Generalized consistent sampling in abstract Hilbert spaces

J. Math. Anal. Appl. 433 (2016) 375–391

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Generalized consistent sampling in abstract Hilbert spaces Kil Hyun Kwon a,1 , Dae Gwan Lee b,∗ a

Department of Mathematical Sciences, KAIST, Daejeon 305-701, Republic of Korea BK21 PLUS SNU Mathematical Sciences Division, Seoul National University, Seoul 151-747, Republic of Korea b

a r t i c l e

i n f o

Article history: Received 20 January 2015 Available online 4 August 2015 Submitted by R. Curto Keywords: Consistent sampling Frame Oblique projection

a b s t r a c t We consider generalized consistent sampling and reconstruction processes in an abstract Hilbert space H. We first study the consistent sampling in H together with its performance analysis. We then study its generalization: partial consistency and quasi-consistency. We give complete characterizations for both of them. We also provide an iterative algorithm to compute the quasi-consistent approximation. An illustrative example is also included. © 2015 Elsevier Inc. All rights reserved.

1. Introduction A fundamental problem of sampling theory concerns the reconstruction of signals from a discrete set of measurements, namely the samples. Classical results on this problem usually focus on perfect reconstruction of signals under suitable assumptions. The most important example is the Shannon sampling theorem [20] which says that any signal f band-limited in [−π, π] can be perfectly reconstructed from its uniform samples  as f (t) = n∈Z f (n) sinc(t − n), where sinc t = sinπtπt is the cardinal sine function. More general approach to the problem is to seek an approximation rather than the perfect reconstruction on some restricted class of signals. As a general abstract formulation, consider a Hilbert space H and a set of measurement vectors {vj }j∈J in H which span a closed subspace V (sampling space). Given any signal f in H, we take measurements of the form {f, vj }j∈J and seek an approximation of f as a linear combination of reconstruction vectors {wk }k∈K which span a closed subspace W (reconstruction space). A natural method for such an approximation is to impose the ‘consistency’, which requires that the approximated signal and the original one yield the same measurements. Consistency was first introduced by Unser and Aldroubi [22] in the setting of shift invariant spaces and later extended significantly by Eldar et al. [7–9,11]. In the beginning, it was assumed [8,9,22] that H = W ⊕ V ⊥ under which we have a unique consistent sampling * Corresponding author. 1

E-mail addresses: [email protected] (K.H. Kwon), [email protected] (D.G. Lee). Tel.: +82 42 350 2716; fax: +82 42 350 5710.

http://dx.doi.org/10.1016/j.jmaa.2015.07.070 0022-247X/© 2015 Elsevier Inc. All rights reserved.

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operator given as the oblique projection onto W along V ⊥ . It is proved that for particular shift-invariant sampling and reconstruction spaces V and W in L2 (R), this consistent sampling performs well as a sampling approximation process [12,13,23,24]. Later, Hirabayashi and Unser [14] treated the consistent sampling when H = W + V ⊥ is finite dimensional and W ∩ V ⊥ may or may not be {0}. Under such assumptions, they proved that there exist infinitely many consistent sampling operators each of which is an oblique projection. Recently, Arias and Conde [2] studied the problem when H  W + V ⊥ in which case consistent sampling operators no longer exist. As a generalization of the consistency, they introduced the ‘quasi-consistency’ which only requires that the measurements of the approximated signal and the original one are as close as possible in l2 sense. In this work, we first study the consistency in an abstract setting on an arbitrary Hilbert space H together with its performance analysis. We then study partial consistency and quasi-consistency [2] when the condition H = W + V ⊥ is violated. In particular, we show that the quasi-consistency can be also interpreted geometrically in terms of oblique projections. We also provide an iterative method to compute the quasi-consistent approximation. This paper is organized as follows. In Section 2, we collect definitions and concepts needed throughout the paper. In Section 3, we give complete characterizations of consistent sampling operators along with their performance analysis. Several useful representations of consistent sampling operators are also provided. Section 4 is devoted to two generalizations of the consistency: partial consistency and quasi-consistency, with more focus on the latter. We give complete characterizations of both partial and quasi-consistencies. We also provide an iterative algorithm to obtain the quasi-consistent approximation. Finally we give an illustrative example. 2. Preliminaries For any countable index set I, let l2 (I) be the set of all complex-valued sequences c = {c(i)}i∈I with  c2 := i∈I |c(i)|2 < ∞. A sequence {φn |n ∈ I} of vectors in a separable Hilbert space H is • a frame of H if there are constants B ≥ A > 0 such that Aφ2H ≤



|φ, φn |2 ≤ Bφ2H ,

φ ∈ H;

n∈I

• a Riesz basis of H if it is complete in H and there are constants B ≥ A > 0 such that Ac2 ≤ 



c(n)φn 2H ≤ Bc2 ,

c = {c(n)}n ∈ l2 (I).

n∈I

For any two Hilbert spaces H and K, let L(H, K) denote the set of all bounded linear operators from H into K, and L(H) := L(H, H). For any T ∈ L(H, K), let R(T ) and N (T ) be the range and the kernel of T respectively. When R(T ) is closed, T † denotes the Moore–Penrose pseudo-inverse of T [1] and T − := {X ∈ L(K, H)| T XT = T } = {T † + Y − T † T Y T T † |Y ∈ L(K, H)} the set of all generalized inverses of T (see Lemma 2.1 below). Lemma 2.1. Let T1 ∈ L(V, K) and T2 ∈ L(H, U) have closed ranges, where H, K, U, V are Hilbert spaces. For any T ∈ L(H, K), the equation T1 XT2 = T

(1)

is solvable for X ∈ L(U, V) if and only if R(T ) ⊆ R(T1 ) and N (T ) ⊇ N (T2 ). In this case, the general solution X is

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X = Xp + Y − T1† T1 Y T2 T2† , where Xp ∈ L(U, V) is any particular solution of (1) (e.g. Xp = T1† T T2† ) and Y ∈ L(U, V) is arbitrary. Proof. See [1] or Theorem 2 in [19].

2

For any closed subspaces A and B of H with H = A ⊕ B, let PA,B be the oblique projection onto A along B defined by PA,B (x) = y for x = y + z, where y ∈ A and z ∈ B. In particular for B = A⊥ , PA,A⊥ := PA is the orthogonal projection onto A. For any closed affine subspace Ah := h + A where h ∈ H, the orthogonal projection onto Ah is defined as PAh (x) := h + PA (x − h). Lemma 2.2. Let A and B be closed subspaces of H. (a) (PB PA )n  = PB PA 2n−1 for any integer n ≥ 1; (b) Assume H = A ⊕ B, and let C ⊆ H. Then PA,B (C) = A ∩ (B + C). In particular, PA (C) = A ∩ (A⊥ + C). Proof. (a) For any T ∈ L(H), we have T ∗ T  = T 2 . If T is normal (i.e. T ∗ T = T T ∗ ), then T n  = T n for any integer n ≥ 1 [15]. Then (PB PA )n 2 = ((PB PA )n )∗ (PB PA )n  = (PA PB )n (PB PA )n  = (PA PB PA )2n−1  = PA PB PA 2n−1 since PA PB PA is self-adjoint. On the other hand, PA PB PA  = (PB PA )∗ PB PA  = PB PA 2 . Thus (PB PA )n  = PB PA 2n−1 . (b) Let a = PA,B (c) (∈ A) where c ∈ C. Then b := a − c = −PB,A (c) ∈ B so that a = b + c ∈ A ∩ (B + C). Conversely, let a = b + c ∈ A ∩ (B + C) where b ∈ B and c ∈ C. Then a = PA,B (a) = PA,B (b + c) = PA,B (c) ∈ PA,B (C). Therefore PA,B (C) = A ∩ (B + C). 2 Lemma 2.3. For any two nontrivial closed subspaces A and B of H, define R(A, B) := inf

a∈A\0

PB a a

and

PB a a∈A\0 a

S(A, B) := sup

so that 0 ≤ R(A, B) ≤ S(A, B) ≤ 1. Then (a) (b) (c) (d) (e) (f)

R(A, B)2 + S(A, B ⊥ )2 = 1; R(A, B) = R(B⊥ , A⊥ ) if A, B = H; S(A, B) = S(B, A) = sup{|a, b| : a ∈ A and b ∈ B with a = b = 1} = PB PA ; A + B is closed and A ∩ B = {0} if and only if S(A, B) < 1 (or R(A, B⊥ ) > 0); H = A ⊕ B if and only if R(A, B ⊥ ) = R(B⊥ , A) > 0; 1 If H = A ⊕ B, then PA,B  = I − PA,B  = R(A,B ⊥) .

