Absolute Boundedness and Absolute Convergence in Sequence Spaces* Martin Buntinas and Naza Tanović Miller

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1 Absolute Boudedess ad Absolute Covergece i Sequece Spaces* Marti Butias ad Naza Taović Miller 1 Itroductio We maily use stadard otatio as give i sectio 2 For a F K space E, various forms of sectioal boudedess ad sectioal covergece have bee show to be equivalet to ivariaces of the form E = D E with respect to coordiatewise multiplicatio by some space D Such statemets show the equivalece of topological properties of E with algebraic properties of E I 1968 Garlig [9] showed that a F K space E has the property of sectioal boudedess AB if ad oly if E is ivariat with respect to the space bv of sequeces of bouded variatio, ad that E has the property of sectioal covergece AK if ad oly if E = bv 0 E I 1970 Butias [4] showed that for a F K space E, Cesàro sectioal boudedess σb is equivalet to ivariace with respect to the space q of bouded quasicovex sequeces ad that Cesàro sectioal covergece σk is equivalet to ivariace with respect to the space q 0 = q c 0 of quasicovex ull sequeces I 1973 results were obtaied for more geeral Toeplitz sectios [5] I 1977, Sember [12] ad Sember ad Raphael [13] showed that for F K spaces, urestricted sectioal boudedess UAB is equivalet to ivariace with respect to the space c of coverget sequeces ad that urestricted sectioal covergece UAK is equivalet to ivariace with respect to the space c 0 of ull sequeces I this paper we study absolute boudedess AB ad absolute covergece AK These coditios are stroger tha UAB ad UAK, respectively However, for F K spaces, we show that the property AK is equivalet to UAK Amog other results, we show i sectio 3 that a F K space E has absolute boudedess if ad oly if it is solid (l ivariat) ad that it has absolute covergece if ad oly if it is c 0 ivariat The itersectio of all solid F K cotaiig a F K space E is called the solid hull of E I sectio 4 we show that it is a F K space ad characterize it as a F K product space We show that the solid hull of a F K space E is related by duality to the AB subspace of E I the last sectio, we give examples ad applicatios to summability theory ad Fourier aalysis * Research partially supported by US Yugoslav Joit Fud (NSF JF 803) 1

2 2 Defiitios Let ω be the space of all real or complex sequeces x = (x ) A F K space is a subspace of ω with a complete metrizable locally covex topology with cotiuous coordiate fuctioals f : x x for all A F K space whose topology is defied by a orm is a Baach space ad is called a BK space Let e be the sequece with 1 i the th coordiate ad 0 elsewhere ad let φ be the liear spa of {e 1, e 2, e 3, } I this paper we cosider oly F K ad BK spaces cotaiig φ, although all the defiitios apply to more geeral K spaces cotaiig φ We use the otatio x y := (x y ) for the coordiatewise product of sequeces x ad y ad, for subsets A ad B of ω, we use A B := {x y x A, y B} If D ω ad F is a F K space, we defie the F dual of D as the multiplier space D F = (D F ) := {y ω x y F for all x D} Let s := e = (1, 1,, 1, 0, ) ad =1 let e := (1, 1, 1, ) be the sequece of all oes The th sectio of a sequece x is s x := s x = (x 1, x 2,, x, 0, ) A sequece x i ω has the property AB of sectioal boudedess i a F K space E if the sectios s x of x form a bouded subset of E ad it has the property AK of sectioal covergece if, i additio, the sectios coverge to x i the topology of E Let H := {h ω h = 1 or h = 0 for all } ad H φ := H φ The ucoditioal (or urestricted ) sectios of a sequece x are the sequeces i the set H φ x The absolute set of x is H x Sice e H, we have x H x Let E be a F K space ad let x ω We say that x has the property UAB of ucoditioal sectioal boudedess i E if H φ x is a bouded subset of E ad we say that x has the property AB of absolute boudedess if H x is a bouded subset of E For each F K space E, we defie the space E AB cosistig of all elemets x of ω with the property AB i E Similarly, for the properties UAB ad AB, we obtai spaces E UAB ad E AB That is, E AB = {x ω {s x} is a bouded subset of E}, =1 E UAB = {x ω H φ x is a bouded subset of E}, E AB = {x ω H x is a bouded subset of E} Each of these spaces is a F K space uder a appropriate topology discussed i sectio 3 These spaces are ot ecessarily subspaces of E as is show by the example (c 0 ) UAB = (c 0 ) AB = l However, E AB is always a subspace of E sice e H We 2

