Stability of Phase Retrievable Frames

Stabilit y of Phase R etriev able F rames Radu Balan Mathematics Departmen t and CSCAM M, Univ ersit y of Maryland, College P ark, MD 20742, US A July 27, 2018 Abstract In this pap er w e study the property of phase retriev ability b y redundant sysems of vectors under p erturbations of the frame set. Sp ecifically w e show that if a set F of m vectors in the complex H ilbert space of d imension n allo ws for vector reconstruction from magnitudes of its co efficients, then there is a p erturbation bound ρ so that any frame s et within ρ f rom F has t he same property . In particular this prov es th e recen t construction in [15] is stable under perturbations. By the same t oke n w e reduce th e critical cardinality conjectured in [11] to proving a stability result for non ph ase-retriev able frames. 1 INTR ODUCTION The ph ase r et rieval problem pr esents itself in man y applications is ph ysics and eng ineering. Recen t pap e rs on t his topic (see [8, 1 8, 6, 7, 1 , 1 1, 51]) present a fu ll list of examples r anging f rom X-Ra y cry stallogra ph y to a udio and image sig nal pro cess ing , cla ssification with deep net works, quantum information theory , and fiber optics data transmission. In this pap er w e consider the complex case, namely the Hilber t space H = C n endow ed wit h the usual Euclidian scalar pro duct h x, y i = P n k =1 x k y k . On H we consider the equiv alence relation ∼ betw een tw o vectors x, y ∈ H defined a s follows; the vectors x and y are similar x ∼ y if and only if there is a complex constant z of unit magnitude, | z | = 1 , so that y = z x . Let ˆ H = H / ∼ b e the quo tien t space. Th us an equiv a lence cla ss (a ra y ) has the for m ˆ x = { e iϕ x , ϕ ∈ [0 , 2 π ) } . A subset F ⊂ H of the Hilb ert space H (reg ardless whether it is finite dimensional or no t) is called fr ame if there are tw o p ositive cons ta n ts 0 < A ≤ B < ∞ (called fr ame b ounds ) so that for any vector x ∈ H , A k x k 2 ≤ X f ∈F |h x, f i| 2 ≤ B k x k 2 (1) In the finite dimensio nal case consider ed in this pap er, the fr ame condition simply re duce s to the spanning condition. Sp ecifically F = { f 1 , . . . , f m } is fra me for H if and only if H = span ( F ). O bviously m ≥ n must hold. When we ca n cho o se A = B t he frame is called tight . If fur thermore A = B = 1 then F is s aid a Parseval fr ame . Co ns ider the fo llowing no nlinear map α : ˆ H → R m , ( α ( ˆ x )) k = |h x, f k i| , 1 ≤ k ≤ m (2) which is well defined on the classe s ˆ x since |h x, f k i| = |h y , f k i| when x ∼ y . 