A generalization of convergent series
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1 Als of the Uiversity of Buchrest (mthemticl series) (Alele Uiversităţii Bucureşti. Mtemtică) 1 (LIX) (2010), A geerliztio of coverget series Io Chiţescu To Professor Io Colojoră o the occsio of his 80th birthdy Abstrct - The otio of coverget series is geerlized, usig fuctiol lysis techiques. The cocept thus obtied, clled here S-coverget series, is lyzed d illustrted by mes of my exmples. Key words d phrses : coverget series, Bch spce, lier d cotiuous opertor, Riesz spce, L p -spce, Bch lgebr, spectrum of opertor, projectio. Mthemtics Subject Clssifictio (2000) : 40A05, 46B25, 46B40, 46B45, 46H10, 47A Itroductio Throughout the pper R will deote the set of rel umbers, C the set of complex umbers d N the set of turl umbers. We shll write K to deote R or C. For y two ormed spces (X, ), (Y, ) d y lier d cotiuous opertor T : X Y, the (opertor) orm of T will be T o = sup{ T (x) x X, x 1}. We shll be cocered with the followig lier spces: 1) The spce of sequeces which coverge to zero: c 0 = {x = (x 0, x 1,..., x,...) x K, x 0} which is Bch spce whe equipped with the usul orm x = sup x. 2) The spce of coverget series. This is the subspce of c 0 defied s follows: { } cc 0 = x = (x 0, x 1,..., x,...) c 0 x is coverget. 47 =0
2 48 Io Chiţescu 3) The spce of bsolutely coverget series (the spce of summble sequeces). This is the subspce of c 0 defied s follows: l 1 = { x = (x 0, x 1,..., x,...) c 0 =0 } x is coverget. 4) The spce of cotiuous fuctios. This is the spce defied s follows: Oe cosiders two rel umbers, b such tht < b. We obti the spce C[, b] = {f : [, b] K f is cotiuous} which becomes Bch spce whe equipped with the orm f = sup{ f(t) t [, b]}. 5) The spce L p (µ) which is defied s follows: Oe cosiders umber p such tht 1 p <. Let us deote by µ the Lebesgue mesure o [0, 1]. We obti the vector spce L p (µ) = {f : [0, 1] K f is µ-mesurble d f p is µ-itegrble} which is semiormed with the semiorm ( N p (f) = ) 1 f p p dµ. The ull spce of L p (µ) is N (µ) = {f L p (µ) N p (f) = 0} = {f : [0, 1] K f(t) = 0 µ-.e.}. The quotiet spce L p (µ) def = L p (µ)/n (µ) is Bch spce, whe equipped with the (quotiet) orm f p def = N p (f) for y represettive f f L p (µ). For geerl Alysis see [7]. For geerl Fuctiol Alysis see [3], [4], [6] d [8]. For Mesure Theory see [1], [2] d [5]. For Spectrl Theory see [3] d [6].
3 A geerliztio of coverget series Results We strt with remrk which motivtes the subsequet fcts. Let us cosider the umericl series x (2.1) =0 with terms i K = R or C. Cuchy s criterio sserts tht the covergece of (2.1) is equivlet to the followig fct: for y ε > 0, there exists turl p(ε) such tht for y turl p p(ε) d y turl, oe hs x p + x p x p+ < ε. (2.2) I order to give ltertive expressio for (2.2), we itroduce the shift opertor S : c 0 c 0 (which is lier, cotiuous d S o = 1) give vi S((x 0, x 1, x 2,..., x,...)) = (x 1, x 2,..., x,...). Let us write S m = S S... S (m times) d S 0 = I = the idetity opertor of c 0. Now, (2.2) becomes: for y ε > 0, there exists turl p(ε) such tht for y turl p p(ε) d y turl, oe hs S p (x ) < ε. (2.3) Here, for x = (x 0, x 1, x 2,..., x,...) c 0 from (2.1), we write x = x + S(x) + S 2 (x) S (x), x 0 = x. Tkig ito ccout the structure of (2.2) d (2.3), we hve the followig result for x c 0 : x cc 0 p S p (x ) = 0 uiformly with respect to N. Hvig this i mid, we shll (from ow o) cosider Bch spce over K (rel or complex) equipped with the orm x x. We shll lso cosider lier d cotiuous opertor S : X X d I : X X is the idetity opertor. For elemet x X, we shll write x 0 = x d, for turl 1: x = x + S(x) + S 2 (x) S (x) = (I + S + S S )(x), where S = S S... S ( times), S 0 = I.
