Orlicz spaces bounds for special classes of hyperbolic sums

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

Orlicz spaces bounds for special classes of hyperbolic sums ✩ Dimitrios Karslidis Department of Mathematics, University of British Columbia, 1984 Mathematics Road, Vancouver, BC, V6T 1Z2, Canada

a r t i c l e

i n f o

Article history: Received 18 March 2016 Available online xxxx Submitted by R.H. Torres Keywords: Small ball inequality Littlewood–Paley inequalities Haar functions Hyperbolic sums Orlicz spaces

a b s t r a c t Let R = R1 × · · · × Rd denote a dyadic rectangle in the unit cube [0, 1]d , d ≥ 3. Let hR be the L∞ -normalized Haar function supported on R. In [10], the conjectured signed small ball inequality,          α h R R   |R|=2−n 

d

 n 2 , where αR ∈ {±1},

∞

was proven under the additional assumption that the coefficients also satisfy the splitting property, αR = αR1 · αR2 ×···×Rd with αR1 , αR2 ×···×Rd ∈ {±1}. We give another proof of this result, using a duality argument. Based on this approach, we also show         αR hR    |R|=2−n 

 n2−a , d

1

2≤a<∞

exp(La )

for any integer n ≥ 1 and any choice of coefficients {αR } ⊂ {−1, 1} which satisfy the splitting property. The above inequality has been conjectured for general coefficients αR ∈ {−1, 1} in d ≥ 3. These bounds are investigated further for more general coefficients {αR } ⊂ {−1, 1}. The proof of the sharpness of the L∞ -lower bound of hyperbolic sums with coefficients satisfying the “splitting property” is also provided. © 2016 Elsevier Inc. All rights reserved.

1. Introduction The problem of obtaining lower bounds of sums of Haar functions supported on dyadic rectangles of fixed volume known as the “Small Ball Inequality”. Linear combinations of Haar functions of the form Hn =  p |R|=2−n αR hR are called hyperbolic sums. The L -behaviour of hyperbolic sums with {αR } ⊂ {−1, 1} is ✩ The author was partially supported by an NSERC Discovery Grant 22R82900 and Foundation for Education and European Culture. E-mail address: [email protected].

http://dx.doi.org/10.1016/j.jmaa.2016.10.051 0022-247X/© 2016 Elsevier Inc. All rights reserved.

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well understood for all 0 < p < ∞ (see Section 3). The sharp bound of L∞ -norm of hyperbolic sums, even in the signed case where the coefficients {αR } ⊂ {−1, 1}, is a challenging problem and remains unsolved. The sharp constant in this inequality, as yet unproven, is of considerable interest due to a variety of applications in probability, approximation theory and discrepancy. We refer the interested reader to the excellent surveys [2,11] for a detailed discussion about the connection of these fields to the “Small Ball Inequality”. In [10], the sharp lower bound was obtained in the signed case under a structural assumption on the coefficients accompanying the Haar functions, {αR } ⊂ {−1, 1}. In this article, we give another proof of this result, using a duality approach. Employed originally by V. Temlyakov to prove the small ball inequality in dimension two, this method uses Riesz products to procure a test function that is able to identify large values of a hyperbolic sum. In higher dimensions, it is a nontrivial matter to extend the Riesz product technique. However, in this paper, our choice of coefficients admits such an extension. Moreover, this method is flexible enough to yield nontrivial bounds on the hyperbolic sums in spaces other than L∞. We have chosen to work with Orlicz spaces in this paper, for two reasons. First, exponential Orlicz spaces lie “between” Lp and L∞ and a family of Orlicz space bounds implies an L∞ bound via a limiting argument. Second, these spaces have already been studied in the context of small ball inequality and irregularities of distributions, making this a natural setting for further exploration. In Section 4, we further study these bounds for more general coefficients {αR } ⊂ {−1, 1}. The sharpness of the L∞ -lower bound is also proven when the coefficients satisfy a structural constrain. 1.1. Preliminaries Let 1I (x) be the characteristic function of the interval I, i.e.  1I (x) =

1, x ∈ I 0, otherwise.

Consider the collection of the dyadic intervals of [0, 1]:  D=

 I=

m m+1 , 2n 2n



: m, n ∈ Z, n ≥ 0, 0 ≤ m < 2n

with (1)

D∗ = D ∪ {[−1, 1]}. If we consider two distinct dyadic intervals, then either one will be strictly contained in the other, or they will be disjoint. Moreover, for every interval I ∈ D, its left and right halves (denoted by Il and Ir respectively) are also dyadic. We define the L∞ -normalized Haar function, hI , corresponding to an interval I as: hI (x) = −1Il (x) + 1Ir (x).

(2)

Haar functions can be easily extended to higher dimensions. In order to do so, we consider the family of dyadic rectangles in dimension d ≥ 2: Dd = {R = R1 × · · · × Rd : Rj ∈ D}, i.e., every R ∈ Dd is a Cartesian product of dyadic intervals. The Haar functions supported on R are defined as a coordinate-wise product of one-dimensional Haar functions:

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hR (x) = hI1 (x1 ) · · · hId (xd ),

3

(3)

where R = I1 × · · · × Id , Ij ∈ D and x = (x1 , . . . , xd ). Haar functions enjoy the following properties: h2R (x) = 1R (x),



(4)

hR (x)hR (x)dx1 . . . dxd = 0,

(5)

hR (x)dx1 . . . dxd = 0, ∀ R ∈ Dd ,

(6)

