Uniform partitions of frames of exponentials into Riesz sequences

Uniform partitions of frames of exponentials into Riesz sequences

J. Math. Anal. Appl. 348 (2008) 739–745 Contents lists available at ScienceDirect Journal of Mathematical Analysis and Applications www.elsevier.com...

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J. Math. Anal. Appl. 348 (2008) 739–745

Contents lists available at ScienceDirect

Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Uniform partitions of frames of exponentials into Riesz sequences Darrin Speegle Department of Mathematics and Computer Science, St. Louis University, 221 North Grand Blvd., St. Louis, MO 63103, USA

a r t i c l e

i n f o

a b s t r a c t The Feichtinger conjecture, if true, would have as a corollary that for each set E ⊂ [0, 1] and Λ ⊂ Z, there is a partition Λ1 , . . . , Λ N of Z such that for each 1  i  N, {exp(2π ixλ): λ ∈ Λi } is a Riesz sequence. In this paper, sufficient conditions on sets E ⊂ [0, 1] and Λ ⊂ R are given so that {exp(2π ixλ)1 E : λ ∈ Λ} can be uniformly partitioned into Riesz sequences. © 2008 Elsevier Inc. All rights reserved.

Article history: Received 6 March 2008 Available online 6 August 2008 Submitted by L. Grafakos Keywords: Beurling density Beurling dimension Riesz sequence Frame of exponentials Feichtinger conjecture

1. Introduction A frame is a collection of elements {e i : i ∈ I } in a Hilbert space H such that there exist positive constants A and B such that for every h ∈ H, A h2 

  h, e i 2  B h2 . i∈ I

A frame {e i : i ∈ I } is bounded if inf e i  > 0. i∈ I

(Note that it is automatic that supi ∈ I e i  < ∞.) A sequence {e i : i ∈ I } is said to be a Riesz sequence if it is a Riesz basis for its closed linear span, i.e., there exist K 1 , K 2 > 0 such that for every finite family of scalars {ai : i ∈ I } K1

 i∈ I

 2      K2 |ai |2   a e |ai |2 . i i   i∈ I

i∈ I

Note that for Riesz basic sequences we always have 0 < inf e i   sup e i  < ∞. i∈ I

i∈ I

With this notation, one can state the Feichtinger conjecture: Conjecture 1 (Feichtinger). Every bounded frame can be written as the finite disjoint union of Riesz basic sequences.

E-mail address: [email protected]. 0022-247X/$ – see front matter doi:10.1016/j.jmaa.2008.07.076

© 2008 Elsevier Inc.

All rights reserved.

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D. Speegle / J. Math. Anal. Appl. 348 (2008) 739–745

The Feichtinger conjecture has been shown to be related to several famous open problems in analysis, and as such, is receiving a fair amount of recent interest [3,6,7,10]. Of particular interest is the special case of the Feichtinger conjecture for frames of exponentials, e.g. frames of the form {exp(2π i λx)1 E : λ ∈ Λ}, where E is measurable subset of [0, 1] with positive measure. The best positive result so far is in [1], where it is shown that in the case Λ = Z, frames of exponentials always contain a Riesz sequence with positive Beurling density. The most natural type of partition of a set indexed by the integers would be a uniform partition, so it is natural to ask which frames can be uniformly partitioned into Riesz sequences. We state this formally as a definition. Definition 2. Let Λ = {· · · < λ−1 < λ0 < λ1 < · · ·} ⊂ R. We say that {e λk : k ∈ Z} can be uniformly partitioned into Riesz sequences if there exists an N such that for 1  J  N, {e λmN + J : m ∈ Z} is a Riesz sequence. Gröchenig [10] showed that if a frame is intrinsically localized, than it can be uniformly partitioned into Riesz sequences. Bownik and the author [3] observed that if E contains an interval a.e., then {exp(2π ixn)1 E : n ∈ Z} can be uniformly partitioned into Riesz sequences. Halpern, Kaftal and Weiss [11] showed that if φ ∈ L ∞ ([0, 1]) is Riemann integrable, then the associated Laurent operator L φ can be uniformly paved. Moreover, to date the primary negative evidence offered against the Feichtinger conjecture is two constructions of frames which can not be uniformly partitioned into Riesz sequences in a strong way, see [3, Theorem 4.13] and [11, Theorem 5.4(b)]. In this paper, we provide a sufficient condition on the pair ( E , Λ), where E ⊂ [0, 1] and Λ ⊂ R, such that {exp(2π i λx)1 E : λ ∈ Λ} can be uniformly partitioned into Riesz sequences. Of particular interest is the application of this condition to the example considered in [2,11], which shows that, perhaps, the proposed counterexample to the paving conjecture in [11], which was proven not to be a counterexample in [2], was not optimally chosen. See Example 12 for details. 2. Beurling dimension In this section, we recall some facts about the Beurling dimension of a subset of Rd , though we will be concerned only with subsets of R. For h > 0, we let Q denote the cube [−1, 1]d and let Q h be the dilation of Q by the factor of h: Q h = h Q = [−h, h]d . For any x = (x1 , . . . , xd ) ∈ Rd we let Q h (x) be the set Q h translated in such a way so that it is “centered” at x, i.e., Q h (x) =

