STARS Citation Lopez, Jerry, "Optimal Dual Frames For Erasures And Discrete Gabor Frames" 2009... Optimal Dual Frames for Erasures and Discrete Gabor Framesby Jerry LopezB.S.. Moreover,
Frames in Hilbert Space
A frame is a broad generalization of a basis in a vector space, particularly in finite-dimensional spaces Unlike a basis, which consists of linearly independent vectors that span the space, a frame can include linearly dependent vectors while still providing a complete spanning set This flexibility makes frames a useful tool in various applications such as signal processing and functional analysis, where redundant representations are advantageous for stability and robustness.
Allowing a spanning set to be linearly dependent offers several benefits, including:
• Relaxed conditions, making it easier to find a spanning set with additional properties (e.g group structure).
For an infinite-dimensional space, the situation is slightly more complicated In- stead of spanning sets there are complete sequences, but not every complete sequence is a frame [18].
Fortunately, there is a definition which is valid for both the finite- and infinite- dimensional cases.
Definition 2.1 Let H be a Hilbert space and {v i }i∈I ⊆ H If there exist constants
A sequence {v_i} is called a frame if it satisfies the inequality |hx, v_i|^2 ≥ Bkxk^2 for some constant B ≥ 2, ensuring stable signal representations The maximal constant A in this context is known as the lower frame bound, guaranteeing a minimum level of stability, while the minimal constant B is referred to as the upper frame bound, ensuring the bounds' consistency Frames play a crucial role in signal processing and functional analysis by providing redundant, yet stable, representations of signals Understanding these bounds helps optimize the robustness and accuracy of signal reconstruction within the frame theory framework.
A frame where \(A = B\) is known as a tight frame, ensuring stable representations in signal processing When \(A = B = 1\), Equation 2.1 reduces to the Parseval identity, making the frame a Parseval tight frame, also called a Parseval frame Additionally, a uniform (or equal-norm) frame is characterized by all vectors having the same norm, which is important for consistent signal analysis and reconstruction.
In finite-dimensional spaces, the condition that a frame spans the entire space is often more practical to verify than checking frame bounds This approach simplifies the concept of frames by providing an alternative definition specifically tailored for finite frames, making it easier to establish complete coverage of the space in a more straightforward manner.
Definition 2.2 Let H be a finite-dimensional Hilbert space and {vi} k i=1 ⊆ H such that span{v i }=H The sequence {v i } is called a frame.
It can be shown that this is equivalent to Definition 2.1 The following proof is adapted from Proposition 3.18 of [18].
Proof: First, suppose that {v i } k i=1 does not span H Then there exists a nonzero vector xsuch that x is in the orthogonal complement of span{v i } k i=1 Thus, for all i, hx, v i i= 0 But then k
Therefore, A = 0 in Equation 2.1 In other words, there is no lower frame bound, and so {v i } k i=1 is not a frame.
Conversely, suppose that {v i } k i=1 violates the lower frame bound condition of Def- inition 2.1 (the upper condition always holds for a finite sequence) Then, for every m∈N, there exists y m ∈H such thatky m k= 1 and k
Since {y m } ∞ m=1 is a bounded sequence, it must have a convergent subsequence {y m j } with limit vector y Thus
|hy, v i i| 2 and y is orthogonal to every v i So either y = 0 or {v i } k i=1 does not span H, but kyk= 1 since every ky m j k= 1 Therefore, span{v i } k i=1 6=H.
It follows from this definition that every basis is also a frame.
Next, we look at a few simple examples of frames.
Example 2.1 The vectors {x i } 3 i=1 given by
It is acceptable to repeat a vector multiple times within a frame, which influences how we define and interpret frames in vector spaces While describing a frame as a set of vectors is convenient in casual discussions, it is more precise to define a frame as an ordered sequence of vectors, though this approach may have limitations depending on the application For instance, in some cases, two frames containing the same vectors but in different sequences can be considered equivalent, highlighting the importance of understanding the order's role An example illustrating this concept will be explored in greater detail in Chapter 3, emphasizing its significance in advanced applications.
Example 2.2 (Mercedes-Benz Frame) The vectors{x i } 3 i=1 given by
Another example shows a simple frame in the infinite-dimensional case.
Example 2.3 Let {e i } ∞ i=1 be an orthonormal basis for the Hilbert space H = ` 2 (N). Then by repeating each element of {e i } ∞ i=1 twice we have
{xi} ∞ i=1 ={e1, e1, e2, e2, } which is a tight frame for H with A=B = 2.
Chapter 5 will continue exploring frames in the infinite-dimensional setting.
Analysis Operator and Frame Operator
This article introduces the foundational concepts of frames by defining key operators linked to arbitrary sequences of vectors It explores the properties of these operators and demonstrates their significance in understanding and characterizing frames By examining how these operators relate to vector sequences, the study provides insights into the structure and applications of frames in various mathematical and engineering contexts.
Definition 2.3 Let H, K be Hilbert spaces, with K of dimension k Let {ei} k i=1 be an orthonormal basis for K, and {v i } k i=1 ⊆H Theanalysis operator is the linear operator Θ :H→K such that Θx k
If K =C k , then this is equivalent to Θx
When dealing with more than one set of frame vectors, it will often be convenient to use a subscript notation to differentiate between their respective analysis operators.
For example, if {v i } k i=1 ⊆H and {w i } k i=1 ⊆H, then Θ v x k
Definition 2.4 The synthesis operatoris the adjoint of the analysis operator.
The analysis operator for \(x \in K\) is defined by \(X i=1 h_x, e_i \), reflecting its fundamental role in decomposing signals into basis components Alternatively, the synthesis operator can be characterized by the relation \(\Theta^* e_i = v_i\), which is derived from the analysis operator's properties and the orthonormal basis, illustrating how signals are reconstructed from their coefficients.
Definition 2.5 The frame operator is the operator Θ ∗ Θ.
The frame operator is often denoted by S.
Definition 2.6 The Grammian operator is the operator ΘΘ ∗
If K =C k , then from the above definitions ΘΘ ∗ x= Θ(Θ ∗ x) k
=Ax where A= (hv j , v i i) That is, ΘΘ ∗ is the matrix ΘΘ ∗
hv 1 , v 1 i hv 2 , v 1 i hv k , v 1 i hv 1 , v 2 i hv 2 , v 2 i hv k , v 2 i . hv 1 , v k i hv 2 , v k i hv k , v k i
Note in particular that the diagonal elements of the Grammian are kv i k 2 See Prop- erty 2.9 for more details.
In addition to standard operators, combining operators associated with different vector sets can enhance functional versatility Creating composite operators, such as Θ ∗ w Θ v x k, enables more complex operations, making them highly useful in advanced mathematical and computational contexts.
We now review some of the basic properties of these operators, beginning with the analysis operator.
Property 2.1 The analysis operator is injective if, and only if, {v i } k i=1 is a frame. Proof: First, suppose {v i } k i=1 is a frame If Θx= 0, then k
X i=1 hx, v i ie i = 0 and sinceei is a basis,hx, vii= 0 for all i Since{vi} k i=1 is a spanning set forH, there existsαi such thatx=Pk i=1αivi, which gives hx, xi=hx, k
If the analysis operator is injective, it implies that the span of the set {v_i}_{i=1}^k must be the entire space H This is because assuming otherwise leads to a contradiction: selecting a non-zero vector y orthogonal to the span would result in Θ y = 0, contradicting the injectivity Therefore, the set {v_i}_{i=1}^k spans H, establishing it as a frame Conversely, when the span of {v_i} is the whole space, the analysis operator is guaranteed to be injective, confirming the fundamental relationship between frames and their analysis operators.