[0,1]d

whenever R = R ,

[0,1]d

and the collection H = {hR }R∈D∗d forms an unconditional basis for the Lebesgue spaces, Lp , 1 < p < ∞. Let d Hdn = {r ∈ Z+ : |r| := r1 + r2 · · · +rd = n}

with Zd+ = {r = (r1 , . . . , rd ) : rj ≥ 0 and rj ∈ Z, j = 1, . . . , d}, and Rr = {R ∈ Dd : Rj ∈ D and |Rj | = 2−rj , j = 1, . . . , d}. In other words, the set Rr consists of all the dyadic rectangles that have the same shape. The rectangles in Rr are disjoint and partition the d-dimensional unit cube, [0, 1)d . A function defined on [0, 1]d of the form fr =



R hR

with R = ±1

R∈R r

is called an r-function with parameter r ∈ Hdn . These functions are also known in the literature as generalized Rademacher functions. It can easily be verified that r-functions are orthonormal with respect to the L2 -norm.  In d = 1, the generalized Rademacher functions, Bk = |I|=2−k αI hI , k = 0, . . . , n, are iid ±1 valued random variables with probability 12 , as shown in [10]. This fact will be used throughout here multiple times. We define the signum function as:  sgn(x) =

1

if x ≥ 0

−1 if x < 0,

and denote Lebesgue measure by | · | in any dimension. 1.2. L∞ conjectured bounds of hyperbolic sums For brevity, let Adn = {R ∈ Dd : |R| = 2−n }, i.e., the set of all dyadic rectangles whose d-dimensional volume is equal to 2−n . Moreover, A  B means that there is a constant K such that A ≥ KB. In our setting, K does not depend on n or {αR }.

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Conjecture 1.1 (The Small Ball Conjecture). In all dimensions d ≥ 2, for any choice of the coefficients αR , and all integers n ≥ 1, one has the following inequality

n

d−2 2

       αR hR     d R∈An



≥ Cd · 2−n ∞

|αR |,

(7)

R∈Ad n

where Cd is a constant that depends only on the dimension d but not on n or the choice of {aR }. The critical feature of this inequality is the precise exponent of n occurring on the left side. If we replace d−1 n with n 2 then we get the so-called “trivial bound”: d−2 2

n

d−1 2

        αR hR   2−n |αR |.    d d R∈An

(8)

R∈An

2

If we compare the conjectured small ball inequality with the inequality (8), but with the L2 -norm replaced by the L∞ -norm, then we can see that the conjectured small ball inequality is better than (8) by a factor √ of n. The proof of the trivial bound (8) can be found in [2]. Also, if we choose each αR in the collection of independent random variables such that αR = ±1, then one verifies that this conjecture is sharp (see [4,14]). 1.3. Recent history The small ball conjecture has been proved in d = 2 by M. Talagrand [13] in 1994. In 1995, V. Temlyakov [14] gave another proof of this inequality. Recently, a new proof of the two-dimensional small ball inequality was provided in [3]. The first improvement over the trivial bound in higher dimensions, specifically in dimension d = 3, by a factor logarithmic in n, was obtained by J. Beck [1]. In 2008, a body of work authored by D. Bilyk, M. Lacey, and A. Vagharshakyan [4,5] made significant progress toward the study of the structure of the small ball inequality. They proved the following theorem: Theorem 1.2. In all dimensions d ≥ 3, there exists 12 > η(d) > 0 such that for all choices of coefficients αR and all non-negative integers n we have the inequality

n

d−1 2 −η(d)

       αR hR     d R∈An

 2−n ∞



|αR |.

(9)

R∈Ad n

The results of [4,5] provide unspecified small values of η(d), whereas the small ball conjecture says that (9) should hold with η = 12 . A complete resolution of the small ball conjecture appears to be a difficult problem, but the following special case, while still unsolved, seems to be more tractable. Conjecture 1.3 (The signed small ball conjecture). If αR = ±1 for every R ∈ Adn , then we have        αR hR     d R∈An

d

 n2 . ∞

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If Hn =





αR hR =

fr ,

 r ∈Hd n

R∈Ad n

then replacing the exponent of the integer n in the above conjecture by on Hn :       d−1  2 . fr    n r∈Hd  n

d−1 2

gives us the trivial bound

(11)

2

This can be obtained using the orthogonality of generalized Rademacher functions and the fact that #Hdn, the cardinality of the set Hdn , is of order nd−1 . The exponent of the integer n appears naturally in the trivial bound (11). This can be explained by observing that “the volume constraint”, |R| = 2−n , on dyadic rectangles, or equivalently |r| = n, reduces the number of “free” parameters in the vector r ∈ Hdn by one. Therefore, the total number of terms in the sum is of order nd−1 while the conjectured signed small ball √ inequality requires the “frozen” parameter to contribute by a factor of n rather than by a factor of 1 as in the trivial estimate (11). In [6], D. Bilyk, M. Lacey, and A. Vagharshakyan quantified explicitly the improvement over the trivial bound, i.e., they proved the signed version of the Theorem 1.2, explicitly providing the range for the values of η. In particular, they showed        αR hR     d R∈An

≥ Cd · n

d−1 1 2 + 8d −

where αR ∈ {±1},

,

(12)

∞

for every  > 0. Under an additional assumption on the length of the first side of a dyadic rectangle R, in [8], D. Bilyk, M. Lacey, I. Parissis and A. Vagharshakyan managed to get a better exponent of the integer n than the one given in the relation (12) when d = 3. Specifically, they showed that     

 −n

|R|=2 |R1 |≥2−n/2

   αR hR  

9

 n8 ,

where αR ∈ {±1}.

