Bounds on moments of weighted sums of finite Riesz products
aa r X i v : . [ m a t h . F A ] M a r BOUNDS ON MOMENTS OF WEIGHTED SUMS OF FINITE RIESZPRODUCTS
ALINE BONAMI, RAFA L LATA LA, PIOTR NAYAR, AND TOMASZ TKOCZ
Abstract.
Let n j be a lacunary sequence of integers, such that n j +1 /n j ≥ r . Weare interested in linear combinations of the sequence of finite Riesz products Q Nj =1 (1 +cos( n j t )). We prove that, whenever the Riesz products are normalized in L p norm ( p ≥ r is large enough, the L p norm of such a linear combination is equivalent tothe ℓ p norm of the sequence of coefficients. In other words, one can describe many waysof embedding ℓ p into L p based on Fourier coefficients. This generalizes to vector valued L p spaces. Primary: 42A55; Secondary: 26D05, 42A05.
Key words.
Riesz products, moment estimates, lacunary sequences, trigonometric polynomials. introduction Let T = R / π Z be the one dimensional torus and m be the normalized Haar measure on T . Let ( n j ) j ≥ be an increasing sequence of positive integers. Riesz products are definedon T by(1) R ≡ R N ( t ) := N Y j =1 (1 + cos( n j t )) for N = 1 , , . . . . To simplify the notation we also put X ≡ X j ( t ) := 1 + cos( n j t ) , j = 1 , , . . . . It was Frigyes Riesz who first realized the usefulness of these objects treated as probabilitymeasures. When n j +1 /n j ≥ j ≥
1, the numbers P Nj =1 ε j n j are all nonzero fornonzero vectors ( ε j ) Nj =1 ∈ {− , , } N , due to the fact that for every l , P lk =1 n k < n l +1 .In particular, the zero mode of R N has Fourier weight 1 and thus R N are densities ofprobability measures µ N . The weak- ∗ limit of ( µ N ) is a singular measure which admitsa number of remarkable Fourier-analytic properties. The reader is referred for instanceto [12] for more information on properties of Riesz products and general trigonometric This material is partially based upon work supported by the NSF grant DMS-1440140, while the authorswere in residence at the MSRI in Berkeley, California, during the fall semester of 2017. P. N. and T. T.were also partially supported by the Simons Foundation. R. L. and P. N. were partially supported by theNational Science Centre Poland grant 2015/18/A/ST1/00553. The research leading to these results is partof a project that has received funding from the European Research Council (ERC) under the EuropeanUnion’s Horizon 2020 research and innovation programme (grant agreement No 637851). olynomials as well as to the short survey [6] of some applications of Riesz products. Wewill always assume that n j +1 /n j ≥ j ≥
1, so that every integer n can be written atmost once as P Nj =1 ε j n j for nonzero vectors ( ε j ) Nj =1 ∈ {− , , } N .In this article we shall study the sum P Nk =0 v k R k where v k are vectors in a normed space( E, k · k ). By the triangle inequality, we trivially have(2) Z T (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X k =0 v k R k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) d m ≤ N X k =0 k v k k . We are interested in the reverse inequality and in L p inequalities. Our interest in thiskind of inequalities comes back to a question of Wojciechowski, who asked for the validityof the reverse bound up to some universal constant (personal communication). He firststudied this problem in the scalar case and in the following probabilistic context. Supposewe replace the functions X , X , . . . appearing in the definition of the Riesz products witha sequence of independent random variables ¯ X , ¯ X , . . . (defined on some probability space(Ω , P )), each having the same distribution as 1 + cos( Y ), where Y is uniform on [0 , π ].We then take ¯ R N = Q Nk =1 ¯ X k and of course ¯ R ≡
1. Note that the functions X j definedon the probability space ( T , m ) have the same distribution as the random variables ¯ X j .Even though the X j are not independent, we shall see that they behave, in many ways,like independent random variables. Capturing this phenomenon in a quantitative way isone of the main difficulties in our investigation.In [11], Wojciechowski showed the existence of universal constants c and C as well as realnumbers a , a , . . . such that for every n , | P ki =0 a i | ≤ C for all k ≤ n and E | P ni =0 a i ¯ R i | ≥ cn . This result was used in [4, 5] in the study of Fourier multipliers on the homogeneousSobolev space ˙ W ( R d ).The reverse of (2) for ¯ R k was proved by the second named author in [7] for general ran-dom variables. More generally, for any sequence ¯ X , ¯ X , . . . of i.i.d. non-negative randomvariables with mean one and such that P ( ¯ X = 1) <
1, we have(3) E (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X k =0 v k ¯ R k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≥ c ¯ X N X k =0 k v k k , for any vectors v i in an arbitrary normed space ( E, k · k ), with a constant c ¯ X dependingonly on the distribution of ¯ X (see Theorem 4 in [7]; see also Theorem 3 therein for nonidentically distributed sequences ( ¯ X i )). This clearly implies Wojciechowski’s result with a i = ( − i (here E = R ). According to a theorem of Y. Meyer (see [8]), under a strongerdivergence of the sequence of modes, namely when P ∞ k =1 n k n k +1 < ∞ , for any real numbers a i , we have Z T (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X k =0 a k R k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≥ c S E (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X k =0 a k ¯ R k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) or a positive constant c S which depends only on the n k . In [7], this principle was combinedwith (3) to show the reverse of (2) in the real case and under the above restrictive conditionon the modes n i .Later the results of [7] have been generalized by Damek et al. in [2], where it was shownthat for any p > X i ), we have(4) 1 C p, ¯ X N X k =0 k v k k p E ¯ R pk ≤ E (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X k =0 v k ¯ R k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) p ≤ C p, ¯ X N X k =0 k v k k p E ¯ R pk N ≥ , with a constant C p, ¯ X depending only on p and the distribution of ¯ X .The aim of this article is to prove the following theorem. Theorem 1.
For every p ≥ there are positive constants d p , c p , C p depending only on p ,such that for any integers n j satisfying n j +1 /n j ≥ d p , j = 1 , , . . . and for any vectors v , v , . . . in a normed space ( E, k · k ) , we have (5) c p N X k =0 k v k k p Z T R pk d m ≤ Z T (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X k =0 v k R k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) p d m ≤ C p N X k =0 k v k k p Z T R pk d m, for any N ≥ , where R k are defined via (1) . The lower bound in the case p = 1 answers the original question of Wojciechowski. Let usalso note that for p >
1, both the upper and the lower bounds are non-trivial (as opposedto the case p = 1 where the upper bound is easy – see (2)). The values of the constants d p , c p and C p that can be obtained from our proofs are far from optimal. In particular, wehave lim p → + d p = ∞ and lim p → + c p = 0, which is inconsistent with the case p = 1. Dueto these blow-ups as p → + , our proof in the case p = 1 is different from the proof for p >
1. It is based on transferring the independent case of [7] using Riesz products. Werestate the result for p = 1 with numerical values of the constants. (For explicit boundson the constants for p >
1, see Remark 24.)
Theorem 2.
