Frames containing a Riesz basis and preservation of this property under perturbations.

Transcription

1 arxiv:math/ v1 [math.fa] 22 Sep 1995 Frames cotaiig a Riesz basis ad preservatio of this property uder perturbatios. Peter G. Casazza ad Ole Christese April 2, 2018 Abstract Aldroubihasshowhowoecacostructayframe{g i } startig with oe frame {f i },usig a bouded operator U o l2 (N). We study the overcompleteess of the frames i terms of properties of U. We also discuss perturbatio of frames i the sese that two frames are close if a certai operator is compact. I this way we obtai a equivalece relatio with the property that members of the same equivalece class have the same overcompleteess. O the other had we show that perturbatio i the Paley-Wieer sese does ot have this property. Fially we costruct a frame which is orm-bouded below, but which does ot cotai a Riesz basis.the costructio is based o the delicate differece betwee the ucoditioal covergece of the frame represetatio, ad the fact that a coverget series i the frame elemets eed ot coverge ucoditioally. 1 Itroductio. The itroductio of frames for a Hilbert space H goes back to the paper [DS] from 1952, where they are used i oharmoic Fourier aalysis. A frame is The first author ackowledges support by NSF Grat DMS ad a grat from the Daish Research Foudatio Mathematics Subject Classificatio: Primary 42C99, 46C99 1

2 a family of elemets i H which ca be cosidered as a overcomplete basis : every elemet i H ca be writte as a liear combiatio of the frame elemets, with square itegrable coefficiets, which do ot eed to be uique. A atural theoretical questio (which is also importat for applicatios, e.g., represetatio of a operator usig a basis) is how far frames are away from bases, i.e., oe may ask questios like 1) does a frame cotai a Riesz basis? 2) which coditios imply that a frame just cosists of a Riesz basis plus fiitely may elemets? 3) what happes with the overcompleteess if the frame elemets are perturbed? The reaso for the iterest i Riesz bases ad ot just bases is that frames ad Riesz bases are closely related: a Riesz bases is just a frame, where the elemets are ω-idepedet. Some aswers has bee foud by Holub [H], who cocetrates o the secod questio. Here we go oe step further, i that we are maily iterested i frames which just cotai a Riesz basis. For such frames oe defies the excess as the umber of elemets oe should take away to obtai a Riesz basis. I the first part of the paper we apply a result of Albroubi [A], explaiig how oe ca map a frame oto aother usig a bouded operator U o l 2. Our results cocer the relatio betwee the frames ivolved ad properties of U. Idepedet of that we costruct a orm-bouded frame ot cotaiig a Riesz basis. I sectio 3 we cocetrate o the third questio. We itroduce the cocept compact perturbatio. This leads to a equivalece relatio o the set of frames, with the property that frames i the same equivalece class have the same overcompleteess properties; this meas, that if a frame cotais a Riesz basis the all members i the class cotai a Riesz basis, ad all those frames have the same excess. Fially we show that perturbatio i the Paley-Wieer sese [C3] ot has this plesat property. 2

3 2 Frames cotaiig a Riesz basis. Let H be a separable Hilbert space. A family {f i } i I is called a frame for H if A,B > 0 : A f 2 < f,f i > 2 B f 2, f H. i I A ad B are called frame bouds. A Riesz basis is a family of elemets which is the image of a orthoormal basis by a bouded ivertible operator. For families with the atural umbers as idex set there is a equivalet characterizatio [Y]: {f i } is a Riesz basis if there exist umbers A,B > 0 such that (1) A c i 2 c i f i 2 B c i 2, for all fiite sequeces c 1,...c. Also, a basis {f i } is a Riesz basis if ad oly if it is ucoditioal (meaig that if c i f i coverges for some coefficiets {c i }, the it actually coverges ucoditioally) ad 0 < if i f i sup i f i <. There is a close coectio betwee frames ad Riesz bases: {f i } is a Riesz basis [{f i } is a frame ad c i f i = 0 c i = 0, i.] If {f i } is a Riesz basis, the the umbers A,B appearig i (1) are actually frame bouds. If {f i } i I is a frame (or if oly the upper frame coditio is satisfied) the we defie the pre-frame operator by T : l 2 (I) H, T { } := {. I The operator T is bouded. Composig T with its adjoit T : H (I), T { = {< {,{ >} I 3

