May 4, 2007 - Conserve School. 5400 N. Black Oak Lake Drive. Land O)Lakes, WI 54540 (USA). E&mail address: [email protected].
Exponential sums with coe cients 0 or 1 and concentrated Lp norms Bruce Anderson J. Marshall Ash Roger Jonesy Daniel G. Rider Bahman Sa ari May 4, 2007
Abstract A sum of exponentials of the form f (x) = exp (2 iN1 x)+exp (2 iN2 x)+ + exp (2 iNm x), where the Nk are distinct integers is called an idempotent trigonometric polynomial (because the convolution of f with itself is f ) or, simply, an idempotent. We show that for every p > 1; and every set E of the torus T = R=Z with jEj > 0; there are idempotents concentrated on E in the Lp sense. More precisely, for each p > 1; there is an explicitly calculated constant Cp > 0 so that for each E with jEj > 0 and > 0 one can nd an idempotent f R R 1=p is greater than Cp . This such that the ratio E jf jp T jf jp is in fact a lower bound result and, though not optimal, it is close to the best that our method gives. We also give both heuristic and computational evidence for the still open problem of whether the Lp concentration phenomenon fails to occur when p = 1: titre francais: Sommes d'exponentielles a coe cients 0 ou 1 et concentration de normes Lp resume francais: Une somme d'exponentielles de la forme f (x) = exp (2 iN1 x) + exp (2 iN2 x) + + exp (2 iNm x), ou les Nk sont des entiers distincts, est appelee un polyn^ ome trigonometrique idempotent (car f f = f ) ou, simplement, un idempotent. Nous prouvons que pour tout reel p > 1, et tout E T = R=Z avec jEj > 0; il existe des idempotents concentres sur E au sens de la norme y
Partially supported by NSF Grant DMS-9707011 Partially supported by NSF Grant DMS-9531526
1
Lp . Plus precisement, pour tout p > 1; nous calculons explicitement une constante Cp > 0 telle que pour tout E avec jEj > 0, et tout reel > 0, on puisse construire un idempotent f tel que le quoR R 1=p tient E jf jp T jf jp soit superieur a Cp . Ceci est en fait un theoreme de minoration qui, bien que non optimal, est proche du meilleur resultat que notre methode puisse fournir. Nous presentons egalement des considerations heuristiques et aussi numeriques concernant le probleme (toujours ouvert) de savoir si le phenomene de concentration Lp a lieu ou non pour p = 1. mots clefs: idempotents, polyn^omes trigonometriques idempotents, normes Lp , noyau de Dirichlet, concentration de normes, sommes d'exponentielles, conjecture de concentration en norme L1 , operateurs faiblement restreints. keywords: idempotents, idempotent trigonometric polynomials, p L norms, Dirichlet kernel, concentrating norms, sums of exponentials, L1 concentration conjecture, weak restricted operators. classi cation code: Primary 42A05; Secondary 42A10, 42A32. running title: Exponential sums with coe cients 0 or 1
1 1.1
Introduction Concentrated Lp norms
Let e (x) := exp (2 ix) : A sum of exponentials of the form f (x) =
m X
e (Nk x)
k=1
(x 2 R) ;
where the Nk are distinct integers is called an idempotent trigonometric polynomial (because the convolution of f with itself is f ) or, simply, an idempotent. In the sequel we adopt the term \idempotent" for brevity, and we denote by } the set of all such idempotents: X } := f e (nx) : S is a nite set of non-negative integersg: n2S
2
The simplest example of an f 2 } is (one form of) the Dirichlet Kernel of length n; de ned by Dn (x) :=
n 1 X
e( x) =
=0
sin (n x) i(n e sin ( x)
1) x
:
Consider any function g 2 Lp (T) ; where T = R=Z, and any set E jEj > 0: (Throughout, jEj denotes the Lebesgue measure of E.) If Z Z 1=p p p jg (x) j dx jg (x) j dx E
(1) T with
(2)
T
(where 1 p < 1 and 0 < < 1); we say that \at least a proportion of the Lp norm of g is concentrated on E" or, equivalently, that \the function g has Lp concentration on E." We will now recall a challenging (and still partially open) problem on idempotents which can be expressed in terms of this notion of Lp concentration. About 25 years ago it was discovered that for any (arbitrarily small ) arc J of the torus T with jJj > 0; there always exists an idempotent f with at least 48% of its L2 norm concentrated on J. The origin of this curious fact occurred about 1977 when one of us (J. M. Ash) was attempting to show that an operator T de ned on L2 (T), commuting with translations, and of restricted weak type (2; 2) is necessarily a bounded operator on L2 (T) : That T is of restricted weak type (2; 2) means that there is a constant C = C (T ) > 0 such that, for every characteristic function of a subset of T; sup >0
