Fractional Biorthogonal wavelets in L 2 (R)
aa r X i v : . [ m a t h . F A ] A ug Fractional Biorthogonal wavelets in L ( R ) Owais Ahmad ∗ , N. A. Sheikh and Firdous A. Shah ∗ Department of Mathematics, National Institute of Technology, Hazratbal, Srinagar -190006, Jammu and Kashmir, India. E-mail: [email protected] Department of Mathematics, National Institute of Technology, Hazratbal, Srinagar -190006, Jammu and Kashmir, India.E-mail: [email protected] Department of Mathematics, University of Kashmir, South Campus, Anantnag-192 101, Jammuand Kashmir, India. E-mail: [email protected]
Abstract.
The fractional Fourier transform (FrFT), which is a generalization of the Fouriertransform, has become the focus of many research papers in recent years because of itsapplications in electrical engineering and optics. In this paper, we introduce the notion offractional biorthogonal wavelets on R and obtain the necessary and sufficient conditionsfor the translates of a single function to form the fractional Riesz bases for their closedlinear span. We also provide a complete characterization for the fractional biorthogonalityof the translates of fractional scaling functions of two fractional MRAs and the associ-ated fractional biorthogonal wavelet families. Moreover, under mild assumptions on thefractional scaling functions and the corresponding fractional wavelets, we show that thefractional wavelets can generate Reisz bases for L ( R ) . Keywords:
Frame; Fractional Biorthogonal wavelets; Fractional MRA; Fractional Fouriertransform.
1. Introduction
The Fourier transform has been used for more than a century in a wide range of appli-cations. However, more recently, it was shown that the Fourier transform is inadequatefor describing some physical applications or dealing with their underlying mathematicalproblems. As a result, some off-shoots of the Fourier transform, such as the windowedFourier transform, the wavelet transform, and the fractional Fourier transform (FrFT)have been introduced to address the shortcoming of the Fourier transform. The FrFT,which is a generalization of the Fourier transform, has gained considerable attention inthe last 20 years or so because of its important applications in signal analysis, optics,and signal recovery and also because of its ability to treat some mathematical problemsthat could not otherwise be handled by the standard Fourier transform [8]. The FrFTappeared implicitly in the work of N. Wiener in 1929 [12] as a way to solve certain typesof ordinary and partial differential equations arising in quantum mechanics. Unaware ofWieners work, V. Namias in 1980 [7] introduced the transform, which he called the FrFT,also to solve ordinary and partial differential equations arising in quantum mechanicsfrom classical quadratic Hamiltonians. His work was later refined by McBride and Kerr[5].Besides lot of advantages, the FrFT has one major drawback due to using global kerneli.e., the fractional Fourier representation only provides such FrFT spectral content witho indication about the time localization of the FrFT spectral components. On the otherhand, the short-time FrFT has rectified almost all the limitations of FrFT, still in somecases short-time FrFT is also not applicable as in the case of real signals having highspectral components for short durations and low spectral components for long durations.Therefore, in order to obtain joint signal representations in both time and FrFT domains,Mendlovic et al. [6] first introduced the fractional wavelet transform (FrWT) in the con-text of time-frequency analysis. The FrWT inherits the excellent mathematical propertiesof wavelet transform and FrFT along with some fascinating properties of its own. Theidea behind this transform is deriving the fractional spectrum of the signal by using theFrFT and performing the wavelet transform of the fractional spectrum. Besides being ageneralization of the wavelet transform, the FrWT can be interpreted as a rotation of thetimefrequency plane and has been proved to relate to other time-varying signal analysistools, which make it as a unified timefrequency transform. In recent years, this transformhas been paid a considerable amount of attention, resulting in many applications in theareas of optics, quantum mechanics, pattern recognition and signal processing. For moreabout fractional wavelet transforms and their applications to signal and image processing,we refer to [10, 4, 6, 9].Along with the study of wavelet transforms, there had been a continuing research ef-fort in the study of biorthogonal wavelets and their promising features in applications haveattracted a great deal of interest in recent years to extensively study them. During thelate 1990s, biorthogonal wavelets brought a major breakthrough into image compression,thanks to their natural feature of concentrating energy in a few transform coefficients.In traditional wavelet theory, biorthogonal wavelets have many advantages over orthogo-nal wavelets, by relaxing orthonormal to biorthogonal, additional degrees of freedom areadded to design problems. Biorthogonal wavelets in L ( R ) were investigated by Bownikand Garrigos [1], Cohen et al. [3], Chui and Wang [2] and many others.Although there are many results for biorthogonal wavelets on the real-line R , thecounterparts on the fractional case are not reported yet in the literature. So this paper isconcerned with the construction of fractional biorthogonal wavelets on R . We introducethe notion of fractional biorthogonal wavelets on R and obtain the necessary and sufficientconditions for the translates of a single function to form the fractional Riesz bases fortheir closed linear span. We also provide a complete characterization for the fractionalbiorthogonality of the translates of fractional scaling functions of two fractional MRAs andthe associated fractional biorthogonal wavelet families. Moreover, under mild assumptionson the fractional scaling functions and the corresponding fractional wavelets, we show thatthe fractional wavelets can generate Reisz bases for L ( R ) . The article is structured in the following manner. In Section 2, we recall the basicdefinitions of fractional Fourier transform and fractional wavelet transform. In Section 3,we establish necessary and sufficient conditions for the translates of a function to forma fractional Riesz basis for its closed linear span. In section 4, we give the definition ofa fractional MRA. We also define the projection operators associated with the fractionalMRAs and show that they are uniformly bounded on L ( R ). In the concluding Section, weshow that the fractional wavelets associated with fractional dual MRAs are biorthogonaland generate Riesz bases for L ( R ) . . Fractional Fourier and wavelet transforms This section gives the basic background to the theory of fractional Fourier and wavelettransforms which is as follows.In 1980, Victor Namias [7] introduced the concept of fractional Fourier transform(FrFT) as a generalization of the conventional Fourier transform to solve certain problemsarising in quantum mechanics. It is also referred as rotational Fourier transform or angularFourier transform since it depends on a parameter α which is interpreted as a rotation byan angle α in the time-frequency plane. Like the ordinary Fourier transform correspondsto a rotation in the time frequency plane over an angle α = 1 × π/
2, the FrFT correspondsto a rotation over an arbitrary angle α = ρ × π/ ρ ∈ R .The fractional Fourier transform with parameter α of function f ( t ) is defined by F α (cid:8) f ( t ) (cid:9) ( ξ ) = ˆ f α ( ξ ) = Z ∞−∞ K α ( t, ξ ) f ( t ) dt, (2 . K α ( t, ξ ) is the so-called kernel of the FrFT given by K α ( t, ξ ) = C α exp n i ( t + ξ ) cot α − itξ csc α o , α = nπ,δ ( t − ξ ) , α = 2 nπ,δ ( t + ξ ) , α = (2 n ± π, (2 . α = ρπ/ F α and C α = (2 πi sin α ) − / e iα/ = r − i cot α π . (2.3)The corresponding inversion formula is given by f ( t ) = Z ∞−∞ K α ( t, ξ ) ˆ f α ( ξ ) dξ, (2 . K α ( t, ξ ) = (2 πi sin α ) / e − iα/ sin α · exp (cid:26) − i ( t + ξ ) cot α itξ csc α (cid:27) = C α exp (cid:26) − i ( t + ξ ) cot α itξ csc α (cid:27) = K − α ( t, ξ ) (2.5)and C α = (2 πi sin α ) / e − iα/ π sin α = r i cot α π = C − α . (2 . Definition 2.1.