4 we get the frame operator S = TT : H H, S{ := I < {,{ > {, which is a bouded ad ivertible operator. This immediately leads to the frame decompositio ; every f H ca be writte as f = i I < f,s 1 f i > f i, where the series coverges ucoditioally. So a frame has a property similar to a basis: every elemet i H ca be writte as a liear combiatio of the frame elemets. For more iformatio about basic properties of frames we refer to the origial paper [DS] ad the research tutorial [HW]. The mai differece betwee a frame {f i } ad a basis is that a frame ca be overcomplete, so it might happe that f H has a represetatio f = c i f i for some coefficiets c i which are differet from the frame coefficiets < f,s 1 f i >. I applicatios oe might wish ot to have too much redudacy. I that spirit Holub [H] discusses ear-riesz bases, i.e. frames {f i } cosistig of a Riesz basis {f i } i N σ plus fiitely may elemets {f i } i σ. The umber of elemets i σ is called the excess. Let us deote the kerel of the operator T by N T. If {f i } is a frame, the {f i } is a ear-riesz basis N T has fiite dimesio {f i } is ucoditioal. The first of the above biimplicatios is due to Holub [H], who also proves the secod uder the assumptio that the frame is orm-bouded below. The geeralizatio above is proved by the authors i [CC]. If the coditios above are satisfied, the the excess is equal to dim(n T ). If dim(n T ) =, two thigs ca happe: {f i } cosists of a Riesz basis plus ifiitely may elemets (i which case we will say that {f i } has ifiiteexcess) or{f i } doesotcotais arieszbasisatall. Ithepreset 4

5 paper we cocetrate o frames which cotai a Riesz basis. Every frame ca be mapped oto such a frame (i fact, oto a arbitrary frame) usig a costructio of Aldroubi [A], which we ow shortly describe. Let {f i } be a frame ad U : l2 (N) l 2 (N) a bouded operator. Let {u i,j } i,j N be the matrix for U with respect to some basis. Defie the family {g i } H by g i = u i,j f j. j=1 Byaabuse ofotatiowe will sometimes write{g i } = U{f i }. Aresult of Aldroubi (differetly formulated) states that {g i } is a frame γ > 0 : UT f γ T f, f H. This is ot too complicated: the boudedess of U implies that {g i } satisfies the upper frame coditio, ad the coditio above is just a differet expressio for the lower coditio. But it is importat that every frame {g i } ca be geerated i this way, i.e., give the frame {g i} we just have to fid the operator U mappig {f i } to {g i }. I coectio with Aldroubis costructio there are (at least) two atural questios related to Holubs work: how is the excess of {g i } related to that of {f i }, ad which coditios imply that {g i} actually is a Riesz basis? We shall give aswers to both questios i this sectio. The defiitio of {g i } immediately shows that {< g i,f >} = U{< f i,f >}, f H; this is true whether or ot {g i } builds a frame. The formula leads to a expressio for the pre-frame operator associated with {g i }. We let UT deote the traspose of U ad U bethe operator correspodig to the matrix where all etries i the matrix of U are complex cojugated. The, give f H,{ } = (N), we have < c i g i,f >= c i < g i,f >= c i < f,g i > 5

6 = < {c i },U{< f,f i >} >=< {c i },UT f >=< TU T {c i },f >. It follows that c i g i = TU T {c i }, {c i} l2 (N). So if {g i } cotais a Riesz basis, the its excess is equal to dim(n TU T). For the calculatio of this umber we eed a lemma. Correspodig to a operator V we deote its rage by R V. Lemma 1: Let X,Y be vector spaces ad V : X Y a liear mappig. Give a subspace Z Y, defie V 1 (Z) := {x X Vx Z}. The dim(v 1 (Z)) = dim(z R V )+dim(n V ). Proof: Let {y i } Y be a basis for Z R V ad take {x i } X such that Vx i = y i. Now, if x X ad Vx Z the we ca fid coefficiets {c i } such that Vx = c i y i = V c i x i, i.e., x spa{x i } + N V. Let ow {z j } be a basis for N V. Correspodig to our elemet x X with Vx Z we ca ow also fid coefficiets {d j } such that x = c i x i + d j z j. So the idepedet set {x i } {z j } spas V 1 (Z). Theorem 2: dim(n TU T) = dim(r U T N T )+dim(r U ). Proof: {{c i } TU T {c i } = 0} = {{c i } U T {c i } N T } = (U T ) 1 (N T ). Now the result follows from Lemma 1 ad the observatio dim(n U T) = dim(n U ) = dim(r U ). So if {g i } actually is a frame cotaiig a Riesz basis, the Theorem 2 6