3 say that a F K space E has the property AB, UAB or AB if E is a subset of E AB, E UAB or E AB, respectively Clearly we have E AB E UAB E AB The coverse iclusios do ot geerally hold For example, the BK space c of all coverget sequeces has the property UAB but it does ot have the property AB The set H φ is a directed set uder the relatio h h defied by h h for all A sequece x i a F K space E cotaiig φ has the property UAK i E if the et h x, where h rages over H φ, coverges to x uder the topology of E We say that x has the property AK of absolute sectioal covergece if H x E ad the et h h x, where h rages over H φ, coverges to h x uiformly i h H uder the topology of E We defie E AK to be the space of all elemets x of E with the property AK i E The same ca be doe for the properties UAK ad AK That is, E AK = {x E lim s x = x}, E UAK = {x E lim h h x = x, h H φ }, E AK = {x E lim h h h x = h x, uiformly i h H, for h H φ } The space E AD is the closure of φ i E Sice φ E, we have the iclusios φ E AK E UAK E AK E AD E If E AD = E, we say that E has the property of sectioal desity AD If y E wheever y x for some x E, we say that E is solid; this is equivalet to l ivariace: E = l E We fiish this sectio with a list of some BK spaces ad their orms The BK spaces l, c ad c 0 are the space of all bouded, coverget ad ull sequeces x, respectively, uder the sup orm x := sup x ; bv is the BK space of all sequeces x of bouded variatio uder the orm x bv := x x +1 + x ; bv 0 = bv c 0 uder the same orm; cs is the BK space of sequeces x with coverget p series uder the orm x bs := sup x ; l, for 1 p <, are the BK spaces of =1 ( ) 1 sequeces x with absolutely p summable series uder the orm x p := x p p ; the mixed l p,q spaces (1 p, 1 q < ) (see [11]) cosist of all x with 3 =1 =1

4 ( )1 x p,q := ( d j x p ) p q j=1 elsewhere; for q =, x p,q := sup j < where d j x = x for 2 j < 2 j+1 ad d j x = 0 d j x p Clearly l p,p = l p 3 Absolute boudedess ad absolute covergece The properties UAB ad UAK were ivestigated by Sember [12] ad Sember ad Raphael [13] The properties AB ad AK cosidered here are related Let E be a BK space uder the orm x E We defie the (exteded) absolute orm o E by x E := sup h x E h H with the covetio that h x E = wheever h x E Clearly x E x E ad E AB = {x E x E < } Similary, if E be a F K space with a icreasig family of semiorms p 1 p 2 p 3 defiig the topology of E, we defie the (exteded) absolute semiorms by p (x) := sup p (h x), = 1, 2, 3, E h H with the covetio that p (h x) = wheever h x E Clearly E AB = {x E p (x) < for = 1, 2, } By Garlig s Theorem [9], (p 998), E E AB is a F K space uder the semiorms p 1, p 1, E p2, p 2, Sice E p p for all, E p 1, p 2, p 3, may be omitted Hece the followig theorem ad corollary hold Theorem 1 Let E be a F K space with defiig semiorms p 1, p 2, p 3, The E AB is a F K space whose topology is defied by the semiorms p 1 E, p2 E, p3 E, Corollary If E is a BK space uder the orm E, the E AB uder the orm E is a BK space Remar It follows that a sequece x i a F K space E has the property AK if ad oly if the et H φ x coverges to x i the topology of E AB Theorem 2 Let E be a F K space cotaiig φ The the followig statemets are equivalet: [a] x E AB ; [b] H x E ; 4