A fractional wavelet is a function ψ ∈ L ( R ) which satisfies the followingcondition: C αψ = Z R (cid:12)(cid:12)(cid:12) F α n e − i ( t − ξ ) / α ψ o ( ξ ) (cid:12)(cid:12)(cid:12) | ξ | dξ < ∞ , (2 . F α denotes the FrFT operator.Analogous to the classical wavelets, the fractional wavelets can be obtained from afractional mother wavelet ψ ∈ L ( R ) by the combined action of translation and dilationsas ψ α,a,b ( t ) = 1 √ a ψ (cid:18) t − ba (cid:19) exp (cid:26) − i ( t − b ) cot α (cid:27) (2 . a ∈ R + and b ∈ R are scaling and translation parameters, respectively. If α = π/ ψ α,a,b reduces to the conventional wavelet basis.Note that if ψ ( t ) ∈ L ( R ), then ψ α,a,b ( t ) ∈ L ( R ), k ψ α,a,b k = | a | − Z ∞−∞ (cid:12)(cid:12)(cid:12)(cid:12) ψ (cid:18) t − ba (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) dt = Z ∞−∞ (cid:12)(cid:12) ψ ( y ) (cid:12)(cid:12) dy = (cid:13)(cid:13) ψ (cid:13)(cid:13) . Moreover, the fractional Fourier transform of ψ α,a,b ( t ) is given by F α (cid:8) ψ α,a,b ( t ) (cid:9) = √ a exp (cid:26) i ( b + ξ ) cot α − ib ξ csc α − ia ξ cot α (cid:27) F α n e − i ( · ) cot α/ ψ o ( aξ ) (2 . f ∈ L ( R ) with re-spect to an analyzing wavelet ψ ∈ L ( R ) is defined as W αψ f ( a, b ) = (cid:10) f, ψ α,a,b (cid:11) = 1 √ a Z ∞−∞ f ( t ) ψ (cid:18) t − ba (cid:19) exp (cid:26) i ( t − b ) cot α (cid:27) dt (2 . ψ α,a,b ( t ) ∈ L ( R ) is given by (2.8).The FrWT (2.10) deals generally with continuous functions, i.e. functions which aredefined at all values of the time t . However, in many applications, especially in signalprocessing, data are represented by a finite number of values, so it is important andoften useful to consider the discrete version of the continuous FrWT (2.10). From amathematical point of view, the continuous parameters a and b in (2.8) can be convertedinto a discrete one by assuming that a and b take only integral values. For a gooddiscritization of the wavelets, we choose a = a − j and b = kb a − j , where a and b arefixed positive constants. Hence, the discritized wavelet family is defined as ψ α,j,k ( t ) = a j/ ψ (cid:0) a j t − kb (cid:1) exp ( − i t − (cid:0) kb a − j (cid:1) α ) (2 . j and k are the controlling factors for the dilation and translation,respectively and are contained in a set of integers. For computational efficiency, thediscrete wavelet parameters a = 2 and b = 1 are commonly used so that equation (2.11)becomes F ψ ( j, k ) := (cid:26) ψ α,j,k ( t ) = 2 j/ ψ (cid:0) j t − k (cid:1) e − i t − ( k − j )22 cot α , j, k ∈ Z (cid:27) . (2 . F ψ ( j, k ) is called a fractional wavelet frame , if there existpositive constants A and B such that A (cid:13)(cid:13) f (cid:13)(cid:13) ≤ X j ∈ Z X k ∈ Z (cid:12)(cid:12)(cid:10) f, ψ α,j,k (cid:11)(cid:12)(cid:12) ≤ B (cid:13)(cid:13) f (cid:13)(cid:13) , (2 . f ∈ L ( R ), and we call the optimal constants A and B the lower framebound and the upper frame bound, respectively. A tight fractional wavelet frame refersto the case when A = B , and a Parseval frame refers to the case when A = B = 1. Onthe other hand if only the right hand side of the above double inequality holds, then wesay F ψ ( j, k ) a Bessel system .
3. Reisz Bases of TranslatesDefinition 3.1.
Let { ψ m : m ∈ Z } and { e ψ n : n ∈ Z } be two collections of functions in L ( R ). We say that they are orthogonal if D ψ n , e ψ m E = δ n,m ∀ m, n ∈ Z . Definition 3.2.
A collection of functions { ψ n : n ∈ Z } in L ( R ) is said to be linearlyindependent if there exists a coefficient sequence a [ n ] ∈ ℓ ( Z ) such that ∞ X n =1 a [ n ] ψ n = 0 in L ( R ) , then a [ n ] = 0 ∀ n ∈ N . Lemma 3.3.
Let { ψ n : n ∈ Z } be a collection of functions in L ( R ). Suppose thatthere is a collection { e ψ n : n ∈ Z } in L ( R ) which is orthogonal to { ψ n : n ∈ Z } .Then { ψ n : n ∈ Z } is linearly independent. Proof.
Let a [ n ] ∈ ℓ ( Z ) be a coefficient sequence satisfying ∞ X n =1 a [ n ] ψ n = 0 in L ( R ) . Then for each m ∈ N , we have 0 = h , e ψ m i = (cid:10) ∞ X n =1 a [ n ] ψ n , e ψ n (cid:11) = ∞ X n =1 a [ n ] h ψ n , e ψ n i = a [ m ] . Hence { ψ n : n ∈ Z } is linearly independent. Definition 3.4.
A collection of functions { g n ( x ) } in L ( R ) is said to form a Reisz basisfor a Hilbert space H ifa) { g n ( x ) } is linearly independent, and(b) there exists positive constants A, B such that A k f k ≤ ∞ X n =1 |h f, g n i| ≤ B k f k ∀ f ∈ H . In the following lemma, we establish a necessary and sufficient condition for the trans-lates of two functions to be biorthogonal in fractional sense.
Lemma 3.5.
