THE NUMBER OF UNIMODULAR ZEROS OF SELF-RECIPROCAL POLYNOMIALS WITH COEFFICIENTS IN A FINITE SET. May 31, 2016

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1 THE NUMBER OF UNIMODULAR ZEROS OF SELF-RECIPROCAL POLYNOMIALS WITH COEFFICIENTS IN A FINITE SET Tamás Erdélyi May 31, 2016 Abstract. Let NZ(T n ) denote the number of real zeros of a trigonometric polynomial T n (t) = n a j,n cos(jt), a j,n C, in a period [a,a + 2π), a R. Let NZ(P n ) denote the number of zeros of an algebraic polynomial n P n (z) = p j,n z j, p j,n C, on the unit circle of C. Let u+k 1 NC k (P n ) := u : 0 u n k +1, p j,n 0. One of the highlights of this paper states that lim n NZ(T n ) = whenever the set is finite and j=u {a j,n : j {0,1,...,n}, n N} [0, ) lim {j {0,1,...,n} : a j,n 0} =. n This follows from a more general result stating that lim n NZ(P 2n ) = whenever P 2n is self-reciprocal, the set {p j,2n : j {0,1,...,2n}, n N} R is finite, and lim n NC k (P 2n ) = for every k N. Key words and phrases. self-reciprocal polynomials, trigonometric polynomials, restricted coefficients, number of zeros on the unit circle, number of real zeros in a period, Conrey s question Mathematics Subject Classifications. 11C08, 41A17, 26C10, 30C15 1 Typeset by AMS-TEX

2 1. Introduction and Notation Research on the distribution of the zeros of algebraic polynomials has a long and rich history. In fact all the papers [1-43] in our list of references are just some of the papers devoted to this topic. The study of the number of real zeros trigonometric polynomials and the number of unimodular zeros (that is, zeros lying on the unit circle of the complex plane) of algebraic polynomials with various constraints on their coefficients are the subject of quite a few of these. We do not try to survey these in our introduction. Let S C. Let Pn c (S) be the set of all algebraic polynomials of degree at most n with each of their coefficients in S. A polynomial (1.1) P n (z) = is called conjugate-reciprocal if n p j,n z j, p j,n C, (1.2) p j,n = p n j,n, j = 0,1,...,n. A polynomial P n of the form (1.1) is called plain-reciprocal or self-reciprocal if (1.3) p j,n = p n j,n, j = 0,1,...,n. If a conjugate reciprocal polynomial P n has only real coefficients, then it is obviously plain-reciprocal. We note also that if P 2n (z) = 2n p j,2n z j, p j,2n C, is conjugate-reciprocal, then there are θ j R, j = 1,2,... n, such that T n (t) := P 2n (e it )e int = p n,2n + n 2 p j,2n cos(jt+θ j ). j=1 If the polynomial P 2n above is plain-reciprocal, then T n (t) := P 2n (e it )e int = p n,2n + n 2p j,2n cos(jt). In this paper, whenever we write P n P c n(s) is conjugate-reciprocal we mean that P n is of the form (1.1) with each p j,n S satisfying (1.2). Similarly, whenever we write P n P c n(s) is self-reciprocal we mean that P n is of the form (1.1) with each p j,n S satisfying (1.3). This is going to be our understanding even if the degree of P n P c n (S) is less than n. Associated with an algebraic polynomial P n of the form (1.1) we introduce the numbers NC(P n ) := {j {0,1,...,n} : p j,n 0}. 2 j=1

