FLATNESS OF CONJUGATE RECIPROCAL UNIMODULAR POLYNOMIALS. June 24, 2015

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1 FLATNESS OF CONJUGATE RECIPROCAL UNIMODULAR POLYNOMIALS Tamás Erdélyi June 24, 2015 Abstract. A polynomial is called unimodular if each of its coefficients is a complex number of modulus 1. A polynomial P of the form P(z) = n j=0 a jz j is called conjugate reciprocal if a n j = a j, a j C for each j = 0,1,...,n. Let D be the unit circle of the complex plane. We prove that there is an absolute constant ε > 0 such that max f(z) (1+ε) 4/3m 1/2, z D for every conjugate reciprocal unimodular polynomial of degree m. We also prove that there is an absolute constant ε > 0 such that M q (f ) exp(ε(q 2)/q) 1/3m 3/2, 1 q < 2, and M q (f ) exp(ε(q 2)/q) 1/3m 3/2, 2 < q, for every conjugate reciprocal unimodular polynomial of degree m, where M q (g) = ( 1 2π 2π 0 g(e it ) q dt) 1/q, q > Introduction LetT n bethesetofallreal trigonometricpolynomialsofdegreeatmost n. Let Pn c bethe set of all algebraic polynomials of degree at most n with complex coefficients. Throughout this paper it will be comfortable for us to denote an appropriate period [a,a+2π) by. Let n A n := Q : Q(t) = cos(jt+γ j ), γ j R j=1 ey words and phrases. Littlewood polynomials; unimodular polynomials; conjugate reciprocal polynomials; flatness properties Mathematics Subject Classifications. 11C08, 41A17 1 Typeset by AMS-TEX

2 and n B n+1/2 := Q : Q(t) = ( ) 2j +1 cos t+γ j, γ j R 2. j=0 We use the notation Q p := ( ) 1/p 1 Q(t) p, p > 0, 2π and Q := max t Q(t). The Bernstein Szegő inequality (see p. 232 in [7], for instance) gives that Q (t) 2 +n 2 Q(t) 2 n 2 Q 2, Q T n, t R. Integrating the left hand side on the period and using Parseval s formula we obtain n(n+1)(2n+1) 12 + n3 2 n2 Q 2, Q A n, and hence (1.1) Q 4/3 n/2, Q A n. One of the highlights of this paper is to improve (1.1) by showing that there is an absolute constant ε > 0 such that Q (1+ε) 4/3 n/2, Q A n. Let n m := P : P(z) = a j z j, a j C, a j = 1, j = 0,1,...,m j=0 be the set of all unimodular polynomials of degree m. Associated with an algebraic polynomial P of the form P(z) = m a j z j, a j C, a m 0, j=0 let m P(z) := a j z j and P (z) := z m P(1/z). j=0 2

3 The polynomial P of degree m is called conjugate reciprocal if P = P. The classes A n, B n+1/2, and m and flatness properties of their elements were studied by many authors, see [1 40], for instance. Let and M q (f) := f(e it ) q = ( 1/q 1 f(e it ) dt) q, q (0, ), 2π M (f) := sup f(e it ). t There is a beautiful short argument to see that (1.2) M (f) 4/3m 1/2 for every conjugate reciprocal unimodular polynomial f m. Namely, Parseval s formula gives ( ) 1/2 m(m+1)(2m+1) M (f ) M 2 (f ) =, f m. 3 Combining this with Lax s Bernstein-type inequality M (f ) m 2 M (f) valid for all conjugate reciprocal algebraic polynomials f Pn c instance), we obtain (see p. 438 in [7], for ( ) 1/2 m(m+1)(2m+1) 4/3m 1/2 M (f) 2 m 3 for all conjugate reciprocal unimodular polynomials f m. One of the highlights of this paper is to improve (1.2) by showing that there is an absolute constant ε > 0 such that M (f) (1+ε) 4/3m 1/2, for every conjugate reciprocal unimodular polynomial f m. We also prove that there is an absolute constant ε > 0 such that M q (f ) exp(ε(q 2)/q) 1/3m 3/2, 1 q < 2, and M q (f ) exp(ε(q 2)/q) 1/3m 3/2, 2 < q, for every conjugate reciprocal unimodular polynomial of degree m. See Theorem