Let the functions φ α ( t − n ) and e φ α ( t − n ) in L ( R ) are given. Then { φ α ( t − n ) : n ∈ Z } is biorthogonal to { e φ α ( t − n ) : n ∈ Z } if and only if X k ∈ Z Θ α ( u + 2 kπ sin α ) e Θ α ( u + 2 kπ sin α ) = 1sin α where Θ α ( u ) is FrFT of φ ( t ). Proof. F α ( φ α,n ( t )) ( u ) = Z ∞−∞ φ ( t − n ) e − j/ t − n − ( t − n ) ) cot α K α ( u, t ) du = A α Z ∞−∞ φ ( t − n ) e j/ t − n ) + u ] cot α − jtu csc α + jn / α du = e jn / α − jnu csc α A α Z ∞−∞ ψ ( t − n ) e j/ t − n ) + u ] cot α − j ( t − n ) u csc α du = e jn / α − jnu csc α Θ α ( u ) . (3.1)Now from the Parseval identity of FrFT, we have D φ α,n ( t ) e φ α,m ( t ) E = hF α ( φ α,n ( t ))( u ) , F α ( φ α,m ( t ))( u ) i = (cid:28) e jn cot α − jnu csc α Θ α ( u ) , e jm cot α − jmu csc α e Θ α ( u ) (cid:29) = Z ∞−∞ e j ( n − m ) cot α − j ( n − m ) u csc α Θ α ( u ) f Θ α ( u ) du. (3.2)Since D φ α,n ( t ) , e φ α,m ( t ) E = δ n,m ∀ n, m (3 . n = n − m, then from (3 .
2) and (3 .
3) it follows that Z ∞−∞ e − jnu csc α Θ α ( u ) e Θ α ( u ) = δ n, (3 . X k ∈ Z Z π sin α e − jn ( u +2 kπ sin α ) csc α Θ α ( u + 2 kπ sin α ) e Θ α ( u + 2 kπ sin α ) du = Z π sin α e − jnu csc α X k ∈ Z Θ α ( u + 2 kπ sin α ) e Θ α ( u + 2 kπ sin α ) du = δ n, . (3.5)et L ( u ) = X k ∈ Z Θ α ( u + 2 kπ sin α ) e Θ α ( u + 2 kπ sin α ) , then we have L ( u + 2 π sin α ) = X k ∈ Z Θ α ( u + 2( k + 1) π sin α ) e Θ α ( u + 2( k + 1) π sin α ) . (2 . k ′ = k + 1 in (3 . L ( u ) = X k ′ ∈ Z Θ α ( u + 2 k ′ π sin α ) e Θ α ( u + 2 k ′ π sin α )= L ( u ) . (3.7)It clearly implies that L ( u ) is 2 kπ sin α periodic function. Therefore from (3 . Z π sin α e − jnu csc α L ( u ) du = δ n, = δ n csc α, . (3 . n ′ = n csc α in (3 . Z π sin α L ( u ) e − jn ′ u du = δ n ′ , , (3 . π sin α Z π sin α L ( u ) e − jn ′ u du = 12 π sin α δ n ′ , , which further implies, L ( u ) = F − (cid:26) π sin α δ n ′ , (cid:27) ( u ) = 1sin α . Thus we have X k ∈ Z Θ α ( u + 2 kπ sin α ) e Θ α ( u + 2 kπ sin α ) = 1sin α . (cid:3) The following result etablishes a sufficient condition for the translates of a function to belinearly independent.
Lemma 3.6.
Let φ α ( t ) ∈ L ( R ). Assume that there exist constants C , C > C ≤ X k ∈ N | Θ α ( u + 2 kπ sin α ) | ≤ C ∀ u ∈ R . (3 . { φ α ( t − n ) : n ∈ Z } is linearly independent. Proof.
By virtue of Lemma 3.3, it suffices to find a function e φ α ( t ) whose translates arebiorthogonal to the translates of φ α . We define e φ α by F α { e φ α ( t ) } ( u ) = Θ α ( u ) X k ∈ Z | Θ α ( u + 2 kπ sin α ) | . y virtue of (3 . X m ∈ N Θ α ( u + 2 mπ sin α ) e Θ α ( u + 2 kπ sin α )= X m ∈ N Θ α ( u + 2 mπ sin α ) Θ α ( u + 2 kπ sin α ) X k ∈ N | Θ α ( u + 2 kπ sin α + 2 mπ sin α ) | = X m ∈ N | Θ α ( u + 2 mπ sin α ) | X ℓ ∈ N | Θ α ( u + 2 ℓπ sin α ) | = 1 . It clearly implies { φ α ( t − n ) } is biorthogonal to { e φ α ( t − n ) } . (cid:3) . Lemmma 3.7.
Suppose that φ α ( t ) satisfies (3 . f in span { φ α ( t − n ) : n ∈ Z } is of the form f = X n ∈ N a [ n ] φ α ( t − n ) , where a [ n ] ∈ ℓ ( Z ) is a finite sequence. Let e a α ( u ) be the discrete FrFT of a [ n ]. Then C Z π sin α | e a α ( u ) | du ≤ k f k ≤ C Z π sin α | e a α ( u ) | du. Proof.
By Plancherel theorem, we have Z ∞−∞ | f ( t ) | dt = Z ∞−∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X n ∈ N a [ n ] φ α ( t − n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) dt = Z ∞−∞ | Θ α ( u ) | | e a α [ u ] | du = Z π sin α X k ∈ N | Θ α ( u + 2 kπ sin α ) e a α [ u ] | du. By invoking (3 . Lemma 3.8.
Let { φ α ( t − n ) : n ∈ Z } be a Reisz basis for its closed linear span.Suppose that there exists a function e φ α such that { e φ α ( t − n ) : n ∈ Z } is biorthogonal to { φ α ( t − n ) : n ∈ Z } . Then(a) for every f ∈ span { φ α ( t − n ) : n ∈ Z } , we have f = X n ∈ Z D f, e φ α ( t − n ) E φ α ( t − n ); (3 . A, B > f ∈ span { φ α ( t − n ) : n ∈ Z } , wehave A k f k ≤ ∞ X n =1 (cid:12)(cid:12)(cid:12)D f, e φ α ( t − n ) E(cid:12)(cid:12)(cid:12) ≤ B k f k . (3 . roof. Since { φ α ( t − n ) : n ∈ Z } forms a Riesz basis for its closed linear span, thenthere exist constants C and C such that (3 .
10) holds. First establish the results for f ∈ span { φ α ( t − n ) : n ∈ Z } and we generalize the established results to span { φ α ( t − n ) : n ∈ Z } .Let f ∈ span { φ α ( t − n ) : n ∈ Z } , then there exists a finite sequence a [ n ] such that f = ∞ X n =1 a [ n ] φ α ( t − n ) . By the definition of biorthogonality, we have D f, e φ α ( t − k ) E = * ∞ X n =1 a [ n ] φ α ( t − n ) , e φ α ( t − k ) + = ∞ X n =1 a [ n ] D φ α ( t − n ) , e φ α ( t − k ) E = a [ k ] . Thus ( a ) is established for f ∈ span { φ α ( t − n ) : n ∈ Z } .Now we proceed to establish ( b ). Since (3 .
