PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 125, Number 11, November 1997, Pages 3363–3370 S 0002-9939(97)04002-1

PERTURBATIONS OF THE HAAR

N. K. GOVIL AND R. A. ZALIK

(Communicated by Palle E. T. Jorgensen)

Abstract. Let m Z+ be given. For any ε>0 we construct a function ε ∈ ε f { } having the following properties: (a) f { } has support in [ ε, 1+ε]. (b) ε m − f { } C ( , ). (c) If h denotes the Haar function and 0 <δ< ,then ε∈ −∞ ∞ δ 1/δ 1/δ ε ∞ f h δ (1 + 2 ) (2ε) .(d)f generates an affine Riesz basis k { }− kL ( ) ≤ {} whose frame boundsR (which are given explicitly) converge to 1 as ε 0. →

Let be a with inner product , and norm := , 1/2.Let + H h· ·i + k·k h· ·i denote the natural numbers. A sequence fn,n Z is called a frame if thereZ are constants A and B such that for every{ f ∈ }⊂H ∈H A f 2 f,f 2 B f 2. k k ≤ |h ni| ≤ k k n Z+ X∈ The constants A and B are called bounds of the frame. If only the right-hand + inequality is satisfied for all f ,then fn,n Z is called a Bessel sequence with bound B. A frame is called∈H exact, or{ a Riesz∈ basis},ifupontheremovalofany single element of the sequence, it ceases to be a frame. However, not every frame + is a Riesz basis: as is well known, a sequence fn,n Z is a Riesz basis if and only if it is the image of an orthonormal{ basis∈ under}⊂H a bounded invertible linear operator U : ([1, 11]). Clearly for an both frame bounds equal 1. H→H In [5] it is shown that adding a Bessel sequence with a small bound to a Riesz basis transforms the original basis into another Riesz basis. It is also shown how frame bounds for the new basis are obtained from frame bounds of the original basis. Given an arbitrary positive integer m, in the present paper we use these results to ε m perturb the Haar function into a function f { } C ( , ) with support in [ ε, 1+ε] (thus having good time and frequency∈ localization)−∞ ∞ that preserves the symmetry− of the Haar wavelet and generates an affine Riesz basis in L2( , ). The lack of precludes the use of the fast .−∞ On∞ the ε other hand the functions f { } are given explicitly in terms of cardinal B-splines. Other do not have a closed form representation and have to be obtained recursively, using the cascade algorithm [4, 10]. Since frame bounds are given

Received by the editors March 18, 1996 and, in revised form, June 21, 1996. 1991 Mathematics Subject Classification. Primary 42C99; Secondary 41A05, 46C99. Key words and phrases. Frames, affine frames, Riesz bases, Haar wavelet, basis perturbations, -bounded mean variation, cardinal splines. ∧ The authors are grateful to Ole Christensen, Sergio J. Favier, Christopher E. Heil, and Luis Miguel Pozo Coronado for their helpful comments.

c 1997 American Mathematical Society

3363

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explicitly, the frame algorithm or other algorithms can be used to reconstruct a function from its wavelet coefficients ([4, 6]). Let denote the integers, therealnumbers,and Z R φ (x)=aj/2φ(ajx bk), j,k − where a>1andb>0. Our construction is based on a sequence of auxiliary propositions, some of which are of independent interest. We begin with:

Lemma 1. Let φ L2( ).If ∈ R φ ,φ B, |h j,k j0,k0 i| ≤ j,k X∈Z for every j0,k0 ,then φj,k j,k is a Bessel sequence with bound B. ∈Z { } ∈Z Proof. From Schur’s lemma ([3, Lemma 5] or [11, p. 159]) we deduce that the 2 2 operator generated by φj,k,φj0,k0 ,j,j0,k,k0 is bounded in ` ( ), and the assertion follows from the{h discussioni in [11, p.∈Z} 159]. Z