4 2. New Results Theorem 2.1. Let Q A n and P = (Q ) 2 +n 2 Q 2. There is an absolute constant δ > 0 such that P 1/2 (1 δ) P 1. Theorem 2.1*. Let Q B n+1/2 and P = (Q ) 2 + (n + 1/2) 2 Q 2. There is an absolute constant δ > 0 such that P 1/2 (1 δ) P 1. Theorem 2.2. Let Q A n and P = (Q ) 2 +n 2 Q 2. There is an absolute constant δ > 0 such that P (1+δ) P 1. Theorem 2.2*. Let Q B n+1/2 and P = (Q ) 2 + (n + 1/2) 2 Q 2. There is an absolute constant δ > 0 such that P (1+δ) P 1. Theorem 2.3. There is an absolute constant δ > 0 such that for every Q A n. Q (1+δ) 4/3 n/2 Theorem 2.3*. There is an absolute constant δ > 0 such that for every Q B n+1/2. Q (1+δ) 4/3 n/2 Let D be the unit circle of the complex plane. Let f be a continuous function on D and let I q (f) := M q (f) q = 1 f(e it ) q dt 2π and M (f) := max z D f(z). Theorem 2.4. There is an absolute constant ε > 0 such that M 1 (f ) (1 ε) 1/3m 3/2 for every conjugate reciprocal unimodular polynomial f m. Theorem 2.5. There is an absolute constant ε > 0 such that M (f ) (1+ε) 1/3m 3/2 for every conjugate reciprocal unimodular polynomial f m. 4

5 Theorem 2.6. There is an absolute constant ε > 0 such that M (f) (1+ε) 4/3m 1/2 for every conjugate reciprocal unimodular polynomial f m. Theorem 2.7. There is an absolute constant ε > 0 such that M q (f ) exp(ε(q 2)/q) 1/3m 3/2, 1 q < 2, and M q (f ) exp(ε(q 2)/q) 1/3m 3/2, 2 < q, for every conjugate reciprocal unimodular polynomial f m. The above results were well known before without the absolute constants δ > 0 and ε > 0, respectively. Remark 2.1. The factor (1+δ) in Theorem 2.2 cannot be replaced by (1+δ)3/2. Remark 2.2. The factor (1+δ) 4/3 in Theorem 2.3 cannot be replaced by (1+δ) 2. Remark 2.3. The factor (1+ε) 1/3 in Theorem 2.5 cannot be replaced by (1+ε) 1/2. Remark 2.4. The factor (1+ε) 4/3 in Theorem 2.6 cannot be replaced by (1+ε) 2. A polynomial f P c m is called skew-reciprocal if f (z) = f( z). A polynomial f P c m is called plain-reciprocal if f = f, that is, f(z) = z m f(1/z) for all z C\{0}. Observe that Corollary 2.8 in [28] may be formulated as follows. Remark 2.5. There is an absolute constant ε > 0 such that max z D f (z) min z D f (z) εm 3/2 for all conjugate reciprocal, plain-reciprocal, and skew-reciprocal unimodular polynomials f m. Observe that for conjugate reciprocal unimodular polynomials Theorem 2.5 is stronger than Remark 2.5. Problem 2.1. Is there an absolute constant ε > 0 such that M (f ) (1+ε) 1/3m 3/2 holds for all plain-reciprocal and skew-reciprocal unimodular polynomials f m? 5

6 Problem 2.2. Is there an absolute constant ε > 0 such that or at least M (f ) (1+ε) 1/3m 3/2 max z D f (z) min z D f (z) εm 3/2 holds for all unimodular polynomials f m? Our methodto provetheorem 2.5doesnot seem towork for allunimodular polynomials f m. In an communication several years ago B. Saffari speculated that the answer to Problem 2.2 is no. However we do not know the answer even to Problem 2.1. Let L m be the collection of all polynomials of degree m with each of their coefficients in { 1,1}. Problem 2.3. Is there an absolute constant ε > 0 such that or at least M (f ) (1+ε) 1/3m 3/2 max z D f (z) min z D f (z) εm 3/2 holds for all Littlewood polynomials f L m? The following problem due to Erdős [29] is open for a long time. Problem 2.4. Is there an absolute constant ε > 0 such that or at least M (f) (1+ε)m 1/2 max f(z) min f(z) εm1/2 z D z D holds for all Littlewood polynomials f L m? The same problem may be raised only for all skew reciprocal Littlewood polynomials f L m and as far as we know, it is also open. 3. Lemmas Let m(a) denote the Lebesgue measure of a measurable set A R. The following lemma is due to Littlewood, see Theorem 1 in [34]. Lemma 3.1. Let R T n be of the form n R(t) = R n (t) = a j cos(jt+γ j ) j=1 6 a j,γ j R, j = 1,2,...,n.