There exists a constant c > · − such that for any positive integers n j satisfying n j +1 /n j ≥ and for any vectors v , v , . . . in a normed space ( E, k · k ) , we have Z T (cid:13)(cid:13)(cid:13) N X j =0 v j R j (cid:13)(cid:13)(cid:13) d m ≥ c N X j =0 k v j k for R k defined in (1) . Theorem 1 was proved in [2] in the real case ( E = R ) under the condition P ∞ k =1 n k n k +1 < ∞ mentioned earlier (again by combining the independent case with the decoupling inequalityof Meyer). It is easy to see that the same proof implies that it is also valid for vector-valued coefficients under the weaker condition P ∞ k =1 (cid:16) n k n k +1 (cid:17) < ∞ , which is known asSchneider’condition [10]. We do it in the next section for completeness. When E = R and p/ n k +1 /n k ≥ p + 1 is sufficient. n general, Theorem 1 cannot be transferred from the independent case by using somegeneralization of Schneider’s condition: L p norms of R k and ¯ R k are not equivalent, as wesee in the next section. So the core of the proof deals directly with Riesz products onthe torus. Many new difficulties appear when compared with the proof for independentfrequencies.We conclude with questions: Is the best constant d p in Theorem 1 an increasing function?Can it be chosen so that it does not depend on p ?The article is organized as follows. First we present those results that are may be ob-tained as consequences of the i.i.d case. This concerns the case when Schneider’s Condition P ∞ k =1 (cid:16) n k n k +1 (cid:17) < ∞ is fulfilled as well as Theorem 2 concerning L norms. The rest of thepaper is devoted to the general case. In Section 4 we give preparatory results. The mainsection is Section 5, which is devoted to the proof of the lower estimate for p >
1. Finally,in Section 6 we give a proof of the upper bound for p > Acknowledgements.
We would like to thank F. Nazarov for stimulating correspon-dence which encouraged us to continue working on this project. We are also indebtedto P. Ohrysko for a helpful discussion.2.
The theorem under Schneider’s Condition
The aim of this section is to prove Theorem 1 under Schneider’s Condition, that is, wehave the following result.
Proposition 3.
Assume that for each j ≥ one has n j +1 /n j ≥ and that, moreover, P (cid:16) n j n j +1 (cid:17) < ∞ . Then the conclusion of Theorem 1 holds: for every p ≥ there arepositive constants c p , C p depending only on p , such that for any vectors v , v , . . . in anormed space ( E, k · k ) , the inequalities (5) hold. To prove this, we proceed as in [7] making a use of Meyer’s condition. Namely, we use thefollowing theorem, which is implicit in the work of Schneider [10]. For an arbitrarily largeinteger N , let us denote by Λ N the set of integers that may be written as P Nj =1 ε j n j , with ε j ∈ {− , , } , for all j ≤ N . The condition n j +1 /n j ≥ T = T N from Λ N to Z N given by T ( P Nj =1 ε j n j ) = ( ε j ) Nj =1 is injective. For a trigonometric poly-nomial P ( x ) = P n ∈ Λ N a n e inx on T with values in E , we define e P ( y ) = P n ∈ Λ N a n e iT ( n ) · y ,which is a trigonometric polynomial on T N with values in E . The next proposition is avariant of results one can find in Meyer’s book [9], Chapter VIII. Proposition 4.
Under the previous assumptions and notations, there exists a constant C which depends only on the sequence ( n j ) such that for all E − valued trigonometric polyno-mials P with frequencies in Λ N and all p ∈ [1 , ∞ ] , (6) C − k ˜ P k L p ( T N ,E ) ≤ k P k L p ( T ,E ) ≤ C k ˜ P k L p ( T N ,E ) . Proposition 4 together with (4) easily implies Proposition 3. We give a complete proofof this proposition for the reader’s convenience. o establish (6), we first consider p = ∞ and E = R and iterate the following simplelemma. Lemma 5.
Let P , P and P be trigonometric polynomials of degree at most d . For aninteger M > d , we let P ( x ) = P ( x ) + P ( x ) e iMx + P ( x ) e − iMx , Q ( x, y ) = P ( x ) + P ( x ) e iMy + P ( x ) e − iMy . Then sup x ∈ T | P ( x ) | ≥ (cid:18) − π d M (cid:19) sup x,y ∈ T | Q ( x, y ) | . Proof.
Let ( x , y ) be a point where | Q | reaches its maximum, which we assume to benonzero. Without loss of generality we may assume that Q ( x , y ) = 1, so that it is alsothe maximum of its real part. This implies in particular that the derivative in the x variableof its real part vanishes at ( x , y ). To conclude it is sufficient to find x ∈ T such thatthe real part of Q ( x , y ) − P ( x ) is smaller than π d M . We take x ∈ T to be such that | x − x | ≤ π/M and exp( iM x ) = exp( iM y ). Then by Taylor’s expansion ℜ ( Q ( x , y ) − P ( x )) = ℜ ( Q ( x , y ) − Q ( x , y )) ≤ π M sup x ∈ T | Q ′′ ( x, y ) | , where Q ′′ stands for the second derivative in the x variable. By Bernstein’s inequality, thissupremum is bounded by d , which allows to conclude. (cid:3) Corollary 6.
There exists a constant C ∞ which depends only on the sequence ( n j ) suchthat for all trigonometric polynomials P with frequencies in Λ N , (7) C − ∞ sup y ∈ T N | ˜ P ( y ) | ≤ sup x ∈ T | P ( x ) | ≤ sup y ∈ T N | ˜ P ( y ) | . Proof.
Let P ( x ) = P n ∈ Λ N a n e inx . Here, for convenience, instead of ˜ P , we shall consider˜ Q ( y ) = P n ∈ Λ N a n e i P j ε j n j y j , y ∈ T N , where ε = T ( n ). Clearly, sup | ˜ Q | = sup | ˜ P | . Theupper bound is obvious because ˜ Q ( x, x, . . . , x ) = P ( x ).We use Lemma 5, with M = n N and d = n + . . . + n N − ≤ n N − (cid:0) + + . . . (cid:1) = n N − . It implies that sup x ∈ T | P ( x ) | ≥ c N sup x,y N ∈ T | ˜ Q ( x, . . . , x, y N ) | , where c N = 1 − π (cid:18) n N − n N (cid:19) . For every fixed y N , ˜ Q ( x, . . . , x, y N ) as a function of x is a trigonometric polynomial withfrequencies in Λ N − and therefore we can iterate the above argument to obtainsup x ∈ T | P ( x ) | ≥ c N · . . . · c N sup x,y N ,...,y N ∈ T | ˜ Q ( x, . . . , x, y N , . . . , y N ) | . bserve that Schneider’s condition implies the existence of N , depending only on thesequence ( n j ), such that c N · . . . · c N ≥ for N ≥ N . Indeed, N Y k = N c k ≥ − π X k ≥ N (cid:18) n k − n k (cid:19) since for every real numbers a , . . . , a l > − Q li =1 (1 + a i ) ≥ P li =1 a i . Therefore there is N depending only on the sequence ( n j ) such that for everypolynomial P , we have(8) sup x ∈ T | P ( x ) | ≥
12 sup x,y N ,...,y N ∈ T | ˜ Q ( x, . . . , x, y N , . . . , y N ) | . Now we handle the first M := N − P M be the space of trigonometricpolynomials on T M spanned by { e i ( P j ≤ M ε j n j y j ) } ε ∈{− , , } M . Any two norms on a finite-dimensional space P M are comparable, in particular there exists δ > x ∈ T | Q ( x, . . . , x ) | ≥ δ sup ( y ,...,y M ) ∈ T M | Q ( y , . . . , y M ) | for Q ∈ P M . The above bound together with (8) implies the lower bound in (7) with C ∞ = 2 δ − . (cid:3) Proof of Proposition 4.
