# New Characterizations of Riesz-Type Frames and Stability of Alternate Duals of Continuous Frames

Authors: Zhong-Qi Xiang
Journal: Advances in Mathematical Physics
 (2013)

---

## Abstract

We give new characterizations of Riesz-type frames, on equivalent
conditions for a continuous frame to be a Riesz-type frame and on equivalency
relations between Riesz-type frames and continuous frames. We characterize also the
Riesz-type frames by using a bounded linear operatorL. Finally, we study the stability of
alternate duals of continuous frames and we prove that if two continuous frames are close
to each other, then we can find alternate duals of them which are close to each other.


---

## Body

## 1. Introduction

Frames were first introduced in 1952 by Duffin and Schaeffer [1]. It seems, however, that Duffin-Scheaffer ideas did not attract much interest outside the realm of nonharmonic Fourier series until the fundamental paper [2] by Daubechies et al. in 1986. During the last 20 years, the theory of frames has been growing rapidly, since several new applications have been developed. For example, besides traditional applications as signal processing, image processing, data compression, and sampling theory, frames are now used to mitigate the effect of losses in packet-based communication systems and hence to improve the robustness of data transmission [3, 4] and to design high-rate constellations with full diversity in multiple-antenna code design [5]. We refer readers to the papers [6–12] for more details on frame theory and its applications.Letℋ be a Hilbert space, and J a set which is finite or countable. A system {fj}j∈J⊆ℋ is called a frame (discrete frame) for ℋ, if there exist two constants C,D>0 such that
(1)C∥f∥2≤∑j∈J‍|〈f,fj〉|2≤D∥f∥2
for all f∈ℋ. The constants C and D are called frame bounds. If C=D, we call this frame a tight frame and if C=D=1 it is called a Parseval frame.The concept of a generalization of frames to a family indexed by some locally compact space endowed with a Radon measure was proposed by Kaiser [13] and independently by Ali et al. [14]. These frames are known as continuous frames. Gabardo and Han in [15] called these frames Frames associated with measurable spaces, and Askari-Hemmat et al. in [16] called them generalized frames and in mathematical physics they are referred to as coherent states [17]. Continuous frames have been widely applied in continuous wavelets transform [17, 18] and the short time Fourier transform [19]. For more details, the interested readers can refer to [20–22].If in the definition of continuous frame, the measure spaceΩ∶=ℕ and μ is the counting measure, then the continuous frame will be a discrete frame and so some known results in frame theory can be generalized to continuous frames. However, the Riesz basis, which is a basic concept in frame theory, has no counterpart in continuous frame theory. This motivates us to pay attention to what kinds of continuous frames that are closer in properties to Riesz bases. We found that the Riesz-type frames introduced by Gabardo and Han [15] are more reasonable. And in the same paper, they made a discussion of some properties of Riesz-type frames in some aspects. In this paper, we give new characterizations of Riesz-type frames, on equivalent conditions for a continuous frame to be a Riesz-type frame, and on equivalency relations between Riesz-type frames and continuous frames. We characterize also the Riesz-type frames by means of a bounded linear operator.The stability of frames is important in practice and is therefore studied widely by many authors [20, 23–25]. The stability of dual frames is also needed in practice. However, most of the known results on this topic are stated about canonical dual; see [26] for frames and [27, 28] for g-frames. Fortunately, Arefijamaal and Ghasemi [29] presented a stability result for alternate duals of g-frames by observing the difference between an alternate dual and the canonical dual. Inspired by their idea we discuss the stability of alternate duals of continuous frames.The paper is organized in the following manner. Section2 investigates some basic properties of Riesz-type frames. The main results of the paper are included in Section 3, where some new characterizations of Riesz-type frames are obtained. Section 4 studies the stability of alternate duals of continuous frames.