Proof. See Theorems 2.1 and 2.3 in [21] for (a)–(e) (also see [22]). For (f), see Theorem 2 in [4].

2

R(A, B) and S(A, B) are called the cosines of the largest and smallest angles from A to B respectively [5,21]. S(A, B) is also called the cosine of the Dixmier angle between A and B. Lemma 2.4. Let H, K1 , K2 be Hilbert spaces and assume that S ∈ L(K1 , H) and T ∈ L(K2 , H) have closed ranges V = R(S) and W = R(T ) in H. Then (a) W ∩ V ⊥ = {0} if and only if N (S ∗ T ) = N (T ); (b) H = W + V ⊥ if and only if R(S ∗ T ) = R(S ∗ ); (c) W + V ⊥ is closed if and only if R(S ∗ T ) is closed.

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Consequently, (d) W + V ⊥ is closed if and only if W ⊥ + V is closed; (e) H = W ⊕ V ⊥ if and only if H = W ⊥ ⊕ V. Proof. First note that R(S ∗ T ) ⊆ R(S ∗ ) and N (S ∗ T ) ⊇ N (T ) hold in general. (a) Assume W ∩V ⊥ = {0} and let x ∈ N (S ∗ T ). Then T (x) ∈ R(T ) ∩N (S ∗ ) = W ∩V ⊥ = {0} so x ∈ N (T ). Therefore N (S ∗ T ) = N (T ). Conversely, assume N (S ∗ T ) = N (T ) and let f ∈ W ∩ V ⊥ = R(T ) ∩ N (S ∗ ). Then f = T (x) for some x ∈ K2 and S ∗ T (x) = S ∗ f = 0. Since N (S ∗ T ) = N (T ), f = T (x) = 0 so W ∩ V ⊥ = {0}. (b) Assume H = W + V ⊥ . Then R(S ∗ ) = S ∗ (H) = S ∗ (W + V ⊥ ) = S ∗ (W) = S ∗ (R(T )) = R(S ∗ T ) since N (S ∗ ) = V ⊥ . Conversely, assume R(S ∗ T ) = R(S ∗ ) and let f ∈ H. Then S ∗ (f ) = S ∗ T (x) for some x ∈ K2 so that h := f − T (x) ∈ N (S ∗ ) = V ⊥ and so f = T (x) + h ∈ W + V ⊥ . Therefore H = W + V ⊥ . (c) See Theorem 22 in [5]. (d) Note that R(T ∗ S) = R((S ∗ T )∗ ) is closed if and only if R(S ∗ T ) is closed. Hence the claim follows immediately from (c) by interchanging the roles of S and T . (e) For any bounded linear operator A, we have N (A)⊥ = R(A∗ ), R(A)⊥ = N (A∗ ), and that R(A∗ ) is closed if and only if R(A) is closed. Then the claim follows easily from (a)–(c). Also see Lemma 2.3(e). 2 3. Consistent sampling We now formulate a sampling problem in an abstract Hilbert space H. Let {vj = 0}j∈J be a set of sampling (or analysis) vectors in H, which forms a frame of the sampling space V := spanj∈J {vj } with pre-frame  operator S : l2 (J) → V given by S(c) = j∈J c(j) vj . Likewise, let {wk = 0}k∈K be a set of reconstruction (or synthesis) vectors in H, which forms a frame of the reconstruction space W := spank∈K {wk } with  pre-frame operator T : l2 (K) → W given by T (d) = k∈K d(k) wk . Then the sampling operator is S ∗ : H  f −→ S ∗ (f ) = {f, vj }j∈J ∈ l2 (J) where c = S ∗ (f ) is the measurement (i.e., generalized samples) of f . Now we look for an operator P : H → W that approximates each signal f ∈ H by f = P(f ) from its measurement c = S ∗ (f ). Specifically, we seek an operator P satisfying: (stability)

(consistency)

P ∈ L(H, W);

(2)

P(f ) = 0 if S ∗ (f ) = 0, i.e., N (S ∗ ) ⊆ N (P);

(3)

P is consistent, i.e., S ∗ (Pf ) = S ∗ (f ) for any f ∈ H.

(4)

We call P satisfying (2)–(4) a consistent sampling operator, and denote the set of all such operators by C(W, V). Note that the conditions (2) and (3) are equivalent to P = T QS ∗

for some Q ∈ L(l2 (J), l2 (K)),

by Lemma 2.1. We call Q a consistent filter if P = T QS ∗ satisfies (4). Note that a filter Q is consistent if and only if S ∗ T QS ∗ = S ∗ . We denote the set of all admissible operators satisfying (2) and (3) and the set of all consistent filters respectively as A(W, V) = {P = T QS ∗ | Q ∈ L(l2 (J), l2 (K))}

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and F(W, V) = {Q ∈ L(l2 (J), l2 (K))| S ∗ T QS ∗ = S ∗ }. Then C(W, V) = T F(W, V) S ∗ and we can easily show that C(W, V) is closed under composition. Throughout the paper we always assume that H0 := W + V ⊥ is closed (or equivalently R(S ∗ T ) is closed so that (S ∗ T )† exists). Proposition 3.1. (See Theorem 1 in [14].) Assume that there exists a consistent sampling operator P˜ . Then H = H0 = L ⊕ V ⊥ and P = PL,V ⊥ where L = R(P) ⊆ W. In general, any set L ⊆ W is a closed complementary subspace of V ⊥ in H0 if and only if it is a closed complementary subspace of W ∩ V ⊥ in W. Let L := {L| L is a closed complementary subspace of W ∩ V ⊥ in W}. Theorem 3.2. The following are equivalent: (a) C(W, V) = ∅ (or equivalently, F(W, V) = ∅); (b) H = H0 ; (c) R(S ∗ T ) = R(S ∗ ). In this case, C(W, V) = {PL,V ⊥ | L ∈ L}. Moreover, F(W, V) = (S ∗ T )− = {(S ∗ T )† + Y − PN (S ∗ T )⊥ Y PR(S ∗ ) | Y ∈ L(l2 (J), l2 (K))}

(5)

so that C(W, V) = {T (S ∗ T )† S ∗ + T PN (S ∗ T ) Y S ∗ | Y ∈ L(l2 (J), l2 (K))}.