7 Let s m := m j=1 a2 j, m = 1,2,...,n, and let µ := R 2, that is, µ 2 = s n. Suppose R 1 cµ, where c > 0 is a constant (necessarily not greater than 1). Suppose also that the coefficients of R satisfy s [n/h] /s n = µ 2 a 2 j 2 9 c 6 1 j n/h for some constant h > 0. Let V = 2 5 c 3. Then there exists a constant B > 0 depending only on c and h such that m({t : v 1 µ R(t) v 2 µ}) B(v 2 v 1 ) 2 for every v 1 and v 2 such that V v 1 < v 2 V. Lemma 3.2. Associated with Q T n we define the sets E δ := {t : Q (t) < δn 3/2 } and We have F δ := {t : Q (t) δ 1/2 n 5/2 }. m(e δ \F δ ) 8δ 1/2. Proof of Lemma 3.2. Observe that E δ \F δ is the union of at most 4n pairwise disjoint open subintervals of the period. Let these intervals be (x j,y j ),j = 1,2,...,µ, where µ 4n. By the Mean Value Theorem we can deduce that there are ξ j (x j,y j ) such that and hence Hence 2δn 3/2 Q (y j ) Q (x j ) = Q (ξ j ) (y j x j ) δ 1/2 n 5/2 (y j x j ), y j x j 2δ 1/2 n 1, j = 1,2,...,µ. m(e δ \F δ ) = µ (y j x j ) µ(2δ 1/2 n 1 ) 8δ 1/2. j=1 Lemma 3.3. Let Q A n, P := (Q ) 2 +n 2 Q 2, δ (0,1), and Suppose G δ := {t : P(t) 1/2 P 1/2 1 δ 1/4 P 1/2 1 }. (3.1) P 1/2 (1 δ) P 1. 7

8 Then m( \G δ ) 4πδ 1/2. Proof of Lemma 3.3. Using (3.1) we have (3.2) P 1/2 1/2 (1 δ)1/2 P 1/2 1 (1 δ) P 1/2 1. Hence P(t) 1/2 P 1/2 1 2 dt = 4π P 1/2 1 ( 1/2 P 1 P 1/2 ) 1/2 4π P 1/2 1 δ P 1/2 1 = 4πδ P 1. Letting a := m( \G δ ) we have and the lemma follows. aδ 1/2 P 1 4πδ P 1, Lemma 3.4. Let Q A n, P := (Q ) 2 + n 2 Q 2, and P 1/ P 1. Then R := n 2 Q 2 +Q satisfies the assumptions of Lemma 3.1 with c := 1/32 and h = for all sufficiently large n. Proof of Lemma 3.4. Let n 3. Observe that n R(t) = a j cos(jt+γ j ), j=1 a j := n 2 j 2, γ j R, j = 1,2,...,n. Then hence Using Parseval s formula, we have s n = µ 2 = R 2 2 = 1 2 (3.3) P 1 = 1 2 n (n 2 j 2 ) 2, j=1 n 5 6 s n = µ 2 n5 2. n j=1 (j 2 +n 2 ) 2n3 3, (3.4) Q 1 Q n j=1 8 j 4 1/2 (1+o(1)) n5/2 10,