1 The frame F = { f 1 , . . . , f m } is called phase r etrievable is the nonlinear map α is injective. Notice that any sig nal x ∈ H is uniquely defined by the magnitudes of its fra me co efficients α ( x ) up to a global phase factor, if and only if F is phase retr iev able. The ma in r esult of this pap er states that the phase retriev able prop erty is stable under s mall p er tur bations o f the frame set. Specifically we show Theorem 1.1 As s ume F = { f 1 , . . . , f m } is a phase r etrievab le fr ame for a c omplex Hilb ert sp ac e H . Then ther e is a ρ > 0 so that any set F ′ = { f ′ 1 , . . . , f ′ m } with k f k − f ′ k k < ρ , 1 ≤ k ≤ m , is also a phase r etrievable fr ame. W e prove this theorem in section 3. The pro of is bas e d on a r e cent necess ary and sufficent condition obtained indep endent ly in [11 ] and [7]. The e xact form of this result is slightly different than the e q uiv alent results s ta ted in the a forementioned pap ers. Conseq uen tly we will provide a direct pro o f. An in teresting problem on phase retriev able fr ames is to find a c ritical cardinal m ∗ ( n ) that has the following pro per ties: (A) F or any m ≥ m ∗ ( n ) the set of phas e r etriev able fra mes is generic with resp ect to the Zar iski top olo gy; (B) I f F is a phas e retriev able frame o f m vectors, then m ≥ m ∗ ( n ). Clearly (B) is equiv alen t to: (C) If m < m ∗ ( n ) there is no frame F of m vectors that is also phas e retriev able. The curr e n t state-o f-the-art on this pr oblem is summar ized by the following sta temen ts: (i) s e e [[8]]. If m ≥ 4 n − 2 then generica lly (with resp ect to the Zar iski top ology) a n y fra me is phase retriev able for C n ; (ii) s e e [[44]]. F or generic 4 n × n unitary matr ices on C n , any subset of m = 4 n − 3 columns for ms a phase retriev able frame; (iii) s e e [[3 3]]. If F is a phase retriev able frame in C n then m ≥ 4 n − 2 − 2 β +    2 if n odd and β = 3 mod 4 1 if n odd and β = 2 mod 4 0 otherw ise where β = β ( n ) is the num ber o f 1’s in the binary expansio n of n − 1 . Hence, if such a critica l car dinal exists, we know 4 n − O ( log ( n )) ≤ m ∗ C ( n ) ≤ 4 n − 2. The author s of [11] conjectured that m ∗ C ( n ) = 4 n − 4. In the case m = 4 n − 4 , Bodmann and H ammen constructed [1 5] a phase retr ie v able frame. In sectio n 4 we r eview their co nstruction and we show it is stable under small per turbations. In section 5 we consider the critical cardinality conjecture a nd show that (C) is equiv alent to a stability result for sets o f fra mes that fail to b e phase retriev able. Note the corresp onding problems fo r the re a l case are completely solved. In fact [8] gives a geo metr ic condition equiv alent to a frame b eing phas e r etriev a ble in R n . That condition (namely , for a n y par tition o f the frame set F = F 1 ∪ F 2 , a t lea s t one of F 1 or F 2 m ust span R n ) is stable under small p erturbations . Additionally that same condition implies that m ∗ R = 2 n − 1 is the critical cardina l in the rea l case. 2 2 Notations In this section we rec all some notations we intro duced in [7] tha t will b e us e d in the fo llowing sections. Let F = { f 1 , . . . , f m } b e a frame in H = C n . Let j : H → R 2 n denote the embedding j ( x ) =  rea l ( x ) imag ( x )  (3) which is a unitary isomor phism j be t ween H seen as a real vector s pa ce endow ed with the re a l inner pro duct h