(13)

∞

1.4. Exponential Orlicz space conjectured lower bounds of hyperbolic sums Conjecture 1.4. If αR = ±1 for every R ∈ Adn , then we have         αR hR    |R|=2−n 

 n 2 − a , 2 ≤ a < ∞, d

1

exp(La )

in all d ≥ 2. In d = 2, the conjectured was solved by V. Temlyakov [14]. 1.5. Main results Let  Asplit = αR : R ∈ Adn , αR = αR1 · αR2 ×···×Rd with αR1 , αR2 ×···×Rd ∈ {±1}

(14)

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be the set of ±1 coefficients which satisfy the splitting property. In this sequel, we show that for any integer n ≥ 1 and any choice of coefficients in Asplit , Conjecture 1.3 and Conjecture 1.4 hold. Theorem 1.5. For all integers n ≥ 1, d ≥ 3 and all choices of coefficients {αR }R∈Adn ⊂ Asplit we have that        αR hR     d R∈An

d

≥ Cd n 2 ,

(15)

∞

where Cd is a constant depending only on the dimension d, not on n or the choice of {αR }. Moreover, the above inequality is sharp. Theorem 1.6. For any integer n ≥ 1 and d ≥ 3, we have        inf  αR hR  Asplit   d R∈An

d

 n2 . ∞

Theorem 1.7. For any integers n ≥ 1, d ≥ 3 and any choice of coefficients αR ∈ Asplit , we have         αR hR    |R|=2−n 

≥ C(d, a) n 2 − a , 2 ≤ a < ∞. d

1

(16)

exp(La )

2. Auxiliary results We write A  B (resp. A  B) if there exist a constant C > 0 such that A ≥ CB (resp. A ≤ CB). The notation A  B means that A  B and A  B. In our context, the implicit constant C does not depend on the choice of coefficients {αR } or the integer n but may depend on some other parameters, such as dimension or a scale of integrability a, etc. 2.1. Khintchine’s inequalities Khintchine’s inequalities allows us to explicitly calculate the p-norms of linear combination of iid random variables up to a constant. Khintchine’s inequalities will be useful in proving Theorem 1.6. Let {Xi , 1 ≤ i ≤ N } be the collection of ±1-valued iid random variables on a probability space (Ω, P) such that P(Xi = 1) = P(Xi = −1) = 12 . Khintchine’s inequalities state that there exist positive constants Ap > 0 and Bp > 0 such that

Ap

 N

|ai |

2

 N  

12 N   2   ≤ ai Xi  ≤ Bp |ai | , 0 < p < ∞,

12

i=1

i=1

p

i=1

for any finite sequence of real numbers {ai }N i=1 . For a proof of these inequalities we refer the interested reader to [9]. A direct application of Khintchine’s inequalities shows that p-norms of a linear combination of Rademacher functions are comparable, i.e., there exist positive constants Ci (p, q), i − 1, 2 such that  N   N   N             C1 (p, q) ak r k  ≤  ak rk  ≤ C2 (p, q) ak r k   , q > p > 0. k=0

p

k=0

q

k=0

p

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 Here, rk = I∈D:|I|=2−k hI are Rademacher functions which are iid random variables on the probability space ([0, 1), | · |). We would like to extend the above inequalities to more general functions. More specifically, we’d like to N   extend them to k=0 |I|=2−k aI hI or, in multivariable case, to R∈Dd :|R|=2−N aR hR , where the coefficients {aI } or {aR } depend not only on the scale of dyadic rectangles, but also on the rectangles themselves. For this reason the Littlewood–Paley inequalities will be useful. 2.2. Littlewood–Paley inequalities To each function of the form f (x) =



R hR (x), where x = (x1 , . . . , xd ) ∈ [0, 1]d ,

(17)

R∈D∗d

we associate the expression ⎡ (Sd f )(x) = ⎣

⎤ 12



|R |2 1R (x)⎦ , with (R )R∈D∗d ⊂ R.

(18)

R∈D∗d

This is called the product dyadic square function of f . The product Littlewood–Paley inequalities state that (Ap )d Sd f p ≤ f p ≤ (Bp )d Sd f p , for p ∈ (1, ∞). The interested reader can find the proof of these inequalities in [2]. Littlewood–Paley inequalities should be viewed as generalizations of Khintchine’s inequalities for Rademacher random variables. Indeed, at any point x ∈ [0, 1], we have S

 ∞

 ak rk (x) =

 ∞



12

|ak | 1I (x) 2

k=0 I∈D:|I|=2−k

k=0

=

 ∞

|ak |

2

12 .

k=0

Therefore, Littlewood–Paley inequalities imply (Ap )

 ∞ k=0

|ak |

2

12

 ∞ 

12  ∞   2   ≤ ak rk  ≤ (Bp ) |ak | , for p > 1 k=0

p

k=0

which are Khintchine’s inequalities. The next proposition, namely Proposition 2.1 will help us in comparing the quantity f p for different values of p in (0, ∞). In particular, when f is a special linear combination of Haar functions, it shows that the Lp -norms, for all p are comparable. Proposition 2.1. Let f be a linear combination of Haar functions, i.e. f (x) =



R hR (x),

R∈Ad n

such that the square function, (Sd f )(·), is a constant on [0, 1]d (i.e. (Sd f )(x) = c(n, d) for every x ∈ [0, 1]d , where the constant c(n, d) depends on the integer n ≥ 1 and on the dimension d). Then there exist a positive constant c2 (p, q, d) such that

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c2 f q ≤ f p ≤ f q

0 < p < q < ∞.

for every

(19)

The Littlewood–Paley inequalities are used in the proof of (19) for the case when q > p > 1. In particular, the assumption the square function is constant on [0, 1]d permits the p-norms of the square function to be comparable. The proof of inequality (19) for the full range of p and q values requires a combination of the previous case and a subtle treatment of Hölder’s inequality. The interested reader can consult [10] for more details. Remark 2.2. It is clear that if 

f (x) =

R hR (x),

R∈Ad n

with (R )R∈Dd :|R|=2−n ⊂ {−1, 1}, then the square function, (Sd f )(·), is independent of x ∈ [0, 1]d . More precisely,  (Sd f )(x) =