Remark. This lemma was implicitly used in the proof of [3, Lemma 9]. See also [7, Proposition 9.10]. The next result is a quantitative version of [2, Lemma 9], and its proof is based on the concept of -bounded mean variation introduced by Shi [8]: ∧ Let := λk,k =1,2,... be a nondecreasing sequence of positive numbers such that∧ {1 = . For any} locally integrable function f and a bounded interval λn ∞ I of length I , define j P| j | 1 f := f(t)dt, Ij I | j | ZIj 1 µ (f):= f(t) f dt. Ij I | − Ij| | j|ZIj The function f is said to be of -bounded mean variation on an interval I,and we write f BMV (I), if ∧ ∈∧ ∞ µ (f) M (f,I)=sup Ij < , ∧ λ ∞ j=1 j X where the supremum is taken over all sequences of nonoverlapping subintervals Ij of I. Let BV ( ) denote the set of functions of bounded variation over .Wehave: R R Lemma 2. Assume that φ satisfies the following conditions: (a) supp φ [c, d]=:I, (b) φ BV⊆( ), (c) ∈φ(t)dtR=0. R If Vφ denotes the total variation of φ on ,then φj,k j,k is a Bessel sequence with BesselR bound R { } ∈Z

√a 1 (a+1) 1 Mφ := φ (d c) [(d c)b− +1]Vφ + b− φ . k k∞ √a 1 − { − √a k k∞} − 

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Proof. Let θj0 ,k0 = φj,k,φj0,k0 .Then j,k |h i| P∈Z θj0,k0 = + =Σ1+Σ2. j j k j

n=1,2,... let αn =infx In φ(x)andβn =supx I φ(x). Then for every ε>0, ∈ ∈ n we can select two points γn and δn in each interval In so that n 1 n 1 (3) φ(γ ) α + ε 2− − ,φ(δ) β ε2−−. n ≤ n n ≥ n− Clearly 1 µ (φ)= φ(x) φ dx β α . In I | − In| ≤ n − n | n|ZIn Hence by (3),

∞ ∞ 1 µ (φ) φ(I0 ) + ε In ≤ | n | 2n n=1 n=1 X X X (4) V + ε, ≤ φ where In0 =[γn,δn]andφ(In0)=φ(γn) φ(δn). Since (4) holds for every ε>0, we obtain− (5) µ (φ) V , In ≤ φ implying that φ BMV (c, d), andX ∈∧ (6) M (φ, I) Vφ, ∧ ≤ with := λk : k =1,2,... and λk =1fork=1,2,.... Thus∧ (2),{ (6) and [3, Lemma} 7] yield µ (φ ) [(d c)b 1 +1]V , Ij j ,k 0,k0 − φ − 0 ≤ − k K(j j ,0,k ) ∈ X− 0 0 and from (1) we conclude that

√a 1 (7) Σ1 φ (d c) [(d c)b− +1]Vφ. ≤k k∞ − √a 1 − −

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From [3, (3.17)]

2 1 (j j0)/2 2 1 (a +1)(d c) (8) Σ2 φ b− (a+1)(d c) a − = φ b− − . ≤k k∞ − k k∞ (√a 1) j

(10) Nm0 (x)=Nm 1(x) Nm 1(x 1), − − − − and

(11) supp Nm =[0,m], whence (a) and (b) follow. Since (11) implies that Nm(m k 1) = 0 for k<0ork>m 2, (c) follows from the identity ([2, Theorem 4.3])− − − (12) N (x k)=1. m − k X∈Z Applying (10) we have: m 2 − g0(x)=χ[0,m 1](x) [Nm 1(x k) Nm 1(x k 1)] (13) − − − − − − − Xk=0 = χ[0,m 1](x)[Nm 1(x) Nm 1(x m +1)], − − − − − and (d) follows from (11).