10) is satisfied, by Lemma 2.7, for every f ∈ span { φ α ( t − n ) : n ∈ Z } , we have C − k f k ≤ Z π sin α | e a α [ u ] | du ≤ C − k f k . By Plancherel formula for the Fourier series and the fact a [ n ] = h f, e φ α ( t − n ) i , we have Z π sin α | e a α [ u ] | du = X n ∈ N | a [ n ] | = X n ∈ N (cid:12)(cid:12)(cid:12)D f, e φ α ( t − n ) E(cid:12)(cid:12)(cid:12) , Thus ( b ) is obtained.Finally we proceed to generalize the results to span { φ α ( t − n ) : n ∈ Z } . we firstestablish ( b ). For f ∈ span { φ α ( t − n ) : n ∈ Z } , there exists a sequence f [ m ] ∈ ℓ ( N ) in span { φ α ( t − n ) : n ∈ Z } such that lim m →∞ f [ m ] = f. Hence for each n ∈ N , we have D f [ m ] , e φ α ( t − n ) E → D f, e φ α ( t − n ) E as m → ∞ . The result holds for each f [ m ]. Hence, N X n = − N (cid:12)(cid:12)(cid:12)D f, e φ α ( t − n ) E(cid:12)(cid:12)(cid:12) = N X n = − N lim m →∞ (cid:12)(cid:12)(cid:12)D f [ m ] , e φ α ( t − n ) E(cid:12)(cid:12)(cid:12) = lim m →∞ N X n = − N (cid:12)(cid:12)(cid:12)D f [ m ] , e φ α ( t − n ) E(cid:12)(cid:12)(cid:12) ≤ B lim m →∞ k f [ m ] k = B k f k . etting N → ∞ in the above expression, we obtain X n ∈ Z (cid:12)(cid:12)(cid:12)D f, e φ α ( t − n ) E(cid:12)(cid:12)(cid:12) ≤ B k f k Hence the upper bound in (3 .
12) holds. Now by the Cauchy Schwarz inequality forsequences, we have for each m ∈ Z (X n ∈ Z (cid:12)(cid:12)(cid:12)D f [ m ] , e φ α ( t − n ) E(cid:12)(cid:12)(cid:12) ) / ≤ (X n ∈ Z (cid:12)(cid:12)(cid:12)D f [ m ] − f, e φ α ( t − n ) E(cid:12)(cid:12)(cid:12) ) / + (X n ∈ Z (cid:12)(cid:12)(cid:12)D f, e φ α ( t − n ) E(cid:12)(cid:12)(cid:12) ) / Since the upper bound in (3 .
12) holds for each f [ m ] − f and the lower bound holds foreach f [ m ], we have A / k f [ m ] k ≤ B / k f [ m ] − f k + (X n ∈ Z (cid:12)(cid:12)(cid:12)D f, e φ α ( t − n ) E(cid:12)(cid:12)(cid:12) ) / Taking limits as m → ∞ , we get A k f k ≤ X n ∈ N (cid:12)(cid:12)(cid:12)D f, e φ α ( t − n ) E(cid:12)(cid:12)(cid:12) which is the upper bound in (3 . a ) for f ∈ span { φ α ( t − n ) : n ∈ Z } . Let ǫ > g ∈ span { φ α ( t − n ) : n ∈ Z } such that k f − g k < ǫ. Since ( a ) holds for every g , therefore forlarge N, M ∈ N , we have f − N X n = − M D f, e φ α ( t − n ) E φ α ( t − n )= f − g + N X n = − M D g, e φ α ( t − n ) E φ α ( t − n ) − N X n = − M D f, e φ α ( t − n ) E φ α ( t − n )= f − g + N X n = − M D g − f, e φ α ( t − n ) E φ α ( t − n ) . Hence, by Cauchy Schwarz Inequality, we have (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) f − N X n = − M D f, e φ α ( t − n ) E φ α ( t − n ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ k f − g k + (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X n = − M D g − f, e φ α ( t − n ) E φ α ( t − n ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ k f − g k + p C ( N X n = − M (cid:12)(cid:12)(cid:12)D g − f, e φ α ( t − n ) E(cid:12)(cid:12)(cid:12) ) / ≤ k f − g k + p C √ B k f − g k < (1 + p C B ) ǫ. ince ǫ is arbitrary, the result follows. (cid:3)
4. Fractional Multiresolution Analysis Associated with Fractional Wavelets
As in case of conventional wavelets, there corresponds a multiresolution analysis. In thesimilar way fractional wavelets give rise to fractional multiresolution analysis.
Definition 4.1.
An MRA associated with the fractional wavelet transform is defined asa sequence of closed subspaces { V αk } ∈ L ( R ) such that(i) V αk ⊆ V αk +1 , k ∈ Z ;(ii) S k ∈ Z V αk is dense in L ( R );(iii) T k ∈ Z V αk = { } ;(iv) f ( t ) ∈ V αk if and only if f (2 t ) e j [(2 t ) − t ] cot α ∈ V αk +1 , k ∈ Z ;(v) there is a function φ ∈ V α called scaling function such that { φ α, ,n = φ ( t − n ) e − j ( tn + n ) cot α : n ∈ Z } is an orthonormal basis of subspace V α .In the above definition, if we assume that the set of functions { φ α, ,n : n ∈ Z } form aReisz basis of V α , then φ ( t ) generates a generalized fractional MRA { V αm } of L ( R ), then φ α,m,n ( t ) = 2 m φ (2 m t − n ) e − j [ t − (2 − m n ) − (2 m t − n ) ] cot α is the orthonormal basis of { V αm } . Theorem 4.2.
Let φ ∈ L ( R ) such that the collection { φ α, ,n ( t ) : n ∈ Z } is a Reisz basisof the space V α = (X n ∈ Z c [ n ] φ α, ,n ( t ) : c [ n ] ∈ ℓ ( Z ) ) of L ( R ) if and only if there exists positive constants A, B such that for all u ∈ I =[0 , π sin α ], we have A ≤ G ( α, φ, u ) ≤ B (4 . G ( α, φ, u ) = s π sin α X k ∈ Z | Θ α ( u + 2 kπ sin α ) | . (4 . Proof.
