Estimates for the Square Variation of Partial Sums of Fourier Series and their Rearrangements

0.41 MB·English

by  Allison Lewko

Estimates for the Square Variation of Partial Sums of Fourier
Series and their Rearrangements
Allison Lewko∗ and Mark Lewko
Abstract
2
We investigate the square variation operator V2 (which majorizes the partial sum max-
1
0 imal operator) on general orthonormal systems (ONS) of size N. We prove that the L2
2 norm of the V2 operator is bounded by O(ln(N)) on any ONS. This result is sharp and
refines the classical Rademacher-Menshov theorem. We show that this can be improved to
n
a O( ln(N)) for the trigonometric system, which is also sharp. We show that for any choice
J of coefficients, this truncation of the trigonometric system can be rearranged so that the
p
1 L2 norm of the associated V2 operator is O( lnln(N)). We also show that for p > 2, a
3 bounded ONS of size N can be rearranged so that the L2 norm of the Vp operator is at
p
most O (lnln(N)) uniformly for all choices of coefficients. This refines Bourgain’s work on
] p
A Garsia’s conjecture, which is equivalent to the V case. Several other results on operators
∞
C ofthisformarealsoobtained. Theproofsrelyoncombinatorialandprobabilisticmethods.
.
h
t 1 Introduction
a
m
[ Let T := [0,1] denote the unit interval with Lebesgue measure dx and let Φ := {φn}n∈N
denoteanorthonormalsystem(ONS)ofrealorcomplexvaluedfunctionsonT. ByanONS,

2 we will always mean the set of orthonormal functions φn n N and the ordering inherited
v from the index set N. For f L2, we let a = f,φ {den}ot∈e the Fourier coefficients of f
n n
1 ∈ h i
with respect to the system Φ. Associated to an ONS is the maximal partial sum operator
7
8
N
0
f(x):=sup a φ (x) .
. n n
6 M N (cid:12) (cid:12)
0 (cid:12)(cid:12)nX=1 (cid:12)(cid:12)
1 ItiswellknownthattheL2boundednessof(cid:12)theoperator(cid:12) impliesthealmosteverwhere
(cid:12) (cid:12)M
1 convergence of the partial sums of the expansion of f L2 in terms of the ONS Φ. Almost
: ∈
v everywhere convergence is known to fail for some ONS, hence the maximal function is
M
i known to be an unbounded operator on L2 for some ONS. There is an optimal estimate
X
known for general ONS.
r
a Theorem 1. (Rademacher-Menshov) Let φn n N =Φ and f L2 be as above. Then,
{ } ∈ ∈
1
2
∞
f a 2ln2(n+1)
L2 n
||M || ≪ | | !
n=1
X
where the implied constant is absolute. Moreover, the function ln2(n + 1) cannot be
replaced with any function that is o(ln2(n+1)).
This last claim is quite deep and is due solely to Menshov.
While this estimate is optimal in general, it can be improved for many specific systems.
For instance, the inequality f f is known to hold when Φ is taken to be the
L2 L2
||M || ≪|| ||
∗Supported bya National Defense Science and Engineering Graduate Fellowship.
1
trigonometric, Rademacher, or Haar systems. We recall the definitions of these systems in
the next section.
Recently, variational norm refinements of the maximal function results stated above
havebeeninvestigated. Tostatetheseresults,wefirstneedtointroducesomenotation. Let
a= a be a sequence of complex numbers. Then we define the r-variation as:
{ n}∞n=1
r 1/r
a := lim sup a ,
Vr n
|| || K (cid:12) (cid:12) !
→∞ PK IX∈PK(cid:12)(cid:12)Xn∈I (cid:12)(cid:12)
(cid:12) (cid:12)
wherethe supremumistakenoverallpartitions K(cid:12)of[K](cid:12)(i.e. allwaysofdividing[K]into
P
disjoint subintervals). When a is a finite sequence of length K, the quantity is defined by
dropping the lim .
K
→∞
Onecaneasilyverifythatthisisanormandisnondecreasingasrdecreases. Nowwewill
denote the sequence a φ (x) by S[f](x). (Note that this is slightly different than the
{ n n }∞n=1
notationusedin[12].) Whenwewrite S[f] Vr(x), wemeanthefunctiononTwhosevalue
|| ||
at x T is obtained by assigning the r-th variation of the sequence S[f](x). Furthermore,
∈
S[f] is the Lp norm of this function. Alternately, we have
Lp(Vr)
|| ||
1/2
K
f (x)=sup sup S [f](x) S [f](x)2 ,
|| ||V2 K n0<...<nK l=1| nl − nl−1 | !
