 # Phase Retrieval using Lipschitz Continuous Maps

In this note we prove that reconstruction from magnitudes of frame coefficients (the so called "phase retrieval problem") can be performed using Lipschitz continuous maps. Specifically we show that when the nonlinear analysis map α: H→R^m is injective, with (α(x))_k=|<x,f_k>|^2, where {f_1,...,f_m} is a frame for the Hilbert space H, then there exists a left inverse map ω:R^m→ H that is Lipschitz continuous. Additionally we obtain the Lipschitz constant of this inverse map in terms of the lower Lipschitz constant of α. Surprisingly the increase in Lipschitz constant is independent of the space dimension or frame redundancy.

## Authors

##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1 Introduction

Assume is a frame (that is a spanning set) for the -dimensional Hilbert space . In this paper can be a real or complex Hilbert space. The result applies to both cases, and the constants are the same.

Let denote the nonlinear map

 α:H→Rm  ,  α(x)=(|⟨x,fk⟩|2)1≤k≤m (1)

On we consider the equivalent replation iff there is a scalar of magnitude one, , so that . Let denote the set of equivalence classes. Note is equivalent to the cross-product between a real or complex projective space of dimension and the positive semiaxis . The nonlinear map is now well defined on that, by abuse of notation, we denote also by . The phase retrieval problem (or the phaseless reconstruction problem) refers to analyzing when is an injective map, and in this case to finding ”good” left inverses.

The frame is said to be phase retrievable if the nonlinear map is injective. In this paper we assume is injective (hence is phase retrievable). The problem is to extend a left inverse of from , the image of through , to the entire space so that it remains Lipschitz continuous. A continuous map , defined between metric spaces and with distances and respectively, is Lipschitz continuous with Lipschitz constant if

 Lip(f):=supx1,x2∈XdY(f(x1),f(x2))dX(x1,x2)<∞

Existing literature (e.g. ) establishes that when the nonlinear map is injective, it is also bi-Lipschitz for metric on and Euclidian norm in . Additionally, the nonlinear map , is bi-Lipschitz in the case of real Hilbert space with distance on and Euclidian distance on . As a consequence of these results we obtain that a left inverse of is Lipschitz when restricted to the image of through . In this paper we show that this left inverse of admits a Lipschitz continuous extension to the entire . Surprisingly we obtain the Lipschitz constant of this extension is just a small factor larger than the minimal Lipschitz constant, a factor that is independent of the dimension

or the number of frame vectors

.

The organization of the paper is as follows. Section 2 introduces notations and presents the main results. Section 3 contains the proof of these results.

## 2 Notations. Statement of Main Results

The nonlinear map naturally induces a linear map between the space of symmetric operators on and :

 A:Sym(H)→Rm  ,  A(T)=(⟨Tfk,fk⟩)1≤k≤m (2)

This linear map has first been observed in  and it has been exploited successfully in various paprs e.g. [5, 17, 6]. Let denote the set of symmetric operators that have at most

strictly positive eigenvalues and

strictly negative eigenvalues. In particular denotes the set of non-negative symmetric operators of rank at most one:

 S1,0(H)={T∈Sym(H) s.t. ∃x∈H∀y∈H , T(y)=⟨y,x⟩x} (3)

In  we studied in more depth geometric and analytic properties of this set. In particular note where

 ⟦x,y⟧=12(⟨⋅,x⟩y+⟨⋅,y⟩x) (4)

denotes the symmetric outer product between vectors and . The map is injective if and only if restricted to is injective.

In previous papers [7, 11] we showed the following necessary and sufficient conditions for a frame to give phase retrieval.

###### Theorem 2.1

[7, 11] The following are equivalent:

1. The frame is phase retrievable;

2. ;

3. There is a constant so that for every

 (5)

On the space we consider two classes of metrics (distances) induced by corresponding distances on and respectively.

The class of vector space norm induced metrics. For every and define

 Dp(^x,^y)=min|a|=1∥x−ay∥p (6)

When no subscript is used, denotes the Euclidian norm, .

The class of matrix norm induced metrics. For every and define

 dp(^x,^y)=∥∥⟦x,x⟧−⟦y,y⟧∥∥p={(∑nk=1(σk)p)1/pfor1≤p≤∞max1≤k≤nσkforp=∞ (7)

where

are the singular values of the matrix

, which is of rank at most 2.

Our choice in (7) corresponds to the class of Schatten norms that extend to ideals of compact operators. In particular corresponds to the operator norm in ; corresponds to the Frobenius norm in ; corresponds to the nuclear norm in :

Note the Frobenius norm induces an Euclidian metric on . In  Lemma 3.7 we computed eplicitely the eigenvalues of . Based on these values, we can easily derive explicit expressions for these distances:

 d2(x,y)=√∥x∥4+∥y∥4−2|⟨x,y⟩|2
 d1(x,y)=√(∥x∥2+∥y∥2)2−4|⟨x,y⟩|2

Since for and , an equivalent condition to (5) is stated as follows:

1. There is a constant so that for every ,

 ∥α(x)−α(y)∥2≥a0(d1(x,y))2 (8)

All metrics and induce the same topology as shown in the following result.