x, y i R = r eal ( h x, y i ) and R 2 n : h x, y i R = real ( h x, y i ) = h j ( x ) , j ( y ) i . (4) F or tw o vectors u, v ∈ R 2 n , J u, v K deno tes the symmetr ic outer p o duct J u, v K = 1 2 ( uv T + v u T ) . (5) and similar ly for tw o vector x, y ∈ C n denote by J x, y K their symmetric outer pro duct defined by J x, y K = 1 2 ( xy ∗ + y x ∗ ) (6) F or each n -vector f k we denote by ϕ k the 2 n r e a l vector, and by Φ k the symmetr ic nonneg ative rank-2, 2 n × 2 n matr ix defined by ϕ k = j ( f k ) =  real( f k ) imag( f k )  , Φ k = J ϕ k , ϕ k K + J J ϕ k , J ϕ k K = ϕ k ϕ k T + J ϕ k ϕ k T J T , where J =  0 − I I 0  (7) Note the following key relations: rea l ( h x, f k i ) = h ξ , ϕ k i (8) |h x, f k i| 2 = h Φ k ξ , ξ i (9) rea l ( h x, f k ih f k , y i ) = h Φ k ξ , η i (10) where ξ = j ( x ) a nd η = j ( y ). F or every ξ ∈ R 2 n set R ( ξ ) = m X k =1 J Φ k ξ , Φ k ξ K (11) Let S 1 , 0 and S 1 , 1 denote the following spaces of s ymmetric op era tors ov er a Hilb ert space K (real or complex) S 1 , 0 ( K ) = { T ∈ S y m ( K ) , rank ( T ) ≤ 1 , λ max ( T ) ≥ 0 = λ min ( T ) } (12) S 1 , 1 ( K ) = { T ∈ S y m ( K ) , rank ( T ) ≤ 2 , S p ( T ) = { λ max , 0 , λ min } , λ max ≥ 0 ≥ λ min } (13) where S y m ( K ) denotes the set of sy mmetric op era tors (ma tr ices) on K , S p ( T ) deno tes the sp ectrum (set of eigenv alues) of T , a nd λ max , λ min denote the larges t, and smalles t eigenv alue of the co r resp onding op erator . Note S 1 , 0 ( K ) = { T = J x, x K , x ∈ K } 3 F or the frame F = { f 1 , . . . , f m } we let A deno te the linear op era to r A : S y m ( H ) → R m , ( A ( T )) k = h T f k , f k i = tr ace ( T J f k , f k K ) (14) Note the frame condition (1) reads equiv alently: A k J x , x K k 1 ≤ kA ( J x, x K ) k 2 ≤ B k J x , x K k 1 (15) where k T k 1 = P r ank ( T ) k =1 | λ k ( T ) | denotes the n uclear norm of op era tor T , that is the s um of its singular v alues , or the s um of magnitudes of its eig enalues when T is symmetric. The upp er b ound is obviously alwa ys true (for an appropria te B ) in the case o f finite fr ames. The lower b ound in (1) or (15) is eq uiv alent to the spanning condition spa n( F ) = H . In turn this spa nning c ondition can b e restated in terms of a null space condition for A . Specifica lly let ker( A ) = { T ∈ S y m ( H ) , A ( T ) = 0 } denote the kernel of the linea r op erator A . Then span( F ) = H (and ther efore F is fra me) if and only if ker( A ) ∩ S 1 , 0 = { 0 } (16) Recall the nonlinear map α in tro duced in (2). W e shall consider a lso the squar e map α 2 defined by: α 2 : H → R m , ( α 2 ( x )) k = |h x, f k i| 2 (17) Of cour s e α is injective if and only α 2 is injective. 