Let µ be a bounded measure on T and ˜ E N be a set of all functionsof the form ˜ P = P n ∈ Λ N a n e iT ( n ) · y and a n ∈ R . We may treat ˜ E N as a subset of thespace of continuous functions C ( T N ). On ˜ E N we define a functional ϕ by the formula ϕ ( ˜ P ) = R P dµ . The upper bound in (7) shows that k ϕ k ≤ k µ k M ( T ) . By the Hahn-Banach theorem we may extend ϕ to C ( T N ) and thus show that there exists a measure˜ µ ∈ M ( T N ) such that k ˜ µ k M ( T N ) ≤ k µ k M ( T ) , by the Riesz-Markov-Kakutani representationtheorem ( k µ k M ( T ) is the total variation of µ ). Moreover, ˆ˜ µ ( T ( n )) = ˆ µ ( n ) for n ∈ Λ N becauseˆ˜ µ ( T ( n )) = Z e − iT ( n ) · y d ˜ µ ( y ) = ˜ ϕ ( e − iT ( n ) · y ) = ϕ ( e − iT ( n ) · y ) = Z e − inx dµ ( x ) = ˆ µ ( n ) . In the same way we show that for any measure ˜ µ ∈ M ( T N ), there exists a measure µ ∈ M ( T ) such that k µ k M ( T N ) ≤ C ∞ k ˜ µ k M ( T ) and the previously stated relation holds.Using these observations for Dirac measures we find for x ∈ T and y ∈ T N measures˜ µ x ∈ M ( T N ) and µ y ∈ M ( T N ) such that k ˜ µ x k ≤ k µ y k ≤ C ∞ and c ˜ µ x ( T ( n )) = e − inx , c µ y ( n ) = e − iT ( n ) · y for n ∈ Λ N . ix now a trigonometric E − valued polynomial ˜ P = P n ∈ Λ N a n e iT ( n ) · y and p ∈ [1 , ∞ ).Observe that for any x ∈ T , k ˜ P k L p ( T N ,E ) = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X n ∈ Λ N a n e inx e iT ( n ) · y ∗ ˜ µ x (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L p ( T N ,E ) ≤ k ˜ µ x k M ( T N ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X n ∈ Λ N a n e inx e iT ( n ) · y (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L p ( T N ,E ) ≤ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X n ∈ Λ N a n e inx e iT ( n ) · y (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L p ( T N ,E ) . Integrating over x ∈ T and changing the order of integration we get k ˜ P k pL p ( T N ,E ) ≤ Z T N Z T (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X n ∈ Λ N a n e inx e iT ( n ) · y (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) p dxdy. However for any y ∈ T N (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X n ∈ Λ N a n e inx e iT ( n ) · y (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L p ( T ,E ) = k P ∗ µ y k L p ( T ,E ) ≤ k µ y k M ( T ) k P k L p ( T ,E ) ≤ C ∞ k P k L p ( T ,E ) . This way we show that k ˜ P k L p ( T N ,E ) ≤ C ∞ k P k L p ( T ,E ) . The case p = ∞ follows by takingthe limit. The upper bound in (6) is shown in an analogous way. (cid:3) In the rest of this section we discuss the question of generalizing this method to sequencesthat do not satisfy Schneider’s condition. It was observed in [1] Chapter I that, as aconsequence of Plancherel’s formula, the double inequality (6) is valid for p an even integerand E = R as soon as n j +1 /n j ≥ p + 1. It means that the conclusions of Theorem 1 arealso valid in this case for scalar functions.For p/ n j +1 /n j ≥ p + 1 is a natural bound for being able totransfer the result for the independent case to the context of the lacunary sequence n j .This is given by the following lemma. Lemma 7.
Let p > be an even integer and n k = p k . Then lim sup k R k k p / k f R k k p = ∞ . Proof.
This comes from a combinatorial argument. We will use the following fact. For asequence of positive integers q , . . . , q k and a trigonometric polynomial g with nonnegativeFourier coefficients, we have(9) Z T | g ( q x ) g ( q x ) · · · g ( q k x ) | dm ( x ) ≥ k g k k and the inequality is strict if and only if there exists two different sequences of integers( m , · · · , m k ) and ( m ′ , · · · , m ′ k ) such that q m + · · · + q k m k = q m ′ + · · · + q k m ′ k while ll Fourier coefficients b g ( m j ) , b g ( m ′ j ) are strictly positive. Indeed, by Plancherel’s formula,the inequality (9) is equivalent to X m X m , ··· ,m k : q m + ··· + q k m k = m b g ( m ) · · · b g ( m k ) ! ≥ X m , ··· ,m k | b g ( m ) | · · · | b g ( m k ) | . This is a direct consequence of the inequality ( P J a j ) ≥ P J a j , while the strict inequalitycomes from the fact that this last inequality is strict whenever a j ’s are positive and J hasmore than one element.Let us come back to the proof of the lemma and prove that k R k k p / k e R k k p tends to ∞ .If we take q = p/ f ( x ) = (1 + cos( x )) q (1 + cos( px )) q , g ( x, y ) = (1 + cos( x )) q (1 + cos( y )) q , then R k ( x ) p = h f ( x ) f ( p x ) ·· · ·· f ( p k − x ) i and we can use the previous fact to prove that k R k k pp ≥ (cid:0)R T f ( x ) d m ( x ) (cid:1) k . Moreover, k g R k k pp = (cid:16)R T × T g ( x, y ) d m ( x )d m ( y ) (cid:17) k . To provethat k R k k pp / k g R k k pp tends to ∞ , it is sufficient to prove that the L norm of f is strictlylarger than the norm of g , that is, to prove that, at least for one value of m , the Fouriercoefficient of b f ( m ) is obtained through different writings of m as a sum of two frequenciesthat belong respectively to the two factors. But, for instance, q = q + 0 = − q + 2 q , whichallows to conclude. (cid:3) The previous lemma allows us to find such examples for other values of p . Namely Lemma 8.
Let q ≥ be an even integer. Except possibly for a discrete set of values of p ∈ (1 , ∞ ) , there exists a sequence n j such that n j +1 /n j ≥ q for all j ≥ and k R k k p / k f R k k p does not remain bounded below or above.Proof. We consider the two quantities k P k pp and k e P k pp , where P and ˜ P are the trigono-metrical polynomials of degree q + 1, respectively on T and T , defined by P ( x ) = (1 + cos( x ))(1 + cos( qx )) , ˜ P ( x, y ) = (1 + cos( x ))(1 + cos( y )) . We have seen in the proof of the previous theorem that k P k pp and k e P k pp differ for p = q .So they differ except on a discrete set of values (this is because k P k pp as a function of p is analytic). Let p be such an exponent and let us construct a sequence n j that satisfiesthe conclusions of the lemma. We let n j = m j and n j +1 = qm j , where the sequence m j increases sufficiently rapidly so that P (cid:16) ( q +1) m j m j +1 (cid:17) < ∞ . The L p ( T k ) norm of e R k is easilyseen to be the k -th power of the norm of ˜ P . We use for P the analog of Proposition 3, butwith the set Λ N defined with ( ε j ) Nj =1 such that ε j ∈ { , ± , · · · ± ( q + 1) } . We deduce thatthe L p ( T ) norm of R k is up to a multiplicative cosntant comparable with the k -th powerof the norm of P . The conclusion that k R k k p / k f R k k p does not remain bounded below orabove follows at once. (cid:3) he last lemma shows that in general Theorem 1 cannot be deduced from the indepen-dent case. We will see that it is nevertheless the case for p = 1, which is not contradictorysince the L norms of R k and e R k are all equal to 1.3. Lower bound for p = 1 Proof of Theorem 2.
We assume n j +1 /n j ≥
3. Then the Fourier expansion of a Rieszproduct Q kj =1 (1+cos( n j x )) has 3 k distinct terms. For a sequence ψ = ( ψ , ψ , . . . ), considerthe Riesz product P ψ ( x ) = ∞ Y j =1 (1 + cos( n j x + ψ j )) . Let ˜ R k ( ψ, x ) = ( P ψ ⋆ R k )( x ) . Then(11) Z T (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X j =0 v j ˜ R k ( ψ, x ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) d m ( x ) ≤ Z T (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X j =0 v j R j ( x ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) d m ( x ) . On the other hand, ˜ R k ( ψ, x ) = k Y j =1 (cid:18) n j x + ψ j ) (cid:19) . We integrate both sides of (11) against d m ( ψ ) and exchange integration. On the left handside we have an i.i.d. sequence (with respect to ψ ) 1 + cos( n j x + ψ j ) (observe also thatthe distribution does not depend on x ), which satisfies conditions of the main theorem of[7]. So we get the desired lower bound . (cid:3) Such techniques involving Riesz products P ψ have been already used in [1]. Unfortu-nately the same argument based on transferring the i.i.d. case from [2] does not seemto work for L p bounds with p >
1. Indeed, the lower bound involves the quantity (cid:0)R T (cid:0) cos( t ) (cid:1) p d m ( t ) (cid:1) k , which is off by an exponential factor (in k ).4. Auxiliary general results
We give here elementary or standard results, which will be our tools in the main proofs.The following simple result will lie in the heart of our induction procedure to obtain thebound below. It is basically [2, Lemma 9].
