WEYL-HEISENBERG FRAMES FOR SUBSPACES OF L 2 (R)

PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 129, Numer 1, Pages 145 154 S 0002-9939(00)05731-2 Article electronically pulished on July 27, 2000 WEYL-HEISENBERG FRAMES FOR SUBSPACES OF L 2 (R) PETER G. CASAZZA AND OLE CHRISTENSEN (Communicated y David R. Larson) Astract. A Weyl-Heisenerg frame {E m T nag} m, = {e 2πim( ) g( na)} m, for L 2 (R) allows every function f L 2 (R) to e written as an infinite linear comination of translated and modulated versions of the fixed function g L 2 (R). In the present paper we find sufficient conditions for {E m T nag} m, to e a frame for span{e m T nag} m,,which,ingeneral,mightjustea suspace of L 2 (R). Even our condition for {E m T nag} m, to e a frame for L 2 (R) is significantly weaker than the previous known conditions. The results also shed new light on the classical results concerning frames for L 2 (R), showing for instance that the condition G(x) := g(x na) 2 >A>0is not necessary for {E m T nag} m, to e a frame for span{e m T nag} m,. Our work is inspired y a recent paper y Benedetto and Li, where the relationship etween the zero-set of the function G and frame properties of the set of functions {g( n)} is analyzed. 1. Preliminaries and notation Let H denote a separale Hilert space with the inner product, linear in the first entry. Let I denote a countale index set. We say that {g i } i I His a frame (for H) if there exist constants A, B > 0 such that A f 2 f,g i 2 B f 2, f H. i I In particular a frame for H is complete, i.e., span{g i } i I = H. In case {g i } i I is not complete, {g i } i I can still e a frame for the suspace span{g i } i I ;inthat case we say that {g i } i I is a frame sequence. ThenumersA, B that appear in the definition of a frame are called frame ounds. Orthonormal ases and, more generally, Riesz ases are frames. Recall that {g i } i I is a Riesz asis for H if span{g i } i I = H and A, B > 0: A c i 2 c i g i 2 B c i 2, {c i } i I l 2 (I). i I i I i I If {g i } i I is a Riesz asis for span{g i } i I,wesaythat{g i } i I is a Riesz sequence. Received y the editors March 10, 1999. 1991 Mathematics Suject Classification. Primary 42C15. The first author was supported y NSF grant DMS 970618 and the second author y the Danish Research Council. The second author also thanks the University of Charlotte, NC, and the University of Missouri-Columia, MO, for providing good working conditions. 145 c 2000 American Mathematical Society License or copyright restrictions may apply to redistriution; see http://www.ams.org/journal-terms-of-use

146 PETER G. CASAZZA AND OLE CHRISTENSEN The present paper deals with frames having a special structure: all elements are translated and/or modulated versions of a single function. Let L 2 (R) denote the Hilert space of functions on the real line which are square integrale with respect to the Leesgue measure. First, define the following operators on functions f L 2 (R): Translation y a R : (T a f)(x) =f(x a), x R. Modulation y R : (E g)(x) =e 2πix f(x), x R. AframeforL 2 (R) oftheform{e m T na g} m, is called a Weyl-Heisenerg frame (or Gaor frame). For a collection of different papers concerning those frames we refer to the monograph [5]. Sufficient conditions for {E m T na g} m, to e a frame for L 2 (R) has een known for aout 10 years. The asic insight was provided y Dauechies [3]. A slight improvement was proved in [6]: Theorem 1.1. Let g L 2 (R) and suppose that (1) A, B > 0: A g(x na) 2 B for a.e. x R, (2) lim T na gt na+ k g =0. 0 Then there exists 0 > 0 such that {E m T na g} m, is a Weyl-Heisenerg frame for L 2 (R) for all ]0; 0 [. The proof of Theorem 1.1 is ased on the following identity, valid for all continuous functions f with compact support whenever g satisfies (1): (3) f,e m T na g 2 m, = 1 f(x) 2 G(x)dx + 1 f(x)f(x k/) g(x na)g(x na k/)dx. An estimate of the second term in (3) now shows that {E m T na g} m, is actually a frame for all values of for which (4) T na gt na+ k g <A. A more recent result can e found in [4]: in Theorem 2.3 it is proved that if (1) is satisfied and there exists a constant D<Asuch that (5) g(x na)g(x na k ) D for a.e. x R, then {E m T na g} m, is a frame for L 2 (R) with ounds A D, B+D. The reader should oserve that [4] does not provide us with a generalization of the results in [3], [6] in a strict sense: there are cases where (5) is satisfied ut (4) is not, and vice versa. The main point is that other conditions (that are easy to check) for {E m T na g} m, to e a frame can e derived from (5); cf. Theorem 2.4 in [4]. License or copyright restrictions may apply to redistriution; see http://www.ams.org/journal-terms-of-use