4 50 Io Chiţescu Defiitio 2.1. A elemet x X is clled S-coverget series if it hs the property tht p S p (x ) = 0 (2.4) uiformly with respect to N. Here (2.4) mes: for y ε > 0, there exists turl p(ε), such tht for y turl p p(ε) d y turl, oe hs S p (x ) < ε. The set of ll S-coverget series will be deoted by C(S). It is esily see tht C(S) is lier subspce of X d C(S) Ker(S). Theorem 2.1. For elemet x X, the followig ssertios re equivlet: 1. Oe hs x C(S). 2. The series S p (x) is coverget (i X). p=0 Proof. The series i the sttemet is coverget if d oly if the sequece ( m ) S p (x) is Cuchy. This is equivlet to the followig fct: for y p=0 m ε > 0, there exists p(ε) N, such tht, for y turl p p(ε) d y turl 1, oe hs p+ S m (x) m=0 p S m (x) = S p+1 (x 1 ) < ε. m=0 The lst ssertio is precisely the ssertio tht p S p (x ) = 0 uiformly with respect to N. Usig Theorem 2.1 we get Propositio 2.1. For elemet x X, the followig ssertios re equivlet: 1. Oe hs x C(S). 2. There exists turl p such tht S p (x) C(S). 3. For y turl p oe hs S p (x) C(S). Cosequetly, C(S) is lier subspce of X, which is ivrit with respect to S (d, of course, Ker(S) C(S)). We shll see (Exmple 3.1) tht, geerlly spekig, C(S) c be ot closed. How lrge is C(S)? The followig theorem swers this questio.
5 A geerliztio of coverget series 51 Theorem 2.2. Either C(S) = X or C(S) is of the first Bire ctegory (i.e. C(S) is meger). Proof. Let us ssume tht C(S) X. We must prove tht C(S) is of the first ctegory. Acceptig the cotrry, we shll rrive t cotrdictio. So let us ssume tht C(S) is of the secod ctegory. We costruct the sequece of lier d cotiuous opertors (T ) give vi T = I + S + S S with T 0 = I. Accordig to the defiitio of C(S), the sequece (T ) coverges poitwise o C(S) which is of the secod ctegory. Usig the Bch- Steihus theorem we get the fct tht the sequece (T ) coverges poitwise o ll of X d the poitwise it is lier d cotiuous opertor (see [8], p. 84 Corollry d p. 86 Corollry or see [3]). I other words, the series S (x) coverges for ll x X, hece C(S) = X, cotrdictio. =0 I order to cotiue, we defie the lier mp T S : C(S) X, give vi T S (x) = S (x). =0 Theorem The mp T S is ijective. 2. For y x C(S), oe hs Cosequetly (I S)(T S (x)) = x. C(S) = (I S)(T S (C(S))) (I S)(X). 3. The followig ssertios re equivlet: ) Oe hs (I S)(X) = C(S). b) For y x X there exists the it S (x). I this cse, for y x X oe hs T S ((I S)(x)) = x S (x). 4. The followig ssertios re equivlet: ) The mp T S is bijectio.
6 52 Io Chiţescu b) For y x X oe hs S (x) = 0. I this cse, cosiderig the mp (I S) 1 : X C(S), give vi (see 3.) (I S) 1 (x) = (I S)(x), oe hs Proof. So 1. Let x C(S) such tht (I S) 1 = T 1 S. T S (x) = 0. 0 = T S (x) = x+s(x+s(x)+s 2 (x)+...+s (x)+...) = x+s(t S (x)) = x+0, hece hece hece 2. Let x C(S). Agi x = 0. T S (x) = x + S(T S (x)), x = T S (x) S(T S (x)) = (I S)(T S (x)). We proved tht y x C(S) hs the form x = (I S)(T S (x)), C(S) (I S)(T S (C(S))). Coversely, let y (I S)(T S (C(S))). We c fid x C(S) such tht d the iclusio y = (I S)(T S (x)) = x (I S)(T S (C(S))) C(S) is lso true. 3. ) b). I view of 2., hypothesis ) mes tht (I S)(X) C(S). Let x X be rbitrry. Becuse (I S)(x) C(S), we c compute T S (x S(x)) = [ (x S(x))+S(x S(x))+S 2 (x S(x))+...+S (x S(x)) ] =
7 A geerliztio of coverget series 53 = (x S +1 (x)), hece S (x) exists d oe hs b) ). We must prove tht T S ((I S)(x)) = x S (x). (2.5) (I S)(X) C(S). Let us choose x X. Usig the hypothesis, there exists ) (x S +1 (x). But we hve see tht x S +1 (x) = (x S(x)) + S(x S(x)) + S 2 (x S(x)) S (x S(x)) which implies tht the series is coverget d so S (x S(x)) =0 x S(x) C(S). The equlity i the sttemet is give by (2.5). 4. ) b). Let y X. We must prove tht p S p (y) = 0. To this ed, let us tke rbitrry ε > 0. Becuse T S is surjective, we fid x C(S) such tht y = T S (x). Hece, for y p N, oe hs S p (y) = S p (T S (x)) = (S p (x) + S p+1 (x) S p+ (x)). Becuse x C(S), the series S (x) =0 is coverget. Usig Cuchy s criterio, we c fid turl p(ε) such tht, for y turl p p(ε) d y turl, oe hs S p (x) + S p+1 (x) S p+ (x) < ε.