 12

 1 = #Hdn 2 ,

|R |2 1R (x)

R∈Ad n

where #Hdn denotes the cardinality of the set Hdn . An immediate consequence of this is that Fr1 1  Fr1 2  (n − r1 )

d−2 2

,

r1 = 0, 1, . . . , n,

(20)

where Fr1 (x ) =



αR hR (x ) with x = (x1 , x ) ∈ [0, 1)d , and

(21)

R ∈Ad−1 n−r1

x = (x2 , x3 , . . . , xd ) ∈ [0, 1)d−1 . Estimate (20) will be crucial in showing Lemma 3.1. In addition, define Ar1 =





αR hR , with (r1 , r ) ∈ Hdn ,

R∈R r  r  ∈Hd−1 n−r1

and r1 = 0, 1, . . . , n. Using the same reasoning as before, we get Ar1 1  Ar1 2  (n − r1 )

d−2 2

,

(22)

r1 = 0, 1, . . . , n. These inequalities will be essential in showing Lemma 4.4. 2.3. Orlicz spaces Orlicz spaces are natural generalizations of Lp -spaces with p ≥ 1, where function sp , entering the definition of Lp , is replaced by a more general convex function. Let φ : R → [0, ∞] be such that φ(·) is a convex, even function with φ(0) = 0 and is different from the constant function 0(s) = 0, s ∈ R. Given a finite measure space (X, M, μ), one defines the Orlicz space as:

 Lφ (X, μ) = f measurable : ∃a > 0, φ(a|f |)dμ < ∞ . X

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We endow the Orlicz space with so-called Luxembourg norm: for any measurable function f ∈ Lφ (X, μ), the norm of f is defined by

 f φ := inf t > 0 : φ(|f |/t)dμ ≤ 1 . X

Note that  Lφ (X, μ) = f measurable : f φ < ∞ . The function φ entering the definition of Orlicz space is increasing on R+ = [0, ∞). Indeed, let 0 ≤ s1 < s2 , then the convexity of φ implies that φ(s1 ) ≤

s2 − s1 s1 φ(0) + φ(s2 ) ≤ φ(s2 ). s2 s2

From the definition we can see that, to understand the composition of Lφ (X, μ), we only need to know how φ(x) grows as |x| → ∞. More specifically, if φ ∼ h as |x| → ∞, i.e., C1 h ≤ φ ≤ C2 h, |x| ≥ M, for some positive constants C1 , C2 , M , then the two quantities f φ and f h are equivalent and we see that it suffices to know the asymptotic behaviour of φ. When φ(x) = |x|p (1 ≤ p < ∞), then Lφ (X, μ) = Lp (X, μ) and f φ = f p . If φ(x) = 0, for |x| ≤ 1 and φ(x) = ∞, otherwise, then Lφ (X, μ) = L∞ (X, μ) and the two norms are equal to each other. We will be interested in the Orlicz spaces which are associated with functions of the form  φa (x) =

e|x| − 1,

|x| ≥ 1

(e − 1)|x|,

otherwise,

a

where a > 0. These spaces are known as exponential Orlicz spaces and are denoted by exp(La ). To prove Theorem 1.7, we will need an Orlicz space version of Hölder’s inequality, i.e., given φ as in the definition of Lφ , we want to find C > 0 and a function φ∗ such that

|f g|dμ ≤ C f φ g φ∗ ,

(23)

for all f ∈ Lφ and g ∈ Lφ∗ . Note that this requires that φ∗ is an even convex function with φ∗ (0) = 0. To prove this generalization of Hölder’s inequality, φ∗ is taken to be the function conjugate to φ, i.e., φ∗ : R → [0, ∞] is the function defined by φ∗ (y) = sup{xy − φ(x) : x ∈ R}.

(24)

It follows from the definition that φ∗ is convex, even and satisfies φ∗ (0) = 0. Another immediate consequence of this definition is Young’s inequality: xy ≤ φ(x) + φ∗ (y) for all x and y. Young’s inequality implies that |f (x)g(x)| ≤ φ(f (x)) + φ∗ (g(x)), for all x ∈ X.

(25)

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We can now see that (23) holds for C = 2. Indeed, it suffices to show (23) for f φ = g φ∗ = 1 and, in such a case, we obtain (23) simply by integrating both sides of inequality (25). This generalization of Hölder’s inequality will be applied to φa and φ∗a to prove the main results of the 1 chapter. Note that, by convention, the Orlicz spaces associated with φ∗a are denoted by L(log L) a , and it 1 can be verified that φ∗a (y) ∼ |y|(log(1 + |y|)) a as |y| → ∞. The next proposition deals with estimating Orlicz space norms of an indicator function. A detailed proof can be found in [12]. Proposition 2.3. Let E ∈ M. Then 1E φ  μ(E)(φ∗ )−1

 1  , μ(E)

where (φ∗ )−1 (y) = inf{x ≥ 0 : φ∗ (x) ≥ y}, y ≥ 0 is the generalized inverse of φ∗ (·). As an application of the above proposition for a probability measure, we have the following estimate 1E

1

1

L(log L) a

 P(E) · (1 − log(P(E)) a ,

(26)

since (φ∗a )∗ = φa . 3. Proofs of the main results  As we have already mentioned, the Lp -bounds of the signed hyperbolic sums, Hn = |R|=2−n αR hR with {αR } ⊂ {−1, 1}, are equivalent to the L2 -bounds. Indeed, applying Proposition 2.1, there exist positive constants Ci (2, p, d), i = 1, 2, such that C1 (2, p, d) Hn 2 ≤ Hn p ≤ C1 (2, p, d) Hn 2 , 0 < p < ∞. d−1

d−1

Using the orthogonality of Haar functions, we obtain Hn 2 = Sd (Hn ) 2  n 2 . Therefore, Hn p  n 2 for all p in (0, ∞), where the implicit constants depend on the value of p and the dimension d. The proof of Theorem 1.5 will use a duality argument. That is, for every collection of coefficients {αR } ⊂ Asplit , we will construct a test function ψ ∈ L1 ([0, 1]d ) such that ψ 1 ≤ C, where C is a constant which does not depend on the choice of coefficients {αR } ⊂ Asplit or the integer n ≥ 1. In addition, the test function ψ will also satisfy

