# On a Generalization of the Hilbert Frame Generated by the Bilinear Mapping

Authors: Migdad Ismailov; Fatima Guliyeva; Yusif Nasibov
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

The concept ofb-frame which is a generalization of the frame in Hilbert spaces generated by the bilinear mapping is considered. b-frame operator is defined; analogues of some well-known results of frame theory are obtained in Hilbert spaces. Conditions for the existence of b-frame in Hilbert spaces are obtained; the relationship between the definite bounded surjective operator and b-frame is also studied. The concept of b-orthonormal b-basis is introduced.


---

## Body

## 1. Introduction

Frames for a Hilbert space were formally defined by Duffin and Schaeffer to study some problems in nonharmonic Fourier series [1]. Frames are applied in various branches of natural sciences, such as in signal processing, in the image processing, and in data compression. More details about the application of frames can be found, for example, in [2–5]. Numerous works have been dedicated to frames (see, e.g., [6–8]). More details about frame theory can be found in the monographs [9, 10] and also in review articles [11, 12]. One of the important facts of the theory of frames is the decomposition of each vector with respect to the frame in Hilbert space. It is a remarkable fact that in this case a sequence of coefficients has a minimal property. In frame theory, an important tool for obtaining frames is the stability of frames. Stability of frames in Hilbert spaces is studied in [13, 14] and so forth. Extension of the concept of frame to Banach spaces belongs to Gröchenig [15]. The concepts of a Banach frame and an atomic decomposition are introduced in [15]. Frames in Banach spaces have been also studied in [16–18].The concept ofg-frame as a generalization of the frame in Hilbert spaces has been considered by Sun in [19]. Analogues of most properties of frames were obtained for the g-frame in [20, 21]. Another generalization of the frame in Hilbert spaces is a concept of a b-frame which is introduced in the works [22, 23] by the bilinear mapping. The concept of t-frame is introduced and studied in [22], where t is tensor mapping. Let us note that the approximate concepts associated with the linear mapping and related results have been introduced in [24, 25]. An atomic decomposition of Lebesgue spaces in the trigonometric systems with degenerate coefficients has been studied in [24, 25].This paper is devoted to the study of the characteristic properties ofb-frame in Hilbert spaces. Analogues of the known results of the frame are established. Namely, a criterion of b-frameness for the given sequence is established. The concept of b-orthonormal b-basis is introduced. In this paper, b-frameness of the sequence, generated by the matrix operator, is studied in Hilbert spaces.


## 2. Some Notations and Auxiliary Facts

We will use the standard notation. LetX and Z be Hilbert spaces with scalar products (
·
,
·
)
X and (
·
,
·
)
Z, respectively. As usual, by L
(
X
,
Z
) we denote the space of all bounded linear operators acting from X to Z. The kernel and the range of the operator T
∈
L
(
X
,
Z
) are denoted by N
(
T
) and R
(
T
), respectively. If R
(
T
)
=
Z and N
(
T
)
=
0, then by Banach theorem ∃
T
-
1
∈
L
(
Z
,
X
). In the case when R
(
T
) is closed, there exists T
+
∈
L
(
Z
,
X
) such that T
T
+
z
=
z, ∀
z
∈
R
(
T
). Indeed, T
+ is a continuation of the operator T
1
-
1
:
R
(
T
)
→
(
N
(
T
)
)
⊥ by zero to all Z, where T
1
=
T
|
(
N
(
T
)
)
⊥. The adjoint of the operator T is denoted by T
∗.Byl
2
(
X
), denote the linear space of sequences of vectors x
→
=
x
k
k
∈
N from X, with coordinatewise linear operations, such that ∑
k
=
1
∞
x
k
X
2
<
+
∞. l
2
(
X
) is a Hilbert space with the scalar product(1)
x
→
′
,
x
→
′
′
l
2
X
=
∑
k
=
1
∞
x
k
′
,
x
k
′
′
X
,
x
→
′
=
x
k
′
k
∈
N
,
x
→
′
′
=
x
k
′
′
k
∈
N
.Obviously, ∀
x
→
=
x
k
k
∈
N
∈
l
2
(
X
) has a decomposition(2)
x
→
=
∑
k
=
1
∞
δ
i
k
x
k
i
∈
N
, where δ
i
k is the Kronecker symbol.LetY be a Banach space with a norm ·
Y. Consider the bilinear mapping b
:
X
×
Y
→
Z, satisfying the condition(3)
∃
M
>
0
:
b
x
,
y
Z
≤
M
x
X
y
Y
,
∀
x
∈
X
,
y
∈
Y
.For each pair z
∈
Z and y
∈
Y, let us consider the functional f
z
,
y in X, defined by the formula f
z
,
y
(
x
)
=
(
b
(
x
,
y
)
,
z
)
Z. Let us show that f
z
,
y
∈
X
∗. Linearity of f
z
,
y is obvious. Further, ∀
x
∈
X; taking into account (3), we have(4)
f
z
,
y
x
=
b
x
,
y
,
z
Z
≤
b
x
,
y
Z
z
Z
≤
M
x
X
y
Y
z
Z
.Hence, f
z
,
y
≤
M
y
Y
z
Z. According to the Riesz theorem, there exists the unique element x
∗
∈
X such that f
z
,
y
(
x
)
=
(
x
,
x
∗
)
X. Further, this element will be called the b-scalar product of z and y and will be denoted by z
,
y. It is clear that z
,
y is linear and continuous for z. It directly follows from z
,
y
X
=
f
z
,
y that(5)
z
,
y
X
≤
M
y
Y
z
Z
.We will also need some concepts and statements from [6, 22, 23].Definition 1.