8 54 Io Chiţescu Let us fix such p p(ε). It is possible to pss to -it i the lst iequlity d we fid S p (y) = S p (x) + S p+1 (x) S p+ (x) ε provig tht p S p (y) = 0. d b) ). Oe must prove tht T S (which is ijective, see 1.) is surjective. Let x X. Accordig to 3. oe hs Agi 3. sys tht Hece (I S)(X) = C(S) y = (I S)(x) C(S). T S (y) = T S ((I S)(x)) = x S (x) = x. x = T S (y) d T S is surjective. Now, cceptig tht T S is bijective, we use gi 3. d otice tht, for y x X, oe hs T S (I S) 1 (x) = x S (x) = x. O the other hd, we c rephrse 2. s follows: for y x C(S), oe hs (I S) 1 T S (x) = x. The lst two equlities show tht (I S) 1 = T 1 S. The results obtied t poit 4. c be rephrsed s follows (we work for K = C): Corollry 2.1. Assume tht for y x S oe hs S (x) = 0. The, we hve three possibilities: 1. If C(S) = X (i.e. (I S)(X) = X) it follows tht 1 is i the resolvet set of S (d T S = (I S) 1 ). 2. If C(S) X, but C(S) = X, it follows tht 1 is i the cotiuous spectrum of S (d the mp T S : C(S) X is ot cotiuous). 3. If C(S) X d C(S) X, it follows tht 1 is i the residul spectrum of S.
9 A geerliztio of coverget series 55 Usig the previous result for opertors o fiite dimesiol spces, we obti Corollry 2.2. Let us ssume tht X is fiite dimesiol vector spce over C. Let lso S : X X be lier opertor. The followig ssertios re equivlet: 1) Oe hs S (x) = 0, for ll x X. 2) For y x X, the series is coverget. S (x) =0 Proof. It is cler tht 2) 1). I order to prove 1) 2) we use Corollry 2.1. Possibilities 2. d 3. cot hppe, becuse the cotiuous spectrum d the residul spectrum of S re empty. It follows tht possibility 1 is vlid, cosequetly C(S) = X, which is precisely 2). Before pssig further, we thik it will be useful to see how Theorem 2.3 works i prticulr cse. Exmple 2.1. We cosider o ull Bch spce X d umber α K. The opertor S : X X will be give vi ) It is see tht S(x) = αx. C(S) = {x X there exists (1 + α + α α )x}. Hece C(S) = b) T S : C(S) X is give vi The equlity (vlid for x C(S)) { {0}, if α 1 X, if α < 1. { 0, if α 1 T S (x) = 1. 1 αx, if α < 1 (I S)(T S (x)) = x
10 56 Io Chiţescu is redily checked. c) For y α K, oe hs (I S)(X) = (1 α)(x) C(S). For α < 1, oe hs (1 α)(x) = X. For α 1, α 1, oe hs (1 α)(x) = X. Filly, for α = 1, oe hs C(S) = {0} d (1 α)(x) = {0}. d) For y x X d N, oe hs S (x) = α x. The it α x exists for y x X if d oly if α < 1 or α = 1. (1 α)(x) = C(S) i ll these cses: It is see tht (1 α)(x) = { X = C(S), if α < 1 {0} = C(S), if α = 1. The equlity (for α < 1 or α = 1) T S ((I S)(x)) = x S (x), vlid for ll x X, is redily checked. Nmely, i cse α < 1 oe hs T S ((I S)(x)) = 1 1 α (1 α)x = x = x α x d i cse α = 1 oe hs T S ((I S)(x)) = T S (x x) = T S (0) = 0 = x 1 x. I cse α = 1 oe c see tht T S : {0} X is ot bijectio. e) Oe hs S (x) = 0 for y x X if d oly if α < 1. I this cse C(S) = X d T S : X X is the bijectio give vi T S (x) = 1 1 α x. I the rest of the prgrph we shll work for K = R d we shll mke the supplemetry ssumptios tht X is Riesz spce d S is positive opertor. Oe c see tht the rel spce c 0 d the shift opertor S : c 0 c 0 fulfill these ssumptios.
11 A geerliztio of coverget series 57 Defiitio 2.2. A elemet x X is clled S-bsolutely coverget series i cse x C(S). The set of S-bsolutely coverget series will be deoted by AC(S). Theorem 2.4. The set AC(S) is lier subspce of C(S) which is ivrit with respect to S. Geerlly spekig, the iclusio AC(S) C(S) is strict d AC(S) my ot be closed (see Exmple 3.1). Proof. ) We prove the iclusio AC(S) C(S). Let x AC(S). For rbitrry ε > 0, we c fid turl p(ε) such tht, for y turl p p(ε) d y turl, oe hs S p ( x ) < ε. We hve successively, for turl p d : S p (x ) = S p (x + S(x) + S 2 (x) S (x)) S p ( x + S(x) + S 2 (x) S (x) ) S p ( x + S( x ) + S 2 ( x ) S ( x )) = S p ( x ). Hece, for p p(ε) d N we get S p (x ) = S p (x ) S p ( x ) < ε. b) It is immedite tht x AC(S) S p (x) AC(S) for y p N, becuse, if q N, oe hs S q ( S p (x) ) S q (S p ( x )) = S q+p ( x ) which implies S q ( S p (x) ) + S q+1 ( S p (x) ) S q+ ( S p (x) ) S p+q ( x ) + S p+q+1 ( x ) S p+q+ ( x ). c) If x, y re i AC(S), we shll prove tht x + y AC(S). Ideed, for y turl : x + y = x + y + S( x + y ) + S 2 ( x + y ) S ( x + y ) x + y +S( x )+S( y )+S 2 ( x )+S 2 ( y )+...+S ( x )+S ( y ) = x + y.