2
measurefx 2 T : jT (x)j > g
C k k2L2 :
Ash[2] was only able to show that this condition was equivalent to there being some positive amount of L2 concentration for idempotents. More explicitly, de ne the absolute constant C2 as the largest real number such that for every arc J T with jJj > 0, we have the inequality Z Z 1=2 2 (3) sup jf (x) j dx jf (x) j2 dx C2 : f 2}
J
T
Thus the issue was whether C2 was 0 or positive. Luckily, at just the same time, Michael Cowling [5] proved, by another method, that every commuting with translations weak restricted type (2; 2) operator is necessarily a 3
bounded operator on L2 : (Actually Cowling [5] proved more. His result allowed the underlying group to be any amenable group, not just T.) This proved, of course, that C2 was indeed positive, but did not give any e ective estimate for it. However, a series of concrete estimates quickly followed. :14, :01, S. Pichorides [14] obtained C2 The referee of [2] obtained C2 H. L. Montgomery [12] and J.-P. Kahane [10] obtained several better lower bounds. (The ideas of H. L. Montgomery were \deterministic" while those of J.-P. Kahane used probabilistic methods from [9].) Finally, in [4], three of us achieved the lower bound 2
sin x = :4802:::; := max p x>0 x
(4)
which, in [6], was proved to be best possible. (See [7] for a more detailed exposition of the contents of [6].) To get a little more feel for what to expect, let be any point of density (also called \Lebesgue point") of a set E T: Then forP every p 2 [1; 1[ the ) e( x); sequence of functions fgn g; where gn (x) := Dn (x ) = n=01 e( p have L concentration tending to 1 as n ! 1: However, the trigonometric polynomials gn are not idempotents, since the non-zero coe cients are not all equal to 1: Note, however, that all the coe cients do have modulus 1: The di culty of the matters studied in [4] and [6], as well as those of the present paper lies precisely in the fact that the trigonometric polynomials f 2 } have all their coe cients equal to 0 or 1; which is a very drastic constraint. At this stage we make an obvious remark: in all the L2 concentration problems on (small) arcs of T studied in [4] and [6] and in all their Lp analogues studied in the present paper, it is equivalent to work on arcs of T or on intervals of [0; 1] : We usually nd it convenient to use \arcs of T" in statements of theorems, but \intervals of [0; 1]" in their proofs!
1.2
The L2 and Lp problems
The results in [4] and [6] were satisfying but, as usual, they led to further questions. The rst two were: a) Can we replace \arc" (or \interval") with \set of positive measure?" b) Can we replace L2 with Lp for any p 4
1?
For each p 2 [1; 1[; de ne Cp as the largest number such that for every set E, E T with jEj > 0, the inequality sup kf kLp ;E /kf kLp := sup f 2}
f 2}
Z
E
p
jf j dx
Z
T
1=p
jf jp dx
Cp
(5)
holds. Similarly, de ne Cp as the largest number such that for every arc J, J T, the inequality Z Z 1=p p sup kf kLp ;J /kf kLp := sup jf j dx jf jp dx Cp (6) f 2}
f 2}
J
T
holds. The de nitions of Cp and Cp are extended to the limit cases p = 1 in the usual way. Obviously, Cp Cp : (7) With regard to question a), although the de nitions allow the possibility for C2 to be smaller than C2 , in [4] it is shown that both are equal to the constant 2 de ned in (4). Whatever the value of p 1, there is no result in this paper which changes when the supremum is taken over all sets of positive measure rather than over all arcs. So we conjecture that inequality (7) is in fact an equality: Cp = Cp ; (8) although we have no proof of this except for p = 2 and p = 1. Question b) is harder. In [6], the constant 2 de ned in (4) is shown to be a lower bound for every Cp when p 2. This is not altogether satisfying for two reasons. First, the cases 1 p < 2 are not addressed. Second, since the constant function 1 is in }, and k1kL1 ;A /k1kL1 = 1 for any non-empty set A T; so that C1 = 1, one might hope to show that limp!1 Cp = 1: In section 1.3 we state new results for the Lp cases, together with some remaining open problems.