3 Here, and in what follows A denotes the number of elements of a finite set A. Let NZ(P n ) denote the number of real zeros (by counting multiplicities) of an algebraic polynomial P n on the unit circle. Associated with a trigonometric polynomial we introduce the numbers T n (t) = n a j,n cos(jt) NC(T n ) := {j {0,1,...,n} : a j,n 0}. Let NZ(T n ) denote the number of real zeros (by counting multiplicities) of a trigonometric polynomial T n in a period [a,a+2π), a R. The quotation below is from [6]. Let 0 n 1 < n 2 < < n N be integers. A cosine polynomial of the form T N (θ) = N j=1 cos(n jθ) must have at least one real zero in a period [a,a + 2π), a R. This is obvious if n 1 0, since then the integral of the sum on a period is 0. The above statement is less obvious if n 1 = 0, but for sufficiently large N it follows from Littlewood s Conjecture simply. Here we mean the Littlewood s Conjecture proved by S. Konyagin [25] and independently by McGehee, Pigno, and Smith [33] in See also [13, pages ] for a book proof. It is not difficult to prove the statement in general even in the case n 1 = 0 without using Littlewood s Conjecture. One possible way is to use the identity n N j=1 T N ((2j 1)π/n N ) = 0. See [26], for example. Another way is to use Theorem 2 of [34]. So there is certainly no shortage of possible approaches to prove the starting observation of this paper even in the case n 1 = 0. It seems likely that the number of zeros of the above sums in a period must tend to with N. In a private communication B. Conrey asked how fast the number of real zeros of the above sums in a period tends to as a function N. In [4] the authors observed that for an odd prime p the Fekete polynomial f p (z) = p 1 k=0( k p) z k (the coefficients are Legendre symbols) has κ 0 p zeros on the unit circle, where > κ 0 > Conrey s question in general does not appear to be easy. Littlewood in his 1968 monograph Some Problems in Real and Complex Analysis [10, problem 22] poses the following research problem, which appears to still be open: If the n m are integral and all different, what is the lower bound on the number of real zeros of N m=1 cos(n mθ)? Possibly N 1, or not much less. Here real zeros are counted in a period. In fact no progress appears to have been made on this in the last half century. In a recent paper [3] we showed that this is false. There exists a cosine polynomials N m=1 cos(n mθ) with the n m integral and all different so that the number of its real zeros in the period is O(N 9/10 (logn) 1/5 ) (here the frequencies n m = n m (N) may vary with N). However, there are reasons to believe that a cosine polynomial N m=1 cos(n mθ) always has many zeros in the period. 3

4 One of the highlights of this paper, Corollary 2.7, shows that the number of real zeros of the sums T N (θ) = N j=1 cos(n jθ) in a period [a,a+2π), a R, tends to whenever 0 n 1 < n 2 < < n N are integers and N tends to, even though the part how fast in Conrey s question remains open. In fact, we will prove more general results of this variety. Let n L n := P : P(z) = p j,n z j, p j,n { 1,1}. Elements of L n are often called Littlewood polynomials of degree n. Let n K n := P : P(z) = p j,n z j, p j,n C, p 0,n = p n,n = 1, p j,n 1. Observe that L n K n. In [10] we proved that any polynomial P K n has at least 8n 1/2 logn zeros in any open disk centered at a point on the unit circle with radius 33n 1/2 logn. Thus polynomials in K n have a few zeros near the unit circle. One may naturally ask how many unimodular roots a polynomial in K n can have. Mercer [34] proved that if a Littlewood polynomial P L n of the form (1.1) is skew reciprocal, that is, p j,n = ( 1) j p n j,n for each j = 0,1,...,n, then it has no zeros on the unit circle. However, by using different elementary methods it was observed in both [18] and [34] that if a Littlewood polynomial P of the form (1.1) is self-reciprocal, that is, p j,n = p n j,n for each j = 0,1,...,n, n 1, then it has at least one zero on the unit circle. Mukunda [35] improved this result by showing that every self-reciprocal Littlewood polynomial of odd degree at least 3 has at least 3 zeros on the unit circle. Drungilas [16] proved that every self-reciprocal Littlewood polynomial of odd degree n 7 has at least 5 zeros on the unit circle and every self-reciprocal Littlewood polynomial of even degree n 14 has at least 4 zeros on the unit circle. In [4] two types of Littlewood polynomials are considered: Littlewood polynomials with one sign change in the sequence of coefficients and Littlewood polynomials with one negative coefficient, and the numbers of the zeros such Littlewood polynomials have on the unit circle and inside the unit disk, respectively, are investigated. Note that the Littlewood polynomials studied in [4] are very special. In [7] we proved that the average number of zeros of self-reciprocal Littlewood polynomials of degree n is at least n/4. However, it is much harder to give decent lower bounds for the quantities NZ n := min P NZ(P), where NZ(P) denotes the number of zeros of a polynomial P lying on the unit circle and the minimum is taken for all self-reciprocal Littlewood polynomials P L n. It has been conjectured for a long time that lim n NZ n =. In this paper we show that lim n NZ(P n ) = whenever P n L n is self-reciprocal and lim n P n (1) =. This follows as a consequence of a more general result, see Corollary 2.3, in which the coefficients of the self-reciprocal polynomials P n of degree at most n belong to a fixed finite set of real numbers. In [6] we proved the following result. 4