12 58 Io Chiţescu This implies, for turl p d : S p ( x + y ) S p ( x ) + S p ( y ).s.o. Now, if x is i AC(S) d α R, we shll prove tht αx AC(S). Ideed, for y turl : d for y turl p d : αx = α x S p ( αx ) = α S p ( x ).s.o. Theorem 2.5. (Geerlized D Alembert Criterio) Let x X. Assume there exists 0 < < 1 such tht there exists M N hvig the property S +1 (x) S (x) for ll M. The x C(S). Proof. Let us tke M. The we hve successively: S +1 (x) S (x) S +1 (x) S (x) S +2 (x) S +1 (x) 2 S (x) S +2 (x) 2 S (x) Hece, oe hs... S +p (x) p S (x) S +p (x) p S (x). S (x) + S +1 (x) + S +2 (x) S +p (x) ( p ) S (x) d the series S (x) coverges bsolutely. =0 I order to itroduce the ext results, it will be ecessry to supplemet our ssumptios with the followig oes: A1. The spce X is lgebr, with multiplictio (x, y) xy d with the property tht, for y x, y i X oe hs xy x y
13 A geerliztio of coverget series 59 d This implies hece X is Bch lgebr. xy y x. xy x y, A2. The opertor S : X X is multiplictive, i.e. S(xy) = S(x)S(y) for y x, y i X. Cosequetly, S is lgebr morphism. A3. For y x X oe hs S (x) = 0. Oe c see tht the rel spce X = c 0, with turl multiplictio ((x 0, x 1,..., x,..), (y 0, y 1,..., y,..)) (x 0 y 0, x 1 y 1,..., x y,..) d the shift opertor S : c 0 c 0 stisfy ssumptios A1, A2 d A3. Theorem 2.6. (Geerlized Abel-Dirichlet Criterio) Assume A1, A2 d A3 re fulfilled. The, for y x C(S) d y X with the property S(), oe hs x C(S) d x C(S). Proof. We begi with lgebric property. Let A be rig d 0, 1,..., +1 ; b 0, b 1,..., b +1, 0, elemets i A. The we hve the idetities d 0 b b b = = ( i i+1 )(b 0 + b b i ) + +1 (b 0 + b b ) i=0 0 b b b = = ( i )(b i b i+1 ) + ( )b +1. i=0 (2.6) (2.7) The proof c be doe by iductio with respect to. We shll tke x C(S) d X such tht S() d we shll prove tht x C(S), usig (2.6). The fct tht x C(S) c be proved i similr wy, usig (2.7). I order to mke thigs shorter, we shll write x + S(x) + S 2 (x) S p (x) = (p),
14 60 Io Chiţescu for y turl p. Let ε > 0. We shll fid turl p(ε) such tht for y turl > m p(ε) oe hs () (m) < ε ( ) d this will prove the covergece of the series which mes x C(S). Usig (2.6) d the fct tht S (x) =0 (p) = x + S()S(x) + S 2 ()S 2 (x) S p ()S p (x) we c write () = (S i () S i+1 ())(x + S(x) S i (x))+ i=0 +S +1 ()(x + S(x) S (x)). Cosequetly, for > m 0 oe hs () (m) = = i=m+1 It follows tht i=m+1 (S i () S i+1 ())(x + S(x) S i (x))+ +S +1 ()(x + S(x) S (x)) S m+1 ()(x + S(x) S m (x)) = (S i () S i+1 ())x i + S +1 ()x S m+1 ()x m. () (m) S +1 ()x + S m+1 ()x m + + (S i () S i+1 ())x i i=m+1 S +1 () x + S m+1 () x m + + x i S i () S i+1 (). i=m+1 (2.8) Due to the fct tht x C(S), we c fid umber M > 0 such tht for y turl p. x p < M, (2.9)
15 A geerliztio of coverget series 61 From (2.8) we obti () (m) S +1 () x + S m+1 () x m + +M S i () S i+1 (). i=m+1 (2.10) But S() implies S i () S i+1 () d (2.10) becomes () (m) S +1 () x + S m+1 () x m + +M i=m+1 (S i () S i+1 ()) = = M(S m+1 () S +1 ()) + S +1 () x + S m+1 () x m M( S m+1 () + S +1 () ) + S +1 () x + S m+1 () x m. Tkig the orms we get: () (m) M( S m+1 () + S +1 () )+ + S +1 () x + S m+1 () x m. Usig (2.9), we get () (m) 2M( S m+1 () + S +1 () ), (2.11) for y > m 0. Becuse of the ssumptio A3, for the lredy tke ε > 0 we c fid turl p(ε) such tht, for y p p(ε) oe hs S p+1 () < ε 4M. Cosequetly, for turl > m p(ε), we obti from (2.11) exctly ( ). The followig result is lso cocered with the multiplictive structure of X. Theorem 2.7. Assume A1 is fulfilled. The AC(S) is bilterl idel i X. Proof. We hve lredy see tht AC(S) is subspce of X. It remis to be proved tht, for y AC(S) d y x X, oe hs x AC(S) d x AC(S). We shll do the proof for x, the proof for x beig similr. For y > m 0 oe hs S m ( x ) + S m+1 ( x ) S ( x ) =
