# An Extension Theorem for a Sequence of Krein Space Contractions

Authors: Gerald Wanjala
Journal: International Journal of Mathematics and Mathematical Sciences
 (2018)

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## Abstract

Consider Krein spacesU and Y and let Hk and Kk be regular subspaces of U and Y, respectively, such that Hk⊂Hk+1 and Kk⊂Kk+1(k∈N). For each k∈N, let Ak:Hk→Kk be a contraction. We derive necessary and sufficient conditions for the existence of a contraction B:U→Y such that BHk=Ak. Some interesting results are proved along the way.


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## Body

## 1. Introduction

A number of extensions and generalizations of classical function theoretic interpolation problems are driven by factors such as applications in systems and control theory. Many such extensions and generalizations make use of the commutant lifting theorem in one way or another. In particular, the commutant lifting theorem in the Hilbert space case which was obtained by Sz.-Nagy and Foias has been used to solve extension problems like the ones of Nevanlinna-Pick, Nudelman, Nehari, and many others. Extensions of this theorem to an indefinite setting are given in [1–5]. In [6] (see also [7, 8]), a time-variant version of the commutant lifting theorem is developed. This time-variant version is called thethree-chain completion theorem and is used to solve a number of nonstationary norm constrained interpolation problems on Hilbert spaces. Recall (see [6]) that the given data for the three-chain completion problem are bounded linear operators(1)Ak:Hk⟶Kk⊖Mkk∈Z,where Hk⊂U and Mk⊂Kk⊂Y for k∈Z are Hilbert spaces satisfying the inclusion relations(2)Hk-1⊂Hk,Mk-1⊂Mk,Kk-1⊂Kkk∈Z.Given operators (1) and tolerance γ the problem is to find an operator B:U→Y such that B≤γ and(3)BHk⊂Kk,I-PMkBHk=Akk∈Z.As far as we know, no extension of this theorem to an indefinite setting has been developed so far.The three-chain theorem mentioned above is the motivation for the extension problem considered in this paper. Fork∈N, consider a sequence of Krein space contractions Ak:Hk→Kk, where Hk and Kk are nested regular subspaces of some Krein spaces U and Y, respectively. The problem is to find a contraction B:U→Y such that (4)Bh=Akhfor all h∈Hk.In order to keep this paper as self-contained as possible, we briefly outline some definitions and some elementary facts about Krein spaces and bounded linear operators defined on them. This is done in Section2. Some important results are stated and proved in Section 3. An extension theorem is considered in Section 4 while Section 5 contains some simple application of the extension theorem discussed in Section 4.