7 gives a recipe for calculatio of the excess. I particular, if {f i } is a ear- Riesz basis ad R U has fiite codimesio, the also {g i } is a ear-riesz basis. Propositio 3: {g i } is a Riesz basis U : R T l 2 (N) is surjective. Proof: The stadig assumptio U bouded implies that {g i } satisfies the upper coditio, so {g i } is a Riesz basis if ad oly if m > 0 : c i 2 m c i g i 2 for all fiite sequeces c 1,...c. By [Y, p.155] this coditio is satisfied if ad oly if the momet problem < f,g i >= c i, i N has at least oe solutio f wheever {c i } l 2 (N),i.e., if ad oly if U{< f,f i >} = {c i } has at least oe solutio wheever {c i} l2 (N). The last coditio meas exactly that U is surjective cosidered as a mappig from the subspace R T of l 2 (N) oto l 2 (N). More geerally oe may wish that the frame at least cotai a Riesz basis. As show i [C2] it is the case for a Riesz frame, which is a frame with the property that every subfamily is a frame for its closed liear spa, with a commo lower boud. It is easy to costruct a frame which does ot cotais a Riesz basis if oe allows a subsequece of the frame elemets to coverge agaist 0 i orm. We ow preset a example showig that the same ca be the case for a frame which is orm-bouded below. Our approach is complemetary to recet work of Seip [Se], who proves that there exist frames of complex expoetials for L 2 ( π,π) which do ot cotai a Riesz basis. While Seip relies o the theory for samplig ad iterpolatio our approach is more elemetary, just usig fuctioal aalysis. Furthermore our costructio puts focus o a differet poit, amely the differece betwee covergece ad ucoditioal covergece of a expasio i the frame elemets. 7

8 Propositio 4: There exists a frame for H made up of orm oe vectors, which has o subsequece which is a Riesz basis. The proof eeds several lemmas, so let us shortly explai the basic idea. As we have see, i I c i f i coverges ucoditioally for every set of frame coefficiets {c i }. But othig guarates that covergece of i I c i f i implies ucoditioal covergece for geeral coefficiets {c i }. Our proof cosists i a costructio of a frame where o total subset is ucoditioal, ad hece ot a Riesz basis. Techically the first step is to decompose H ito a direct sum of fiite dimesioal subspaces of icreasig dimesio. The idea behid the proof might be useful i other situatios as well. Lemma 5: Let {e i } be a orthoormal basis for a fiite dimesioal space H. Defie f j = e j 1 e i for j = 1,.. The +1 j=1 f +1 = 1 e i. < f,f j > 2 = f 2, f H. Proof: Give f H, write f = a i e i, a i =< f,e i >. If we let P 1 deote the orthogoal projectio oto the uit vector e i, the Pf = 1 < f, e i > e i = a i 1 e i. Therefore Pf = a i 2 = < f,f +1 > 2. Also j=1 (I P)f 2 = f Pf 2 a j = a i e i e i 2 8

9 j=1 a j j=1 = (a i )e i 2 a j = a i 2 = < f,f i > 2. Puttig the two results together we obtai ad the proof is complete. f 2 = Pf 2 + (I P)f 2 = +1 < f,f i > 2 Give a sequece {g i } i I H its ucoditioal basis costat is defied as the umber sup{ σ i c i g i c i g i = 1 ad σ i = ±1, i}. i I i I As show i [Si], a total family {g i } i I cosistig of o-zero elemets is a ucoditioal basis for H if ad oly if it has fiite ucoditioal basis costat. Lemma 6: Defie {f 1,...f +1 } as i Lemma 5. Ay subset of {f 1,f 2,..f +1 } which spas H has ucoditioal basis costat greater tha or equal to 1 1. Proof: Sice f i = 0, ay subset of {f 1,..f +1 } which spas H must cotai 1 elemets from {f 1,...f } plus f +1. By the symmetric costructio it is eough to cosider the family {f 1,..f 1,f +1 }. We have 1 f i = 1 e i 1 e i = (1 1 1 e i 1 = 1 1 e i 1 e = 1 + ( 1)2 2 2 = 1 ( 1) 1. e Nowcosider 1 ( 1) f i ; ifisoddthisumberisequalto 1 ( 1) e i = 1, ad if is eve it is equal to 1 ( 1) i e i 1 1 e i ( 1) i e i 1 e i

10 That is, i all cases, 1 ( 1) f i 1 1. Combiig this with the orm estimate 1 f i 1 it follows that the ucoditioal basis costat of {f 1,...f 1 } is greater tha or equal to 1 1, so clearly the same is true for {f1,...f 1,f +1 }. Now we are ready to do the costructio for Propositio 4. Let {e i } be a orthoormal basis for H ad defie H := spa{e( 1) 2 +1,e( 1) 2 +2,...e( 1) 2 + }. So H 1 = spa{e 1 }, H 2 = spa{e 2,e 3 },H 3 = spa{e 4,e 5,e 6 },... By costructio, H = ( H\ ) H. \= That is, g H } = \= } \,} \ H \, ad g 2 = =1 g 2. We refer to [LT] for details about such decompositios. ForeachspaceH wecostructthesequece {fi } +1 asilemma5, startig with the orthoormal basis {e( 1) }. Specifically, give N, +1,...e( 1) f i = e( 1) 2 +i 1 j=1 f+1 = 1 e( 1) 2 +j, 1 i j=1 e( 1) 2 +j. Lemma 7: {f i }+1,,=1 is a frame for H, with bouds A = B = 1. Proof: Write g H as g = =1 g, g H. Give N it is clear that < g,f i >=< g,f i > for i = 1,

11 From this calculatio it follows that +1 =1 < g,f i > 2 = where we have used Lemma =1 < g,fi > 2 = g 2 = g 2, =1 Lemma 8: No subsequece of {f i }+1,,=1 is a Riesz basis for H. Proof: Ay subsequece of {fi } +1,,=1 which spas H must cotai elemets from {fi }+1 ad so by Lemma 6, its ucoditioal basis costat is greater tha or equal to 1 1 for every. That is, the ucoditioal basis costat is ifiite, hece the subsequece ca ot be a ucoditioal basis for H. Lemma 7 ad Lemma 8 proves Propositio 4. It would be iterestig to determie whether Propositio 4 still holds if oe oly cosiders classes of frames with a special structure, for example Weyl-Heiseberg frames, wavelet frames, or frames cosistig of traslates of a sigle fuctio. Remark: The projectio method developed i [C1, C2] ca be used to calculate the frame coefficiets if a certai techical coditio is satisfied. The block structure of the frame {fi } +1,,=1 costructed here shows that the projectio method ca be used. As show i [C2] the method ca also be used for every Riesz frame. The two questios, i.e. the questio about cotaimet of a Riesz basis ad the questio whether the projectio method works, do t seem to be strogly related. 3 Excess preservig perturbatio. At several places i the followig we eed results for perturbatio of frames ad Riesz bases. We deote the frames by {f i },{g i }, usually with the covetio that {f i } is the frame we begi with, ad {g i} is the perturbed family. Commo for all the result is that they ca be formulated usig the perturbatio operator K mappig a sequece {c i } of umbers to ci (f i g i ). 11