For any f ( t ) ∈ V α , we have f ( t ) = X n ∈ Z c [ n ] φ α, ,n ( t ) (4 . c [ n ] ∈ ℓ ( Z ).On taking FrFT on both sides of (4 . F α { f ( t ) } ( u ) = √ π e c α ( u )Θ α ( u ) (4 . e c α ( u ) denotes the discrete FrFT of c [ n ]. By using Parseval formula of the FrFT,we have k f ( t ) k L ( R ) = kF α { f ( t ) } ( u ) k L ( R ) = Z ∞−∞ π | e c α ( u ) | | Θ α ( u ) | du = X k ∈ Z Z π sin α π | e c α ( u + 2 kπ sin α ) | | Θ α ( u + 2 kπ sin α ) | du = Z π sin α | e c α ( u ) | G ( α, φ, u ) du, (4.5)Further, Parsevals formula for discrete FrFT yields k c [ n ] k ℓ ( Z ) = X n ∈ Z | c [ n ] | = Z π sin α | e c α ( u ) | du. (4 . . , (4 .
2) and (4 .
3) yields A k c [ n ] k ℓ ( Z ) ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)X n ∈ Z c [ n ] φ α, ,n ( t ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ B k c [ n ] k ℓ ( Z ) . (4 . .
7) and Definition 3.4. that { φ α, ,n ( t ) : n ∈ Z } is a Reisz basis for V α .In particular { φ α, ,n ( t ) : n ∈ Z } is an orthonormal basis for V α if and only if A = B = 1 . Lemma 4.3.
Let φ be the scaling function for fractional MRA { V αj : j ∈ Z } . Then foreach j ∈ Z , { φ α,j,k : k ∈ Z } is a Reisz basis for V αj . Proof.
Define e φ by F α { e φ ( t ) } ( u ) = Θ α ( u ) q π sin α P k ∈ Z | Θ α ( u + 2 kπ sin α ) | , then by the same arguments as in the proof of Lemma 3.4, { e φ α ( t − k ) : k ∈ Z } isbiorthogonal to { φ α ( t − k ) : k ∈ Z } Hence, D φ α,j,n , e φ α,j,m E = D δ j φ α ( t − n ) , δ j e φ α ( t − m ) E = D φ α ( t − n ) , e φ α ( t − m ) E = δ n,m that is, n e φ α,j,k : k ∈ Z o is biorthogonal to { φ α,j,k : k ∈ Z } for every j ∈ Z . Therefore byLemma 3.3. { φ α,j,k : k ∈ Z } is linearly independent.Now we need to show the collection { φ α,j,k : k ∈ Z } satisfies the frame condition. Forany f ∈ V αj , we have k ∈ Z |h f, φ α,j,k i| = X k ∈ Z |h f, δ j φ α ( t − k ) i| = X k ∈ Z |h δ − j f, φ α ( t − k ) i| . Since { φ α ( t − n ) : k ∈ Z } is a Riesz basis for V α and δ − j f ∈ V α , there exist constants A, B > f ∈ V αj , A k δ − j f k ≤ X k ∈ Z |h δ − j f, φ α ( t − k ) i| ≤ B k δ − j f k . This is equivalent to A k f k ≤ X k ∈ Z |h f, φ α ( t − k ) i| ≤ B k f k . Hence, { φ α,j,k : k ∈ Z } satisfies the frame condition. Lemma 4.4.
Suppose that (cid:8) V αj : j ∈ Z (cid:9) is a fractional MRA with scaling function φ .Then there exists a sequence { h [ n ] } n ∈ Z in ℓ ( Z ) called the scaling filter such that φ ( t ) = X n ∈ Z h [ n ] √ φ (2 t − n ) e − j [ t − ( n ) − (2 t − n ) ] cot α and a 2 kπ sin α -periodic function Λ α called the auxilliary function such thatΘ α ( u ) = Λ α (cid:16) u (cid:17) Θ α (cid:16) u (cid:17) . Proof.
By Lemma 4.3, { φ α, ,n : n ∈ Z } is a Riesz basis for V α . Since φ α, , ( t ) ∈ V α ⊆ V α , therefore by virtue of (3 .
11) there must exist a sequence { h [ n ] } n ∈ Z in ℓ ( Z ) such that φ α, , = X n ∈ Z h [ n ] φ α, ,n ( t )which can be simplified as φ ( t ) = X n ∈ Z h [ n ] √ φ (2 t − n ) e − j [ t − ( n ) − (2 t − n ) ] cot α (4 . h [ n ] = √ Z ∞−∞ φ ( t ) φ ∗ (2 t − n ) e j [ t − ( n ) − (2 t − n ) ] cot α dt. (4 . α ( u ) = X n ∈ Z h [ n ] √ A α Z ∞−∞ φ (2 t − n ) e j [ t − ( n ) − (2 t − n ) ] cot α − jtu csc α dt = 1 √ e ju cot α X n ∈ Z h [ n ] e jn cot α − jnu csc α × A α Z ∞−∞ φ (2 t − n ) e j [ t − ( n ) − (2 t − n ) ] cot α − j (2 t − n ) u csc α d (2 t − n )= 1 √ e ju cot α X n ∈ Z h [ n ] e jn cot α − jnu csc α Θ α (cid:16) u (cid:17) = 1 √ e ju cot α D α (cid:16) u (cid:17) Θ α (cid:16) u (cid:17) , (4.10)where D α ( u ) = X n ∈ Z h [ n ] e jn cot α − jnu csc α . By defining Λ α ( u ) = 1 √ e ju cot α D α ( u )= 1 √ X n ∈ Z f [ n ] A α e jn cot α − jnu csc α . Eq.(4.10) can be written as Θ α ( u ) = Λ α (cid:16) u (cid:17) Θ α (cid:16) u (cid:17) . It is to be noted that Λ α ( u ) is a 2 kπ sin α -periodic function since we haveΛ α ( u + 2 kπ sin α ) = 1 √ X n ∈ Z f [ n ] A α e jn cot α − jn ( u +2 kπ sin α ) csc α = 1 √ X n ∈ Z f [ n ] A α e jn cot α − jnu csc α = Λ α ( u ) . Definition 4.5.
A pair of fractional MRAs (cid:8) V αj : j ∈ Z (cid:9) and n e V αj : j ∈ Z o with scalingfunctions φ and e φ respectively are said to be dual to each other if { φ α ( t − k ) : k ∈ Z } and n e φ α ( t − k ) : k ∈ Z o are biorthogonal. Definition 4.6.
Let φ and e φ be scaling functions for dual MRAs. For each j ∈ Z wedefine the operators P α,j , e P α,j on L ( R ) by P α,j f = X k ∈ Z D f, e φ α,j,k E φ α,j,k , P α,j = X k ∈ Z h f, φ α,j,k i e φ α,j,k . Lemma 4.7.
The operators P α,j , e P α,j are uniformly bounded. Proof.
