Phase retrieval in infinite-dimensional Hilbert spaces 6 1 Jameson Cahill1, Peter G. Casazza 2, and Ingrid Daubechies 3 0 ∗ † 2 1Department of Mathematical Sciences, New Mexico State University n u 2Department of Mathematics, University of Missouri J 7 3Department of Mathematics, Duke University 2 ] A June 28, 2016 F . h t a m Abstract [ The main result of this paper states that phase retrieval in infinite-dimensional 2 Hilbert spaces is never uniformly stable, in sharp contrast to the finite dimensional v 1 setting inwhich phase retrieval is always stable. This leads us to derive stability results 1 for signals depending on how well they are approximated by finite expansions. 4 6 0 . 1 0 6 1 Introduction 1 : v i X Given a separable Hilbert space , phase retrieval deals with the problem of recovering H r an unknown f from a set of intensity measurements ( f,ϕ ) for some countable a n n∈I ∈ H |h i| collection Φ = ϕ . Note that if f = αg with α = 1 then f,ϕ = g,ϕ for n n∈I n n { } ⊆ H | | |h i| |h i| every n I regardless of our choice of Φ; we say Φ does phase retrieval if the converse of ∈ this statement is true, i.e., if the equalities f,ϕ = g,ϕ for every n imply that there n n |h i| |h i| is a unimodular scalar α so that f = αg. We will generally assume that Φ forms a frame for , i.e., there are positive constants H 0 < A B < so that ≤ ∞ A f 2 f,ϕ 2 B f 2 n k k ≤ |h i| ≤ k k n∈I X ∗The second author was supported by NSF DMS 1609760;NSF ATD 1321779;and ARO W911NF-16-1- 0008 †The third authorwassupportedby AFOSRgrant00002113-02;ONRgrantN00014-11-1-0714-06-7;and NSF grant DMS-1516988 1 for every f in . We call the operator T : ℓ2(I) given by Φ H H → T (f) = ( f,ϕ ) Φ n n∈I h i the analysis operator of Φ. We denote by : ℓ2(I) the nonlinear mapping given by Φ A H → (f) = ( f,ϕ ) , Φ n n∈I A |h i| so that Φ does phase retrieval if and only if is injective on / where f g if f = αg Φ A H ∼ ∼ with α = 1. | | Definition 1.1. We say a frame ϕ for a Hilbert space has the complement property n n∈I { } H if for every subset S I we have ⊆ span ϕ = or span ϕ = . n n∈S n n6∈S { } H { } H Theorem 1.2. (a) Let be a separable Hilbert space and let Φ be a frame for . If Φ does H H phase retrieval then Φ has the complement property. (b) Let be a separable Hilbert space over the real numbers and let Φ be a frame for . If H H Φ has the complement property then Φ does phase retrieval. Proof. (a) Suppose Φ does not have the complement property and find S I so that neither ⊆ ϕ nor ϕ spans . Then we can find nonzero f,g so that f,ϕ = 0 for n n∈S n n6∈S n { } { } H ∈ H h i all n S and g,ϕ = 0 for all n S. Also, since Φ is a frame we know that f = λg n ∈ h i 6∈ 6 for any scalar λ, so in particular we know f + g = 0 and f g = 0. It now follows that 6 − 6 f +g,ϕ = f g,ϕ for all n I but f +g = λ(f g) for any scalar λ, so Φ does n n |h i| |h − i| ∈ 6 − not do phase retrieval. (b) Suppose Φ does not do phase retrieval and find nonzero f,g so that f,ϕ = n ∈ H |h i| g,ϕ for every n I, but f = g. Since is a real Hilbert space this means that n |h i| ∈ 6 ± H f,ϕ = g,ϕ , so let S = n I : f,ϕ = g,ϕ . Then f g = 0 but f g,ϕ = 0 n n n n n h i ±h i { ∈ h i h i} − 6 h − i for every n S so span ϕ = , and similarly f +g = 0 but f +g,ϕ = 0 for every n n∈S n ∈ { } 6 H 6 h i n S so span ϕ = , which means Φ does not have the complement property. n n6∈S 6∈ { } 6 H This theorem was originally proved in [3] where it was only stated in the finite-dimensional case, but the proof still holds in infinite dimensions without any modifications. The original proof of part (a) presented in [3] did not give the correct conclusion in the case where is H a Hilbert space over the complex numbers. This was observed by the authors of [5] where they presented a much more complicated proof for this case. It turns out that the proof presented in [3] does hold in this case with only minor modifications, which is the proof presented above. We remark here that recently several papers have been devoted to showing that certain frames do phase retrieval for infinite-dimensional spaces over both the real and complex numbers, so by Theorem 1.2 all of these frames have the complement property. For instance, in [10] it is shown that a real-valued band-limited signal can be recovered up to sign from 2 the absolute values of its samples at any rate greater than twice the Nyquist rate. A similar result for complex valued band-limited signals was shown in [9] which required a minimal oversampling rate of four times the Nyquist rate. In [8] the authors study an instance of the phase retrieval problem using the Cauchy wavelet transform to recover analytic functions in L2(R,C) that have compactly supported Fourier transforms. In that paper they observe that although they are able to show that is Φ A injective (for the particular choice of and Φ) there is an inherent lack of robustness in H the sense that arbitrarily small perturbations of the measurements (f) can result in large Φ A errors in the reconstructed signal (see sections 4.1 and 4.2 in [8]). The main result of the present paper states that this type of lack of robustness is unavoidable when doing phase retrieval in an infinite-dimensional Hilbert space. In this paper, we restrict ourselves to the case of countably infinite frames in Hilbert spaces; in work extending the present results, [1] proves similar lack of robustness for phase retrieval in infinite-dimensional Banach spaces with infinite frames that need not be countable. One way to quantify the robustness of the phase retrieval process for a given frame Φ is in terms of the lower Lipschitz bound of the map with respect to some metric on the space Φ A / . A natural choice of metric is the quotient metric induced by the metric on given H ∼ H by ˜ d(f,g˜) = inf f αg . |α|=1k − k We would like to find a positive constant C (depending only on Φ) so that for every f,g ∈ H inf f αg C (f) (g) . (1.1) Φ Φ |α|=1k − k ≤ kA −A k In [5] the authors introduced a numerical version of the complement property as a means of quantifying the constant C in (1.1): Definition 1.3. We say a frame ϕ has the σ-strong complement property if for every n n∈I { } subset S I either ϕ or ϕ is a frame for with lower frame bound at least σ. n n∈S n n6∈S ⊆ { } { } H In [5] it is shown that when = RM the lower Lipschitz bound of is precisely controlled Φ H A by the largest σ for which Φ has the σ-strong complement property (see also [4]). Although this result does not apply to the complex case, much like the complement property cannot be used to determine whether a given frame does phase