ψ, Hn  :=

d

ψ(x)Hn (x)dx1 . . . dxd  n 2 , [0,1]d

 where Hn = |R|=2−n αR hR with {αR } ⊂ Asplit . Theorem 1.5 follows from the above two inequalities by a simple application of Hölder’s inequality to ψ, Hn . We will prove the two inequalities in Lemma 3.1. This lemma is also essential in proving Theorem 1.7. The construction of our test function will be similar to the one used by Temlyakov in the proof of the conjectured small ball inequality when d = 2. Lemma 3.1. For any n ≥ 1, d ≥ 3, and any choice of coefficients {αR }R∈Adn ⊂ Asplit , there exists a function ψ ∈ L1 ([0, 1]d ) with ψ 1 = 1 such that

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d

ψ, Hn   n 2 .

11

(27)

Here, the implicit constant depends only on the dimension d and not on the collection of coefficients {αR }R∈Adn ⊂ Asplit or the integer n ≥ 1. Proof. First, we rewrite the function Hn (x) =



αR hR (x)

(28)

Br1 (x1 )Fr1 (x ),

(29)

R∈Ad n

in a more convenient way for us as: n 

Hn (x) =

r1 =0

where 

Br1 =

αR1 hR1

(30)

|R1 |=2−r1

and 

Fr1 =

αR hR

(31)

|R |=2−(n−r1 )

with r1 = 0, . . . , n and x = (x1 .x ) ∈ [0, 1)d . Define the test function ψ as the Riesz product ψ(x) =

n  

 1 + Br (x1 )sgn(Fr (x )) .

(32)

r=0

Here, we have replaced r1 with r for our convenience. We claim that ψ has the property that ψ 1 = 1

(33)

and satisfies (27). Indeed, we first observe that, for any x ∈ [0, 1]d , ψ(x) ≥ 0, since each factor in the product forming ψ is nonnegative. Expanding the product in (32), we get ψ = 1 + ψ1 + ψ2 ,

(34)

with ψ1 (x) =

n 

Br (x1 )sgn(Fr (x )),

(35)

r=0

and ψ2 (x) =

n 



k 

Bsj (x1 )sgn(Fsj (x )).

(36)

k=2 0≤s1 <···
  Now, it is easy to see that [0,1]d ψ1 (x)dx1 . . . dxd = [0,1]d ψ2 (x)dx1 . . . dxd = 0 since {Br }nr=0 are iid  ±1-valued random variables. Hence, ψ 1 = [0,1]d ψ(x)dx1 . . . dxd = 1.

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12

To show that ψ satisfies (27), we first use (34) to get

ψ, Hn  =

Hn (x)dx1 . . . dxd + ψ1 , Hn  + ψ2 , Hn .

(37)

[0,1]d

Now, since {Br }nr=0 are of mean zero, we get that we show that

 [0,1]d

Hn (x)dx1 . . . dxd = 0. Therefore, to conclude (27),

d

ψ1 , Hn   n 2

(38)

ψ2 , Hn  = 0.

(39)

and

Since {Br }nr=0 are ±1-valued independent random variables of mean zero, using (20), we get that

n 

ψ1 , Hn  = [0,1]d−1

=

n 

r,r  =0

Fr 1 

r=0



n 

 sgn(Fr )Fr

 Br Br

[0,1] n 

Fr 2

(40)

r=0

(n − r)

d−2 2

d

 n2 ,

r=0

which proves (38). Again, using the fact that {Br }nr=0 are ±1-valued independent random variables of mean zero once more, from the definitions of the functions ψ2 and Hn we get that

ψ2 , Hn  = [0,1]d−1

n  n 



r=0 k=2 0≤s1 <···
Fr

k 

sgn(Fsj )

j=1

[0,1]

Br

k 

Bsj

j=1

= 0,

(41)

which completes the proof. 2 3.1. Proof of Theorem 1.5 Proof. Lemma 3.1 ensures the existence of the function ψ with ψ 1 = 1 for which

d

ψ(x)Hn (x)dx1 . . . dxd  n 2 .

(42)

[0,1]d d

Applying Hölder’s inequality in (42), we get that Hn ∞  n 2 and the proof is complete. 2 3.2. Proof of Theorem 1.6 In this section, we prove that there exists a choice of coefficients {αR }R∈Adn ⊂ {−1, 1} satisfying the “splitting property” for which

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     αR hR   R∈Ad n

d

∞

 n2 ,

13

(43)

for all d ≥ 3. The proof of Theorem 1.6 relies on two lemmas: the first establishes the independence of a product of random variables, and the second estimates the expected value of the point-wise maximum of a collection of random variables. Lemma 3.2. Fix an integer l ≥ 1. Let {Xi , Yji : 1 ≤ i ≤ m, 1 ≤ j ≤ l} be a collection of independent random variables taking the values ±1 with probability 12 . Then the random variables {Xi Yji : 1 ≤ i ≤ m, 1 ≤ j ≤ l} are independent. − → Proof. LetW i = (Xi Y1i , . . . , Xi Yli ), i = 1, . . . , m. It is easy to see that the probability distribution function − → − → of W i is P W i = i = 21l for i ∈ {−1, 1}l and i = 1, . . . , m. To prove the independence of {Xi Yji : 1 ≤ i ≤ m, 1 ≤ j ≤ l}, we need to show that −   1 m → − → P W 1 = 1 , . . . , W m = m = l . 2  = (X1 , . . . , Xm ) and Y l,i = (Y i , . . . , Y i ), i = 1, . . . , m. The independence Let b = (b1 , . . . , bm ) ∈ {−1, 1}m , X 1 l of the random variables {Xi , Yji : 1 ≤ i ≤ m, 1 ≤ j ≤ l} gives us that −  → − → P W 1 = 1 , . . . , W m = m =

 b∈{−1,1}m

=

 1 m 2l

.