Sequencey
k
k
∈
N
⊂
Y is said to be b-orthonormal in Z, if (6)
b
x
,
y
k
,
y
j
=
δ
k
j
x
,
∀
x
∈
X
,
k
,
j
∈
N
.ByL
b
(
y
k
k
∈
N
), denote the set of all finite sums of the form ∑
k
=
1
n
b
(
x
k
,
y
k
), x
k
∈
X.Definition 2.
Sequencey
k
k
∈
N
⊂
Y is said to be b-complete in Z, if the closure L
b
(
y
k
k
∈
N
)
¯ coincides with Z.Let us provide the following criterion forb-completeness of the sequence.Theorem 3.
Sequencey
k
k
∈
N
⊂
Y is b-complete in Z if and only if ∀
z
∈
Z the equality z
,
y
k
=
0, ∀
k
∈
N, implies z
=
0.Proof.
Let the sequencey
k
k
∈
N
⊂
Y be b-complete in Z. Let us show that, ∀
z
∈
Z from the equality z
,
y
k
=
0, ∀
k
∈
N it follows that z
=
0. Assume the contrary that is let there exists an element z
∈
Z such that z
,
y
k
=
0, ∀
k
∈
N, but z
≠
0. Then, ∀
x
∈
X and ∀
k
∈
N, we have(7)
0
=
x
,
z
,
y
k
X
=
b
x
,
y
k
,
z
Z
. Thus, z is orthogonal to L
b
(
y
k
k
∈
N
), and hence, by its continuity, z will be orthogonal to L
b
(
y
k
k
∈
N
)
¯. Since L
b
(
y
k
k
∈
N
)
¯ coincides with Z, then, in particular, (
z
,
z
)
Z
=
0, and therefore z
=
0, which contradicts the assumption.
Conversely, let the sequencey
k
k
∈
N
⊂
Y be such that, from the equation z
,
y
k
=
0, ∀
k
∈
N it follows that z
=
0. Let us prove its b-completeness in Z. If L
b
(
y
k
k
∈
N
)
¯ does not coincide with Z, then there exists a nonzero element z
0
∈
Z such that (
z
,
z
0
)
Z
=
0, ∀
z
∈
L
b
(
y
k
k
∈
N
)
¯. Then, ∀
x
∈
X, we have(8)
0
=
b
x
,
y
k
,
z
0
Z
=
x
,
z
0
,
y
k
X
,
∀
k
∈
N
, and, consequently, z
0
,
y
k
=
0, ∀
k
∈
N. Then, z
0
=
0, and this gives a contradiction. Therefore, y
k
k
∈
N is b-complete in Z.
The theorem is proved.Definition 4.
The sequencey
k
k
∈
N
⊂
Y is said to be b-basis in Z, if ∀
z
∈
Z uniquely represented in the form (9)
z
=
∑
k
=
1
∞
b
x
k
,
y
k
,
x
k
∈
X
.In the case ofb-basis, y
k
k
∈
N is an orthonormal; then, it is called a b-orthonormal b-basis in Z.It is clear that if the systemy
k
k
∈
N forms a b-orthonormal b-basis for Z, then ∀
z
∈
Z has a unique representation(10)
z
=
∑
k
=
1
∞
b
z
,
y
k
,
y
k
. Indeed, taking into account the continuity of b-scalar product z
,
y with respect to z, from (9) we obtain(11)
z
,
y
j
=
lim
n
→
∞
⁡
∑
k
=
1
n
b
x
k
,
y
k
,
y
j
=
x
j
,
∀
j
∈
N
.The following theorem is an analogue of the criterion for an orthonormal basis.Theorem 5.