Property 2.2 If {v i } k i=1 is a Parseval frame, then the analysis operator is an isom- etry.
Where the last equality follows from Equation 2.1, with A =B = 1, since {v i } k i=1 is a Parseval frame Therefore, the analysis operator is an isometry.
Property 2.3 LetT :H →H be a linear operator so that the set of vectors{T v i } k i=1 has analysis operator Θ T v Then Θ T v x= Θ v T ∗ x.
Property 2.4 Let α be a scalar so that the set of vectors {αv i } k i=1 has analysis op- erator Θ αv Then Θ αv =αΘ v
Proof: This can be shown from the definition, or simply by using Property 2.3 with
Next, we give some of the properties of the frame operator.
Property 2.5 The frame operator is invertible if, and only if, {vi} k i=1 is a frame. Proof: If S −1 exists, then for allx∈H x k
X i=1 hx, S −1 viivi by Proposition 2.2 Thus{v i } k i=1 spans H, and so is a frame.
Conversely, suppose {v i } k i=1 is a frame with analysis operator Θ If x 6= 0 ∈ H, then Θx6= 0∈Θ(H), by Property 2.1 SinceK = ker Θ ∗ ⊕Range Θ, ify 6= 0∈Θ(H), then Θ ∗ y6= 0 Thus, Θ ∗ (Θx) = Θ ∗ Θx6= 0 Therefore, Θ ∗ Θ is invertible.
Property 2.6 The frame operator is self-adjoint.
Property 2.7 The frame operator is the identity operator I if, and only if, {v i } k i=1 is a Parseval frame.
This follows from the reconstruction formula See Section 2.3 for more details.
Property 2.8 The frame operator is a scalar multiple of the identity operator, λI, if, and only if {v i } k i=1 is a tight frame with frame bound λ.
Proof: Let {v i } k i=1 be a tight frame with frame bound λ >0, so that for all x λkxk 2 k
Thus the set of vectors {λ −1/2 v i } k i=1 is a Parseval frame From Property 2.7, this frame has frame operator I, and so by Property 2.4
Therefore the frame operator is a scalar multiple of the identity operator.
Finally, we show a useful property of the Grammian operator.
Property 2.9 For a frame{v i } k i=1 withΘits analysis operator, tr(ΘΘ ∗ ) =Pk i=1kv i k 2
The analysis operator Θ : H → K associated with the frame {v_i}_i=1^k maps elements from the Hilbert space H to the finite-dimensional space K The Grammian operator ΘΘ* acts on K and plays a crucial role in understanding the frame's properties If {e_i}_i=1^k forms an orthonormal basis for K, then the trace of the Grammian operator, tr(ΘΘ*), provides valuable information about the frame's tightness and energy distribution This relationship highlights the significance of the analysis and synthesis operators in frame theory and their applications in signal processing and functional analysis.
Parseval Frames
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It turns out that there are sets of vectors other than orthonormal bases which exhibit this extremely useful reconstruction property.
Definition 2.7 Let {v i } k i=1 be a set of vectors in H, and x ∈ H Then the recon- struction formula is x k
This is equivalent to the equation x= Θ ∗ v Θ v x In other words, the frame operator is the identity operator This leads to the following theorem
Theorem 2.1 A set of vectors {v i } k i=1 ⊆ H is a Parseval frame if, and only if, it satisfies the reconstruction formula (Equation 2.3).
Proof: Suppose {v i } k i=1 satisfies the reconstruction formula Then kxk 2 =hx, xi
Therefore, the Parseval identity is satisfied for all x∈H, and so{v i } k i=1 is a Parseval frame.
Conversely, suppose {v i } k i=1 is a Parseval frame By Property 2.2, Θ v is an isome- try, and so it also preserves inner products Let {u i } n i=1 be an orthonormal basis for
H and {e i } k i=1 be an orthonormal basis for K Thus x n
X m=1 hhx, v j ie j ,hu i , v m ie m iu i n
Therefore, the reconstruction formula is satisfied, as required.
Parseval frames are an important class of frames with many useful results As was just shown in Theorem 2.1, Parseval frames satisfy the same reconstruction formula as orthonormal bases, x k
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Parseval frames offer the advantage of redundancy, ensuring stability and robustness in signal representation For example, Theorem 2.4 demonstrates that, under specific conditions, removing a single vector from a Parseval frame still results in a collection of vectors that continue to form a frame and span the entire space This redundancy property makes Parseval frames valuable in applications requiring reliable and resilient signal processing.
What follows are some of the basic results for Parseval frames, beginning with a very important theorem which shows that every frame has a Parseval frame associated with it.
Theorem 2.2 For any frame {v i } k i=1 , the set {S −1/2 v i } k i=1 is a Parseval frame. Proof: It is enough to show that {S −1/2 v i } k i=1 satisfies the reconstruction formula x= (S −1/2 SS −1/2 )x
X i=1 hx, S −1/2 v i iS −1/2 v i since S, and thus S −1/2 , is self-adjoint Therefore, {S −1/2 v i } k i=1 satisfies the recon- struction formula and is a Parseval frame.
Proposition 2.1 If{v i } k i=1 is a Parseval frame, then for alli, kv i k ≤1, with equality if, and only if, v i is orthogonal to all v j where j 6=i.
Proof: From the Parseval identity, kv j k 2 k
For equality, suppose kv j k= 1 Then
|hv j , v i i| 2 which implies that hv j , v i i= 0 for all i6=j.
Conversely, suppose v j is orthogonal to all j 6=i Then kv j k 2 k
|hv j , v i i| 2 =|hv j , v j i| 2 =kv j k 4 so that kvjk= 1 ifvj 6= 0.
This leads to the following Corollary.
Corollary 2.1 Let{v i } k i=1 be a Parseval frame Then{v i } k i=1 is an orthonormal basis if, and only if, every v i is a unit vector.
In fact, a more general proposition proved identically to Proposition 2.1 giveskv i k 2 ≤
Afor a tight frame with frame boundA, with equality only forv i which are orthogonal to all vj such thatj 6=i.
For a general frame with upper frame bound B
|hv j , v i i| 2 ≥ |hv j , v j i| 2 =kv j k 4 so that kv i k 2 ≤B for all v i 6= 0.
Lemma 2.1 Let H be a Hilbert space of dimension n ≤ k If {v i } k i=1 is a uniform Parseval frame, then kv i k 2 = n k for all i.
Proof: Since the frame is Parseval, Θ ∗ Θ =I n by Property 2.7 By the definition of uniform, kvik=kvjk for all i, j Combining this with Property 2.9 gives, for any j kv j k 2 = 1 k k
Theorem 2.3 Let H be a Hilbert space of dimension n < k If {v i } k i=1 ⊆ H is a uniform Parseval frame, then for any index j, {v i }i6=j is still a frame.
|hx, v i i| 2 by Cauchy-Schwarz Thus kxk 2 (1− kv j k 2 )≤ k
Therefore, provided (1−kv j k 2 )>0, there is a lower frame bound and so the remaining set of vectors is a frame But by Lemma 2.1,kv j k 2 = n k for all j, and since k > n
1−n k >0 the lower frame bound exists, and the set of vectors is a frame.
In fact this is a special case of a more general theorem which makes use of Proposi- tion 2.1.