## 2. The Definitions and Some Basic Results

In the following we briefly recall some definitions and results about continuous frames and Riesz-type frames in Hilbert spaces.We first give some notations which we will need later. Throughout this paper,ℋ and 𝒦 are complex Hilbert spaces and ℒ(ℋ,𝒦) is the collection of all bounded linear operators from ℋ into 𝒦. For T∈ℒ(ℋ,𝒦), we use ℛ(T) and 𝒩(T) to denote the range and the null space of T, respectively. As usual we use Iℋ to denote the identity operator on ℋ.Definition 1.
Let(Ω,μ) be a measure space with positive measure μ. A mapping F:Ω→ℋ is called a continuous frame with respect to (Ω,μ) for ℋ if it is weakly measurable and if there exist two constants A, B>0 such that
(2)A∥f∥2≤∫Ω‍|〈f,F(ω)〉|2dμ(ω)≤B∥f∥2,∀f∈ℋ.
The constants A and B are called the lower and upper continuous frame bounds, respectively. We call F a tight continuous frame if A=B and a Parseval continuous frame if A=B=1. The mapping F is called Bessel if the right-hand inequality in (2) holds. In this case, B is called the Bessel bound.Example 2.
LetD be a bounded, Lebesgue measurable subset of ℝn and let ℋ=L2(D) and Ω=ℝn. Define F(ω)(t)=e2πi〈ω,t〉χD(t), for each t∈ℝn. Clearly, for each f∈ℋ, 〈f,F(ω)〉=f^(ω), for ω∈Ω, where f^ denotes the Fourier transform of f. By the Plancherel’s theorem, we have
(3)∫Ω‍|〈f,F(ω)〉|2dμ(ω)=∫Ω‍|f^(ω)|2dμ(ω)=∫ℝn‍|f^(ω)|2dμ(ω)=∥f∥2,
which shows that F is a Parseval continuous frame with respect to (Ω,μ) for ℋ, where μ is the Lebesgue measure.By definition, ifF is a Bessel mapping with respect to (Ω,μ) for ℋ, then TF:ℋ→L2(Ω,μ) defined by
(4)(TFf)(ω)=〈f,F(ω)〉,ω∈Ω
is a bounded linear operator, which is called the analysis operator. It follows immediately from the observation that for all f∈ℋ, φ∈L2(Ω,μ),
(5)〈φ,TFf〉=∫Ω‍φ(ω)〈F(ω),f〉dμ(ω)=〈∫Ω‍φ(ω)F(ω)dμ(ω),f〉,
the adjoint operator TF*:L2(Ω,μ)→ℋ is weakly given by
(6)TF*φ=∫Ω‍φ(ω)F(ω)dμ(ω),∀φ∈L2(Ω,μ).
We call TF* the synthesis operator. By composing TF* and TF, we obtain the frame operator SF:ℋ→ℋ. Note that SF is a positive, self-adjoint operator, and it is invertible if and only if F is a continuous frame with respect to (Ω,μ) for ℋ. If F is a continuous frame, then every f∈ℋ has a representation of the form
(7)f=∫Ω‍〈f,SF-1F(ω)〉F(ω)dμ(ω)=∫Ω‍〈f,F(ω)〉SF-1F(ω)dμ(ω).
Remember that the representations above have to be interpreted in the weak sense.We can also characterize the continuous frames in terms of the associated analysis and synthesis operators as in frame theory.Proposition 3.
LetF:Ω→ℋ be a weakly measurable vector-valued function on Ω. Then the following conditions are equivalent.(1)
Fis a continuous frame with respect to(Ω,μ)  forℋ.(2)
The analysis operator  TF  is bounded and injective with closed range.(3)
The synthesis operator  TF*  is bounded, linear, and onto.Proof.
This claim holds in an analogous way as in frame theory.The definitions of similar and unitary equivalent frames give rise to the definitions of similar and unitary equivalent continuous frames.Definition 4.
LetF and G be two continuous frames with respect to (Ω,μ) for ℋ and 𝒦, respectively.(1)
They are said to besimilar or equivalent if there is a bounded invertible operator T:ℋ→𝒦 such that G=TF.(2)
They are said to beunitary equivalent if there exists a unitary linear operator U:ℋ→𝒦 such that G=UF.It is easily shown that a Bessel mappingF with respect to (Ω,μ) for ℋ is a continuous frame if and only if there exists a Bessel mapping G with respect to (Ω,μ) for ℋ such that