Lety
k
k
∈
N
⊂
Y be a b-orthonormal sequence. Then, the following conditions are equivalent:(1)
y
k
k
∈
N is b-orthonormal b-basis in Z.
(2)
y
k
k
∈
N is b-complete in Z.
(3)
∀
z
∈
Z, the b-Parseval identity z
Z
2
=
∑
k
=
1
∞
z
,
y
k
X
2 holds.Proof.
Let condition(
1
) hold. Then, it is clear that L
b
(
y
k
k
∈
N
)
¯ coincides with Z; that is, condition (
2
) is fulfilled. The equivalence of conditions (
1
) and (
3
) follows from the relation(12)
z
-
∑
k
=
1
n
b
z
,
y
k
,
y
k
Z
2
=
z
Z
2
-
∑
k
=
1
n
z
,
y
k
Z
2
. It remains to show that condition (
3
) follows from condition (
2
). For every x
1
,
x
2
,
…
,
x
n
∈
X, we have(13)
z
-
∑
k
=
1
n
b
x
k
,
y
k
Z
2
=
z
-
∑
k
=
1
n
b
x
k
,
y
k
,
z
-
∑
k
=
1
n
b
x
k
,
y
k
Z
=
z
Z
2
-
∑
k
=
1
n
z
,
y
k
,
x
k
X
-
∑
k
=
1
n
x
k
,
z
,
y
k
X
+
∑
k
=
1
n
x
k
X
2
=
z
Z
2
+
∑
k
=
1
n
z
,
y
k
-
x
k
X
2
-
∑
k
=
1
n
z
,
y
k
X
2
≥
z
Z
2
-
∑
k
=
1
n
z
,
y
k
X
2
.Hence, it is easy to see that if the system y
k
k
∈
N is b-complete in Z, then condition (
3
) is true.
The theorem is proved.In theory of frames, often use the following theorem, which describes some properties of the adjoint operator.Theorem 6.
LetT
∈
L
(
X
,
Z
). Then,(1)
T
∗
∈
L
(
Z
,
X
) and T
∗
=
T;
(2)
R
(
T
) is closed if and only if R
(
T
∗
) is closed;
(3)
T is surjective if and only if ∃
A
>
0
:
T
∗
z
Z
≥
c
z
Z, ∀
z
∈
Z.


## 3. Main Results

LetX and Z be a Hilbert space with scalar products (
·
,
·
)
X and (
·
,
·
)
Z, respectively, and let Y be a Banach space with the norm ·
Y. Suppose that b
:
X
×
Y
→
Z is bilinear mapping and satisfies condition (3). Consider a sequence of vector y
k
k
∈
N from Y.Definition 7.
Sequencey
k
k
∈
N is called a b-frame in Z, if there exist constants A
,
B
>
0 such that (14)
A
z
Z
2
≤
∑
k
=
1
∞
z
,
y
k
X
2
≤
B
z
Z
2
,
∀
z
∈
Z
.ConstantsA and B are called the bounds of b-frame. When the right-hand side of (14) is fulfilled, then the sequence y
k
k
∈
N is called b-Besselian in Z with a bound B.Let us provide some examples tob-frames.Example 8.
LetX, Z be Hilbert spaces and Y
=
L
(
X
,
Z
). Assume b
(
x
,
Λ
)
=
Λ
∗
(
x
), where x
∈
X and Λ
∈
Y. Let us recall that (see. [16]) the sequence Λ
k
k
∈
N
⊂
L
(
X
,
Z
) is called g-frame in Z, if there exist constants A
,
B
>
0 such that(15)
A
z
Z
2
≤
∑
k
=
1
∞
Λ
k
z
X
2
≤
B
z
Z
2
,
∀
z
∈
Z
. Then, b-frame Λ
k
k
∈
N
⊂
Y in Z is g-frame in Z.Example 9.
LetX, Y be Hilbert spaces and Z
=
X
⊗
Y is their Hilbert tensor product. Assume that b
(
x
,
y
)
=
x
⊗
y, where x
⊗
y is a tensor product of elements x
∈
X and y
∈
Y. Then, b-frame y
k
k
∈
N
⊂
Y in Z forms a t-frame for Z [19].Let us provide a characteristic property ofb-Besselian sequence in Z.Theorem 10.
Sequencey
k
k
∈
N
⊂
Y is b-Besselian in Z with a bound B if and only if the bounded operator T
:
l
2
(
X
)
→
Z is defined:(16)
T
x
k
k
∈
N
=
∑
k
=
1
∞
b
x
k
,
y
kand T
≤
B. Moreover, an adjoint operator T
∗
:
Z
→
l
2
(
X
) is determined by T
∗
(
z
)
=
z
,
y
k
k
∈
N.Proof.