Theorem 2.4 Let H be a Hilbert space of dimension n ≤ k If {v i } k i=1 ⊆ H is a
Parseval frame, then for any index j where v j is not orthogonal to every v i such that i6=j, {v i }i6=j is still a frame.
The proof closely follows the approach of Theorem 2.3, utilizing Proposition 2.1 to establish that kvjk 0, thereby confirming the existence of a positive lower frame bound for the remaining vectors.
In an orthonormal basis, each vector is orthogonal to every other vector, meaning no vector is orthogonal to all others simultaneously This ensures the basis vectors are linearly independent and collectively span the entire space Removing one vector from an orthonormal basis results in a set that no longer spans the space, highlighting the fundamental importance of each vector in maintaining the basis's completeness.
General Reconstruction Formula and Dual Frames
For general frames, there is also a reconstruction formula similar to Equation 2.3 Let {v i } k i=1 be a frame for H Then, there exists a frame {w i } k i=1 such that every x∈H can be reconstructed with the formula x k
Definition 2.8 Let{v i } k i=1 be a frame for H Then any frame{w i } k i=1 which satisfies Equation 2.4 is called a dual frame for {vi} k i=1
The frame {S −1 v i } k i=1 always satisfies this reconstruction formula and is called the canonical (or standard)dual frame Any other frame which satisfies the equation is called an alternate dual frame.
Proposition 2.2 The set of vectors{S −1 v i } k i=1 is a dual frame for {v i } k i=1
Proof: Substituting S −1 x forx in Equation 2.2 gives
X i=1 hx, S −1 v i iv i since S, and thusS −1 , is self-adjoint Similarly, applying S −1 to both sides of Equa- tion 2.2 gives
X i=1 hx, viiS −1 vi (2.5) as required.
The next result gives a characterization for all of the alternate dual frames of a given frame.
Proposition 2.3 Every dual frame of {v i } is of the form w i =S −1 v i +h i , where k
Proof: Let {w i } be a dual frame for {v i }, and define h i =w i −S −1 v i k
Conversely, suppose {w i } k i=1 is a set of vectors such that w i = S −1 v i +h i , where
Thus{w i }satisfies the dual reconstruction formula, and so it is a dual frame for{v i }. The proof for Pk i=1hx, h i iv i = 0 is similar.
The condition that Pk i=1hx, v i ih i = 0, or in operator notation Θ ∗ h Θ v = 0, is related to the concept of orthogonal frames See Section 2.5 for more details.
Next, we review some properties of the canonical dual frame.
Let {v i } k i=1 be a frame with frame operator S and canonical dual {S −1 v i } k i=1 Property 2.10 If {v i } k i=1 is a Parseval frame, then it is its own canonical dual.
This follows from Property 2.7, that the frame operator of a Parseval frame is the identity operator S=I, so that {S −1 v i } k i=1 ={v i } k i=1
Property 2.11 If {v i } k i=1 is a tight frame with frame bound λ, then {λ −1 v i } k i=1 is its canonical dual.
This follows from Property 2.8, that the frame operator of aλ-tight frame is S =λI, so that {S −1 v i } k i=1 ={λ −1 v i } k i=1
The next two properties together show that a frame and its canonical dual frame are actually canonical duals of each other.
Property 2.12 The frame operator for {S −1 v i } k i=1 is S −1
Proof: Let Θ S −1 v be the analysis operator for the canonical dual, so that T Θ ∗ S −1 vΘ S −1 v is its frame operator Then by Property 2.3
Property 2.13 The canonical dual of {S −1 vi} k i=1 is {vi} k i=1
The duality between the sequences {S⁻¹ v_i} and {v_i} is evident from the definitions of dual frames and Equation 2.5 Additionally, {S⁻¹ v_i} serves as the canonical dual frame of {v_i}, which follows from Property 2.12, given that T⁻¹ = S.
Orthogonal Frames
Let {vi} k i=1 ⊂ H and {wi} k i=1 ⊂ K with Θv : H → C k and Θw : K → C k their respective analysis operators Then {v i } and {w i } are orthogonal sequences if the range space of Θ v is orthogonal to the range space of Θ w , that is, Θ v (H) ⊥Θ w (K).
In addition, if {v i } and {w i } are frames for their respective spaces, they are called orthogonal frames.
Orthogonal frames are essential in applications like data multiplexing, and their significance will be further explored in Chapter 5 Additionally, orthogonal frames play a crucial role in characterizing all possible alternate dual frames, as detailed in Proposition 2.3.
Proposition 2.4 Θ v (H)⊥Θ w (K) if and only if Θ ∗ v Θ w = 0
Proof: First, suppose Θ ∗ v Θ w = 0 Then for all a∈H, b∈K hΘ v a,Θ w bi=ha,Θ ∗ v Θ w bi
Now, suppose that Θ v (H)⊥Θ w (K) Then for alla∈H, b ∈K
0 =hΘva,Θwbi=ha,Θ ∗ v Θwbi and since this must hold for all a ∈H, Θ ∗ v Θ w b= 0 And since this must hold for all b∈K, Θ ∗ v Θ w = 0.
Similar Frames
Definition 2.9 Let {v i } k i=1 and {w i } k i=1 be frames Then these frames are said to be similar if there exists an invertible operator T such that T v i = w i for i = 1, , k.
If T is a unitary operator, the frames are said to be unitarily equivalent.
Similar frames are denoted as{v i } k i=1 ∼ {w i } k i=1 Clearly every frame is similar to its canonical dual frame More importantly, from Theorem 2.2 every frame is similar to a Parseval frame.
Property 2.14 Similarity (and consequently unitary equivalence) is an equivalence relation.
Proof: Let {v i },{w i },and {z i } be frames LetT,S be invertible operators.
• Reflexive: For alli, Iv i =v i , so that {v i } ∼ {v i }.
• Symmetric: For all i, if T v i =w i , then v i =T −1 w i
• Transitive: For all i, if T v i = w i and Sw i = z i , then z i = Sw i = S(T v i ) (ST)v i , with (ST) invertible.
Proposition 2.5 If {v i } and {w i } are similar, Range(Θ v ) =Range(Θ w ).
Proof: Suppose Av i =w i , for all i, where A is invertible Lety ∈Range(Θ v ) Then there exists x such that y= Θ v x= Θ A −1 w x= Θ w (A −1 ) ∗ x so that y∈Range(Θ w ) Thus Range(Θ v )⊆Range(Θ w ).
Similarly, if y ∈Range(Θ w ), then there exists x such that y= Θ w x= Θ Av x= Θ v A ∗ x so that y∈Range(Θ v ) Thus Range(Θ w )⊆Range(Θ v ).
Note that if {v i } and {w i }are unitarily equivalent, this becomes Θ v x= Θ w Ax and Θ w x= Θ v A ∗ x
Operator Trace Using Dual Frames
In a finite-dimensional Hilbert space, the trace of a linear operator T is defined as the sum of inner products over an orthonormal basis, expressed as tr(T) = ∑_{i=1}^n ⟨T e_i, e_i⟩ The standard proof demonstrating that the trace is independent of the chosen orthonormal basis primarily relies on the reconstruction property of these bases To extend this concept, recent generalizations explore the use of dual frames, broadening the applicability of the trace definition beyond orthonormal bases and emphasizing its robustness in various frame settings within Hilbert spaces.
Theorem 2.5 Let T be a linear operator on H having dimension n Let k ≥ n and ` ≥ n If {e i } k i=1 and {f i } ` i=1 are frames, with dual frames {v i } k i=1 and {w i } ` i=1 respectively, then k
Corollary 2.2 For any Parseval frame {f i } k i=1 , tr(T) =Pk i=1hT f i , f i i.