5 [c] l x E Proof We have [a] [b] by defiitio of AB Suppose [b] The H ({x} E) The multiplier space ({x} E) i a F K space [10], (p 229) By Beett ad Kalto [3], H is a subset of a F K space if ad oly if l is a subset Thus l ({x} E), or l x E Thus [b] [c] Fially suppose [c] Let T x be the multiplier map from l to E defied by T x (y) = x y By the Closed Graph Theorem, all multiplier maps betwee F K spaces are cotiuous [20] Sice H is a bouded subset of l, T x (H) = x H is a bouded subset of E Thus x has the property AB i E Sice e l, we have (l E) E Aderso ad Shields [1] have observed that (l E) is the largest solid subspace of E By [c] above we obtai the followig Corollary 1 Let E be a F K space cotaiig φ The E AB = (l E) This is the largest solid subspace of E Corollary 2 A F K space has the property AB if ad oly if it is solid Corollary 3 For ay F K space E, E UAB is solid Proof If E is a F K space uder the semiorms p, = 1, 2, 3,, the E UAB is a F K space uder the semiorms sup p (h x) Sice H φ = H φ H φ = H φ H, h H φ it follows that H φ x is a bouded subset of E if ad oly if H φ x is a bouded subset of E UAB This is true if ad oly if H x is a bouded subset of E UAB This is, E UAB = (E UAB ) UAB = (E UAB ) AB, which is solid The space E UAB ca be characterized as follows Theorem 3 Let E be a F K space cotaiig φ The E UAB = (c 0 E) Proof By Theorem 4 of [12] c 0 E UAB E Thus E UAB (c 0 E) Coversely, suppose c 0 x E Defie the map T x : c 0 E by T x (y) = x y Sice T x is cotiuous, T x taes bouded subsets of c 0 ito bouded subsets of E Let U be the uit sphere of c 0 The U x is bouded i E By Theorem 3 of [12], we have x E UAB 5

6 Corollary Let E be a F K space cotaiig φ The E E UAB = (c E) I the same way, we ca use the results i [9] to show that E AB = (bv 0 E) ad E E AB = (bv E) Although E UAB is solid, E UAB E eed ot be solid This is the case whe E = c Thus E AB is geerally a proper subspace of E UAB E Also if E E UAB, the space E UAB eed ot be the smallest solid space cotaiig E The space (c 0 ) UAB = l provides a example Theorem 4 Let E be a F K space If E UAB E, the E AB = E UAB Proof Clearly E AB E UAB By Theorem 2 Corollary 3, E UAB is the largest solid subspace of E, the statemet follows is solid Sice E AB Theorem 5 Let E be a F K space cotaiig φ The the followig statemets are equivalet: [a] E is solid ad has the property AD ; [b] E is solid ad has the property AK ; [c] E = c 0 E ; [d] E has the property UAK ; [e] E has the property AK Proof The equivalece of [a], [b] ad [c] was proved by Garlig [9], (p 1007) [b] [e]: Fremli ad Garlig showed that a solid F K space E is locally solid [9], (p 1006); that is, the topology of E is defied by semiorms p with the property p(d x) p(x) for all sequeces d i the uit sphere of l Let p be such a cotiuous semiorm, let x E ad let ɛ > 0 The p E = p Suppose p(s x x) < ɛ 2 ad let h H such that h s The s = h s Hece p E (h x x) p E (h x s x)+p E (s x x) = p E (h x h s x) + p E (s x x) = p(h (x s x)) + p(s x x) 2p(s x x) < ɛ This shows that the et H x coverges to x uder the topology of E AB [e] [d] is immediate from the defiitios [d] [b]: The property UAK clearly implies AK sice s H φ for all Sember ad Raphael [13], (Corollary 32) have show that is solid Sice E = E UAK, it follows that E is solid E UAK 6