Since the translates of φ and e φ form Riesz basis for their closed linear spans,therefor by Lemma 4.2, there exist constants C and C such that c ≤ X k ∈ Z | Θ α ( u + 2 kπ sin α ) | ≤ C and c ≤ X k ∈ Z (cid:12)(cid:12)(cid:12) e Θ α ( u + 2 kπ sin α ) (cid:12)(cid:12)(cid:12) ≤ C . For a sequence { c [ k ] } k ∈ Z ∈ ℓ ( Z ), there exists B > (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)X k ∈ Z c [ k ] φ α, ,k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ X k ∈ Z | c [ k ] | . Now for any f ∈ L ( R ), we have X k ∈ Z |h f, φ α, ,k i| = X k ∈ Z (cid:12)(cid:12)(cid:12)(cid:12)Z ∞−∞ F α { f ( t ) } ( u )Θ α ( u ) e jn cot α − jnu csc α du (cid:12)(cid:12)(cid:12)(cid:12) = Z π sin α (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X ℓ ∈ Z F α { f ( t ) } ( u + 2 πℓ sin α )Θ α ( u + 2 πℓ sin α ) du (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Z π sin α (X ℓ ∈ Z |F α { f ( t ) } ( u + 2 πℓ sin α ) | ) (X ℓ ∈ Z | Θ α ( u + 2 πℓ sin α ) | ) du ≤ Z π sin α X ℓ ∈ Z |F α { f ( t ) } ( u + 2 πℓ sin α ) | du = C Z ∞−∞ |F α { f ( t ) } ( u ) | du = C k f k . Similar estimates hold for e φ . Hence for f ∈ L ( R ), we have kP α, f k = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)X k ∈ Z D f, e φ α, ,k E φ α, ,k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ B X k ∈ Z (cid:12)(cid:12)(cid:12)D f, e φ α, ,k E(cid:12)(cid:12)(cid:12) ≤ BC k f k . hus P α, is a bounded operator on L ( R ) with norm at most √ BC = C, (say). Sincethe dilation operators are unitary and since P α,j f = X k ∈ Z D f, e φ α,j,k E φ α,j,k = X k ∈ Z D δ − j f, e φ α, ,k E δ − j φ α, ,k , we conclude that the operator norm of P α,j is at most C . Similar arguments work for e P α,j . This finishes the proof of the lemma.Now we proceed to prove some useful properties of the operators P α,j and e P α,j . Lemma 4.8.
The operators P α,j and e P α,j satisfy the following properties(a) P α,j f = f if and only if f ∈ V αj and e P α,j = f if only if f ∈ e V αj ;(b) lim j →∞ kP α,j f − f k = 0 and lim j →−∞ kP α,j f k = 0 for every f ∈ L ( R ) . Proof. (a) P α,j f = f if only if f = P n ∈ Z D f, e φ α,j,n E φ α,j,n . Since { φ α,j,n : n ∈ Z } is aRiesz basis for V αj and n e φ α,j,n : n ∈ Z o is biorthogonal to { φ α,j,n : n ∈ Z } . By Lemma3.8, f = P n ∈ Z D f, e φ α,j,n E φ α,j,n if and only if f ∈ span { φ α,j,n : n ∈ Z } = V αj . Similarargument applies for e P α,j f .(b) It is straight forward.
5. Biorthogonality of fractional waveletsDefinition 5.1.
Let φ and e φ be the scaling functions for dual fractional MRAs. Definethe fractional wavelet ψ and and e ψ by ψ ( t ) = X n ∈ Z g [ n ] √ φ (2 t − n ) e − j [ t − ( n/ − (2 t − n ) ] cot α (5 . e ψ ( t ) = X n ∈ Z e g [ n ] √ e φ (2 t − n ) e − j [ t − ( n/ − (2 t − n ) ] cot α . (5 . Lemma 5.2
Let ψ and e ψ be the wavelet and dual wavelet corresponding to the fractionalMRA’s { V αj } with the scaling function φ and { e V αj } with the scaling function e φ . Then thefollowing hold(a) ψ ∈ V α and e φ ∈ e V α .(b) n e ψ α, ,n : n ∈ Z o is bi orthogonal to { ψ α, ,n : n ∈ Z } .(c) For all m, n ∈ Z , we have D ψ α, ,n , e φ α, ,m E = D e ψ α, ,n , φ α, ,m E = 0 . roof. (a) This clearly follows from the definition of ψ and e ψ .(b) Taking the FrFT on both sides of (5 .
1) and (5 .
2) gives F α { ψ ( t ) } ( u ) = Γ α (cid:16) u (cid:17) Θ α (cid:16) u (cid:17) ; (5 . F α { e ψ ( t ) } ( u ) = e Γ α (cid:16) u (cid:17) e Θ α (cid:16) u (cid:17) , (5 . α ( u ) = e − jπ ( u +2 kπ sin α ) e Λ α ( u + 2 kπ sin α ); (5 . e Γ α ( u ) = e − jπ ( u +2 kπ sin α ) Λ α ( u + 2 kπ sin α ) . (5 . { φ α, ,n : n ∈ Z } is biorthogonal to { e φ α, ,n : n ∈ Z } ,we have1sin α = X n ∈ Z Θ α ( u + 2 nπ sin α ) e Θ α ( u + 2 nπ sin α )= X n ∈ Z Λ α (cid:16) u kπ sin α (cid:17) Θ α (cid:16) u nπ sin α (cid:17) × e Λ α (cid:16) u kπ sin α (cid:17) e Θ α (cid:16) u nπ sin α (cid:17) = Λ α (cid:16) u (cid:17) e Λ α (cid:16) u (cid:17) X k ∈ Z Θ α (cid:16) u kπ sin α (cid:17) e Θ α (cid:16) u kπ sin α (cid:17) + Λ α (cid:16) u kπ sin α (cid:17) e Λ α (cid:16) u kπ sin α (cid:17) X k ∈ Z Θ α (cid:16) u k + 1) π sin α (cid:17) × e Θ α (cid:16) u k + 1) π sin α (cid:17) = Λ α (cid:16) u (cid:17) e Λ α (cid:16) u (cid:17) α + Λ α (cid:16) u kπ sin α (cid:17) e Λ α (cid:16) u kπ sin α (cid:17) α . (4.7)Combining (5.5), (5.6) and (5.7) givesΓ α (cid:16) u (cid:17) e Γ α (cid:16) u (cid:17) α + Γ α (cid:16) u kπ sin α (cid:17) e Γ α (cid:16) u kπ sin α (cid:17) α = 1 . (5 . X n ∈ Z F α { ψ } ( u + 2 nπ sin α ) F α { e ψ } ( u + 2 nπ sin α )= Γ α (cid:16) u (cid:17) e Γ α (cid:16) u (cid:17) α + Γ α (cid:16) u kπ sin α (cid:17) e Γ α (cid:16) u kπ sin α (cid:17) α = 1sin α . (5.9)herefore by Lemma 3.5., { ψ α, ,n : n ∈ Z } is biorthogonal to n e ψ α, ,n : n ∈ Z o .(c) For fixed m, n ∈ Z , we have by Plancherel’s formula, we have D ψ α, ,n , e φ α, ,m E = Z ∞−∞ F α { ψ } ( u ) e Θ α ( u ) e j ( n − m ) cot α − j ( n − m ) u csc α du = Z ∞−∞ Γ α (cid:16) u (cid:17) e Λ α (cid:16) u (cid:17) Θ α (cid:16) u (cid:17) e Θ α (cid:16) u (cid:17) e j ( n − m ) cot α − j ( n − m ) u csc α du = 1sin α Z π sin α e j ( n − m ) cot α − j ( n − m ) u csc α X k ∈ Z Γ α (cid:16) u kπ sin α (cid:17) × e Λ α (cid:16) u kπ sin α (cid:17) Θ α (cid:16) u kπ sin α (cid:17) e Θ α (cid:16) u kπ sin α (cid:17) du = 1sin α Z π sin α e j ( n − m ) cot α − j ( n − m ) u csc α ( Γ α (cid:16) u (cid:17) e Λ α (cid:16) u (cid:17) × X ℓ ∈ Z Θ α (cid:16) u ℓπ sin α (cid:17) e Θ α (cid:16) u ℓπ sin α (cid:17) + Γ α (cid:16) u ℓπ sin α (cid:17) e Λ α (cid:16) u ℓπ sin α (cid:17) × X ℓ ∈ Z Θ α (cid:16) u ℓ + 1) π sin α (cid:17) e Θ α (cid:16) u ℓ + 1) π sin α (cid:17)) du = 1sin α Z π sin α e j ( n − m ) cot α − j ( n − m ) u csc α (cid:26) Γ α (cid:16) u (cid:17) e Λ α (cid:16) u (cid:17) + Γ α (cid:16) u ℓπ sin α (cid:17) e Λ α (cid:16) u ℓπ sin α (cid:17)(cid:27) du = 0 . Similarly, we can show that
D e ψ α, ,n , φ α, ,n E = 0 for all n, m ∈ Z . Now our main objective is to show that the wavelets associated with the dual frac-tional MRAs are biorthogonal and also they form Riesz basis for L ( R ) . For that thefollowing proposition is very useful.