retrieval for a complex space, we still have the following result in the finite-dimensional case: Proposition 1.4. If is a finite-dimensional Hilbert space and Φ = ϕ does phase n n∈I H { } retrieval for then has a positive lower Lipschitz bound, i.e., satisfies (1.1) for some Φ Φ H A A C < . ∞ Proof. Since this result is already known if is a real Hilbert space we will prove it for H the case where = CN. Note that the inequality (1.1) is homogeneous so without loss of H generality we can assume that f = 1 and g 1. k k k k ≤ 3 Let H denote the space of N N Hermitian matrices equipped with the Hilbert-Schmidt N × inner product X,Y = Trace(XY). Because of the restriction to Hermitian matrices, this is h i a Hilbert space over the reals (of dimension M2), and no adjoint is necessary in the definition of X,Y . Define the linear mapping 2 : H ℓ2(I) by h i AΦ N → 2(X) = ( X,ϕ ϕ∗ ) = ( Xϕ ,ϕ ) , AΦ |h n ni| n∈I |h n ni| n∈I where we denote by gg∗ the rank one operator that maps h CN to h,g g. (Note that ∈ h i if X is rank 1, i.e. X = ff∗, then 2(X) = ( f,ϕ 2) , hence the notation 2.) AΦ |h ni| i∈I AΦ It is well known that Φ does phase retrieval if and only if ker( 2) does not contain any AΦ matrix of rank 1 or 2 (see Lemma 9 in [5]). This, together with the compactness of the set S = X H : rank(X) 2, X = 1 (since H is finite-dimensional), implies that N N { ∈ ≤ k k } min 2(X) = c > 0 X∈SkAΦ k where X denotes the operator norm (however, we can choose any norm on H and this N k k will still be true). For f,g CN, ff∗ gg∗ is rank 1 or 2, so we have ∈ − 1 ff∗ gg∗ 2 2(ff∗ gg∗) 2 k − k ≤ c2kAΦ − k 1 = 2(ff∗) 2(gg∗) 2 c2kAΦ −AΦ k 1 = ( f,ϕ 2 g,ϕ 2)2. (1.2) c2 |h ni| −|h ni| n∈I X Furthermore, since we are assuming f = 1 and g 1 we have k k k k ≤ ( f,ϕ 2 g,ϕ 2)2 = ( f,ϕ g,ϕ )2( f,ϕ + g,ϕ )2 n n n n n n |h i| −|h i| |h i|−|h i| |h i| |h i| n∈I n∈I X X ( f,ϕ g,ϕ )2(2 ϕ )2 n n n ≤ |h i|−|h i| k k n∈I X (4max ϕ 2) ( f,ϕ g,ϕ )2. (1.3) n n n ≤ n∈I k k |h i|−|h i| n∈I X Since we are assuming f g a direct computation shows that the largest (in absolute k k ≥ k k value) eigenvalue of ff∗ gg∗ is − 1 ( f 2 g 2+(( f 2 + g 2)2 4 f,g 2)1/2). 2 k k −k k k k k k − |h i| Therefore, we have that 1 ff∗ gg∗ (( f 2 + g 2)2 4 f,g 2)1/2 k − k ≥ 2 k k k k − |h i| 1 = ( f 2 + g 2 2 f,g )1/2( f 2 + g 2+2 f,g )1/2 2 k k k k − |h i| k k k k |h i| 1 = ( inf f αg )( f 2+ g 2 +2 f,g )1/2, 2 |α|=1k − k k k k k |h i| 4 and since f = 1 this says k k inf f αg 2 ff∗ gg∗ . (1.4) |α|=1k − k ≤ k − k Finally, combining (1.2),(1.3), and (1.4) yields (1.1). 2 Main results Before stating the main result we first remark that, when viewed as a subset of CM×N, the set of frames ϕ N that do phase retrieval for CM is an open subset for each N, see [2, 7]. { n}n=1 In fact, in [7] it is shown that if N 4M 4 then this set is open and dense, and it is clear ≥ − that it must be empty if N is sufficiently small. At this time it is not known if there exists a pair (M,N) where the set of frames consisting of N vectors which do phase retrieval for CM is nonempty but not dense (see [11]), but in any case, the set of frames which do not do phase retrieval is never dense unless it is all of CM×N. The next statement says that this situation is reversed when we consider frames for an infinite-dimensional space. Proposition 