X
where S [f](x)= nl a φ (x) is the n -th partial sum.
nl n=1 n n l
We note that the function S[f] (x) is essentially the maximal function. More pre-
V∞
P || ||
cisely, f(x) S[f] (x) f(x). Since the quantity a is nondecreasing as r
V∞ Vr
M ≪ || || ≪ M || ||
decreases, we see that S[f] (x) majorizes the maximal function whenever r < . In
Vr
|| || ∞
[12], the following is proved for the trigonometric system e2πinx :
{ }∞n=1
Theorem 2. Let r >2 and r <p< , where 1 + 1 =1. Then
′ ∞ r r′
S[f] C f ,
Lp(Vr) p,r Lp
|| || ≤ || ||
where C is a constant depending only on p and r.
p,r
Thisresultisratherdeep,beingastrengthenedversionofthecelebratedworkofCarleson
and Hunt on the almost everywhere convergence of Fourier series. The analogous inequal-
ities were previously obtained in [9] in the simpler situation of Ces`aro partial sums of the
trigonometricsystem. Moreover,the aboveinequality is knownto holdfor the Haar system
and more generally for martingale differences by Lepingles inequality, a variational variant
of Doob’s maximal inequality. In [12], it is shown that the condition r > 2 is necessary in
case of the trigonometric system. Our focus here will be to study the case p = r = 2 for
general ONS. In this direction, we prove (closely following the classical proof):
Theorem 3. Let Φ be an ONS. Then
1/2
∞
S[f] a 2ln2(n+1) . (1)
L2(V2) n
|| || ≪ | | !
n=1
X
If f ∆(N) f for all f = N a φ for somerealvaluedfunction ∆(N), then
||M ||L2 ≪ || ||L2 n=1 n n
P 1/2
N
f ∆(n)ln(n+1)a 2 . (2)
L2(V2) n
||| || ≪ | | !
n=1
X
Interestingly, the first inequality strengthens the Rademacher-Menshov theorem stated
above, since the right sides are the same (up to implicit constants), yet we have replaced
the maximal function with the square variation operator V2 on the left side. Since the V2
operator dominates the maximal operator, this implies the Rademacher-Menshov theorem
2
and the claim that this result is sharp follows from the sharpness of Rademacher-Menshov.
This might lead one to think that the two operators behave similarly, however we will see
that the V2 operator is much larger than the maximal operator for the classical systems.
Theorem 3 can be refined further for certain classes of ONS, see Section 7 for discussion of
this.
We can apply (2) to the trigonometric system with ∆(N) = O(1), the Carleson-Hunt
inequality, and obtain the following corollary:
Corollary 4. Let e2πinx be the trigonometric system. We then have
{ }∞n=1
1/2
∞
S[f] a 2ln(n+1) . (3)
L2(V2) n
|| || ≪ | | !
n=1
X
Moreover, the function ln(n+1) cannot be replaced by a function that is o(ln(n+1)).
ThelowerboundcanbeobtainedbyconsideringtheDirichletkernelD (x)= N e2πinx.
N n=1
A proof of this is contained in Section 2 of [12]. Strictly speaking, they work with the de la
P
Vallee-Poussin kernel there, but the same proof works for the Dirichlet kernel.
As we will see below, it is easy to construct an infinite ONS such that S[f]
L2(V2)
|| || ≪
f holds, bychoosingthe basis functions φ (x) to havedisjointsupports. However,this
L2 n
|| ||
is a very contrived ONS, and it is then natural to ask if there exists a complete ONS such
that S[f] f . This is not possible. In fact, we show slightly more:
L2(V2) L2
|| || ≪|| ||
Theorem 5. Let φ be a complete orthogonal system. There exists a L function such
n ∞
{ }
that S[f] (x)= for almost every x.