(8)f=∫Ω‍〈f,G(ω)〉F(ω)dμ(ω),∀f∈ℋ.
In this case, we call G a dual frame for F and (F,G) a dual pair. If SF is the frame operator for F, a continuous frame with respect to (Ω,μ) for ℋ, then it is easy to check that SF-1F is a dual of F, and it is called the canonical dual, but a dual which is not the canonical dual is called an alternate dual, or simply a dual. It is certainly possible for a continuous frame F to have only one dual, and if this is the case, then we call F a Riesz-type frame.The following proposition is a useful tool and will be used frequently in the rest of the paper. The reader can find a proof in [15].Proposition 5.
LetF be a continuous frame with respect to (Ω,μ) for ℋ with analysis operator TF, and then F is a Riesz-type frame if and only if ℛ(TF)=L2(Ω,μ).The following is an immediate consequence of Propositions3 and 5.Proposition 6.
A weakly measurable vector-valued functionF:Ω→ℋ is a Riesz-type frame with respect to (Ω,μ) for ℋ if and only if the operator TF defined by (4) is well defined, bounded, and invertible.The definition shows that ifF is a continuous frame with respect to (Ω,μ) for ℋ, then F is complete in ℋ, that is, span¯{F(ω)}ω∈Ω=ℋ. We say that F is a continuous frame sequence of ℋ if it is a continuous frame for span¯{F(ω)}ω∈Ω.Proposition 7.
LetF and G be two continuous frame sequences for ℋ, and let us denote X=span¯{F(ω)}ω∈Ω and Y=span¯{G(ω)}ω∈Ω and assume that ℋ=X⊕Y⊥. Then F is a Riesz-type frame with respect to (Ω,μ) for X if and only if G is a Riesz-type frame with respect to (Ω,μ) for Y.Proof.
Denote byTF and TG the analysis operators of F and G, respectively, and let T=TGTF*. Let φ∈ℛ(TF) and let f=TF*φ. If Tφ=0, then 0=(TGf)(ω)=〈f,G(ω)〉 for each ω∈Ω, and thus f∈Y⊥. Since f∈X∩Y⊥ and ℋ=X⊕Y⊥, it follows that f=0. Hence φ∈𝒩(TF*). Since 𝒩(TF*)=(ℛ(TF))⊥ we know that φ=0, which implies that T is injective. Now let ψ∈ℛ(TG). Then there is some g∈X such that TGg=ψ. Let SF be the frame operator of F and let ϕ={〈SF-1g,F(ω)〉}ω∈Ω, and then ϕ∈L2(Ω,μ) and
(9)ψ=TGg=TG∫Ω‍〈SF-1g,F(ω)〉F(ω)dμ(ω)=TGTF*ϕ=Tϕ.
Thus, we conclude that T is an invertible operator from L2(Ω,μ) onto ℛ(TG) and hence by Proposition 5, it follows that F is a Riesz-type frame with respect to (Ω,μ) for X if and only if G is a Riesz-type frame with respect to (Ω,μ) for Y.LetF:Ω→ℋ be a weakly measurable vector-valued function and φ∈L2(Ω,μ). It is natural to ask whether we can find f∈ℋ such that
(10)〈f,F(ω)〉=φ(ω),∀ω∈Ω.
A problem of this type is called a moment problem. It is clear that if the moment problem (10) has a solution, and it is unique if and only if  span¯{F(ω)}ω∈Ω=ℋ. Now an alternative formulation of Proposition 5 is that a continuous frame F is a Riesz-type frame if and only if the moment problem (10) has a unique solution for all φ∈L2(Ω,μ), and furthermore we have the following.Proposition 8.
Suppose thatF is a Riesz-type frame with respect to (Ω,μ) for ℋ, and let φ∈L2(Ω,μ). Then the unique solution for the moment problem (10) is exact the vector f=SF-1TF*φ, where SF and TF denote the frame operator and analysis operator of F, respectively.Proof.
Letf be an element of ℋ such that 〈f,F(ω)〉=φ(ω) for each ω∈Ω, and let P be the orthogonal projection from L2(Ω,μ) onto ℛ(TF)=L2(Ω,μ). A simple calculation shows that
(11)P(φ)(ω)=∫Ω‍φ(ν)〈SF-1F(ν),F(ω)〉dμ(ν),∀φ∈L2(Ω,μ),ω∈Ω.
Hence
(12)〈f,F(ω)〉=φ(ω)=∫Ω‍φ(ν)〈SF-1F(ν),F(ω)〉dμ(ν)=〈SF-1TF*φ,F(ω)〉,∀ω∈Ω,
and so f=SF-1TF*φ as desired.