3 Stabilit y of Phase Retriev a ble F rames W e star t by pr esenting tw o lemmas reg arding the o b jects we introduced earlier. Lemma 3.1 Fix a Hilb ert sp ac e K . (i) A s sets , S 1 , 1 ( K ) = S 1 , 0 ( K ) − S 1 , 0 ( K ) . Explicitely t his me ans: ∀ T ∈ S 1 , 1 ∃ T 1 , T 2 ∈ S 1 , 0 s.t. T = T 1 − T 2 ; Conv ersely : ∀ T 1 , T 2 ∈ S 1 , 0 , T 1 − T 2 ∈ S 1 , 1 ; (ii) F or any T ∈ S 1 , 1 ( K ) ther e ar e u, v ∈ K s o that T = J u, v K ; (iii) F or any u, v ∈ K , J u, v K ∈ S 1 , 1 ( K ) ; (iv) S 1 , 1 ( K ) = { T = J u, v K , u, v ∈ K } . The pr o of of this lemma ca n b e found in [7] Lemma 3.7 . Lemma 3.2 The fol lowing ar e e quivalent: (i) The nonline ar map α is inje ct ive; (ii) ker( A ) ∩ S 1 , 1 = { 0 } (iii) Ther e is a c onstant a 0 > 0 so that m X k =1   |h x, f k i| 2 − |h y , f k i| 2   2 ≥ a 0  k x − y k 2 k x + y k 2 + 4( imag ( h x, y i )) 2  (18) for any x, y ∈ H ∈ C n ; 4 (iv) Ther e is a c onstant a 0 > 0 so that for al l ξ ∈ R 2 n , λ 2 n − 1 ( R ( ξ )) ≥ a 0 k ξ k 2 (her e, λ 2 n − 1 ( T ) denotes t he 2 n − 1 th lar gest eigenvalue of T ); (v) Ther e is a c onstant a 1 > 0 so that for al l ξ ∈ R 2 n , L ( ξ ) := R ( ξ ) + J J ξ , ξ K J T = m X k =1 Φ k ξ ξ T Φ k + J ξ ξ T J T ≥ a 1 k ξ k 2 1 R 2 n (19) wher e t he ine quality is in t he sense of quadr atic forms; (vi) F or every ξ ∈ R 2 n , dim ker R ( ξ ) = 1 ; (vii) F or every ξ ∈ R 2 n , r ank ( R ( ξ )) = 2 n − 1 ; (viii) F or every ξ ∈ R 2 n − 1 , ker( R ( ξ )) = { c J ξ , c ∈ R } ; (ix) Ther e is a c onstant a 0 > 0 so that for al l ξ ∈ R 2 n , R ( ξ ) ≥ a 0 k ξ k 2 (1 − P J ξ ) ( 20) wher e P J ξ = 1 k ξ k 2 J J ξ , J ξ K is the ort ho gonal pr oje ction ont o J ξ . Remark 3. 3 Bef or e pr esenting t he pr o of, note the c onstants a 0 at (iii), (iv) and (ix) c an b e chosen to b e e qu al. Henc e the optimal (i.e. the lar gest) a 0 is given by a 0 = min k ξ k =1 λ 2 n − 1 ( R ( ξ )) (21) A ddi tional ly, the c onstant a 1 at (v) c an b e chosen as a 1 = min (1 , a 0 ) . Pro of o f Lemm a 3 .2 Claims (i),(ii),(iv),(vi)-(ix) are known in literature - see for instance [7] Theo rem 2.2 and the bibliograph- ical indications - and Theor em 3.1 o f [7]. Claim (v) follows from (ix) by adding k ξ k 2 P J ξ on b oth sides. Claim (iii) follows from Theorem 3.1 (2) of [7], where w e set u = x − y and v = x + y and b y remarking imag ( h u, v i ) = 2 i mag ( h x, y i ) and r eal ( h u, f k ih f k , v i ) = r eal ( |h x, f k i| 2 − |h y , f k i| 2 ). ✷ Recall t wo frames F = { f 1 , . . . , f m } and G = { g 1 , . . . , g m } fo r the s ame Hilb ert space H a re said e qu ivalent if there is a n inv ertible op erato r T : H → H so that g k = T f k , for all 1 ≤ k ≤ m (see [3, 3 2]).The pr op erty of b eing phase retriev able is inv ar iant amo ng equiv a lent fr ames, as the following lemma shows. Lemma 3.4 Ass u me F = { f 1 , . . . , f m } is a phase r etrievable fr ame for H . Then (i) F or any invertible op er ator T : H → H and n on-zer o sc alars z 1 , . . . , z m ∈ K , the fr ame G = { g 1 , . . . , g m } define d by g k = z k T f k , 1 ≤ k ≤ m , is also phase r etrievable; (ii) F or any invertible op er ator T : H → H , the e quivalent fr ame G = { g 1 , . . . , g m } define d by g k = T f k , 1 ≤ k ≤ m is also phase r etrievable; (iii) The c anonic al dual fr ame ˜ F = { ˜ f 1 , . . . , ˜ f m } is also phase r etrievable, wher e ˜ f k = S − 1 f k , 1 ≤ k ≤ m ; 5 (iv) The asso cia te d Parseval fr ame F # = { f # 1 , . . . , f # m } is also phase r etrievable, wher e f # k = S − 1 / 2 f k , 1 ≤ k ≤ m ; (v) A ny finite set of ve ct ors G ⊂ H so that F ⊂ G is a phase r etrievable fr ame; (vi) If G ⊂ H is not a phase r etrievable fr ame then any subset H ⊂ G is also not a phase r etrievable fr ame. Pro of o f Lemm a 3 .4 (i) Note that each z k 6 = 0 and hence G is also frame. Let α G : ˆ H → R m be the nonlinear map a sso ciated to G , ( α G ( x )) k = |h x, g k i| 2 . If x, ∈ ˆ H ar e so that α G ( x ) = α G ( y ) then α ( T ∗ x ) = α ( T ∗ y ). Since F is phas e retriev able it follows T ∗ x = T ∗ y and hence x = y . (Note a n y o per ator R : H → H lifts to a unique op era tor R : ˆ H → ˆ H that is denoted using the same letter). (ii)-(iv) follows fro m (i). Claims (v) a nd (vi) are obvious. ✷ Remark 3. 5 While the claim (vi) in pr evio us L emma is obvious, the fol lowing quest ion is not so obvious. Assume F is a phase r etrievabl e fr ame in the r e al c ase (that is H = R n ). We know m ≥ 2 n − 1 . Assu me m > 2 n − 1 . Do es ther e always exist a su bset G ⊂ F so that G is a phase r etrievable fr ame? In ter estingly the answer to this question is ne gative. The fol lowing example shows this phenomenon (similar example was pr op ose d by [24]). Example 3. 6 Co nsider H = R 3 and m = 6 wher e t he 6 ve ct ors ar e: f 1 =   1 0 0   , f 2 =   0 1 0   , f 3 =   0 0 1   , f 4 =   1 1 0   , f 5 =   1 0 1   , f 6 =   0 1 1   (22) The asso ciate d r ank-1 op er ators F k = f k f T k , 1 ≤ k ≤ 6 , b elong to the line ar sp ac e of symmetric 3 × 3 matric es S y m ( R 3 ) . Note the S y m ( R 3 ) is a r e al ve ctors sp ac e of dimension 6. The Gr am matrix G (2) asso ciate d to { F 1 , . . . , F 6 } is a 6 × 6 symmetric matrix of ent r ies G (2) k,l = h F k , F l i = |h f k , f l i| 2 , which ar e the s qu ar e of the entries of Gr am matrix asso ciate d to F . E xplicitely G (2) is given by G (2) =         1 0 0 1 1 0 0 1 0 1 0 1 0 0 1 0 1 1 1 1 0 4 1 1 1 0 1 1 4 1 0 1 1 1 1 4         (23) Its determinant is det ( G (2) ) = 8 . Henc e { F 1 , . . . , F 6 } is a b asis for S y m ( R 3 ) and thus k er ( A ) = { 0 } which implies F is a phase r etrievable fr ame. On the other hand c onside r any subset G of 5 ve ctors of F . It is e asy to che ck G is a fr ame for R 3 . Howeverfor e ach G ther e is a s u bset of 3 elements that is not line arly indep endent, henc e c annot sp an R 3 . This fact to gether with Cor ol lary 2.6 fr om [8] pr oves t hat G is n ot phase r etrieva ble. Thus we c onstructe d a fr ame F of 6 ve ctors (which is mor e than the critic al c ar dinal 2 n − 1 = 5 ) so that any subset is not phase r etrievable. 