V2
|| || ∞
In general, this divergence cannot be made quantitative. We show that for any function
w(n) , there exists a complete ONS such that S[f] w(N) f whenever
L2(V2) L2
→ ∞ || || ≪ || ||
f(x) = N a φ (x). However, a quantitative refinement is possible if we restrict our
n=1 n n
attention to uniformly bounded ONS:
P
Theorem 6. In the case of a uniformly bounded ONS, it is not possible for w(N) =
o( lnln(N)). However,theredoexistuniformlyboundedONSsuchthatw(N)=O( lnln(N)).
pTheRademachersystemprovidesanexampleofthesecondclaim. SeeTheoremp9below.
RecallthatwedefinedanONStobeasequenceoforthonormalfunctionswithaspecified
ordering. This is essential since the behavior of the maximal and variational operators
depend heavily on the ordering. For instance, the Carleson-Hunt bound on the maximal
function forthe trigonometricsystemmakesessentialuseofthe orderingofthe system,and
the result is known to fail for other orderings. It is thus natural to ask what one can say
abouttheV2 operatorforreorderingsofthetrigonometricsystem. Surprisingly,itturnsout
that the O( ln(N)) bound can be improved to O( lnln(N)) for any choice of coefficients
by reordering the system. More generally:
p p
Theorem 7. Let φ N be an ONS such that φ (x) = 1 for all x and n, and let
{ n}n=1 | n |
f(x)= N a φ (x). Then there exists a permutation π :[N] [N] such that
n=1 n n →
P
f lnln(N) f
L2(V2) L2
|| || ≪ || ||
holds (for sufficiently large N) with respect topthe rearranged ONS ψ N , where ψ (x):=
{ n}n=1 n
φ (x).
π(n)
This is perhaps the most technically interesting part of the paper. This result should
be compared to Garsia’s theorem [7], which states that the Fourier series of an arbitrary
function with respect to an arbitrary ONS can be rearrangedso that the maximal function
is bounded on L2. Garsia’s proof proceeds by selecting a uniformly random permutation,
and arguing that it will satisfy the claim with positive probability. In our case, however,
we randomize over a subset of all permutations. This subset is chosen based on structural
information about the Fourier coefficients of the function. It is unclear if this restriction
3
is necessary or an artifact of our proof techniques. It would be interesting to extend this
resulttomoregeneralONS.WenotethatitcanbeseenfromtheworkofQian[16](seealso
our refinement [11]) that N r Nlnln(N) = lnln(N) N r ,
|| n=1 n||L2(V2) ≫ || n=1 n||L2
regardlessofthe orderingofthe Rademacherfunctions r , hence the lnln(N) terminthe
P p n p P
statement of the theorem is sharp. A similar result can be obtained for general ONS when
p
the coefficients are multiplied by random signs:
Theorem 8. Let φ N be an ONS and f(x) = N a φ (x). Then there exists a
{ n}n=1 n=1 n n
sequence of signs ǫ such that
n
P
g lnln(N) g
L2(V2) M L2
|| || ≪ || ||
p
holds, where g(x)= N ǫ a φ (x).
n=1 n n n
This easily followPs from the following inequality:
Theorem9. Let r N beasequenceofuniformlyboundedindependentrandomvariables.
{ n}n=1
Then
1/2
N N
a r lnln(N) a2 .
(cid:12)(cid:12) n n(cid:12)(cid:12) ≪ n!
(cid:12)(cid:12)nX=1 (cid:12)(cid:12)L2(V2) p nX=1
(cid:12)(cid:12) (cid:12)(cid:12)
In particular, combinin(cid:12)g(cid:12) this with(cid:12)(cid:12)Theorem 6, we see that the L2 norm of the V2 operator
(cid:12)(cid:12) (cid:12)(cid:12)
for the Rademacher system grows like lnln(N).
Finally, we prove that the Vp normpof some systems can be improved uniformly for all
choices of coefficients by a rearrangement,for p>2.
Theorem 10. Let φ N be an ONS such that φ M for each n, and let p >2.
{ n}n=1 || n||L∞ ≤
There exists a permutation π : [N] [N] such that the orthonormal system φ N
→ { π(n)}n=1
satisfies
S[f] lnln(N) f (4)
L2(Vp) M,p L2
|| || ≪ || ||
for all f = N a φ .
n=1 n n
The maxPimal V∞ version of this result is due to Bourgain [1] and represents the best
progressknowntowardsGarsiaandKolmogorov’srearrangementconjectures. Ourmethods
rely heavily on those developed in that paper. This also leads us to perhaps the most
interesting open problem relating to V2 operators:
Question 11. Does there exist a permutation π :[N] [N] such that the L2 norm of the
→
associated V2 operator on the trigonometric system grows like o( ln(N))?