## 3. New Characterizations of Riesz-Type Frames

In this section, we get several characterizations of Riesz-type frames. We start with the following equivalent conditions under which a continuous frame is a Riesz-type frame.Theorem 9.
LetF be a continuous frame with respect to (Ω,μ) for ℋ with analysis operator TF and the frame operator SF. Then the following are equivalent.(1)
F is a Riesz-type frame.(2)
There exist constantsA,B>0 such that
(13)A∥φ∥2≤∥TF*φ∥2≤B∥φ∥2,∀φ∈L2(Ω,μ).(3)
If(14)∫Ω‍φ(ω)〈F(ω),f〉dμ(ω)=0for someφ∈L2(Ω,μ) and for any f∈ℋ, then φ=0.Proof.
(1)⇒(2). Assume that F is a Riesz-type frame with frame bounds A and B, and then the right-hand inequality in (13) holds. To complete the proof, it remains to establish the left-hand inequality. By Proposition 5, we know that ℛ(TF)=L2(Ω,μ) and, hence, for any φ∈L2(Ω,μ), there is f∈ℋ such that φ(ω)=〈f,F(ω)〉 for each ω∈Ω. Then, we compute that
(15)∥φ∥4=(∫Ω‍|〈f,F(ω)〉|2dμ(ω))2=(〈SFf,f〉)2≤∥SFf∥2∥f∥2≤1A∥SFf∥2∫Ω‍|〈f,F(ω)〉|2dμ(ω),
and hence
(16)A∥φ∥2=A∫Ω‍|〈f,F(ω)〉|2dμ(ω)≤∥SFf∥2=∥TF*TFf∥2=∥TF*φ∥2,
as desired.
(2)⇒(3). It follows immediately from the left-hand inequality of (13).
(3)⇒(1). In order to show that F is a Riesz-type frame, it is sufficient to show, by Proposition 5, that TF is surjective. The assumption implies that TF* is injective, and TF* is also surjective by using Proposition 3. Now the conclusion follows from the fact that SF=TF*TF is invertible.The following perturbation result for Riesz-type frames is due to Gabardo and Han in [15]. Here, as an application of Theorem 9, we present a new approach to its proof.Proposition 10.
LetF be a Riesz-type frame with respect to (Ω,μ) for ℋ with frame bounds A, B and the analysis operator TF. Let G:Ω→ℋ be a weakly measurable vector-valued function and assume that there exist constants λ1, λ2, μ≥0 such that max{λ1+μ/A,λ2}<1 and
(17)∥∫Ω‍φ(ω)(F(ω)-G(ω))dμ(ω)∥≤λ1∥∫Ω‍φ(ω)F(ω)dμ(ω)∥+λ2∥∫Ω‍φ(ω)G(ω)dμ(ω)∥+μ∥φ∥
for all φ∈L2(Ω,μ) with μ({ω∈Ω:φ(ω)≠0})<∞. Then G is also a Riesz-type frame with frame bounds
(18)A[1-(λ1+(μ/A))1+λ2]2,B[1+λ1+(μ/B)1-λ2]2.Proof.
It follows from [15, Theorem 1.2] that G is a continuous frame with respect to (Ω,μ) for ℋ. Let φ be an arbitrary element of L2(Ω,μ), then
(19)∥∫Ω‍φ(ω)G(ω)dμ(ω)∥≤∥∫Ω‍φ(ω)(F(ω)-G(ω))dμ(ω)∥+∥∫Ω‍φ(ω)F(ω)dμ(ω)∥≤(1+λ1)∥∫Ω‍φ(ω)F(ω)dμ(ω)∥+λ2∥∫Ω‍φ(ω)G(ω)dμ(ω)∥+μ∥φ∥.
Since ∥∫Ω‍φ(ω)F(ω)dμ(ω)∥=∥TF*φ∥≤B∥φ∥, we get
(20)∥∫Ω‍φ(ω)G(ω)dμ(ω)∥≤(1+λ1)B1-λ2∥φ∥+μ∥φ∥=[(1+λ1)B+μ1-λ2]∥φ∥,
and hence
(21)∥∫Ω‍φ(ω)G(ω)dμ(ω)∥2≤[(1+λ1)B+μ1-λ2]2∥φ∥2.
Similarly, we obtain
(22)∥∫Ω‍φ(ω)G(ω)dμ(ω)∥≥∥∫Ω‍φ(ω)F(ω)dμ(ω)∥-∥∫Ω‍φ(ω)(F(ω)-G(ω))dμ(ω)∥≥(1-λ1)∥∫Ω‍φ(ω)F(ω)dμ(ω)∥-λ2∥∫Ω‍φ(ω)G(ω)dμ(ω)∥-μ∥φ∥.
Since ∥∫Ω‍φ(ω)F(ω)dμ(ω)∥≥A∥φ∥, we have
(23)∥∫Ω‍φ(ω)G(ω)dμ(ω)∥2≥[(1-λ1)A-μ1+λ2]2∥φ∥2.
If we denote by TG the analysis operator of  G, then by combing (21) with (23) we have
(24)[(1-λ1)A-μ1+λ2]2∥φ∥2≤∥TG*φ∥2≤[(1+λ1)B+μ1-λ2]2∥φ∥2,∀φ∈L2(Ω,μ),
and so the result follows from Theorem 9.A similar result was proved in [15]. Our next proposition points out that if μ=0 in the condition (17) of Proposition 10 and if F is a continuous frame, then G is a Riesz-type frame if and only if F is a Riesz-type frame.Proposition 11.
LetF be a continuous frame with respect to (Ω,μ) for ℋ and let G:Ω→ℋ be a weakly measurable vector-valued function and assume that there exist constants λ1,λ2∈[0,1) such that
(25)∥∫Ω‍φ(ω)(F(ω)-G(ω))dμ(ω)∥≤λ1∥∫Ω‍φ(ω)F(ω)dμ(ω)∥+λ2∥∫Ω‍φ(ω)G(ω)dμ(ω)∥