5mm W e ar e now ready to pre s ent the pro of o f Theorem 1.1. Pro of o f Theorem 1.1 6 Assume F is a phase r etriev a ble fra me. Then equa tio n (19) is satisfied for so me a 1 > 0. Let B b e the upper frame b ound for F . Then set: ρ = min( 1 √ m , a 1 4(3 B + 2 ) 3 / 2 ) (24) W e will find a ρ > 0 so that (19) is satisfied for any set F ′ = { f ′ 1 , . . . , f ′ m } with k f k − f ′ k k < ρ . Let 0 < A ≤ B < ∞ b e the fra me b ounds of F and let L ′ ( ξ ) denote the r ight hand side in (19 ) asso c ia ted to F ′ . W e compute |h L ( ξ ) η , η i − h L ′ ( ξ ) η , η i| ≤ m X k =1 | |h Φ k ξ , η i| 2 − |h Φ ′ k ξ , η i| 2 | ≤ m X k =1 ( |h Φ k ξ , η i| + |h Φ ′ k ξ , η i| ) |h (Φ k − Φ ′ k ) ξ , η i| ≤ m X k =1 |h Φ k ξ , η i| + m X k =1 |h Φ ′ k ξ , η i| ! max 1 ≤ k ≤ m |h (Φ k − Φ ′ k ) ξ , η i| Fix ξ ∈ R 2 n with k ξ k = 1 . Then max k η k =1 m X k =1 |h Φ k ξ , η i| = m X k =1 h Φ k ξ , ξ k ξ k i ≤ B k ξ k Thu s for any ξ , η ∈ R 2 n , m X k =1 |h Φ k ξ , η i| ≤ B k ξ k k η k , m X k =1 |h Φ ′ k ξ , η i| ≤ B ′ k ξ k k η k where B ′ is the upp er frame b ound of F ′ . On the other ha nd we b ound |h (Φ k − Φ ′ k ) ξ , η i| ≤ k Φ k − Φ ′ k k k ξ k k η k According to Lemma 3.12 (4 ) from [7], k Φ k − Φ ′ k k = k F k − F ′ k k , where F k = f k f ∗ k and F ′ k = f ′ k f ′∗ k . Note F k − F ′ k ∈ S 1 , 1 ( C n ) a nd F k − F ′ k = J f k − f ′ k , f k + f ′ k K . Now using Lemma 3.7 (4) from [7], we obtain k F k − F ′ k k ≤ k F k − F ′ k k 1 = q k f k − f ′ k k 2 k f k + f ′ k k 2 + ( imag ( h f k − f ′ k , f k + f ′ k i )) 2 ≤ √ 2 k f k − f ′ k k k f k + f ′ k k where k T k 1 is the nuclear nor m (the sum o f its singular v alues) of T . Next notice k f k + f ′ k k ≤ k f k k + k f ′ k k ≤ √ B + √ B ′ ≤ p 2( B + B ′ ). Putting all the estimates to g ether we obtain: |h L ( ξ ) η , η i − h L ′ ( ξ ) η , η i| ≤ 2( B + B ′ ) 3 / 2  max 1 ≤ k ≤ m k f k − f ′ k k  k ξ k 2 k η k 2 Thu s L ′ ( ξ ) ≥ ( a 1 − 2( B + B ′ ) 3 / 2 ρ ) k ξ k 2 1 R 2 n Finally we obtain an estimate of B ′ in ter ms o f B , ρ and m . This estimate can b e further r e fined, but we do not need such a r e finmen t for this pro of. Let δ k = f ′ k − f k . Then m X k =1 |h x, f ′ k i| 2 = m X k =1 |h x, f k i + h x, δ k i| 2 ≤ 2 m X k =1 |h x, f k i| 2 + m X k =1 |h x, δ k i| 2 ! = 2( B + m max k k δ k k 2 ) k x k 2 (25) 7 Due to (24) we obtain B ′ = sup k x k =1 P m k =1 |h x, f ′ k i| 2 ≤ 2( B + 1). This b ound implies that L ′ ( ξ ) ≥ a 1 2 k ξ k 2 1 R 2 n and hence F ′ is phase retriev able. ✷ 4 Critical case m = 4 n − 4 This se ction co mmen ts o n the recent construction by Bo dma nn a nd Hammen [15] of a 4 n − 4 phase retr ie v able frame in C n . Their co nstruction is as follows. Fix a ∈ R \ π Q , an irra tional multiple of π . The fra me set F is given by a union of tw o sets, F = F 1 ∪ F a 2 , where F 1 constains the following 2 n − 3 vectors: F 1 = { f 1 k =  1 e 2 π i ( k +1) / (2 n − 1) e 2 π i ( k +1)2 / (2 n − 1) · · · e 2 π i ( k +1)( n − 1) / (2 n − 1)  T , 1 ≤ k ≤ 2 n − 3 } (26) and F a 2 contains the following 2 n − 1 vectors: F a 2 = { f 2 k =  1 z k z 2 k · · · z n − 1 k  T , 1 ≤ k ≤ 2 n − 1 } (27) where z k = sin  π 2 n − 1  sin ( a ) e i k − 1 2 n − 1 − e i ( π 2 n − 1 − a 