12 Theorem 9: Let {f i },{g i} H. a) If {f i } is a frame for H ad K is compact as a operator from l 2 (N) ito H, the {g i } is a frame for its closed liear spa. b) Suppose {f i } is a frame for H with bouds A,B. If there exist umbers λ,µ 0 such that λ+ µ A < 1 ad c i (f i g i ) λ c i f i +µ ci 2 for all fiite sequeces {c i }, the {g i } A(1 (λ+ µ A )) 2,B(1+λ+ µ B ) 2. is a frame for H with bouds For the proofs we refer to [C3, CH]. Several variatios are possible. If i b) we just assume that {f i } is a Riesz basis for spa{f i}, the {g i} is also a Riesz basis for its closed liear spa. Also observe, that if {f i } is a frame ad σ N is fiite, the {f i } i N σ is a frame for spa{f i } i N σ ; this is a cosequece of a). The ext result coects Theorem 9 with the questio about overcompleteess. Theorem 10: Suppose that {f i } is a frame cotaiig a Riesz basis, that {g i } is total, ad that K is compact as a mappig from l 2 (N) ito H. The {g i } is a frame for H cotaiig a Riesz basis, ad the frames {f i } ad {g i} have the same excess. Proof: First assume that {f i } has fiite excess equal to. By chagig the idex set we may write {f i } = {f i } {f i } i=+1, where {f i } i=+1 is a Riesz basis for H. Let A be a lower frame boudfor {f i } i=+1 ad choose µ < A. By compactess there exists a umber m > such that c i (f i g i ) µ c i 2 for all sets of sequeces {c i } l 2 (N). So by the remark after Theorem 9, {g i } is a Riesz basis for spa{g i}. If we defie the operator T o H by Tf i = f i, < i m, Tf i = g i, i m+1, 12

13 (exteded by liearity) the we have a ivertible operator o H. The argumet is that every f H has a represetatio f = i=+1 c i f i, leadig to (I T)f = c i (f i g i ) µ As a cosequece, c i 2 µ A c i f i = µ A f. codim(spa{g i } ) = codim(spa{f i} ) = m. Take m idepedet elemets {g ik } m k=1 outside spa{g i }. The {g ik } m k=1 {g i} is a frame for spa{{g i k } m k=1 {g i} } = H, sice oly fiitely may elemets have bee take away from the frame {g i }. If ow m k=1 c kg ik + c i g i = 0, the all coefficiets are zero; first, m k=1 c k g ik = c i g i = 0 (if the sums were ot equal to zero we could delete a elemet g ik ad still have a frame for H cotradictig the fact that codim(spa{g i } ) = m ) ad sice {g ik } m k=1 is a idepedet set ad {g i } a Riesz basis, all coefficiets must be zero. So {g ik } m k=1 {g i} is a Riesz basis, i.e., {g i } also has excess. Now suppose that {f i } has ifiite excess. Let {f i} i I be a subset which is a Riesz basis. The the correspodig set {g i } i I spas a space of fiite codimesio, i.e., codim(spa{g i } i I ) <. This follows by the same compactess argumet as we used i the fiite excess case, which shows that there exist fiitely may f i,i I with the property that if we take them away the we obtai a family which spas a space with the same codimesio as the correspodig space of g i s. Now take a fiite family {g i } i J such that {g i } i I J is total. Sice {f i } i I J is a frame with fiite excess, the fiite excess result gives that {g i } i I J is a frame cotaiig a Riesz basis, implyig that {g i } has ifiite excess. 13

14 We ca express the result i the followig way: defie a equivalece relatio o the set of frames for H by {f i } {g i} K is compact as a operator from l2 (N) ito H. The equivalece relatio partitios the set of frames ito equivalece classes. If a frame cotais a Riesz basis, the every frame i its equivalece class cotais a Riesz basis, ad the frames have the same excess. Let us go back to Theorem 10. If {g i } is ot total, the rest of the assumptios implies that {g i } is a frame for its closed spa.now the proof of Theorem 10 shows that {g i } cotais a Riesz basis for spa{g i}, ad that the excess referig to this space is equal to the excess of {f i } as a frame for H plus the dimesio of the orthogoal complemet of spa{g i } i H. Now we wat to study the excess property of perturbatios i the sese of Theorem 9 b). We eed a result, which might be iterestig i itself. To motivate it, cosider a ear-riesz basis {f i } cotaiig a Riesz basis {f i } i I. Ufortuately, the lower boud for {f i } i I ca be arbitrarily small compared to the lower boud A of {f i }. Our result states, that if we are willig to delete sufficietly (still fiitely) may elemets, the we ca obtai a family which is a Riesz basis for its closed spa, ad which has a lower boud so close to A as we wat: Propositio 11: Let {f i } be a ear-riesz basis with lower boud A. Give ǫ > 0, there exists a fiite set J N such that {f i } i N J is a Riesz basis for its closed spa, with lower boud A ǫ. Proof: As i the proof of Theorem 10, write {f i } = {f i} {f i} i=+1, where {f i } i=+1 is a Riesz basis for H. Let d(, ) deote the distace iside H (i.e., d(f,e) = if g E f g for f H,E H) ad choose a umber m > such that d(f j,spa{f i } m i=+1 ) < ǫ, j = 1,... Wewattoshowthat{f i } isarieszbasisforitsclosedspa, withlower boud A ǫ. Let P deote the orthogoal projectio oto spa{f i } m i=+1. 14