7 Remar A BK space has the property AK if ad oly if, for all x E, s x x E 0 (as ) Moreover, i this case x E = sup s x E A similar statemet ca be made about F K spaces Theorem 6 If E is a F K space cotaiig φ ad E AD is solid, the E AD = E AK = E UAK = E AK = c 0 E Proof Clearly E AK E UAK E AK E AD Sice E AD is a closed subspace of E, it is a F K space uder the subspace topology Hece E AK = (E AD ) AK By Theorem 5 ([a] [e]) we have (E AD ) AK = E AD Corollary 1 If E is a F K space cotaiig φ with the property UAB, the E AD = E AK = E UAK = E AK = c 0 E Proof If E has the property UAB, the by Sember ad Raphael [13], (Theorem 4) E AD = E UAK = c 0 E Thus E AD is solid ad satisfies the coditios of Theorem 6 If E is solid, the E has the property UAB sice E = E AB E UAB Corollary 2 If E is a solid F K space cotaiig φ, the E AD = E AK = E UAK = E AK = c 0 E Corollary 3 For ay F K space E cotaiig φ, E AK = c 0 E AB Proof By defiitio E AK = (E AB ) UAK Sice E AB = c 0 E AB by Corollary 2 is solid, we have (E AB ) UAK Theorem 7 Let E be a F K space The E UAK = E AK Proof Let x E The statemet x E UAK meas that the et H φ x coverges to x i the topology of E The statemet x E AK meas that the et H φ x coverges to x i the topology of E AB ; that is, E AK = (E AB ) UAK Also E UAK = c 0 E UAB = c 0 (E UAB ) UAB = (E UAB ) UAK It remais to be show that (E AB ) UAK = (E UAB ) UAK, which by Theorem 6 Corollary 2 is equivalet to (E AB ) AK = (E UAB ) AK Sice E AB E UAB, we have the iclusio (E AB ) AK (E UAB ) AK Coversely suppose x (E UAB ) AK The for each cotiuous semiorm p o E, sup p(h (s x h H φ s m x)) 0 as, m Sice s x s m x φ, we have sup p(h (s x s m x)) 0 h H as, m ; that is, x (E AB ) AK 7

8 4 The solid hull of a F K space For a F K space E, the solid hull E is the itersectio of all solid F K spaces cotaiig E It is clearly solid The solid hull was ivestigated by Aderso ad Shields i [1] We show that the solid hull is a F K product space ad we fid a dual relatioship betwee E AB ad E The F K product E F of two F K spaces E ad F was defied i [6] ad [7] ad was characterized as the smallest F K space cotaiig the coordiate product E F If E ad F are BK spaces, the E F turs out to be a BK space Theorem 8 Let E be a F K space The solid hull of E is the F K space l E Proof If F is a solid F K space cotaiig E, the F = l F l E E Thus F l E E But l E is itself solid, sice l (l E) = (l l ) E = l E Thus it is the smallest solid F K space cotaiig E Theorem 43 of [7] states that (E F ) AD = (E F ) AK = E AK E E AB We obtai the followig F, wheever Corollary Let E be a F K space cotaiig φ The (l E) AD = (l E) AK = c 0 E From this Corollary we see that if E is solid ad has the property AD, the E = c 0 E This is Theorem 5 ([a] [c]) Other parts of Theorems 5 ad 6 ca also be obtaied The ext theorem exhibits a dual relatioship betwee the space E AB solid hull E ad the Theorem 9 Let E ad F be F K spaces The ( E F ) = (E F ) AB That is, the F dual of the solid hull of E is the largest solid subspace of the F dual of E Proof By Theorem 2 Corollary 1, (E F ) AB = (l (E F )) By (56) of [7], (l (E F )) = ((l E) F ) For example, let the α ad β duals of E be defied by E α = (E l 1 ) ad E β = (E cs), respectively We have E α = (E α ) AB = E α Also (E β ) AB = (l E β ) = (E ββ l 1 ) = E ββα = E α ([7], Theorem (51); [8], Theorem 1) Similarly ( E ) β = E α 8