Our Theorems 7 and 10 may be viewed as evidence that thips may in fact be possible.
It is consistent with our knowledge that one could get growth as slow as lnln(N). It is
known that purely probabilistic techniques in the maximal (V ) case can only go as far as
∞ p
Bourgain’sboundoflnln(N)(seeRemark2of[1]). Thus,findingapermutationthatreduces
the growth further (Garsia’s conjecture is the assertion that there exists a rearrangement
thatgetstoO(1))wouldrequirefundamentallynewideas. However,itisconsistentwithour
current knowledge that the purely probabilistic techniques could get one down to lnln(N)
in the V2 case. If true, this will certainly require a much more delicate analysis than the
methods used here. Theorem 3 combined with the V case of the previous theorem does
∞
give a bound of ln(N)lnln(N) for general bounded ONS for the V2 operator. This is a
nontrivial improvement for some systems, but not the most interesting classical systems.
p
2 Notation and General Remarks
WewillworkwithONSdefinedontheunitintervalT. TheunderlyingspaceTplaysalmost
noroleinourproofs(the roleissimilartothatofaprobabilityspaceinprobabilitytheory),
and one could replace it with an abstract probability space.
4
We assume that the ONS is real valued in most of our results. In these cases, one can
obtain the same results for complex valued ONS by splitting into real and imaginary parts
andapplying the argumentsto each. The details areroutineso weomitthem. The proofof
Theorem 7 is the one place where this requires some care, and thus we work with complex
valued functions directly there.
Wedefinethetrigonometricsystemtobethesystemofcomplexexponentials e2πinx .
{ }∞n=1
Typicallythetrigonometricsystemisdefinedtobethedoublyinfinitesystem e2πinx
andthemaximalandvariationaloperatorsaredefinedwithrespecttothesym{metric}p∞na=r−ti∞al
sums. However,we find it more convenientto define the trigonometricsystem this way and
avoidhavingto state allofthe following resultsfor both singly anddoubly infinite systems.
All of our results can easily be transferred to the doubly infinite setting (using symmetric
partialsums)bysplittingtheFourierseriesofafunctionf L2(T)withrespecttoadoubly
∈
infinite systeminto twofunctions withsingly infinite Fourierseriesandapplyingthe results
in this setting. For instance, note that
N 0 N
f(x):=sup a φ (x) sup a φ (x) +sup a φ (x) .
n n n n n n
M N (cid:12) (cid:12)≪ N (cid:12) (cid:12) N (cid:12) (cid:12)
(cid:12)(cid:12)nX=−N (cid:12)(cid:12) (cid:12)(cid:12)n=X−N (cid:12)(cid:12) (cid:12)(cid:12)nX=1 (cid:12)(cid:12)
Thusitfollowsthat(cid:12)theL2bounded(cid:12)nessoft(cid:12)hemaximalope(cid:12)ratora(cid:12)ssociatedtot(cid:12)hesystem
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
e2πinx implies the L2 boudedness of the symmetric maximal operator associated to
{ }∞n=1
e2πinx , and similarly for the Vp operators.
{ The}H∞n=a−a∞r system, which we denote by , is a complete ONS comprised of the
{Hn}∞n=0
following functions. For k N and 1 j 2k, we define by
k,j
∈ ≤ ≤ {H }
√2k x∈ j2−k1,j−21k/2 ,
Hk,j(x)=−0√2k xot∈he(cid:16)(cid:16)rwj−i2s1ke/.2,2jk(cid:17)(cid:17),
We formthe system byorderingthe basisfunctions firstbythe parameterk and
n k,j
H {H }
then by the parameter j, or = for n=2k+j. Lastly, we set H =1.
n j,k 0
H H
The Rademacher system, denoted r (x) , is defined by
{ n }∞n=1
r (x)=signsin(2nπx).
n
TheRademachersystemcanalsobethoughtofasindependentrandomvariableswhichtake
each of the values 1,1 with probability 1/2.
{− }
3 Variational Rademacher-Menshov-Type Results
We start by giving a proof of Theorem 3.