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Finally, integrating by parts, and using (11), (10) and [2, (3.2.16)] we have: m 1 − ixt √2πgˆ(x)= e− g(t)dt 0 Z m 1 itx m 1 − itx =(i/x) g(t)e− − (i/x) e− g0(t)dt 0 − Z0 ( m 1)xi =(i/x)e − − g(m 1) − m 1 − itx (i/x) e− [Nm 1(t) Nm 1(t m +1)]dt − 0 − − − − Z m 1 (m 1)xi − itx =(i/x)e− − (i/x) e− Nm 1(t)dt − − Z0 (m 1)xi ˆ =(i/x)e− − (√2πi/x)Nm 1(x) − − m 1 (m 1)xi (m 1)xi/2 sin x/2 − =(i/x)e− − (i/x)e− − − x/2   m 1 (m 1)xi/2 (m 1)xi/2 sin x/2 − =(i/x)e− − e− − . − x/2 "   # Separating the expression in brackets into real and imaginary parts, passing m 1 to the limit and noting thatg ˆ(0) = 1 − g(t)dt is real, we conclude that √2π 0 √2πgˆ(0) = m 1 , i.e. g(t)dt = m 1 . 2− 2− R R Lemma 6. Let m R2 and ≥ m 2 1 − q(t):=g(t m+1) N (t k). − −2 m − kX=0 Then (a) supp q [0, 2m 2], m⊆2 − (b) q C − ( , 2m 2], (c) q(0)∈ = 0, −∞ − (d) q(2m 2) = 1, dk − dk (e) dxk q(0) = dxk q(2m 2) = 0, 1 k m 2, (f) q(t)dt =0, − ≤ ≤ − R (g) q BV ( ) with Vq 2(m 1), (h) R q∈ R1. ≤ − k k∞ ≤

Proof. (a), (b), (c), (d) and (e) follow from (10), (11) and Lemma 5. To prove (f), note that m 2 1 − q(t)dt = g(t m +1)dt N (t k)dt − − 2 m − ZR ZR Xk=0 ZR m 2 m 1 1 − = − N (t k)dt (by Lemma 5(e)) 2 − 2 m − Xk=0 ZR m 1 m 1 = − − (by [2, Theorem 4.3(i)]) 2 − 2 =0.

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To prove (g), note that on [0,m 1],g(t m+1)= 0, which implies that − − g0(t m +1)=0for0 t m 1. Hence, − ≤ ≤ − m 2 − q0(t)= (1/2) Nm0 (t k)= (1/2)[Nm 1(t) Nm 1(t m +1)]. − − − − − − − kX=0 1 m 1 Thus for 0 t m 1, q0(t) ,which implies that V [0,m 1] − . ≤ ≤ − | |≤ 2 q − ≤ 2 On [m 1, 2m 2], we have − − m 2 1 − q0(t)=g0(t m+1) N 0 (t k) − −2 m − kX=0 1 =[Nm1(t m+1) Nm 1(t 2m+2)]+ [Nm 1(t m+1) Nm 1(t)]. − − − − − 2 − − − − 3 Thus, (m 1) t 2(m 1) implies that q0(t) ,and so V [m 1, 2m 2] − ≤ ≤ − | |≤ 2 q − − ≤ 3 (m 1). Hence, 2 −

