On Lipschitz Analysis and Lipschitz Synthesis for the Phase Retrieval Problem

Transcription

1 On Lipschitz Analysis and Lipschitz Synthesis for the Phase Retrieval Problem Radu Balan a, Dongmian Zou b, a Department of Mathematics, Center for Scientific Computation and Mathematical Modeling, University of Maryland, College Park, MD 20742, USA b Department of Mathematics, University of Maryland, College Park, MD 20742, USA Abstract We prove two results with regard to reconstruction from magnitudes of frame coefficients the so called phase retrieval problem First we show that phase retrievable nonlinear maps are bi-lipschitz with respect to appropriate metrics on the quotient space Specifically, if nonlinear analysis maps α, β : Ĥ R m are injective, with αx = x, f k m k=1 and βx = x, f k 2 m k=1, where {f 1,, f m } is a frame for a Hilbert space H and Ĥ = H/T 1, then α is bi-lipschitz with respect to the class of natural metrics D p x, y = min ϕ x e iϕ y p, whereas β is bi-lipschitz with respect to the class of matrix-norm induced metrics d p x, y = xx yy p Second we prove that reconstruction can be performed using Lipschitz continuous maps That is, there exist left inverse maps synthesis maps ω, ψ : R m Ĥ of α and β respectively, that are Lipschitz continuous with respect to appropriate metrics Additionally, we obtain the Lipschitz constants of ω and ψ in terms of the lower Lipschitz constants of α and β, respectively Surprisingly, the increase in both Lipschitz constants is a relatively small factor, independent of the space dimension or the frame redundancy Keywords: Frames, Lipschitz maps, Stability, Phase retrieval 2010 MSC: 15A29, 65H10, 90C26 1 Introduction Let H be an n-dimensional real or complex Hilbert space On H we consider the equivalence relation defined by x y iff there is a scalar a of magnitude one, a = 1, for which y = ax Let Ĥ = H/ denote the collection of the equivalence classes We use ˆx to denote the equivalence class of x in Ĥ When there is no ambiguity, we also use x in place of ˆx for simplicity Corresponding author addresses: rvbalan@mathumdedu Radu Balan, zou@mathumdedu Dongmian Zou Preprint submitted to Linear Algebra and its Applications November 11, 2015

2 Assume that F = {f 1, f 2,, f m } is a frame that is, a spanning set for H Let α and β denote the nonlinear maps and α : Ĥ Rm, αx = x, f k 1 k m, 1 β : Ĥ Rm, βx = x, f k 2 1 k m 2 The phase retrieval problem, or the phaseless reconstruction problem, refers to analyzing when α or equivalently, β is an injective map, and in this case to finding good left inverses The frame F is said to be phase retrievable if the nonlinear map α or β is injective In this paper we assume α and β are injective maps hence F is phase retrievable The problem is to analyze the stability properties of phaseless reconstruction We explore this problem by studying Lipschitz properties of these nonlinear maps A continuous map f : X, d X Y, d Y, defined between metric spaces X and Y with distances d X and d Y respectively, is Lipschitz continuous with Lipschitz constant Lipf if Lipf := d Y fx 1, fx 2 sup x 1,x 2 X d X x 1, x 2 < Further, the map f is called bi-lipschitz with lower Lipschitz constant a and upper Lipschitz constant b if for every x 1, x 2 X, a d X x 1, x 2 d Y fx 1, fx 2 b d X x 1, x 2 Obviously the smallest upper Lipschitz constant is b = Lipf If f is bi-lipschitz then f is injective The space Ĥ admits two classes of inequivalent distances We introduce and study them in detail in section 2 In particular, consider the following two distances: D 2 x, y = min x e iϕ y ϕ 2 = x 2 + y 2 2 x, y, and d 1 x, y = xx yy 1 = x 2 + y x, y 2 When the frame is phase retrievable the nonlinear maps α : Ĥ, D 2 R m, 2 and β : Ĥ, d 1 R m, 2 are shown to be bi-lipschitz This statement was previously known for the map β in the real and complex case see [2, 3, 6], and for the map α in the real case only see [13, 6, 8] In this paper we prove this statement for α in the complex case In general, noisy measurements are not in the image of the analysis map αĥ or βĥ In this paper we prove that the unique left inverses of α and β can be extended from αĥ and βĥ, respectively, to the entire space Rm while the extended maps remain to be Lipschitz 2

3 continuous Specifically, there exist two Lipschitz continuous maps ω : R m, 2 Ĥ, D 2 and ψ : R m, 2 Ĥ, d 1 so that ωαx = x and ψβx = x for every x H Consider one of the maps α and β, say α a similar discussion works for β Assume an additive noise model y = αx + ν, where ν R m is the noise For a signal x 0 Ĥ, and noise ν 1 R m, let y 1 = αx 0 + ν 1 R m be the measurement vector, and let x 1 = ωy 1 be the reconstructed signal We have d 1 x 0, x 1 = d 1 ωαx 0, ωy 1 Lipω αx 0 y 1 = Lipω ν 1 Figure 1 is an illustration of this model In fact, we have stability in a stronger sense If we have two noisy measurements y 1 = αx 0 + ν 1 and y 2 = αx 0 + ν 2 of the signal x 0, then d 1 x 1, x 2 = d 1 ωy 1, ωy 2 Lipω y 1 y 2 = Lipω ν 1 ν 2 Figure 1: Illustration of the noisy measurement model Denote by a α and a β the lower Lipschitz constants of α and β respectively In this paper we prove also that the upper Lipschitz constants of these maps obey Lipω 825 a α and Lipψ 825 a β Surprisingly, this shows the Lipschitz constant of these left inverses are just a small factor larger than the minimal Lipschitz constants Furthermore this factor is independent of dimension n or number of frame vectors m The organization of this paper is as follows Section 2 introduces notations and presents the results for bi-lipschitz properties Section 3 presents the results for the extension of the left inverse Section 4 contains the proof of these results 2 Bi-Lipschitz Properties for the Analysis Map 21 Notations To study the bi-lipschitz properties, we need to choose an appropriate distance on Ĥ We consider two classes of metrics distances, respectively: 3