2 ) sin  π 2 n − 1 − a 2  sin ( a ) (28) The pro of that F is a pha se retrie v able fra me is based o n a result by P . J amming from [37]. O ur Theorem 1.1 pr ov es that, in fact, F rema ins phase retr iev able for a small perturba tion. Since f 2 k depe nds contin uously on a , it follows that the set R \ π Q ca n b e replaced by a m uch larg er set o f r e a l nu mbers that includes most of rational multiples of π . Going thr ough the pro o f o f Theo rem 2.3 in [15] , a nd in par ticular o f Le mma 2 .2, the only requirement on a is that, any set of 2( n − 1 ) complex num b e r s cannot b e simultaneoulsy symmetric with r esp e c t to the real line a nd to a line of angle a pas s ing throug h the origin. This phenomenon happ ens for any n when a is an irr ational multiple of π . How ever, for a fixed n , only finitely ma n y v alues of a may allow such a symmetry . In fact when such a symmetric set of 2 ( n − 1) complex num bers ex ists, a = π p q for some q ≤ 2( n − 1). Thus the frame set a bove F = F 1 ∪ F a 2 remains phase retr iev able for all v alues of a except a finite set o f the form { π p q , 0 ≤ p ≤ 2 q ≤ 4( n − 1) } . 5 Non Phase-Retriev able F rames Consider now the case when m is ” s mall”. The conjecture in [1 1] reads that for m < 4 n − 4 there is no phase re tr iev able frame. In this section we co mmen t on a pa rtial res ult supp orting this conjecture. The main res ult of this section is the following Prop ositio n 5. 1 Fix the n -dimensional Hilb ert sp ac e H . Denote by m ∗ ( n ) the critic al c ar dinal m ∗ ( n ) = 4 n − 4 . Assume the fol lowing statement holds tru e for some m < m ∗ ( n ) : (O) F or any fr ame set F = { f 1 , . . . , f m } ⊂ H that is not a phase r etrievable fr ame for H ther e exists a ρ > 0 so that any other set F ′ = { f ′ 1 , . . . , f ′ m } with k f k − f ′ k k < ρ , 1 ≤ k ≤ m , is also not phase re trievable. Then any subset G = { g 1 , . . . , g m } ⊂ H of m ve ctors in H is not a phase r etrievabl e fr ame. 8 Remark 5. 2 Bef or e pr esenting its pr o of, we make the fol lowing r emark. While not pr oving the fu l l l c on- je ctu r e, this r esult r e duc es the pr o of of the 4 n − 4 c onje ct u r e to a stability r esult for n on phase r etrievable fr ames. A dd itional ly the r esult holds true even if the critic al c ar dinal is not 4 n − 4 . However t he aut hor r e c o gnizes this is just a p artial r esult that do es not pr ove the ful l c onje ctur e. Remark 5. 3 Note t he set F is supp ose d to b e fr ame. If (O) holds for any set of m ve ctors of H , one c an use the t rivial set { 0 , 0 , . . . , 0 } of m ve ctors. Then if (O) holds for this sp e cial set, then any set of m ve ctors whose norms ar e less than some ρ > 0 is non phase r etrievable fr ame. By sc aling we obtin imme diately that any m -set of ve ctors is not phase r etrievable fr ame. Pro of o f Prop os ition 5.1 First note that if m < 2 n any set F of m elements canno t b e a pha se r etriev able frame (confor m [28]). Hence we