WEYL-HEISENBERG FRAMES FOR SUBSPACES OF L 2 (R) 147 Define the Fourier Transform F(f) = ˆf of f L 1 (R) y ˆf(y) = f(x)e 2πiyx dx. As usual we extend the Fourier Transform to an isometry from L 2 (R) ontol 2 (R). We denote the inverse Fourier transformation of g L 2 (R) yf 1 g or ǧ. It is important to oserve the following comutator relations, valid for all a R: FT a = E a F, FE a = T a F. We need a result from [2]. The asic insight was provided y Benedetto and Li [1], who treated the case a =1. Theorem 1.2. Let g L 2 (R). Then {T na g} is a frame sequence with ounds A, B if and only if 0 <aa ĝ( x + n a ) 2 ab for a.e. x for which ĝ( x + n a ) 2 0. In that case {T na g} is a Riesz sequence if and only if the set of x for which ) 2 =0has measure zero. ĝ( x+n a Theorem 1.2 leads immediately to an equivalent condition to (1). Define the function G and its kernel N G y G : R [0, ], G(x) := g(x na) 2, N G = {x R G(x) =0}. Corollary 1.3. {E n a g} is a frame sequence with ounds A, B if and only if 0 < A a g(x na) 2 B a for a.e. x R N G. In that case {E n g} is a Riesz sequence iff N a G has measure zero. Proof. The inequality 0 < A a g(x na) 2 B a for a.e. x R N G holds if and only if (6) 0 < A a g([x n]a) 2 B a for a.e. x R N G. By Theorem 1.2, (6) is equivalent to {T n ǧ} eing a frame sequence with ounds a A, B. Applying the Fourier transformation this is equivalent to {E n g} eing a a frame sequence with ounds A, B. 2. The results From now on we concentrate on Weyl-Heisenerg frames {E m T na g} m,.our first result gives a sufficient condition for {E m T na g} m, to e a frame sequence. It can e considered as a suspace version of a result y Ron and Shen; cf. [7]. Our condition for {E m T na g} m, to e a frame for L 2 (R) is significantly weaker than the conditions mentioned in section 1. Let L 2 (R N G ) denote the set of functions in L 2 (R) thatvanishesatn G. License or copyright restrictions may apply to redistriution; see http://www.ams.org/journal-terms-of-use

148 PETER G. CASAZZA AND OLE CHRISTENSEN Theorem 2.1. Let g L 2 (R), a, > 0 and suppose that (7) A := inf x [0,a] N G g(x na) 2 > 0, g(x na)g(x na k ) (8) B := sup x [0,a] k Z g(x na)g(x na k ) <. Then {E m T na g} m, is a frame for L 2 (R N G ) with ounds A, B. Proof. First, oserve that span{e m T na g} m, L 2 (R N G ). Now consider a function f L 2 (R N G ) which is ounded and has support in a compact set. The Heil-Walnut argument (3) is valid under the assumption (8) and it gives that (3) f,e m T na g 2 m, = 1 f(x) 2 g(x na) 2 dx + 1 f(x)f(x k/) g(x na)g(x na k/)dx. We want to estimate the second term aove. For k Z, define H k (x) := T na g(x)t na+k/ g(x). First, oserve that T k/ H k (x) = T k/ T na g(x)t na+k/ g(x) = T na k/ g(x)t na g(x) = T na+k/ g(x)t na g(x) = T na+k/ g(x)t na g(x) = H k (x). License or copyright restrictions may apply to redistriution; see http://www.ams.org/journal-terms-of-use