Lemma 9.
Let µ be a measure on X and let f, g : X → E be measurable functions. Supposethat for some p > and γ > , we have Z X k g k p − k f k d µ ≤ γ Z X k f k p d µ. hen, Z X k f + g k p d µ ≥ (cid:18) p − pγ (cid:19) Z X k f k p d µ + Z X k g k p d µ. Proof.
For any real numbers a, b we have | a + b | p ≥ | a | p − p | a | p − | b | . If, additionally, | a | ≤ | b | , then | a + b | ≥ | b | − | a | ≥ | a | + | b | and thus | a + b | p ≥ | a | p + p | b | p . Taking a = k g k , b = −k f k and using the inequality k f + g k ≥ |k f k − k g k| , we obtain Z X k f + g k p d µ = Z X k f + g k p {k g k≤ k f k} d µ + Z X k f + g k p {k g k > k f k} d µ ≥ Z X k g k p {k g k≤ k f k} d µ + 13 p Z X k f k p {k g k≤ k f k} d µ + Z X k g k p {k g k > k f k} d µ − p Z X k g k p − k f k {k g k > k f k} d µ = Z X k g k p d µ + 13 p Z X k f k p (1 − {k g k > k f k} )d µ − p Z X k g k p − k f k {k g k > k f k} d µ. Note that Z X (cid:18) p k f k p + p k g k p − k f k (cid:19) {k g k > k f k} d µ ≤ (cid:18)
13 + p (cid:19) Z X k g k p − k f k d µ ≤ pγ Z X k f k p d µ. Therefore, Z X k f + g k p d µ ≥ Z X k g k p d µ + 13 p Z X k f k p d µ − pγ Z X k f k p d µ. (cid:3) The next lemma gives a comparison between explicit constants that we will need.
Lemma 10.
For k, p ≥ , Z T | cos( t ) | p | sin( t ) | kp d m ≤ kp + 1 Z T | cos( t ) | p d m Z T | sin( t ) | kp d m. Proof.
All integrals are equal to beta functions. So the ratio between the left and the righthand sides is equal toΓ( p + 1)Γ( kp + 1)Γ( kp + p + 1) = p Γ( p )Γ( kp + 1)Γ( kp + p + 1) = p Z x p − (1 − x ) kp d x ≤ p Z x p − d x Z (1 − x ) kp d x = 1 kp + 1 , where we have used the continuous version of Chebyshev’s sum inequality. (cid:3) Our next lemma concerns exact algebraic factorization for integrals of products oftrigonometric polynomials and is also standard. emma 11. Suppose that g , . . . , g N − are trigonometric polynomials of degree at most d , g N is an arbitrary continuous function on T and n j +1 /n j ≥ d + 1 for j ≥ . Then Z T N Y j =1 g j ( n j t )d m = N Y j =1 Z T g j ( n j t )d m. Proof.
Indeed the left hand side is the sum of products of Fourier coefficients b g j ( l j ), with P Nj =1 l j n j =0, | l j | ≤ d for j ≤ N −
1. This only occurs when all l j are zero, which allows toconclude. (cid:3) Even if exact factorization does not hold, one can establish approximate factorization inthe presence of a highly oscillating factor. This idea is quantified in the following lemma.
Lemma 12.
Suppose that f is a Lipschitz function on T and g is an integrable functionon T . Then for any integer n ≥ , we have (cid:12)(cid:12)(cid:12) Z T f ( t ) g ( nt )d m − Z T f d m Z T g ( nt )d m (cid:12)(cid:12)(cid:12) ≤ πn Z T | f ′ ( t ) | d m Z T | g ( nt ) | d m. Proof.
Let I k = [ kn π, k +1 n π ] for k = 0 , , . . . , n −
1. Observe that for any k , R T g ( nt )d m = | I k | R I k g ( nt )d t , hence (cid:12)(cid:12)(cid:12) Z I k f ( t ) (cid:16) g ( nt ) − Z T g ( ns )d m ( s ) (cid:17) d t (cid:12)(cid:12)(cid:12) = 1 | I k | (cid:12)(cid:12)(cid:12) Z I k × I k ( f ( t ) − f ( s )) g ( nt )d t d s (cid:12)(cid:12)(cid:12) ≤ sup t,s ∈ I k | f ( t ) − f ( s ) | Z I k | g ( nt ) | d t ≤ Z I k | f ′ ( u ) | d u Z I k | g ( nt ) | d t = 2 πn Z I k | f ′ ( u ) | d u Z T | g ( nt ) | d m. Summing the above estimate over 0 ≤ k ≤ n − (cid:3) In the context of trigonometric polynomials, in the above lemma we can pass from thebound in terms of f ′ to the bound in terms of the original factor f . Namely, we havethe following lemma. Its first part is the classical Bernstein inequality for vector valuedtrigonometric polynomials. Lemma 13.
Suppose that f is a vector-valued trigonometric polynomial of order at most d . Then (12) Z T k f ′ k p d m ≤ d p Z T k f k p d m. Moreover, for any integrable (complex valued) function h on T , we have (13) (cid:12)(cid:12)(cid:12)(cid:12)Z T k f ( t ) k p h ( nt )d m − Z T k f k p d m Z T h ( nt )d m (cid:12)(cid:12)(cid:12)(cid:12) ≤ π pdn Z T k f k p d m Z T | h ( nt ) | d m. roof. Formula (3.11) in [12, Chapter X] gives f ′ ( t ) = P dk =1 b k f ( t + t k ), where P dk =1 | b k | = d and t k = d ( k − ) π . Thus k f ′ k p ≤ P dk =1 | b k |k f k p = d k f k p and (12) follows.To show (13), take g = k f k p . Then | g ′ | ≤ p k f k p − k f ′ k ( g is in fact almost everywheredifferentiable) and Z T | g ′ | d m ≤ p (cid:18)Z T k f k p d m (cid:19) ( p − /p (cid:18)Z T k f ′ k p d m (cid:19) /p ≤ pd Z T k f k p d m, by H¨older’s inequality and estimate (12). Thus Lemma 12 yields (13). (cid:3) Lemma 14.
Let f and f be vector-valued trigonometric polynomials of degree at most d .Then for n ≥ d , we have Z T k f + f cos( nt ) k p d m ≥ p Z T k f k p d m. Proof.