It suffices to assume that N is a power of 2, say N =2ℓ. For all i,k such that 0 i ℓ
≤ ≤
and 0 k 2ℓ i 1, we consider the collection of intervals I :=(k2i,(k+1)2i].
− k,i
≤ ≤ −
Lemma12. AnysubintervalofS [0,2ℓ]canbeexpressedasthedisjoint unionofintervals
⊂
of the form I , such as
k,i
S = I (5)
km,im
m
[
where at most two of the intervals I in the union are of each size, and where the union
km,im
consists of at most 2ℓ intervals.
Proof. Let S = [a,b] and set i := max i. It follows that there are at most two
′ Ik,i⊆S
intervals of the form I contained in S (otherwise S would contain an interval of the
k,i′
form I ). Let r denote the right-most element of the interval with the largest k value
k,i′+1
satisfying I S. Now b r has a unique binary expansion. It easily follows from this
k,i′
⊆ −
that (r,b] can be written as [r,b] = I where the union contains only one interval
m km,im
S
5
of the form I of any particular size, and these intervals are disjoint. An analogous
km,im
argument allows us to obtain a decomposition of this form also for [a,r ], where r is the
′ ′
left-most element of an interval with the smallest k value satisfying I S. The lemma
k,i′
⊆
follows by taking the union of these two decompositions.
We now prove
Lemma 13. In the notation above, we have that
1/2
∞
S[f] ln(N) a 2 . (6)
L2(V2) n
|| || ≪ | | !
n=1
X
Proof. Byroundinguptothenearestpoweroftwo,wecanassumewithoutlossofgenerality
that N =2ℓ for some positive integer ℓ (this change will only affect the constants absorbed
by the notation). Now, for each x, we have some disjoint intervals J ,...,J [N] such
1 b
≪ ⊆
that:
2
b
||S[f]||V2(x)=vu  anφn(x) .
uuXj=1 nX∈Jj
t  
It is important to note that these intervals depend on x.
By Lemma 12, each J can be decomposed as a disjoint union of the form (5). In this
j
disjoint union of intervals I , each value of i appears at most twice. For each j and i,
km,im m
we let Ij denote the union of the (at most two) intervals in the decomposition of J which
i j
are of length 2i. We then have:
2
b ℓ
||S[f]||V2(x)=vu  anφn(x) .
uuXj=1 Xi=0nX∈Iij
t  
Applying the triangle inequality for the ℓ2 norm, this is:
2
ℓ b
v anφn(x) .
≤ u  
Xi=0uuXj=1 nX∈Iij
t  
Now, since each Ij is a union of at most two intervals, this implies:
i
2
ℓ 2ℓ−i 1
−
||S[f]||V2(x)≪ vu  anφn(x) . (7)
Xi=0uu kX=0 nX∈Ik,i
t  
Notice that we are now summing over all intervals I for each i, regardless of the value of
k,i
x.
We take the L2 norm of both sides of (7), and apply the triangle inequality to obtain:
2
ℓ 2ℓ−i 1
−
||S[f]||L2(V2) ≪ (cid:12)(cid:12)(cid:12)(cid:12)vu  anφn(x) (cid:12)(cid:12)(cid:12)(cid:12) . (8)
Xi=0(cid:12)(cid:12)(cid:12)(cid:12)uu kX=0 nX∈Ik,i (cid:12)(cid:12)(cid:12)(cid:12)
(cid:12)(cid:12)t   (cid:12)(cid:12)L2
(cid:12)(cid:12) (cid:12)(cid:12)
By linearity of the integral and Par(cid:12)s(cid:12)eval’s identity, we have that(cid:12)(cid:12)
(cid:12)(cid:12) (cid:12)(cid:12)
2ℓ−i 1 2 2ℓ−i 1 12 N 21
− −
(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)vuuu Xk=0 nX∈Ik,ianφn(x) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = kX=0 nX∈Ik,ia2n = nX=1a2n! ,
(cid:12)(cid:12)t   (cid:12)(cid:12)L2  
(cid:12)(cid:12) (cid:12)(cid:12)
(cid:12)(cid:12) (cid:12)(cid:12)
(cid:12)(cid:12) (cid:12)(cid:12)
6
for each i. Combining this with (8) and noting that there are lnN values of i, we have:
≪
1/2
∞
S[f] ln(N) a 2 .
L2(V2) n
|| || ≪ | | !
n=1
X
Wenowdefineavariantofthefunction S[f] (x)whichwewilldenoteby S [f] (x).