5 Theorem 1.1. If the set {a j : j N} R is finite, the set {j N : a j 0} is infinite, the sequence (a j ) is not eventually periodic, and then lim n NZ(T n ) =. T n (t) = n a j cos(jt), In [6] Theorem 1.1 is stated without the assumption that the sequence (a j ) is not eventually periodic. However, as the following example shows, Lemma 3.4 in [6], dealing with the case of eventually periodic sequences (a j ), is incorrect. Let n 1 T n (t) :=cost+cos((4n+1)t)+ (cos((4k+1)t) cos((4k +3)t)) = 1+cos((4n+2)t) 2cost k=0 +cost. It is easy to see that T n (t) 0 on [,π]\{/2,π/2} and the zeros of T n at /2 and π/2 are simple. Hence T n has only two (simple) zeros in the period. So the conclusion of Theorem 1.1 above is false for the sequence (a j ) with a 0 := 0, a 1 := 2, a 3 := 1, a 2k := 0, a 4k+1 := 1, a 4k+3 := 1 for every k = 1,2,... Nevertheless, Theorem 1.1 can be saved even in the case of eventually periodic sequences (a j ) if we assume that a j 0 for all sufficiently large j. See Lemma So Theorem 1 in [6] can be corrected as Theorem 1.2. If the set {a j : j N} R is finite, a j 0 for all sufficiently large j, and then lim n NZ(T n ) =. T n (t) = n a j cos(jt), It was expected that the conclusion of the above theorem remains true even if the coefficients of T n do not come from the same sequence, that is T n (t) = n a j,n cos(jt), where the set S := {a j,n : j {0,1,...,n}, n N} R is finite and lim {j {0,1,...,n}, a j,n 0} =. n The purpose of this paper is to prove such an extension of Theorem 1.1. This extension is formulated as Theorem 2.1, which is the main result of this paper. The already mentioned Littlewood Conjecture, proved by Konyagin [25] and independently by McGehee, Pigno, and B. Smith [33], plays an key role in the proof of the main results in this paper. This states the following. 5

6 Theorem 1.3. There is an absolute constant c > 0 such that 2π m a j e iλ jt dt cγlogm 0 j=1 whenever λ 1,λ 2,...,λ n are distinct integers and a 1,a 2,...,a m are complex numbers of modulus at least γ > 0. This is an obvious consequence of the following result a book proof of which has been worked out by Lorentz in [13, pages ]. Theorem 1.4. If λ 1 < λ 2 < < λ m are integers and a 1,a 2,...,a m are complex numbers, then 2π m a j e iλ jt dt 1 m a j. 30 j let 0 j=1 2. New Results j=1 Associated with an algebraic polynomial n P n (z) = p j,n z j, p j,n C, Recall that if is self-reciprocal, then It is also clear that u+k 1 NC k (P n ) := u : 0 u n k +1, p j,n 0. P 2n (z) = 2n j=u p j,2n z j, p j,2n R, T n (t) := P 2n (e it )e int = p n,2n + n 2p j,2n cos(jt). j=1 NZ(P 2n ) = NZ(T n ). Theorem 2.1. If S R is a finite set, P 2n P2n c (S) are self-reciprocal polynomials, and T n (t) := P 2n (e it )e int, (2.1) lim n NC k(p 2n ) = for every k N, then (2.2) lim n NZ(P 2n) = lim n NZ(T n) =. 6

7 Corollary 2.2. If S R is a finite set, P 2n P c 2n(S) are self-reciprocal polynomials, and T n (t) := P 2n (e it )e int, (2.3) lim n P 2n(1) =, then (2.2) lim n NZ(P 2n) = lim n NZ(T n) =. Our next result is slightly more general than Corollary 2.2, and it follows from Corollary 2.2 simply. Corollary 2.3. If S R is a finite set, P n Pn c (S) are self-reciprocal polynomials, and (2.4) lim n P n(1) =, then (2.5) lim n NZ(P n) =. We say that S R has property (2.6) if (for every k N) (2.6) s 1 +s 2 + +s k = 0, s 1,s 2,...,s k S, implies s 1 = s 2 = = s k = 0, that is, any sum of nonzero elements of S is different from 0. Corollary 2.4. If the finite set S R has property (2.6), P 2n P c 2n(S) are self-reciprocal polynomials, T n (t) := P 2n (e it )e int, and (2.7) lim n NC(P 2n) =, then (2.2) lim n NZ(P 2n) = lim n NZ(T n) =. Our next result is slightly more general than Corollary 2.4, and it follows from Corollary 2.4 simply. 7