## 2. Preliminaries

In this section we recall some definitions and basic notions of indefinite metric space theory. More details can be obtained from [2, 9–12].LetK be a linear space and let ·,· be a sesquilinear form on K. This sesquilinear form is called an indefinite metric on K. If K admits a direct orthogonal sum decomposition(5)K=K+⊕K-in which K±,±·,· are Hilbert spaces, then K (or K,·,·) is called aKrein space. Decomposition (5), which is not unique in general, is called afundamental decomposition and gives rise to orthogonal projections from K onto K±, which we denote by P±, respectively. The self-adjoint and unitary operator J on K defined by J=P+-P-, is called afundamental symmetry on K.The spaceK with the inner product(6)x,y=Jx,y=x+,y+-x-,y-,x±,y±∈K±is a Hilbert space. This inner product is used to define the norm of an element x of the Krein space K by(7)x2=x,x.Topological notions such as convergence and continuity are understood to be with respect to this norm topology. The inner product ·,· in (6) depends on decomposition (5), as does the norm · in (7), but the norms generated by different decompositions of K are equivalent. We will denote the norm · in (7) by ·J where clarity is needed.Anorthogonal projection in a Krein space is a bounded self-adjoint operator R in K such that R2=R. Note that the norm of an orthogonal projection in a Krein space need not be less than 1. The range M of an orthogonal projection R in a Krein space K is a closed subspace of K and the space K can be decomposed as(8)K=M⊕M⊥,where M⊥=l∈K:l,m=0∀m∈M. On the other hand, given a closed subspace M of K such that decomposition (8) holds, then M is the range of an orthogonal projection in K. In this case, M,·,· is again a Krein space. Unlike in the Hilbert space case, the decomposition (8) need not hold for a given closed subspace. Closed subspaces for which this decomposition holds are referred to asregular subspaces.LetH and K be Krein spaces and let T:H→K be a bounded linear operator. We say that T is acontraction if for all x∈H, Tx,TxK≤x,xH. If both T and its Krein space adjoint T∗ are contractions, then T is called abicontraction.LetH and K be Krein spaces and let T∈B(H,K), the space of bounded linear operators from H into K. A column extension of T is an operator of the form (9)C=TE:H⟶KE,where E is a Krein space and E∈B(H,E). By a row extension of T we shall mean an operator of the form (10)L=TE:HE⟶K,where E is a Krein space and E∈B(E,K). It is shown in [2] (see also [3]) that if T is a contraction, then there exist contractive row and column extensions of T where the extension space is a Hilbert space. Contractive 2×2 matrix extensions of a contractive operator T:H→K where the extending space E is a Hilbert space are thoroughly discussed in [2] where Lemma 1, Theorem 2, and Lemma 3 can be found. Results more general than those provided by Lemma 1 and Theorem 2 can be found in [1] while Lemma 3 can also be found in [4].Lemma 1.
LetH and K be Krein spaces and let T∈BH,K be a contraction. Let(11)C=TE~∗∈BH,K⊕E~ be a contractive column extension of T with E~ a Hilbert space. If norms are computed relative to some fixed fundamental decompositions of H and K and the induced fundamental decomposition of K⊕E~, then(12)C2≤1+2T2.The above norm estimate enables one to fix a bound for the norm of the operatorC. The following lemma is helpful in finding 2×2 matrix extensions of a contractive operator T. See [13] for a similar result in a Hilbert space setting.Theorem 2.
LetH, K, and E be Krein spaces and let E~ be a Hilbert space. Assume that (13)C11C12:HE⟶K,C11C21:H⟶KE~ are contractions. Then there exists an operator C22:E→E~ such that (14)C=C11C12C21C22:HE⟶KE~ is a contraction. If (15)C11C12 is a bicontraction, C22 may be chosen such that C is a bicontraction.We conclude this section by stating the following lemma.Lemma 3.
LetN1,N2,… be regular subspaces of a Krein space H such that N1⊃N2⊃⋯. Suppose that the projections P1,P2,… of H onto the subspaces N1,N2,… are uniformly bounded. Then(16)N=⋂n=1∞Nn,M=span¯⁡Nn⊥,n=1,2,… are regular subspaces of H with N = M⊥. If P is the projection of H onto N, then P=limn→∞⁡Pn with convergence in the strong operator topology.