d 

[xi − h, xi + h].

i =1

Employing these notions we will first define a generalization of Beurling density. Definition 3. Let Λ ⊂ Rd and r > 0. Then the lower Beurling density of Λ with respect to r is defined by

Dr− (Λ) = lim inf inf

#(Λ ∩ Q h (x))

h→∞ x∈Rd

hr

,

and the upper Beurling density of Λ with respect to r is defined by

Dr+ (Λ) = lim sup sup h→∞

x∈Rd

#(Λ ∩ Q h (x)) hr

.

If Dr− (Λ) = Dr+ (Λ), then we say that Λ has uniform Beurling density with respect to r and denote this density by Dr (Λ). With this definition in hand, we can define the upper and lower Beurling dimensions of subsets of Rd . Definition 4. Let Λ ⊂ Rd . Then the lower dimension of Λ ⊂ Rd is defined by



dim− (Λ) = inf r > 0: Dr− (Λ) < ∞ and the upper dimension of Λ ⊂ Rd is







dim+ (Λ) = sup r > 0: Dr+ (Λ) > 0 . When these two quantities are equal, we refer to the Beurling dimension of Λ, and we denote it by dim(Λ). We note here that the upper Beurling dimension is a base point independent version of the upper mass dimension considered in [4], see [8,9] for details. One result on Beurling dimension that we will use in the main section of this paper is the following [8,9].

D. Speegle / J. Math. Anal. Appl. 348 (2008) 739–745

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Proposition 5. Let Λ ⊂ Rd . (i) The following conditions are equivalent. (a) Dd+ (Λ) < ∞. (b) There exists some h > 0 such that supx∈Rd #(Λ ∩ Q h (x)) < ∞. (c) For all h > 0, supx∈Rd #(Λ ∩ Q h (x)) < ∞. (d) Λ is relatively uniformly discrete. (e) For all h > 0, supx∈Rd #{λ ∈ Λ: x ∈ Q h (λ)} < ∞. (ii) Also the following conditions are equivalent. (a) Dd− (Λ) > 0. (b) There exists some h > 0 such that infx∈Rd #(Λ ∩ Q h (x)) > 0. (c) Λ contains a subsequence of positive uniform density. (d) There exists some h > 0 such that Λ is h-dense. 3. Results For E a measurable subset of [0, 1] and Λ ⊂ R, we will say that ( E , Λ) can be uniformly partitioned into Riesz sequences if the frame {exp(2π i λx)1 E : λ ∈ Λ} can be uniformly partitioned into Riesz sequences, see Definition 2. Characterizing sets E such that ( E , Z) can be uniformly partitioned into Riesz sequences is related (but not equivalent) to characterizing the Laurent operators which can be uniformly paved, which was characterized in [11] as those Laurent operators whose symbol is Riemann integrable. In this article, we consider general sets Λ ⊂ R. Our main tool is a theorem due to Montgomery and Vaughan: [12, Theorem 1, Chapter 7], [13]. Theorem 6. Suppose that λ1 , . . . , λ N are distinct real numbers, and suppose that δ > 0 is chosen so that |λn − λm |  δ whenever n = m. Then, for any coefficients a1 , . . . , a N , and any T > 0,

( T − 1/δ)