8 Corollary 2.5. If the finite set S R has property (2.5), P n P c n(s) are self-reciprocal polynomials, and (2.8) lim n NC(P n) =, then (2.5) lim n NZ(P n) =. Our next result is an obvious consequence of Corollary 2.2. Corollary 2.6. If where the set is finite and then T n (t) = n a j,n cos(jt), S := {a j,n : j {0,1,...,n}, n N} R lim n n a j,n =, lim NZ(T n) =. n Our next result is an obvious consequence of Corollary 2.6. Corollary 2.7. If where the set is finite, and then T n (t) = n a j,n cos(jt), S := {a j,n : j {0,1,...,n}, n N} [0, ) lim NC(T n) =, n lim NZ(T n) =. n 8

9 3. Lemmas Let Pn c denote the set of all algebraic polynomials of degree at most n with complex coefficients. Lemma 3.1. If S C is a finite set, P 2n P2n c (S), and H Pc m is a polynomial of minimal degree m such that (3.1) supnc(p 2n H) <, then each zero of H is a root of unity, and each zero of H is simple. Proof. Let H Pm c satisfy the assumptions of the lemma and suppose to the contrary that H(α) = 0, where 0 α C is not a root of unity. Let G Pm 1 c be defined by (3.2) G(z) := H(z) z α. Let S n be the set of the coefficients of P 2nG, and let S := S n. As P 2n P 2n (S) and the set S is finite, the set S is also finite. Let (3.3) (P 2n H)(z) = 2n+m a j,n z j and (P 2n G)(z) = Note that b 2n+m,n = 0. Due to the minimality of H we have (3.4) supnc(p 2n G) =. Observe that (3.2) implies 2n+m (3.5) a j,n = b j 1,n αb j,n, j = 1,2,...,2n+m. Let A n := {j : 1 j 2n+m, b j 1,n αb j,n }. Combining (3.1), (3.3), and (3.5), we can deduce that µ := sup A n <. b j,n z j. Hence we have A n = {j 1,n < j 2,n < < j un,n}, 9

10 where u n µ for each n N. We introduce the numbers j 0,n := 1 and j un +1,n := 2n+m. As α C is not a root of unity, the inequality for some l = 0,1,...,u n implies j l+1,n j l,n S b j,n = 0, j = j l,n, j l,n +1, j l,n +2,..., j l+1,n 1. But then b j,n 0 is possible only for (µ + 1) S values of j = 1,2,...,2n + m, which contradicts (3.4). This finishes the proof of the fact that each zero of H is a root of unity. Now we prove that each zero of H is simple. Without loss of generality it is sufficient to prove that H(1) = 0 implies that H (1) 0, the general case can easily be reduced to this. Assume to the contrary that H(1) = 0 and H (1) = 0. Let G 1 P c m 1 and G 2 P c m 2 be defined by (3.6) G 1 (z) := H(z) z 1 respectively. Let and G 2 (z) := H(z) (z 1) 2 = G 1(z) z 1, (3.7) (P 2n H)(z) = (P 2n G 1 )(z) = 2n+m 2n+m a j,n z j, b j,n z j and (P 2n G 2 )(z) = Due to the minimality of the degree of H we have (3.8) supnc(p 2n G 1 ) =. Observe that (3.6) implies (3.9) a j,n = b j 1,n b j,n, j = 1,2,...,2n+m, and (3.10) b j,n = c j 1,n c j,n, j = 1,2,...,2n+m. Combining (3.1), (3.7), and (3.9), we can deduce that 2n+m (3.11) µ := sup j : 1 j 2n+m, b j 1,n b j,n <. c j,n z j. By using (3.8) and (3.11), for every N N there are n N and L N such that 0 b := b L,n = b L+1,n = = b L+N,n. 10