Lety
k
k
∈
N be b-Besselian in Z with a bound B. The series ∑
k
=
1
∞
b
(
x
k
,
y
k
) converges for every x
→
=
x
k
k
∈
N
∈
l
2
(
X
). Indeed, for m
>
n, we have(17)
z
m
-
z
n
Z
=
∑
k
=
n
+
1
m
b
x
k
,
y
k
Z
=
s
u
p
z
=
1
∑
k
=
n
+
1
m
b
x
k
,
y
k
,
z
Z
=
s
u
p
z
=
1
∑
k
=
n
+
1
m
x
k
,
z
,
y
k
X
≤
s
u
p
z
=
1
∑
k
=
n
+
1
m
x
k
X
z
,
y
k
X
≤
s
u
p
z
=
1
∑
k
=
n
+
1
m
z
,
y
k
X
2
1
/
2
∑
k
=
n
+
1
m
x
k
X
2
1
/
2
≤
B
∑
k
=
n
+
1
m
x
k
X
2
1
/
2
.
Since the right-hand side tends to zero, asn
→
∞, then the sequence z
n
n
∈
N satisfies the Cauchy criterion and also converges. So the operator T
(
x
k
k
∈
N
)
=
∑
k
=
1
∞
b
(
x
k
,
y
k
) is defined. It is easy to show that(18)
T
x
k
k
∈
N
Z
≤
B
x
k
k
∈
N
l
2
X
. Hence, we get T
≤
B.
Conversely, suppose that the bounded operatorT
:
l
2
(
X
)
→
Z, T
(
x
k
k
∈
N
)
=
∑
k
=
1
∞
b
(
x
k
,
y
k
) is defined, and T
≤
B. Let us prove that the system y
k
k
∈
N is b-Besselian in Z with a bound B. For arbitrary x
→
=
x
k
k
∈
N
∈
l
2
(
X
) and z
∈
Z, consider (
T
(
x
k
k
∈
N
)
,
z
)
Z. We have(19)
T
x
k
k
∈
N
,
z
Z
=
∑
k
=
1
∞
b
x
k
,
y
k
,
z
Z
=
∑
k
=
1
∞
x
k
,
z
,
y
k
X
=
x
k
k
∈
N
,
z
,
y
k
k
∈
N
X
.Hence, we directly obtain an expression for the adjoint operator(20)
T
∗
z
=
z
,
y
k
k
∈
N
,
z
∈
Z
. We have(21)
∑
k
=
1
∞
z
,
y
k
X
2
=
T
∗
z
l
2
X
2
≤
T
∗
2
z
Z
2
=
T
2
z
Z
2
≤
B
z
Z
2
.
The theorem is proved.Let the sequencey
k
k
∈
N
⊂
Y form a b-frame for Z with the bounds A and B. Then, a bounded operator S
:
Z
→
Z is defined by the following formula:(22)
S
z
=
∑
k
=
1
∞
b
z
,
y
k
,
y
k
,
∀
z
∈
Z
. Operator S is called b-frame operator for y
k
k
∈
N. Many of the properties of the ordinary frames are valid in this case.Theorem 11.
Let the sequencey
k
k
∈
N
⊂
Y form a b-frame for Z with bounds A, B and with b-frame operator S. Then S is a positive, self-adjoint, bounded invertible operator and B
-
1
≤
S
-
1
≤
A
-
1.Proof.
For everyz
∈
Z, we have(23)
S
z
,
z
Z
=
∑
k
=
1
∞
b
z
,
y
k
,
y
k
,
z
Z
=
∑
k
=
1
∞
z
,
y
k
X
2
; that is, S is a positive operator. Let the operator T
:
l
2
(
X
)
→
Z be defined by the expression T
(
x
k
k
∈
N
)
=
∑
k
=
1
∞
b
(
x
k
,
y
k
). Then, by Theorem 10, it directly follows that T
∗
(
z
)
=
z
,
y
k
k
∈
N, and therefore S
=
T
T
∗. Hence, we obtain that S is a self-adjoint operator. Then, for every z
∈
Z, the following inequality holds:(24)
A
z
Z
2
≤
∑
k
=
1
∞
z
,
y
k
X
2
=
S
z
,
z
Z
≤
S
z
Z
z
Z
, and consequently A
z
Z
≤
S
z
Z. Thus, as surjective and injective operator, by Banach theorem, S is a bounded invertible operator. Since A
≤
S
≤
B holds, then from the relation(25)
A
S
-
1
z
Z
≤
S
S
-
1
z
Z
=
z
Z
,
z
Z
=
S
S
-
1
z
Z
≤
B
S
-
1
z
Z it follows that B
-
1
≤
S
-
1
≤
A
-
1.