WEYL-HEISENBERG FRAMES FOR SUBSPACES OF L 2 (R) 149 Now, y a slight modification of the argument in [4], Theorem 2.3, f(x)f(x k/) g(x na)g(x na k/)dx f(x) T k/ f(x) H k (x) dx = f(x) H k (x) T k/ f(x) H k (x) dx ( ) 1/2 ( f(x) 2 H k (x) dx ) 1/2 T k/ f(x) 2 H k (x) dx f(x) 2 H k (x) dx 1/2 T k/ f(x) 2 H k (x) dx 1/2 = f(x) 2 H k (x) dx 1/2 f(x) 2 T k/ H k (x) dx 1/2 = f(x) 2 H k (x) dx. Note that H k(x) = T nag(x)t na+k/ g(x) is a periodic function with period a. By (3) and the assumption (7) we now have f,e m T na g 2 m, 1 f(x) 2 g(x na) 2 g(x na)g(x na k ) dx A f 2. Similary, y (3) and (8), f,e m T na g 2 m, 1 f(x) 2 g(x na) 2 + g(x na)g(x na k ) dx = 1 f(x) 2 g(x na)g(x na k ) k Z B f 2. Since those two estimates holds on a dense suset of L 2 (R N G ), they hold on L 2 (R N G ). Thus {E m T na g} m, is a frame for L 2 (R N G ) with the desired ounds. License or copyright restrictions may apply to redistriution; see http://www.ams.org/journal-terms-of-use

150 PETER G. CASAZZA AND OLE CHRISTENSEN The advantage of Theorem 2.1 compared to the results in section 1 is that we compare the functions g(x na) 2 and H k(x) pointwise rather than assuming that the supremum of H k(x) is smaller than the infimum of g(x na) 2. It is easy to give concrete examples where Theorem 2.1 shows that {E m T na g} m, is a frame for L 2 (R) ut where the conditions in section 1 are not satisfied: Example. Let a = = 1 and define 1+x if x [0, 1[, 1 g(x) = 2x if x [1, 2[, 0 otherwise. For x [0, 1[ we have and G(x) = g(x n) 2 = g(x) 2 + g(x +1) 2 = 5 (x +1)2 4 g(x n)g(x n k) =(1+x) 2, so y Theorem 2.1 {E m T n g} m, is a frame for L 2 (R) with ounds A = 1 4,B =9. But inf x R G(x) = 5 4 and T n gt n+k g =4, so the condition (4) is not satisfied. (5) is not satisfied either. Remark. It is well known that G eing ounded elow is a necessary condition for {E m T na g} m, to e a frame for L 2 (R); cf. [3]. Theorem 2.1 shows that this condition is not necessary for {E m T na g} m, to e a frame sequence. However, it is implicit in (7) that G has to e ounded elow on R N G in order for Theorem 2.1 to work, and an easy modification of the proof in [3] shows that this is actually a necessary condition for {E m T na g} m, to e a frame for L 2 (R N G ). We shall later give examples of frame sequences for which G is not ounded elow on R N G. In case g has support in an interval of length 1 an equivalent condition for {E m T na g} m, to e a frame sequence can e given. First, oserve that y (3) this condition on g implies that for all continuous functions f with compact support, we have f,e m T na g 2 = 1 f(x) 2 G(x)dx. m, It is not hard to show that this actually holds for all f L 2 (R); cf. [6]. Corollary 2.2. Suppose that g L 2 (R) has compact support in an interval I of length I 1/. Then {E m T na g} m, is a frame sequence with ounds A, B if and only if 0 <A g(x na) 2 B, for a.e. x R N G. In that case {E m T na g} m, is actually a frame for L 2 (R N G ). License or copyright restrictions may apply to redistriution; see http://www.ams.org/journal-terms-of-use