Proof: For any orthonormal basis{e i } n i=1 ofH, using{e i } n i=1 ={v i } n i=1 and{f i } k i=1 {wi} k i=1 in the above theorem gives tr(T) n
UsingT =S =In in the above Corollary gives an alternate proof that the sum of the frame vector norms is the dimension of H, without using the Grammian matrix.
Corollary 2.3 For any Parseval frame, {f i } k i=1 , on a Hilbert space H of dimension n, Pk i=1kf i k 2 =n.
The following Corollaries for general frames mirror the ones for Parseval frames
Corollary 2.4 For any frame{f i } k i=1 with dual frame{w i } k i=1 , tr(T) = Pk i=1hT f i , w i i.
Proof: For any orthonormal basis{e i } n i=1 of H, using{e i } n i=1 ={v i } n i=1 in the above theorem gives tr(T) n
UsingT =I n in the above Corollary
Corollary 2.5 For any frame, {fi} k i=1 with dual frame {wi} k i=1 , on a Hilbert space
In addition, if the dual used is the canonical dual
Corollary 2.6 For any frame, {f i } k i=1 with frame operator S and canonical dual frame {S −1 f i } k i=1 , tr(S) =Pk i=1kf i k 2 and tr(S −1 ) =Pk i=1kS −1 f i k 2
Finally, the independence of the choice of Parseval frame in the trace shows that no Parseval frame can have a Parseval frame as a proper subset (with non-zero vectors omitted).
Corollary 2.7 If {f i } ` i=1 is a Parseval frame for a Hilbert space H of dimension n, then there is no Parseval frame {fi} k i=1 , with n ≤ k < `, with some fj 6= 0 for k < j ≤`.
Proof: The proof is by contradiction Let{f i } k i=1 ⊂ {f i } ` i=1 both be Parseval frames, with f j 6= 0 for some k < j ≤` Then by the trace,
But this is impossible if f j 6= 0 for some k < j ≤ ` Therefore, there is no suchParseval frame with a Parseval frame (or orthonormal basis) as a proper subset.
Finite frames can possess special structures that enhance their properties, with one notable example being group-structured frames Incorporating group structures like cyclic and abelian groups into frame design introduces unique characteristics, enabling more efficient and robust applications in signal processing and data analysis Understanding how these specific group types influence frame properties is essential for optimizing their performance in various computational contexts.
Unitary Representations
Let G be a group and H a Hilbert space The set B(H) consists of all bounded, linear operators acting on H An operator T within B(H) is considered unitary if it satisfies the condition T* = T^{-1}, meaning its adjoint operator T* is equal to its inverse.
Definition 3.1 The set of all unitary operators of H form a group, and a group homomorphism π from G into this group of unitary operators is called a unitary representation.
Definition 3.2 For any unitary representation π, thecommutant is the set π(G) 0 ={T ∈B(H) | T π(g) = π(g)T,∀g ∈G}
The following properties hold for any unitary representation.
Property 3.1 For any unitary representation π, π(e) =I.
Proof: Since π is a homomorphism π(e) =π(gg −1 )
Property 3.2 For any unitary representation π, π(g) ∗ =π(g −1 ).
Property 3.3 For any unitary representation π, kπ(g)ξk=kξk for all g ∈G. Proof: Since π(g) is unitary kπ(g)ξk 2 =hπ(g)ξ, π(g)ξi
Frame Representations
A frame representation is a unitary representation on a Hilbert space H where there exists a vector ξ ∈ H such that the set {π(g)ξ | g ∈ G} forms a frame for H The vector ξ is referred to as a frame vector for the representation π, which is said to admit a frame This concept is fundamental in understanding how group actions can generate frames within Hilbert spaces, providing valuable insights for applications in signal processing and harmonic analysis.
For frame representations, most definitions follow directly from those of a general frame Let{π(g)ξ}g∈G be a collection of vectors inH, and{χ g }an orthonormal basis for K =` 2 (G).
Definition 3.4 The analysis operator is the operator Θ :H →K defined by Θx=X g∈G hx, π(g)ξiχ g
A given frame representation can admit multiple frame vectors, and so the subscripted notation will be useful Θ ξ =X g∈G hx, π(g)ξiχ g and Θ η =X g∈G hx, π(g)ηiχ g
This notation differs from that in Chapter 2 for general frames, as here the subscript indicates the frame vector of the representation rather than the vectors themselves For example, when applying a linear operator, the notation ΘT ξ is used to represent Θ T ξ = X_{g∈G} ⟨x, π(g)T ξ⟩ χ_g, highlighting the distinction from previous conventions.
In general, these two are different unless T ∈π(G) 0
Definition 3.5 The synthesis operatoris the adjoint of the analysis operator. Θ ∗ x=X g∈G hx, χ g iπ(g)ξ or Θ ∗ χ g =π(g)ξ
Definition 3.6 The frame operator is the operator Θ ∗ Θ. Θ ∗ Θx=X g∈G hx, π(g)ξiπ(g)ξ
Definition 3.7 The Grammian operator is the operator ΘΘ ∗
In addition, it will also be useful to take compositions of operators associated with different frame vectors for a given representation That is, to create operators of the form Θ ∗ η Θ ξ =X g∈G hx, π(g)ξiπ(g)η
The first result is a basic property of frame representations.
Proposition 3.1 If π(g) admits a frame, it is a uniform frame.
Proof: This is an immediate consequence of Property 3.3, since every vector in the frame has the same norm, the norm of the frame vector ξ.
In addition to the properties for the analysis and frame operators given in Chap- ter 2, we have the following additional property for the frame operator of a group representation frame.
Property 3.4 The frame operator is in the commutant of π(g).
Proof: This follows fromS = Θ ∗ ξ Θξ and Lemma 3.3 See Section 3.3 for more details.
Consequently, S −1 is also in the commutant of π(g) This leads to the following
Proposition 3.2 If ξ is a frame vector so that {π(g)ξ}g∈G is a frame for H, then
Proof: From Proposition 2.2, {S −1 π(g)ξ}g∈G is a dual frame for {π(g)ξ}g∈G, and since S −1 commutes withπ(g), this becomes{π(g)S −1 ξ}g∈G.
We say that (ξ, S −1 ξ) form a dual pair.
Commutant of Group Representation Frames
Lemma 3.2 If π(g) admits a Parseval frame vector ξ, then π(G) 0 ⊆ {Θ ∗ y Θ z | y, z ∈ H}
Proof: Let T ∈π(G) 0 Then from Lemma 3.1
Proof: If T = Θ ∗ y Θ z for some y, z ∈H, then for anyx∈H
Theorem 3.1 Ifπ(g)admits a Parseval frame vector ξ, thenπ(G) 0 ={Θ ∗ ξ Θ η |ξ, η ∈ H}
Proof: Follows from Lemma 3.2 and Lemma 3.3.
In fact, a more general result is given by the following.
Unitary Equivalence
Recall that if two frames are unitarily equivalent, denoted {π(g)ξ}g∈G ∼ {σ(g)η}g∈G, then there exists a unitary operator U such that for all g ∈G
In addition, if two unitary representations are unitarily equivalent, denoted π ∼ σ, then there exists a unitary operator U such that for all g ∈G
Lemma 3.4 If {π(g)ξ} ∼ {σ(g)η} as frames, then U ξ =η.