## 3. Some Preliminary Results

In this section we prove some useful results regarding sequences of regular subspaces. See [4, Corollary 3.3 and Lemma 3.4] for closely related results.Theorem 4.
Forn≥1, let Ln be a sequence of regular subspaces of a Krein space H such that Ln⊂Ln+1, n∈N. For each n∈N, let Pn be the projection of H onto Ln. If the projections Pn are uniformly bounded then L=span¯⁡Ln is a regular subspace of H.Proof.
SetN1=L1⊥, N2=L2⊥,…. Then N1,N2,… is a sequence of regular subspaces of H such that N1⊃N2⊃⋯. Let Q1,Q2,… be the projections of H onto the subspaces N1,N2,…. Then Qn=1-Pn and so Qn≤1+Pn. Since the projections Pn are uniformly bounded we see that the projections Qn are also uniformly bounded. Hence L is a regular subspace of H by Lemma 3.Theorem 5.
Forn≥1, let Ln be regular subspaces of a Krein space K such that Ln⊂Ln+1. Let Pn be the projection of K onto Ln. If Ln=Ln+⊕Ln- is a fundamental decomposition of Ln, we denote by Qn± the projection of Ln onto Ln±. Let (17)L=span¯⁡Ln. Then the following are equivalent: (1)
L is regular and there exists a fundamental symmetry J on L such that J∣Ln is a fundamental symmetry on Ln.(2)
There exist fundamental decompositionsLn=Ln+⊕Ln- such that
(a)
Ln±⊂Ln+1±,(b)
supQn±PnJ~<∞,whereJ~ is any fundamental symmetry on K.Proof.
Suppose that (1) holds and letL=L+⊕L- be the fundamental decomposition of L which gives rise to the fundamental symmetry J=Q+-Q- with the stated property, where Q±L=L±. Then J∣Ln=Qn+-Qn- where Qn±Ln=Ln±. To show that (a) holds, we let x∈Ln+. Then (18)x=Qn+x-Qn-x=Jx=Qn+1+x-Qn+1-x. Hence x=Qn+1+x-Qn+1-x and so I-Qn+1+x=-Qn+1-x. This means that Qn+1-x=-Qn+1-x and that Qn+1-x=0. Therefore x=Qn+1+x∈Ln+1+. Similarly, we have that Ln-⊂Ln+1-. Hence (a) holds. To show that (b) also holds, we let (19)y∈K=Ln⊕L⊖Ln⊕L⊥. Let Jn=J∣Ln, J^n=J∣(L⊖Ln) and let J⊥ be a fundamental symmetry on L⊥. Then J~≔J+J⊥ is a fundamental symmetry on K. Now, (20)yK,J~2=J~y,y=J~xn+yn+zn,xn+yn+zn=Jnxn,xn+J^nyn,yn+J⊥zn,zn, where y=xn+yn+zn with xn∈Ln, yn∈L⊖Ln and zn∈L⊥. Therefore (21)Qn+PnyLn,Jn2=Qn+xnLn,Jn2=JnQn+xn,Qn+xn=Qn+xn,Qn+xn=Qn+xn,xn=Qn+-Qn-xn,xn+Qn-xn,xn≤Qn+-Qn-xn,xn=Jnxn,xn≤Jnxn,xn+J^nyn,yn+J⊥zn,zn=yK2. Hence Qn+Pn≤1. Similarly Qn-Pn≤1.
Conversely, assume that (2) holds. We start by showing thatL is regular. To do this we note that Pn=Qn++Qn-Pn. So Pn≤Qn+Pn+Qn-Pn≤M1 for some constant M1>0. Hence L=span¯⁡Ln is regular by Theorem 4.
Next setNn=Ln+⊥ and let Rn be the projection of K onto Nn=Ln+⊥. Then Rn=I-Qn+Pn. Hence Rn≤1+Qn+Pn≤M2 for some constant M2>0 since (2) holds. Hence N+≔⋂Nn and M+≔span¯⁡Ln+ are regular subspaces by Lemma 3. Let x∈M+. Then there exists a sequence xn∈span⁡Ln+ such that xn→x. Hence xn,xn→x,x. This means that x,x≥0 since xn,xn≥0. Hence M+ is regular and nonnegative and so it is a uniformly positive subspace of L. Similarly M-≔span¯⁡Ln- is a uniformly negative subspace of L. We now show that (22)L=M+⊕M-. Let x∈M+. Then there