  1 m     P Yl,1 = , . . . , Yl,m = ,X = b b1 bm 2

Lemma 3.3. Let Y1 , . . . , YN be random variables on a probability space (Ω, P) and φ : [0, ∞) → R be a convex and strictly increasing function. If Eφ(|Yk |) ≤ C, k = 1, . . . , N , then we have that 

E

 sup |Yk | ≤ φ−1 (N ), 1≤k≤N

where C is an absolute constant and φ−1 is the inverse function of φ. The proof of this lemma, which we omit here, can be found in M. Lacey’s excellent survey [11] on the small ball inequality and discrepancy theory. Proof of Theorem 1.6. Suppose that the coefficients {αR1 ×R } in Asplit are chosen randomly and independently, i.e., consider a probability space (Ω, P) where {αR1 } and {αR } are two families of ±1 valued iid random variables which are independent of each other. To prove (43), it suffices to show      E αR hR  R∈Ad n

d

∞

 n2 ,

(44)

where E denotes the expected value with respect to the probability P. Therefore, we concentrate on the proof of (44). Let Qk be a dyadic cube in Dd of side-length 2−(n+1) , k = 1, . . . , 2(n+1)d . Note that all these dyadic cubes partition the unit cube, [0, 1)d . Set Xk =

 |R|=2−n

αR hR |Qk =





 r ∈Hd n

Qk ⊂R∈R r

αR1 αR hR ,

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14

k = 1, . . . , 2(n+1)d . We can see that Xk is constant on the dyadic cube Qk . Note also, that #{R ∈ Adn , Qk ⊂ R} = #Hdn . Using this fact and Lemma 3.2, Xk is the sum of #Hdn many iid ±1 valued random variables. Applying Khintchine’s inequality, we obtain Cd−1 n Set Yk =

|Xk | E|Xk |A

d−1 2

≤ E(|Xk |) ≤ Cd n

d−1 2

, k = 1, . . . , 2(n+1)d .

(45)

2

and φ(t) = exp( t2 ), where A is a large positive constant and t > 0. Then we have that

Eφ(Yk )  1, as shown in [2]. Appealing to Lemma 3.3, we get (44). Indeed, set N = 2(n+1)d and Lemma 3.3 implies that E( sup |Yk |) 

√

n.

(46)

k≤N

Combining (45) and (46), we get      E αR hR  R∈Ad n

 ∞

=E

 sup |Xk | k≤N

n

d−1 2

 E

 sup |Yk | k≤N

d 2

2

=n . 3.3. Proof of Theorem 1.7

Before we proceed to the proof of Theorem 1.7, note that it follows from the definition of exponential Orlicz spaces that, for each 1 < p < ∞ and a > 0, we have continuous embeddings L∞ ⊂ exp(La ) ⊂ Lp . Therefore, n

d−1 2

 Hn 2  Hn exp(La )

(47)

holds for any choice of coefficients {αR } ⊂ Asplit . In fact, (47) holds for any collection {αR } ⊂ {−1, 1}. We can see that the lower bound in (47) is better than the one given in Theorem 1.7 when a < 2, but when 2 ≤ a < ∞, inequality (16) from the statement of Theorem 1.7 gives us an improvement over the lower 1 1 bound in (47) by a factor of n 2 − a for any collection of coefficients {αR } ⊂ Asplit . It is for this reason that Theorem 1.7 is stated only for a ≥ 2. As we will soon see, the proof of Theorem 1.7 works for all a > 0. Proof of Theorem 1.7. From Lemma 3.1 there is a positive function ψ such that ψ 1 = 1 and d

ψ, Hn   n 2 .

(48)

Also note that this test function can be written as ψ = 2n+1 1E , where  E = x = (x1 , x ) ∈ [0, 1]d : Br (x1 ) · sgn(Fr (x )) = 1 for all r = 0, . . . , n . Furthermore, |E| = 2−(n+1) since to (48), we get

 [0,1]d

ψ(x)dx1 . . . dxd = 1. Applying Hölder’s inequality for Orlicz spaces

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D. Karslidis / J. Math. Anal. Appl. ••• (••••) •••–•••

2n+1 1E Combining (26) with the fact that |E| =

d

1 2n+1 ,

2n+1 1E

1

L(log L) a

Hn exp(La )  n 2 .

15

(49)

we obtain 1

1

L(log L) a

1

≤ Cd 2 a n a , n > 1.

(50)

Now, using (50), (49) becomes Hn exp(La ) ≥ (Cd )−1 2− a n 2 − a , 1

d

1

which completes the proof. 2 From the proof of Theorem 1.7 we can see that the constant C(d, a) appearing in (16) must be equal 1 to (Cd )−1 2− a . Taking the limit of both sides of inequality (16) as the scale of integrability a approaches infinity, we get the signed small ball inequality when {αR } ⊂ Asplit . It is worth mentioning that bounds of this type have already appeared in the field of irregularities of distribution of points (see [7]). In particular, Bilyk et al. [7] showed that, in d = 2, the discrepancy function,   DN (x1 , x2 ) := # PN ∩ [0, x1 ) × [0, x2 ) − N x1 x2 , associated with the N -point set PN ⊂ [0, 1]2 satisfies the following lower bound 1

DN exp(La )  (log N )1− a ,

2 ≤ a < ∞.