Proposition 5.3.
Let φ and e φ be the scaling functions for dual fractional MRAs and ψ, e ψ be the associated wavelets satisfying the matrix condition M α ( u ) f M α ( u ) = I (5 . M α ( u ) = (cid:20) Λ α ( u ) Λ α ( u + 2 π sin α )Γ α ( u ) Γ α ( u + 2 π sin α (cid:21) (5 . ψ = φ and e ψ = e φ . Then for every f ∈ L ( R ), we have P α, f = P α, f + X k ∈ Z D f, e ψ α, ,k E ψ α, ,k (5 . e P α, f = e P α, f + X k ∈ Z h f, ψ α, ,k i e ψ α, ,k (5 . L ( R ) Proof.
We will prove only (5.12) as the proof of (5.13) follows in the similar manner.Further it suffices to prove (5.12) in the weak sense, that is, for all f, g ∈ L ( R ), hP α, f, g i = hP α, f, g i + X k ∈ Z D f, e ψ α, ,k E h g, ψ α, ,k i = X k ∈ Z D f, e ψ α, ,k E h g, ψ α, ,k i . e have X k ∈ Z D f, e ψ α, ,k E h g, ψ α, ,k i = X k ∈ Z (cid:26)Z ∞−∞ F α { f ( t ) } ( u ) F α { e ψ } ( u ) e jn cot α − jnu csc α du (cid:27) × (cid:26)Z ∞−∞ F α { g ( t ) } ( u ) F α { ψ } ( u ) e − jn cot α + jnu csc α du (cid:27) = X k ∈ Z (Z π sin α X ℓ ∈ Z F α { f ( t ) } ( u + 2 ℓπ sin α ) × F α { e ψ } ( u + 2 ℓπ sin α ) e jn cot α − jn ( u +2 ℓπ sin α ) csc α du (cid:27) × (Z π sin α X ℓ ′ ∈ Z F α { g ( t ) } ( u + 2 ℓ ′ π sin α ) × F α { ψ } ( u + 2 ℓ ′ π sin α ) e − jn cot α + jn ( u +2 ℓ ′ ψ sin α ) csc α du (cid:27) = X k ∈ Z (Z π sin α X ℓ ∈ Z F α { f ( t ) } ( u + 2 ℓπ sin α ) F α { e ψ } ( u + 2 ℓπ sin α ) du ) × (Z π sin α X ℓ ′ ∈ Z F α { g ( t ) } ( u + 2 ℓ ′ π sin α ) F α { ψ } ( u + 2 ℓ ′ π sin α ) du ) = Z π sin α X ℓ ∈ Z F α { f ( t ) } ( u + 2 ℓπ sin α ) e Γ α (cid:16) u ℓπ sin α (cid:17) e Θ α (cid:16) u ℓπ sin α (cid:17) × X ℓ ′ ∈ Z F α { g ( t ) } ( u + 2 ℓ ′ π sin α ) e Γ α (cid:16) u ℓπ sin α (cid:17) Θ α (cid:16) u ℓπ sin α (cid:17) du = Z π sin α X ℓ ∈ Z X ℓ ′ ∈ Z F α { f } ( u + 2 ℓπ sin α ) e Θ α (cid:16) u ℓ sin α (cid:17) × F α { g } ( u + 2 ℓ ′ π sin α )Θ α (cid:16) u ℓ ′ π sin α (cid:17) du. (5.14)In the similar lines, we can obtain X k ∈ Z D f, e φ α, ,k E h g, φ α, ,k i = Z π sin α X s ∈ Z X s ′ ∈ Z F α { f } ( u + 2 sπ sin α ) , e Θ α (cid:16) u sπ sin α (cid:17) × F α { g } ( u + 2 s ′ π sin α )Θ α (cid:16) u s ′ π sin α (cid:17) du. (5.15)Since the right hand sides of (5 .
14) and (5 .
15) are same. This completes the proof. (cid:3) ombining the Proposition 5.3. and Lemma 4.8., we have the following proposition.
Proposition 5.4.
Let φ, e φ and ψ, e ψ be defined as above. Then for every f ∈ L ( R ) , wehave f = X j ∈ Z X j ∈ Z D f, e ψ α,j,k E ψ α,j,k = X j ∈ Z X j ∈ Z h f, ψ α,j,k i e ψ α,j,k (5 . Theorem 5.5.
Let φ and e φ be the scaling functions for Dual fractional MRAs and ψ, e ψ bethe associated wavelets as in the Proposition 4.3. Then the collections { ψ α,j,k : j, k ∈ Z } and { e ψ α,j,k : j, k ∈ Z } are biorthogonal. Further, if | Θ α ( u ) | ≤ C (1 + | u | ) − / − ǫ , | e Θ α ( u ) | ≤ C (1 + | u | ) − / − ǫ , (5 . |F α { ψ } ( u ) | ≤ C | u | , and (cid:12)(cid:12)(cid:12) F α { e ψ } ( u ) (cid:12)(cid:12)(cid:12) ≤ C | u | (5 . C > , ǫ > u ∈ R , then the collections { ψ α,j,k : j, k ∈ Z } and { e ψ α,j,k : j, k ∈ Z } form Riesz basis for L ( R ). Proof.