1.3
Statement of the result
This paper is devoted to proving one single theorem, Theorem 1 below. It was announced in the Comptes Rendus note [1] (in a weaker form, and presented as two distinct results). The aim of this paper is to supply the proofs of [1], to strengthen the rst result thereof, and to unify the results in the form of a single theorem (which is valid for all sets of positive measure and not just 5
for all arcs). Our theorem is stated in terms of the \constants" cp and cp ; de ned as follows: cp :=
sin ( !) = ( !)
sup 0
21+1=p
b1=!c + 1 +
1 p 1
3 p 8
1=p
;
b1=!c
where for a real number r; dre = ceiling of r = the smallest integer greater than or equal to r, brc = oor of r = the greatest integer less than or equal to r; and cp :=
2 p+1
Z
0
1
p
sin x dx x
1=p
sin ( !) : 1 ! 1 1=p
max
0 !
As p increases from 1 to +1 (resp. from 2 to +1), cp (resp. cp ) increases from 0 to :5 (resp. from 2 = 0:48 : : : to 1). That cp tends to 1 as p ! 1 follows from an easy calculation, which is done in Remark 2 for the reader's convenience. Theorem 1 Whenever 1 < p < 1; we have the estimate 8 if 1 < p 2 < cp : Cp : if 2 p < 1 cp
In other words, if p > 1 and > 0 are given, then for each set E T with jEj > 0, there is a nite set of integers S = S(E; p; ) such that 8 p ,Z p p Z X if 1 < p 2 < cp X e(nx) dx e(nx) dx : : p E n2S T n2S cp if 2 p < 1 (9) Furthermore, cp (and a fortiori Cp ) tends to 1 as p tends to in nity. For a slightly larger lower estimate of Cp ; see inequality (28) at the end of Section 2 (where Part I of Theorem 1 is proved).
6
Remark 2 The estimate in Part II of Theorem 1, which is quite good although not optimal, does have two virtues. First, it is sharp when p = 2; since Z 2 2 1 sin x dx = 1: x 0 Second, it implies that limp!1 Cp = 1. Indeed lim inf Cp p!1
lim inf cp p!1
Z 1 1=p p 1=p sin ! 2 sin x = lim inf max lim dx lim p!1 p!1 p!1 0 ! 1 ! 1 1=p x 0 sin ( =p) sin (1=p) sin x 1 Essup = lim lim (1=p)1=p = 1: lim 1 1=p p!1 p!1 p!1 x ( =p) (1=p)
To prove Theorem 1, it obviously su ces to prove the inequalities Cp cp (for all p > 1) and Cp cp (for all p 2). These two inequalities will be proved, respectively, in the next sections 2 and 3. As for the open problems pertaining to the case p = 1 (conjectures of non-concentration in the L1 sense), we shall state them in Section 4 at the end of the paper.