for all φ∈L2(Ω,μ) with μ({ω∈Ω:φ(ω)≠0})<∞. Then F is a Riesz-type frame with respect to (Ω,μ) for ℋ if and only if G is a Riesz-type frame with respect to (Ω,μ) for ℋ.Proof.
By Theorem  1.2 in [15], we know that G is also a continuous frame with respect to (Ω,μ) for ℋ. We denote by TF and TG the analysis operators of F and G, respectively. We now show that ℛ(TF)=ℛ(TG). If φ∈𝒩(TF*), then we have ∥TG*φ∥≤λ2∥TG*φ∥, which leads to φ∈𝒩(TG*) and so 𝒩(TF*)⊆𝒩(TG*). In the same manner, we can show that 𝒩(TG*)⊆𝒩(TF*), and thus 𝒩(TG*)=𝒩(TF*). It follows from Proposition 3 that both ℛ(TF) and ℛ(TG) are closed. Now applying the fact
(26)𝒩(TF*)⊕ℛ(TF)=𝒩(TF*)⊕(𝒩(TF*))⊥=L2(Ω,μ)=𝒩(TG*)⊕(𝒩(TG*))⊥=𝒩(TG*)⊕ℛ(TG),
we see that ℛ(TF)=ℛ(TG). Then by Proposition 5, we can infer that F is a Riesz-type frame with respect to (Ω,μ) for ℋ if and only if G is a Riesz-type frame with respect to (Ω,μ) for ℋ.In the following theorem we give necessary and sufficient conditions on Bessel mappingsF and G and bounded linear operators L1,L2:ℋ→ℋ so that L1F+L2G is a Riesz-type frame with respect to (Ω,μ) for ℋ.Theorem 12.
LetF and G be two Bessel mappings with respect to (Ω,μ) for ℋ such that
(27)f=∫Ω‍〈f,F(ω)〉G(ω)dμ(ω)
for all f∈ℋ. Let TF and TG be the analysis operators of F and G, respectively, and assume that L1,L2:ℋ→ℋ are bounded linear operators so that L1L2*=Iℋ. Then the following statements are equivalent.(1)
L1F+L2G is a Riesz-type frame with respect to (Ω,μ) for ℋ.(2)
The operatorTFL1*+TGL2* is surjective.(3)
There exists a constantM>0 such that
(28)M∥φ∥2≤∥(L1TF*+L2TG*)φ∥2,∀φ∈L2(Ω,μ).Proof.
(1)⇔(2). We first show that L1F+L2G is a continuous frame with respect to (Ω,μ) for ℋ. Let f be an arbitrary element of ℋ. Then we have
(29)∫Ω‍|〈f,L1F(ω)+L2G(ω)〉|2dμ(ω)=∥{〈f,L1F(ω)+L2G(ω)〉}ω∈Ω∥2=〈{〈f,L1F(ω)+L2G(ω)〉}ω∈Ω,{〈f,L1F(ω)+L2G(ω)〉}ω∈Ω〉=∫Ω‍|〈f,L1F(ω)〉|2dμ(ω)+∫Ω‍|〈f,L2G(ω)〉|2dμ(ω)+∫Ω‍〈f,L1F(ω)〉〈L2G(ω),f〉dμ(ω)+∫Ω‍〈f,L2G(ω)〉〈L1F(ω),f〉dμ(ω)=∫Ω‍|〈f,L1F(ω)〉|2dμ(ω)+∫Ω‍|〈f,L2G(ω)〉|2dμ(ω)+〈L1*f,L2*f〉+〈L2*f,L1*f〉=∫Ω‍|〈f,L1F(ω)〉|2dμ(ω)+∫Ω‍|〈f,L2G(ω)〉|2dμ(ω)+2∥f∥2.
Let us denote by DF and DG the Bessel bounds of F and G, respectively. For all f∈ℋ we have
(30)2∥f∥2≤∫Ω‍|〈f,L1F(ω)+L2G(ω)〉|2dμ(ω)≤(2+DF∥L1∥2+DG∥L2∥2)∥f∥2.
By Proposition 5, L1F+L2G is a Riesz-type frame with respect to (Ω,μ) for ℋ if and only if its analysis operator T is surjective where
(31)Tf={〈f,L1F(ω)+L2G(ω)〉}ω∈Ω={〈f,L1F(ω)〉}ω∈Ω+{〈f,L2G(ω)〉}ω∈Ω=TFL1*f+TGL2*f.
(1)⇔(3). From the proof of (1)⇔(2) we know that L1F+L2G is a continuous frame with respect to (Ω,μ) for ℋ. Now the result follows immediately from Theorem 9.We now turn our attention to the characterizations of the equivalency relations between Riesz-type frames and continuous frames. To prove our first result, we need the following.Lemma 13.
LetF and G be two Parseval continuous frames with respect to (Ω,μ) for ℋ, with analysis operators TF and TG, respectively. Then ℛ(TG)⊆ℛ(TF) if and only if there exists an isometry U:ℋ→ℋ such that G=U*F.Proof.
“⇒”. Suppose that ℛ(TG)⊆ℛ(TF). We already know that ℛ(TF) and ℛ(TG) are closed subspaces of L2(Ω,μ). Let us denote P=TFTF* and Q=TGTG*. It is easy to see that ℛ(P)=ℛ(TF) and ℛ(Q)=ℛ(TG) since, by Proposition 3, both ℛ(TF*) and ℛ(TG*) are surjective. Since F and G are Parseval continuous frames of ℋ, it follows that P and Q are orthogonal projections from L2(Ω,μ) onto ℛ(TF) and ℛ(TG), respectively. Let U=TF*TG, and then, for an arbitrary element f of ℋ, recalling that TGf∈ℛ(TF), we have
(32)U*Uf=TG*TFTF*TGf=TG*TGf=f.
Thus, U is isometric. Also
(33)U*f=TG*TFf=∫Ω‍〈f,F(ω)〉G(ω)dμ(ω),
and so