N  n=1

2 T   N N    2π i λn t  |an |   an e |an |2 .  dt  ( T + 1/δ)   2

n=1

0

(3.1)

n=1

We will also need two lemmas concerning partitioning subsets of R with finite upper dimension. Lemma 7. Let Λ = {· · · < λ−1 < λ0 < λ1 < · · ·} ⊂ R such that dim+ (Λ)  1, and K ∈ N. There exists N ∈ N such that whenever |i − j | > N, |λi − λ j | > K . Proof. This is just a restatement of Proposition 5. For Λ ⊂ R,

2

#(Λ∩ Q r (x)) α  0 and r > 0, we define D + : x ∈ R}. α ,Λ (r ) = sup{ rα

Lemma 8. Let Λ = {· · · < λ−1 < λ0 < λ1 < · · ·} ⊂ R be such that dim+ (Λ) < β  1, and > 0. There exist N , R ∈ N such that for −β + , where Λ ( N ) = {λ each 1  j  N and r  R, D + j mN + j : m ∈ Z}. β,Λ ( N ) (r )  2R j

Proof. Choose R such that for r  R, sup x∈R

#(Λ ∩ Q r (x)) rβ

 sup

#(Λ ∩ Q R (x)) Rβ

x∈R

+ < ∞.

Choose N = supx∈R #(Λ ∩ Q R (x)). Now, fix r  R and 1  j  N. Then, D+ β,Λ j ( N ) (r ) = sup

#(Λ j ( N ) ∩ Q r (x)) rβ

x∈R

 sup

#(Λ ∩ Q r (x))/ N + 1 rβ

x∈R

 sup x∈R

#(Λ ∩ Q R (x)) Rβ N

 2R −β + .

+ + R −β

2

Using Theorem 6, we obtain the following improvement to Lemma 5.1 in [5].

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D. Speegle / J. Math. Anal. Appl. 348 (2008) 739–745

Lemma 9. Suppose Λ ⊂ R. If I is an interval contained in [0, 1], then for any sequence of numbers {aλ }λ∈Λ in 2 ,

  2

  −1 aλ e 2π i λξ   2 ( I ) D + .  0,Λ ( I )

(3.2)

I

Proof. First note that the interval [0, T ] in Theorem 6 can be replaced by any interval of length T , which we set to be ( I ). −1 ) subsets Λ such that if λ , λ are in Λ , then |λ − λ | > ( I )−1 . It follows that We can partition Λ into D + 1 2 1 2 i i 0,Λ ( ( I )

  2 1/2     aλ e 2π i λξ    i

I



 2 1/2   2π i λξ   aλ e   λ∈Λi

I

 1/ 2  (2 ( I ))1/2 |aλ |2 λ∈Λi

i

1/2

1/ 2 +

1/2    2 ( I ) D 0,Λ ( I )−1 |aλ |2 i



1/ 2 +

1/2 = 2 ( I ) D 0,Λ ( I )−1



λ∈Λi

1/ 2

|aλ |2

,

λ∈Λ

where the second inequality is from Theorem 6 and the third inequality is the generalized mean inequality with −1 ) terms. 2 D+ 0,Λ ( ( I ) We recall for hypotheses in the following theorem that the essential α -Hausdorff measure of a set E

motivation ofthe ∞ is H α ( E ) = inf{ ( I n )α : E ⊂ n=1 I n ∪ J , | J | = 0}. The following theorem is related to the essential α -Hausdorff measure of E in Corollary 11. Theorem 10. Let E ⊂ [0, 1] have positive measure, Λ = {< · · · < λ1 < λ0 < λ1 < · · ·} ⊂ R and 0 < α < 1. If there exists a sequence of intervals { E n : n ∈ N} of nonincreasing length, an integer Z and 0 < α < 1 such that (i) (ii)

∞

n=1

Z

E n ⊃ E, and

n=1 | E n | +



n= Z +1 | E n |

α < 1,

and dim+ (Λ) < 1 − α , then ( E , Λ) can be uniformly partitioned into Riesz sequences. Proof. Define F = (i)

3 4

∞

n=1

E n , and note that | F | < |[0, 1]|, and normalize |[0, 1]| = 1. Choose

> 0 satisfying

+ 14 | F | < 1 − and

(ii) | F | + 2 <

1 2

+ | F |.