16 62 Io Chiţescu = S m ( x ) + S m+1 ( x ) S ( x ) S m ( x ) + S m+1 ( x ) S ( x ) = Becuse the series = x (S m ( ) + S m+1 ( ) S ( )). (2.12) S ( ) =0 coverges, we fid p(ε) such tht, for y > m p(ε) oe hs x S m ( ) + S m+1 ( ) S ( ) < ε. Let us tke > m p(ε). Usig (2.12), we get S m ( x ) + S m+1 ( x ) S ( x ) x S m ( ) + S m+1 ( ) S ( ) < ε d this proves tht the series coverges.s.o. S ( x ) =0 3. Exmples Exmple 3.1. (Semil Exmple) Actully, we beg the precedig prgrph with this exmple, which motivtes the preset pper. We tke X = c 0 d S : X X is the shift opertor, give vi where, if oe hs S(x) = y, x = (x 0, x 1,..., x,..) y = (x 1, x 2,..., x,..). It is esy to see tht S is lier, cotiuous d S o = 1. Rephrsig the cosidertios from the begiig of the precedig prgrph, we get C(S) = cc 0 = the coverget series. At the sme time (workig for the rel cse), we hve AC(S) = l 1 = the bsolutely coverget series. We hve lredy see tht (i the rel cse) the spce X = c 0 d the shift opertor S fulfill ssumptios A1, A2 d A3.
17 A geerliztio of coverget series 63 ) We prove tht C(S) = cc 0 is dese i X = c 0 (hece C(S) is ot closed). The sme proof shows (i cse K = R) tht AC(S) = l 1 is dese i c 0 (hece AC(S) is ot closed). To this ed, we pick rbitrry x = (x ) c 0. The sequece (y()) will be costructed s follows: y(0) = (x 0, 0, 0,..., 0,..) y(1) = (x 0, x 1, 0,..., 0,..)... y() = (x 0, x 1,..., x, 0, 0,..., 0,...). The y() cc 0 d (i cse K = R) y() l 1. O the other hd, oe c see tht, for y : thus fiishig the proof. y() x = sup x p 0, p> b) We shll compute the vlues of T S : cc 0 c 0. Let us tke rbitrry elemet x = (x 0, x 1,..., x,..) c 0 d write d for y N. We shll prove tht t(x) = =0 x σ = x 0 + x x, T S (x) def = x = (t(x), t(x) x 0, t(x) (x 0 + x 1 ),..., t(x) σ 1,..). Becuse it remis to be proved tht T S (x) = p x p p x p x = 0. The -th compoet of x p is equl to x + x x +p
18 64 Io Chiţescu d the -th compoet of x is equl to t(x) σ 1, hece the -th compoet of x p x is equl to x + x x +p t(x) + σ 1 = σ +p t(x). It follows tht, for y p oe hs x p x = sup σ +p t(x). Tkig rbitrry ε > 0, we c fid p(ε) N such tht, for y q p(ε) oe hs σ q t(x) < ε d this implies for y p p(ε). x p x ε, c) It is obvious tht, for y x X = c 0, oe hs S (x) = 0. Cosequetly, Theorem 2.3 implies the fct tht (I S) 1 : X C(S) is bijective. Hece I S : c 0 c 0 is ijective d (I S)(c 0 ) = cc 0 which is dese i c 0. Therefore 1 is i the cotiuous spectrum of S d T S is discotiuous (Corollry 2.1). d) The discotiuity of T S c be proved i other wy. Nmely, we shll prove tht the fuctiol L : cc 0 K, give vi L(x) = t(x) is discotiuous. Acceptig this fct, the discotiuity of T S follows from the reltio L = π 0 T S, where π 0 : c 0 K is the (lier d cotiuous) projectio umber zero give vi π 0 ((x ) ) = x 0. Now, let us tur to the proof of the fct tht L is discotiuous. We must prove the existece of ε 0 > 0 hvig the property tht, for y δ > 0 oe c fid x cc 0 with x δ d yet L(x) > ε 0.
19 A geerliztio of coverget series 65 To this ed we c begi with rbitrry ε 0 > 0. Let lso δ > 0 be rbitrrily tke. We cosider y = (y ) semicoverget series with rel terms, e.g. ) ( ( 1) y = Oe c fid (δ) N such tht, for y (δ) oe hs y < δ. We costruct the semicoverget series z = (z ) 0, where z 0 = y (δ), z 1 = y (δ)+1,..., z = y (δ)+,... d, of course, z δ. Usig Riem s permuttio theorem, we c fid permuttio π : N N such tht the series x = (z π(0), z π(1),..., z π(),...) hs the sum equl to ε Hece x δ d L(x) = L(x) = ε 0 +1 > ε 0. e) Becuse T S : cc 0 c 0 is bijectio, s we hve see, we hve o cc 0 two orms. The first orm o cc 0 is the orm iduced by the orm of c 0, hvig the lyticl expressio x = sup x, where x = (x 0, x 1,..., x,...). With respect to this orm, cc 0 is ot Bch spce (becuse cc 0 is ot closed i c 0 ). The secod orm o cc 0 is obtied vi T S -trsport from the orm of c 0. For x cc 0, this orm will be x def = T S (x). Hece, the lyticl expressio of this orm will be x = sup{ t(x), t(x) σ 0, t(x) σ 1,..., t(x) σ,...} with previous ottios. With respect to this orm, cc 0 is Bch spce. f) Now we shll see tht Theorem 2.5 geerlizes D Alembert s criterio. Nmely, we cosider series x = (x ) cc 0 d ssume tht the hypothesis of Theorem 2.5 is fulfilled. We hve, for y turl : S +1 (x) = (x +1, x +2,...) d S (x) = (x, x +1,...).