15 Sice c i f i c i (I P)f i for all sequeces, it suffices to show that {(I P)f i } satisfies the lower Riesz basis coditio with boud A ǫ. Let f (I P)H. The < f,(i P)f i > 2 = < f,(i P)f i > 2 < f,(i P)f i > 2 A f 2 f 2 (I P)f i 2 (A ǫ) f 2. Now we oly have to show that {(I P)f i } is ω-idepedet. But if c i (I P)f i = 0, the c i f i = P c i f i, implyig that both sides are equal to zero, sice P c i f i spa{f i } m i=+1 ad {f i } i=+1 is idepedet. Therefore c i = 0 for all i. Theorem 12: Let {f i } be a frame for H with bouds A,B. Let {g i} H ad assume that there exist λ,µ 0 such that λ+ µ A < 1 ad c i (f i g i ) λ c i f i +µ ci 2 for all fiite sequeces {c i }. The {f i } is a ear-riesz basis {g i } is a ear-riesz basis, i which case {f i } ad {g i } have the same excess. Proof: First assume that {f i } is a ear-riesz basis with excess. Let m be choose as i the proof of Propositio 11, correspodig to a ǫ satisfyig the coditio λ + µ A ǫ < 1. Let Q deote the orthogoal projectio oto spa{f i }. The every elemet f H ca be writte f = (I Q)f +Qf = (I Q)f + c i f i, for some coefficiets c i. Now defie a operator T : H H by Tf = f, f spa{f i }, Tf i = g i, i m+1. T is bouded. Give f H we choose a represetatio as above. The (I T)f = c i (f i g i ) λ c i f i +µ c i 2 15

16 µ (λ+ ) A ǫ c i f i = (λ+ µ µ ) Qf (λ+ ) f. A ǫ A ǫ It follows, that T is a isomorphism of H oto H. So {g i } is a Riesz basis for its closed spa, ad dim(spa{g i } ) = dim(spa{f i } ). As a cosequece, {f i } ad {g i} have the same excess. Now assume that {g i } is a ear-riesz basis. By reidexig we may agai assume that {g i } i=+1 is a Riesz basis for H. Defie a bouded operator U : H H by Uf := < f,s 1 f i > g i. The as i the origial proof from [C3] oe proves that U is a isomorphism of H oto H. If we defie U : H H by U f = i=+1 < f,s 1 f i > g i, the this operator has a rage with fiite codimesio i H, which we will write as codim H (R U ) <. Now let {e i } be the atural basis for l2 (N), i.e., e i is the sequece with 1 i the i th etry, otherwise 0. There exists a bouded ivertible operator V : H \{ } =\+ such that Vg i = e i for i +1, ad clearly codim spa{ei } i=+1 (R VU ) <. Observe that VU f = i=+1 < f,s 1 f i > e i = {< f,s 1 f i >} i=+1. So (VU ) {c i } = i=+1 c i S 1 f i = S 1 i=+1 c i f i. Sice R VU = N (V U) has fiite dimesio, also {c i} i=+1 i=+1 c i f i has a fiite dimesioal kerel. Therefore T : l 2 (N) H, T { } = = { has a fiite dimesioal kerel, ad ow the theorem of Holub implies that {f i } is a ear-riesz basis. By the first part of the Theorem the two frames 16 =