exists a sequence xn∈span⁡Ln+ such that xn→x. If y∈M- then there exists a sequence yn∈span⁡Ln- such that yn→y. Hence 〈x,y〉 = lim⁡〈xn,yn〉 = 0. This implies that M+⊥M-. Next, assume that u∈L⊖M+⊕M-. Then u⊥M+ and u⊥M-. This in turn means that u⊥Ln+ and u⊥Ln- and so u⊥Ln. Let v∈L. Then there exists a sequence vn∈span⁡Ln such that vn→v. Hence u,v=lim⁡u,vn⁡=0 for all v∈L and so v=0. Hence L=M+⊕M-. This implies that M+ is maximal positive and M- is maximal negative. Hence M+ is a maximal uniformly positive space and M- is a maximal uniformly negative space. Since M+⊥M-, M+ is uniformly positive and M- is uniformly negative, there exists a fundamental decomposition L=L+⊕L- such that M±⊂L±. But M± maximal implies M±=L±. Hence L=M+⊕M- is a fundamental decomposition of L. This decomposition gives rise to a fundamental symmetry J=Q+-Q- where Q±L=M±.
We show thatJ∣Ln=Qn+-Qn-. Let x∈Ln⊂L. Then x=x++x-, x±∈Ln±⊂span¯⁡Ln±=M±. Hence Q+x=Q+x++x-=x+=Qn+x. Similarly, Q-x=Qn-x and so J∣Ln=Qn+-Qn-.Corollary 6.
Forn≥1, let K, Ln, L, and Pn be as in Theorem 5. Then there exists a fundamental symmetry J on K which commutes with the projections Pn onto Ln if and only if L is regular and there exists a fundamental symmetry J′ on L such that J′∣Ln is a fundamental symmetry on Ln.Proof.
Suppose thatL is regular and J′ exists which has the stated property. Let L=L+⊕L- be the fundamental decomposition which gives rise to J′. Since L is regular there exists a fundamental decomposition K=K+⊕K- such that L±⊂K±. This gives rise to J on K such that J∣Ln is a fundamental symmetry on Ln. Let f∈K. Then f=fn+fn⊥, where fn∈Ln and fn⊥∈Ln⊥. Now, JPnf=Jfn=fn+-fn- (where fn±∈Ln±,Ln=Ln+⊕Ln-). On the other hand,(23)PnJf=PnJfn+fn⊥=PnJfn++fn-+fn⊥++fn⊥-=PnJfn++fn-+Jfn⊥++fn⊥-=Pnfn+-fn-+Pnfn⊥+-fn⊥-=fn+-fn-. Hence JPn=PnJ.
Conversely, assume that there exists a fundamental symmetryJ on K that commutes with the projections Pn. To show that Pn’s are uniformly bounded we consider the fundamental decomposition K=K+⊕K- which gives rise to the fundamental symmetry J. Let(24)Pn=A11nA12nA21nA22n:K+K-⟶Ln+Ln- be the matrix representation of Pn with respect to these decompositions. Then the commutativity condition(25)100-1A11nA12nA21nA22n=A11nA12nA21nA22n100-1 implies that A21n=A12n=0 and so the matrix representation of Pn is diagonal; that is to say,(26)Pn=A11n00A22n:K+K-⟶Ln+Ln-. Since A11n≤1 and A22n≤1 for all n∈N, we see that Pn’s are uniformly bounded. The uniform boundedness of Pn’s implies that L is regular. From the matrix representation of Pn above we see that A11n=Qn+Pn and A22n=Qn-Pn and so supQn±Pn<∞. Hence condition (b) in Theorem 5 holds. Since, for each n∈N, Ln is a regular subspace and Ln⊂Ln+1, we have that, for any fundamental decomposition Ln=Ln+⊕Ln-, there exist a fundamental decomposition (27)Ln+1=Ln+1+⊕Ln+1- such that Ln±⊂Ln+1±. Hence condition (a) in Theorem 5 also holds. This shows that part (1) of Theorem 5 holds and this completes the proof.