9 Corollary For ay F K space E, we have E αα = E AB = E UAB Proof By Theorem 9 we have E α = (E α ) AB Also (E α ) AB = E α sice E α is solid Thus E αα = E αα But E αα = E AB by [ 5 ], Theorem 4 ad [ 8 ], Remar (6) As oted after Theorem 3, E AB = (bv 0 E ) This is a subset of (c 0 E ), which is E UAB by Theorem 3 That is, E AB E UAB, ad thus E AB = E UAB 5 Examples ad applicatios The properties AB ad AK are strog properties of F K spaces We have the followig list l AK = c 0, l AB = l = l ; c AK = c AB = c 0, c = l ; cs AK = cs AB = l 1, cs = c 0 ; bv AK = bv AB = l 1, bv = l ; l p AK = l p AB = l p = l p (1 p < ) ; l p,q AK = l p,q AB = l p,q = l p,q (1 p <, 1 q ) Give a ifiite matrix T = (t ) of real or complex umbers, let c T deote the covergece field of T, ie, c T = {x ω : T x c} By the above list we have cs AB = l 1 We will ow show that this ca be exteded to covergece fields c T of all series sequece regular matrices T (ie, c T cs ad lim t x = x for all x cs ) Theorem 10 If T is a matrix with lim t = 1 for each, the (c T ) AB l 1 Proof The space c T is a F K space uder the semiorms p (x) = x ( = 1, 2, 3, ), q m (x) = sup t x ( = 1, 2, 3, ) ad r(x) = sup t x, m =1 [ 20 ], [ 10 ] By Theorem 1, (c T ) AB is a F K space uder the semiorms p ct, q, ad r c T c T Let x = sup t x It ca be easily verified that (c T ) AB is a BK space uder the orm sice 2r ct r ct = sup Furthermore x x for all x (c T ) AB q c T sup p c T 9

10 Every series sequece regular matrix T satisfies the coditios of Theorem 10 This ca be show by cosiderig the sequeces e, = 1, 2, 3, Corollary If T is a series sequece regular matrix, the (c T ) AB = l 1 We ow apply the cocepts cosidered i this paper to the spaces of Fourier coefficiets of some classes of fuctios Let L p (p 1) be the Baach space of all real or complex valued 2π periodic itegrable fuctios with the orm f L p = ( ) 1 2π f p 1 p, where the itegral is tae over ay iterval of legth 2π Let C be the Baach space of all cotiuous real or complex valued 2π periodic fuctios with the orm f C = sup f(x) x If f L 1, let f(),, deote the th complex Fourier coefficiet of f, f = ( f()) ad let s f, = 0, 1, deote the th partial sum of the Fourier series of f If E is a subspace of L 1, let Ê deote the class of all sequeces of Fourier coefficiets of fuctios i E, ie, Ê = { f : f E} Although the results i the precedig sectios are for spaces of oe way sequeces, they ca be easily exteded to the classes Ê of two way sequeces If E is a Baach space the Ê is a Baach space uder the iduced orm f Ê := f E ad coversely Give a Baach space E cotaied i L 1 we ca determie the correspodig subspaces of absolutely bouded ad absolutely coverget Fourier series, i the topology of E, by determiig the spaces Ê AB ad Ê AK We shall also cosider the correspodig solid hull Ê Two classical spaces of fuctios i Fourier aalysis, determied by the poitwise covergece, ordiary I ad absolute I, are the spaces of uiformly ad absolutely coverget Fourier series: U = {f C : s f f I uiformly} ad A = {f C : s f f I ae} They are Baach spaces, uder the orms: f U := sup s f C ad f A := f() = f l 1 It is well ow that A U C L properly, where L is the correspodig space of essetially bouded measurable fuctios We shall also cosider the Baach space M of 2π periodic Rado measures, uder the orm f M = sup 1 +1 s f L 1 10 =0