4 1 the class of natural metrics For every 1 p and x, y H, we define D p ˆx, ŷ = min a =1 x ay p When no subscript is used, denotes the Euclidean norm, = 2 2 the class of matrix norm induced metrics For every 1 p and x, y H, we define d p ˆx, ŷ = xx yy p = { n k=1 σ k p 1/p for 1 p < max 1 k n σ k for p =, 3 where σ k 1 k n are the singular values of the operator xx yy, which is of rank at most 2 Here x denotes the adjoint of x see [3] for a detailed discussion, which is the transpose conjugate of x if H = R n or C n Our choice in 3 corresponds to the class of Schatten norms In particular, d corresponds to the operator norm op in SymH = {T : H H, T = T }; d 2 corresponds to the Frobenius norm F r in SymH; d 1 corresponds to the nuclear norm in SymH Specifically, we have d x, y = xx yy op, d 2 x, y = xx yy F r, d 1 x, y = xx yy Note that the Frobenius norm T F r = tracet T induces the Euclidean distance on SymH As a consequence of Lemma 38 in [3], we have: d x, y = 1 2 x 2 y x 2 + y x, y 2, d 2 x, y = x 4 + y 4 2 x, y 2, d 1 x, y = x 2 + y x, y 2 To study the above distances it is important to study eigenvalues of symmetric matrices Let S p,q H denote the set of symmetric operators that have at most p strictly positive eigenvalues and q strictly negative eigenvalues In particular, S 1,0 H is the set of nonnegative symmetric operators of rank at most one: S 1,0 H = { xx, x H } 4 If H = R n or C n, then SymH is the set of n-dimensional Hermitian matrices For a matrix X SymR n or SymC n, we use λ 1 X, λ 2 X,, λ n X to denote its eigenvalues These eigenvalues are real numbers and we arrange them to satisfy λ 1 X λ 2 X λ n X To analyze the bi-lipschitz properties, we define the following three types of Lipschitz bounds for α Note that the Lipschitz constants are square-roots of those bounds 4

5 i The global lower and upper Lipschitz bounds, respectively: αx αy 2 2 A 0 = inf, x,y Ĥ D 2 x, y 2 αx αy 2 2 B 0 = sup ; x,y Ĥ D 2 x, y 2 ii The type I local lower and upper Lipschitz bounds at z Ĥ, respectively: Az = lim r 0 Bz = sup r 0 inf x,y Ĥ D 2 x,z<r D 2 y,z<r inf x,y Ĥ D 2 x,z<r D 2 y,z<r αx αy 2 2 D 2 x, y 2, αx αy 2 2 D 2 x, y 2 ; iii The type II local lower and upper Lipschitz bounds at z Ĥ, respectively: Ãz = lim Bz = sup inf r 0 x Ĥ D 2 x,z<r inf r 0 x Ĥ D 2 x,z<r αx αz 2 2 D 2 x, z 2, αx αz 2 2 D 2 x, y 2 Similarly, we define the three types of Lipschitz constants for β i The global lower and upper Lipschitz bounds, respectively: βx βy 2 2 a 0 = inf, x,y Ĥ d 1 x, y 2 βx βy 2 2 b 0 = sup ; x,y Ĥ d 1 x, y 2 ii The type I local lower and upper Lipschitz bounds at z Ĥ, respectively: az = lim r 0 bz = lim r 0 inf x,y Ĥ d 1 x,z<r d 1 y,z<r sup x,y Ĥ d 1 x,z<r d 1 y,z<r 5 βx βy 2 2 d 1 x, y 2, βx βy 2 2 d 1 x, y 2 ;

6 iii The type II local lower and upper Lipschitz bounds at z Ĥ, respectively: ãz = lim inf r 0 x Ĥ d 1 x,z<r βx βz 2 2 d 1 x, z 2, bz = lim r 0 sup x Ĥ d 1 x,z<r βx βz 2 2 d 1 x, z 2 Due to homogeneity we have A 0 = A0, B 0 = B0, a 0 = a0, b 0 = b0 Also, for z 0, we have Az = Az/ z, Bz = Bz/ z, az = az/ z, bz = bz/ z We analyze the bi-lipschitz properties of α and β by studying these constants 22 Bi-Lipschitz Properties for α The real case H = R n is studied in [6] We summarize the results as a theorem Recall that F = {f 1,, f m } is a frame in H if there exist positive constants A and B for which m A x 2 x, f k 2 B x 2 5 k=1 We say A resp, B is the optimal lower resp, upper frame bound if A resp, B is the largest resp, smallest positive number for which the inequality 5 is satisfied For any index set I {1, 2,, m}, let F[I] = {f k, k I} denote the frame subset indexed by I Also, let σ1[i] 2 and σn[i] 2 denote the upper and lower frame bound of set F[I], respectively It is straightforward to see that they respectively correspond to the largest and smallest eigenvalues of k I f kfk, that is, σ 2 1[I] = λ max k I f k f k and σ 2 n[i] = λ min k I f k f k Theorem 21 [6] Let F R n be a phase retrievable frame for R n Let A and B denote its optimal lower and upper frame bound, respectively Then i For every 0 x R n, Ax = σ 2 nsuppαx where suppαx = {k, x, f k 0}; ii For every x R n, Ã = A; iii A 0 = A0 = min I {1,2,,m} σ 2 n[i] + σ 2 n[i c ]; iv For every x R n, Bx = Bx = B; v B 0 = B0 = B0 = B Now we consider the complex case H = C n We analyze the complex case by doing a realification first Consider the R-linear map j : C n R 2n defined by [ ] realz jz = imagz 6

7 This realification is studied in detail in [3] We call jz the realification of z For simplicity, in this paper we will denote ξ = jx, η = jy, ζ = jz, ϕ = jf, δ = jd, respectively For a frame set F = {f 1, f 2,, f m }, define the symmetric operator Φ k = ϕ k ϕ T k + Jϕ k ϕ T k J T, k = 1, 2,, m where is a matrix in R 2n 2n Also, define S : R 2n SymR 2n by { Sξ = J = k:φ k ξ 0 [ 0 I I 0 We have the following result proved in Section 4: ] 0, if ξ = 0 1 Φ k ξ,ξ kξξ T Φ k, if ξ 0 6 Theorem 22 Let F C n be a phase retrievable frame for C n Let A and B denote its optimal lower and upper frame bound, respectively For any z C n, let ζ = jz be its realification Then i For every 0 z C n, Az = λ 2n 1 Sζ ; ii A 0 = A0 > 0 ; iii For every z C n, Ãz = λ 2n 1 Sζ + k: z,f k =0 Φ k ; iv Ã0 = A ; v For every z C n, Bz = Bz = λ 1 Sζ + k: z,f k =0 Φ k vi B 0 = B0 = B0 = B ; 23 Bi-Lipschitz Properties for β The nonlinear map β naturally induces a linear map between the space SymH of symmetric operators on H and R m : A : SymH R m, AT = T f k, f k 1 k m This linear map has first been observed in [5] and it has been exploited successfully in various papers eg [1, 11, 2] Note that the map β is injective if and only if A restricted to S 1,0 H is injective In previous papers [3, 6], the authors establish global bi-lipschitz results for phaseretrievable frames We summarize them as follows: Theorem 23 [3], [6] Let F be a phase retrievable frame for H = C n Then i the global lower Lipschitz bound a 0 > 0; 7