(6)

Remark 3.3. Most parts of Theorem 3.2 are known and can be found in the literature. See e.g., Proposition 1 in [7], Theorem 1 in [8], and Theorem 3.1 in [2]. However, these works do not give statements in a complete form as presented in Theorem 3.2. Below we provide its short proof only for self-containedness. Proof. (a) ⇒ (b): Let P = T QS ∗ be a consistent sampling operator and f ∈ H. Then h := f − P(f ) ∈ V ⊥ since S ∗ (h) = S ∗ (f ) − S ∗ P(f ) = 0 so that f = P(f ) + h ∈ W + V ⊥ = H0 . Thus, H = H0 . (b) ⇔ (c): See Lemma 2.4(b). (c) ⇒ (a): Assume that R(S ∗ T ) = R(S ∗ ). Then by Lemma 2.1, there exists Q ∈ L(l2 (J), l2 (K)) such that S ∗ T QS ∗ = S ∗ . Therefore C(W, V) = ∅. Now assume that any one of the equivalent conditions (a)–(c) holds. Then for any P ∈ C(W, V), (3) and (4) yield that P2 = T Q(S ∗ P) = T QS ∗ = P and N (P) = N (S ∗ ) = V ⊥ . Hence P = PL,V ⊥ where L := R(P) = N (I − P) ∈ L. Conversely, for any L ∈ L, let P := PL,V ⊥ . Then P : H → L ⊆ W is a bounded linear operator with N (P) = V ⊥ = N (S ∗ ). For any f = g + h ∈ H = L ⊕ V ⊥ where g ∈ L and h ∈ V ⊥ , S ∗ (P(f )) = S ∗ (g) = S ∗ (g + h) = S ∗ (f ) so that P ∈ C(W, V). Thus, C(W, V) = {PL,V ⊥ | L ∈ L}. Note that since R(S ∗ T ) = R(S ∗ ), any Q ∈ L(l2 (J), l2 (K)) satisfies S ∗ T QS ∗ = S ∗ if and only if S ∗ T QS ∗ T = S ∗ T . This yields (5) and (6) by Lemma 2.1. 2

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Corollary 3.4. (See Theorem 1 and Proposition 1 in [8].) There is a unique consistent sampling operator P if and only if H = W ⊕ V ⊥ . In this case, P = PW,V ⊥ = T (S ∗ T )† S ∗ . When both index sets J and K are finite, it is convenient to use the following equivalent expression for F(W, V): F(W, V) = {Q ∈ L(l2 (J), l2 (K))| B ◦ Q = I on R(S ∗ )} where B = [wk , vj ]j∈J,k∈K is the matrix representation of S ∗ T ∈ L(l2 (K), l2 (J)) called the input–output cross correlation matrix (or the Gram matrix). Corollary 3.5. Assume that the index sets J and K are finite and S ∗ is surjective, i.e., {vj }j∈J is a Riesz basis of V. Let M and N denote the cardinality of J and K respectively. Then the following are equivalent. (a) C(W, V) = ∅ (or equivalently, F(W, V) = ∅); (b) H = H0 ; (c) B : CN → CM is surjective (so M ≤ N ). Moreover in this case, F(W, V) = {B † + (I − B † B)Y | Y ∈ L(CM , CN )}. Furthermore, if M = N , then the above conditions are also equivalent to: (d) H = W ⊕ V ⊥ (so C(W, V) = {PW,V ⊥ }); (e) det(B) = 0. In this case, {wk }k∈K is also a Riesz basis of W. Proof. Assume that J and K are finite and S ∗ is surjective, i.e., R(S ∗ ) = CM . Then Q ∈ F(W, V) if and only if B ◦ Q = I, which implies that F(W, V) = ∅ if and only if B is surjective. Therefore (a)–(c) are all equivalent. In this case, F(W, V) = {Q ∈ L(CM , CN )| B ◦ Q = I} = {B † + (I − B † B)Y | Y ∈ L(CM , CN )}. Now assume that M = N . Then (c) ⇔ (e) and (d) ⇒ (b) hold trivially. (a) ⇒ (d): Let Q ∈ F(W, V), i.e., Q ∈ L(CM , CM ) such that B ◦ Q = I. Note that Q : CM → CM is an isomorphism. Then P := T QS ∗ is a consistent sampling operator with R(P) = R(T ) = W since S ∗ and Q are surjective. Therefore H = W ⊕ V ⊥ and P = PW,V ⊥ by Proposition 3.1. Finally, let d ∈ CN be such that T (d) = 0. Since S ∗ and Q are surjective, d = QS ∗ (f ) for some f ∈ H so that PW,V ⊥ (f ) = T QS ∗ (f ) = T (d) = 0. Then f ∈ V ⊥ = N (S ∗ ) so d = QS ∗ (f ) = 0. Therefore N (T ) = {0}, which means that {wk }k∈K is a Riesz basis of W. Hence (a)–(e) are all equivalent, in which case {wk }k∈K is a Riesz basis of W. 2  := T (S ∗ T )† S ∗ ∈ C(W, V) whenever H = H0 so that C(W, V) = ∅. It is therefore From (6), we have U  more closely. worthwhile to observe U  := T (S ∗ T )† S ∗ ∈ L(H, W) and U 0 := U  |H . Then the following hold. Lemma 3.6. Let U 0  ) = R(T (S ∗ T )† ) = R(T T ∗ S) ⊆ W along N (U  ) = N (PR(S ∗ T ) S ∗ ) ⊇  is the oblique projection onto R(U (a) U  ) ⊕ N (U  )). V ⊥ (so H = R(U ⊥   ) = R(U 0 ) and V ⊥ = N (U 0 ) so U 0 = P  ⊥ . (b) H0 = R(U ) ⊕ V where R(U R(U ),V  ) = W if and only if W ∩ V ⊥ = {0}. (c) R(U  ) = V ⊥ if and only if H = H0 . (d) N (U  = PW,V ⊥ if and only if R(U  ) = W and N (U  ) = V ⊥ if and only if H = W ⊕ V ⊥ . (e) U

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 = PR(T T ∗ S),V ⊥ when H = H0 . This can be Remark 3.7. It was shown in (Lemmas 2 and 3, [14]) that U seen immediately from (a) and (d).  2 = T (S ∗ T )† S ∗ T (S ∗ T )† S ∗ = T PN (S ∗ T )⊥ (S ∗ T )† S ∗ = U  so that Proof. (a) Since R((S ∗ T )† ) = N (S ∗ T )⊥ , U ∗ † ∗ ∗ † ∗ † ∗ ⊥   U is a projection. R(U ) = R(T (S T ) S ) ⊆ R(T (S T ) ) = T (R((S T ) )) = T (N (S T ) ) = T (R(T ∗ S)) =  ). R(T T ∗ S) = R(T PR(T ∗ S) T ∗ S) = R(T PN (S ∗ T )⊥ T ∗ S) = R(T (S ∗ T )† (S ∗ T )T ∗ S) ⊆ R(T (S ∗ T )† S ∗ ) = R(U ∗ † ∗ ∗ ∗ ∗   Hence R(U ) = R(T (S T ) ) = R(T T S) ⊆ W. Since S U = PR(S ∗ T ) S , N (U ) ⊆ N (PR(S ∗ T ) S ). Conversely,  ). Then U  (f ) ∈ R(U  ) ∩ N (S ∗ ) ⊆ R(U  ) ∩ N (U  ) = {0} so that f ∈ N (U  ). let f ∈ N (PR(S ∗ T ) S ∗ ) = N (S ∗ U ∗ ⊥  Therefore N (U ) = N (PR(S ∗ T ) S ) ⊇ V .  and R(U  ) ⊆ W ⊂ H0 , we have U 0 so that H0 = R(U 0 ) ⊕ N (U 0 ) and U 2 = U 2 = U 0 = (b) Since U 0 2 0 ) = R(U  ) since R(U 0 ) ⊆ R(U  ) = R(U  )=U  (R(U  )) ⊆ U  (H0 ) = R(U 0 ). Let PR(U0 ),N (U0 ) . Note that R(U ⊥ ∗ ∗ ∗ 0 ) where g = T (d) ∈ W, d ∈ l2 (K) and h ∈ V . Then S (f ) = S T (d) = PR(S ∗ T ) S T (d) = f = g+h ∈ N (U ∗  (f ) = 0 so that f ∈ V ⊥ . Since V ⊥ ⊆ N (U  ) ∩ H0 = N (U 0 ), we have N (U 0 ) = V ⊥ . PR(S ∗ T ) S (f ) = S ∗ U ⊥ ⊥  ) = W, then H0 = W ⊕ V by (b) so that W ∩ V = {0}. Conversely, if W ∩ V ⊥ = {0}, then (c) If R(U  ) ⊕ V ⊥ where R(U  ) ⊆ W, and thus R(U  ) = W. H0 = W ⊕ V ⊥ = R(U ⊥ ⊥    ) ⊕N (U ) = (d) If N (U ) = V , then H = R(U ) ⊕V = H0 by (a) and (b). Conversely, if H = H0 , then R(U ⊥ ⊥ ⊥  ) ⊕ V where N (U  ) ⊇ V , and thus N (U ) = V . R(U (e) It follows immediately from (a), (c) and (d). 2 Below we let L0 := W ∩ (W ∩ V ⊥ )⊥ (∈ L) be the orthogonal complementary subspace of W ∩ V ⊥ in W. Note that if W ∩ V ⊥ = {0}, then L0 = W is the unique element of L.  = PL ,V ⊥ . Proposition 3.8. Assume H = H0 . Then PL0 U 0 Proof. We always have (when H0 ⊆ H): W ∩ V ⊥ = R(T PN (S ∗ T ) ).