20 66 Io Chiţescu Hece, the hypothesis sys tht there exist 0 < < 1 d M N such tht A +1 = sup p +1 x p sup x p = A, p for ll M. I cse A +1 = A it follows tht A = A +1 = 0 for ll M, hece x = 0 for ll M d the series x (3.1) =0 coverges (trivilly). So, the cse whe some A = 0 is trivil d let us see wht hppes whe A > 0 for ll. It follows tht, for ll M, oe hs 0 < A +1 A, hece d, cosequetly 0 < A +1 < A, A = sup p The, for ll M, oe hs which implies x p = sup x p = x. p= sup x p x p +1 x +1 x (3.2) d the series (3.1) coverges. I cse ll the x re o ull, we c write (3.2) i the form: for ll M d this implies (of course) sup x +1 x x +1 x (3.3) < 1. (3.4) It is redily see tht, i cse (3.4) is fulfilled, oe c fid turl M, such tht, for ll M, oe hs (3.3). This completes the proof of the fct tht Theorem 2.5 is exctly D Alembert s criterio i this specil cse.
21 A geerliztio of coverget series 67 g) At this poit we begi with the remrk tht Theorem 2.7 is pplicble. Hece l 1 is bilterl idel i c 0. Filly, we shll be cocered with Theorem 2.6 i this prticulr cse. Iterpretig the result we obti the followig FACT. Assume = ( ) is decresig sequece which coverges to zero. Let (x ) be sequece such tht the series is coverget. The, the series is coverget. =0 x x =0 This Fct is equivlet to the ppretly more geerl followig result: Abel-Dirichlet Criterio Let (b ) be decresig sequece which is bouded. Let (x ) be sequece such tht the series is coverget. The, the series is coverget. =0 x b x =0 Ideed, ssumig the vlidity of the Fct, let us write, for b = b : b = + b, where = b b. The ( ) is decresig d coverges to 0. It follows tht the series x coverges. Becuse the series =0 bx =0
22 68 Io Chiţescu coverges too, it follows tht the sum of the precedig series, i.e. the series b x =0 is coverget. We proved tht Theorem 2.6 geerlizes the Abel-Dirichlet Criterio. Exmple 3.2. We cosider rbitrry Bch spce X. Let X 1 d X 2 be two closed subspces of X which re complemetry (i.e. ech elemet x X c be writte uiquely i the form x = x 1 + x 2, with x 1 X 1 d x 2 X 2 ). We cosider the lier d cotiuous projectios P i : X X ssocited with X i, i = 1, 2 (i.e. P i is give vi P i (x) = x i, for i {1, 2}). Let us tke S = P 1. The I S = P 2. Becuse S = S for y turl 1, we use Theorem 2.3 d get For y x X 2 = C(S), oe hs for y turl 1, hece C(S) = (I S)(X) = P 2 (X) = X 2. S (x) = 0, T S (x) = S (x) = x = x =0 S (x). Exmple 3.3. Let us cosider the rel umbers < b. The Bch spce will be X = C[, b]. The opertor S : X X will be defied s follows: where g : [, b] K is give vi g(x) = S(f) = g, x f(t)dt. It is see tht S is lier, cotiuous d S o = b.
23 A geerliztio of coverget series 69 The lst ssertios follow from the iequlity x b f(t)dt f(t) dt (b ) f, which is vlid for y x [, b], d from the fct tht S(u) = b, where u is the costt fuctio equl to 1. It is kow tht, for y f X d y turl 1, oe hs where g : [, b] K is give vi S (f) = g, g(x) = x (x t) 1 f(t)dt. ( 1)! ) We c tht oe hs (see lso Corollry 2.1, 1.) C(S) = X. Ideed, for y f X, y 1 d y x [, b], we hve successively S (f)(x) = x (x t) 1 ( 1)! b f(t)dt x t 1 ( 1)! f dt b Becuse the series b 1 ( 1)! f dt = =1 b ( 1)! f =. is coverget (with sum (b )e b f ) it follows tht the series of fuctios S (f) =0 coverges uiformly d bsolutely, hece coverges i X. Workig for K = R, it is see tht AC(S) = X.