Theorem is proved.Theorem 12.
Sequencey
k
k
∈
N
⊂
Y forms a b-frame for Z if and only if the bounded surjective operator,(26)
T
:
l
2
X
⟶
Z
,
T
x
k
k
∈
N
=
∑
k
=
1
∞
b
x
k
,
y
k
, is defined.Proof.
Let the sequencey
k
k
∈
N form a b-frame for Z and let S be its b-frame operator. Then, it is a b-Besselian in Z and therefore, by Theorem 10, the bounded operator T
:
l
2
(
X
)
→
Z, T
(
x
k
k
∈
N
)
=
∑
k
=
1
∞
b
(
x
k
,
y
k
), is defined. It remains to show that R
(
T
)
=
Z. Since T
T
∗
=
S and R
(
S
)
=
Z, then R
(
T
)
=
Z.
Conversely, suppose that a bounded surjective operatorT
:
l
2
(
X
)
→
Z, T
(
x
k
k
∈
N
)
=
∑
k
=
1
∞
b
(
x
k
,
y
k
) is defined. Let us show that y
k
k
∈
N forms a b-frame for Z. According to Theorem 10, the sequence y
k
k
∈
N is b-Besselian in Z. Let us take an arbitrary z
∈
Z. So z
=
T
T
+
z; then supposing T
+
z
=
t
k
k
∈
N, we have(27)
z
Z
4
=
z
,
z
Z
2
=
T
t
k
k
∈
N
,
z
Z
2
=
∑
k
=
1
∞
b
t
k
,
y
k
,
z
Z
2
=
∑
k
=
1
∞
t
k
,
z
,
y
k
X
2
≤
∑
k
=
1
∞
t
k
X
z
,
y
k
X
2
≤
∑
k
=
1
∞
t
k
X
2
∑
k
=
1
∞
z
,
y
k
X
2
=
T
+
z
Z
2
∑
k
=
1
∞
z
,
y
k
X
2
≤
T
+
2
z
Z
2
∑
k
=
1
∞
z
,
y
k
X
2
.Thus,(28)
1
T
+
2
z
Z
2
≤
∑
k
=
1
∞
z
,
y
k
X
2
.
The theorem is proved.Let us provide the criterion ofb-frameness with bounds A and B.Theorem 13.
Sequencey
k
k
∈
N
⊂
Y forms a b-frame for Z with the bounds A and B if and only if the following conditions are fulfilled:(1)
y
k
k
∈
N is b-complete in Z.
(2)
The bounded operator,(29)
T
:
l
2
X
⟶
Z
,
T
x
k
k
∈
N
=
∑
k
=
1
∞
b
x
k
,
y
k
, is defined, such that (30)
A
∑
k
=
1
∞
x
k
X
2
≤
T
x
k
k
∈
N
Z
2
≤
B
∑
k
=
1
∞
x
k
X
2
,
∀
x
k
k
∈
N
∈
N
T
⊥
.Proof.
Lety
k
k
∈
N form a b-frame for Z with bounds A and B. Let us prove the b-completeness of the sequence y
k
k
∈
N in Z. Assume the contrary; that is, the sequence y
k
k
∈
N is not b-complete in Z. Then, by Theorem 3, there exists a nonzero element z
0
∈
Z such that z
0
,
y
k
=
0, ∀
k
∈
N. According to (14), we obtain that z
0
=
0, and hence the assumption is incorrect. Then, from the closedness of R
(
T
), closedness of R
(
T
∗
) follows. Therefore, R
(
T
∗
)
=
(
N
(
T
)
)
⊥. Let us take an arbitrary element x
k
k
∈
N
∈
(
N
(
T
)
)
⊥. Let z
∈
Z, such that x
k
k
∈
N
=
T
∗
z
=
z
,
y
k
k
∈
N. We have(31)
∑
k
=
1
∞
z
,
y
k
X
2
2
=
T
T
∗
z
,
z
Z
2
≤
T
T
∗
z
Z
2
z
Z
2
≤
1
A
T
T
∗
z
Z
2
∑
k
=
1
∞
z
,
y
k
X
2
.Hence,(32)
A
∑
k
=
1
∞
z
,
y
k
X
2
≤
T
T
∗
z
Z
2
. On the other hand,(33)
T
T
∗
z
Z
2
≤
T
2
T
∗
z
l
2
X
2
≤
B
∑
k
=
1
∞
z
,
y
k
X
2
.