2
Proof of Theorem 1; Part I: Cp p > 1)
cp (for all
2.1. Since the proof is quite technical and computational, before giving the full proof we start by sketching a (heuristic) outline of the beginning of the proof. Outline of (the beginning of ) the proof. Let q be a large odd positive integer and let m be an integer at least as large. We begin with the special 1 1 1 ; q + mq ]; where m q. The idea of the proof is to think case of J = [ 1q mq of T (or rather of some suitable interval of length 1; which is more convenient in the proofs) as \partitioned" (except for common endpoints) into q conh i 2 1 2 +1 gruent arcs of the form 2q ; 2q ; with centers at q ; ( = 0; 1; : : : ; q 1)
and common length 1q : The idempotent Dmq (qx); where Dn (x) is the Dirichlet kernel de ned by the relation (1) of the introduction, has period 1=q 7
h i 11 11 and when restricted to behaves approximately like the Dirac mea; 2q 2q sure. Now consider the idempotent D!q (x); where ! 2]0; 1=2[ is chosen to make o This, when restricted to a small neighborhood of the n !q an integer. q 1 set 0q ; 1q ; 2q ; :::; q ; behaves roughly (as far as its modulus is concerned) like the function 1=x: Thus the idempotent $(x) = Dmq (qx)D!q (x) satis es R1 R Pq p p j$(x)jp dx P 1 p : Concentration at 1=q j=1 1=j and J j$(x)j dx 0 q follows since j=1 j p is bounded. This outline will be made rigorous in the proof below. We will also have to treat the problem of concentration at points which are not of the form 1=q: 2.2. To prove Theorem 1 we need the following lemma, which will be used in the proof of Part II as well. Lemma 3 Let DN (x) be the Dirichlet Kernel de ned by (1), so that jDN (x)j = jsin ( N x) = sin ( x)j ; and let p be greater than 1: Then Z 1 jDN (x)jp dx = p N p 1 + op (N p 1 ); N !1 0
where p
=
2
Z
1
0
p
sin u du u
and op is the \little o" notation of Landau modi ed to emphasize the dependence of the associated constant on p: More precisely, Z
0
1
jDN (u)jp du =
pN
where the error term Rp (N ) satis es: 8 < Op (N p 3 ) O (log N ) Rp (N ) = : Op (1)
p 1
+ Rp (N ) ;
if p > 3 if p = 3 : if 1 < p < 3
Proof. This result is classical, but we give the full proof for the reader's R 1=2 convenience. Since DN is even, we need only estimate 2 0 jDN (x)jp dx: By 8
the triangle inequality, this di ers from Z 1=2 sin N u p j Ap (N ) := 2 j du u 0 by at most Ep (N ) := 2
Z
0
1=2
Substituting x := N u yields Z N =2 2 sin x p Ap (N ) = j j dx N 0 x=N Z 2 1 sin x p j j dx N p = x 0 =
since
Z
pN
p 1
1
N
1
j sin N ujp
sin
1
p
2
1 ( u)p
u
N
p 1
Z
1
1
N =2
+ Op (1) ;
sin x p j j dx < x =2
Z
du:
j
sin x p j dx x
p+1 p 1
x p dx =
N =2
2 p 1
N
(p 1)
:
So proving the lemma reduces to proving that 8 if p > 3 < Op (N p 3 ) O (log N ) if p = 3 Ep (N ) = : : Op (1) if 1 < p < 3 For 0 < u
1=2, the inequality
1 p sin ( u)
1 ( u)p
( u)p sinp ( u) ( u)p sinp ( u) p up 1 (1 + O (u2 )) = Op (1) = Op u2 u2p
p
immediately leads to the two estimates j sin N ujp
1 sin
p
u
1 ( u)p
= (N u)p Op u2
and 9
p
= Op N p u2
(10)
1
j sin N ujp
sin
p
1 ( u)p
u
= 1 Op u2
p
p
= Op u2
:
(11)
If 1 < p < 3; from estimate (11) we have Z
Ep (N )
1=2
Op u2
p
du = Op (1) ;
0
if p = 3; from estimates (10) and (11) we have Z
E3 (N )
Z
1=N 3 2
O N u
du +
0
1=2
O u2
3
du
1=N
O N 3N
3
+ O (log N ) = O (log N ) ;
and if p > 3; again from estimates (10) and (11) we have Ep (N )
Z
1=N p 2
Op N u
du +
0
Z
1=2 p
Op u2
du
1=N
= Op N p N
3
+ Op N
(3 p)
= Op N p
3
:
Thus the lemma is proved. 2.3. We now proceed to prove Theorem 1 in detail. We nd it convenient to split the proof into eight steps. The rst three steps deal with concentration at 1=q and the remaining ve steps with the general case. First Step. Concentration at 1=q: lower estimation of numerator. Let ! = !(q) be a constant in (0; 1=2) such that !q is an integer. We will estimate Z Z 1=p
1
p
p
J
j$(x)j dx
0
j$(x)j dx
;
(12)
where $(x) := Dmq (qx)D!q (x):
We begin with the numerator, N; of (12). Suppose that j 1q bounded derivative,
xj
jD!q (x)j = jD!q (
1 q
1 ; mq
and let
)j =
:=
1 q
x: Then, since sin x has a
sin ! + O ( q) sin ! = + O (1) sin q + O ( ) sin q 10
=
sin !