This is an easy consequence of the use of de la Vall´ee Poussin kernel V d (see, e.g. [3,2.13, p. 16]), which is a trigonometric polynomial of degree 2 d − − d and d equal to 1. The L norm of V d is bounded by 3 /
2. If g ( t ) = 2 e int V d ( t ),then e int f coincides with the convolution of f + f cos( nt ) with g . The result follows atonce. (cid:3) Lower bound for p > f = v N R N − yield k v N kk R N k p ≤ k v N kk R N − ( t ) cos( n N t ) k p ≤ (cid:13)(cid:13)(cid:13) N X k =0 v k R k (cid:13)(cid:13)(cid:13) L p ( T ,E ) under the condition that n k +1 ≥ n k . But we are far from having the possibility of aninduction from this. Our first step will concern this inequality, but for a family of weightedmeasures on the torus.Let ϕ k ( t ) = ( − cos t ) k . For k, l ≥
1, we say that a function g on T belongs to family ofweights F pk,l if it has the form g ( t ) := l Y j =1 h j ( n j t ) , where h j ∈ (cid:26) , ϕ pk , − ϕ pk (cid:27) for j = 1 , . . . , l. We also set F pk, := { } . With a slight abuse of notation we will say that a measure µ on T belongs to F pk,l if it has the form d µ = g d m for some g ∈ F pk,l .We will approximate these weights by trigonometric polynomials. We start with thenext lemma, which is a rather standard application of Bernstein polynomials. We prove itfor completeness. emma 15. Let p > and f p ( t ) = (1 − t p ) /p , t ∈ [0 , . For any ε > , there exists apolynomial w ε,p of degree at most ⌈ ε − ⌉ such that f p ( t ) ≤ w ε,p ( t ) ≤ (1 + ε ) f p ( t ) for t ∈ [0 , . Proof.
We have | f ′ p ( t ) | = t p − (1 − t p ) /p − ≤ − /p ≤
1, so f p is 1-Lipschitz. Let S n,t have the binomial distribution with parameters n and t and define ˜ w n,p ( t ) := E f p ( n S n,t ).Then ˜ w n,p is a polynomial of degree at most n and | ˜ w n,p ( t ) − f p ( t ) | ≤ E (cid:12)(cid:12)(cid:12)(cid:12) f p (cid:18) n S n,t (cid:19) − f p ( t ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ E (cid:12)(cid:12)(cid:12)(cid:12) n S n,t − t (cid:12)(cid:12)(cid:12)(cid:12) ≤ n (cid:0) E | S n,t − nt | (cid:1) / = 1 n p nt (1 − t ) ≤ √ n . Define w ε,p = ˜ w n,p + √ n , where n = ⌈ ε − ⌉ . Observe that f p ( t ) ≤ w ε,p ( t ) ≤ f p ( t ) + 1 √ n ≤ f p ( t ) + ε ≤ (1 + ε ) f p ( t ) . (cid:3) Let us now approximate the weights by trigonometric polynomials.
Lemma 16.
Suppose that n j +1 /n j ≥ for all j ≥ and let k ≥ , l ≥ . Then for any g ∈ F pk,l , there exists a trigonometric polynomial h of degree at most C ( p ) n l k such that g ≤ h p ≤ g .Proof. There exist disjoint I , I ⊂ { , . . . , l } such that g := 2 −| I | Y j ∈ I ϕ pk ( n j t ) Y j ∈ I (cid:18) − ϕ pk ( n j t ) (cid:19) . Let ε j := ln 2 p j − l − for j ∈ I and h := 2 − | I | p Y j ∈ I ϕ k ( n j t ) Y j ∈ I w ε j ,p ( ϕ k ( n j t )) , where w ε j ,p are polynomials given by Lemma 15. Then h is a trigonometric polynomial ofdegree at mostdeg( h ) ≤ X j ∈ I n j k + X j ∈ I ⌈ ε − j ⌉ n j k ≤ p ln l X j =1 l +1 − j n j k ≤ p ln n l k. Moreover, g ≤ h p ≤ g Y j ∈ I (1 + ε j ) p ≤ e p P j ∈ I ε j g ≤ e ln 2 P lj =1 j − l − g ≤ g. (cid:3) The following lemma will comprise a first step in our main inductive argument. emma 17. Suppose that k ≥ , l ≥ and n j +1 /n j ≥ C ( p ) k for j ≥ . Then for any µ ∈ F pk,l and any vectors v , . . . , v l +1 in a normed space ( E, k · k ) , we have Z T (cid:13)(cid:13)(cid:13) l +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ ≥ c ( p ) k v l +1 k p Z T R pl +1 d µ. Proof.
We may assume that C ( p ) ≥
8. Let g = d µ d m and h be a trigonometric polynomialgiven by Lemma 16. We have Z T (cid:13)(cid:13)(cid:13) l +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ ≥ Z T (cid:13)(cid:13)(cid:13) l +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p h p d m. Observe that l +1 X j =0 v j R j h = f h + v l +1 cos( n l +1 t ) R l h, where f is a vector-valued trigonometric polynomial. Moreover,max { deg( R l h ) , deg( f h ) } ≤ deg( h ) + l X j =1 n j ≤ ( C ( p ) + 2) n l k and the assertion easily follows by Lemma 14. (cid:3) Lemma 18.
For any p > , there exists a real polynomial w p such that x p − ≤ w pp ( x ) for x ∈ [0 , and λ ( p ) := R T w pp ( X )d m (cid:0)R T X p d m (cid:1) ( p − /p < . Proof.
By the Weierstrass approximation theorem, for any ε >
0, there exists a polynomial w p such that x ( p − /p ≤ w p ( x ) ≤ x ( p − /p + ε for x ∈ [0 , (cid:18)Z T w pp ( X )d m (cid:19) /p ≤ (cid:18)Z T X p − d m (cid:19) /p + ε, (cid:18)Z T X p d m (cid:19) /p > (cid:18)Z T X p − d m (cid:19) / ( p − . The assertion follows by taking ε sufficiently small. (cid:3) We are now in position to give the main ingredients for the induction procedure.
Lemma 19.
For p > , there exist constants C ( p ) , C ( p ) , C ( p ) and λ ( p ) < with thefollowing property. If n j +1 /n j ≥ C ( p ) k for j ≥ , k ≥ , l ≥ , then for any µ ∈ F pk,l , any N ≥ l + 1 and any vector valued polynomial f of order at most n l , we have (14) Z T k f k p ϕ pk ( n l +1 t )d µ ≥ Z T k f k p d µ Z T ϕ pk d m nd Z T k f k R p − N ϕ pk ( n l +1 t )d µ ≤ C ( p ) k ( p − /p λ ( p ) N − l − (cid:18)Z T k f k p d µ (cid:19) /p (cid:18)Z T ϕ pk d m (cid:19) (cid:18)Z T R pN d µ (cid:19) ( p − /p . (15) Moreover for any v l +1 , . . . , v N we have Z T k f k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X j = l +1 v j R j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) p − ϕ pk ( n l +1 t )d µ ≤ C ( p ) k ( p − /p (cid:18)Z T k f k p d µ (cid:19) /p (cid:18)Z T ϕ pk d m (cid:19) N X j = l +1 λ ( p ) j − l − k v j k p Z T R pj d µ ( p − /p . (16) Proof.