Conversely, suppose that conditions(
1
) and (
2
) hold. Let us show that T is a surjective operator. Indeed, let z
∈
R
(
T
)
¯ and z
n
∈
R
(
T
) such that z
n
→
z as n
→
∞. Then, there exists x
→
n
∈
(
N
(
T
)
)
⊥ such that z
n
=
T
x
→
n. From (
2
), it is easy to see that the sequence x
→
n is fundamental and therefore converges to x
→
∈
(
N
(
T
)
)
⊥. Considering continuity of an operator T, we obtain z
n
=
T
x
→
n
→
T
x
→
=
z; that is, R
(
T
) is closed. Future, from b-completeness of y
k
k
∈
N in Z and from L
b
(
y
k
k
∈
N
)
⊂
R
(
T
), it follows that R
(
T
)
=
Z. As(34)
∀
x
→
∈
l
2
X
;
T
+
T
x
→
∈
N
T
⊥
, then, from (
2
), we have(35)
A
T
+
T
x
→
l
2
X
2
≤
T
T
+
T
x
→
Z
2
=
T
x
→
Z
2
. Consequently, T
+
≤
1
/
A. We have(36)
R
T
∗
+
=
R
T
+
∗
=
N
T
+
⊥
=
R
T
=
Z
, and thus ∀
z
∈
Z we obtain z
=
(
T
∗
)
+
T
∗
z and(37)
z
Z
2
=
T
∗
+
T
∗
z
Z
2
≤
T
∗
+
2
T
∗
z
l
2
X
2
=
T
+
∗
2
T
∗
z
l
2
X
2
=
T
+
2
T
∗
z
l
2
X
2
≤
1
A
T
∗
z
l
2
X
2
=
1
A
∑
k
=
1
∞
z
,
y
k
X
2
.
The theorem is proved.The following property of frames is a generalization of the similar properties with respect to the Riesz basis.Theorem 14.
Let the sequencee
k
k
∈
N
⊂
Y form a b-orthonormal b-basis for Z. Then, the system y
k
k
∈
N
⊂
Y forms a b-frame for Z if and only if there exists a bounded surjective operator U
:
Z
→
Z such that U
(
b
(
x
,
e
k
)
)
=
b
(
x
,
y
k
), ∀
x
∈
X and ∀
k
∈
N.Proof.
Lety
k
k
∈
N form a b-frame for Z. By Theorem 12, the bounded surjective operator T
:
l
2
(
X
)
→
Z is defined:(38)
T
x
k
k
∈
N
=
∑
k
=
1
∞
b
x
k
,
y
k
. Consider the operator Φ
:
Z
→
l
2
(
X
) by the formula Φ
(
z
)
=
z
,
e
k
k
∈
N. It is clear that Φ
(
b
(
x
,
e
j
)
)
=
δ
j
k
x
k
∈
N. Let us define the operator U
:
Z
→
Z as follows: U
=
T
Φ. It is obvious that U is a bounded surjective operator and(39)
U
b
x
,
e
k
=
T
Φ
b
x
,
e
k
=
T
δ
j
k
x
k
∈
N
=
b
x
,
y
k
,
∀
x
∈
X
,
∀
k
∈
N is fulfilled.
Conversely, suppose that there is a bounded surjective operatorU
:
Z
→
Z, such that U
(
b
(
x
,
e
k
)
)
=
b
(
x
,
y
k
), ∀
x
∈
X, ∀
k
∈
N. Let us show that y
k
k
∈
N is a b-frame in Z. Let z
∈
Z be an arbitrary element. We have(40)
z
,
y
k
=
U
∗
z
,
e
k
.Indeed, ∀
x
∈
X we have(41)
x
,
z
,
y
k
X
=
b
x
,
y
k
,
z
Z
=
U
b
x
,
e
k
,
z
Z
=
b
x
,
e
k
,
U
∗
z
Z
=
x
,
U
∗
z
,
e
k
X
,and therefore holds (40). Then,(42)
∑
k
=
1
∞
z
,
y
k
X
2
=
∑
k
=
1
∞
U
∗
z
,
e
k
X
2
=
U
∗
z
l
2
X
2
.By virtue of the surjectivity of U, there exists A
>
0 such that U
∗
z
l
2
(
X
)
≥
A
z
Z. On the other hand, U
∗
z
l
2
(
X
)
≤
U
z
Z. Thus, from (42) it follows that y
k
k
∈
N forms a b-frame for Z.