11 Combining this with (3.10), we get Hence c j 1,n = c j,n +b, j = L,L+1,...,L+N. (3.12) sup max c j,n =.,1,...,2n+m On the other hand P 2n P c 2n(S) together with the fact that the set S is finite implies that the set { c j,n : j {0,1,...,2n+m}, n N} is also finite. This contradicts (3.12), and the proof of the fact that each zero of H is simple is finished. Lemma 3.2. If S C is a finite set, P 2n P c 2n (S), H(z) := zk 1, (3.13) µ := supnc(p 2n H) <, then there are constants c 1 > 0 and c 2 > 0 depending only on µ, k, and S and independent of n and δ such that δ δ P 2n (e it ) dt > c 1 log(nc k (P 2n )) c 2 δ 1 for every δ (0,π), and hence assumption (2.1) implies for every δ (0,π). Proof. We define δ lim P 2n (e it ) dt = n δ k 1 G(z) := so that H(z) = G(z)(z 1). Let S n be the set of the coefficients of P 2nG. We define S := z j Sn. n=1 As P 2n P c 2n (S) and the set S is finite, the set S is also finite. So by Theorem 1.3 there is an absolute constant c > 0 such that (3.14) 2π 0 (P 2n G)(e it ) dt cγlog(nc(p 2n G)) cγlog(nc k (P 2n )), n N, 11

12 with Observe that γ := min z S \{0} z. (P 2n G)(e it ) = 1 e it 1 (P 2nH)(e it ) µm e it 1 = µm sin(t/2) πµm t, t (,π), where µ is defined by (3.13) and M := max{ z : z S } depends only on the set S, and hence M > 0 depends only on k and the set S. It follows that (3.15) [,π]\[ δ,δ] Now (3.14) and (3.15) give (P 2n G)(e it ) dt 2π πµm 2δ = π2 µm δ. δ δ δ P 2n (e it ) dt 1 (P 2n G)(e it ) dt k δ ( = 1 2π (P 2n G)(e it ) dt k 0 1 k cγlog(nc k(p 2n )) π2 µm kδ [,π]\[ δ,δ]. (P 2n G)(e it ) dt Lemma 3.3. If S R is a finite set, P 2n P c 2n (S) are self-reciprocal, H(z) := zk 1, (3.13) µ := supnc(p 2n H) <, and 0 < δ (2k) 1, then T n (t) := P 2n (e it )e int, R n (x) := sup max R n(x) <. x [ δ,δ] x 0 T n (t)dt, ) Proof. Let n T n (t) = a 0,n + 2a j,n cos(jt), a j,n S. j=1 Observe that (3.13) implies that (3.16) sup {j : k j n, a j k,n a j,n } µ := supnc(p 2n H) <. 12

13 We have Now (3.16) implies that where R n (x) = a 0,n x+ n j=1 R n (x) = a 0,n x+ F m,k,n (x) := n m 2a j,n j u n m=1 sin(jx). F m,n (x), 2A m,k,n sin((j m +jk)x) j m +jk with some A m,k,n S, m = 1,2,...,u n, j m N, and n m N, where supu n kµ < (we do not know much about j m and n m ). Since the set S R is finite, and hence it is bounded, it is sufficient to prove that max F m,k,n(x) M, x [ δ,δ] where M is a uniform bound valid for all n N, j m N, n m N, m = 1,2,...,u n, that is, it is sufficient to prove that if then F(x) := ν sin((j 0 +jk)x) j 0 +jk (3.17) max F(x) = max F(x) M, x [ δ,δ] x [0,δ] where M isauniform bound validfor all ν N and j 0 N. Notethatthe equalityin(3.17) holds as F is odd. To prove the inequality in (3.17) let x (0,δ], where 0 < δ (2k) 1. We break the sum as (3.18) F = R+S, where and R(x) := S(x) := ν j 0 +jk x 1 ν x 1 <j 0 +jk 13 sin((j 0 +jk)x) j 0 +jk sin((j 0 +jk)x) j 0 +jk.

14 Here (3.19) R(x) ν j 0 +jk x 1 sin((j 0 +jk)x) j 0 +jk (x 1 +1) x 1+ x 1+δ = 1+(2k) 1 3 2, where each term in the sum in the middle is estimated by sin((j 0 +jk)x) j 0 +jk (j 0 +jk)x) j 0 +jk = x, and the number of terms in the sum in the middle is clearly at most x Further, using Abel rearrangement, we have S(x) = B v(x) j 0 +vk + B u(x) ν ( ) 1 j 0 +uk + B j (x) j 0 +jk 1 j 0 +(j +1)k with x 1 <j 0 +jk B j (x) := B j,k (x) := j sin((j 0 +hk)x) and with some u,v N 0 for which x 1 < j 0 +(u+1)k and x 1 < j 0 +(v +1)k. Hence, (3.20) S(x) B v (x) j 0 +vk + B u (x) ν ( ) 1 j 0 +uk + B j (x) j 0 +jk 1. j 0 +(j +1)k h=0 x 1 <j 0 +jk Observe that x (0,δ], 0 < δ (2k) 1, x 1 < j 0 +(w+1)k, and w N 0 imply and and hence (3.21) x 1 < j 0 +(w+1)k < 2(j 0 +wk) if w 1, 2k δ 1 x 1 < j 0 +k if w = 0, 1 j 0 +wk 2x, w N 0. Observe also that x (0,δ] and 0 < δ (2k) 1 imply that 0 < x < πk 1. Hence, with z = e ix we have ( j 1 B j (x) = 2 Im z 0+hk) j 1 j z j 0+hk 2 = 1 j (3.22) z hk 2 h=0 h=0 h=0 = 1 1 z (j+1)k 2 1 z k z(j+1)k 1 z k 1 1 z k 1 sin(kx/2) π kx. 14