Let g = d µ d m and h be a trigonometric polynomial given by Lemma 16. Notice that hf is a vector-valued trigonometric polynomial with degree at most ( C ( p ) + 2) n l k . Thusby (13) we have for sufficiently large C ( p ), Z T k f k p ϕ pk ( n l +1 t )d µ ≥ Z T k f h k p ϕ pk ( n l +1 t )d m ≥ Z T k f h k p d m Z T ϕ pk d m ≥ Z T k f k p d µ Z T ϕ pk d m. To establish (15), let us define d˜ µ = h p ( t ) ϕ pk ( n l +1 t )d m . By H¨older’s inequality, we have Z T k f k R p − N ϕ pk ( n l +1 t )d µ ≤ Z T k f k R p − N d˜ µ ≤ (cid:18)Z T k f k p R p − l +2 ,N d˜ µ (cid:19) /p (cid:18)Z T R pl +1 R p − l +2 ,N d˜ µ (cid:19) ( p − /p . We have used the notation, for 1 ≤ l ≤ N, (17) R l,N = N Y j = l X j . et w p be given by Lemma 18 and ε = ε p be a small positive number to be chosen later.By (13), if C ( p ) is sufficiently large, we have Z T k f k p R p − l +2 ,N d˜ µ ≤ Z T k f k p N Y j = l +2 w pp ( X j )d˜ µ ≤ (1 + ε ) Z T k f k p N − Y j = l +2 w pp ( X j )d˜ µ Z T w pp ( X N )d m ≤ . . . ≤ (1 + ε ) N − l − Z T k f k p d˜ µ N Y j = l +2 Z T w pp ( X j )d m ≤ (1 + ε ) N − l Z k f h k p d m Z T ϕ pk d m N Y j = l +2 Z T w pp ( X j )d m ≤ ε ) N − l λ ( p ) N − l − Z T k f k p d µ Z T ϕ pk d m N Y j = l +2 (cid:18)Z T X pj d m (cid:19) ( p − /p . In the same way we show that Z T R pl +1 R p − l +2 ,N d˜ µ ≤ ε ) N − l λ ( p ) N − l − Z T R pl d µ Z T X pl +1 ϕ pk ( n l +1 t )d m N Y j = l +2 (cid:18)Z T X pj d m (cid:19) ( p − /p . The above estimates together with Lemma 10 yield Z T k f k R p − N ϕ pk ( n l +1 t )d µ ≤ (cid:18) kp + 1 (cid:19) ( p − /p (1 + ε ) N − l λ ( p ) N − l − (cid:18)Z T k f k p d µ (cid:19) /p × (cid:18)Z T ϕ pk d m (cid:19) Z T R pl d µ N Y j = l +1 Z T X pj d m ( p − /p . Estimate (13) implies however that for sufficiently large C ( p ), Z T R pN dµ ≥
12 (1 − ε ) Z T R pN − h p d m Z T X pN d m ≥ . . . ≥
12 (1 − ε ) N − l Z T R pl g d m N Y j = l +1 Z T X pj d m. To derive (15) we choose ε = ε p in such a way that λ ( p ) := (1 + ε )(1 − ε ) (1 − p ) /p λ ( p ) < . o show (16) we consider two cases. First assume that 1 < p ≤
2. By (15), we have Z T k f k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X j = l +1 v j R j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) p − ϕ pk ( n l +1 t )d µ ≤ Z T k f k N X j = l +1 k v j R j k p − ϕ pk ( n l +1 t )d µ ≤ C ( p ) k ( p − /p (cid:18)Z T k f k p d µ (cid:19) /p (cid:18)Z ϕ pk d m (cid:19) N X j = l +1 λ ( p ) j − l − k v j k p − (cid:18)Z T R pj d µ (cid:19) ( p − /p . However N X j = l +1 λ ( p ) j − l − k v j k p − (cid:18)Z T R pj d µ (cid:19) ( p − /p ≤ N X j = l +1 λ ( p ) j − l − /p N X j = l +1 λ ( p ) j − l − k v j k p Z T R pj d µ ( p − /p ≤ (1 − λ ( p )) − /p N X j = l +1 λ ( p ) j − l − k v j k p Z T R pj d µ ( p − /p , which concludes for this case.Finally, if p >
2, we have by the triangle inequality in L p − and (15) Z T k f k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N X j = l +1 v j R j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) p − ϕ pk ( n l +1 t )d µ ≤ N X j = l +1 k v j k (cid:18)Z T k f k R p − j ϕ pk ( n l +1 t )d µ (cid:19) / ( p − p − ≤ C ( p ) k ( p − /p (cid:18)Z T k f k p d µ (cid:19) /p (cid:18)Z T ϕ pk d m (cid:19) N X j = l +1 k v j k λ ( p ) ( j − l − / ( p − (cid:18)Z T R pj d µ (cid:19) /p p − . o finish the proof of (16) in this case it is enough to observe that by H¨older’s inequality N X j = l +1 k v j k λ ( p ) ( j − l − / ( p − (cid:18)Z T R pj d µ (cid:19) /p ≤ N X j = l +1 λ ( p ) ( j − l − / ( p − ( p − /p N X j = l +1 λ ( p ) j − l − k v j k p Z T R pj d µ /p ≤ (cid:16) − λ ( p ) / ( p − (cid:17) (1 − p ) /p N X j = l +1 λ ( p ) j − l − k v j k p Z T R pj d µ /p . (cid:3) Proposition 20. If k ≥ , N ≥ l ≥ , n j +1 /n j ≥ max { C ( p ) , C ( p ) , } k for j ≥ , thenfor any µ ∈ F pk,l and any vectors v , v , . . . , v N in a normed space ( E, k · k ) we have Z T (cid:13)(cid:13)(cid:13) N X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ ≥ α p Z T (cid:13)(cid:13)(cid:13) l X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ + N X j = l +1 ( β p − c p,j − l ) k v j k p Z T R pj d µ, where α p = 116 · p Z ϕ pk d m, β p = c ( p )2 α p , γ p = (16 p p C ( p )) pp − α p k and c p,j = γ p j − X i =0 λ ( p ) i . Proof.
We proceed by induction on N − l . If N − l = 0 the assertion is obvious, since α p ≤
1. To show the induction step we may assume that l is fixed and we increased N .We consider two cases. Case 1. α p R T (cid:13)(cid:13)(cid:13) P lj =0 v j R j (cid:13)(cid:13)(cid:13) p d µ ≤ γ p P N +1 j = l +1 λ ( p ) j − l − k v j k p R T R pj d µ . y the induction assumption (applied to N + 1 and l + 1), we have Z T (cid:13)(cid:13)(cid:13) N +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ ≥ α p Z T (cid:13)(cid:13)(cid:13) l +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ + N +1 X j = l +2 ( β p − c p,j − l − ) k v j k p Z T R pj d µ ≥ β p k v l +1 k p Z T R pl +1 d µ + N +1 X j = l +2 ( β p − c p,j − l − ) k v j k p Z T R pj d µ ≥ α p Z T (cid:13)(cid:13)(cid:13) l X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ − γ p N +1 X j = l +1 λ ( p ) j − l − k v j k p Z T R pj d µ + β p k v l +1 k p Z T R pl +1 d µ + N +1 X j = l +2 ( β p − c p,j − l − ) k v j k p Z T R pj d µ = α p Z T (cid:13)(cid:13)(cid:13) l X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ + N +1 X j = l +1 ( β p − c p,j − l ) k v j k p Z T R pj d µ, where the second inequality follows by Lemma 17. Case 2. α p R T (cid:13)(cid:13)(cid:13) P lj =0 v j R j (cid:13)(cid:13)(cid:13) p d µ > γ p P N +1 j = l +1 λ ( p ) j − l − k v j k p R T R pj d µ .Let d µ = (cid:18) − ϕ pk ( n l +1 t ) (cid:19) d µ and d µ = 12 ϕ pk ( n l +1 t )d µ. The induction assumption applied to l + 1 and N + 1 with the measure µ ∈ F pk,l +1 yields Z T (cid:13)(cid:13)(cid:13) N +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ ≥ α p Z T (cid:13)(cid:13)(cid:13) l +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ + N +1 X j = l +2 ( β p − c p,j − l − ) k v j k p Z T R pj d µ . Since 1 − ϕ pk ≥ , we get by Lemma 17 Z T (cid:13)(cid:13)(cid:13) l +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ ≥ Z T (cid:13)(cid:13)(cid:13) l +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ ≥ c ( p ) k v l +1 k p Z T R pl +1 d µ, hence(18) Z T (cid:13)(cid:13)(cid:13) N +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ ≥ β p k v l +1 k p Z T R pl +1 dµ + N +1 X j = l +2 ( β p − c p,j − l − ) k v j k p Z T R pj d µ . Define f = l X j =0 v j R j and g = N +1 X j = l +1 v j R j . stimate (14) and the assumptions of Case 2 yield Z T k f k p d µ ≥ Z T k f k p d µ Z T ϕ pk d m ≥ (cid:18)Z T k f k p d µ (cid:19) /p (cid:18)Z T ϕ pk d m (cid:19) γ p α p N +1 X j = l +1 λ ( p ) j − l − k v j k p Z T R pj d µ ( p − /p . On the other hand, by (16) we get Z T k f kk g k p − d µ ≤ C ( p )2 k ( p − /p (cid:18)Z T k f k p d µ (cid:19) /p (cid:18)Z T ϕ pk d m (cid:19) N +1 X j = l +1 λ ( p ) j − l − k v j k p Z T R pj d µ ( p − /p . Thus Z T k f kk g k p − d µ ≤ p p Z T k f k p d µ and Lemma 9 gives Z T (cid:13)(cid:13)(cid:13) N +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ = Z T k f + g k p d µ ≥ · p Z T k f k p d µ + Z T k g k p d µ . Inequality (14) gives12 · p Z T k f k p d µ ≥ · p Z T k f k p d µ Z ϕ pk d m = α p Z T (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) l X j =0 v j R j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) p d µ. The induction assumption applied to l + 1 , N + 1 and measure µ ∈ F pk,l +1 yields Z T k g k p d µ ≥ α p Z T k v l +1 R l +1 k p d µ + N +1 X j = l +2 ( β p − c p,j − l − ) k v j k p Z T R pj d µ . Thus(19) Z T (cid:13)(cid:13)(cid:13) N +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ ≥ α p Z T (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) l X j =0 v j R j (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) p d µ + N +1 X j = l +2 ( β p − c p,j − l − ) k v j k p Z T R pj d µ . dding (18) and (19), we obtain Z T (cid:13)(cid:13)(cid:13) N +1 X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ ≥ α p Z T (cid:13)(cid:13)(cid:13) l X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ + β p k v l +1 k p Z T R pl +1 d µ + N +1 X j = l +2 ( β p − c p,j − l − ) k v j k p Z T R pj d µ ≥ α p Z T (cid:13)(cid:13)(cid:13) l X j =0 v j R j (cid:13)(cid:13)(cid:13) p d µ + N +1 X j = l +1 ( β p − c p,j − l ) k v j k p Z T R pj d µ. (cid:3) Proof of Theorem 1 (lower bound).