## 4. An Extension Theorem for a Sequence of Contractions

In this section, we formulate and give a proof of an extension theorem for a sequence of contractions defined on a nested sequence of regular subspaces. Please refer to [14, Lemma 3.1] and [5] for closely related results.Theorem 7.
LetU and Y be Krein spaces and let Hk⊂U and Kk⊂Y,k∈N, be sequences of regular subspaces satisfying(28)Hk⊂Hk+1,Kk⊂Kk+1.Let PHk and PKk be the orthogonal projections of U onto Hk and Y onto Kk, respectively, and let H~=span¯⁡Hk and K~=span¯⁡Kk. Assume that there exist fundamental symmetries J on U and J′ on Y such that J commutes with PHk and J′ commutes with PKk and that U⊖H~, Y⊖K~, Hk+1⊖H1, and Kk+1⊖K1 are all Hilbert space.
For eachk∈N, let(29)Ak:Hk⟶Kkbe a contraction. Then there exists a contraction B:U→Y such that(30)BHk=Ak,k∈N,if and only if(31)Ak+1Hk=Ak,supk∈N⁡Ak<∞.Proof.
First, let us assume that such an extensionB∈BU,Y exists and take hk∈Hk. Since Hk⊂Hk+1 and (30) holds, we have that(32)Ak+1hk=Bhk=Akhk and therefore Ak+1Hk=Ak. Hence the first condition in (31) holds. To show that the second condition also holds, we first note that since J and J′ commute with PHk and PKk, respectively, Corollary 6 ensures that the subspaces H~ and K~ are regular and that there exist fundamental symmetries JH~ on H~ and JK~′ on K~ such that JH~∣Hk is a fundamental symmetry on Hk and JK~′∣Kk is a fundamental symmetry on Kk. Let h∈Hk. Then we have that AkhBh. Hence, (33)AkhK~=AkhKk=JK~′Akh,Akh=JK~′Bh,Bh=BhK~=BhH~. Hence, Ak≤B for all k∈N and so the norms of Ak’s are uniformly bounded.
Conversely, letAk:Hk→Kk, k∈N, be contractions satisfying both conditions in (31). Since Ak+1Hk=Ak, we can decompose the operator (34)Ak+1Hk:Hk⟶KkKk+1⊖Kk as Ak+1Hk=Ak0. The fact that Ak is a contraction and Hk+1⊖Hk is a Hilbert space implies that the operator (35)Ak0:HkHk+1⊖Hk⟶Kk is a contraction. Since Ak+1Hk=Ak0 is a contraction, Theorem 2 implies that there exists a bounded operator c22:Hk+1⊖Hk→Kk+1⊖Kk such that(36)Ck+1=Ak00c22:HkHk+1⊖Hk⟶KkKk+1⊖Kkis a contraction. For each k=0,1,2,…, the operator Ck+1 is clearly a contractive extension of Ak.
We now show that, fork=0,1,2,…, the contractive liftings Ck+1’s of Ak’s are uniformly bounded. Let M be the bound for the norm of Ak’s. Then Ak0=Ak≤M and so Lemma 1 implies that Ck+12≤1+2Ak2≤1+2M2. Hence Ck+1’s are uniformly bounded.
Define an operatorC∞′:span⁡Hk→span⁡Kk by C∞′x=Ck+1x for x∈Hk. The operator C∞′ is well defined. To see this, assume that x∈Hk and x∈Hl. Then C∞′x=Ck+1x and C∞′x=Cl+1x. For k<l, Cl+1Hk=Al|Hk=Ak=Ck+1Hk. The operator C∞′ is bounded since the contractions Ck+1 are uniformly bounded. To see that this is the case, consider x∈Hk. Then (37)x,xH~=JH~x,xH~=JH~x,xHk=JHkx,xHk=x,xHk. Hence for x∈span⁡Hk, we see that(38)C∞′xK~=Ck+1xK~=Ck+1xKk≤Ck+1xHk=Ck+1xH~ and so C∞′ is a bounded operator. Since it is defined on a dense set we can extend it by continuity to a bounded operator C∞:H~→K~. The operator C∞ is clearly an extension of Ak for each k≥0 since Ck+1 is an extension of Ak for each k≥0. To show that C∞ is a contraction, let x∈H~. Then there exists a sequence xn∈span⁡Hk such that xn→x. Since the inequality (39)C∞′xn,C∞′xn≤xn,xn holds for each n we have (40)C∞x,C∞x=limn→∞⁡C∞′xn,C∞′xn≤limn→∞⁡xn,xn=x,x. Hence C∞:H~→K~ is a contraction.
Define a new operatorB:U→Y by the matrix(41)B≔C∞E0X:H~U⊖H~⟶K~Y⊖K~, where(42)C∞E is any contractive row extension of C∞ and X∈BU⊖H~,Y⊖K~. Since Y⊖K~ is a Hilbert space, Theorem 2 guarantees the existence of X such that B is a contraction. Note that since U⊖H~ is also a Hilbert space we may set E=0, the zero operator, in the matrix representation of B and the above result still holds.
Clearly,B∣Hk=Ak and so B is the required extension.