(7)

 ) ⊕ (W ∩ V ⊥ ) = R(T ) = T (R(T ∗ S) ⊕ Trivially R(T PN (S ∗ T ) ) ⊆ W ∩ V ⊥ . On the other hand, R(U  ) + R(T PN (S ∗ T ) ) so that R(T ∗ S)⊥ ) = T (R(T ∗ S)) + T (R(T ∗ S)⊥ ) = R(T T ∗ S) + T (N (S ∗ T )) = R(U ⊥  =U −  = P(W∩V ⊥ )⊥ U  = (I −PW∩V ⊥ )U W ∩V ⊆ R(T PN (S ∗ T ) ). Hence (7) holds. Now we have by (7), PL0 U † ∗ † ∗     PR(T PN (S∗ T ) ) U = U − T PN (S ∗ T ) [(T PN (S ∗ T ) ) T (S T ) ]S . So PL0 U ∈ C(W, V) by (6). Since R(PL0 U ) ⊆ L0 ,  = PL ,V ⊥ by Proposition 3.1. 2 PL U 0

0

Lemma 3.9. Assume H = H0 and let L ∈ L. Then PV |L : L → V and PL |V : V → L are isomorphisms. In particular, PL0 |V = PW |V . Proof. Assume H = H0 and let L ∈ L. Then H = L ⊕ V ⊥ = L⊥ ⊕ V so that PV (L) = V ∩ (L ⊕ V ⊥ ) = V and PL (V) = L∩(L⊥ ⊕V) = L by Lemma 2.2(b). If f ∈ L is such that PV (f ) = 0, then f ∈ L∩V ⊥ = {0} so f = 0. If g ∈ V is such that PL (g) = 0, then g ∈ L⊥ ∩ V = {0} so g = 0. Hence PV |L : L → V and PL |V : V → L are isomorphisms. Finally, PL0 |V = PL0 PW |V = PW |V since L0 := W ∩ (W ∩ V ⊥ )⊥ = W ∩ (W ⊥ + V) = PW (V) by Lemma 2.2(b). 2 We now give another expression for consistent sampling operators in terms of orthogonal projections. This expression is motivated by Theorem 1 in [7], which concerns characterizing the set of consistent approximations (also see Proposition 3.1 in [2]).

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Theorem 3.10. Assume H = H0 . Then for any L ∈ L, PL,V ⊥ = FL−1 PV = PL G−1 L PV , where FL := I −PV ⊥ PL and GL := I − PV PL⊥ . In particular, PL0 ,V ⊥ = F † PV = PW G−1 PV , where F := I − PV ⊥ PW and G := I − PV PW ⊥ . Proof. Assume H = H0 and let L ∈ L. Then H = L ⊕ V ⊥ so that PV ⊥ PL  < 1 by Lemma 2.3. Thus,  FL := I − PV ⊥ PL =

PV

on L,

I

on L⊥

is invertible with its inverse FL−1

 =

(PV |L )−1

on V,

I

on L⊥ .

Similarly, GL and G are invertible since PV PL⊥  < 1 and PV PW ⊥  < 1 respectively (see Lemmas 2.3 and 2.4). Note that FL PL = (I − PV ⊥ PL )PL = (I − PV ⊥ )PL = PV PL = PV (I − PL⊥ ) = −1 −1 −1 PV (I − PV PL⊥ ) = PV GL . Then P := FL−1 PV = FL−1 (PV GL )G−1 L PV = FL (FL PL )GL PV = PL GL PV . −1 −1 ⊥ Since P|L = FL PV |L = (PV |L ) PV |L = I on L and N (P) ⊇ V , we have P = PL,V ⊥ . Therefore PL,V ⊥ = FL−1 PV = PL G−1 L PV . Similarly, F PW = (I − PV ⊥ PW )PW = (I − PV ⊥ )PW = PV PW = PV (I − PW ⊥ ) = PV (I − PV PW ⊥ ) = PV G so that P := F † PV = F † (PV G)G−1 PV = F † (F PW )G−1 PV = PN (F )⊥ PW G−1 PV . Note that G|V : V → V is an isomorphism since GPV = (I − PV PW ⊥ )PV = PV − PV PW ⊥ PV = PV (I − PW ⊥ )PV = PV PW PV = PV PL0 PV by Lemma 3.9. Then R(PW G−1 PV ) = PW (V) = PL0 (V) = L0 ⊂ (W ∩V ⊥ )⊥ = N (F )⊥ so that P = PW G−1 PV . On the other hand, F |L0 = PV |L0 = FL0 |L0 is an isomorphism from L0 onto V so that F † |V = (PV |L0 )−1 = FL−1 | . Thus, P = F † PV = FL−1 PV = PL0 ,V ⊥ . This completes 0 V 0 the proof. 2 Lemma 3.11. Assume H = H0 . Then PL0 ,V ⊥ (f ) ≤ PL,V ⊥ (f ) for any L ∈ L and any f ∈ H. Hence PL0 ,V ⊥  ≤ PL,V ⊥  for any L ∈ L. Proof. Let L ∈ L, f ∈ H, g = PL,V ⊥ (f ), and g0 = PL0 ,V ⊥ (f ). Then h := g − g0 ∈ W ∩ V ⊥ since PV (h) = PV (g) − PV (g0 ) = PV (f ) − PV (f ) = 0. Note that g0 ⊥ h since L0 ⊥ (W ∩ V ⊥ ). Therefore g0 2 ≤ g0 2 + h2 = g2 , thus, PL0 ,V ⊥ (f ) ≤ PL,V ⊥ (f ). 2 The following theorem was first given in [2] in a form of minimization problem. Its motivation comes from the work of Eldar and Dvorkind [10], which finds approximations of a given input signal that minimize the worst error or the worst regret based on minimax formulations. Here we consider the case in which consistent sampling operators exist (H = H0 ). Its simple proof can be deduced from Lemma 2.3(f) and Lemma 3.11. The general case (H = H0 ) will be discussed later in Theorem 4.13. Theorem 3.12. (See Theorem 5.4 in [2].) Assume H = H0 . Then for any L ∈ L, I − PL0 ,V ⊥  ≤ I − PL,V ⊥ 

(8)

PW − PL0 ,V ⊥  ≤ PW − PL,V ⊥ .

(9)

and

Theorem 3.12 naturally leads to the following question: when is PL,V ⊥ the closest to its corresponding orthogonal projection PL ? This question was first treated in [13] when V and W are shift-invariant subspaces of H = L2 (R).