V2 L V2
|| || || ||
For each x, we define S [f](x) to be the sequence of differences of lacunary partial sums
L
of f at x, i.e. S [f](x) := S [f](x),S [f](x) S [f](x),S [f](x) S [f](x),... . As
L 20 21 20 22 21
{ − − }
usual, we let S [f] (x) denote the 2-variation of this function.
L V2
|| ||
Lemma 14. In the notation above we have that
1/2
∞
SL[f] L2(V2) ln(n+1)an 2 .
|| || ≪ | | !
n=1
X
Proof. We will need the inequality a2 2a b2 +2b2 for any real numbers a,b. For
| | ≤ | − | | |
each x, there exists some sequence m (x),m (x),m (x),... such that:
0 1 2
S [f] 2 (x)= S [f](x)2+ ∞ S [f](x) S [f](x) 2. (9)
|| L ||V2 | 2m0(x) | 2mi(x) − 2mi−1(x)
i=1
X(cid:12) (cid:12)
(cid:12) (cid:12)
Setting a:=S [f](x) S [f](x)andb:=f(x) S [f](x),we canapply
2mi(x) − 2mi−1(x) − 2mi−1(x)
the inequality above to obtain:
S [f](x) S [f](x) 2 2 S [f](x) f(x)2+2 S [f](x) f(x) 2
2mi(x) − 2mi−1(x) ≤ | 2mi(x) − | 2mi−1(x) −
(cid:12) (cid:12) (cid:12) (cid:12)
fo(cid:12)r each i 1. Combining this w(cid:12)ith (9), we have: (cid:12) (cid:12)
≥
S [f] 2 (x) S [f](x)2+ ∞ S [f](x) f(x)2+ S [f](x) f(x) 2
|| L ||V2 ≪ | 2m0(x) | | 2mi(x) − | 2mi−1(x) −
i=1
X (cid:12) (cid:12)
(cid:12) (cid:12)
S [f](x)2+ ∞ S [f](x) f(x)2
≪ | 2m0(x) | | 2mi(x) − |
i=0
X
S [f](x)2+ ∞ S [f](x) f(x)2.
≪ | 2m0(x) | | 2m − |
m=0
X
Note that in this last quantity, we are always summing over all values of m, instead of
summing over a subsequence dependent on x.
This gives us
1
2
S [f] (x) S [f](x)2+ ∞ S [f](x) f(x)2 .
|| L ||V2 ≪ | 2m0(x) | | 2m − | !
m=0
X
Now we take the L2 norm of both sides of this inequality to obtain:
1
2
∞
S [f] ln(n+1)a2 .
|| L ||L2(V2) ≪ n!
n=1
X
To see this, note that |S2m[f](x)−f(x)| = ∞n=2m+1anφn(x) and each n is greater than
2m for ln(n) values of m. The result then follows from Parseval’s identity.
≪ (cid:12)P (cid:12)
(cid:12) (cid:12)
We now combine these two results to prove the following theorem.
7
Theorem 15. For an arbitrary ONS, in the notation above, we have
1
2
∞
S[f] ln2(n+1)a2 .
|| ||L2(V2) ≪ n!
n=1
X
Proof. We write U (x) := 2k a φ (x) (when k = 0, U (x) := a φ (x).). We claim
k n=2k−1+1 n n 0 1 1
that
P
∞
S[f] 2 S [f] 2 (x)+ U 2 (x) dx.
|| ||L2(V2) ≪ T || L ||V2 || k||V2 !
Z k=0
X
To see this, note that any interval [a,b] can be decomposed as the disjoint union of at
most three intervals I ,I ,I , where I = (2k,2k′] and I (2k 1,2k] and I (2k′,2k′+1)
l c r c l − r
(here, 2k can be set as the smallest integral power of 2⊆contained in [a,b],⊆and 2k′ can
be set as the largest integral power of 2 contained in [a,b]). Now, S [f] 2 (x)dx
T|| L ||V2 ≪
∞n=1ln(n+1)|an|2fromthepreviouslemma,whichisclearlyboundedRby ∞n=1ln2(n+1)a2n.
By Lemma 13, we have
P P
2k 2k
U 2 (x)dx ln2(2k+1) a2 ln2(n+1)a2.