(34)∫Ω‍〈Uf,F(ω)〉〈G(ω),f〉dμ(ω)=〈∫Ω‍〈Uf,F(ω)〉G(ω)dμ(ω),f〉=〈U*Uf,f〉=∥f∥2.
Note that
(35)∫Ω‍|〈f,G(ω)〉|2dμ(ω)=∥f∥2,∫Ω‍|〈Uf,F(ω)〉|2dμ(ω)=∥Uf∥2=∥f∥2,
and it follows that
(36)∫Ω‍|〈Uf,F(ω)〉-〈f,G(ω)〉|2dμ(ω)=∫Ω(〈Uf,F(ω)〉-〈f,G(ω)〉)‍×(〈F(ω),Uf〉-〈G(ω),f〉)dμ(ω)=∫Ω‍|〈Uf,F(ω)〉|2dμ(ω)-∫Ω‍〈Uf,F(ω)〉〈G(ω),f〉dμ(ω)-∫Ω‍〈f,G(ω)〉〈F(ω),Uf〉dμ(ω)+∫Ω‍|〈f,G(ω)〉|2dμ(ω)=∥f∥2-∥f∥2-∥f∥2+∥f∥2=0.
Hence 〈f,U*F(ω)-G(ω)〉=0, for all f∈ℋ, and so G(ω)=U*F(ω), for all ω∈Ω, and G=U*F as a consequence.
“⇐”. It is straightforward.Proposition 14.
LetF and G be two Parseval continuous frames with respect to (Ω,μ) for ℋ, with analysis operators TF and TG, respectively, and assume besides that F is Riesz-type. Then G is Riesz-type if and only if F and G are unitary equivalent.Proof.
Assume first thatG is a Riesz-type Parseval continuous frame with respect to (Ω,μ) for ℋ, and then ℛ(TF)=ℛ(TG). From the proof of Lemma 13, we know that there is a bounded linear operator U such that U*U=Iℋ and G=U*F, and hence TG=TFU. Since both TF and TG are invertible, it follows that U is a unitary operator and so F and G are unitary equivalent.
For the other implication, letU:ℋ→ℋ be a unitary linear operator such that G=UF, and then ℛ(TF)=ℛ(TG). By Proposition 5, G is Riesz-type.Furthermore, we have the following.Proposition 15.
LetF be a Riesz-type frame with respect to (Ω,μ) for ℋ and G a continuous frame with respect to (Ω,μ) for ℋ. Then G is also a Riesz-type frame with respect to (Ω,μ) for ℋ if and only if F and G are similar.Proof.
Suppose first thatG is a Riesz-type frame with respect to (Ω,μ) for ℋ. Let SF and SG be the frame operators of F and G, respectively. It is clear that F~=SF-1/2F and G~=SG-1/2G are both Riesz-type Parseval continuous frames. By Proposition 14, we know that there is a unitary operator U such that G~=UF~, that is, SG-1/2G=USF-1/2F and so G=SG1/2USF-1/2F, and consequently, F is similar to G.
The opposite implication is obvious.As an application of Proposition15, we have the following equivalent condition for a continuous frame to be a Riesz-type frame.Theorem 16.
LetF be a Riesz-type frame with respect to (Ω,μ) for ℋ and G is a continuous frame with respect to (Ω,μ) for ℋ. Then the following conditions are equivalent.(1)
G is a Riesz-type frame with respect to (Ω,μ) for ℋ.(2)
There exists a constantM>0 such that for all φ∈L2(Ω,μ),
(37)∥∫Ω‍φ(ω)F(ω)dμ(ω)-∫Ω‍φ(ω)G(ω)dμ(ω)∥2≤M·min{∥∫Ω‍φ(ω)F(ω)dμ(ω)∥2,∥∫Ω‍φ(ω)G(ω)dμ(ω)∥2}.Proof.
(1)⇒(2). Suppose that G is a Riesz-type frame with respect to (Ω,μ) for ℋ. By Proposition 15, we know that there is a bounded and invertible operator T:ℋ→ℋ such that G=TF. If f=∫Ω‍φ(ω)F(ω)dμ(ω) for certain φ∈L2(Ω,μ), then we have
(38)〈Tf,g〉=∫Ω‍φ(ω)〈F(ω),T*g〉dμ(ω)=〈∫Ω‍φ(ω)TF(ω)dμ(ω),g〉,
for all g∈ℋ. Hence, Tf=∫Ω‍φ(ω)TF(ω)dμ(ω)=∫Ω‍φ(ω)G(ω)dμ(ω). Therefore,
(39)∥f-Tf∥2=〈f-Tf,f-Tf〉=〈f,f〉+〈Tf,Tf〉+〈(-T*-T)f,f〉≤∥f∥2+∥T∥2∥f∥2+∥T*+T∥∥f∥2≤(1+∥T∥)2∥f∥2.
On the other hand,
(40)∥f-Tf∥2=〈f-Tf,f-Tf〉=〈f,f〉+〈Tf,Tf〉+〈(-T*-T)f,f〉≤(∥T-1∥2+1)∥Tf∥2+〈(T-1)*(-T*-T)T-1Tf,Tf〉≤(∥T-1∥2+1)∥Tf∥2+∥(T-1)*(-T*-T)T-1∥∥Tf∥2≤(1+∥T-1∥)2∥Tf∥2.
Hence (37) holds.
(2)⇒(1). For each f=∫Ω‍φ(ω)F(ω)dμ(ω)∈ℋ, we define an operator T:ℋ→ℋ as follows:
(41)Tf=T(∫Ω‍φ(ω)F(ω)dμ(ω))=∫Ω‍φ(ω)G(ω)dμ(ω).
It is clear that T is well defined and bounded. Let TG be the analysis operator of G, and since TG* is surjective by Proposition 3, it follows that T is surjective. And (37) implies that T is injective and thus T is invertible. In order to show that G is a Riesz-type frame, it suffices to show, by Proposition 15, that G=TF. To see this, let f∈ℋ, φ∈L2(Ω,μ), we have