Theorem is proved.Let the sequencey
k
k
∈
N
⊂
Y form a b-frame for Z and consider the operator-matrix U
=
(
u
l
k
), i
,
k
∈
N, where u
l
k
∈
L
(
X
,
X
). Assume that the system φ
l
l
∈
N
⊂
Y satisfies the following condition:(43)
z
,
φ
l
=
∑
k
=
1
∞
u
l
k
z
,
y
k
.The following theorem is true, which of an independent interest.Theorem 15.
Let the sequencey
k
k
∈
N
⊂
Y form a b-frame for Z with bounds A, B and a b-frame operator S; φ
l
l
∈
N
⊂
Y forms a b-frame for Z with bounds A
1 and B
1. If u
l
k
(
x
)
=
S
-
1
(
b
(
x
,
y
k
)
)
,
φ
l, i
,
k
∈
N, then U
=
(
u
l
k
) is a bounded operator in l
2
(
X
).Proof.
Take∀
x
→
=
x
k
k
∈
N
∈
l
2
(
X
).
We have(44)
U
x
→
l
2
X
2
=
∑
l
=
1
∞
∑
k
=
1
∞
u
l
k
x
k
X
2
=
∑
l
=
1
∞
∑
k
=
1
∞
S
-
1
b
x
k
,
y
k
,
φ
l
X
2
=
∑
l
=
1
∞
∑
k
=
1
∞
S
-
1
b
x
k
,
y
k
,
φ
l
X
2
≤
B
1
∑
k
=
1
∞
S
-
1
b
x
k
,
y
k
Z
2
=
B
1
s
u
p
z
=
1
∑
k
=
1
∞
S
-
1
b
x
k
,
y
k
,
z
Z
2
=
B
1
s
u
p
z
=
1
∑
k
=
1
∞
x
k
,
S
-
1
z
,
y
k
X
2
≤
B
1
s
u
p
z
=
1
∑
k
=
1
∞
x
k
X
S
-
1
z
,
y
k
X
2
≤
B
1
s
u
p
z
=
1
∑
k
=
1
∞
x
k
X
2
∑
k
=
1
∞
S
-
1
z
,
y
k
X
2
≤
B
1
S
-
1
2
x
→
l
2
X
2
.Thus, the operator U is bounded and U
≤
B
1
S
-
1.
The theorem is proved.The following theorem in a certain sense is the inverse of Theorem15.Theorem 16.
Let the sequencey
k
k
∈
N
⊂
Y form a b-frame for Z with bounds A and B, let U
=
(
u
l
k
) be a bounded operator in l
2
(
X
), and the sequence φ
l
l
∈
N
⊂
Y is such that the relation (43) holds. Then, φ
l
l
∈
N forms a b-frame for Z if and only if (45)
∃
c
>
0
:
∑
l
=
1
∞
z
,
φ
l
X
2
≥
c
∑
k
=
1
∞
z
,
y
k
X
2
,
∀
z
∈
Z
.Proof.
Letφ
l
l
∈
N be a b-frame in Z with the bounds A
1 and B
1. Then, ∀
z
∈
Z, we have(46)
∑
l
=
1
∞
z
,
φ
l
X
2
≥
A
1
z
Z
2
≥
A
1
B
∑
k
=
1
∞
z
,
y
k
X
2
.
Conversely, suppose that inequality (45) holds. Let us show that φ
l
l
∈
N is b-frame in Z. It is clear that(47)
U
z
,
y
k
k
∈
N
=
z
,
φ
l
l
∈
N
,
∀
z
∈
Z
. Consequently,(48)
∑
l
=
1
∞
z
,
φ
l
X
2
=
U
z
,
y
k
k
∈
N
l
2
X
2
≤
U
2
∑
k
=
1
∞
z
,
y
k
X
2
≤
U
2
B
z
Z
2
.Then, according to (45) and (14), we obtain(49)
∑
l
=
1
∞
z
,
φ
l
X
2
≥
c
∑
k
=
1
∞
z
,
y
k
X
2
≥
c
A
z
Z
2
; that is, φ
l
l
∈
N forms a b-frame for Z.
The theorem is proved.The following theorem establishes a relationship between the frames and systems of systemsy
k
k
∈
N and φ
l
l
∈
N.Theorem 17.
Let the sequencey
k
k
∈
N
⊂
Y form a b-frame for Z with bounds A and B, the operators u
l
k
∈
L
(
X
,
X
), i
,
k
∈
N, are such that the inequality u
l
k
(
x
)
X
≥
a
l
k
x
X is valid, a
l
k
≥
0, ∀
x
∈
X, and the following conditions are fulfilled:(50)
b
=
sup
k
⁡
∑
j
=
1
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
<
+
∞
;
a
=
inf
k
⁡
∑
l
=
1
∞
a
l
k
2
-
∑
j
≠
k
∑
l
=
1
∞
u
l
j
∗
u
l
k
>
0
. Then, φ
l
l
∈
N is a b-frame in Z with bounds a
A and b
B.Proof.
