﬁlter≥ on M admits a cluster point in M. − −

Proof. N.C.] Obvious. S.C.] Let amaximumK convex filter of M. For any j 1,πX( )is F − ≥ j F amaximumK convex filter of πX (M) (proposition 2), so πX ( )converges − j j F to xj in (X, τj). is then TK convergent, so it admits a cluster point in M, and hence Fconverges in (−E(X),τ) (Proposition 1). F Proposition 16. Let M a K convex subset of E(X); if M is C compact, any K convex and TK convergent− filter on M has a unique cluster− point in M. − −

Proof. Let a K convex and TK convergent ﬁlter on M. For all F − X − j 1letxj X such that π ( ) converges to xj in (X, τj). admits ≥ ∈ j F F at least one cluster point (zn)inM. For all j 1,zj is a cluster point of X ≥ πj ( )in(X, τj), and then xj = zj. So (xj) is the only cluster point of in M.F F

5. AK completion and completion − Let M a subset of E(X)andτ a topology on E(X), we put:

[n] n SM = x M/x →∞ xin(E(X),τ) . ∈ −→ n o If M is a subspace of E(X), we say that M is an AK space if SM = M. − Proposition 17. Let τ a polar topology of convergence on E(X); (E(X),τ) is AK complete. A − [n] Proof. Let x =(xn) E(X) such that (x ) is a Cauchy sequence ∈ [n] [m] in (E(X),τ). For all A there exists n0 1 such that x x A◦ ∈ A [n] ≥ − [n] n ∈ for all n m n0, and then x x A◦ for all n n0, then x →∞ x in (E(X)≥,τ). ≥ − ∈ ≥ −→ 118 R.AmezianeHassani,A.ElamraniandM.Babahmed

Corollary 6. Let M a subset of E(X).M is AK complete if and only if M contains every element x of E(X) such that (x[n]−) is the Cauchy sequence in M.

Corollary 7. Let τ alocallyK convex topology on E(X) coarser than 0 − τ; any AK complete subset of (E(X),τ0 ) is complete in (E(X),τ). −

Proof. Let M an AK complete subset of (E(X),τ0), and either x E(X) such that (x[n])isaCauchysequencein(− M,τ), (x[n])isa ∈ Cauchy sequence in (M,τ0), so x M and hence M is AK complete in (E(X),τ), (Corollary 6). ∈ − ψx : E(Y ) c0(K) For all x =(xn) E(X), we put −→ ∈ (yn) ( xn,yn )n −→ h i ψx is a linear map.

Lemma 1. For any x E(X),ψx is (σ(E(Y ),E(X)),σ(c0(K),m(K))) continuous. ∈ −

β Proof. c0(K) = m(K)and c0(K),m(K) is a separating dual- h i ity. Let (αn) m(K); E(X)issolid,then(αnxn) E(X), and we have ∈ ∈ ψx( (αn xn) ◦) (αn) ◦ . { } ⊂ { } Proposition 18. (E(X),σ(E(X),E(Y ))) is an AK space. −

Proof. Let x =(xn) E(X). For all y =(yn) E(Y ), ( xn,yn ) ∈ ∈ [i] h i ∈ c0(K); there exists i0 1 such that sup xn,yn 1, then x x y ◦ ≥ n i0 |h i| ≤ − ∈ { } ≥ [i] i for all i i0, and then x →∞ x in (E(X),σ(E(X),E(Y ))). ≥ −→ Proposition 19. Suppose that K be local, and let τ a (E(X),F(Y )) compatible topology on E(X); if τ is solid, (E(X),τ) is an AK space. − − Proof. Let afamilyofσ(F (Y ),E(X)) compacts and absolutely K convex subsetsA of F (Y ) such that τ be a polar− topology of convergence − A− ([1] , theorem 4.5.) Let x =(xn) E(X); for all A ,ψx(A)issolid ∈ [i] i∈ A and σ(c0(K),m(K)) compact in c0(K). Then z →∞ z uniformly on − −→ z ψx(A)in(c0(K),σ(c0(K),m(K))) (theorem 7); there exists i0 1 ∈ [i] ≥ such that z z,e 1 for all i i0 and for all z ψx(A), then − ≤ ≥ ∈ [i] ¯D E¯ [i] i x x A¯◦ for all i ¯ i0, and so x →∞ xin(E(X),τ). − ∈ ¯ ≥¯ −→ We have the following result which is a kind of reciprocal of theorem 1: Polar topologies on sequence spaces in non-archimedean analysis 119