11 For the spaces E = L p (p 1) ad L, the questios of determiig the largest solid space cotaied i Ê, ad the smallest solid space cotaiig Ê, have already bee cosidered i [ 1 ] Slightly expadig those results i view of the cocepts of this paper, we ca write the followig theorem where the stadard sequece spaces are to be iterpreted as the spaces of two way sequeces Theorem 11 [a] If E is a Baach space ad L 2 E L 1, the Ê AK = Ê AB = l2 ad l 2 = L 2 Ê L 1 = c 0 Moreover if 1 < p 2, the L p l q,2 where 1/p + 1/q = 1 [b] If p > 2 ad 1/p + 1/q = 1, the l q,2 L p AK ad L p = l 2 [c] If E is a Baach space ad A E L, the Ê AK = Ê AB = l1 ad l 1 = Â Ê L l 2 [d] M AB = L 1 = AB l2 ad M = l Proof [a]: It was poited out i [ 1 ] that L 1 AB = l2 = L 2 AB Thus Ê AB = l 2 ad sice l 2 has AD, by Theorem 5 we have that Ê AK = Ê AB The correspodig statemets about solid hulls were discussed i [ 1 ], ad the last statemet is a corollary of a result i [ 11 ] [b]: The iclusio follows from [ 11 ] ad the equality L p = l 2 for p > 2 was also discussed i [ 1 ] [c]: Sice  = l1 is solid,  AB = l 1 The equality L = AB l1 was explaied i [ 1 ] Thus Ê AB = l1 ad by Theorem 5, Ê AK = Ê AB The iclusio about solid hulls is obvious [d]: Clearly L 1 M AB AB ad by [a] L1 = AB l2 Hece l 2 M AB Coversely, let f M AB The f <, ie, sup 1 M +1 s f < Sice =0 L 1 L 1 = l 2 we have sup 1 AB +1 s f < ad therefore 2 l 1 2 sup ( [/2] =0 f() ) ( sup ( 1 +1) f() 2 )12 < Thus f l 2 This proves the first iequality To show that M = l, we first ote that e M, so that e l = l M But M l implies M l Thus M = l 11

12 Corollary  AB = Û AB = Ĉ AB = L = AB l1 =  ad the same strig of equalities is true for AK We cosider ow some ewer classes of fuctios itroduced i Fourier aalysis They are determied by other types of poitwise covergece, amely strog covergece of idex p 1, [ I ] p, ad absolute covergece of idex p 1, I p The latter exteds the cocept of absolute covergece I i the sese that a sequece s s I p if ad oly if s s I ad p 1 s s 1 p < The strog covergece [ I ] p lies betwee the absolute I p ad the ordiary covergece I, that is, I p [ I ] p I, see [ 14 ] or [ 16 ] ad the refereces cited there These otios were applied to trigoometric ad Fourier series i a series of recet papers, [ 16 ] through [ 18 ], which led to the study of the related spaces of fuctios, [ 14 ], [ 15 ], [ 19 ] : S p = {f L 1 : s f f [ I ] p ae}, p = {f C : s f f [ I ] p uiformly}, A p = {f L 1 : s f f I p ae}, A p = {f C : s f f I p uiformly} For p = 1 we write simply S,, A ad A, respectively They have may iterestig properties: p S p L r properly, but S p L ; the classes p ad S p decrease 1<r< with p icreasig while the classes A p are icomparable ad the same is true for A p ; A = A ; A p p U ad A p S p L p properly From the results i [ 14 ], [ 15 ] ad [ 17 ] they ca be described as follows { For p 1, let s p 1 := x : 2+1 L 1 s 1 ad Ŝ s1 properly, Ŝ p = s p = {x : p 1 } p x p = o(1) ( ) The Ŝ = p > 1, ad p = Ĉ Ŝp for p 1 For p 1,  p = a p := } x p = o(1) ( ) { x : for } p 1 x p < ad Âp = Ĉ Âp They are Baach spaces uder the correspodig orms: f S = f L 1 + f [1] ; f S p = f [p] for p > 1 ; f p = f U + f [p] for p 1 ; f A p = f p ad f A p = f U + f p for p 1, where f [p] = sup ( ( + 1) p f() p )1 p ad 12