T|| k||V2 ≪ n ≪ n
Z n=2Xk−1+1 n=2Xk−1+1
Combining these estimates completes the proof.
Next we show that these estimates can be improved if one has additional information
regarding the ONS. In particular,if the partialsum maximal operator associatedto the
system is bounded then one can replace the ln2(n) above with an ln(n)M.
1/2
Theorem16. Letf(x)= N a φ (x)andassumethat f ∆(N) N a2
n=1 n n ||M ||L2 ≪ n=1 n
for any choice of f. Then (cid:16) (cid:17)
P P
1/2
N
f ∆(N) ln(N) a2
|| ||L2(V2) ≪ n!
n=1
p X
and
1/2
N
f ∆(n)ln(n+1)a2 .
|| ||L2(V2) ≪ n!
n=1
X
In particular, if the quantity on the right is finite, then the variational operator applied to f
must be finite almost everywhere.
Proof. As before, without loss of generality, we may assume that N =2ℓ for some positive
integer ℓ. And we consider the collection of dyadic subintervals of [1,N] of the form I =
k,i
(k2i,(k+1)2i] for each 0 i ℓ, 0 k 2ℓ i 1. We will refer to intervals of this form
−
≤ ≤ ≤ ≤ −
as admissible intervals.
NowwenotethatanarbitraryintervalJ =[a,b] [N]canbewrittenasadisjointunion
⊆
J = J J , where J I and J I and J 1 I and J 1 I . We
l ∪ r r ⊆ kr,ir l ⊆ kl,il | l| ≥ 2| kl,il| | r| ≥ 2| kr,ir|
allow one of the intervals to be empty if needed, although in the following we will always
assume that the intervals are not empty, since estimating the contribution from an empty
intervalis trivial. Thatis, we canwrite anarbitraryinterval J asthe unionoftwointervals
which are contained within admissible intervals and the intersection with the admissible
intervals is a constant fraction of the the admissible interval.
For J [N], let S := a φ (x). We now claim the pointwise inequality
⊆ J n∈J n n
P
f 2 (x) S (x)2.
|| ||V2 ≪ |M Ik,i |
0≤Xi≤ℓ 0≤k≤X2ℓ−i−1
8
Note that the sum on the right is only over all admissible intervals. To see that this
inequality holds, let J m be a partition of [N] that maximizes the square variation
{ i}i=1
(at x). From the discussion above, we can associate disjoint Jl and Jr to J such that
i i i
J Jl Jr. Moreover,wecanfinddisjointadmissibleintervalsIl andIr suchthatJs Is
i ⊂ i∪ i i i i ⊆ i
and Js 1 Is (s r,l ).
| i|≥ 2| i| ∈{ }
We observe that |SJi(x)|2 ≪ |MSIil(x)|2 +|MSIir(x)|2. Moreover, any particular ad-
missible interval I will be associated to at most two intervals in the partition J since
i
{ }
the intervals in the partition are disjoint and have at least half the length of the associated
admissible interval. The pointwiseinequality abovenow follows. Nowintegratingeachside,
applying the hypothesized inequality S 2 ∆2(N) a2, and noting that every
||M J||L2 ≪ n J n
point in [N] is in O(ln(N)) admissible intervals, we have that ∈
P
f 2 dx S (x)2dx
T|| ||V2 ≪ T|M Ik,i |
Z 0≤Xi≤ℓ 0≤k≤X2ℓ−i−1Z
N
∆2(N)ln(N) a2.
≪ n
n=1
X
Taking the square rootof eachside completes the the proof of the first inequality in the
theorem statement. The second statement follows from the first via the argument used to
prove Theorem15. Note that we obtained a bound on the lacunary partialsums in Lemma
14 of the order ln(n). This estimate was better than we needed for the proof of Theorem
15, however is exactly the order we need here.
p
This completes the proof of Theorem 3 and Corollary 4 follows.
4 Lower bounds
In this section, we prove:
Theorem 5. Let φ (x) be a complete ONS.Then thereexists a function f L (T) such
n ∞
{ } ∈
that for almost every x T
∈
f (x)= . (10)
V2
|| || ∞
1/2
Here,asbefore, f (x)=sup sup K S [f](x) S [f](x)2 where
|| ||V2 K n0<...<nK l=1| nl − nl−1 |
S [f](x)= nl a φ (x) is the n -th partial sum(cid:16). (cid:17)
nl n=1 n n l P
Using Lemma 17 below and properties of the Dirichlet kernel, Jones and Wang showed
P
(10) for the trigonometric system. In the case of general orthonormal systems, we do not
have analytic information regarding the partial summation operator and need to proceed
differently. We start by establishing the result for the Haar system.