## 5. Applications

In this section we use Theorem7 to solve a nonstationary extension problem in a Krein space setting. Let Uk and Yjj,k∈N be Krein spaces with fixed fundamental decompositions(43)Uk=Uk+⊕Uk-,Yj=Yj+⊕Yj-and let(44)fj,k:Uk⟶Yjbe bounded operators with matrix representations(45)fj,k=fj,k11fj,k21fj,k21fj,k22:Uk+Uk-⟶Yj+Yj-with fj,k11=0 for j>k and fj,k21=0 for j>0. Define the Krein spaces U→ and Y→ by (46)U→≔U0⊕U1⊕U2⊕⋯,Y→≔Y0⊕Y1⊕Y2⊕⋯and let G:U→ to Y→ be a bounded operator such that(47)PYjGUk=fj,k.We use Theorem 7 to establish conditions under which such an operator G is a contraction. Equality (47) implies that the operator G is of the form(48)G=⋯f4,0f3,0f2,0f1,0f0,0⋯f4,1f3,1f2,1f1,1f0,1⋯f4,2f3,2f2,2f1,2f0,2⋯f4,3f3,3f2,3f1,3f0,3⋰⋮⋮⋮⋮⋮:⋮U4U3U2U1U0⟶Y0Y1Y2Y3Y4⋮.Consider the decompositions (49)U→=⋯⊕U1-⊕U0-⊕U0+⊕U1+⊕⋯,Y→=⋯⊕Y1-⊕Y0-⊕Y0+⊕Y1+⊕⋯.With the above decompositions, the operator G takes the form(50)
G
=
⋱
⋮
⋮
⋮
⋮
⋮
⋮
⋮
⋮
⋰
⋯
f
2,3
11
f
2,2
11
0
0
0
0
0
0
⋯
⋯
f
1,3
11
f
1,2
11
f
1,1
11
0
0
0
0
0
⋯
⋯
f
0,3
11
f
0,2
11
f
0,1
11
f
0,0
11
f
0,0
12
f
0,1
12
f
0,2
12
f
0,3
12
v
k
⋯
⋯
f
0,3
21
f
0,2
21
f
0,1
21
f
0,0
21
f
0,0
22
f
0,1
22
f
0,2
22
f
0,3
22
v
k
⋯
⋯
f
1,3
11
f
1,2
11
f
1,1
11
f
1,0
11
f
1,0
22
f
1,1
22
f
1,2
22
f
1,3
22
⋯
⋰
⋮
⋮
⋮
⋮
⋮
⋮
⋮
⋮
⋮
:
⋮
U
2
+
U
1
+
U
0
+
U
0
-
U
1
-
U
2
-
⋮
⟶
⋮
Y
2
+
Y
1
+
Y
0
+
Y
0
-
Y
1
-
Y
2
-
⋮
.For eachk∈N, let Ak denote the operator(51)
A
k
=
f
k
,
k
11
0
⋯
0
0
0
0
0
0
⋯
⋮
⋱
⋮
⋮
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⋮
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f
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,
k
11
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11
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k
11
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11
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k
11
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f
0,2
11
f
0,1
11
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11
f
0,0
12
f
0,1
12
f
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12
f
0,3
12
v
k
⋯
f
0
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k
21
⋯
f
0,2
21
f
0,1
21
f
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21
f
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22
f
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k
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1
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22
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:
U
k
⋮
U
2
+
U
1
+
U
0
+
U
0
-
U
1
-
U
2
-
⋮
⟶
Y
k
⋮
Y
2
+
Y
1
+
Y
0
+
Y
0
-
Y
1
-
Y
2
-
⋮
.With these notations, we are in a position to state the following theorem.Theorem 8.
LetUk and Yj be Krein spaces with decompositions (43) with operators fj,k:Uk→Yj in (44). Then there exists a contraction G:U→→Y→ satisfying (47) if and only if, for each k∈N, the operator Ak in (51) defines a contraction whose norm has an upper bound independent of k.Proof.
Assume that a contractionG:U→→Y→ satisfying (47) exists and consider the operator matrices (50) and (51). We see that, for k∈N,(52)Ak=PKkGHk,where(53)
H
k
=
U
k
⋮
U
1
+
U
0
+
U
0
-
U
1
-
⋮
,
K
k
=
Y
k
⋮
Y
1
+
Y
0
+
Y
0
-
Y
1
-
⋮
.Since Y→⊖Kk is a Hilbert space, we see that PKk≤1 and so (54)Ak≤PKkG≤G. This shows that the operators Ak are uniformly bounded. Since G is a contraction and Y→⊖Kk is a Hilbert space, (52) implies that Ak is a contraction for each k∈N.
To prove the reverse implication, we assume that for eachk∈N the operator Ak:Hk→Kk is a contraction of norm at most γ where 0<γ<∞. Since span⁡Hk=U→ and span⁡Kk=Y→, it follows that span¯⁡Hk and span¯⁡Kk are regular subspaces. By construction, the subspaces Hk and Kk satisfy the conditions of Corollary 6 and hence do satisfy all the conditions specified in Theorem 7. Since, for each k∈N, the operators (55)Ak:Hk⟶Kk are defined by (51) and are assumed to be uniformly bounded, condition (31) in Theorem 7 is also fulfilled (it can be easily seen that the first condition in (31) is satisfied by the operators Ak). Since (31) is satisfied for all k∈N, it follows that there exists a contraction B satisfying (30) and therefore must be of the form (50). This concludes the proof.



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*Source: 5178454-2018-03-01.xml*