17 {f i } ad {g i} ow have the same excess, ad the proof is complete. Ufortuately, the requiremet that {f i } has fiite excess is eeded i Theorem 12. I fact we are able to costruct examples, where {f i } has ifiite excess ad {g i } does ot cotai a Riesz basis, but where the perturbatio coditio is satisfied. Let us shortly describe how oe ca do this. Defie {fi }+1,,=1 as i Lemma 7. Give ǫ > 0, let g i = e( 1) 2 +i 1 ǫ j=1 g+1 = 1 Now, give a sequece {c i} we have j=1 e( 1) 2 +j, 1 i e( 1) 2 +j. c i (f i gi ) = ǫ [ =1 c i ]1 j=1 ǫ =1 c i e( 1) 2 +j 1 2 ǫ c i 2. (1) By Lemma 5, {fi },,=1 is a frame with bouds 1. If we choose ǫ < 1, the the perturbatio coditio is satisfied with λ = 0,µ = ǫ, implyig that {gi }+1,,=1 is a frame with bouds (1 ǫ) 2,(1+ǫ) 2. Claim: {g i },,=1 is a Riesz basis for H. We oly eed to prove that {gi },,=1 satisfies the lower Riesz basis coditio. Give a sequece {c i } we have c igi c (1 ǫ) ie( 1) ( +i 2 =1 =1 =1 c i) 1 e( 1) +i 2 c i 2 (1 ǫ) =1 c i 1 2 ǫ c i 2. 17

18 So actually we have a example where {f i }+1,,=1 does ot cotais a Riesz basis but the perturbed family does. To obtai the example we were lookig for, we use that {g i } has the lower boud (1 ǫ) 2. By (1) above we ca cosider {fi } as a perturbatio of {g i } if ǫ < 1, i.e., if ǫ < 1. So we get 1 ǫ 2 our example by choosig ǫ < 1 ad switchig the roles of {fi } ad {gi }. Ackowledgemets: The secod author would like to thak Chris Heil for fruitful discussios o the subject. Refereces: [A] Aldroubi,A.: Portraits of frames. Proc. Amer. Math. Soc., 123 (1995), p [CC] Casazza, P.G. ad Christese, O.: Hilbert space frames cotaiig a Riesz basis ad Baach spaces which have o subspace isomorphic to c 0. Submitted, [C1] Christese, O.: Frames ad the projectio method. Appl. Comp. Harm. Aal. 1 (1993), p [C2] Christese, O.: Frames cotaiig Riesz bases ad approximatio of the frame coefficiets usig fiite dimesioal methods. Accepted for publicatio by J. Math. Aal. Appl. [C3] Christese, O.: A Paley-Wieer theorem for frames. Proc. Amer.Math. Soc. 123 (1995 ), p [CH] Christese, O. ad Heil, C.: Perturbatio of Baach frames ad atomic decompositios. Accepted for publicatio by Math. Nach.. [DS] Duffi, R.J. ad Schaeffer, A.C.: A class of oharmoic Fourier series. Tras. Amer. Math. Soc. 72 (1952) p [HW] Heil, C. ad Walut, D.: Cotiuous ad discrete wavelet trasforms. SIAM Review 31 (1989), p

19 [H] Holub, J.: Pre-frame operators, Besselia frames ad ear-riesz bases. Proc. Amer. Math. Soc. 122 (1994), p [LT] Lidestrauss, J. ad Tzafriri, L.: Classical Baach spaces 1. Spriger [Se] Seip, K.: O the coectio betwee expoetial bases ad certai related sequeces i L 2 ( π,π). J. Fuct. Aal. 130 (1995), p [Si] Siger, I.: Bases i Baach spaces 1. Spriger New York, [Y] Youg, R.M.: A itroductio to oharmoic Fourier series. Academic Press, New York, Peter G. Casazza Departmet of Mathematics Uiversity of Missouri Columbia, Mo USA. pete@casazza.cs.missouri.edu Ole Christese Mathematical Istitute Buildig 303 Techical Uiversity of Demark 2800 Lygby Demark. olechr@mat.dtu.dk 19