They used a duality argument in the setting of exponential Orlicz spaces in their proof which, in fact, provided the motivation for the proof of Theorem 1.7. 4. Study of signed hyperbolic sums with free “splitting property” coefficients In this section, we are concerned with exploring L∞ and exp(La )-lower bounds of a linear combination of Haar functions with signed coefficients αR ∈ {−1, 1}. To motivate our discussion, we first restate the signed small ball conjecture and Conjecture 1.4 in an equivalent form.  = (μ0 , . . . , μn ) ∈ Conjecture 4.1 (Restatement of the signed small ball conjecture). There exists a vector μ {−1, 1}n+1 such that, for any choice of coefficients {βR } ⊂ {−1, 1} and any integer n ≥ 1, we have the inequality       μr1 βR hR    r ∈Hd n

R∈R r

∞

d

 n2 ,

d ≥ 3.

(51)

By choosing μr1 = 1, r1 = 0, 1, . . . , n and βR = αR , R ∈ Adn , it is immediate that the small ball conjecture implies (51). To get the reverse implication, (51) implies the small ball conjecture, simply set βR = μr1 αR for R = R1 × R ∈ Adn with |R1 | = 2−r1 , r1 = 0, . . . , n.  = (μ0 , . . . , μn ) ∈ {−1, 1}n+1 such Conjecture 4.2 (Restatement of Conjecture 1.4). There exists a vector μ that, for any choice of coefficients {βR } ⊂ {−1, 1} and any integer n > 1, we have the inequality

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D. Karslidis / J. Math. Anal. Appl. ••• (••••) •••–•••

      μr1 βR hR    r ∈Hd n

R∈R r

 n2−a , d

exp(La )

1

2 ≤ a < ∞,

(52)

for all d ≥ 3. The equivalence can be obtained by methods similar to those presented before. Using the techniques developed in the previous section, we will show the existence of (n + 1)-dimensional vectors μ  and ν ∈ {−1, 1}n+1 which satisfy inequalities (51) and (52) respectively, though these vectors may depend on the choice of coefficients {αR } ⊂ {−1, 1}. Specifically, we show: Theorem 4.3. For any integer n ≥ 1, d ≥ 3 and any choice of coefficients {αR } ⊂ {−1, 1} there exist (n + 1)-dimensional vectors μ  = (μ0 , . . . , μn ) and ν = (ν0 , . . . , νn ) ∈ {−1, 1}n+1 such that       μr1 αR hR    r ∈Hd n

R∈R r

d

∞

 n2 ,

(53)

and       νr1 αR hR    r ∈Hd n

R∈R r

 n2−a , d

exp(La )

1

2 ≤ a < ∞.

(54)

This theorem will be proved by employing a duality argument which uses a test function in the form of a Riesz product. Instead of working with the function 

μr1

 r ∈Hd n



αR hR ,

R∈R r

we let μr1 : Ω → {−1, 1} be iid random variables with P(μr1 = ±1) = 12 , r1 = 0, 1, . . . , n and consider the function   Hμ (ω, x) = μr1 (ω) αR hR (x)  r ∈Hd n

R∈R r

  defined on the probability space Ω × [0, 1)d , P × | · | . Then, for a particular choice of ω ∈ Ω, Hμ (ω, ·) takes the form of the function inside the L∞ and exp(La )-norm in (51) and (52) respectively. The usage of these random variables combined with (22) will be crucial in getting the lower bounds in (53) and (54). The idea of introducing random coefficients is not new. For instance, it has been used to prove sharpness of the conjectured lower bound in the small ball inequality [2]. However, the randomization used in [2] required each coefficient in {αR } to be iid ±1-valued random variables, and there were about nd−1 2n iid random coefficients needed for that proof, whereas our randomization is considerably milder, employing just n + 1 iid random variables. Just as we proved the lower bound appearing in Theorem 1.5 separately in Lemma 3.1, we will prove the lower bound appearing in Theorem 4.3 in the following lemma. Lemma 4.4. There exists a positive test function ψμ ∈ L1 (Ω × [0, 1]d ) such that ψμ 1 = 1, and

(55)

d

Eμ E(ψμ Hμ )  n 2 , where Eμ is the expected value with respect to the probability measure P and E(ψμ Hμ ) = dxd is the expected value with respect to the Lebesgue measure.

(56)  [0,1]d

(ψμ Hμ )dx1 . . .

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17

Proof. We will find a test function ψμ ∈ L1 (Ω × [0, 1]d ) such that ψμ 1 = 1

(57)

and d

Eμ E(ψμ Hμ )  n 2 .

(58)

The function ψμ (·) takes the form of a Riesz product and is defined as: ψμ (ω, x) =

n 

(1 + μr (ω)sgn(Ar (x))) ,

(59)

r=0

where we replaced r1 with r for our convenience. We claim that the function ψμ (·) satisfies (57) and (58). Indeed, first we observe that ψμ (ω, x) ≥ 0 ∀(ω, x) ∈ Ω × [0, 1]d ,

(60)

since each factor in the product is positive, and expanding the product in (32), we get ψμ = 1 + ψ1 + ψ2 ,

(61)

with ψ1 (ω, x) =

n 

μr (ω)sgn(Ar (x)),

(62)

r=0

and ψ2 (ω, x) =

n 



k 

μsj (ω)sgn(Asj (x)).

(63)

k=2 0≤s1 <···
It is easy to verify that ψμ 1 = EEμ ψμ = 1 + EEμ ψ1 + EEμ ψ2 . Hence, to complete the proof of (57), it is sufficient to prove Eμ ψ1 = 0,

(64)

Eμ ψ2 = 0.