Proof: Let{π(g)ξ} ∼ {σ(g)η}as frames Then, by definition, there exists a unitary operator U such that U π(g)ξ =σ(g)η, ∀g ∈G Thus, by Property 3.1
Lemma 3.5 If {π(g)ξ} ∼ {σ(g)η} as frames, then π∼σ.
Proof: Let{π(g)ξ} ∼ {σ(g)η}as frames Then, by definition, there exists a unitary operator U such that U π(g)ξ =σ(g)η, ∀g ∈G Thus, by Lemma 3.4
Abelian and Cyclic Groups
An abelian group is one where the group operation is commutative.
A cyclic group G = {a^i | i ∈ Z} is a mathematical structure generated by a single element a, where a^0 is the identity element and the group operation satisfies a^{i + j} = a^i a^j An important property of cyclic groups is that they are always abelian, meaning their elements commute under the group operation Additionally, frame representations induced by abelian and cyclic groups possess unique characteristics, highlighting their significance in group theory and applications.
The Mercedes-Benz frame from Example 2.2 exemplifies a cyclic group frame, where the group structure is modeled by the 3rd roots of unity Each vector in this frame is generated through a group operation involving rotation by 120 degrees, showcasing the symmetry and cyclic nature of the frame Specifically, the vectors {x_i} for i = 1 to 3 are constructed based on these rotational transformations, highlighting the role of cyclic groups in frame design and their applications in advanced signal processing.
For one particular cyclic group, the K-th roots of unity, the following lemma will be useful.
Lemma 3.6 If e 2πi K A is a K-th root of unity that is not equal to 1, then
The following results are summarized without proof from Han and Larson The first is Proposition 1.14 in [19], which says that distinct alternate duals for a given frame are never similar.
Lemma 3.7 Suppose that {x n } is a frame and {y n } is an alternate dual for {x n }.
If T ∈ B(H) is an invertible operator such that {T y n } is also an alternate dual for {x n }, then T =I.
The next result is Corollary 3.14 in [19], adapted for our notation.
Lemma 3.8 states that for an abelian group G with a representation π on a Hilbert space H, if ξ is a Parseval frame vector for π, then any frame vector η in H can be expressed uniquely as η = Vξ, where V is an invertible operator in π(G) These findings demonstrate that all frame vectors for H are interconnected through invertible operators in the group representation, providing a foundational understanding of the structure of frames in this context.
Proposition 3.3 Let{π(g)ξ}g∈G be a frame, withGan abelian group Then the only dual frame for{π(g)ξ}g∈G with the same group structure is the canonical dual frame.
Orthogonal Group Frames and Super-Frames
In Section 2.5, we define two sequences as orthogonal when their analysis operator range spaces are orthogonal, which is equivalent to the condition Θ₂* Θ₁ = 0 When these sequences are also frames, they are referred to as orthogonal frames or strongly disjoint frames, highlighting their mutually exclusive analysis properties.
In terms of group representation frames, suppose that ξ and η are frame vectors for π Then {π(g)ξ}g∈G and {π(g)η}g∈G are orthogonal frames if Θ ∗ η Θ ξ = 0.
Proposition 3.4 Ifη andζ generate two dual frames for {π(g)ξ} g∈G , thenu=η−ζ generates a sequence which is orthogonal to {π(g)ξ}g∈G.
We say that (ξ, u) form an orthogonal pair.
Orthogonal frames have applications in multiplexing, and we briefly mention some of the ideas used later Let {φ (`) j }j∈J be Parseval frames for Hilbert spaces H`,
A disjoint k-tuple, consisting of collections {φ(1)j}, {φ(2)j}, , {φ(k)j}, is characterized by the property that their direct sum, {φ(1)j ⊕ ⊕ φ(k)j}, forms a frame for the orthogonal direct sum space H₁ ⊕ ⊕ Hₖ When this combined collection is a Parseval frame for the entire space, the k-tuple is considered strongly disjoint, also known as a superframe of length k These concepts are fundamental in frame theory, especially in the study of signal decomposition and reconstruction within Hilbert spaces.
Lemma 3.9 (Han-Larson) {φ (1) j ⊕ .⊕φ (k) j }j∈J is a superframe for H⊕ .⊕H if and only if all of the following hold
(i) Each {φ (`) j } is a Parseval frame for H.
(ii) {φ (m) j } and {φ (n) j } are orthogonal when m6=n
Introduction
To add redundancy to transmitted data, frames can be employed effectively For example, consider a scenario where H = C² and K = C³, utilizing a uniform Parseval frame {v_i} for encoding The analysis operator Θ_v facilitates this process, transforming the data vector x = (x₁, x₂) into a redundant format Encoding is achieved by computing Θ_v x, which results in a combination of inner products hx, v_i, providing robustness against data loss This approach enhances data transmission reliability by leveraging frame theory and redundancy in communication systems.
The coefficients of Θ v x can then be transmitted to a receiver, who recovers x by computing Θ ∗ v (Θ v x) Θ ∗ v (Θvx) = (Θ ∗ v Θv)x=Ix=x
Suppose, however, that one of the coefficients is lost in tranmission, and the receiver only receives
Even if a transmitted data vector is lost, the receiver can still potentially recover the original information According to Theorem 2.3, a uniform Parseval frame that loses one vector remains a frame, ensuring that specific vectors such as v1 and v3 continue to span the space This enables the existence of coefficients α and β such that v2 can be expressed as a linear combination of v1 and v3, facilitating data reconstruction through the relation x = Θ* Θ v x.
=hx, v 1 iv 1 +hx, v 2 iv 2 +hx, v 3 iv 3
=hx, v 1 iv 1 +hx, αv 1 +βv 3 iv 2 +hx, v 3 iv 3
=hx, v 1 iv 1 + αhx, v 1 i+βhx, v 3 i v 2 +hx, v 3 iv 3 where all of the inner products in the last equality were received in transmission. Thus the receiver can reconstruct x exactly.
This simplified example highlights the redundant nature of frames but also underscores their limitations, as successful reconstruction requires precise knowledge of which coefficient was lost—a detail often unavailable in real-world applications As the number of vectors in a frame increases, the computational complexity of recovering the original signal skyrockets, rendering exact reconstruction impractical in many cases Fortunately, many applications do not require perfect recovery; instead, they focus on achieving optimal estimations of the original signal despite transmission losses.
Optimal Frames for Erasures
What follows is an overview of the typical method of dealing with erasures, as from
In a Hilbert space of dimension n, a Parseval frame {v_i}₁^k with analysis operator Θ enables efficient signal representation The original vector x is encoded as Θx and transmitted to the receiver Decoding is achieved by applying the synthesis operator Θ* to Θx, resulting in (Θ*Θ)x, which simplifies to the identity operator times x, ensuring perfect reconstruction.
When certain components of the vector Θx are lost, corrupted, or delayed during transmission, the received vector can be modeled as EΘx, where E is a diagonal matrix with m zeros and k−m ones The zeros in the matrix E represent the components of Θx that have been "erased" or failed to reach the receiver, illustrating how data loss or corruption affects the transmitted information.
One option to recover the original data is to attempt to compute a left inverse for
An alternative approach is to continue using Θ* for reconstruction, acknowledging that the recovered data is only an approximation of the original x The reconstruction error can be expressed as x−Θ*EΘx, which simplifies to Θ*DΘx, where D is a diagonal matrix with m ones representing the lost coordinates and k−m zeros corresponding to the received coordinates.