Choose M  Z ∈ N such that (i) | E M |1−α < 4 ∞



(ii) 4 | E M |

1−α

1

n = M +1 | E n |

+



α

,

n= M +1 | E n |

α < , and

(iii) | E M |−1 > R, where R is chosen from Lemma 8 with β = 1 − α and

as above.

Choose K ∈ N such that M / K < . By Lemma 7, there exists L ∈ N such that |λ j − λk | > K whenever | j − k| > L. By α −1 +  2| E |1−α + for all r  | E |−1 and 1  i  J . Finally, Lemma 8, there exists J ∈ N such that D + M M 1−α ,Λi ( J ) (r )  2R let N be the larger of J and L. Fix 1  j  N, and {aλ : λ ∈ 2 (Λ j ( N ))}. We compute

2  ∞    M     2π i λξ   a e = λ   n=1 E

n

λ∈Λ j ( N )

n=1 E

+

n

  2   2π i λξ   a e λ   λ∈Λ j ( N )

∞ 

2      2π i λξ   a e λ   := S 1 + S 2 .

n= M +1 E

n

λ∈Λ j ( N )

D. Speegle / J. Math. Anal. Appl. 348 (2008) 739–745

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Moreover, by Theorem 6 and our choice of K ,

 S1 

 M  

 |aλ |2  | F | + |aλ |2 . | E n | + 1/ K n=1

λ∈Λ j ( N )

(3.3)

λ∈Λ j ( N )

We also have that ∞ 

S2  2

n= M +1

∞ 

=2

n= M +1

∞ 

2



 −1

( E n ) D + |aλ |2 0,Λ j ( N ) ( E n ) λ∈Λ j ( N )



 −1

( E n ) ( E n )α −1 D + |aλ |2 α ,Λ j ( N ) ( E n ) λ∈Λ j ( N )





( E n )α 2| E M |1−α + |aλ |2

n= M +1

λ∈Λ j ( N )



 4 | E M |1−α +

∞ 



( E n )α

n= M +1

|aλ |2 .

λ∈Λ j ( N )

In particular,



S2 <

|aλ |2 .

(3.4)

λ∈Λ j ( N )

Therefore, combining (3.3) and (3.4), we get that

2 ∞      

 2π i λξ   a e |aλ |2 λ    | F | + 2 n=1 E

n

λ∈Λ j ( N )

λ∈Λ j ( N )



< 1 + | F | /2 |aλ |2 .

λ∈Λ j ( N )

Therefore, denoting P = P



3+| F | , 4

we have



|aλ |2  (1 − 1/ K )

λ∈Λ j ( N )

|aλ |2

λ∈Λ j ( N )

2      2π i λξ   a e λ   dξ



[0,1]

λ∈Λ j ( N )



=

+ Ec

 

Ec

2      2π i λξ   a e λ   dξ λ∈Λ j ( N )

E

2      2π i λξ   + a e λ  dξ  

En

λ∈Λ j ( N )

2      

  aλ e 2π i λξ  dξ + 1 + | F | /2 |aλ |2 . Ec

λ∈Λ j ( N )

It follows that





1 − | F | /4

 λ∈Λ j ( N )

|aλ |2 <

λ∈Λ j ( N )

2      2π i λξ   a e λ   , Ec

λ∈Λ j ( N )

i.e. {e 2π i λξ 1 E c : λ ∈ Λ} has a lower Riesz bound. Since each Λ j ( N ) is separated, it also has an upper Riesz bound, which completes the proof. 2 Corollary 11. Let E ⊂ [0, 1] have positive measure, Λ = {< · · · < λ1 < λ0 < λ1 < · · ·} ⊂ R and 0 < dim+ (Λ) < 1 − α , then ( E , Λ) can be uniformly partitioned into Riesz sequences.