24 70 Io Chiţescu As cosequece (see Theorem 2.3), we hve the lier d cotiuous opertors I S : X X d T S : X X d T S = (I S) 1. b) It is possible to express T S more cocretely. To this ed, we tke rbitrry f X d rbitrry x [, b]. Due to the uiform covergece, we get = f(x) + x =1 T S (f)(x) = f(x) + S (f)(x) = =1 (x t) 1 x f(t)dt = f(x) + ( 1)! x = f(x) + e x t f(t)dt. ( (x t) )f(t)dt =! =0 We used the fct tht, for fixed x [, b], the series of fuctios =1 coverges uiformly o [, x], where ϕ x, (t) = ϕ x, (x t) 1 f(t) ( 1)! ( becuse ϕ x, (t) (x ) ) 1 f. ( 1)! Tkig ito ccout the previous fcts, we proved tht T S = (I S) 1 = I + V. Here S : X X is the lier d cotiuous itegrl opertor with kerel Q : [, b] 2 K give vi Q(x, t) = (it is cler tht, for y x [, b], oe hs S(f)(x) = { 0, if x < t 1, if x t b Q(x, t)f(t)dt).
25 A geerliztio of coverget series 71 Also V : X X is the lier d cotiuous itegrl opertor with kerel P : [, b] 2 K give vi (mely, for y x [, b], oe hs { 0, if x < t P (x, t) = e x t, if x t V (f)(x) = b P (x, t)f(t)dt). Remrk 3.1. Oe c obti T S = (I S) 1 i other wy. Let us give rbitrry f X = C(S). We wt to obti g X such tht g = T S (f), i.e. f = (I S)(g). Writig y : [, b] K, y(x) = x g(t)dt (i.e. y = S(g)) it is see tht y is the solutio of the Cuchy problem y = g d y() = 0. The reltio f = (I S)(g) becomes the differetil equtio (Cuchy problem) y y = f with the iitil coditio y() = 0. The solutio of the homogeeous equtio is of the form y y = 0 y(x) = Ce x. Usig the vritio of costts method, we write y(x) = C(x)e x, which implies Hece y (x) y(x) = C (x)e x = f(x). x C(x) = f(t)e t dt + A.
26 72 Io Chiţescu Becuse y() = 0, oe must hve C() = 0, cosequetly A = 0. So x y(x) = e x f(t)e t dt, for y x [, b], which implies x g(x) = y (x) = e x f(t)e t dt + e x f(x)e x = We got gi x = f(x) + f(t)e x t dt. g(x) = (I + V )(f)(x). c) I view of the existece of the bijectios I S : X X d I + V = (I S) 1 : X X we obti vi trsport two ew Bch spces orms o X. These orms re give s follows: d d f def = (I S)(f) = sup x [,b] f def = (I + V )(f) = sup x [,b] x f(x) f(t)dt x f(x) + e x t f(t)dt. It is see tht, for y x [, b] d y f X, we hve: x x f(x) f(t)dt f(x) + f(t) dt f + (b ) f x x f(x) + e x t f(t)dt f(x) + e b f(t) dt f + e b (b ) f, hece d f (1 + (b )) f f (1 + e b (b )) f.
27 A geerliztio of coverget series 73 It follows tht the orms, d re equivlet. Problem. Fid (s smll s possible) costts A > 0 d B > 0 such tht f A f d for y f X. f B f, Remrk 3.2. Workig for K = C, we kow tht the spectrl rdius of S is equl to 0, hece the spectrum of S reduces to {0} = the residul spectrum of S d (of course) 1 is i the resolvet set of S. Exmple 3.4. We shll cosider the spce X = L p (µ) where 1 p < d µ is the Lebesgue mesure o [0, 1]. The lier d cotiuous opertor S : L p (µ) L p (µ) will be give vi where S( f) = g, g(t) = tf(t) for ll t [0, 1]. It is see tht (i cse K = R) the opertor S is positive. For y turl, oe hs S ( f) = g, where for ll t [0, 1]. ) Oe c prove tht g(t) = t f(t), S ( f) = 0, for y f X. Ideed, let f L p (µ) = X d tke represettive f f. The, for y t [0, 1), oe hs t f(t) = 0, hece t f(t) = 0 µ-.e. d t f(t) f(t),
28 74 Io Chiţescu for ll t [0, 1]. Usig Lebesgue s domited covergece theorem (see [5]) we get f 0 i L p (µ), where f (t) = t f(t) for ll t [0, 1]. Hece S ( f) 0. b) Usig the result from ), we c prove tht the followig ssertios re equivlet for mesurble fuctio f : [0, 1] K: 1. f C(S). 2. There exists u L p (µ) such tht 3. Oe hs g L p (µ), where f(t) = (1 t)u(t) µ-.e.. g(t) = { 1 1 t 4. f AC(S) (for K = R). f(t), if 0 t < 1. 0, if t = 1 The equivlece 1 2 follows from Theorem 2.3, poits 3. d 4. d from ): we hve C(S) = (I S)(X) d the elemets f (I S)(X) hve represettives give by f(t) = (1 t)u(t), with u L p (µ). 2 3 is obvious. The implictio 3 1 (which will be proved immeditely) estblishes the equivlece of ssertios 1, 2, 3. We cosider the sequece (f ) i L p (µ), give vi { (1 + t + t f (t) = t )f(t), if 0 t < 1. 0, if t = 1 It is see tht for y d t oe hs f (t) g(t) d f (t) g(t) µ-.e..