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Lemma 3.13. Let T ∈ L(H, K) be such that T = 0 on a closed subspace G of H. Then T  = T |G ⊥ . Proof. T  = supg∈G,h∈G ⊥ ,g+h≤1 T (g+h) = supg∈G,h∈G ⊥ ,g2 +h2 ≤1 T (h) ≤ suph∈G ⊥ ,h≤1 T (h) = T |G ⊥ . Hence T  = T |G ⊥ . 2 Theorem 3.14. (Cf. Theorem 4.1 in [13].) Assume H = H0 . Then for any L ∈ L, PL0 − PL0 ,V ⊥  ≤ PL − PL,V ⊥ . Proof. First note that PL0 ,V ⊥ |W = PL0 ,W∩V ⊥ = PL0 |W since W = L0 ⊕ (W ∩ V ⊥ ) is an orthogonal direct sum. So, PL0 −PL0 ,V ⊥  = (PL0 −PL0 ,V ⊥ )|W ⊥  by Lemma 3.13. Note that for any L ∈ L, PL |W ⊥ = 0 since L ⊂ W. Then (PL0 − PL0 ,V ⊥ )|W ⊥  = PL0 ,V ⊥ |W ⊥  ≤ PL,V ⊥ |W ⊥  = (PL − PL,V ⊥ )|W ⊥  ≤ PL − PL,V ⊥  by Lemma 3.11. Hence PL0 − PL0 ,V ⊥  ≤ PL − PL,V ⊥ . 2 Theorem 3.15. (Cf. Theorems 4.2 and 4.3 in [13].) Assume H = H0 and let L ∈ L. Then for any f ∈ / L, 0 < R(L, V) ≤

f − PL (f ) ≤ S(V ⊥ , L⊥ ) ≤ 1 f − PL,V ⊥ (f )

and 0 ≤ R(V ⊥ , L) ≤

PL (f ) − PL,V ⊥ (f ) ≤ S(L, V ⊥ ) < 1. f − PL,V ⊥ (f )

Proof. Let f ∈ / L so that g := f − PL,V ⊥ (f ) = PV ⊥ ,L (f ) ∈ V ⊥ \0. Then f − PL (f ) = PL⊥ (f ) = PL⊥ PV ⊥ ,L (f ) = PL⊥ (g) and PL (f ) −PL,V ⊥ (f ) = PL (f −PL,V ⊥ (f )) = PL (g) so that and

PL (f )−PL,V ⊥ (f ) f −PL,V ⊥ (f )

=

PL (g) g .

f −PL (f ) f −PL,V ⊥ (f )

Since H = L ⊕ V ⊥ , the claims follow by Lemma 2.3.

=

PL⊥ (g) g

2

4. Partial consistency and quasi-consistency Recall that C(W, V) = ∅ if and only if H = H0 := W + V ⊥ (see Theorem 3.2). So if H0  H, then there is no consistent sampling operator, that is, the consistency condition (4) cannot hold for any P ∈ A(W, V). In this case, we are interested in some relaxed properties such as partial consistency or quasi-consistency. Given a closed subspace M of H, P ∈ A(W, V) is said to be partially consistent on M if S ∗ (Pf ) = S ∗ (f )

for any f ∈ M.

We denote the set of all such operators as PC M . Theorem 4.1. Let M be a closed subspace of H. Then PC M = ∅ if and only if M ⊆ H0 . Moreover in this case, PC M = {T QS ∗ | Q ∈ L(l2 (J), l2 (K)), S ∗ T QS ∗ PM = S ∗ PM }. If W ⊆ M ⊆ H0 , then PC M = {T QS ∗ | Q ∈ (S ∗ T )− }. Proof. Assume that P = T QS ∗ ∈ PC M and let f ∈ M. Then S ∗ (f ) = S ∗ P(f ) = S ∗ T QS ∗ (f ) ∈ R(S ∗ T ) so that S ∗ (f ) = S ∗ (g) for some g ∈ W. Then h := f −g ∈ N (S ∗ ) = V ⊥ so f = g +h ∈ W +V ⊥ = H0 . Therefore M ⊆ H0 . Conversely, assume M ⊆ H0 . Then R(S ∗ PM ) = S ∗ (M) ⊆ S ∗ (H0 ) = R(S ∗ T ) so the equation S ∗ T QS ∗ PM = S ∗ PM is solvable for Q ∈ L(l2 (J), l2 (K)). Hence there exists P = T QS ∗ ∈ A(W, V) which is partially consistent on M. Moreover, {T QS ∗ | Q ∈ L(l2 (J), l2 (K)), S ∗ T QS ∗ PM = S ∗ PM } is precisely the set of all operators that are partially consistent on M. Finally, assume W ⊆ M ⊆ H0 .

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Then Q ∈ L(l2 (J), l2 (K)) satisfies S ∗ T QS ∗ PM = S ∗ PM if and only if S ∗ T QS ∗ T = S ∗ T . Hence PC M = {T QS ∗ | Q ∈ (S ∗ T )− }. 2 A more interesting relaxation of the consistency is the ‘quasi-consistency’ introduced by Arias and Conde [2]. We call P ∈ A(W, V) a quasi-consistent sampling operator if S ∗ P(f ) − S ∗ (f ) ≤ S ∗ P(f ) − S ∗ (f ) for any P ∈ A(W, V) and any f ∈ H. When P = T QS ∗ is a quasi-consistent sampling operator, we call Q a quasi-consistent filter. We denote the set of all quasi-consistent sampling operators (quasi-consistent filters, resp.) by QC(W, V) (QF(W, V), resp.) so that QC(W, V) = T QF(W, V) S ∗ . If H = H0 , then quasi-consistency coincides with consistency so that QC(W, V) = C(W, V) and QF(W, V) = F(W, V). In fact, the converse also holds, i.e., if QF(W, V) = F(W, V) (and thus, QC(W, V) = C(W, V)), then H = H0 . As in Section 3, we always assume that H0 := W + V ⊥ is closed (so that R(S ∗ T ) is closed and (S ∗ T )† exists). We further assume that S ∗ T ≡ 0 (otherwise, QC(W, V) = A(W, V) which is not interesting). Proposition 4.2. (See Theorems 5.1 and 5.2 in [2].) QF(W, V) = {Q ∈ L(l2 (J), l2 (K))| S ∗ T QS ∗ = PR(S ∗ T ) S ∗ } = {(S ∗ T )† + Y − PN (S ∗ T )⊥ Y PR(S ∗ ) | Y ∈ L(l2 (J), l2 (K))}

(10)

and QC(W, V) = {T (S ∗ T )† S ∗ + T PN (S ∗ T ) Y S ∗ | Y ∈ L(l2 (J), l2 (K))}.

(11)

Let QF p (W, V) := {Q ∈ L(l2 (J), l2 (K))| S ∗ T QS ∗ T = S ∗ T, (S ∗ T Q)∗ = S ∗ T Q}. Then QF p (W, V) = {Q ∈ L(l2 (J), l2 (K))| S ∗ T Q = PR(S ∗ T ) } ⊆ QF(W, V) ⊆ (S ∗ T )− ⊆ {Q ∈ L(l2 (J), l2 (K))| S ∗ T Q is a projection onto R(S ∗ T )}. Moreover, (a) QF(W, V) = (S ∗ T )− if and only if H = H0 . (b) QF(W, V) = QF p (W, V) if and only if S ∗ is surjective, i.e., {vj | j ∈ J} is a Riesz basis of V. We remark that the formula (10) follows from Lemma 2.1 and the observation that (S ∗ T )† PR(S ∗ ) = (S ∗ T )† PR(S ∗ T ) PR(S ∗ ) = (S ∗ T )† PR(S ∗ T ) = (S ∗ T )† since R(S ∗ T ) ⊆ R(S ∗ ). Note that the formulas (10) and (11) coincide exactly with (5) and (6) respectively; the only difference is that we require H = H0 for (5)  := T (S ∗ T )† S ∗ ∈ QC(W, V), whereas U  ∈ C(W, V) if and only if H = H0 . and (6). Here we always have U We point out that like C(W, V), QC(W, V) is also closed under composition.  is the unique quasi-consistent sampling operator, that is, Corollary 4.3. (See Theorem 5.1 in [2].) U ⊥  QC(W, V) = {U } if and only if W ∩ V = {0}.  is a projection (cf. Lemma 3.6), but in general, P ∈ QC(W, V) may not be a projection. Note that U However, we have:

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Proposition 4.4. Let P ∈ QC(W, V). Then P 3 = P2 and so P2 is a projection. Proof. Let P = T QS ∗ ∈ QC(W, V). Then S ∗ P = PR(S ∗ T ) S ∗ by Proposition 4.2 so P3 = (T QS ∗ )P2 = T QPR(S ∗ T ) S ∗ P = T QPR(S ∗ T ) S ∗ = T QS ∗ P = P2 . Hence (P2 )2 = P 2 , so P2 is a projection. 2 Theorem 4.5. The set of all projections in QC(W, V) is  + T PN (S ∗ T ) Y S ∗ | T PN (S ∗ T ) Y PR(S ∗ T )⊥ S ∗ = 0}. {P ∈ QC(W, V)| P2 = P} = {U +R  ∈ QC(W, V) where R  := T PN (S ∗ T ) Y S ∗ and Y ∈ L(l2 (J), l2 (K)). Then P2 = Proof. Let P = U 2 2        2 = 0. Therefore P 2 = P if and only if R U  =R  if and only   (U + R) = U + RU since U = U and U R = R ∗ ∗ ∗ −R U  = T PN (S ∗ T ) Y S − T PN (S ∗ T ) Y PR(S ∗ T ) S = T PN (S ∗ T ) Y PR(S ∗ T )⊥ S = 0. 2 if R Corollary 4.6. QC(W, V) consists of projections if and only if either W ∩ V ⊥ = {0} or H = H0 . Proof. Note that (i) W ∩ V ⊥ = {0} ⇔ N (S ∗ T ) = N (T ) ⇔ T PN (S ∗ T ) = 0 and (ii) H = H0 ⇔ R(S ∗ T ) = R(S ∗ ) ⇔ PR(S ∗ T )⊥ S ∗ = 0. Hence if either W ∩ V ⊥ = {0} or H = H0 , then T PN (S ∗ T ) Y PR(S ∗ T )⊥ S ∗ = 0 for any Y ∈ L(l2 (J), l2 (K)) so that any P in QC(W, V) is a projection. Conversely, assume that W ∩ V ⊥ = {0} and H = H0 . Then T PN (S ∗ T ) = 0 and PR(S ∗ T )⊥ S ∗ = 0 so there exists some Y ∈ L(l2 (J), l2 (K)) such that  + T PN (S ∗ T ) Y S ∗ ∈ QC(W, V) is not a projection. This completes the T PN (S ∗ T ) Y PR(S ∗ T )⊥ S ∗ = 0. Then U proof. 2 We now give a geometric interpretation of QC(W, V), which extends the characterization of C(W, V) in Theorem 3.2. Let L := {L| L is a closed complementary subspace of W ∩ V ⊥ in W} and QC r (W, V) :=  = PR(S ∗ T ) S ∗ |H⊥ }.  ∈ L(H0⊥ , W)| S ∗ R {R 0

Proposition 4.7. Let P ∈ QC(W, V) and P0 := P|H0 ∈ L(H0 , W). Then P0 = PR(P0 ),V ⊥ is a projection and R(P0 ) = N (I − P) ∈ L. Proof. Let P = T QS ∗ ∈ QC(W, V) and f = g+h ∈ H0 , where g = T (d) ∈ W, d ∈ l2 (K), and h ∈ V ⊥ . Then P2 (f ) = T QS ∗ P(f ) = T QPR(S ∗ T ) S ∗ (f ) = T QPR(S ∗ T ) S ∗ T (d) = T QS ∗ T (d) = P(g) = P(f ) so that P02 = P0 . Note that N (P0 ) = N (P) ∩ H0 ⊇ V ⊥ . Further, if f ∈ N (P0 ), then S ∗ (f ) = S ∗ (g) = PR(S ∗ T ) S ∗ T (d) = PR(S ∗ T ) S ∗ (f ) = S ∗ P(f ) = 0. So N (P0 ) = V ⊥ . Therefore P0 = PR(P0 ),V ⊥ where H0 = R(P0 )⊕V ⊥ . Moreover, since R(P0 ) ⊆ W ⊂ H0 , we have (I − P)P0 = (I − P0 )P0 = 0 so that R(P0 ) ⊆ N (I − P). Conversely if x ∈ N (I − P), then x = P(x) ∈ W ⊂ H0 so that x = P0 (x) ∈ R(P0 ). Therefore R(P0 ) = N (I − P) ∈ L. This completes the proof. 2 Proposition 4.8. P ∈ QC(W, V) if and only if P =



PL,V ⊥  R

on H0 , on H0⊥

 ∈ QC r (W, V). where L ∈ L and R ∈ Proof. Let P ∈ QC(W, V). Then P|H0 = PL,V ⊥ where L = R(P|H0 ) ∈ L by Proposition 4.7 and P|H⊥ 0  ∈ QC r (W, V) and QC r (W, V) since S ∗ P|H⊥ = PR(S ∗ T ) S ∗ |H⊥ by Proposition 4.2. Conversely, let L ∈ L, R 0

0

P :=



PL,V ⊥  R

on H0 , on H0⊥ .

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386

Then P ∈ L(H, W), N (P) ⊇ V ⊥ , and R(P) = P(H0 ⊕ H0⊥ ) = P(H0 ) + P(H0⊥ ) ⊆ L + W = W so that P ∈ A(W, V) by Lemma 2.1. Let f = g + h + e ∈ H where g = T (d) ∈ L, d ∈ l2 (K), h ∈ V ⊥ , and e ∈ H0⊥ .  Then PR(S ∗ T ) S ∗ (f ) = PR(S ∗ T ) S ∗ (T (d) + e) = PR(S ∗ T ) S ∗ T (d) + PR(S ∗ T ) S ∗ (e) = S ∗ T (d) + S ∗ R(e) = ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ S (g) + S R(e) = S (PL,V ⊥ (g + h)) + S R(e) = S P (g + h) + S P (e) = S P (f ), that is, S P = PR(S ∗ T ) S ∗ . Hence P ∈ QC(W, V). 2 p ∈ QC r (W, V), QC r (W, V) = R p + L(H⊥ , W ∩ V ⊥ ). Lemma 4.9. For any R 0 p + A) = p ∈ QC r (W, V). If A ∈ L(H⊥ , W ∩ V ⊥ ), then R p + A ∈ L(H⊥ , W) and S ∗ (R Proof. Fix any R 0 0 ∗ ∗   −R p . S Rp = PR(S ∗ T ) S |H⊥ so that Rp + A ∈ QC r (W, V). Conversely, let R ∈ QC r (W, V) and A := R 0 ⊥ ∗ ∗  ⊥ ⊥ p ) = 0 so that A ∈ L(H , W ∩ V ). Hence QC r (W, V) = Then A ∈ L(H0 , W) and S A = S (R − R 0 ⊥ ⊥  Rp + L(H , W ∩ V ). 2 0

 |H⊥ ∈ QC r (W, V), we have QC r (W, V) = U  |H⊥ + L(H⊥ , W ∩ V ⊥ ), which leads to the following: Since U 0 0 0 Theorem 4.10. There exists a one-to-one correspondence between QC(W, V) and L × L(H0⊥ , W ∩ V ⊥ ).  )|H⊥ ). Then F(P) ∈ L × L(H⊥ , W ∩ V ⊥ ) by Proof. For any P ∈ QC(W, V), let F(P) = (R(P|H0 ), (P − U 0 0 Proposition 4.7 and Lemma 4.9. Conversely, for any L ∈ L and A ∈ L(H0⊥ , W ∩ V ⊥ ), let  G((L, A)) =