α < 1. If H α ( E ) < ∞ and

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D. Speegle / J. Math. Anal. Appl. 348 (2008) 739–745

Proof. If H α ( E ) < ∞, then we can find intervals { E n : n ∈ N} such that E ⊂ Choose an integer Z such that ∞ 

∞

n=1

and



n=1 | E n |

α < ∞.

| E n |α < 1 − α .

n= Z

Let

 F=E∩

∞ 

 En .

n= Z

Then, Theorem 10 yields that

 2π i λ  e 1 F λ∈Λ

can be uniformly partitioned into Riesz basic sequences given by, say, {Λ j } Kj=1 with lower Riesz bounds {C j } Kj=1 . Now, for any j = 1, . . . , K , and any scalars {αλ : λ ∈ Λ j },

 2  2  2          2π i λ x 2π i λ x 2π i λ x   =  +  a e 1 a e 1 a e 1 λ E λ F λ E \ F       λ∈Λ j

λ∈Λ j

 2    2π i λ x   a e 1 λ F 

λ∈Λ j

λ∈Λ j

Cj



|aλ |2 ,

λ∈Λ j

as desired.

2

Example 12. An example considered by Bourgain–Tzafriri [2] and Halpern–Kaftal–Weiss [11]. Let {rn : n ∈ N} be a partition of the rational numbers in [0, 1]. Foreach n, let E n be an interval centered at rn with length | E n | < 2−n . Let φ be the ∞ indicator function supported on E = n=1 E n . In [11], it was shown that the Laurent operator with symbol φ , L φ , cannot be uniformly paved (in fact, something stronger was shown), while in [2], it was shown that L φ can still be paved. In particular, this implies that there is a partition of the integers Λ1 , . . . , Λ N such that for each 1  i  N, {1 E c exp(λ): λ ∈ Λi } is a Riesz sequence. (Note that ( E c , Z) cannot be uniformly partitioned into Riesz sequences.) However, by Theorem 10, whenever dim+ (Λ) < 1, ( E , Λ) can be uniformly partitioned into Riesz sequences.

So, perhaps a better candidate for a counterexample to the Feichtinger conjecture would have been to have a series an that converges more slowly to a number less than 1 and to let | E n | < an . We end this paper by presenting an example of a set E ⊂ [0, 1] and a set Λ ⊂ Z such that dim+ (Λ) < 1 yet ( E , Λ) can not be uniformly partitioned into Riesz sequences. We begin by recalling the following theorem. Theorem 13. (See [3].) There exists a set E ⊂ [0, 1] such that whenever K ⊂ Z is such that for all δ > 0, there exist M , N ∈ Z such that (i) N −1/2 log3 N < δ , and (ii) { M , M + , . . . , M + N } ⊂ K, then {exp(λ): λ ∈ K} is not a Riesz sequence in L 2 ( E ). Corollary 14. There exists a set E ⊂ [0, 1] such that for each 1 > β > 2/3 there is a set Λ ⊂ Z such that dim+ (Λ) = β and ( E , Λ) cannot be uniformly partitioned into Riesz sequences. Proof. Let E be the set guaranteed to exist from Theorem 13. Let 2/3 < β < 1. Let

∞

γ =

1−β

β . Define

Λ =

γ 2j j =1 { Q j + α j : 0  α  j }, where the Q j ’s are chosen to be some rapidly increasing sequence such as 2 . It follows that

supx∈R |#(Λ ∩ Q j γ +1 (x))| ≈ j, and so dim+ (Λ) = 1+1γ = β . Now, let N be a positive integer and write Λ = {λn : n ∈ N} where λi < λ j when i < j. Let Λ N = {λ Nn : n ∈ N}. We show that Λ N satisfies (i) and (ii) of the hypotheses of Theorem 13, which completes the proof of this corollary. Let δ > 0. For each j, let k j be the smallest nonnegative inteer such that Q j + k j j γ  ∈ Λ N . We have that









Q j + (α N + k j ) j γ : 0  α <  j / N  ⊂ Λ N .

D. Speegle / J. Math. Anal. Appl. 348 (2008) 739–745

Since

745

γ < 1/2, we can find j such that    j / N −1/2 N j γ log3  j / N  < δ,

which finishes the proof.

2

Acknowledgments The author wishes to thank Pete Casazza who suggested the improved form of Corollary 11 appearing in this paper and who made several other suggestions for improvement. The author also thanks Dick Gundy, whose insightful questions at a seminar at Washington University led to some improvements to this paper, as well as Wojtek Czaja and Gitta Kutyniok for their encouragement to pursue this line of thought. The author is partially supported by NSF-DMS 0354957.

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