(42)∫Ω‍φ(ω)〈TF(ω),f〉dμ(ω)=〈T∫Ω‍φ(ω)F(ω)dμ(ω),f〉=〈∫Ω‍φ(ω)G(ω)dμ(ω),f〉=∫Ω‍φ(ω)〈G(ω),f〉dμ(ω)
that is,
(43)0=∫Ω‍φ(ω)〈G(ω)-TF(ω),f〉dμ(ω)=〈φ,{〈f,G(ω)-TF(ω)〉}ω∈Ω〉.
Hence G(ω)=TF(ω) for each ω∈Ω, and consequently G=TF, as desired.Next we will introduce a bounded linear operatorL about two Bessel mappings from Ω to ℋ. The idea is derived from the operator SVW which was considered for fusion frames by Găvruţa in [30]. In this paper, we will use the operator L to characterize the Riesz-type frames further.LetF and G be two Bessel mappings with respect to (Ω,μ) for ℋ, and we introduce the operator L:ℋ→ℋ in weak sense by
(44)Lf=∫Ω‍〈f,F(ω)〉G(ω)dμ(ω),∀f∈ℋ.
It is obvious that L is well defined and bounded, and a direct calculation yields
(45)L*f=∫Ω‍〈f,G(ω)〉F(ω)dμ(ω),∀f∈ℋ.Proposition 17.
LetF be a continuous frame with respect to (Ω,μ) for ℋ with analysis operator TF. Suppose that G is a Bessel mapping with respect to (Ω,μ) for ℋ with analysis operator TG. If L is surjective, then G is a continuous frame for ℋ.
On the contrary, ifF is a Riesz-type frame for ℋ, then L is surjective.Proof.
Suppose first thatF is a continuous frame for ℋ. It is clear that
(46)Lf=∫Ω‍〈f,F(ω)〉G(ω)dμ(ω)=TG*TFf
for any f∈ℋ, namely, L=TG*TF. Since L is surjective, then for any f∈ℋ, there exists g∈ℋ such that f=Lg=TG*TFg, and TFg∈L2(Ω,μ), and it follows that TG* is surjective. By Proposition 3, we know that G is a continuous frame with respect to (Ω,μ) for ℋ.
On the contrary, suppose thatF is a Riesz-type frame with respect to (Ω,μ) for ℋ, and G is a continuous frame with respect to (Ω,μ) for ℋ. By Proposition 6, TF is invertible. By Proposition 3, TG* is surjective, and thus L=TG*TF is surjective.Theorem 18.
Suppose thatF is a Riesz-type frame with respect to (Ω,μ) for ℋ, and G is a continuous Bessel mapping with respect to (Ω,μ) for ℋ. Then L is invertible if and only if G is a Riesz-type frame with respect to (Ω,μ) for ℋ.Proof.
We first suppose thatL is invertible. Let TF and TG be the analysis operators of F and G, respectively. By Proposition 6 we know that TF is invertible. From the proof of Proposition 17, we know that L=TG*TF. Since L is invertible, so is TG*. Again by Proposition 6, we can infer that G is a Riesz-type frame with respect to (Ω,μ) for ℋ.
Now we letF and G be two Riesz-type frames with respect to (Ω,μ) for ℋ. By Proposition 6, both TF and TG are invertible and thus L=TG*TF is invertible.Theorem 19.
LetF be a continuous frame with respect to (Ω,μ) for ℋ, with frame bounds A,B and the frame operator SF, and G is a Bessel mapping with respect to (Ω,μ) for ℋ. Suppose that there exists a number λ with 0<λ<A such that for all f∈ℋ,
(47)∥Lf-SFf∥≤λ∥f∥.
Then F is a Riesz-type frame with respect to (Ω,μ) for ℋ if and only if G is a Riesz-type frame with respect to (Ω,μ) for ℋ.Proof.
For anyf∈ℋ, we have
(48)∥Lf∥=∥Lf-SFf+SFf∥≥∥SFf∥-∥Lf-SFf∥≥(A-λ)∥f∥.
So L is bounded below and is thus injective with closed range. On the other hand, since
(49)∥L*f-SFf∥≤∥L*-SF∥∥f∥=∥(L-SF)*∥∥f∥≤λ∥f∥,∀f∈ℋ,
by the above result L* is also injective with closed range. Thus, by Proposition 3, both L and L* are invertible. Let TF and TG be the analysis operators of F and G, respectively, and we already know that L=TG*TF. Hence TF is invertible if and only if TG is invertible, and consequently F is a Riesz-type frame with respect to (Ω,μ) for ℋ if and only if G is a Riesz-type frame with respect to (Ω,μ) for ℋ.