We letE :L1 L1 denotethe conditionalexpectationoperatordefinedasfollows. For
k
→
x [l2 k,(l+1)2 k), 0 l <2k, l N we define
− −
∈ ≤ ∈
(l+1)2−k
E f(x)= f(x)dx.
k
Zl2−k
Using a probabilistic result of Qian [16], Jones and Wang [9] showed that:
Lemma 17. (Proposition 8.1 of [9]) There exists f L (T) such that
∞
∈
1/2
K
sup sup E f(x) E f(x)2 =
K n0<...<nK ℓ=1| nℓ − nℓ−1 | ! ∞
X
almost everywhere.
9
If we let S [f] denote the partial summation operator with respect to the Haar system,
n
theniteasilyfollowsthatE f(x)=S [f](x) S [f](x)forsomesequence n . There-
k nk+1 − nk { k}
fore, there exists f L∞(T) such that f V2(x) = for almost every x T, where the
∈ || || ∞ ∈
operator V2 is associated to the Haar system. For future use, let us define b to be the
n
{ }
Haar coefficients of the function f, that is
b = f(x), (x) . (11)
n n
h H i
We will also need a theorem of Olevskii (see [13] Chapter 3), which requires that we
introduce some additional notation. Let g and f be two sequences of real-valued
n n
{ } { }
measurable functions on T. We say that they are weakly isomorphic if for each n N there
∈
exists an invertible measure-preserving mapping T : T T that is one-to-one on a set of
n
→
full measure and satisfies
f (T x)=g (x)
k n k
for all 1 k n.
≤ ≤
Theorem 18. (Olevskii) Let φ be a complete real-valued orthonormal system. There
{ n}∞n=1
exists an orthonormal system H that is weakly isomorphic to the Haar system, and a
{ k}∞k=1
sequence n such that
{ k}∞k=1
nk+1
H ,φ φ (x) 2 k j
j i i − −
(cid:12)(cid:12) h i (cid:12)(cid:12) ≤
(cid:12)(cid:12)i=Xnk+1 (cid:12)(cid:12)L2
(cid:12)(cid:12) (cid:12)(cid:12)
(cid:12)(cid:12) (cid:12)(cid:12)
whenever j =k. (cid:12)(cid:12) (cid:12)(cid:12)
6
We now set f˜(x) := ∞n=1bnHn(x), for bn defined in (11). Using the fact that the
(finite) partial sums of the series defining f˜(x) are weakly isomorphic to the partial sums
P
of the Haar expansion of f, it follows that the partial sums of the function f˜are uniformly
bounded, hence f˜ L (T).
∞
∈
Lemma 19. For f˜defined as above, we set c := f˜,φ . It follows that
n n
D E
nk+1
c φ (x)=b H (x)+e (x),
n n k k k
n=Xnk+1
where e (x) < for almost every x.
k| k | ∞
Proof.PSince f˜(x)= ∞j=1bjHj(x), we have
nPk+1 nk+1
∞
c φ (x)= b H (x),φ (x) φ (x)
n n j j n n
* +
nXk+1 n=Xnk+1 Xj=1
nk+1 nk+1
= b H (x),φ (x) φ (x)+ b H (x),φ (x) φ (x).
k k n n j j n n
h i * +
n=Xnk+1 n=Xnk+1 Xj6=k
By applying the triangle inequality, we obtain:
nk+1
b H (x) c φ (x) b H (x),φ (x) φ (x)
(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k k −nXk+1 n n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Ln2k≤+1| k|(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)n∈/[nkX+1,nk+1]h k n i n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)L2
+ b H (x),φ (x) φ (x) .
j j n n
| |(cid:12)(cid:12) h i (cid:12)(cid:12)
Xj6=k (cid:12)(cid:12)(cid:12)(cid:12)n=Xnk+1 (cid:12)(cid:12)(cid:12)(cid:12)L2
(cid:12)(cid:12) (cid:12)(cid:12)
(cid:12)(cid:12) (cid:12)(cid:12)
10