29 A geerliztio of coverget series 75 Agi Lebesgue s domited covergece theorem implies tht i L p (µ). But d we obti the fct tht f g f = f + S( f) S ( f) f g i L p (µ), which c be rephrsed s follows: the series S ( f) =1 coverges to g i L p (µ). Hece f C(S). I order to fiish the proof we must prove tht 1 4. So, let us tke f C(S). Becuse 1 3, we obti (tkig represettive f f) the fuctio g L p (µ) costructed with the id of f s i the sttemet of 3. It follows tht g L p (µ), hece the fuctio u : [0, 1] K give vi u(t) = { 1 1 t f (t), if 0 t < 1 0, if t = 1 is i L p (µ). From 3 1 we get f C(S), i.e. f AC(S). c) We cosider the subspce H X which cosists of costt fuctios. More precisely H = { f X f is costt}. We shll prove tht H C(S) = {0} (d this shows tht C(S) X). Ideed, ssume by bsurd, the existece of 0 f H C(S). Let 0 α K be such tht f(t) = α µ-.e.. Usig 3, from b) we obti the fuctio g L p (µ) give s follows g(t) = { α 1 t, if 0 t < 1. 0, if t = 1
30 76 Io Chiţescu A simple computtio shows tht g p dµ = d we got cotrdictio. d) Now we cosider the subspce Y = { f L p (µ) there exists 0 < < 1 such tht f(t) = 0 for ll t [, 1]}. (of course, chges with f). We shll prove tht Y is dese i X. To this ed, we tke first elemet u L p (µ) such tht u(t) 0 µ-.e.. We defie the sequece (u ) i L p (µ) such tht u (t) = { u(t), if 0 t 1 1 0, if 1 1 < t 1, for 2. The u Y d for ll. We hve lso u = u u, u (t) = u(t) µ-.e.. Lebesgue s domited covergece theorem sys tht i L p (µ) = X. u u For rbitrry rel f : [0, 1] R, f X, we write f = u v with u, v : [0, 1] R, u d v beig positive fuctios i L p (µ). We cosider the sequeces ( u ) d ( v ) i Y such tht u u d v v i X d where u v Y. u v f, For complex (i cse K = C) f : [0, 1] C, f X, we write f = g + i h d fid rel g, h i Y such tht g g d h h i Y. The g + i h Y d g + i h f.
31 A geerliztio of coverget series 77 e) Now we prove tht Y C(S). Let us tke f i Y d prove tht f C(S). Accordig to b) 3, we must prove tht (tkig represettive f f) the fuctio g : [0, 1] K, give vi { f(t) g(t) = 1 t, if 0 t < 1 0, if t = 1 is i L p (µ). There exists 0 < < 1 such tht f(t) = 0 for t 1. It follows tht { f(t) g(t) = 1 t, if 0 t < 0, if t 1. For 0 t <, oe hs Cosequetly for ll t [0, 1] d g L p (µ). f) We coclude tht 0 < 1 1 t < 1 1. g(t) 1 1 f(t), C(S) = AC(S) = (I S)(X). Accordig to poit c, it follows tht C(S) X d C(S) is meger. Accordig to poit d) d poit e), it follows tht C(S) is dese i X. Theorem 2.3 sys tht (usig poit )) we hve the mutully iverse bijectios (I S) 1 : X C(S) d T S : C(S) X. Hece 1 is i the cotiuous spectrum of S d T S is discotiuous. Notice tht T S : C(S) X cts s follows: T S ( f) = g where g(t) = { 1 1 t f(t), if 0 t < 1 0, if t = 1 for y represettive f f C(S). Remrk 3.3. Actully, the spectrum of S is equl to [0, 1] d coicides with the cotiuous spectrum of S.
32 78 Io Chiţescu Refereces [1] I. Chiţescu, Spţii de fucţii, Ed. Ştiiţifică şi Eciclopedică, Bucureşti, [2] I. Chiţescu d N.-A. Secele, Elemete de teori măsurii şi itegrlei, Ed. Fudţiei Româi de Mâie, Bucureşti, [3] I. Colojoră, Elemete de teorie spectrlă, Ed. Acdemiei R.S.R., Bucureşti, [4] R. Cristescu, Aliză Fucţiolă, Ed. Didctică şi Pedgogică, Bucureşti, ed. III, [5] N. Diculeu, Vector Mesures, Veb Deutscher Verlg der Wisseschfte, Berli, [6] N. Duford d J.T. Schwrtz, Lier Opertors, prt I, Itersciece Publishers, Ic., New York, [7] M. Nicolescu, N. Diculeu d S. Mrcus, Aliză Mtemtică, vol. I, Ed. Didctică şi Pedgogică, Bucureşti, ed. V, [8] H.H. Schfer, Topologicl Vector Spces, Spriger Verlg, third pritig corrected, Io Chiţescu Uiversity of Buchrest, Fculty of Mthemtics d Computer Sciece 14 Acdemiei Street, Buchrest, Romi E-mil: iochitescu@yhoo.com
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