The primary goal is to identify the "best" frames in scenarios involving data erasures, by selecting frames that minimize the norm of the error operator regardless of which coordinates are erased These optimal frames ensure minimal reconstruction error across all possible m-erasures, making them ideal for reliable data transmission and signal processing By focusing on error operator minimization, we can develop robust frames that enhance resilience against data loss, ensuring accurate signal reconstruction even in the presence of multiple erasures.
To ensure the independence of erased coordinates, it is essential to assign each analysis operator, and consequently each Parseval frame, a measure of the maximum error introduced by m erasures This is achieved by defining the worst-case norm of the error operator across all potential erasure patterns Specifically, we consider the set Dm of all k×k diagonal matrices with m ones and k–m zeros, representing different erasure scenarios The quantity dm(Θ) is then defined as the maximum of the operator norms of kΘ* D Θk over all D in Dm, providing a rigorous means to evaluate the robustness of the frame against m coordinate erasures.
Minimizing d_m(Θ) across all possible Θ can be considered optimal in theory, but it is more desirable for a frame that is optimal for erasures to also be effective for m or fewer erasures To achieve this, we develop a decreasing family of frames that ensures robustness even with reduced erasures, enhancing the reliability and efficiency of frame-based systems in data transmission and signal processing.
The equation E m (k, n) = min Θ∈E m−1 (k,n) d m (Θ) represents the minimal measure over the set of all Parseval frames of k vectors in F n, denoted by F(k, n) The set F(k, n) includes all frames where Θ belongs to the union of sets E i for 1 ≤ i ≤ m, indicating that frames in E m are suitable for m-erasure scenarios Optimal frames designed for m-erasures are characterized by their analysis operators residing within the set E m, commonly referred to as m-erasure frames, which are crucial for enhancing robustness against data loss in frame theory applications.
Optimal frames have been extensively studied, with uniform Parseval frames identified as ideal for single erasures, ensuring robust signal reconstruction For scenarios involving two erasures, equiangular, uniform Parseval frames are considered optimal, providing maximum resilience and minimal coherence These specialized frames, known as Grassmannian frames, are particularly effective in minimizing worst-case coherence in signal processing and communication systems, highlighting their significance in designing robust frame structures.
Optimal Dual Frames for Erasures
Existence of Optimal Dual Frames
An (n, k)-frame X = {xi} for a Hilbert space H, with frame operator S, provides a robust representation of data An optimal dual frame Y for X minimizes the reconstruction error when subjected to 1-erasure, satisfying d₁(X, Y) = min{d₁(X, Z) : Z is a dual frame for X} Similarly, for m-erasures, an optimal dual frame Y is designed to minimize errors, with dₘ(X, Y) = min{dₘ(X, Z) : Z is a dual frame for X}, ensuring the best possible stability and reconstruction quality despite data losses.
From Proposition 2.3, Y = {yi} n i=1 is a dual frame for X if and only if Y S −1 X+U for some U ={u i } n i=1 such that n
In the context of dual frames and m-erasures, the condition Xi=1 hx, xiiui=0 for all x∈H implies that Θ*UΘX=0, defining the set NX of all U satisfying this condition An optimal dual frame for X in the presence of m-erasures is characterized as the one that minimizes the worst-case reconstruction error, expressed as minimizing the distance dm(X, S^(-1)X + U) over all U in NX This approach ensures robust recovery of signals even when up to m frame elements are erased, optimizing the performance of dual frames under erasure scenarios.
We first prove the existence of an optimal dual frame for one erasure.
Let x, y ∈ H We will use x⊗ y to denote the rank-one operator defined by (x⊗y)(v) =hv, yix for all v ∈H.
Note that if D ∈ D 1 and Y = {S −1 x i +u i } n i=1 with U ={u i } n i=1 ∈ N X , then we have
||Θ ∗ Y DΘ X ||=||(S −1 x i +u i )⊗x i ||=||(S −1 x i +u i )|| ã ||x i || for some 1≤i≤ n Therefore when we consider 1-erasure optimal dual frames, it is reasonable to assume that x i 6= 0 for all 1≤i≤n Thus the function defined by
The function F(U) = d₁(X, S⁻¹X + U) measures the maximum norm difference between the transformed points and their original positions, defined as max{||S⁻¹xᵢ + uᵢ|| vs ||xᵢ|| for 1 ≤ i ≤ n} This function is continuous with respect to U and exhibits unbounded growth, meaning F(U) approaches infinity as ||U|| approaches infinity Here, U is considered as a vector in the orthogonal direct sum Hilbert space H(n) By restricting our analysis to a bounded subset of N×X, we can effectively determine the minimum of F(U) within this confined region.
This leads to the following:
Lemma 4.1 Let X = {x i } n i=1 be a frame for H with x i 6= 0 for all i Then optimal dual frames of X exist for1-erasure Moreover, the set of all the optimal dual frames of X for m-erasures form a convex, closed and bounded subset of H (n)
To demonstrate the proof, it is essential to establish the convexity of the set of optimal dual frames Specifically, if Y(1) and Y(2) are two optimal dual frames of X for m-erasure, then both achieve the minimal dual frame distance, satisfying dm(X, Y(1)) = dm(X, Y(2)) = min{dm(X, Z) : Z is a dual frame for X} This confirms that the set of optimal dual frames forms a convex set, ensuring the robustness and stability of dual frame solutions in the context of m-erasure.
Let Y =λY (1) + (1−λ)Y (2) for λ ∈ [0,1] Clearly, Y is a dual of X It remains to show that d m (X, Y) = d m (X, Y (1) ) =d m (X, Y (2) ) In fact, for any D∈ D m we have
Thus d m (X, Y)≤d m (X, Y (1) ) = d m (X, Y (2) ) and so the equality holds.
Following from the above lemma and using the induction argument we obtain:
Corollary 4.1 Let X ={x i } n i=1 be a frame forH with x i 6= 0 for alli Then optimal dual frames of X exist for any m-erasures Moreover, the set of all the optimal dual frames of X for m-erasures form a convex and closed subset of H (n)
The next two sections show that for some cases, the canonical dual frame is the unique optimal dual frame.
Optimal Dual Frame for a Uniform, Tight Frame
Proposition 4.1 For one erasure, the unique optimal dual frame for a tight frame with uniform length is the canonical dual frame.
Proof: Let{x i } n i=1 be a tight frame with equal norms,kx i k qλk n,∀i ThenS =λI for some λ6= 0, and so S −1 = λ 1 I Thus, the canonical dual frame is{ λ 1 x i } Suppose {y i }is a dual of {x i } We need to show
Now, by property of the trace n
Thus,Pn i=1ky i k ≥q nk λ Now, suppose that
< rnk λ which is a contradiction Therefore, the canonical dual is an optimal dual for a uniform tight frame.
In fact, it is the unique optimal dual Suppose {z i } is an optimal dual frame. Then
1≤i≤nmax{kz i k} r k λn so kzik ≤ qk λn,∀i If kzjk< qk λn for some j, then n
X i=1 kzik< rnk λ which is the same contradiction as above So, if{zi}is optimal,kzik q k λn for alli.
Thus Pn i=1kh i k 2 = 0, and so h i = 0 for all i Therefore, the canonical dual is the unique optimal dual.
Optimal Dual Frame for a Group Representation Frame
LetG be a group and H a Hilbert space, with {π(g)ξ:g ∈G} a frame forH.
First, note that for an abelian group, {π(g)S −1 ξ} is the only dual with the same group structure, so the problem is only interesting for non-abelian groups.