V ( )=V [0, 2m 2] 2(m 1). q R q − ≤ − Finally, to prove (h) note that for t , ∈R m 2 1 − 1 0 g(t m+1) 1and0 N (t k) , ≤ − ≤ ≤2 m − ≤ 2 Xk=0 implying that m 2 1 − q(t) = g(t m +1) N (t k) 1, | | | − −2 m − |≤ Xk=0 and the assertion follows. Let ε 2m 2 ε 2m 2 p{ }(x):=q − x+ε ,p{}(x):=q − 1+ε x 1 ε { } 2 ε { − } ε 2m 2 1  ε 2m 2 1  p{ }(x):=q − x ,p{}(x):=q − x , 3 ε {2− } 4 ε { −2} ε 1 1  A{ } := [ ε, 0) ( ε, + ε) (1, 1+ε], − ∪ 2 − 2 ∪ ε ε ε ε ε ε p{ }(x):=p{ }(x) p{ }(x)+p{ }(x) p{ }(x) ψ{ }(x). 1 − 2 3 − 4 − We have: Lemma 7. Let ε>0.Then ε ε (a) supp p{ } A{ }, ε m⊆2 ε (b) p{ } C − (A{ }), ε ∈ ε 1 ε 1 + ε ε (c) p{ }( ε)=p{ }(( 2 ε)−)=p{ }(( 2 + ε) )=p{ }(1 + ε)=0,p{ }(0−)=1, ε + − − p{ }(1 )= 1, (d) For 1− k m 2 we have ≤ ≤ − k k k k d ε d ε d ε 1 + d ε 1 p{ }( ε)= p{ }(0−)= p{ }(( ε) )= p{ }( ) dxk − dxk dxk 2 − dxk 2 dk 1 dk dk = p ε (( + ε) )= p ε (1+)= p ε (1 + ε)=0, dxk { } 2 − dxk { } dxk { }

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ε (e) p{ } j,k is a Bessel sequence with bound { j,k } ∈Z

√2 ε 3 1 √2ε 1 2 4[ (ε + 1)(2m 2) + ] 2 +[ 8ε +4+3/√2 ]2 . { (√2 1){ − √2} (√2 1){ } } − − Proof. (a), (b), (c) and (d) follow from Lemma 6. By Lemmas 3 and 6, each of the ε functions p{ }(x), 1 i 4, generates a Bessel sequence with bound i ≤ ≤ √2 ε (ε + 1)(2m 2) + 3/√2 . (√2 1){ − } − Applying Lemma 4 and Minkowski’s inequality, (e) follows.

Theorem. Let m 2 be an integer and let ε>0be such that ≥

√2 ε 3 1 √2ε 3 1 2 B := 4[ (ε + 1)(2m 2) + ] 2 +[ 8ε +4+ ]2 <1. { (√2 1){ − √2} (√2 1){ √2} } − − ε ε Let h := χ 1 χ 1 and f { } := h + p{ }.Then [0, 2 ] [ 2 ,1] ε −m 2 (a) f { } C − ( , ), ∈ ε −∞ ∞ (b) supp f { } [ ε, 1+ε], ε ⊆ − 2 2 (c) f { } j,k is a Riesz basis of L ( ) with frame bounds (1 √B) and { j,k } ∈Z R − (1 + √B)2, ε δ 1/δ 1/δ (d) If 0 <δ< ,then f{ } h Lδ( ) (1+2 ) (2ε) . ∞ k − k R ≤

Proof. (a) and (b) follow from Lemma 7. Since the Haar system hj,k j,k is an orthogonal basis of L2( ) ([2, 4]), (c) follows from Lemma 7 and{ [5, Theorem} ∈Z 5]. R ε ε ε To prove (d), note that f { } h = p{ }. Since the functions pi{ },1 i 4, have − ε ≤ε ≤ disjoint support, Lemma 6 and the definition of p{ } imply that p{ }(x) 1on ε ε | |≤ [ ε, 0] [1, 1+ε], p{ }(x) 2on[1/2 ε, 1/2+ε], and p{ }(x) = 0 elsewhere. − ∪ | |≤ − | |

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10. G. Strang and T. Nguyen, “Wavelets and Filter Banks”, Wellesley–Cambridge Press, Welles- ley, Massachussetts, 1996. CMP 97:02 11. R. M. Young, “An Introduction to Nonharmonic Fourier Series”, Academic Press, New York, 1980. MR 81m:42027

Department of Mathematics, Auburn University, Auburn, Alabama 36849–5310 E-mail address: [email protected] E-mail address: [email protected]

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