13 f p = ( f(0) p +, 0 p 1 f() p )1 p Theorem 12 [a] Ŝ p AK = Ŝp AB = sp = Ŝp = Ŝp for p > 1 [b] Ŝ AK = Ŝ AB = l2 s 1 ( l 2 ad 1 are icomparable) [c] p AK = p AB = l1 s p (l 1 s p except for p = 1 ad s p l 1 ) [d] p = s p = Ŝp for p > 1 ad l 2 s 1 = Ŝ AB Proof [a]: By the above remars Ŝp = s p for p > 1 Sice s p is solid ad has the property AD, the statemet follows from Theorem 5 [b]: Ŝ = L 1 s 1 ad cosequetly Ŝ AB L 1 AB s1 sice s 1 is solid By Theorem 11 [a], L 1 = AB l2 ad therefore Ŝ AB l2 s 1 Coversely, l 2 s 1 L 1 AB s1 Ŝ AB Moreover, by Theorem 5, Ŝ AK = Ŝ AB [c]: By the above remars p = C S p ad by the Corollary of Theorem 11 Ĉ AB = l1 Hece by statemet [a], p AB l1 s p ad coversely l 1 s p Ĉ AB Ŝp p AB The equality p = p is clear by Theorem 5 AK AB [d]: Sice p = C S p, clearly p Ŝp = s p AB for all p 1 To show the coverse iclusio for p > 1 we refer to a result due to Salem ( [ 2 ], vol 1, p 335), otig first that x s p, implies that =2 1 ( x i 2)1 2 log i < ( ) Sice s p s 2 for p > 2 it clearly suffices to assume that 1 < p 2 Taig 1 < p 2 for x s p we have ( i x i 2)1 2 ( i x i p)1 p = O ( 1 1/q ) where 1/p + 1/q = 1, from which it follows that ( ) is satisfied We ow show that s p l Ŝp For x s p let x = x r + x l where x r = x for 0, x r = 0 for < 0 ad xl = 0 for 0, x l = x for < 0 By the above argumet, both x r ad x l satisfy ( ) Cosequetly by Salem s theorem, there exists a sequece (α r ) =0 x r cos (t α) r =0 13 such that the series

14 coverges uiformly ad is therefore the Fourier series of its sum fuctio g C Hece ĝ c () = x r cos αr ad ĝ s() = x r si αr Expressed i the complex form, g is the uiform sum of the series ĝ() e it where ĝ() = 1 2 xr e iαr, ĝ( ) = 1 2 xr e iαr Cosequetly, defiig a two way sequece y, by y = 2e iαr for 0 ad y = 0 for < 0, we have x r = y ĝ where y l But clearly ĝ s p = Ŝp ad therefore x r = y ĝ l S p I the same way we ca show that x l l Ŝp Cosequetly x l Ŝp Ŝp The correspodig properties for the spaces Âp ad that a p s p so that x a p implies ( ) Âp are proved similarly, otig Theorem 13 [a]  p AK = Âp AB = ap = Âp = Âp for p 1 [b]  p AK = Âp AB = l1 a p for p 1 [c] Âp = a p = Âp for p > 1 ad  = l1 REFERENCES 1 J M Aderso ad A L Shields, Coefficiet multipliers of Bloch fuctios, Tras Amer Math Soc 224 (1976), N Bary, A treatise o trigoometric series, vols 1 ad 2, Pergamo Press, New Yor, G Beett ad N J Kalto, Iclusio theorems for K spaces, Ca J Math 25 (1973), M Butias, Coverget ad bouded Cesàro sectios i FK spaces, Math Z 121 (1971), M Butias, O Toeplitz sectios i sequece spaces, Math Proc Cambridge Philos Soc 78 (1975), M Butias, Products of sequece spaces, Aalysis 7 (1987), M Butias ad G Goes, Products of sequece spaces ad multipliers, Radovi Mat 3 (1987),

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