Proposition 4.2 For one erasure, the optimal dual frame with the same group struc- ture is the canonical dual frame.
It follows, by setting x=S −3/2 ξ and taking the inner product with S −1/2 ξ, that hX g∈G hS −3/2 ξ, π(g)ξiπ(g)h, S −1/2 ξi= 0
Then, since S −1 is in the commutant of π(G) and π(g) is an isometry minh max g∈G kS −1 π(g)ξ+π(g)hk= min h max g∈G kπ(g) S −1 ξ+h k
In this analysis, it is shown that the sum of the squared norms kS^−1 ξ + hk 2 equals the sum of the squared norms kS^−1 ξk 2 and khk 2, due to the orthogonality between S^−1 ξ and h The optimal solution occurs when khk 2 equals zero, which implies h = 0, indicating that the canonical dual provides the optimal dual solution with the same group structure in the case of a single erasure.
Proposition 4.3 For one erasure, the canonical dual frame is the unique optimal dual for a group representation frame.
Standard Dual Frame as the Unique Optimal Dual Frame
The results of the two previous sections can actually be shown to be corollaries of a more general result We require the following lemma.
X i=1 hx, x i ih i = 0 then Pn i=1hS −1 x i , h i i= 0.
Proof: In operator notation, Θ ∗ h Θ S −1 x = 0, since for any x inH n
Therefore, by the property of the trace n
Theorem 4.1 For one erasure, the canonical dual frame is the unique optimal dual frame for any frame where kS −1 x i k ã kx i k is a constant for all i.
Let {x i } be a frame with kS −1 x i k ã kx i k = c, a constant for all i Let {y i } {S −1 x i +h i }be an optimal dual frame Then maxi kS −1 x i +h i k ã kx i k ≤max i kS −1 x i k ã kx i k
Thus maxi ky i k ã kx i k ≤ kS −1 x j k ã kx j k, ∀j kyik ã kxik ≤ kS −1 xjk ã kxjk, ∀i, j ky j k ã kx j k ≤ kS −1 x j k ã kx j k, ∀j ky j k ≤ kS −1 x j k, ∀j
Now, for all i ky i k 2 =kS −1 x i +h i k 2
=kS −1 x i k 2 +kh i k 2 + 2RehS −1 x i , h i i kh i k 2 + 2RehS −1 x i , h i i=ky i k 2 − kS −1 x i k 2 kh i k 2 + 2RehS −1 x i , h i i ≤0
X i=1 kh i k 2 ≤0 and sohi = 0 for alli Therefore,{yi}is the canonical dual frame, and so the optimal dual frame is unique.
Corollary 4.2 For one erasure, the canonical dual frame is the unique optimal dual frame for a tight frame with uniform length.
This follows with kS −1 x i k ã kx i k q k λn.
Corollary 4.3 For one erasure, the canonical dual frame is the unique optimal dual frame for a group representation frame.
This follows since π(g) is an isometry for all g, withkπ(g)S −1 ξk ã kπ(g)ξk=kS −1 ξk ã kξk.
An optimal dual frame, defined inductively as one that remains optimal for m-erasures, must also be optimal for (m−1)-erasures Since the dual frame uniquely optimal for a 1-erasure has no alternative choices, it is inherently optimal for any number of erasures Therefore, the key result is that a dual frame optimal for single erasures extends its optimality to multiple erasures, ensuring robust performance in signal reconstruction scenarios.
Theorem 4.2 Let {x i } n i=1 be a frame for a k-dimensional Hilbert space H and S be its frame operator If kS −1 x i k ã kx i k = c is a constant for all i, then the canonical dual frame is the unique optimal dual frame for any m-erasures.
In Theorem 4.2, the question arises whether the sequence {y_i}_i=1^n is an optimal dual of {x_i}_i=1^n when the norms satisfy ||y_i|| ≈ ||x_i|| as a constant However, as demonstrated by Corollary 4.3, the answer to this query is negative, indicating that equal norm conditions alone do not ensure optimal duality.
Proposition 4.4 There exists a group frame{π(g)ϕ}g∈G such that it admits a dual frame of the form{π(g)η}g∈Gthat is not the canonical dual, and consequently{π(g)η}g∈G is not optimal.
Proof: Let π be a group representation on H and {π(g)ϕ}g∈G is a Parseval frame for H with the property that there exists g 1 , g 2 ∈Gsuch that hπ(g1)ϕ, π(g2)ϕi 6= hπ(g2)ϕ, π(g1)ϕi.
Then by the main result on the uniqueness of dual frame generators in [11], there existsη∈Hsuch thatη6=S −1 ϕand{π(g)η}g∈Gis a dual frame of{π(g)ϕ}g∈G, where
S is the frame operator for {π(g)ϕ}g∈G Since π is an isometry, kπ(g)ηk ã kπ(g)ξk is a constant for all g ∈G However, by Corollary 4.3, {π(g)η}g∈G is not optimal.
One further generalization of Theorem 4.2 requires an additional definition Let
X = {x i } n i=1 be a sequence We say that a decomposition Sm j=1I j = {1,2, , n} isX-linearly independentifM 1 + .+M m is a direct sum, whereM j = span{x i |i∈I j }.
Theorem 4.3 Let{I 1 , , I m }be anX-linearly independent decomposition of{1, , n}. Suppose that kS −1 x i k ã kx i k=c j for all i∈I j Then
(ii) Assume c m = c m−1 = = c k > c k−1 ≥ c k−2 ≥ ≥ c 1 Then {x i } n i=1 has a unique 1-optimal dual if and only if {x i } i∈ S k−1 j=1 I j is linearly independent.
Proof: Let {yi} n i=1 be 1-optimal, whereyi =S −1 xi+ui, and n
Thus, by the X-linearly independent decomposition, for every 1≤j ≤m
Letc j 0 = max{c j : 1≤j ≤m} Then, for all 1≤i≤n ky i k ã kx i k ≤c j 0
In particular, for all i∈I j 0 kyik ã kxik ≤cj 0 =kS −1 xik ã kxik ky i k ≤ kS −1 x i k
Thus, just as in Theorem 4.1, ui = 0 for all i ∈ Ij 0 Therefore, yi = S −1 xi for all i∈I j 0 Consequently max{kS −1 x i k ã kx i k: 1≤i≤n}=c j 0 = max{ky i k ã kx i k: 1≤i≤n}
Therefore {S −1 x i } n i=1 is optimal for 1-erasure as claimed.
For part (ii), assume that {x i } i∈ S k−1 j=1 I j is linearly independent, and let {y i } n i=1 be a 1-optimal dual for{x i } n i=1 , with y i =S −1 x i +u i and n
In our analysis, we establish that \( x_i = 0 \) for all \( x \in H \), indicating that the inner product \( \langle x, u_i \rangle = 0 \) for every \( x \in H \) From previous results, we know that \( y_i = S^{-1} x_i \), meaning \( u_i = 0 \) for all \( i \) in the union of the index sets \( I_k \cup I_{k+1} \cup \dots \cup I_m \) The critical step is to verify that \( u_i = 0 \) for all \( i < k \), which follows immediately due to the linear independence of the relevant vectors This confirms that the orthogonality condition holds across all indices, ensuring the completeness of the proof.
X i=1 hx, u i ix i = X i∈ S k−1 j=1 I j hx, u i ix i implies that hx, uii = 0 for all x Therefore, ui = 0 for all i and the optimal dual frame is unique.