8 ii the global upper Lipschitz bound b 0 <, and b 0 = max x = y =1 = max x =1 m real x, f k f k, y 2 k=1 m x, f k 4 k=1 = T 4 BH,l 4 m, where T : H C m is the analysis operator defined by x x, f k m k=1, and l4 m := C m, 4 Remark 24 An upper bound of b 0 is given by 2 b 0 B max f k B 2, 1 k m where B is the upper frame bound of F by We give an expression of the local Lipschitz bounds as well Define R : R 2n SymR 2n Rξ = m Φ k ξξ T Φ k k=1 Theorem 25 Let F be a phase retrievable frame for H = C n For every 0 z H, let ζ = jz denote the realification of z Then i az = ãz = λ 2n 1 Rζ/ ζ 2 ; ii bz = bz = λ 1 Rζ/ ζ 2 ; iii see [3] a0 = a 0 = min ζ =1 λ 2n 1 Rζ; iv ã0 = min x =1 m k=1 x, f k 4 ; v b0 = b0 = b 0 3 Extension of the Inverse Map The results in this section work for both H = R n and C n First we show that all metrics D p and d p defined in Section 2 induce the same topology in the following result Proposition 31 We have the following statements regarding D p and d p : i For each 1 p, D p and d p are metrics distances on Ĥ ii D p 1 p are equivalent metrics, that is each D p induces the same topology on Ĥ as D 1 Additionally, for every 1 p, q the embedding i : Ĥ, D p Ĥ, D q, ix = x, is Lipschitz with Lipschitz constant L D p,q,n = max1, n 1 q 1 p 7 8

9 iii For 1 p, q, d p 1 p are equivalent metrics, that is each d p induces the same topology on Ĥ as d 1 Additionally, for every 1 p, q the embedding i : Ĥ, d p Ĥ, d q, ix = x, is Lipschitz with Lipschitz constant L d p,q,n = max1, 2 1 q 1 p 8 iv The identity map i : Ĥ, D p Ĥ, d p, ix = x, is continuous with continuous inverse However it is not Lipschitz, nor is its inverse v The metric space Ĥ, D p is Lipschitz isomorphic to S 1,0 H endowed with Schatten norm p The isomorphism is given by the map κ α : Ĥ S1,0 H, κ α x = { 1 x xx if x 0 0 if x = 0 9 The embedding κ α is bi-lipschitz with the lower Lipschitz constant and the upper Lipschitz constant min p, n 1 p maxn p, 2 1 p 1 2 In particular, for p = 2, the lower Lipschitz constant is 1 and the upper Lipschitz constant is 2 vi The metric space Ĥ, d p is isometrically isomorphic to S 1,0 H endowed with Schatten norm p The isomorphism is given by the map κ β : Ĥ S1,0 H, κ β x = xx 10 In particular the metric space Ĥ, d 1 is isometrically isomorphic to S 1,0 H endowed with the nuclear norm 1 vii The nonlinear map ι : Ĥ, D p Ĥ, d p defined by { x x if x 0 ιx = 0 if x = 0 is bi-lipschitz with the lower Lipschitz constant min p, n p 1 2 and the upper Lipschitz constant 2 maxn p, 2 p 1 2 Remark 32 i Note that the Lipschitz bound L D p,q,n is equal to the operator norm of the identity between C n, p and C n, q : L D p,q,n = I l p n ln q ii Note the equality L d p,q,n = L D p,q,2 9

10 The results in Section 2, together with the previous proposition, show that if the frame F is phase retrievable, then the nonlinear map α resp, β is bi-lipschitz between the metric spaces Ĥ, D p resp, Ĥ, d p and R m, q Recall that the Lipschitz constants between Ĥ, D 2 resp, Ĥ, d 1 and R m, = 2 are given by A 0 resp, a 0 and B 0 resp, b0 : A0 D 2 x, y αx αy B 0 D 2 x, y, 11 a0 d 1 x, y βx βy b 0 d 1 x, y 12 Clearly the inverse map defined on the range of α resp, β from metric space αĥ, resp, βĥ, to Ĥ, D 2 resp, Ĥ, d 1: ω : αĥ Rm Ĥ, ωc = x if αx = c ; 13 ψ : βĥ Rm Ĥ, ψc = x if βx = c 14 is Lipschitz with Lipschitz constant 1/ A 0 resp, 1/ a 0 We prove that both ω and ψ can be extended to the entire R m as a Lipschitz map, and its Lipschitz constant is increased by a small factor The precise statement is given in the following Theorem, which is the main result of this paper Theorem 33 Let F = {f 1,, f m } be a phase retrievable frame for the n dimensional Hilbert space H, and let α, β : Ĥ R m denote the injective nonlinear analysis maps as defined in 1 and 2 Let A 0 and a 0 denote the positive constants as in 11 and 12 Then i there exists a Lipschitz continuous function ω : R m Ĥ so that ωαx = x for all x Ĥ For any 1 p, q, ω has an upper Lipschitz constant Lipω p,q between R m, p and Ĥ, D q bounded by: Lipω p,q A0 2 1 q max1, m 2 1 p for q 2; q A0 n q max1, m 2 1 p for q > 2 Explicitly this means: for q 2 and for all c, d R m : D q ωc, ωd A0 2 1 q 1 2 max1, m p c d p, 16 whereas for q > 2 and for all c, d R m : D q ωc, ωd q n q max1, m 2 1 p c d p 17 A0 In particular, for p = 2 and q = 2 its Lipschitz constant Lipω 2,2 is bounded by a0 : D 2 ωc, ωd a0 c d 18 10

11 ii there exists a Lipschitz continuous function ψ : R m Ĥ so that ψβx = x for all x Ĥ For any 1 p, q, ψ has an upper Lipschitz constant Lipψ p,q between R m, p and Ĥ, d q bounded by: Lipψ p,q a0 2 1 q 1 2 max1, m p for q 2; Explicitly this means: for q 2 and for all c, d R m : q a0 max1, m p for q > 2 19 d q ψc, ψd a0 2 1 q 1 2 max1, m p c d p, 20 whereas for q > 2 and for all c, d R m : d q ψc, ψd q max1, m p c d p 21 a0 In particular, for p = 2 and q = 1 its Lipschitz constant Lipψ 2,1 bounded by a0 : d 1 ψc, ψd a0 c d 22 The proof of Theorem 33, presented in Section 4, requires the construction of a special Lipschitz map We believe this particular result is interesting in itself and may be used in other constructions This construction is given in [7] for the case p = 2 Here we consider a general p and give a better bound for the Lipschitz constant We state it as a lemma Lemma 34 Consider the spectral decomposition of any self-adjoint operator A in SymH, say A = d k=1 λ mkp k, where λ 1 λ 2 λ n are the n eigenvalues including multiplicities, and P 1,,P d are the orthogonal projections associated to the d distinct eigenvalues Additionally, m1 = 1 and mk + 1 = mk + rk, where rk = rankp k is the multiplicity of eigenvalue λ mk Then the map satisfies the following two properties: π : SymH S 1,0 H, πa = λ 1 λ 2 P 1 23 i for 1 p, π is Lipschitz continuous from SymH, p to S 1,0 H, p with Lipschitz constant Lipπ p ; ii πa = A for all A S 1,0 H Remark 35 Numerical experiments suggest that the Lipschitz constant of π is smaller than 5 for p = On the other hand it cannot be smaller than 2 as the following example shows 11