2.1. Let be an infinite-dimensional separable Hilbert space and suppose H ϕn n∈N does phase retrieval. For every ǫ > 0 there is another frame ψn n∈N which does { } { } not do phase retrieval and satisfies ϕ ψ 2 < ǫ. n n k − k n∈N X Proof. Let en n∈N be an orthonormal basis for and choose k N so that { } H ∈ ∞ e ,ϕ 2 < ǫ. 1 n |h i| n=k+1 X For n k let ψ = ϕ and for n > k let n n ≤ ∞ ψ = ϕ ,e e . n n i i h i i=2 X Now we have that ∞ ϕ ψ 2 = e ,ϕ 2 < ǫ. n n 1 n k − k |h i| n∈N n=k+1 X X Also, it is clear that ψ k cannot span , and for every n > k we have that e ,ψ = 0, { n}n=1 H h 1 ni so {ψn}∞n=k+1 does not span H either. Therefore {ψn}n∈N does not have the complement property and so by Theorem 1.2 does not do phase retrieval. Furthermore, for ǫ sufficiently small ψn n∈N is still a frame. { } 5 The above proposition suggests an infinite-dimensional space is fundamentally different from a finite-dimensional setting when doing phase retrieval. In particular, since any frame can be perturbed by an arbitrarily small amount to arrive at a frame that does not do phase retrieval,itsuggeststhatphaseretrievalforinfinite-dimensionalspacesisinherentlyunstable. We now state the main result, confirming this intuition. Theorem 2.2. Let be an infinite-dimensionalseparable Hilbert space and let Φ = ϕn n∈N H { } be a frame for with frame bounds 0 < A B < ; further suppose that ϕ c > 0 n H ≤ ∞ k k ≥ for every n N. Then, for every δ > 0, there exist f,g so that inf f αg 1 but ∈ ∈ H |α|=1k − k ≥ (f) (g) < δ. Φ Φ kA −A k Before proving the theorem we need a lemma. Lemma 2.3. Let be an infinite-dimensional separable Hilbert space and let Φ = ϕn n∈N H { } be a frame for with frame bounds 0 < A B < . For every ǫ > 0 and every N N H ≤ ∞ ∈ there is a k > N and a m > k so that ϕ ,ϕ 2 < ǫ. k n |h i| n6∈{N+1,...,m} X Proof. Fix ǫ > 0 and N N. Let V = span ϕ ,...,ϕ and let P denote the orthogonal 1 N V ∈ { } projection onto V. Let e L be an orthonormal basis for V and note that { ℓ}ℓ=1 L P ϕ 2 = ϕ ,e 2 V n n ℓ k k |h i| n∈N ℓ=1 n∈N X XX L B e 2 = BL. ℓ ≤ k k ℓ=1 X ǫ So since P ϕ 2 < we can find k > N so that P ϕ 2 < . Then V n V k k k ∞ k k 2B n∈N X N N ϕ ,ϕ 2 = ϕ ,P ϕ 2 k n k V n |h i| |h i| n=1 n=1 X X N = P ϕ ,ϕ 2 V k n |h i| n=1 X P ϕ ,ϕ 2 V k n ≤ |h i| n∈N X ǫ ǫ B P ϕ 2 < B = . V k ≤ k k 2B 2 6 Now observe that ∞ ϕ ,ϕ 2 ϕ ,ϕ 2 k n k n |h i| ≤ |h i| n=k+1 n∈N X X B ϕ 2 < , k ≤ k k ∞ so there is an m > k so that ∞ ǫ ϕ ,ϕ 2 < . k n |h i| 2 n=m+1 X Therefore N ∞ ϕ ,ϕ 2 = ϕ ,ϕ 2 + ϕ ,ϕ 2 k n k n k n |h i| |h i| |h i| n6∈{N+1,...,m} n=1 n=m+1 X X X ǫ ǫ < + = ǫ. 2 2 Note that one consequence of the above lemma is that no frame for an infinite-dimensional Hilbert space can have the σ-strong complement property, regardless of how small one picks σ > 0. Proof of Theorem 2.2. We use the Lemma to construct f and g explicitly. Pick ǫ = c2δ2/4 and N N, and determine k and m as in the Lemma, for these choices of ∈ ǫ,N. Next, pick ψ so that it is orthogonal to the finite-dimensional span of ϕ , ...,ϕ , N+1 m and set f = ϕ ϕ −1 + ψ, and g = ϕ ϕ −1 ψ. k k k k k k k k − For n N +1,...,m we have f,ϕ = g,ϕ , so that n n ∈ { } h i h i ( f,ϕ g,ϕ )2 = ( f,ϕ g,ϕ )2 . n n n n |h i|−|h i| |h i|−|h i| n∈Z n6∈{N+1,...,m} X X The triangle inequality implies that z +z z z 2 z for all z ,z C; applying 