(65)

and

Implementing the fact that {μr }nr=0 are iid ±1-valued random variables, we see that (64) and (65) hold. Now, we turn our attention to the proof of (58). Using (61), we have Eμ E(ψμ Hμ ) = Eμ EHμ + Eμ E(ψ1 Hμ ) + Eμ E(ψ2 Hμ ), and, since {μr (·)}nr=0 are iid random variables, we get Eμ Hμ = 0. Therefore, it suffices to show

(66)

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18

d

Eμ E(ψ1 Hμ )  n 2 ,

(67)

Eμ E(ψ2 Hμ ) = 0.

(68)

and

Thus, appealing to the fact that {μr (·)}nr=0 are iid random variables and using (22), we obtain (67) as follows: Eμ E(ψ1 Hμ ) = E

  sgn(Ar )Ar Eμ μr μr

n  r,r  =0

=

n 

Ar 1

r=0



n 

(69) Ar 2

r=0



n  d−2 d (n − r) 2  n 2 . r=0

Therefore (67) holds. Using the definition of the functions ψ2 and Hμ , and employing the fact that {μr (·)}nr=0 are iid random variables one more time, it is clear that we have

Eμ E(ψ2 Hμ ) = E

n  n 



Ar

r=0 k=2 0≤s1 <···
k 

sgn(Asj )Eμ μr

j=1

k 

μsj

(70)

j=1

= 0, and the proof is complete. 2 We show that Theorem 4.3 holds. Proof of Theorem 4.3. To prove (53), we apply Hölder’s inequality in (58) to get d

Hμ ∞  n 2 . Therefore, there exists ω0 ∈ Ω such that Hμ (ω0 , ·) ∈ L∞ ([0, 1]d ) with Hμ (ω0 , ·) ∞  n 2 and this shows the truth of (53). To prove (54), we note first that the test function can also take the following form: d

ψμ = 2n+1 1E , where E =

 (ω, x) ∈ Ω × [0, 1]d : μr (ω) · sgnAr (x) = 1 for all r = 0, . . . , n and (P × | · |)(E) = 2−(n+1) , 1

1

1 ≤ 2 a n a . Applying Hölder’s inequality for Orlicz spaces, we since Eμ Eψμ = 1. In addition, ψμ L(log L) a get a lower bound on Hμ :

Hμ exp(Laω,x ) ≥ C(d, a)n 2 − a , d

1

(71)

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19

where C(d, a) is a positive constant depending on the dimension d, and the scale of integrability a, and · exp(Laω,x ) denotes a norm in exponential Orlicz space with respect to the measure P × | · |. We claim that there exists ω0 ∈ Ω such that Hμ (ω0 , ·) exp(La ) ≥ 12 C(d, a)n 2 − a , where the norm in Orlicz space is considered with respect to the Lebesgue measure only, and this fact will imply the truth of (54). Indeed, using the definition of exponential Orlicz space, (71) implies d





 φa

Ω [0,1]d



|Hμ (ω, x)| (2)−1 C(d, a)n

1

d 1 2−a

dPdx1 . . . dxd > 1,

and this shows that there exists ω0 ∈ Ω such that

 φa

[0,1]d

|Hμ (ω0 , ·)| (2)−1 C(d, a)n

which means Hμ (ω0 , ·) exp(La ) ≥ 12 C(d, a)n 2 − a . d

1

d 1 2−a

 dx1 . . . dxd > 1

2

Acknowledgments I would like to thank my advisor Malabika Pramanik for introducing me to this problem and the numerous discussions that led to this paper. I am grateful to the anonymous referee for valuable comments concerning the content of this article. I am also grateful to Professors Richard Froese, Philip Loewen and Dmitriy Bilyk for their valuable comments. References [1] J. Beck, A two-dimensional van Aardenne-Ehrenfest theorem in irregularities of distribution, Compos. Math. 72 (3) (1989) 269–339. [2] D. Bilyk, On Roth’s orthogonal function method in discrepancy theory, Unif. Distrib. Theory 6 (1) (2011). [3] D. Bilyk, N. Feldheim, A new proof of the two-dimensional small ball inequality, submitted for publication, available online at: arXiv:1511.07326. [4] D. Bilyk, M. Lacey, On the small ball inequality in three dimensions, Duke Math. J. 143 (1) (2008) 81–115. [5] D. Bilyk, M. Lacey, A. Vagharshakyan, On the small ball inequality in all dimensions, J. Funct. Anal. 254 (9) (2008) 2470–2505. [6] D. Bilyk, M. Lacey, A. Vagharshakyan, On the signed small ball inequality, Online J. Anal. Comb. 3 (2008). [7] D. Bilyk, M. Lacey, I. Parissis, A. Vagharshakyan, Exponential squared integrability of the discrepancy function in two dimensions, Matematika 55 (1–2) (2009) 1–27. [8] D. Bilyk, M. Lacey, I. Parissis, A. Vagharshakyan, A three-dimensional small ball inequality, in: Dependence in Probability Analysis and Number Theory, Walter Philip memorial volume, Kendrick Press, Heber City, UT, 2010, pp. 73–78. [9] L. Grafakos, Classical Fourier Analysis, third edition, Springer, 2014. [10] D. Karslidis, On the signed small ball inequality with restricted coefficients, Indiana Univ. Math. J. 65 (2016) 797–812. [11] M. Lacey, Small ball and discrepancy inequalities, arXiv:math/0609816, 2006. [12] M.M. Rao, Z.D. Ren, Theory of Orlicz Spaces, Taylor & Francis, 1991. [13] M. Talagrand, The small ball problem for the Brownian sheet, Ann. Probab. 22 (3) (1994) 1331–1354. [14] V.N. Temlyakov, Some inequalities for multivariate Haar polynomials, East J. Approx. 1 (1) (1995) 61–72.