9 and (3.5) Q n j=1 j 2 1/2 (1+o(1)) n3/2 6, and hence (3.6) nq 1 =(2π) 1 (2π) 1 P(t) Q (t) 2 1/2 dt P(t) 1/2 dt (2π) 1 Q (t) dt ( ) 1/2 = P 1/2 31 1/2 Q 1 P 1/2 1 Q n3/2 (1+o(1)) 6 n3/2. Combining (3.5) and (3.4) we conclude R 1 = n 2 Q+Q 1 n 2 Q 1 Q n5/2 (1+o(1)) 1 32 n5/2 6 n5/2 (1+o(1)) n5/2 10 forallsufficientlylargen. Also,s [n/h] (n/h)n 4 = n 5 /h,hences [n/h] /s n 1/h. Therefore the assumptions of Lemma 3.1 are satisfied with c := 1/32 and h = Proof of the Theorems Proof of Theorem 2.1. Let Q A n, P = (Q ) 2 +n 2 Q 2, and R := n 2 Q 2 +Q. Let δ (0,1). Suppose P 1/2 (1 δ) P 1. Lemma 3.4 states that if 0 < δ 1/32 then R satisfies the assumptions of Lemma 3.1 with c = 1/32 and h = Now let and E δ := {t : Q (t) < δn 3/2 }, F δ := {t : Q (t) δn 5/2 }, G δ := {t : P(t) 1/2 P 1/2 1 δ 1/4 P 1/2 1 }, Recall that by the Parseval formula we have H γ := {t : γn 5/2 R(t) < 2γn 5/2 }. (4.1) P 1 = n3 2 + n(n+1)(2n+1). 12 9

10 Hence, if t G δ E δ F δ and the absolute constant δ > 0 is sufficiently small, then that is, R(t) n 2 Q(t) Q (t) = n(p(t) Q (t) 2 ) 1/2 Q (t) ( n n ((1 δ 1/4 ) n(n+1)(2n+1) ) ) 1/2 δ 2 n 3 δn 5/ n5/2, (4.2) R(t) 1 2 n5/2, t G δ E δ F δ. By Lemma 3.2 we have (4.3) m(e δ \F δ ) 8δ 1/2. By Lemma 3.3 we have (4.4) m( \G δ ) 4πδ 1/2. Observe that if 0 < γ < 1/4 then (4.2) implies that H γ \(G δ E δ F δ ), hence H γ E δ (E δ \G δ ) (E δ \F δ ). Therefore, by (4.3) and (4.4) we can deduce that (4.5) m(h γ E δ ) m(e δ \G δ )+m(e δ \F δ ) 4πδ 1/2 +8δ 1/2. By Lemmas 3.1 and 3.4 there are absolute constants 0 < γ < 1/4 and B > 0 such that (4.6) m(h γ ) Bγ 2. It follows from (4.5) and (4.6) that (4.7) m(h γ \E δ ) 1 2 Bγ2 for all sufficiently small absolute constants δ > 0. Observe that (4.8) 2Q (t)r(t) 2δn 3/2 γn 5/2 = 2γδn 4, t H γ \E δ, and (4.9) P (t) = 2Q (t)r(t). 10

11 Combining (4.7), (4.8), and (4.9), we obtain m({t : P (t) 2γδn 4 }) 1 2 Bγ2, and hence (4.10) P (t) dt 1 2 Bγ2 (2γδn 4 ) = Bγ 3 δn 4. Now let P := P 2π P 1 T 2n. Then (4.10) can be rewritten as P (t) dt Bγ 3 δn 4, and by Bernstein s inequality in L 1 (see p. 390 of [7], for instance), we have (4.11) 2π P 1 = P(t) dt 1 2 Bγ3 δn 3. Observe that (4.12) 2π P 1 = = ( P(t) dt P(t) P 1 dt (P(t) 1/2 P 1/2 1 =2π ( 2 P 1/2 1 ( P 1/2 1 P 1/2 4π P 1/2 1 Combining (4.11), (4.12), and (4.1), we conclude ( P(t) 1/2 P 1/2 )( 1 P(t) 1/2 + P 1/2 ) dt 1 ) 1/2 ( ) 2 dt (P(t) 1/2 + P 1/2 1/2 2 1 dt )) 1/2 ( 1/2 1/2 2 P 1 ( ) P 1 P ( 1/2 ) 1/2 = 4πn 3/2 1/2. P 1 P 1/2 P 1 P 1/2 ( 2π P 1 4πn 3/2 δ P 1 with an absolute constant δ > 0. ( P 1/2 1 + P 1/2 ) 2 ( Bγ 3 δn 3/2 ) 2 δ n 3 8π 1/2 )) 1/2 Proof of Theorem 2.2. By Theorem 2.1 there is an absolute constant δ > 0 such that P 1 = 2π P(t) dt = 2π P(t) 1/2 P(t) 1/2 dt 2π P(t) 1/2 P 1/2 dt P 1/2 1/2 P 1/2 (1 δ)1/2 P 1/2 1 P 1/2. 11