Theorem 9. Suppose that K be local, and let τ a polar and solid topology β on E(X) for separating duality E(X),E(X) . If τj is (X, Y ) compatible − for all j 1,τis (E(X),E(X)Dβ) compatible.E ≥ − Proof. E(X)β =(E(X),σ(E(X),E(X)β))0 (E(X),τ)0 . Let f ⊂ ∈ (E(X),τ)0 and x =(xn) E(X). (E(X),τ)isanAK space (proposi- i ∈ − [i] tion 19), therefore x[i] →∞ x in (E(X),τ), and then f(x)=limf(x )= −→ i X X 0 foδj (xj). For all j 1,foδj (X, τj) = Y ; therefore f(x)= j ≥ ∈ X X β xj,yj , with yj = foδj for all j 1. Hence (yj) E(X) , and so j h i ≥ ∈ X (E(X),τ) E(X)β. 0 ⊂ Let a family of subsets of F (Y ) such that: 1. Cis the right filtering for inclusion; C 2. There exist λ0 K, λ0 > 1 such that λ0A for all A ; Y ∈ | | ∈ C ∈ C 3. πj (A)isσ(Y,X) bounded for all j 1andforallA 4. The subspace of E−(Y ) generated by ≥ A/A contains∈ C ϕ(Y ). ∪ { ∈ C} (X)= (x ) ω(X)/ sup x ,y < for all A We put: n n n ⎧ C ( ∈ (yn) A ¯ n h i¯ ∞ ∈ C) ∈ ¯X ¯ ⎨⎪ (Y )=subspace generated by¯ A/A ¯ . C ¯ ∪ { ∈¯C} If is the family of all finite subsets of F (¯Y ), (X)=¯ F (Y )β. ⎩⎪ ϕ(CX) (X)and (X), (Y ) is a separatingC duality defined by the bilinear form:⊂ C hC C i (xn), (yn) = xn,yn for all (xn) (X)andforall(yn) (Y ). h i n h i ∈ C ∈ C If τ is the polarX topology of convergence of E(X), ( (X),τ )is defined, where τ is the polar topologyA− defined on (X)bythefamilyA A , and we have: A A A 1. E(X) (X) F (Y )β; ⊂ A ⊂ 2. τ /E(X) = τ. A Proposition 20. Let τ a polar topology of convergence on E(X). A− 1. S( (X),τ ) E(X), 2. ( A(X),τA )⊂is AK complete. A A −

[i] i [i] Proof. 1. Let x =(xn) S( (X),τ ); x →∞x (τ ), therefore (x ) ∈ A A −→ A is Cauchy sequence in (E(X),τ)(τ = τ /E(X)), and then x E(X) (proposition 17). A ∈ 120 R.AmezianeHassani,A.ElamraniandM.Babahmed

2. Let (x[i])aCauchysequencein( (X),τ ); for all A , there exists A jA ∈ A i 1 such that for all i, j i sup x ,y /(y ) A 1. 0 0 ⎧¯ n n ¯ n ⎫ ≥ ≥ ¯n=i+1 h i¯ ∈ ≤ ⎨¯ X ¯ ⎬ ¯ ¯ ⎩¯ ¯ ⎭ We have on the one hand, sup xn¯,yn /(yn) A¯ 1, therefore ⎧¯ h i¯ ∈ ⎫ ≤ ¯n>i0 ¯ ⎨¯ X ¯ ⎬ ¯ ¯ ¯ ¯ sup xn,yn /(yn) A <⎩¯ (ϕ(X) ¯ (X)), and⎭ then x (¯ h i¯ ∈ ) ∞ ⊂ A ∈ A ¯ n ¯ ¯X ¯ ¯ ¯ ∞ (X); on¯ the other¯ hand, for all i i sup x ,y /(y ) A 1, 0 ⎧¯ n n ¯ n ⎫ ≥ ¯n=i+1 h i¯ ∈ ≤ ⎨¯ X ¯ ⎬ therefore ¯ ¯ [i] ⎩¯ [i] i ¯ ⎭ sup x x, (yn) /(yn) A 1, and¯ then x →∞ x¯ (τ ). − ∈ ≤ −→ A n¯D E¯ o ¯ ¯ Theorem¯ 10. Let τ¯ a solid and polar topology of convergence on E(X). For E(X) is a closed subspace of ( (X),τ ) it isA necessary− and suﬃ- cient that any Cauchy net TK convergentA of E(XA ) converges in (E(X),τ). −