## 4. Stability of Duals of Continuous Frames

We begin with showing that the difference between an alternate dual and the canonical dual can be considered as a bounded operator.Lemma 20.
LetF be a continuous frame with respect to (Ω,μ) for ℋ with analysis operator TF and frame bounds A and B. Then there exists a one-to-one correspondence between duals of F and operators S∈ℒ(ℋ,L2(Ω,μ)) such that TF*S=0.Proof.
Assume first thatG is a dual of F with frame bounds A1 and B1 and let SF be the frame operator of F. Define S:ℋ→L2(Ω,μ),f↦Sf by
(50)(Sf)(ω)=〈f,G(ω)-SF-1F(ω)〉,f∈ℋ,ω∈Ω.
Then S is bounded. Indeed,
(51)∥Sf∥=(∫Ω‍|(Sf)(ω)|2dμ(ω))1/2=(∫Ω‍|〈f,G(ω)-SF-1F(ω)〉|2dμ(ω))1/2≤(∫Ω‍|〈f,G(ω)〉|2dμ(ω))1/2+(∫Ω‍|〈f,SF-1F(ω)〉|2dμ(ω))1/2≤B1∥f∥+1A∥f∥=(B1+1A)∥f∥
for all f∈ℋ. Moreover, by (50) we have
(52)TF*Sf=∫Ω‍(Sf)(ω)F(ω)dμ(ω)=∫Ω‍〈f,G(ω)-SF-1F(ω)〉F(ω)dμ(ω)=∫Ω‍〈f,G(ω)〉F(ω)dμ(ω)-∫Ω‍〈f,SF-1F(ω)〉F(ω)dμ(ω)=f-f=0.
Conversely, letS∈ℒ(ℋ,L2(Ω,μ)) and TF*S=0. Let G:Ω→ℋ be a weakly measurable vector-valued function such that for any f∈ℋ,
(53)〈f,G(ω)〉=(Sf)(ω)+〈f,SF-1F(ω)〉,f∈ℋ,ω∈Ω.
It is easy to check that G is a Bessel mapping with respect to (Ω,μ) for ℋ. Furthermore,
(54)∫Ω‍〈f,G(ω)〉F(ω)dμ(ω)=∫Ω‍(Sf)(ω)F(ω)dμ(ω)+∫Ω‍〈f,SF-1F(ω)〉F(ω)dμ(ω)=TF*Sf+f=f.
This follows immediately that G is a dual of F.Theorem 21.
LetF and G be two continuous frames with respect to (Ω,μ) for ℋ with frame bounds A1,B1 and A2,B2, respectively. Also let F′ be a fixed dual for F. If F-G is a Bessel mapping with sufficiently small bound ϵ>0, then there exists a dual G′ for G such that F′-G′ is also a Bessel mapping.Proof.
LetTF, TG and SF, SG be the analysis operators and frame operators of F and G, respectively. It follows from Lemma 20 that there exists S∈ℒ(ℋ,L2(Ω,μ)) such that TF*S=0 and
(55)〈f,F′(ω)〉=〈f,SF-1F(ω)〉+(Sf)(ω),f∈ℋ,ω∈Ω.
Let M:Ω→ℋ be a weakly measurable vector-valued function such that for any f∈ℋ,
(56)〈f,M(ω)〉=〈f,SG-1G(ω)〉+(Sf)(ω),ω∈Ω.
It is easy to see that M is a Bessel mapping with Bessel bound (1/A2+∥S∥)2. If we denote the analysis operator of M by TM, then
(57)∥f-TG*TMf∥=∥f-∫Ω‍〈f,M(ω)〉G(ω)dμ(ω)∥=∥f-∫Ω‍〈f,SG-1G(ω)〉G(ω)dμ(ω)-∫Ω‍(Sf)(ω)G(ω)dμ(ω)∥=∥TG*Sf∥=∥TG*Sf-TF*Sf∥≤∥TG-TF∥∥S∥∥f∥≤ϵ∥S∥∥f∥.
Hence ∥Iℋ-TG*TM∥≤ϵ∥S∥ and so TG*TM is invertible since, by assumption, ϵ is sufficiently small. Therefore, every f∈ℋ can be represented by
(58)f=TG*TM(TG*TM)-1f=∫Ω‍〈(TG*TM)-1f,M(ω)〉G(ω)dμ(ω)=∫Ω‍〈f,(TM*TG)-1M(ω)〉G(ω)dμ(ω),
showing that G′=(TM*TG)-1M is a dual for G. In what follows, we will show that G′ is the desired continuous frame. First note that
(59)∥SF-SG∥=∥TF*TF-TF*TG+TF*TG-TG*TG∥≤∥TF-TG∥(∥TF∥+∥TG∥)≤ϵ(B1+B2).
Let TF′ be the analysis operator of F′, and then by combining (55) with (56) we have
(60)〈TF′*φ,f〉-〈TM*φ,f〉=∫Ω‍φ(ω)〈F′(ω),f〉dμ(ω)-∫Ω‍φ(ω)〈M(ω),f〉dμ(ω)=〈TF*φ,SF-1f〉-〈TG*φ,SG-1f〉=〈TF*φ,SF-1f〉-〈TF*φ,SG-1f〉+〈TF*φ,SG-1f〉-〈TG*φ,SG-1f〉=〈TF*φ,(SF-1-SG-1)f〉+〈(TF*φ-TG*φ),SG-1f〉
Therefore,
(61)∥TF′*φ-TM*φ∥=sup∥f∥=1|〈TF′*φ-TM*φ,f〉|=sup∥f∥=1|〈TF*φ,(SF-1-SG-1)f〉+〈(TF*φ-TG*φ),SG-1f〉|≤∥SF-1-SG-1∥∥TF*φ∥+∥SG-1∥∥TF*φ-TG*φ∥≤∥SG-1∥∥SF-SG∥∥SF-1∥∥TF*φ∥+∥SG-1∥∥TF*-TG*∥∥φ∥≤1A2ϵ(B1+B2)1A1B1∥φ∥+1A2ϵ∥φ∥=ϵ(A1+B1+B1B2)A1A2∥φ∥.
If we takeT=(TG*TM)-1, then
(62)∥T∥≤11-∥Iℋ-T-1∥≤11-ϵ∥S∥,
and so
(63)∥Iℋ-T∥≤∥T∥∥Iℋ-T-1∥≤ϵ∥S∥1-ϵ∥S∥.
Consequently,
(64)∥TF′*φ-TG′*φ∥=sup∥f∥=1|〈TF′*φ-TG′*φ,f〉|≤sup∥f∥=1(|〈(Iℋ-T*)TF′*φ,f〉|+|〈T*(TF′*-TM*)φ,f〉|)≤∥Iℋ-T∥∥TF′*φ∥+∥T∥∥(TF′*-TM*)φ∥≤ϵ1-ϵ∥S∥×(∥S∥B′+A1+B1+B1B2A1A2)∥φ∥,
where B′ is the upper frame bound of F′. The proof is completed.



---

*Source: 298982-2013-06-09.xml*