12 Example 36 If A =, B =, then πa = and πb = Here we have πa πb = 2 and A B = 1 Thus for this example we have πa πb = 2 A B It is unlikely to obtain an isometric extension in Theorem 33 Kirszbraun theorem [14] gives a sufficient condition for isometric extensions of Lipschitz maps The theorem states that isometric extensions are possible when the pair of metric spaces satisfy the Kirszbraun property, or the K property: Definition 37 The Kirszbraun Property K Let X and Y be two metric spaces with metric d x and d y respectively X, Y is said to have Property K if for any pair of families of closed balls {Bx i, r i : i I}, {By i, r i : i I}, such that d y y i, y j d x x i, x j for each i, j I, it holds that Bx i, r i By i, r i If X, Y has Property K, then by Kirszbraun s Theorem we can extend a Lipschitz mapping defined on a subspace of X to a Lipschitz mapping defined on X while maintaining the Lipschitz constant Unfortunately, if we consider X, d X = R m, and Y = Ĥ, Property K does not hold for either D p or d p Example 38 Property K does not hold for Ĥ with norm D p Specifically, R m, R n / does not have Property K We give a counterexample for m = n = 2, p = 2: Let ỹ 1 = 3, 1, ỹ 2 = 1, 1, ỹ 3 = 0, 1 be the representatives of three points y 1, y 2, y 3 in R 2 / Then D 2 y 1, y 2 = 2 2, D 2 y 2, y 3 = 1 and D 2 y 1, y 3 = 3 Consider x 1 = 0, 0, x 2 = 0, 2 2, x 3 = 1, 2 2 in R 2 with the Euclidean distance, then we have x 1 x 2 = 2 2, x 2 x 3 = 1 and x 1 x 3 = 3 For r 1 = 6, r 2 = 2 2, r 3 = 6 3, we see that 1 2, i=1 Bx i, r i but 3 i=1 By i, r i = To see 3 i=1 By i, r i =, it suffices to look at the upper half plane in R 2 If we look at the upper half plane H, then By 1, r 1 becomes the union of two parts, namely Bỹ 1, r 1 H and B ỹ 1, r 1 H, and By i, r i becomes Bỹ i, r i for i = 2, 3 But Bỹ 1, r 1 H Bỹ 2, r 2 = and B ỹ 1, r 1 H Bỹ 3, r 3 = So we obtain that 3 i=1 By i, r i = The following example is given in [7] Example 39 Property K does not hold for Ĥ with norm d p Specifically, R m, C n / does not have Property K Let m be any positive integer and n = 2, p = 2 We want to show that X, Y = R m, C n / does not have Property K Let ỹ 1 = 1, 0 and ỹ 2 = 0, 3 be representitives of y 1, y 2 Y, respectively Then d 1 y 1, y 2 = 4 Pick any two points x 1, x 2 in X with x 1 x 2 = 4 Then Bx 1, 2 and Bx 2, 2 intersect at x 3 = x 1 + x 2 /2 X It suffices to show that the closed balls By 1, 2 and By 2, 2 have no intersection in H Assume on the contrary that the two balls intersect at y 3, then pick a representive of y 3, say ỹ 3 = a, b where a, b C It can be computed that d 1 y 1, y 3 = a 4 + b 4 2 a b a 2 b 2 + 1, 24 12

13 and d 1 y 2, y 3 = a 4 + b a 2 6 b a 2 b Set d 1 y 1, y 3 = d 1 y 2, y 3 = 2 Take the difference of the right hand side of 24 and 25, we have b 2 a 2 = 1 and thus b 2 1 However, the right hand side of 24 can be rewritten as a 2 + b b 2, so d 1 y 1, y 3 = 2 would imply that b 2 1/2 This is a contradiction Remark 310 Using nonlinear functional analysis language [9], Lemma 34 can be restated by saying that S 1,0 H is a 5-Lipschitz retract in SymH Remark 311 The Lipschitz inversion results of Theorem 33 can be easily extended to systems of quadratic equations, not necessarily of rank-1 matrices from the phase retrieval model considered in this paper 4 Proof of the Results 41 Proof of Theorem 22 i First we prove the following lemma Lemma 41 Fix x C n and z C n Let ξ = jx and ζ = jz be their realifications, respectively Let ξ 0 ˆξ := {j x R 2n : x ˆx} be a point in the equivalency class that satisfies D 2 x, z = ξ 0 ζ Then it is necessary that ξ 0, Jζ = 0 26 and where J is defined as in 6 ξ 0, ζ 0, 27 Proof For θ [0, 2π define Then it is easy to compute that Therefore, D 2 x, z = Uθ := cosθi + sinθj je iθ x = Uθξ min Uθξ θ [0,2π ζ 2 = ξ 2 + ζ 2 2 max Uθξ, ζ θ [0,2π If Uθξ, ζ is constantly zero, then we are done Otherwise, note that max Uθξ, ζ = ξ, ζ 2 + Jξ, ζ θ [0,2π 13

14 and the maximum is achieved at θ = θ 0 if and only if cosθ 0 = ξ, ζ ξ, ζ 2 + Jξ, ζ and Now we can compute sinθ 0 = Jξ, ζ ξ, ζ 2 + Jξ, ζ ξ 0, Jζ = Uθ 0 ξ, Jζ = cosθ 0 ξ, Jζ + sinθ 0 Jξ, Jζ = = ξ, ζ ξ, ζ 2 + Jξ, ζ ξ, ζ ξ, ζ 2 + Jξ, ζ = 0 So we get is obvious ξ, Jζ + Jξ, ζ + Now we come back to the proof of the theorem Denote Jξ, ζ ξ, ζ 2 + Jξ, ζ Jξ, ζ ξ, ζ 2 + Jξ, ζ Jξ, Jζ ξ, ζ px, y := αx αy 2 D 2 x, y 2, x, y C n, ˆx ŷ 28 We can represent this quotient in terms of ξ and η It is easy to compute that px, y = P ξ, η := m k=1 Φ kξ, ξ + Φ k η, η 2 Φ k ξ, ξ Φ k η, η 29 ξ 2 + η 2 2 ξ, η 2 + ξ, Jη 2 Fix r > 0 Take ξ, η R 2n that satisfy D 2 x, z = ξ ζ < r and D 2 y, z = η ζ < r Let µ = ξ + η/2 and ν = ξ η/2 Then ν < r Note that for r 14