Take∀
z
∈
Z. Firstly, let us prove that φ
l
l
∈
N is b-Besselian in Z with a bound b
B. We have(51)
∑
l
=
1
∞
z
,
φ
l
X
2
=
∑
l
=
1
∞
∑
k
=
1
∞
u
l
k
z
,
y
k
X
2
=
∑
l
=
1
∞
∑
k
=
1
∞
u
l
k
z
,
y
k
,
∑
j
=
1
∞
u
l
j
z
,
y
j
Z
=
∑
l
=
1
∞
∑
k
=
1
∞
∑
j
=
1
∞
u
l
j
∗
u
l
k
z
,
y
k
,
z
,
y
j
Z
=
∑
k
=
1
∞
∑
j
=
1
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
z
,
y
k
,
z
,
y
j
Z
≤
∑
k
=
1
∞
∑
j
=
1
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
z
,
y
k
X
z
,
y
j
X
≤
∑
k
=
1
∞
∑
j
=
1
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
z
,
y
k
X
2
1
/
2
∑
k
=
1
∞
∑
j
=
1
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
z
,
y
j
X
2
1
/
2
=
∑
k
=
1
∞
∑
j
=
1
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
z
,
y
k
X
2
≤
b
∑
k
=
1
∞
z
,
y
k
X
2
≤
b
B
z
Z
2
.Now, let us establish for φ
l
l
∈
N the left-hand side of inequality (14). We obtain(52)
∑
l
=
1
∞
z
,
φ
l
X
2
=
∑
l
=
1
∞
∑
k
=
1
∞
u
l
k
z
,
y
k
X
2
=
∑
l
=
1
∞
∑
k
=
1
∞
u
l
k
z
,
y
k
,
∑
j
=
1
∞
u
l
j
z
,
y
j
Z
=
∑
l
=
1
∞
∑
k
=
1
∞
u
l
k
z
,
y
k
X
2
+
∑
l
=
1
∞
∑
k
=
1
∞
∑
j
≠
k
∞
u
l
j
∗
u
l
k
z
,
y
k
,
z
,
y
j
Z
=
I
1
+
I
2
.Consider I
1 and I
2, separately. We have(53)
I
1
≥
∑
l
=
1
∞
∑
k
=
1
∞
a
l
k
2
z
,
y
k
X
2
,
I
2
=
∑
k
=
1
∞
∑
j
≠
k
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
z
,
y
k
,
z
,
y
j
Z
≤
∑
k
=
1
∞
∑
j
≠
k
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
z
,
y
k
X
z
,
y
j
X
≤
∑
k
=
1
∞
∑
j
≠
k
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
z
,
y
k
X
2
1
/
2
∑
k
=
1
∞
∑
j
≠
k
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
z
,
y
j
X
2
1
/
2
=
∑
k
=
1
∞
∑
j
≠
k
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
z
,
y
k
X
2
.Consequently,(54)
∑
l
=
1
∞
z
,
φ
l
X
2
≥
∑
l
=
1
∞
∑
k
=
1
∞
a
l
k
2
z
,
y
k
X
2
-
∑
k
=
1
∞
∑
j
≠
k
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
z
,
y
k
X
2
≥
a
∑
k
=
1
∞
z
,
y
k
X
2
≥
a
A
z
Z
2
.
The theorem is proved.In particular, in the case ofg-frame, we obtain the following result.Theorem 18.
Let the sequenceΛ
k
k
∈
N
⊂
L
(
X
,
Z
) form a g-frame for Z with bounds A and B, the operators u
l
k
∈
L
(
X
,
X
), i
,
k
∈
N, are such that u
l
k
(
x
)
X
≥
a
l
k
x
X, a
l
k
≥
0, ∀
x
∈
X, and the following condition is fulfilled:(55)
b
=
sup
k
⁡
∑
j
=
1
∞
∑
l
=
1
∞
u
l
j
∗
u
l
k
<
+
∞
;
a
=
inf
k
⁡
∑
l
=
1
∞
a
l
k
2
-
∑
j
≠
k
∑
l
=
1
∞
u
l
j
∗
u
l
k
>
0
. Then, the bounded operators Φ
l
=
∑
k
=
1
∞
u
l
k
Λ
k are defined in X and Φ
l
l
∈
N are g-frames in Z with bounds a
A and b
B.



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*Source: 9516839-2016-04-05.xml*