Conversely, suppose that {x i } i∈ S k−1 j=1 I j is linearly dependent Then there exists u i not all 0 such that
X i∈ S k−1 j=1 I j hx, u i ix i = 0 for all x ∈ H Let u i = 0 when i ∈ Sm j=kI j Then {u i } n i=1 is a non-zero finite sequence such that n
The equation \( \sum_{i=1}^x \langle h x, t u_i \rangle x_i = 0 \) holds for every \( x \in H \), where \( t \neq 0 \) is any constant Based on this, the dual frame \( Y_t = \{ S^{-1} x_i + t u_i \}_{i=1}^n \) is established for the original frame \( \{ x_i \}_{i=1}^n \) Since the norm \( \| S^{-1} x_i \| \) is bounded by a constant \( c_m \), there exists a \( \delta > 0 \) such that for all \( t \) with \( |t| \leq \delta \), the inequality \( \| S^{-1} x_i + t u_i \| < c_m \) holds for every \( i \) in the index set Consequently, \( Y_t \) is 1-erasure optimal whenever \( |t| \leq \delta \), indicating that the original frame \( \{ x_i \}_{i=1}^n \) possesses infinitely many 1-erasure optimal dual frames, owing to the non-zero finite nature of the sequence \( \{ u_i \}_{i=1}^n \).
Examples
One example of a frame where the canonical dual is optimal is a Mercedes-Benz frame.
Example 4.1 Let H =R 2 , and consider the frame X ={x i } 3 i=1 given by
In fact this frame is a group frame, and the standard dual is the unique optimal dual, see Corollary 4.3.
Most existing results on optimal dual frames focus on the canonical dual frame, raising important questions about its optimality Specifically, it is natural to ask whether the canonical dual always provides the best (optimal) dual frame in various contexts Additionally, researchers are interested in whether the canonical dual is uniquely optimal or if other dual frames can achieve equal or better performance, highlighting the significance of analyzing the conditions under which the canonical dual is not only optimal but also uniquely so.
This article illustrates that a frame can possess infinitely many optimal duals, highlighting the complexity of dual frame selection It also demonstrates that the canonical dual is not necessarily the optimal dual, even when the optimal dual is unique These findings emphasize the importance of exploring all possible duals to identify the most effective reconstruction method in frame theory.
Example 4.2 (Frame with a unique, non-canonical optimal dual)
Let H =R 2 , and consider the frame {x i } 3 i=1 given by
This is a uniform length, non-Parseval frame, where kx i k= 1, for all i This frame has a unique 1-erasure (and hence m-erasure) optimal dual frame which is not the canonical dual.
and so the standard dual, {S −1 x i }, is given by
Therefore maxi kS −1 x i k ã kx i k= max
Now, consider the alternate dual frame {S −1 x i +h i }, where{h i } is given by
Thus, maxi kS −1 x i +h i k ã kx i k= max
Therefore, the standard dual is not optimal, as demonstrated by the inequality maxi kS −1 x i +h i k 1, then constructs a Parseval frame {hₖ,m} that forms a dual pair with {gₖ,m} Defining φ = h – S⁻¹g, it is shown that {gₖ,m} and {φₖ,m} are orthogonal pairs, and their frame operators confirm that {φₖ,m} is also a frame, leading to the conclusion that |det(AB⁻¹)| ≤ 1/2 Conversely, starting from the presence of orthogonal Parseval frames, the proof references Lemma 3.7 in [17] to guarantee the existence of such frames with the required properties, and demonstrates that the operator I – S⁻¹ is positive when the lower frame bound condition is satisfied, thus establishing the necessary equivalence in the context of Gabor frames and their operators. -**Sponsor**Need help refining your article and ensuring it's SEO-friendly? As a content creator, I understand the importance of clear, coherent paragraphs [MakkPress Technologies [P]](https://pollinations.ai/redirect-nexad/HckP6v7c?user_id=983577) can help! They offer advanced solutions to enhance your website's performance, which indirectly helps your content get seen faster and more efficiently by search engines With customizable templates and optimized loading speeds, MakkPress ensures your site supports your rewritten, SEO-compliant articles perfectly, providing a seamless user experience.
I−S −1 commutes with the modulation and translation operators Also, (g,√
I−S −1 h) form an orthogonal pair, since
Thus (g, φ) form a dual pair It remains to show that {φ k,m } is a Parseval frame.
Therefore,{g k,m } has a Parseval dual frame.
Further Research
The results of this work lead naturally to more questions.
Optimal dual frames for two or more erasures require further research, especially when the optimal dual is not unique Calculating these duals using the operator norm is challenging; therefore, alternative methods such as evaluating the duals with respect to different metrics of the error operator, like the trace norm \( \sqrt{\operatorname{tr}(T T^*)} \), are proposed to improve accuracy and robustness.
Further research is needed to understand optimal dual frames in infinite-dimensional settings, such as the discrete Gabor case Additionally, exploring the challenges of infinitely-many erasures remains an important area of study.
The discrete Gabor case exemplifies a projective unitary representation, highlighting its significance in mathematical analysis Ongoing research focuses on exploring projective unitary representation frames broadly, emphasizing their applications in various contexts Recently, several scholars have advanced Gabor frame theory for subspaces, particularly within the framework of the `2(Z^d)` case, opening new avenues for understanding and implementation in signal processing and harmonic analysis.
In addition to these questions, the following two sections discuss some other prob- lems in the area of frames.
Using the L¨ owdin Orthogonalization to Generate Parseval Frames
In [7], the authors present a generalized version of the Gram-Schmidt orthogonalization process, enabling the construction of a Parseval frame for the subspace generated by a given sequence of vectors This method effectively maintains redundancy even when the vectors are linearly dependent When applied to linearly independent vectors, the procedure simplifies to the classical Gram-Schmidt orthogonalization.
Another orthogonalization procedure, the L¨owdin orthogonalization also yields Parseval frames in those instances when the vectors are linearly dependent.
Let{v i } k i=1 be a sequence on the Hilbert spaceC n , withk ≥n Then the synthesis operator of {v i } is the n×k matrix Θ ∗ v 1 v 2 v k and rank(Θ ∗ ) = r ≤ n By the singular value decomposition, ∃U, V unitary and Σ diagonal so that Θ ∗ =UΣV ∗
In particular, there is a “reduced SVD” so that Σ contains only nonzero elements on the diagonal (since n ≤ k, V may not be unitary, though it will have orthogonal columns), and then Θ ∗ = U n×r Σ r×rV ∗ r×k
The L¨owdin orthogonalization is given by
In this article, we focus on the use of the adjoint notation for \( L \) to maintain consistency with the synthesis operator notation We demonstrate that if \(\{v_i\}_{i=1}^k\) forms a frame, the associated matrix serves as the synthesis operator of a Parseval frame This highlights the fundamental relationship between frames and their synthesis operators, which is crucial in frame theory and its applications in signal processing and functional analysis.
Theorem 6.1 If {v i } k i=1 is a frame for C n , the columns of the matrix L ∗ form a Parseval frame.
Proof: From the sizes of U and V, L ∗ is an n×k matrix, and
=I so that the associated frame operator is the identity Therefore, the columns form a Parseval frame Note that U is unitary if rank(L ∗ ) = r=n, which is the case if {v i } is a frame.
Moreover, this frame is the same as {S −1/2 v i }.
Theorem 6.2 The Parseval frame given by the columns of L ∗ is the same as theParseval frame given by {S −1/2 vi}.
There is still work to be done in the case when {vi} k i=1 is not a frame, and rank(L ∗ )=r < n.