15 small enough we have that µ > ν and that Φ k ζ 0 Φ k µ 0 Thus m P ξ, η = Φ k µ + ν, µ + ν + Φ k µ ν, µ ν k=1 2 Φ k µ + ν, µ + ν Φ k µ ν, µ ν 1 µ + ν 2 + µ ν 2 2 µ + ν, µ ν 2 + µ + ν, Jµ ν 2 m = Φ k µ, µ + Φ k ν, ν k=1 Φ k µ, µ + Φ k ν, ν 2 4 Φ k µ, ν 2 1 µ 2 + ν 2 µ 4 + ν 4 2 µ 2 ν µ, Jν 2 Φ k µ, µ + Φ k ν, ν Φ k µ, µ + Φ k ν, ν 2 4 Φ k µ, ν 2 k:φ k ζ 0 = 1 2 ν 2 µ 2 + ν 2 k:φ k ζ 0 = 1 2 ν 2 k:φ k ζ 0 Φ k µ, µ = 1 2 ν 2 k:φ k ζ 0 Φ k µ, µ = 1 2 ν 2 = k:φ k ζ 0 k:φ k ζ 0 Φ k µ, µ µ 4 + ν 4 2 µ 2 ν 2 1 Φ k µ, µ + Φ k ν, ν Φ k µ, µ + Φ k ν, ν 1 + Φ 2 kν, ν 4 Φ kµ, ν 2 Φ k µ, µ Φ k µ, µ 2 Φ k µ, µ + Φ k ν, ν Φ k µ, µ + Φ k ν, ν 2 4 Φ k µ, ν Φ kν, ν Φ k µ, µ + Φ kν, ν 2 Φ k µ, µ 2 4 Φ kµ, ν 2 Φ k µ, µ 2 Φ k µ, µ + Φ k ν, ν 1 + Φ kν, ν Φ k µ, µ 2 Φ kµ, ν 2 Φ k µ, µ 2 Φ k µ, ν 2 Φ k µ, µ ν 2 + O ν 2 = 1 ν 2 Sµν, ν + O ν O ν 4

16 Note that Jµ, ν = Jµ, ν Jζ, ν Jµ Jζ ν = µ ζ ν 30 since Jζ, ν = 0 by Lemma 41 Also, µ ζ < r Therefore, P Jµ ν = Jµ, ν Jµ = Jµ, ν µ r ν µ and thus As a consequence, we have P Jµ ν 2 1 r2 µ 2 ν 2 P ξ, η = 1 SµP ν 2 Jµ ν, PJµν + O ν 2 1 SµP PJµ ν 2 Jµ ν, PJµν 1 r2 µ 2 + Or 2 1 r2 µ 2 λ 2n 1 Sµ + Or 2 Take r 0, by the continuity of eigenvalues with respect to matrix entries we have that Az λ 2n 1 Sζ 31 On the other hand, take E 2n 1 to be the unit-norm eigenvector correspondent to λ 2n 1 Sζ For each r > 0, take ξ = ζ + r E 2 2n 1 and η = ζ r E 2 2n 1 Then Hence Together with 31 we have px, y = P ξ, η = λ 2n 1 Sζ Az λ 2n 1 Sζ Az = λ 2n 1 Sζ ii Assume on the contrary that A 0 = 0, then for any N N, there exist x N, y N H for which px N, y N = αx N αy N 2 D 2 x N, y N 2 1 N 32 Without loss of generality we assume that x N y N for each N, for otherwise we can just swap the role of x N and y N Also due to homogeneity we assume x N = 1 By compactness of the closed ball B 1 0 = {x H : x 1} in H = C n, there exist convergent subsequences of {x N } N N and {y N } N N, which to avoid overuse of notations we still denote as {x N } N N x 0 H and {y N } N N y 0 H 16

17 Since x 0 = 1 we have from i that Ax 0 > 0 Note that D 2 x N, y N x N + y N 2, so by 32 we have αx N αy N 0 That is, αx 0 αy 0 = 0 By injectivity we have x 0 = y 0 in Ĥ By Proposition 22i, px N, y N Ax 0 1/N > 1/N for N large enough This is a contradiction with 32 iii The case z = 0 is an easy computation We now present the proof for z 0 First we consider px, z = P ξ, ζ as defined in 29 Fix r > 0 Take ξ R 2n that satisfy D 2 x, z = ξ ζ < r Let d = x z and δ = jd = ξ ζ Note that P ξ, ζ = m k=1 Φ kξ, ξ + Φ k ζ, ζ 2 Φ k ξ, ξ Φ k ζ, ζ ξ 2 + ζ 2 2 ξ, ζ 2 + ξ, Jζ 2 We can compute its numerator = m Φ k ξ, ξ + Φ k ζ, ζ 2 Φ k ξ, ξ Φ k ζ, ζ k=1 m Φ k ζ, ζ + 2 Φ k ζ, δ + Φ k δ, δ + Φ k ζ, ζ k=1 = k:φ k ζ 0 = k:φ k ζ 0 2 Φ k ζ, ζ + 2 Φ k ζ, δ + Φ k δ, δ Φ k ζ, ζ 2 Φ k ζ, ζ + 2 Φ k ζ, δ + Φ k δ, δ + 2 Φ k ζ, ζ 1 + Φ kζ, ζ Φ k ζ, δ + 1 Φ 2 kζ, ζ Φ k δ, δ Φ k ζ, ζ Φ kζ, ζ 2 Φ k ζ, δ 2 Φ k ζ, ζ 4 + O δ 3 Φ k ζ, δ 2 Φ k ζ, ζ + k:φ k ζ=0 and its denominator ξ 2 + ζ 2 2 ξ, ζ 2 + ξ, Jζ 2 = 2 ζ 2 + δ ζ, δ 2 ζ 2 1+ = δ 2 + O δ 3 Φ k δ, δ + O δ 3 ; + k:φ k ζ=0 Φ k δ, δ ζ 2 ζ, δ + 1 ζ, δ + 1 Jζ, δ ζ 4 4 ζ 4 ζ, δ 2 8 ζ 8 + O δ 3 We used Lemma 41 to get Jζ, δ = 0 in the above 17