c a n a ssume m ≥ 2 n . Let H m = H × H × · · · × H deno te the m -pro duct space endow ed with the top ology induced by the no rm k x k H m = max 1 ≤ k ≤ m   x k   , x = ( x 1 , . . . , x m ) ∈ H m Note that H m is ho meomorphic with C nm endow ed with the usual Euclidian no rm. Let F m n denote the set o f frames fo r H with m elemen ts. F m n is a n o pen set in H m since each frame set is sta ble under s ma ll per turbations: for instance this c a n b e seen us ing an es timate similar to equation (25) that bounds below P m k =1 |h x, f ′ k i| 2 : m X k =1 |h x, f ′ k i| 2 ≥ m X k =1 |h x, f k i| 2 −   2 √ m m X k =1 |h x, f k i| 2 ! 1 / 2  max k k f k − f ′ k k  k x k + m  max k k f k − f ′ k k  2 k x k 2   ≥ ( A − 2 √ mB ρ − mρ 2 ) k x k 2 Thu s for s o me sufiiciently small ρ , A − 2 √ mB ρ − mρ 2 > 0 and { f ′ 1 , . . . , f ′ m } is also frame when k f k − f ′ k k < ρ , 1 ≤ k ≤ m . Assume the hypo thesis (O ) holds true. Let N m n denote the set o f non phase- r etriev able frames of m vectors in H . Thus N m n ⊂ F m n ⊂ H m is an op en set in H by hypothesis (O). On the other hand the complement Γ m n := F m n \ N m n represents the set of phase re tr iev able frames. Theorem 1 .1 shows Γ m n is op en in H m . Now let us show the set of fra mes F m n is connected in H m . Firstly tw o equiv alen t fr ames are connected by path as sho wn e.g. in [3]. W e will sho w that any tw o f rames of m elements for the n dimensiona l Hilber t spa c e H can b e connected by a contin uous path (in fact t wo seg men ts of line), when m ≥ 2 n . Let F 1 = { f 1 1 , . . . , f 1 m } and F 2 = { f 2 1 , . . . , f 2 m } be t w o m -frames. Let I = { k 1 , . . . , k n } be the n - set so that F 1 [ I ] = { f 1 k 1 , . . . , f 1 k n } is a linear ly independent subset of F 1 . Let J = { j 1 , . . . , j m − n } be a m − n -s et so that F 2 [ J ] = { f 2 j , j ∈ J } is a frame for H . Let γ : I c → J and δ : J c → I be t w o bijective maps where I c = { 1 ≤ k ≤ m } \ I and J c = { 1 ≤ j ≤ m } \ J are the complement sets o f I a nd J resp ectively . W e build a piecewise linear path β : [ − 1 , 1] → H m connecting F 1 to F 2 as follows: F or − 1 ≤ t ≤ 0 , ( β ( t )) k =  f 1 k if k ∈ I − tf 1 k + ( t + 1) f 2 γ ( k ) if k ∈ I c F o r 0 ≤ t ≤ 1, ( β ( t )) k =  f 2 j if j ∈ J tf 1 δ ( j ) + (1 − t ) f 2 j if j ∈ J c 9 One can eas ilt y chec k that β ( t ) is a fr ame for each − 1 ≤ t ≤ 1, and β ( − 1) = F 1 , β (1) = F 2 . This prov es the se t of frames F m n is path c onnected, hence connected. W e obtained that the connected set F m n can b e par titioned into tw o o pens se ts Γ m n and N m n . It follows that o ne of the tw o sets must b e the empt y set. How ev er we c a n a lw ays co ns truct a non phase retr iev able frame, for instance F = { e 1 , . . . , e n , e n , . . . , e n } where { e 1 , . . . , e n } is a basis of H and the vector e n is rep eated a total of m − n + 1 times. This shows Γ m n m ust b e empty . 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