WEYL-HEISENBERG FRAMES FOR SUBSPACES OF L 2 (R) 151 Proof. Suppose that g has support in an interval I of length I 1.If0<A G(x) B for a.e. x R N G, it follows from Theorem 2.1 that {E m T na g} m, is a frame sequence with the desired ounds. Now suppose that {E m T na g} m, is a frame sequence with ounds A, B. Then, for every interval I of length I =1/ and every function f L 2 (I), f,e m T na g 2 = 1 f(x) 2 G(x)dx B f 2. m,n But this is clearly equivalent to R G(x) = g(x na) 2 B a.e. To prove the lower ound for G we proceed y way of contradiction. Suppose that for some ɛ>0wehave0<g(x) (1 ɛ)a on a set of positive measure. In this case there is a set of positive measure and supported in an interval of length 1 so that 0 <G(x) (1 ɛ)a on. Then, for any function f L 2 (R) supported on, we have f,e m T na g 2 = 1 f(x) 2 G(x)dx m,n (1 ɛ)a f(x) 2 dx =(1 ɛ)a f 2. R Since G(x) > 0on,thereisak Z so that χ T ka g is not the zero function. With := Supp(T ka g)wehave f := χ T ka g span{e m T ka g} span{e m T na g} m,, so the aove calculation shows that the lower ound for {E m T na g} m, is at most (1 ɛ)a, which is a contradiction. Thus G(x) A for a.e. x R N G. In case the condition in Corollary 2.2 is satisfied, it follows from Theorem 2.1 that {E m T na g} m, is a frame for L 2 (R N G ). For functions g with the property that the translates T na g, n Z, have disjoint support we can give an equivalent condition for {E m T na g} m, to e a frame sequence. Define the function G(x) :R [0, ], G(x) = g(x + m ) 2. R Proposition 2.3. Let g L 2 (R),a,>0 and suppose that (9) supp(g) supp(t na g)=, n Z {0}. Then {E m T na g} m, is a frame sequence with ounds A, B if and only if there exist A, B > 0 such that A g(x + m ) 2 B for a.e. x R N G. In that case, {E m T na g} m, is a Riesz sequence iff N G has measure zero. License or copyright restrictions may apply to redistriution; see http://www.ams.org/journal-terms-of-use

152 PETER G. CASAZZA AND OLE CHRISTENSEN Proof. Because of the support condition (9), it is clear that {E m g} is a frame sequence iff {E m T na g} m, is a frame sequence, in which case the sequences have the same frame ounds. But y Corollary 1.3 {E m g} is a frame sequence with ounds A, B iff A g(x + m ) 2 B for a.e. x R N G. Also, {E m T na g} m, is a Riesz sequence iff {E m g} is a Riesz sequence, which, y Corollary 1.3, is the case iff N G has measure zero. We are now ready to show that G eing ounded elow on R N G (y a positive numer) is not a necessary condition for {E m T na g} m, to e a frame sequence. Example. Let a, > 0 and suppose that 1 a / N. Choseɛ>0 such that [0,ɛ]+na [ 1, 1 + ɛ] =, n Z. This implies that ɛ<min(a, 1 ). Define x if x [0,ɛ], g(x) := 1 (x 1 )2 if x [ 1, 1 + ɛ], 0 otherwise. Then the condition (9) in Proposition 2.3 is satisfied. Also, for x [0,ɛ], G(x) = g(x + m ) 2 = g(x) 2 + g(x +1) 2 =1 and for x ]ɛ, 1 ], we have G(x) =0. Thus, y Proposition 2.3 {E m T na g} m, is a frame sequence. But for x [0,ɛ], G(x) = g(x na) 2 = x 2. Thus G is not ounded elow y a positive numer on R N G. By the remark after Theorem 2.1 this implies that span{e m T na g} m, L 2 (R N G ). For a > 1 it is even possile to construct an orthonormal sequence having all the features of the aove example. For example, let a =2,=1and x if x [0, 1], g(x) := 2x x 2 if x ]1, 2], 0 otherwise. Since g(x + m ) 2 =1, x, it follows y Proposition 2.3 that {E m T 2n g} m, is a Riesz sequence with ounds A = B = 1, which implies that {E m T 2n g} m, is an orthonormal sequence. But G(x) = g(x 2n) 2 is not ounded elow on R N G. G eing ounded aove is still a necessary condition for {E m T na g} m, to e a frame sequence (repeat the argument in Corollary 2.2). G also has to e ounded aove: License or copyright restrictions may apply to redistriution; see http://www.ams.org/journal-terms-of-use