18 Take r 0, we see that Ãz λ 2n 1 Sζ + k: z,f k =0 Let Ẽ2n 1 be the unit-norm eigenvector correspondening to Sζ + Note that λ 2n 1 k: z,f k =0 Φ k Φ k Jζ, Ẽ2n 1 = 0 since SζJζ = 0 and Φ k Jζ = JΦ k ζ = 0 for each k with z, f k = 0 Take ξ = ζ + r for each r, we again also have 2Ẽ2n 1 Ãz λ 2n 1 Sζ + Therefore Ãz = λ 2n 1 Sζ + k: z,f k =0 k: z,f k =0 iv Take z = 0 in iii v Bz can be computed in a similar way as in iii in particular, the expansion for P ξ, ζ is exactly the same We compute Bz B0 is computed in [8], Lemma 16 Now we consider z 0 Use the same notations as in 29 Fix r > 0 Again, take ξ, η R 2n that satisfy D 2 x, z = ξ ζ < r and D 2 y, z = η ζ < r Let µ = ξ + η/2 and ν = ξ η/2 Also let δ 1 = ξ ζ and δ 2 = η ζ Recall that Φ k Φ k P ξ, η = = m k=1 Φ kξ, ξ + Φ k η, η 2 Φ k ξ, ξ Φ k η, η ξ 2 + η 2 2 ξ, η 2 + ξ, Jη 2 m Φ k ξ, ξ + Φ k η, η 2 Φ k ξ, ξ Φ k η, η k=1 ξ 2 + η 2 2 ξ, η 2 + ξ, Jη 2 Now we compute it as m k=1 = k:φ k ζ 0 + k:φ k ζ=0 Again, k:φ k ζ 0 = k:φ k ζ 0 Φ k ξ, ξ + Φ k η, η 2 Φ k ξ, ξ Φ k η, η ξ 2 + η 2 2 ξ, η 2 + ξ, Jη 2 Φ k µ, µ + Φ k ν, ν Φ k µ, µ + Φ k ν, ν 2 4 Φ k µ, ν 2 µ 2 + ν 2 µ 4 + ν 4 2 µ 2 ν µ, Jν

19 Using the same computation as in i, we get that the numerator is Φ k µ, µ + Φ k ν, ν Φ k µ, µ + Φ k ν, ν 2 4 Φ k µ, ν 2 k:φ k ζ 0 = 2 Sµν, ν + O ν 4 Since µ 0, the denominator is µ 2 + ν 2 µ 4 + ν 4 2 µ 2 ν µ, Jν 2 = µ 2 + ν 2 µ ν 4 µ 4 2 ν 2 4 µ, Jν 2 µ 2 + µ 4 = µ 2 + ν 2 µ 2 1 ν 2 2 µ, Jν µ µ 4 + O ν 4 34 = 2 ν 2 2 Jµ, ν 2 µ 2 + O ν 4 = 2 ν 2 + O ν 4 by 30 Also we can compute using the denominator as above note that ν = δ 1 δ 2 /2 that k:φ k ζ=0 = k:φ k ζ=0 Φ k ξ, ξ + Φ k η, η 2 Φ k ξ, ξ Φ k η, η ξ 2 + η 2 2 ξ, η 2 + ξ, Jη 2 Φ 1/2 k δ 1 Φ 1/2 k δ 2 2 δ 1 δ O ν 4 35 Now put together 33, 34 and 35, we get P ξ, η = Sµν, ν + O ν 4 ν 2 + O ν 4 + k:φ k ζ=0 Φ 1/2 k δ 1 Φ 1/2 k δ 2 δ 1 δ O ν 4 2 Note that Φ 1/2 since it is equivalent to k δ 1 Φ 1/2 k δ 2 2 Φk δ 1 δ 2, δ 1 δ 2 Φ k δ 1, δ 1 Φ k δ 2, δ 2 Φ k δ 1, δ 2 2, 36 which is the Cauchy-Schwarz inequality Therefore, we have that Sµ + k:φ k ζ=0 Φ kν, ν + O ν 4 P ξ, η ν 2 + O ν 4 λ 1 Sµ + 19 k:φ k ζ=0 Φ k + Or 2

20 Take r 0 we have that Bz λ 1 Sζ + Φ k k:φ k ζ=0 Again we get the other direction of the above inequality by taking ξ = ζ + r E 2 1 and η = ζ r E 2 1 for each r > 0 where E 1 is the unit-norm eigenvector correspondent to λ 1 Sζ + k: z,f k =0 Φ k Note that for each r, the equality in 36 holds for this pair of ξ and η vi Take z = 0 in v 42 Proof of Theorem 25 Only the first two parts are nontrivial We prove them as follows Fix z C n Take x = z + d 1 and y = z + d 2 with d 1 < r and d 2 < r for r small Let u = x + y = 2z + d 1 + d 2 and v = x y = d 1 d 2 Let µ = 2ζ + δ 1 + δ 2 R 2n and ν = δ 1 δ 2 R 2n be the realification of u and v, respectively Define ρx, y = βx βy 2 d 1 x, y 2 By the same computation as in [3], Section 41, we get ρx, y = Qζ; δ 1, δ 2 := R2ζ + δ 1 + δ 2 δ 1 δ 2, δ 1 δ 2 2ζ + δ 1 + δ 2 2 P J2ζ+δ 1 +δ 2 δ 1 δ 2, δ 1 δ 2 Since J2ζ + δ 1 + δ 2 ker R2ζ + δ 1 + δ 2, we have R2ζ + δ 1 + δ 2 PJ2ζ+δ 1 +δ 2 δ 1 δ 2, PJ2ζ+δ 1 +δ 2 δ 1 δ 2 Qζ; δ 1, δ 2 = 2ζ + δ 1 + δ 2 2 PJ2ζ+δ 1 +δ 2 δ 1 δ 2, δ 1 δ 2 Now let δ = δ 1 + δ 2 and ν = δ 1 δ 2 Note the set inclusion relation {δ 1, δ 2 R 2n : δ < r 2, ν < r 2, ν J2ζ + δ } Thus we have inf δ <2r ν <2r ν J2ζ+δ { δ 1, δ 2 R 2n : δ 1 < r, δ 2 < r, ν J2ζ + δ } { δ 1, δ 2 R 2n : δ < 2r, ν < 2r, ν J2ζ + δ } Qζ; δ 1, δ 2 inf δ 1 <r δ 2 <r ν J2ζ+δ Qζ; δ 1, δ 2 inf δ <r/2 ν <r/2 ν J2ζ+δ Qζ; δ 1, δ 2 20

21 That is, λ 2n 1 R2ζ + δ inf δ <2r 2ζ + δ 2 inf δ 1 <r δ 2 <r ν J2ζ+δ λ 2n 1 R2ζ + δ Qζ; δ 1, δ 2 inf δ <r/2 2ζ + δ 2 Take r 0, by the continuity of eigenvalues with respect to the matrix entries, we have That is, Now consider λ 2n 1 Rζ/ ζ 2 az λ 2n 1 Rζ/ ζ 2 az = λ 2n 1 Rζ/ ζ 2 ρx, z = For simplicity write δ = δ 1 We can compute that Note that ρx, z = Qζ; δ = inf Qζ; δ inf δ <r δ J2ζ+δ Take r 0 we have that βx βz 2 d 1 x, z 2 R2ζ + δδ, δ 2ζ + δ 2 P J2ζ+δ δ, δ = inf σ <r δ <r δ J2ζ+δ R2ζ + δp J2ζ+δ δ, P J2ζ+δ δ 2ζ + δ 2 P J2ζ+δ δ, δ Qζ; δ = inf σ <r λ 2n 1R2ζ + δ ãz λ 2n 1 R2ζ/ 2ζ 2 = λ 2n 1 Rζ/ ζ 2 On the other hand, take ẽ 2n 1 to be a unit-norm eigenvector correspondent to λ 2n 1 R2ζ Then by the continuity of eigenvalues with respect to the matrix entries, for any ε > 0, there exists t > 0 so that δ = tẽ 2n 1 satisfy R2ζ + δδ, δ P J2ζ+δ δ, δ λ 2n 1 R2ζ + ε and from there we have Therefore, ãz λ 2n 1 R2ζ/ 2ζ 2 = λ 2n 1 Rζ/ ζ 2 ãz = λ 2n 1 Rζ/ ζ 2 In a similar way replacing infimum by supremum we also get bz and bz as stated in the theorem 21