+ O (1) ;
q
since m q: Using this and Minkowski's inequality in the form kF + GkLp kF kLp kGkLp ; we have Z
N
J
jDmq (qx)jp
Z
1=p
p
q sin !
dx
O
J
1=p
jDmq (qx)jp dx
!
:
(13)
Substitute u = qx to get sin !
N
De ne
:=
R1 0
Z
q1
1=p
O q
Z
1=p
1=m
p
jDmq (u)j du =
p
2
Z
N
p
(14)
1=2
1=m
Use the estimates jsin mq uj 1 and jsin uj gral, thereby obtaining the estimate
By the lemma,
:
: Since Dmq is even, we may write
1=m
Z
jDmq (u)jp du
!1=p
1=p
jDmq (u)jp du
1=m
1=m
jDmq (u)jp du: 2u to control the last inte-
1=m 1=m
'
pq
jDmq (u)jp du =
p 1
sin !
p
O mp
1
:
mp 1 ; which together with (14) implies q1
1=p
O q
1=p
1
O q
1+1=p
;
(15)
or, more simply, N
sin !
q1
1=p
f1
o(1)g :
(16)
Second Step. Concentration at 1=q: upper estimation of denominator.
11
Passing now to the estimate of the denominator, D; of (12), we have Z 1 p D = jDmq (qx)jp jD!q (x)jp dx: (17) 0
We now estimate this in great detail. We decompose q 1 Z j+1 X q p D = jDmq (qx)jp jD!q (x)jp dx: j q
j=0
Let x = y + qj ; dy = dx; to get p
D =
q 1Z X
1 q
jDmq (qy + j)jp D!q y +
0
j=0
p
j q
dy:
Since jDmq (qy + j)j = jDmq (qy)j for any integer j; p
D =
q 1Z X
1 q
jDmq (qy)j D!q
0
j=0
p
j y+ q
p
t j + q q
p
t j + q q
p
dy:
Let t = qy to get q 1
1X D = q j=0 p
Z
0
1
jDmq (t)jp D!q
Interchange sum and integral: Z 1 q 1 X p 1 p D = jDmq (t)j D!q q j=0 0
dt:
dt:
(18)
Replace the sum by its supremum over all t 2 [0; 1] which is the same as sup
q 1 X
D!q x +
x2[0; 1q ) j=0
so recalling that
p
=
D
p
R1 0
j q
p
;
jDmq (t)jp dt; we have p1
q
sup
q 1 X
x2[0; 1q ) j=0
12
D!q
j x+ q
p
:
Since D!q is even and has period 1; q 1
D
p
2
p1
q
2 X
sup
p
j x+ q
D!q
x2[0; 1q ) j=0
:
Break the sum into two pieces using the standard estimates jD!q (x + j=q)j !q when j b1=!c and j sup D!q (x + ) q x2[0; 1 )
j q
1= sin
q
q 2j
when j 2 [b1=!c + 1; (q 1)=2] : We have 8 9 1 q 1 > > bX !c p 2 < = X q p p p1 D 2 (!q) + q> 2j > : j=0 ; 1 j=b ! c+1 ( ) Z 1 p 1 1 q dx 2 p + 1 (!q)p + p q ! 2 b !1 c x !p ) ( p 1 1 1 1 ! = 2 pqp 1!p +1 + ! 1 ! p 1 2 ! ( ) 1 1 = 2 pqp 1!p +1+ ! ; ! p 1 where
p
1 1=! 2 b1=!c
=
:
Third Step. Concentration at 1=q: conclusion. Now combine this estimate with estimate (16) to get sin !
N D
2
sin ! !
2
1 !
1=p
1 !