WEYL-HEISENBERG FRAMES FOR SUBSPACES OF L 2 (R) 153 Proposition 2.4. If {E m T na g} m, is a frame sequence with upper ound B, then g(x + m ) 2 B a.e. Proof. If {E m T na g} m, is a frame sequence, then {F 1 E m T na g} m, = {T m E na ǧ} m, is a frame sequence with the same ounds. In particular the sequence {T m ǧ} m, has the upper frame ound B. By Theorem 1.2 (or, more precisely, the proof of it in [2]) it follows that It follows that g(x + m ) 2 B a.e. g( x + m ) 2 B for a.e. x. Remark. Recall that a wavelet frame for L 2 (R) has the form { 1 a g( x n/2 a n m)} m,, where a>1,>0andg L 2 (R) arefixed. As well as Weyl-Heisenerg frames, wavelet frames play a very important role in applications. The theory for the two types of frames was developed at the same time, with the main contriution due to Dauechies. Several results for Weyl- Heisenerg frames have counterparts for wavelet frames. For example, Theorem 5.1.6 in [6] gives sufficient conditions for { 1 g( x a n/2 a m)} n m, to e a frame ased on a calculation similar to (3). Also our results for Weyl-Heisenerg frames have counterparts for wavelet frames. The ideas in the proof of Theorem 2.1 can e used to modify [6], Theorem 5.1.6, which leads to the following: Theorem 2.5. Let a>1,>0 and g L 2 (R) e given. Let { } N := γ [1,a] ĝ(a n γ) 2 =0 and suppose that A := inf γ [1,a] N B := sup γ [0,a] k, ĝ(a n γ) 2 ĝ(a n γ)ĝ(a n γ + k/) > 0, ĝ(a n γ)ĝ(a n γ + k/) <. Then { 1 a n/2 g( x a n m)} m, is a frame sequence with ounds A, B. Acknowledgment The authors would like to thank Christopher Heil for discussions and Baiqiao Deng for giving us access to the preprint [4]. License or copyright restrictions may apply to redistriution; see http://www.ams.org/journal-terms-of-use

154 PETER G. CASAZZA AND OLE CHRISTENSEN References [1] Benedetto, J. and Li, S.: The theory of multiresolution analysis frames and applications to filter anks. Appl. Comp. Harm. Anal. 5 (1998), 389 427. MR 99k:42054 [2] Casazza, P.G., Christensen, O. and Kalton, N.: Frames of translates, Collectanea Mathematica, accepted for pulication. [3] Dauechies, I.: The wavelet transformation, time-frequency localization and signal analysis. IEEE Trans. Inform. Theory 36 (1990), p. 961 1005. [4] Deng, B., Schempp, W., Xiao, C. and Wu, Z.: On the existence of Weyl-Heisenerg and affine frames. Preprint, 1997. [5] Feichtinger, H.G. and Strohmer, T., (Eds.): Gaor analysis and algorithms: Theory and applications. Birkhäuser, 1998. MR 98h:42001 [6] Heil, C. and Walnut, D.: Continuous and discrete wavelet transforms. SIAM Review 31 (1989), p.628 666. MR 91c:42032 [7] Ron, A. and Shen, Z.: Weyl-Heisenerg systems and Riesz ases in L 2 (R), Duke Math. J. 89 (1997), 237 282. MR 98i:42013 [8] Young, R.: An introduction to nonharmonic Fourier series. Academic Press, New York, 1980. MR 81m:42027 Department of Mathematics, University of Missouri, Columia, Missouri 65211 E-mail address: pete@casazza.math.missouri.edu Department of Mathematics, Technical University of Denmark, Building 303, 2800 Lyngy, Denmark E-mail address: ole.christensen@mat.dtu.dk License or copyright restrictions may apply to redistriution; see http://www.ams.org/journal-terms-of-use