22 43 Proof of Proposition 31 i Obviously D p ˆx, ŷ 0 for any ˆx, ŷ Ĥ and D pˆx, ŷ = 0 if and only if ˆx = ŷ Also D p ˆx, ŷ = D p ŷ, ˆx since x ay p = y a 1 x p for any x, y H, a = 1 Moreover, for any ˆx, ŷ, ẑ Ĥ, fix D pˆx, ŷ = x ay p, D p ŷ, ẑ = z by, then D p ˆx, ẑ x ab 1 z p = bx az p bx aby p + aby az p = D p ˆx, ŷ + D p ŷ, ẑ Therefore D p is a metric d p is also a metric since p in the definition of d p is the standard Schatten p-norm of a matrix ii For p q, by Hölder s inequality we have for any x = x 1, x 2,, x n H that n i=1 x i p n 1 p 1 q n i=1 x i q p q Thus x p n 1 p 1 q x q Also, since p is homogeneous, we can assume x p = 1 Then n i=1 x i q n i=1 x i p = 1 Thus x q x p Therefore, we have D q ˆx, ŷ = x a 1 y q n 1 p 1 q x a 1 y p n 1 p 1 q D p ˆx, ŷ and D p ˆx, ŷ = x a 2 y p x a 2 y q D q ˆx, ŷ for some a 1, a 2 with magnitude 1 Hence D q ˆx, ŷ D p ˆx, ŷ n 1 p 1 q D q ˆx, ŷ We see that D p 1 p are equivalent The second part follows then immediately iii The proof is similar to ii Note that there are at most 2 σ i s that are nonzero, so we have 2 1 p 1 q instead of n 1 p 1 q iv To prove that D p and d q are equivalent, we need only to show that each open ball with respect to D p contains an open ball with respect to d p, and vice versa By ii and iii, it is sufficient to consider the case when p = q = 2 First, we fix x H = C n, r > 0 Let R = min1, rn 2 2 x Then for any ŷ such that D 2 ˆx, ŷ < R, we take y such that x y < R, then 1 i, j n, x i x j y i y j = x i x j y j + x i y i y j < x i R + R x i + R = R2 x i + R R2 x i + 1 r Hence d n 2 2 ˆx, ŷ = xx yy 2 < n 2 r = r n 2 On the other hand, we fix x H = C n, R > 0 Let r = R 2 / 2 Then for any ŷ such that d 2 ˆx, ŷ < r, we have But we also have so d 2 ˆx, ŷ 2 = x 4 + y 4 2 x, y 2 < r 2 = R4 2 D 2 ˆx, ŷ 2 = min a =1 x ay 2 = x, y x x, y y 2 = x 2 + y 2 2 x, y, D 2 ˆx, ŷ 4 = x 4 + y x 2 y 2 4 x 2 + y 2 x, y + 4 x, y 2 22

23 Since x, y x y x 2 + y 2 /2, we can easily check that D 2 ˆx, ŷ 4 2d 2 ˆx, ŷ 2 < R 4 Hence D 2 ˆx, ŷ < R Thus D 2 and d 2 are indeed equivalent metrics Therefore D p and d q are equivalent Also, the imbedding i is not Lipschitz: if we take x = x 1, 0,, 0 C n, then D 2 ˆx, 0 = x 1, d 2 ˆx, 0 = x 1 2 v First, for p = 2, for ˆx ŷ in Ĥ {0}, we compute the quotient ρx, y = κ αx κ α y 2 D 2 x, y 2 x 1 xx y 1 yy 2 = x 2 + y 2 2 x, y = xx 2 y 2 + x 2 yy 2 2 x y tracexx yy x 4 y 2 + x 2 y 4 2 x 2 y 2 x y = x y x y x y tracexx yy x 4 y 2 + x 2 y 4 2 x 2 y 2 x y = x y x y tracexx yy x 3 y + x y 3 2 x y x y, where we used xx = x 2 For simplicity write a = x, b = y and t = x, y x y 1 We have a > 0, b > 0 and 0 t 1 Now ρx, y = 1 + 2abt abt2 a 2 + b 2 2abt Obviously ρx, y 1 Now we prove that ρx, y 2 Note that but 1 + 2abt abt2 a 2 + b 2 2abt 2 a2 + b 2 4abt + 2abt 2 0, a 2 + b 2 4abt + 2abt 2 2ab 4abt + 2abt 2 = 2abt 1 2 0, so we are done Note that take any x, y with x, y = 0 we would have ρx, y = 1 On the other hand, taking x = y and let t 1 we see that ρx, y = 2 ε is achievable for any small ε > 0 Therefore the constants are optimal The case where one of x and y is zero would not break the constraint of these two constants Therefore after taking the square root, we get lower Lipschitz constant 1 and upper Lipschitz constant 2 For other p, we use the results in ii and iii to get that the lower Lipschitz constant for κ α is min p, n p and the upper Lipschitz constant is 2 maxn p, 2 p 1 2 vi This follows directly from the construction of the map vii This follows directly from v and vi 23