pqp 1!p
or N D
q1
+1+
f1
+1+ 13
f1
o(1)g p 1
o(1)g b !1 c
p 1
1=p
b !1 c
1=p
:
The numbers ! = ! (q) appearing in the above two steps are, by construction, rational numbers. However it is clear, from their de nition, that as q varies these ! (q) are everywhere dense in [0; 1=2] : (Cf. also eighth step of this proof.) As will be made explicit below, this estimate, when extended to general intervals, is su cient to prove Theorem 1. (Notice that < (3=8)p since ! < 1=2.) Fourth Step. General case: heuristic search for the concentrated exponential sum. From now on, our goal is to extend the above estimate (of the third step) to any interval. This fourth step is purely heuristic. Given any interval J ]0; 1[; we can nd q (and in fact in nitely many 1 k 1 ; + qm ] J; where such q's) so that for some integer k 2 [1; q[; [ kq qm q m := q. (This choice of m is for technical reasons that will become clear in the eighth step of the proof.) It is convenient to pick q prime, which implies that k and q are relatively prime. Hence there exists a unique pair (a; b) of integers such that ak
bq = 1;
(0 < a < q
and
0 < b < k)
(19)
i.e., (the conjugate class of) a is the multiplicative inverse of k in the nite eld Zq = GF (q) : Multiplication by a; when reduced modulo q; de nes a bijection from f0; 1; :::; q 1g to itself. Furthermore, (k) = 1: For the sake of this heuristic argument, temporarily suppose that (as previously) ! is chosen so that !q is an integer. (In fact, in the rigorous argument below, we shall choose ! according to another criterion, so that !q will not be an integer, but instead of !q we shall use the integer d!qe : Now the idempotent D!q (ax) behaves very much like the idempotent D!q (x); except that the former does at k=q what o at 1=q. Since Dmq (qx) is constant n the latter does
and large on the set 0q ; 1q ; 2q ; :::; q q 1 ; and D!q (ax) restricted to this set takes on the same set of values as D!q (x) did, but has its maximum at k=q (instead of at 1=q); it seems reasonable that, the idempotent Dmq (qx)D!q (ax) should work here. The de nition of the idempotent G (x) analyzed below was motivated by these considerations. Fifth Step. General case: the desired concentrated exponential sum. Let E be a subset of T of positive measure and let > 0 be given. First we will nd an integer Q, and an = (Q; ) > 0; then an interval J of 14
h i 1 k 1 the form J = kq mq so that jJ \ Ej > (1 ) jJj; and nally we ; q + mq will de ne an idempotent f depending on k and Q which is -close to being su ciently concentrated on rst J and then E. Actually in what follows we will always take m to be equal to q, but we leave m in the calculations since some increase of the concentration constants may be available by taking other values of m. First we de ne . The function f will have the form mQ 1
f (x) = G (x)
X
e (nqx) .
n=0
Since G will turn out to be a sum of q exponentials (see (22), kGk1 so the decomposition J = (J r E) [ (J \ E) allows Z jf jp dx jJj (kf k1 )p
q;
JnE
2 (qmQ)p mq = ( Q) 2q p 1 (mQ)p
In the sixth step below, we will get Z p sin ! jf jp dx qp 1
p
(mQ)p
1
1
:
+ o qp
1
(mQ)p
1
J
as long as q is large enough. Pick (Q; ) so small that from this will follow Z Z Z p sin ! p p p p jf j dx = jf j dx jf j dx (1 ) q p 1 p (mQ)p 1 : J\E
J
JnE
(20) Now that we know how to choose , we show how to nd J = J (E; ). We will use m = q while choosing J. Almost every point has the property that there are in nitely many primes q and integers k for which k 1 < 2 . (See [8].) q q
(21)
We may assume that E has an irrational point of density for which inequality (21) holds for in nitely many primes q. Suppose that the prime q 15
i + q22 satis es jK Ej 2 jKj and such that i h condition (21) holds. With J = kq q12 ; kq + q12 , we have is so large that K =
h
2 ; q2
jJ Ej
jK Ej 2
jKj =
4 2 = 2 = jJj ; 2 2q q
so that jJ \ Ej > (1 ) jJj, as required. Let a be (uniquely) de ned by (19), which in turn uniquely de nes the bijection from f0; 1; :::; q 1g into itself (reduction modulo q of multiplication by a). We have (r) = ra sq where s is the largest (necessarily nonnegative) integer such that sq ra: This leads us to consider the following sets Ej ; j = 1; 2; :::. For each integer j 0 let Ej denote the set of those r 2 f0; 1; :::; q 1g such that (r) = ra jq: A priori the Ej are pairwise disjoint, and it is straightforward to check that 8 if 0 j < a < fr 2 N : tj r < tj+1 g Ej = ; : ; if j a
where ftj g is the (strictly increasing) nite sequence of integers de ned by t0 := 0; tj = djq=ae if 0 < j < a; and tj = q if j = a: Thus fEj g0 j