PL,V ⊥  +A U

on H0 , on H0⊥

which belongs in QC(W, V) by Proposition 4.8 and Lemma 4.9. Then it is easy to see that (G ◦ F)(P) = P for any P ∈ QC(W, V) and (F ◦ G)((L, A)) = (L, A) for any (L, A) ∈ L × L(H0⊥ , W ∩ V ⊥ ). This completes the proof. 2 Theorem 4.10 implies that if W ∩ V ⊥ = {0}, then QC(W, V) contains exactly one element, in fact,  }. This matches perfectly with Corollary 4.3. QC(W, V) = {U As generalizations of Proposition 3.8, Lemma 3.11 and Theorem 3.12, we have the following.  =P  Lemma 4.11. PL0 U  ) ∈ QC(W, V) and PL0 U |H0 = PL0 ,V ⊥ . L0 ,N (U  = U  − T PN (S ∗ T ) [(T PN (S ∗ T ) )† T (S ∗ T )† ]S ∗ so that Proof. As in the proof of Proposition 3.8, PL0 U  ∈ QC(W, V) by (11). Moreover, PL U  is a projection by Theorem 4.5. Note that R(PL U  ) = L0 P L0 U 0 0    since R(PL0 U |H0 ) ⊆ R(PL0 U ) ⊆ L0 and R(PL0 U |H0 ) ∈ L. Since N (PL0 ) ∩ W = N (S ∗ ) ∩ W,  ) = {f ∈ H| U  f ∈ N (PL ) ∩ W} = {f ∈ H| U  f ∈ N (S ∗ ) ∩ W} = N (S ∗ U  ) so that N (PL0 U 0 ∗ ∗    N (PL0 U ) = N (S U ) = N (PR(S ∗ T ) S ) = N (U ) by Lemma 3.6(a). Hence PL0 U = PL0 ,N (U ) . Finally,  ). 2  |H = PL ,V ⊥ since V ⊥ ⊆ N (U PL U 0

0

0

 (f ) ≤ P(f ) for any P ∈ QC(W, V) and any f ∈ H. Hence PL U   ≤ P for any Lemma 4.12. PL0 U 0  P ∈ QC(W, V).  + T PN (S ∗ T ) [Y − (T PN (S ∗ T ) )† T (S ∗ T )† ]S ∗ = PL U  + T PN (S ∗ T ) Y S ∗ Proof. For any P ∈ QC(W, V), P = U 0 where Y ∈ L(l2 (J), l2 (K)) by (11). Note that R(T PN (S ∗ T ) ) = W ∩ V ⊥ (cf. (7)) and L0 ⊥ W ∩ V ⊥ . Then for any f ∈ H,  (f ) + T PN (S ∗ T ) Y S ∗ (f )2 P(f )2 = PL0 U

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 (f )2 + T PN (S ∗ T ) Y S ∗ (f )2 = PL0 U  (f )2 ≥ PL0 U   ≤ P. so that PL0 U

2

Theorem 4.13. (See Theorem 5.4 in [2].) For any P ∈ QC(W, V),   ≤ I − P I − PL0 U

(12)

  ≤ PW − P. PW − PL0 U

(13)

and

Remark 4.14. Theorem 4.13 was first given in (Theorem 5.4, [2]) but with some unnecessary zero terms. Precisely, using our notations, PR(T PN (S∗ T ) ) (S ∗ )† S ∗ = PR(T PN (S∗ T ) ) PW (S ∗ )† S ∗ = 0 by (7). Then the oper . Another proof of ator given in (Theorem 5.4, [2]) can be reduced into PR(T P ∗ )⊥ T (S ∗ T )† S ∗ = PL U 0

N (S T )

Theorem 4.13 can be deduced from Lemmas 2.3, 4.11, and 4.12. Finally we introduce an iterative algorithm for quasi-consistent approximations. We first need a simple generalization of Lemma 3.9. Lemma 4.15. Let L ∈ L and V0 := V ∩ H0 . Then PV |L : L → V0 and PL |V0 : V0 → L are isomorphisms. In particular, PL0 |V0 = PW |V0 . Proof. Let L ∈ L. Since V0⊥ = (V ∩ H0 )⊥ = V ⊥ ⊕ H0⊥ , H = (L ⊕ V ⊥ ) ⊕ H0⊥ = L ⊕ V0⊥ so that H = L⊥ ⊕ V0 by Lemma 2.4(e). Then PV (L) = V ∩ (L ⊕ V ⊥ ) = V0 and PL (V0 ) = L ∩ (L⊥ ⊕ V0 ) = L by Lemma 2.2(b). Since L ∩ V ⊥ = L⊥ ∩ V0 = {0}, PV |L : L → V0 and PL |V0 : V0 → L are isomorphisms. Finally, note that V0 = V ∩ H0 = (V + H0⊥ ) ∩ H0 = PH0 (V) since H0⊥ = W ⊥ ∩ V ⊂ V. Then L0 = W ∩ (W ⊥ + V) = PW (V) = PW PH0 (V) = PW (V0 ) so that PL0 |V0 = PL0 PW |V0 = PW |V0 . 2 Theorem 4.16. Let P ∈ QC(W, V) and L := R(P|H0 ). Then for any f ∈ H0 , P(f ) = lim fn

(14)

n→∞

and PV (f − fn ) ≤ PL,V ⊥ (f ) − fn  ≤

α2n−1 PV (f ) 1−α

(15)

where α = PV ⊥ PL  and f1 := PL PV (f )

and

fn := f1 + PL PV ⊥ (fn−1 ),

n ≥ 2.

(16)

Proof. First note that since H0 = L ⊕ V ⊥ is closed, α = PV ⊥ PL  = S(L, V ⊥ ) < 1 by Lemma 2.3. Hence (cf. Theorem 3.10)  FL := I − PV ⊥ PL = is invertible with its inverse (cf. Lemma 4.15)

PV

on L,

I

on L⊥

388

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FL−1

 =

(PV |L )−1

on V0 ,

I

on L⊥ .

Moreover, PL FL−1 PV = FL−1 PV

on H0

(17)

since FL−1 PV (H0 ) = FL−1 PV (L) = FL−1 (V0 ) = L by Lemma 4.15. Now the iterations of (16) leads to fn = [I + PL PV ⊥ + · · · + (PL PV ⊥ )n−1 ]PL PV f = PL [I + PV ⊥ PL + · · · + (PV ⊥ PL )n−1 ]PV f,

n ≥ 1.

(18)

−1 −1 −1 ˜ Then {fn }∞ n=1 converges in L to f := limn→∞ fn = PL FL PV (f ) = FL PV (f ) by (17). Since FL PV |L = −1 −1 −1 ⊥ (PV |L ) PV |L = I on L and N (FL PV |H0 ) = N (PV |H0 ) = V , FL PV = PL,V ⊥ on H0 . Therefore

f˜ = FL−1 PV (f ) = PL,V ⊥ (f ) = P(f ) by Proposition 4.7 so (14) holds. Finally, for any integers m > n ≥ 1, fm − fn = PL (PV ⊥ PL )n [I + · · · + (PV ⊥ PL )m−n−1 ]PV f by (18). Then by letting m → ∞, we have PL,V ⊥ (f ) − fn = PL (PV ⊥ PL )n FL−1 PV (f ) = (PL PV ⊥ )n FL−1 PV (f ) by (17) so that PL,V ⊥ (f ) − fn  ≤ (PL PV ⊥ )n  · FL−1  · PV (f ). Since (PL PV ⊥ )n  = α2n−1 by Lemma 2.2(a) 1 and FL−1  ≤ 1−α , the last inequality of (15) holds. The first inequality of (15) follows immediately from observing that PV (f − fn ) = PV (PL,V ⊥ (f ) − fn ). This completes the proof. 2 Corollary 4.17. Let P ∈ QC(W, V) be such that R(P|H0 ) = L0 . Then for any f ∈ H0 , P(f ) = lim fn n→∞

where f1 := PW PV (f )

and

fn := f1 + PW PV ⊥ (fn−1 ),

n ≥ 2.