We begin the proof by proving that the collections { ψ α,j,k : j, k ∈ Z } and n e ψ α,j,k : j, k ∈ Z o are biorthogonal to each other. First we will show that, for j ∈ Z D ψ α,j,k , e ψ α,j,k ′ E = δ k,k ′ . We have already proved it for j = 0, by Lemma 5.2. (b). For j = 0, we have D ψ α,j,k , e ψ α,j,k ′ E = D δ − j ψ α, ,k , δ − j e ψ α, ,k ′ E = D ψ α, ,k , e ψ α, ,k ′ E = δ k,k ′ . Let k, k ′ ∈ Z be fixed and Let j, j ′ ∈ Z . Assume that j < j ′ , we will show that D ψ α,j,k , e ψ α,j ′ ,k ′ E = 0 . It can be shown that ψ α, ,k ∈ V α . Hence, ψ α,j,k = δ − j ψ α, ,k ∈ V αj +1 ⊆ V αj ′ . Therefore, itwill be enough to show that e ψ α,j ′ ,k ′ is orthogonal to every element of V αj ′ . Let f ∈ V αj ′ .By Lemma 4.3, { φ α,j ′ ,k ′ : k ∈ Z } is a Riesz basis for V αj ′ . Hence, there exists a sequence c [ k ] ∈ ℓ ( Z ) such that f = X k ∈ Z c [ k ] φ α,j ′ ,k ′ in L ( R ) . By Lemma 5.2 (c), we have
D e ψ α,j ′ ,k ′ , φ α,j ′ ,k E = D δ − j ′ e ψ α, ,k ′ , δ − j ′ φ α, ,k E = D e ψ α, ,k ′ , φ α, ,k E = 0 . ence, D e ψ α,j ′ ,k ′ , f E = * e ψ α,j ′ ,k ′ , X k ∈ Z c [ k ] φ α,j ′ ,k ′ + = X k ∈ Z c [ k ] D e ψ α,j ′ ,k ′ , φ α,j ′ ,k E = 0 . In order to show that these two collections form Riesz bases for L ( R ), we must verify thatthey are linearly independent and satisfy the frame condition. Since they are biorthogonalto each other,therefore by Lemma 3.3, both the collections are linearly independent.In order to show the frame condition, we must show that there exist constants A, B, e A, and e B > f ∈ L ( R ) , we have A k f k ≤ X j ∈ Z X k ∈ Z |h f, ψ α,j,k i| ≤ B k f k , (5 . , and e A k f k ≤ X j ∈ Z X k ∈ Z (cid:12)(cid:12)(cid:12)D f, e ψ α,j,k E(cid:12)(cid:12)(cid:12) ≤ e B k f k , (5 . . We first establish the existence of upper bounds in (5.18) and (5.19). we have X k ∈ Z |h f, ψ α,j,k i| = X k ∈ Z (cid:12)(cid:12)(cid:12)(cid:12)Z ∞−∞ F α { f } ( u ) F α { ψ } (2 j u ) du (cid:12)(cid:12)(cid:12)(cid:12) = X k ∈ Z (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z π sin α X m ∈ Z F α { f } ( u + 2 j πm sin α ) F α { ψ } (2 j u + 2 mπ sin α ) du (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Z ∞−∞ |F α { f } ( u ) | (cid:12)(cid:12) F α { ψ } (2 j u ) (cid:12)(cid:12) δ X n ∈ Z (cid:12)(cid:12) F α { ψ } (2 j u + 2 nπ sin α ) (cid:12)(cid:12) − δ ) du. We have assumed that | Θ α ( u ) | ≤ C (1 + | u | ) − − ǫ , hence we have |F α { ψ } ( u ) | ≤ C (cid:0) | u | (cid:1) − − ǫ .Therefore, P n ∈ Z |F α { ψ } (2 j u + 2 nπ sin α ) | − δ ) is uniformly bounded if δ < ǫ (1 + 2 ǫ ) − .Hence, there exists C > X j ∈ Z X k ∈ Z |h f, ψ α,j,k i| ≤ C Z ∞−∞ |F α { f } ( u ) | (cid:12)(cid:12) F α { ψ } (2 j u ) (cid:12)(cid:12) δ du ≤ C sup (X j ∈ Z (cid:12)(cid:12) F α { ψ } (2 j u ) (cid:12)(cid:12) δ : u ∈ [0 , π sin α ] ) k f k . urther for 1 < | u | ≤ α , we have X j = −∞ (cid:12)(cid:12) F α { ψ } (2 j u ) (cid:12)(cid:12) δ ≤ ∞ X j =0 C δ (1 + | j − u | ) δ (1+2 ǫ ) ≤ ∞ X j =0 C δ ( j − δ (1+2 ǫ ) = C δ δ (1+2 ǫ ) − − δ (1+2 ǫ ) . Also ∞ X j =1 (cid:12)(cid:12) F α { ψ } (2 j u ) (cid:12)(cid:12) δ ≤ ∞ X j =1 (cid:0) C − j | u | (cid:1) δ ≤ C δ ∞ X j =1 ( − j +1)2 δ = C δ − − δ These two estimates show that sup nP j ∈ Z |F α { ψ } (2 j u ) | δ : u ∈ [0 , π sin α ] o is finite. Hence,there exists B > f ∈ L ( R ), then wehave f = X j ∈ Z X k ∈ Z D f, e ψ α,j,k E ψ α,j,k = X j ∈ Z X k ∈ Z h f, ψ α,j,k i e ψ α,j,k . Therefore, we have k f k = h f, f i = *X j ∈ Z X k ∈ Z D f, e ψ α,j,k E ψ α,j,k , f + = X j ∈ Z X k ∈ Z D f, e ψ α,j,k E h ψ α,j,k , f i≤ (X j ∈ Z X k ∈ Z (cid:12)(cid:12)(cid:12)D f, e ψ α,j,k E(cid:12)(cid:12)(cid:12) ) / (X j ∈ Z X k ∈ Z |h f, ψ α,j,k i| ) / ≤ p e B k f k (X j ∈ Z X k ∈ Z |h f, ψ α,j,k i| ) / . Hence, 1 e B k f k ≤ X j ∈ Z X k ∈ Z |h f, ψ α,j,k i| . imilarly we can show that 1 B k f k ≤ X j ∈ Z X k ∈ Z (cid:12)(cid:12)(cid:12)D f, e ψ α,j,k E(cid:12)(cid:12)(cid:12) . This completes the proof of the theorem.
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