Proof. N.C.] A is solid for all A , therefore A◦ =[A ϕ(X)]◦ . i ∈ A ∩ Let (x )i I aCauchyandTK convergent net in (E(X),τ). For all ∈ i − j 1, let xj X such that (xj)i I converges in (X, τj)toxj.τj is the ≥ ∈ Y ∈ polar topology of π ( ) convergence on X. Let A , there exists k0 I j A − ∈ A ∈ N such that for all r, s k xr xs,y 1 for all N 1andforall 0 ¯ j j j ¯ ≥ ¯j=1 − ¯ ≤ ≥ ¯X D E¯ ¯ ¯ r y A. There exists kj I¯ such that for all¯ r kj, xj xj,yj 1 ∈ ∈ ¯ ¯ ≥ − ≤ for all (y ) A. Let r =max k ,k ,...,k for all¯Dr r weE¯ have: n 0 0 1 N ¯ 0 ¯ N ∈ { } ¯ ≥ ¯ r r xj xj,yj max xj xj,yj 1forall(yn) A. ¯ − ¯ ≤ 1 j N − ≤ ∈ ¯j=1 D E¯ ≤ ≤ ¯D E¯ ¯X ¯ ¯ ¯ ¯ N ¯ ¯ ¯ ¯ s ¯ ¯ x xj,y¯ j 1 for all (yn) A and for all s r0;therefore ¯ j − ¯ ≤ ∈ ≥ ¯j=1 D E¯ s ¯X ¯ s s x ¯x [A ϕ(X)]◦¯for all s r0. Furthermore, x = x (x x) (X). −¯ ∈ ∩i ¯ ≥ − − ∈ A Therefore¯ (x )i I converges¯ to x in ( (X),τ ), and then x E(X)and i ∈ A A ∈ (x )i I converges to x in (E(X),τ). ∈ i i S.C.] Let (x )i I anetinE(X) which converges to x in ( (X),τ ). (x )i I ∈ A A ∈ is a Cauchy and TK convergent net in (E(X),τ)(τ = τ /E(X)), therefore i − A (x )i I converges to x in (E(X),τ). ∈ Polar topologies on sequence spaces in non-archimedean analysis 121

Lemma 2. Let L and M two K vector spaces, τ a topology on L, L π δ − −→ M L two linear maps such as πoδ = idM , and τδ the inverse image topology−→ of τ by δ on M. The application ψ :(M,τδ) (δ(M),τ),x δ(x), is an homeomor- phism. −→ −→

1 Proof. If is a F.S.N of 0 for τ;aF.S.Nof0forτδ is δ− ( )= δ 1(U)/U ,Uand we have: ψ 1(U δ(M)) = δ 1(U) for all U U . − ∈ U − ∩ − ∈ U © ª Theorem 11. Let τ a polar and solid topology of convergence on E(X); (E(X),τ) is complete if and only if: A− (i.) (X, τj) is complete for all j 1; ≥ (ii.) E(X) is a closed subspace of ( (X),τ ). A A X Proof. N.C.] δ is (τ,τj) closed for all j 1 (proposition 13), j − ≥ X X therefore δj (X) is a closed subspace of (E(X),τ), hence δj (X),τ is X complete. Now (δj (X),τ) (X, τj) (lemma 2), therefore (³X, τj)iscom-´ plete. Furthermore E(X) is' a closed subspace of ( (X),τ ) (theorem 10). i A A i S.C.] Let (x )i I aCauchynetin(E(X),τ). For j 1, (xj)i I is Cauchy ∈ i ≥ ∈ in (X, τj) so it converges, and then (x )i I is TK convergent in (E(X),τ) so it converges in (E(X),τ), (theorem 10).∈ −