(19)

Proof. Note that f1 ∈ PW PV (H0 ) = PW PV (L0 ) = PW (V0 ) = L0 by Lemma 4.15. Assume that fn ∈ L0 for some n ≥ 1. Then fn+1 = f1 + PW (fn ) − PW PV (fn ) = f1 + fn − PW PV (fn ) ∈ L0 since PW PV (L0 ) = L0 . Therefore fn ∈ L0 for any n ≥ 1 by induction. Since PL0 PW = PL0 , f1 = PL0 (f1 ) = PL0 PV (f ) and fn = PL0 (fn ) = f1 + PL0 PV ⊥ (fn−1 ), n ≥ 2. So, fn ’s defined by (19) are the same as (16) with L = L0 . Hence the claim follows from Theorem 4.16. 2 Remark 4.18. The above algorithms can be interpreted using MAP (method of alternating projections), which was first proposed by von Neumann [17]. In its simplest form, the MAP is described as follows: for any closed subspaces (or affine subspaces) M1 and M2 of a Hilbert space H, lim (PM2 PM1 )n (f ) − PM1 ∩M2 (f ) = 0,

n→∞

f ∈ H.

See [3,6,16] for more details on MAP and its generalizations such as POCS (projection onto convex sets method). We now show the convergence of iteration (16) using MAP. Let L ∈ L, f ∈ H0 , and M := {h ∈ H| S ∗ (h) = S ∗ (f )} = {h ∈ H| PV (h) = PV (f )}. Note that M = PV (f ) + V ⊥ is a closed affine subspace of H so that PM (h) = PV (f ) + PV ⊥ (h), h ∈ H. Then (16) becomes f1 = PL PV (f ) and fn = PL PV (f ) +

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389

PL PV ⊥ (fn−1 ) = PL [PV (f ) + PV ⊥ (fn−1 )] = PL PM (fn−1 ), n ≥ 2. Thus, MAP guarantees that limn→∞ fn = PL∩M (f1 ). Since H0 = L ⊕ V ⊥ , we have L ∩ M = L ∩ (PV (f ) + V ⊥ ) = PL,V ⊥ (PV (f )) = PL,V ⊥ (f ) by Lemma 2.2. Hence limn→∞ fn = PL,V ⊥ (f ). As an illustration of Theorem 4.16 and Corollary 4.17, we revisit the example introduced in [8] with some minor modifications. Example 4.19 (Band-limited sampling of time-limited sequences). Let J0 , J, K, and N be positive integers such that J = 2J0 + 1 < N and K < N , and let H be the space of all complex-valued sequences x = {x(n)}n∈Z such that x(n) = 0 for n < 0 and n ≥ N . Let VJ = {x ∈ H| x ˆ(n) = 0 for J0 < n < N − J0 } and WK = {x ∈ H| x(n) = 0 for n ≥ K}, where x ˆ(·) denotes the N point DFT (discrete Fourier transform) of x. Then the sequences in VJ and K−1 WK are band-limited and time-limited respectively. Let {vj }J−1 j=0 and {wk }k=0 be the bases for VJ and i2π(j−J0 )n/N WK respectively, given by vj (n) = e and wk (n) = δk,n for 0 ≤ n < N . Then the pre-frame J−1 K operators S : CJ → VJ of {vj }J−1 and T : C → WK of {wk }K−1 j=0 j=1 c(j) vj and k=0 are given by S(c) = K−1 ∗ T (d) = k=1 d(k) wk respectively. For any x ∈ H, the measurements c = S (x) of x are c(j) = x, vj  =

N −1 

x(n) vj (n) =

n=0

N −1 

x(n) e−i2π(j−J0 )n/N = x ˆ[((j − J0 ))N ],

0 ≤ j < J,

n=0

where ((·))N means modulo N . That is, the measurements c(j)’s are precisely the J lowpass DFT coefficients ˜ of x in WK has the same lowpass DFT coefficients of the N point DFT of x. So, a consistent approximation x N −1 as x. On the other hand, wk , vj  = n=0 wk (n) vj (n) = vj (k) = (e−i2π/N )jk (ei2πJ0 /N )k for 0 ≤ j < J and 0 ≤ k < K so that the Gram matrix B (the matrix representation of S ∗ T ) is ⎛

1

⎜ ⎜1 ⎜ ⎜ B = ⎜1 ⎜. ⎜. ⎝. 1

1

1

···

1

z

z2

···

z K−1

z2 .. .

z4 .. .

··· .. .

z 2(K−1) .. .

z J−1

z 2(J−1)

⎞ ⎟ ⎟ ⎟ ⎟ ⎟D ⎟ ⎟ ⎠

· · · z (J−1)(K−1)

where z := e−i2π/N , d := ei2πJ0 /N and D = diag(1, d1 , . . . , dK−1 ). Note that B always has full rank. By Lemma 2.4, B is injective (surjective, resp.) if and only if WK ∩ VJ⊥ = {0} (H = WK + VJ⊥ , resp.). ˜= If J = K (perfectly determined), then B is an invertible square matrix so that H = WK ⊕ VJ⊥ and x PWK ,VJ⊥ (x) = T (S ∗ T )−1 S ∗ (x) = T B −1 (c) is the unique consistent approximation of x (see Corollaries 3.4 and 3.5). In this case, any sequence x in WK can be perfectly reconstructed from its lowpass DFT coefficients c = S ∗ (x). As noted in [8], this corresponds to a discrete time version of Papoulis’ theorem [18] which says that a time-limited signal can be recovered from a lowpass segment of its Fourier transform. If J > K (over-determined), then B is injective but not surjective. Then H  WK ⊕VJ⊥ so that consistent sampling operators do not exist, i.e., C(WK , VJ ) = ∅, and there is no consistent approximation for general  := T (S ∗ T )† S ∗ ∈ QC(WK , VJ ) (see Proposition 4.2), x  (x) = T B † (c) is ˜ = U x ∈ H. However, since U a quasi-consistent approximation of x; in fact, it is the unique quasi-consistent approximation of x by

390

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Fig. 1. Absolute value of DFT (x − x5 ).

Corollary 4.3. Note that, as in the perfectly determined case, any x ∈ WK can be perfectly reconstructed  |W ⊕V ⊥ (x) = PW ,V ⊥ (x). This can be also understood from the ˜ = T B † (c) = U from c = S ∗ (x) since x K K J J following perspective. Since we have more than enough number of measurements, some measurement values may be neglected. For example, when K = 2K0 + 1 for some integer K0 , we neglect the first J0 − K0 and the last J0 − K0 entries of c = S ∗ (x) so that only the K lowpass entries of c are left. Then as in the perfectly determined case, any x ∈ WK can be perfectly recovered from the remaining K entries. In fact, neglecting the measurement values in such a way corresponds to reducing the subspace VJ (thus enlarging VJ⊥ ) so as to meet the condition H = WK ⊕ VJ⊥ . If J < K (under-determined), then B is surjective but not injective. Then H = WK + VJ⊥ with WK ∩ VJ⊥ = {0} so that there are infinitely many consistent sampling operators {PL,VJ⊥ | L ∈ L} where L := {L| L is a closed complementary subspace of WK ∩ VJ⊥ in WK }. For a fixed L ∈ L, any sequence x in L ( WK ) can be perfectly reconstructed from c = S ∗ (x). For L0 := WK ∩ (WK ∩ VJ⊥ )⊥ ∈ L, the approximation PL0 ,VJ⊥ (x) of x ∈ H can be easily computed by the iteration (19). As an illustration, let J = 31 (J0 = 15), K = 99, N = 100, and x ∈ H an arbitrary sequence with x = 1. Then PL0 ,VJ⊥ (x) can be obtained from the following iteration:

x1 = PWK PVJ (x), xn+1 = x1 + PWK PVJ⊥ (xn ),

n ≥ 1.

(20)

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