1 2 1 2 1 1 2 || |−| − || ≤ | | ∈ this to each term in the right hand side of the inequality, setting z = ϕ ,ϕ ϕ −1 and 1 k n k h ik k z = ψ,ϕ , leads to 2 n h i 4ǫ ( f,ϕ g,ϕ )2 4 ϕ ,ϕ 2 ϕ −2 = δ2, |h ni|−|h ni| ≤ |h k ni| k kk ≤ c2 n∈Z n6∈{N+1,...,m} X X or (f) (g) δ. Φ Φ kA − A k ≤ On the other hand, because ψ and ϕ −1ϕ are orthogonal unit vectors, we have that, for k k k k all α C with α = 1, ∈ | | f αg 2 = (1 α) ϕ −1ϕ + (1+α)ψ 2 = 1 α 2 + 1+α 2 = 4, k k k − k k − k k k | − | | | so that inf f αg = 2. |α|=1k − k 7 Remark 2.4. Although it is not important here, it may be interesting to note that, regardless of how small δ is, the functions f, g constructed in the proof lie within the closed bounded ball with radius √2 (in fact f = g = √2). k k k k Since Theorem 2.2 says that we can never do phase retrieval in a robust way for an infinite- dimensionalspace, butTheorem1.4sayswecanbasicallyalwaysdoitforafinite-dimensional space, it seems natural to try to use finite-dimensional approximations when working in an infinite-dimensional setting. Our next theorem is a first step in that direction; again, we first establish a lemma. Lemma 2.5. Let be a separable Hilbert space and let Φ be a frame for with frame H H bounds 0 < A B < , then for every f,g we have ≤ ∞ ∈ H (f) (g) B1/2 inf f αg . Φ Φ kA −A k ≤ |α|=1k − k Proof. First note that f,ϕ g,ϕ f,ϕ g,ϕ n n n n ||h i|−|h i|| ≤ |h i−h i| by the reverse triangle inequality. This means that (f) (g) T f T g B1/2 f g . Φ Φ Φ Φ kA −A k ≤ k − k ≤ k − k Since (αg) = (g) for any unimodular scalar α, we have Φ φ A A (f) (g) = inf (f) (αg) B1/2 inf f αg . Φ Φ Φ Φ kA −A k |α|=1kA −A k ≤ |α|=1k − k Remark 2.6. Since Theorem 2.2 says that can never have a positive lower Lipschitz Φ A bound when Φ is a frame it may seem tempting to ask whether we can achieve a positive lower bound for a set that does not form a frame, i.e., a sequence Φ that does not have an upper frame bound. While this might be possible, Lemma 2.5 tells us that in this case Φ A cannot have a finite upper Lipschitz bound. To see this take g = 0 in the proof of the Lemma so that (f) (g) = (f) = T (f) for every f , then use the fact that Φ Φ Φ Φ Φ kA −A k kA k k k ∈ H does not have a finite upper frame bound to produce a sequence of unit vectors fn n∈N with { } T (f ) . Φ n k k → ∞ Theorem 2.7. Let be an infinite-dimensionalseparable Hilbert space and let Φ = ϕn n∈N H { } be a frame for with frame bounds 0 < A B < . For each m N let V be a finite- m H ≤ ∞ ∈ dimensional subspace of so that dim(V ) > dim(V ). Suppose there is an increasing m+1 m H function G(m), with lim G(m) = , so that the following holds for every m: for every m→∞ ∞ f,g V m ∈ inf f αg G(m) (f) (g) . Φ Φ |α|=1k − k ≤ kA −A k 8 For γ > 1 and R > 0 define (R) = f : f P f G(m+1)−γR f for every m N γ m B { ∈ H k − k ≤ k k ∈ } where P denotes the orthogonal projection onto V . Then for every f,g (R) we have m m γ ∈ B inf f αg C ( f + g )1/γ Φ(f) Φ(g) γ−γ1 (2.1) |α|=1k − k ≤ k k k k kA −A k where C depends on only B,R,γ and G(1). Proof. Let f,g (R). We start by proving an equality of the type γ ∈ B γ−1 inf f αg C′(1+ f + g ) (f) (g) γ ; (2.2) Φ Φ |α|=1k − k ≤ k k k k kA −A k equation (2.1) will then follow by an amplification trick. If (f) (g) RB1/2G(1)−γ, then it follows from Φ Φ kA −A k ≥ f αg f + g k − k ≤ k k k k γ−1 that (2.2) is satisfied for C′ = C := R−1B−1/2G(1)γ γ . 