12 Hence and the result follows. P 1/2 1 (1 δ) 1/2 P 1/2, Proof of Theorem 2.3. The Bernstein Sze go inequality(see p. 232 of[7],for instance) yields hence if P = (Q ) 2 +n 2 Q 2, then Q (t) 2 +n 2 Q(t) 2 n 2 Q 2, t R, Q A n T n, P n 2 Q. Hence, using Theorem 2.1 and Parseval s formula we can deduce that Q 2 n 2 P n 2 (1+δ) P 1 = n 2 (1+δ) ( ) Q 2 2 +n2 Q 2 2 ( ) n(n+1)(2n+1) =n 2 (1+δ) + n3 (1+δ)(4/3)(n/2) 12 2 with an absolute constant δ > 0 and the theorem follows. The proofs of Theorems 2.1*, 2.2*, and 2.3* are similar to those of Theorems 2.1 and 2.2, respectively. The modifications required in the proofs of Theorems 2.1* and 2.2* are straightforward for the experts and we omit the details. Proof of Theorem 2.4. First assume that m = 2n is even and f m is a conjugate reciprocal unimodular polynomial. Let f(z) = m j=0 a jz j, where a j C and a j = 1 for each j = 0,1,...m. As f is conjugate reciprocal, we have a m j = a j, j = 0,1,...,m, and a n { 1,1}, in particular. Let Q A n be defined by 2Q(t) = e int f(e it ) a n. Then hence the triangle inequality implies that ie it f (e it ) = e int (2Q (t)+in(2q(t)+a n )), f (e it ) 2 e int (Q (t)+inq(t)) + e int ina n = 2 Q (t)+inq(t) +n = 2 P(t) 1/2 +n, where P := (Q ) 2 + n 2 Q 2 is the same as in Theorem 2.1, and the theorem follows from Theorem 2.1 as M 1 (f ) 2 P 1/2 1/2 +n 2(1 δ)1/2 P 1/2 1 +n ( ) 1/2 n(n+1)(2n+1) 2(1 δ) 1/2 + n3 +n 12 2 ( ) 2(n+1) 2(1 δ) 1/2 3 1/2 +n 3 (1 δ) 1/2 1/3m 3/2 +o(m 3/2 ). 12

13 Now assume that m = 2n+1 is odd and f m is a conjugate reciprocal unimodular polynomial. Let Q B m/2 be defined by 2Q(t) = e imt/2 f(e it ). Then implies that ie it f (e it ) = 2e imt/2 (Q (t)+(im/2)q(t)) f (e it ) = 2 e imt/2 (Q (t)+(im/2)q(t)) = 2 Q (t)+(im/2)q(t) = 2 P(t) 1/2, where P := (Q ) 2 +(n+1/2) 2 Q 2 is the same as in Theorem 2.1*, and the theorem follows from Theorem 2.1* as M 1 (f ) 2 P 1/2 1/2 2(1 δ)1/2 P 1/2 1 ( (n+1)(n+2)(2n+3) 2(1 δ) 1/2 12 ( ) 2(n+1) 2(1 δ) 1/2 3 1/2 3 (1 δ) 1/2 1/3m 3/2 +o(m 3/2 ). ) 1/2 + (n+1)3 2 Proof of Theorem 2.5. Let f m be a conjugate reciprocal unimodular polynomial. By Theorem 2.4 there is an absolute constant ε > 0 such that m(m+1)(2m+1) 6 = ( M 2 (f ) ) 2 = 2π 2π M 1 (f )M (f ) f (e it ) max τ f (e iτ ) dt (1 ε) 1/3m 3/2 M (f ). f (e it ) 2 dt = 2π f (e it ) f (e it ) dt Hence 1/3m 3/2 (1 ε)m (f ), and the theorem follows. Proof 1 of Theorem 2.6. First assume that m = 2n is even and f m is a conjugate reciprocal unimodular polynomial. Let f(z) = m j=0 a jz j, where a j C and a j = 1 for each j = 0,1,...m. As f is conjugate reciprocal, we have a m j = a j, j = 0,1,...,m, 13