We now conclude the proof of Theorem 2. We recallthat kγ p is uniformly bounded. Let k = k ( p ) be the smallest integer such that γ p − λ ( p ) ≤ β p .Then c p,m ≤ γ p − λ ( p ) ≤ β p and thus applying Proposition 20 with l = 0 and µ = m yieldsthe result with the constant c p = min( α p , β p / (cid:3) Proof of the upper bound
We first remark that, using the Minkowski inequality, we can replace the vector-valuedcoefficients v j by their norms. So it is sufficient to prove the inequality v j ’s are real positivecoefficients.All the integrals over the one dimensional torus T appearing in this section are withrespect to its (normalised) Haar measure m . We shall need three preparatory facts. Thefirst two are immediate corollaries to Lemma 13. Corollary 21.
For p ≥ and a nonzero integer n , we have (20) (cid:12)(cid:12)(cid:12)(cid:12)Z T R k ( t ) p e int (cid:12)(cid:12)(cid:12)(cid:12) ≤ πp deg R k | n | Z T R pk , k ≥ . Corollary 22.
Let p ≥ , d ≥ πp + 1 and n j +1 /n j ≥ d , j ≥ . For positive integers k < l , we have (21) 1 − πpd − ≤ R T R pk,l X pl +1 R T R pk,l R T X pl +1 ≤ πpd − . In particular, for k ≥ , l ≥ , (22) (cid:18) − πpd − (cid:19) l − ≤ R T X pk +1 . . . X pk + l R T X pk +1 . . . R T X pk + l ≤ (cid:18) πpd − (cid:19) l − . Proof.
Note that deg( R k,l ) n l +1 = n k + . . . + n l n l +1 ≤ d l − k +1 + . . . + 1 d < d − , ence applying Lemma 13 for f = R k,l , h ( t ) = (1 + cos t ) p and n = n l +1 gives (cid:12)(cid:12)(cid:12)(cid:12)Z T R pk,l X pl +1 − Z T R pk,l Z T X pl +1 (cid:12)(cid:12)(cid:12)(cid:12) ≤ πpd − Z T R pk,l Z T X pl +1 . This is (21). Iterating (21) yields (22). (cid:3)
Lemma 23.
Let p ≥ , d > p + 1 and n j +1 /n j ≥ d , j ≥ . Then for every k ≥ , m ≥ and non negative integers l , . . . , l m ≤ p , we have (23) Z T R pk X l k +1 . . . X l m k + m ≤ (1 + ǫ ) Z T R pk Z T X l k +1 . . . X l m k + m , where ǫ = πdd − p (2 p +1) d − p − .Proof. For any t , (cid:18) e it + e − it (cid:19) l = 12 l (cid:16) e it/ + e − it/ (cid:17) l = l X j = − l l (cid:18) lj + l (cid:19) e itj . Thus,(24) f = X l k +1 . . . X l m k + m = X j " l (cid:18) l j + l (cid:19) . . . l m (cid:18) l m j m + l m (cid:19) e itN j , where the sum is over all vectors j = ( j , . . . , j m ) ∈ X ms =1 {− l s , . . . , , . . . , l s } and N j = n k +1 j + . . . + n k + m j m .A standard computation shows that if d > p + 1, then the mapping j N j is injective.Let us write f = b + X j ∈ Comb b j e itN j , where b j = l (cid:0) l j + l (cid:1) . . . lm (cid:0) l m j m + l m (cid:1) and Comb denotes the set X ms =1 {− l s , . . . , l s }\{ (0 , . . . , } of all nonzero vectors j . Since all the Fourier coefficients b j are positive, they are all upper-bounded by the first one b = R T f . Applying Corollary 21 yields Z T R pk f = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z T R pk b + X j ∈ Comb b j Z T R pk e itN j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Z T R pk b + X j ∈ Comb b j πp deg R k | N j | Z T R pk ≤ (cid:18)Z T R pk Z T f (cid:19) πp deg R k X j ∈ Comb | N j | . o deal with the sum over j , we break Comb into the sets
Comb r , r = 1 , . . . , m , of thevectors j for which the largest index of a nonzero coordinate is r . We thus get X j ∈ Comb | N j | ≤ m X r =1 X j ∈ Comb r n k + r − p ( n k +1 + . . . + n k + r − ) ≤ m X r =1 | Comb r | n k + r (cid:16) − pd − (cid:17) ≤ m X r =1 (2 p + 1) r n k d r (cid:16) − pd − (cid:17) ≤ n k (cid:16) − pd − (cid:17) p + 1 d − p − < n k p + 1 d − p − . Plugging this back into the previous estimate and noticing that (deg R k ) /n k ≤ ( n + . . . + n k ) /n k ≤ /d k − + . . . + 1 < d/ ( d −
1) yields (23). (cid:3)
Proof of the upper bound of Theorem 1.