1 Now assume that Φ(f) Φ(g) <(cid:0) RB1/2G(1)−γ. (cid:1)Find then m so that kA −A k RB1/2G(m+1)−γ (f) (g) RB1/2G(m)−γ (2.3) Φ Φ ≤ kA −A k ≤ (we can always do this since G is increasing and lim G(m) = ). We have that m→∞ ∞ f αg f P f + g P g + P f αP g , m m m m k − k ≤ k − k k − k k − k and since P f,P g V we also have that m m m ∈ inf P f αP g G(m) (P f) (P g) m m Φ m Φ m |α|=1k − k ≤ kA −A k G(m)( (f) (P f) + (f) (g) + (g) (P g) ) Φ Φ m Φ Φ Φ Φ m ≤ kA −A k kA −A k kA −A k G(m) B1/2 f P f + (f) (g) +B1/2 g P g m Φ Φ m ≤ k − k kA −A k k − k (cid:0) (cid:1) so that inf f αg 1+G(m)B1/2 ( f P f + g P g ) + G(m) (f) (g) . m m Φ Φ |α|=1k − k ≤ k − k k − k kA −A k (cid:0) (cid:1) (2.4) Because f and g are both in (R), we have γ B f P f + g P g G(m+1)−γR( f + g ) B−1/2 (f) (g) ( f + g ); m m Φ Φ k − k k − k ≤ k k k k ≤ kA −A k k k k k 9 on the other hand, using (f) (g) < RB1/2G(1)−γ and (2.3), we derive Φ Φ kA −A k 1 1 1 1 1 1 1+G(m)B1/2 Rγ B2γ G(1)−1 Φ(f) Φ(g) −γ + B1/2Rγ B2γ Φ(f) Φ(g) −γ ≤ kA −A k kA −A k 1 1 1 Rγ B2γ G(1)−1 +B1/2 Φ(f) Φ(g) −γ . ≤ kA −A k Finally, we also have, using (cid:0)(2.3) again, (cid:1) 1 G(m) (f) (g) (f) (g) −1RB1/2 γ (f) (g) Φ Φ Φ Φ Φ Φ kA −A k ≤ kA −A k kA −A k 1 1 γ−1 = (cid:0)Rγ B2γ Φ(f) Φ(g) γ (cid:1) kA −A k Substituting all this into (2.4), we obtain 1 1 γ−1 inf f αg Rγ B2γ B−1/2G(1)−1 +1 ( f + g ) + 1 Φ(f) Φ(g) γ , |α|=1k − k ≤ k k k k kA −A k (cid:2)(cid:0) (cid:1) (cid:3) 1 1 which does indeed imply the inequality (2.2), with C′ = C2 := Rγ B2γ B−1/2G(1)−1 +1 . It thus follows that, for all f,g (R), (2.2) holds for C′ = max(C ,C ); C′ is completely γ 1 (cid:0)2 (cid:1) ∈ B determined by R,B,γ and G(1). This inequality can be sharpened further by exploiting its non-homogeneous nature. The set (R) is invariant under scaling: if f (R), then so are all multiples of f. If, given γ γ B ∈ B f,g (R), we write the inequality (2.2) for f′ = Mf, g′ = Mg, where M R is to be γ + ∈ B ∈ fixed below, then we find, upon dividing both sides by M = 0, 6 γ−1 γ−1 inf f αg C′M−1 [1 + M ( f + g )]M γ (f) (g) γ Φ Φ |α|=1k − k ≤ k k k k kA −A k = C′M−γ1 [1 + M ( f + g )] Φ(f) Φ(g) γ−γ1 . k k k k kA −A k Since this inequality holds for all M R , it holds in particular for the value M = + ∈ [(γ 1)( f + g )]−1 that minimizes the right hand side. We obtain − k k k k 1−γ 1 γ−1 inf f αg C′γ(γ 1) γ ( f + g )γ (f) (g) γ Φ Φ |α|=1k − k ≤ − k k k k kA −A k Remark 2.8. We note that although (2.1) describes a type of Hölder continuity (and thus uniform continuity) for , when restricted to (R), with Hölder exponent (γ 1)/γ, it Φ γ A B − does not establish Lipschitz continuity (which would require Hölder exponent 1). So far, most papers on the stability of phase retrieval have focused on showing Lipschitz continuity; we do not know whether Lipschitz bounds are possible within our framework, or whether these weaker Hölder-type bounds are the strongest possible here. This theorem complements Theorem 2.2: even though uniformly stable phase retrieval is never possible in infinite-dimensional , Theorem 2.7 establishes that stable phase retrieval H 10

Phase retrieval in infinite-dimensional Hilbert spaces

Similar Phase retrieval in infinite-dimensional Hilbert spaces