14 and a n { 1,1}, in particular. Let Q A n be defined by 2Q(t) = e int f(e it ) a n. Observe that max f(z) 2Q z D 1, hence the theorem follows from Theorem 2.3. Now assume that m = 2n + 1 is odd and f m is a conjugate reciprocal unimodular polynomial. Let Q B n/2 be defined by 2Q(t) = e imt/2 f(e it ). Observe that max f(z) = 2Q, z D hence the theorem follows from Theorem 2.3*. Proof 2 of Theorem 2.6. It is well known (see p. 438 of [7], for instance) that if f is a conjugate reciprocal unimodular polynomial of degree m then f = (m/2) f. Hence the theorem follows from a combination of this and Theorem 2.5. Proof of Theorem 2.7. Let f m be conjugate reciprocal. Observe that by Parseval s formula we have ( ) 1/2 m(m+1)(2m+1) (4.13) M 2 (f ) = = (1+o(1)) m3/ As we will see, both inequalities of the theorem follow from Theorem 2.4, (4.13), and the following convexity property of the function h(q) := qlog(m q (f)) on (0, ). Let f be a continuous function on D and let I q (f) := M q (f) q = 1 2π f(e it ) q dt. Then h(q) := log(i q (f)) = qlog(m q (f)) is a convex function of q on (0, ). This is a simple consequence of Hölder s inequality. For the sake of completeness, before we apply it, we present the short proof of this fact. We need to see that if q < r < p, then that is, (4.14) To see this let I r (f) I p (f) r q p q Iq (f) p r p q, ( p q ( ) r q ( p r f(e it ) dt) r f(e it ) p dt f(e it ) dt) q. 2π 2π 2π α := p q r q, hence 1/α+1/β = 1 and γ +δ = r. Let p q β := p r, γ := p α, δ := q β, F(t) := f(e it ) γ = f(e it ) p(r q) p q, 14

15 and Then by Hölder s inequality we conclude G(t) := f(e it ) δ = f(e it ) q(p r) p q. ( F(t)G(t)dt ) 1/α ( 1/β F(t) α dt G(t) dt) β, and (4.14) follows. Let q [1,2). Then, using the convexity property of the function h(q) := qlog(m q (f)) on (0, ), we obtain 2logM 2 qlogm q 2 q 2logM 2 logm Combining this with the Theorem 2.4 and (4.13) gives theorem. Now let q (2, ). Then, using the convexity property of the function h(q) := qlog(m q (f)) on (0, ), we obtain qlogm q 2logM 2 q 2 2logM 2 logm Combining this with the Theorem 2.4 and (4.13) gives theorem. Proof of Remark 2.1. Let (f n ) be anultraflat sequence of unimodular polynomialsf n n satisfying M (f n ) (1+ε n ) n with a sequence (ε n ) of numbers ε n > 0 converging to 0. It is shown in [7] that such a sequence (f n ) exists. Let g n (z) = zf n 1 (z). Let Q n A n be defined by Q n (t) := Re(g n (e it )). Then the Bernstein Szegő inequality (see p. 232 in [7], for instance) gives that P n := (Q n) 2 +n 2 Q 2 n satisfy while by Parseval s formula we have P n n 2 Q n 2 (1+ε n) 2 n 3, P n 1 = n3 2 + n(n+1)(2n+1) 12 2n3 3. Proof of Remark 2.2. Let Q n A n be the same as in the proof of Remark 2.1. Then Q n P n 1/2 (1+ε n)n 3/2. Proof of Remark 2.4. Let f n n and g n (z) = zf n 1 (z) be the same as in the proof of Remark 2.1. For m = 2n we define h m m by h m (z) := z n (g n (z)+g n (1/z)+1. 15

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