###### Proposition 2.2

1. For each , and are metrics (distances) on .

2. are equivalent metrics, that is each induces the same topology on as . Additionally, for every the embedding , , is Lipschitz with Lipschitz constant

 LDp,q,n=max(1,n1q−1p). (9)

3. For , are equivalent metrics, that is each induces the same topology on as . Additionally, for every the embedding , , is Lipschitz with Lipschitz constant

 Ldp,q,n=max(1,21q−1p). (10)

4. The metrics and are equivalent, that is they produce the same topology on . However neither the embedding nor is Lipschitz, where .

5. The metric space is isometrically isomorphic to endowed with Schatten norm . The isomorphism is given by the map

 κ:^H→S1,0(H)  ,  x↦⟦x,x⟧. (11)

In particular the metric space is isometrically isomorphic to endowed with the nuclear norm .

###### Remark 2.3

1. Note the Lipschitz bound is equal to the operator norm of the identity between and : .

2. Note the equality .

Theorem 2.1 together with the previous proposition show that if the frame is phase retrievable then the nonlinear map (1) is bi-Lipschitz between metric spaces and . In particular, the Lipschitz constants between and are given by and :

 √a0d1(x,y)≤∥α(x)−α(y)∥≤√b0d1(x,y) (12)

Clearly the inverse map defined on the range of from metric space to :

 ~ω:α(^H)⊂Rm→^H  ,  ~ω(c)=x  if α(x)=c (13)

is Lipschitz with Lipschitz constant . In this paper we prove that can be extended to the entire as a Lipschitz map with Lipschitz constant that increases by a small factor.

The precise statement is given in the following Theorem which is the main result of this paper.

###### Theorem 2.4

Let be a phase retrievable frame for the dimensional Hilbert space , and let denote the injective nonlinear analysis map . Let denote the positive constant introduced in (5) and (8). Then there exists a Lipschitz continuous function so that for all . For any , has an upper Lipschitz constant between and bounded by:

 Lip(ω)p,q≤⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩3+2√2√a021q−12max(1,m12−1p)% for q≤23+21+1q√a0max(1,m12−1p)for q>2 (14)

Explicitly this means: for and for all :

 dq(ω(c),ω(d))≤3+2√2√a021q−12max(1,m12−1p)∥c−d∥p (15)

whereas for and for all :

 dq(ω(c),ω(d))≤1√a0(3+21+1q)max(1,m12−1p)∥c−d∥p (16)

In particular, for and its Lipschitz constant bounded by :

 d1(ω(c),ω(d))≤4+3√2√a0∥c−d∥ (17)

The proof of Theorem 2.4, presented in Section 3, requires construction of a special Lipschitz map. We believe this particular result is interesting in itself and may be used in other constructions. Due to its importance we state it here:

###### Lemma 2.5

Consider the spectral decomposition of any self-adjoint operator in , , where are the eigenvalues including multiplicities, and ,…, are the orthogonal projections associated to the distinct eigenvalues. Additionally, and , where is the multiplicity of eigenvalue . Then the map

 π:Sym(H)→S1,0(H)  ,  π(A)=(λ1−λ2)P1 (18)

satisfies the following two properties: (1) for , it is Lipschitz continuous from to with Lipschitz constant less than or equal to ; (2) for all .

###### Remark 2.6

Numerical experiments suggest the Lipschitz constant of is smaller than 5 for . On the other hand it cannot be smaller than 2 as the following example shows.

###### Example 2.7

If , , then and . Here we have and . Thus for this example .

It is unlikely to obtain an isometric extension in Theorem 2.4. Kirszbraun theorem  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 2.8

The Kirszbraun Property (K): Let and be two metric spaces with metric and respectively. is said to have Property (K) if for any pair of families of closed balls , , such that for each , it holds that .

If has Property (K), then by Kirszbraun’s Theorem we can extend a Lipschitz mapping defined on a subspace of to a Lipschitz mapping defined on while maintaining the Lipschitz constant. Unfortunately, if we consider and , Property (K) does not hold for either or .

Property (K) does not hold for with norm . Specifically, does not have Property K.

###### Example 2.9

We give a counterexample for : Let , , be the representatives of three points , , in . Then , and . Consider , , in with the Euclidean distance, then we have , and . For , , , we see that but . To see , it suffices to look at the upper half plane in . If we look at the upper half plane , then becomes the union of two parts, namely and , and becomes for , . But and . So we obtain that .