Remark 2. We can replace (ii) of theorem 11 by: (ii) Any Cauchy TK convergent net in (E(X),τ) converges in (E(X),τ). − Corollary 8. Let τ a polar and solid topology of convergence on E(X). If E(X) is a closed subspace of ( (X),τ ); (E(XA),τ− ) is sequentially com- A A plete if and only if (X, τj) is sequentially complete for all j 1. ≥

Lemma 3. Let τ a vector topology on E(X); if τ is solid, SE(X) is the closure of ϕ(X) in (E(X),τ).

Proof. SE(X) ϕ(X). Let x =(xn) ϕ(X)andU a solid neighbor- ⊂ ∈ [i] hood of 0, it is z =(zn) ϕ(X)asx z U. Since U is solid x x U ∈[i] i − ∈ − ∈ for i large enough, then x →∞ x in (E(X),τ) and hence x S . −→ ∈ E(X) Proposition 21. Let τ a solid and polar topology of convergence on A− E(X); if (X, τj) is complete for all j 1, (S ,τ) is complete. ≥ E(X) 122 R.AmezianeHassani,A.ElamraniandM.Babahmed

Proof. SE(X) = ϕ(X) (lemma 3), therefore (SE(X),τ)isaclosed subspace of ( (X),τ ), andthen(SE(X),τ) is complete. Application:A LetA (X, . )an.a Banach space, we consider m(X)en- k k dowed with the n.a. norm . . We have c0(X)=Sm(X), and . deﬁnes k k∞ k k∞ a polar and solid topology on m(X), therefore (c0(X), . ) is complete. k k∞

Theorem 12. Let τ a solid and polar topology of convergence on A− E(X); if E(X) is an AK space, (E(X),τ) is complete if and only if (X, τj) is complete for all j 1.− ≥ Proof. N.C.] Obvious. S.C.] E(X)isanAK space, therefore E(X)=S(E(X),τ). Now S( (X),τ ) − A A ⊂ E(X) (proposition 20) and S(E(X),τ) S( (X),τ ) , therefore E(X)= ⊂ A A S(E(X),τ) = S( (X),τ ), and then E(X) is a closed subspace of ( (X),τ ). Hence (E(X),τA ) is completeA (theorem 11). A A

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[7] ;O.Toeplitz,LineareRaüme mit unendlich vielen Koordi- naten und Ringe unendlicher Matrizen, J. reine angew. Math. 171, pp. 193-226, (1934). [8] G. Matthews, Generalised Rings of infinite matrices, Ned. Akad. Wet. Proc. 61, pp. 298-306 (1958). [9] A.F.Monna, Analyse non-archimédienne, Springer-Verlag Berlin New York Heidelberg (1970). [10] H.H. Schaefer, Topological vector spaces, Springer-Verlag Berlin New york Heidlberg, (1971). [11] W. H. Schikhof, Locally convex spaces over nonspherically complete valued fieldI,II.Bull.Soc.Math.Belg.Sér. B. 38, pp. 187-224, (1986). [12] J. Van Tiel, Espaces localement K-convexes I-III, Indag. Math. 27, pp. 249-289 (1965). R. Ameziane Hassani Département de Mathématiques Faculté des Sciences Dhar El Mehraz Université Sidi Mohamed Ben Abdellah B. P. 1796 FES - MAROC e-mail : [email protected]

A. El Amrani Département de Mathématiques Faculté des Sciences Dhar El Mehraz Université Sidi Mohamed Ben Abdellah B. P. 1796, FES - MAROC e-mail : [email protected]

and

M. Babahmed Département de Mathématiques Faculté des Sciences de Meknès Université Moulay Ismail B. P. 11201 Zitoune MEKNES - MAROC e-mail : [email protected]