We want to show that, for ( a k ), a sequence ofnonnegative real numbers, we have(25) Z T N X k =0 a k R k ! p ≤ C p N X k =0 a pk Z T R pk . For N = 0 this is obvious. When 0 < p ≤ x + y ) p ≤ x p + y p , x, y ≥ C p = 1). Let N ≥
1. Suppose that for some integer m ≥
1, (25) holds when m − < p ≤ m and we want to show it when m < p ≤ m + 1.Iterating the inequality ( x + y ) p ≤ x p + 2 p ( yx p − + y p ), x, y ≥ Z T N X k =0 a k R k ! p ≤ a pN Z T R pN + 2 p N − X k =0 a k Z T R k N X i = k +1 a i R i ! p − + N − X k =0 a pk Z T R pk . The challenge is to deal with the mixed term N − X k =0 a k Z T R k N X i = k +1 a i R i ! p − = N − X k =0 a k Z T R pk F p − k , where F k = N X i = k +1 a i R k +1 ,i , k ≥ . We shall make several observations. Firstly, take α, β > /α + 1 /β = 1 and useH¨older’s inequality, Z T R pk F p − k = Z T R p/αk (cid:16) R p/βk F p − k (cid:17) ≤ (cid:18)Z T R pk (cid:19) /α (cid:18)Z T R pk F ( p − βk (cid:19) /β which holds trivially when β = 1). Choosing β so that ( p − β = ⌈ p ⌉ − m gives usthe natural power at F k . Then brutally expanding yields Z T R pk F ( p − βk = Z T R pk N X i = k +1 a i R k +1 ,i ! m = X m k +1 + ... + m N = m (cid:18) mm k +1 , . . . , m N (cid:19) Z T R pk N Y i = k +1 a m i i R m i k +1 ,i . The integral R T R pk Q Ni = k +1 R m i k +1 ,i is of the form R T R pk X l k +1 . . . X l N N with the nonnegativeinteger powers l k +1 , . . . , l N not exceeding m < p . Therefore we can apply Lemma 23 tofactor R pk out, Z T R pk N Y i = k +1 R m i k +1 ,i ≤ (1 + ǫ ) Z T R pk Z T N Y i = k +1 R m i k +1 ,i , provided that d > p + 1, and then use the multinomial formula again to get back to F mk , Z T R pk F mk ≤ (1 + ǫ ) Z T R pk Z T F mk . Recall that ǫ = πdd − p (2 p +1) d − p − . We choose d p large enough to assure that for d ≥ d p we have ε <
1. By the inductive assumption, Z T F mk ≤ C m N X i = k +1 a mi Z T R mk +1 ,i with C m ≥
1, provided that d ≥ d m . We finally get N − X k =0 a k Z T R pk F p − k ≤ N − X k =0 a k (cid:18)Z T R pk (cid:19) /α Z T R pk · C m N X i = k +1 a mi Z T R mk +1 ,i ! /β ≤ C m N X k =0 N X i = k +1 a k a p − i Z T R pk (cid:18)Z T R mk +1 ,i (cid:19) /β . Lastly, notice that we have R k +1 ,i to the power of m but we want the p -th power. Since m < p , there is some room. Introduce the constant λ p = ( R T X m ) /m ( R T X p ) /p ! p − < . y (22) we obtain (cid:18)Z T R mk +1 ,i (cid:19) /β ≤ (cid:18) πpd − (cid:19) i − k Z T X mk +1 . . . Z T X mi ! /β ≤ (cid:18) πpd − (cid:19) i − k (cid:16) λ m/ ( p − p (cid:17) i − k (cid:18)Z T X pk +1 . . . Z T X pi (cid:19) m/p ! /β = " (cid:18) πpd − (cid:19) /β λ p i − k (cid:18)Z T X pk +1 . . . Z T X pi (cid:19) ( p − /p ≤ η i − kp (cid:18)Z T X pk +1 . . . Z T X pi (cid:19) ( p − /p , where η p = (cid:16) πpd − (cid:17) λ p . Therefore, N − X k =0 a k Z T R pk F p − k ≤ C m N X k =0 N X i = k +1 η i − kp (cid:18)Z T R pk (cid:19) · a k a p − i (cid:18)Z T X pk +1 . . . Z T X pi (cid:19) ( p − /p ≤ C m N X k =0 N X i = k +1 η i − kp (cid:18)Z T R pk (cid:19) · (cid:18) p a pk + p − p a pi Z T X pk +1 . . . Z T X pi (cid:19) . Provided that η p <
1, the first bit can be easily estimated as desired, N X k =0 N X i = k +1 η i − kp (cid:18)Z T R pk (cid:19) · p a pk ≤ η p p (1 − η p ) N X k =0 a pk Z T R pk . The second one requires some more work. With the aid of (21) with k = 1 and (22), Z T R pk Z T X pk +1 . . . Z T X pi ≤ (cid:18) − πpd − (cid:19) − ( i − k ) Z T R pi , so, provided that η p < − πpd − , that is λ p (cid:16) πpd − (cid:17) < (cid:16) − πpd − (cid:17) , we obtain N X k =0 N X i = k +1 η i − kp (cid:18)Z T R pk (cid:19) · p − p a pi Z T X pk +1 . . . Z T X pi ≤ p − p N X i =1 a pi Z T R pi i − X k =0 " η p − πpd − i − k ≤ p − p − n p − πpd − ! − N X i =1 a pi Z T R pi . Putting everything together, N − X k =0 a k Z T R k N X i = k +1 a i R i ! p − ≤ C N X k =0 a pk Z T R pk , here C = 2 C m η p p (1 − η p ) + p − p − n p − πpd − ! − . Thus, Z T N X k =0 a k R k ! p ≤ p (1 + C ) N X k =0 a pk Z T R pk , which completes the proof. (cid:3) Remark . Even though we have not kept track of the values of the constants d p , c p and C p in our arguments, with some extra work it can be shown that for the upper bound inTheorem 1 one can take d (upper) p = 80 p and C p = (16 p ) p +1 , p > , whereas for the lower bound it is enough to have d ( lower ) p = (cid:16) p − (cid:17) p − c p = (cid:16) p − (cid:17) p − , p ∈ (1 ,
2] and d ( lower ) p = 10 p c p = 10 − p , p > . References [1] A. Bonami, ´Etude des coefficients de Fourier des fonctions de L p ( G ). Ann. Inst. Fourier (Grenoble) L p -norms of combinations ofproducts of independent random variables, Stochastic Process. Appl. 125 (2015), 1688–1713.[3] Y. Katznelson, An introduction to harmonic analysis. Third edition. Cambridge Mathematical Library. Cambridge University Press , Cambridge, 2004.[4] K. Kazaniecki, M. Wojciechowski, Ornstein’s non-inequalities Riesz product approach, preprint,arXiv:1406.7319[5] K. Kazaniecki, M. Wojciechowski, On the continuity of Fourier multipliers on the homogeneous Sobolevspaces ˙ W ( R d ), Annales de l’Institut Fourier 66(3) (2016), 1247–1260.[6] Keogh, F., Riesz products, Proc. London Math. Soc. (3) 14a 1965, 174–182.[7] R. Lata la, L -norm of combinations of products of independent random variables, Israel J. Math. 203(2014), 295–308.[8] Y. Meyer, Endomorphismes des id´eaux ferm´es de L ( G ), classes de Hardy et s´eries de Fourier lacunaires,Ann. Sci. Ecole Norm. Sup. 1(4) (1968) 499–580.[9] Y. Meyer, Algebraic numbers and harmonic analysis. Vol. 2. North-Holland Publishing Co., New York,1972.[10] R. Schneider, Some theorems in Fourier analysis on symmetric sets. Pacific J. Math. 31 (1969), 175–196.[11] M. Wojciechowski, On the strong type multiplier norms of rational functions in several variables, IllinoisJ. Math. 42 (1998), 582–600.[12] A. Zygmund, Trigonometric series, Vol. II, Third edition, Cambridge Mathematical Library, CambridgeUniversity Press, Cambridge, 2002. A. B.) Institut Denis Poisson, CNRS-UMR 2013, Universit´e dOrl´eans, France
E-mail address : [email protected] (R.L.) Institute of Mathematics, University of Warsaw, Banacha 2, 02-097, Warsaw, Poland
E-mail address : [email protected] (P. N.) Institute of Mathematics, University of Warsaw, Banacha 2, 02-097, Warsaw, Poland E-mail address : [email protected] (T. T.) Department of Mathematical Sciences, Carnegie Mellon University, Pittsburgh,PA 15213, USA E-mail address : [email protected]@math.cmu.edu