Property (K) does not hold for with norm . Specifically, does not have Property K.

###### Example 2.10

We give a counterexample for : Let , be the representatives of two points , in . Then . Consider and in with the Euclidean distance, then . For , we see that . On the other hand, since can be isometrically embedded in but is not in the map.

###### Remark 2.11

Using nonlinear functional analysis language () Lemma 2.5 can be restated by saying that is a 5-Lipschitz retract in .

## 3 Proofs of results

We start by proving proposition 2.2.

Proof of Proposition 2.2

1. For obviously we have and if and only if . We also have since for any , , . Also, for any , , , if is achieved by , is achieved by , then . So is a metric.

Since in the definition of is the standard Schatten p-norm of a matrix, is also a metric.

2. For , by Hölder’s inequality we have for any that . Thus . Also since is homogeneous, if we assume we have . Thus . Therefore, we have and for some , with magnitude . Hence

 Dq(^x,^y)⩽Dp(^x,^y)⩽n(1p−1q)Dq(^x,^y)

We see that are equivalent. The second part follows then immediately.

3. The proof is similar to 2. Note that there are at most 2 ’s that are nonzero, so we have instead of .

4. To prove that and are equivalent, we need only to show that each open ball with respect to contains an open ball with respect to , and vise versa. By 2 and 3, it is sufficient to consider the case when .

First, we fix , . Let . Then for any such that , we take such that , then , . Hence .

On the other hand, we fix , . Let . Then for any such that , we have

 (19)

But we also have

 (D2(^x,^y))2=min|a|=1∥x−ay∥2=∥∥∥x−⟨x,y⟩|⟨x,y⟩|y∥∥∥2=∥x∥2+∥y∥2−2|⟨x,y⟩| (20)

So

 (21)

Since , we can easily check that . Hence .

Thus and are indeed equivalent metrics. Therefore and are equivalent. Also, the imbedding is not Lipschitz: if we take , then , .

5. This follows directly from the construction of the map.

Q.E.D.

Next we prove Lemma 2.5.

Proof of Lemma 2.5

(2) follows directly from the expression of . We prove (1) below.

Let , where and . We now show that

 (22)

Assume . Otherwise switch the notations for and . If then and the inequality (22) is satisfied. Assume now . Thus is of rank 1 and therefore for all . First note that

 π(A)−π(B)=(λ1−λ2)P1−(μ1−μ2)Q1=(λ1−λ2)(P1−Q1)+(λ1−μ1−(λ2−μ2))Q1 (23)

Here . Therefore we have since , . From that we have .

Also, by Weyl’s inequality (see  III.2) we have for each . Apply this to , we get . Thus .

Let , , then apply the above inequality to (23) we get

 ∥π(A)−π(B)∥p⩽g∥P1−Q1∥p+2δ⩽21pg+2δ (24)

If , then and we are done.

Now we consider the case where . We use holomorphic functional calculus and put

and

 Q1=−12πi∮γRBdz (26)

where , , and is the contour given in the picture below and used also by .

Therefore we have

 (27)

Now we have

 (RA−RB)(z)=RA(z)−(I+RA(z)(B−A))−1RA(z)=∑n⩾1(−1)n(RA(z)(B−A))nRA(z) (28)

since for large we have , where denotes the spectrum of A.

Therefore we have

 ∥(RA−RB)(γ(t))∥p (29)

since for each t for large . Here we used the fact that if we order the singular values of any matrix such that , then for any we have , and thus for two operators , , we have .

Hence by (27) and (29) we have

 ∥P1−Q1∥p⩽(21p+2−1)∥A−B∥pπ∫I1dist2(γ(t),ρ(A))|γ′(t)|dt (30)

By evaluating the integral and letting approach infinity for the contour, we have as in 

 ∫I1dist2(γ(t),ρ(A))|γ′(t)|dt=2∫∞01t2+(g2)2dt=[4garctan(2tg)]∞0=2πg (31)

Hence

 ∥P1−Q1∥p⩽(21p+2−1)∥A−B∥pπ⋅2πg=(21+1p+1)δg (32)

Thus by the first inequality in (24) and (32) we have .

We have proved that . That is to say, is Lipschitz continuous with Lipschitz constant less than or equal to .

Q.E.D.

Now we are ready to prove Theorem 2.4.

Proof of Theorem 2.4

We construct a map so that for all , and is Lipschitz continuous. We prove the Lipschitz bound (14) which implies (17) for and .

Set . By hypothesis, there is a map that is Lipschitz continuous and satisfies for all . Additionally, the Lipschitz bound between (that is, with Euclidian distance) and is given by .

First we change the metric on from to and embed isometrically into with Frobenius norm (i.e. Euclidian metric):

 (M,∥⋅∥2)~ω1⟶(^H