24 44 Proof of Lemma 34 ii follows directly from the expression of π We prove i below Let A, B SymH where A = d k=1 λ mkp k and B = d k =1 µ mk Q k We now show that πa πb p p A B p 37 Assume λ 1 λ 2 µ 1 µ 2 Otherwise switch the notations for A and B If µ 1 µ 2 = 0 then πa = πb = 0 and the inequality 37 is satisfied Assume now µ 1 µ 2 > 0 Thus Q 1 is of rank 1 and Q 1 p = 1 for all p First we consider the case λ 1 λ 2 > 0 In this case P 1 is of rank 1, and we have πa πb = λ 1 λ 2 P 1 µ 1 µ 2 Q 1 = λ 1 λ 2 P 1 Q 1 +λ 1 µ 1 λ 2 µ 2 Q 1 38 Here P 1 = Q 1 = 1 Therefore we have P 1 Q 1 1 since P 1, Q 1 0 From that we have P 1 Q 1 p 2 1 p Also, by Weyl s inequality we have λ i µ i A B for each i Apply this to i = 1, 2 we get λ 1 µ 1 λ 2 µ 2 λ 1 µ 1 + λ 2 µ 2 2 A B Thus λ 1 µ 1 + λ 2 µ 2 2 A B 2 A B p Let g := λ 1 λ 2, δ := A B p, then apply the above inequality to 38 we get πa πb p g P 1 Q 1 p + 2δ 2 1 p g + 2δ 39 If 0 < g p δ, then πa πb p p + 3δ and we are done Now we consider the case where g > p δ Note that in this case we have δ < g/2 Thus we have λ 1 µ 1 < g/2 and λ 2 µ 2 < g/2 That means µ 1 > λ 1 + λ 2 /2 and µ 2 < λ 1 + λ 2 /2 Therefore, we can use holomorphic functional calculus and put and P 1 = 1 2πi γ R A dz Q 1 = 1 R B dz 2πi γ where R A = A zi 1, R B = B zi 1, and γ = γt is the contour given in Figure 2 note that γ encloses µ 1 but not µ 2 and used also by [15] Therefore we have P 1 Q 1 p 1 R A R B γt 2π p γ t dt 40 Now we have I R A R B z = R A z I + R A zb A 1 R A z = n 1 1 n R A zb A n R A z, 24 41

25 Figure 2: Contour for the integrals since for large L we have R A zb A R A z B A p 2 < 1, where σa denotes the spectrum of A Therefore we have p δ distz,σa 2δ g < R A R B γt p n 1 R A γt n+1 A B n p = R Aγt 2 A B p 1 R A γt A B p < 42 A B p dist 2 γt, σa p + 1, since distγt, σa g/2 for each t for large L Here we used the fact that if we order the singular values of any matrix X such that σ 1 X σ 2 X, then for any i we have σ i XY σ 1 Xσ i Y, and thus for two operators X, Y SymH, we have XY p X Y p Hence by 40 and 42 we have P 1 Q 1 p 2 1 p A B p π I 1 dist 2 γt, σa γ t dt 43 By evaluating the integral and letting L approach infinity for the contour, we have as in [15] 1 [ ] 1 4 2t I dist 2 γt, σa γ t dt = 2 0 t 2 + g dt = 2 g arctan = 2π g 2 0 g 44 Hence P 1 Q 1 p 2 1 p A B p π 2π g = p + 1 δ g 45 Thus by the first inequality in 39 and 45 we have πa πb p p δ 25

26 Now we are left with the case λ 1 λ 2 = 0 < µ 1 µ 2 Note that in this case we have that πa πb = µ 1 µ 2 Q 1 = λ 1 µ 1 λ 2 µ 2 Q 1, and therefore πa πb p 2 A B p < p A B p We have proved that πa πb p p A B p That is to say, π : SymH, p S 1,0 H, p is Lipschitz continuous with Lipπ p Now we are ready to prove Theorem Proof of Theorem 33 The proof for α and β are the same in essence For simplicity we do it for β first We need to construct a map ψ : R m, p Ĥ, d q so that ψβx = x for all x Ĥ, and ψ is Lipschitz continuous We prove the Lipschitz bound 15, which implies 14 for p = 2 and q = 1 Set M = βĥ Rm By the result in Section 23, there is a map ψ 1 : M Ĥ that is Lipschitz continuous and satisfies ψ 1 βx = x for all x Ĥ Additionally, the Lipschitz bound between M, 2 that is, M with Euclidean distance and Ĥ, d 1 is given by 1/ a 0 First we change the metric on Ĥ from d 1 to d 2 and embed isometrically Ĥ into SymH with Frobenius norm ie the Euclidean metric: M, 2 ψ 1 Ĥ, d 1 i 1,2 Ĥ, d 2 κ β SymH, F r, 46 where i 1,2 x = x is the identity of Ĥ and κ β is the isometry 10 We obtain a map ψ 2 : M, 2 SymH, F r of Lipschitz constant Lip ψ 2 Lip ψ 1 Lipi 1,2 Lipκ β = 1 a0, where we used Lipi 1,2 = L d 1,2,n = 1 by 8 Kirszbraun Theorem [14] extends isometrically ψ 2 from M to the entire R m with Euclidean metric Thus we obtain a Lipschitz map ψ 2 : R m, SymH, F r of Lipschitz constant Lipψ 2 = Lip ψ 2 1 a0 so that ψ 2 βx = xx for all x Ĥ The third step is to piece together ψ 2 with norm changing identities For q 2 we consider the following maps: R m, p j p,2 R m, 2 ψ 2 SymH, F r π S 1,0 H, F r κ 1 β Ĥ, d 2 i 2,q Ĥ, d q, 47 where j p,2 and i 2,q are identity maps on the respective spaces that change the metric and π is the map defined in Eq 23 The map ψ claimed by Theorem 33 is obtained by composing: ψ : R m, p Ĥ, d q, ψ = i 2,q κ 1 β π ψ 2 j p,2 26

27 Its Lipschitz constant is bounded by Lipψ p,q Lipj p,2 Lipψ 2 LipπLipκ 1 β Lipi 2,q max1, m p q 1 2 a0 Hence we obtained 20 The other equation 14 follows for p = 2 and q = 1 For q > 2 we use: R m, p j p,2 R m, 2 ψ 2 SymH, F r I 2,q SymH, q π S 1,0 H, q κ 1 β Ĥ, d q, 48 where j p,2 and I 2,q are identity maps on the respective spaces that change the metric The map ψ claimed by Theorem 33 is obtained by composing: ψ : R m, p Ĥ, d q, ψ = κ 1 β π I 2,q ψ 2 j p,2 Its Lipschitz constant is bounded by Lipψ p,q Lipj p,2 Lipψ 2 LipI 2,q LipπLipκ 1 β max1, m p 1 a q 1 Hence we obtained 21 Replace β by α, ψ by ω, and κ β by κ α in the proof above, using the Lipschitz constants for κ α in Proposition 31, we obtain 16 and 17 Acknowledgements The authors were supported in part by NSF grant DMS and DMS The first author acknowledges fruitful discussions with Krzysztof Nowak and Hugo Woerdeman both from Drexel University who pointed out several references, with Stanislav Minsker Duke University for pointing out [15] and [12], and Vern Paulsen University of Houston, Marcin Bownick University of Oregon and Friedrich Philipp University of Berlin We also thank the reviewers for their constructive comments and suggestions References [1] R Balan, On Signal Reconstruction from Its Spectrogram, Proceedings of the CISS Conference, Princeton NJ, May 2010 [2] R Balan, Reconstruction of Signals from Magnitudes of Redundant Representations, available online arxiv: v1 [mathfa] 4 July 2012 [3] R Balan, Reconstruction of Signals from Magnitudes of Redundant Representations: the Complex Case, to appear in Foundations of Computational Mathematics 2015 [4] R Balan, P Casazza, D Edidin, On signal reconstruction without phase, ApplComputHarmonAnal ,

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