15 Combining (3.20), (3.21), and (3.22), we conclude (3.23) S(x) π kx 2x+ π kx 2x+ π kx 2x 6π k. Now (3.18), (3.19), and (3.23) give the inequality in (3.17) with M := 6π/k 6π. Our next lemma is well known and may be proved simply by contradiction. Lemma 3.4. If R is a continuously differentiable function on the interval [ δ,δ], δ > 0, δ δ R (x) dx = L and max x [ δ,δ] R(x) = M, then there is an η [ M,M] such that R η has at least L(2M) 1 zeros in [ δ,δ]. Lemma 3.5. If S R is a finite set, P 2n P c 2n(S) are self-reciprocal, H(z) := z k 1, T n (t) := P 2n (e it )e int, (3.13) µ := supnc(p 2n H) <, and (2.1) lim n NC k(p 2n ) =, then (2.2) lim n NZ(T n) =. Proof. Let 0 < δ (2k) 1. Let R n be defined by R n (x) := x 0 T n (t)dt. Observe that T n (x) = P 2n (e ix ) for all x R. By Lemmas 3.2 and 3.3 we have (3.24) lim n and δ δ R n(x) dx = lim n δ δ T n (x) dx = lim n (3.25) supx max R n(x) <. [ δ,δ] δ δ P 2n (e ix ) dx = (Note that to obtain (3.24) from Lemma 3.2 we use the second statement of Lemma 3.2, which is valid under the assumption (2.1), that is why (2.1) is also assumed in this lemma.) Therefore, by Lemma 3.4 there are c n R such that lim NZ(R n c n ) =. n However, T n (x) = (R n c n ) (x) for all x R, and hence lim NZ(T n) =. n Our next lemma follows immediately from Lemmas 3.1 and

16 Lemma 3.6. If S R is a finite set, P 2n P c 2n(S) are self-reciprocal, T n (t) := P 2n (e it )e int, (2.1) lim n NC k(p 2n ) =, and there is a polynomial H P m such that (3.13) µ := supnc(p 2n H) <, then (2.2) lim n NZ(T n) = Moreover, we have the following observation. Lemma 3.7. Let (n ν ) be a strictly increasing sequence of positive integers. If S R is a finite set, P 2nν P c 2n ν (S) are self-reciprocal, T nν (t) := P 2nν (e it )e in νt, lim NC k(p 2nµ ) = µ for every k N, and there is a polynomial H P m such that supnc(p 2nν H) <, ν N then lim NZ(T n ν ν ) =. Proof. Without loss of generality we may assume that 0 S. We define the self-reciprocal polynomials P 2n P2n c (S) by and apply Lemma 3.6. P 2n (z) := z n n ν P 2nν (z), n ν n < n ν+1, The next lemma is straightforward consequences of Theorem 1.4. Lemma 3.8. Let λ 0 < λ 1 < < λ m be nonnegative integers and let m Q m (t) = A j cos(λ j t), A j R, j = 0,1,...,m. Then π Q m (t) dt 1 60 m A m j j +1. We will also need the lemma below in the proof of Theorem

17 Lemma 3.9. Let λ 0 < λ 1 < < λ m be nonnegative integers and let m Q m (t) = A j cos(λ j t), A j R, j = 0,1,...,m. Let A := max,1,...,m A j. Suppose Q m has at most K 1 zeros in the period [,π). Then π m 1 Q m (t) dt 2KA π + 2KA(5+logm). λ j=1 j Proof. We may assume that λ 0 = 0, the case λ 0 > 0 can be handled similarly. Associated with Q m in the lemma let R m (t) := A 0 t+ m A j λ j sin(λ j t). Clearly m max R 1 m(t) A π +. t [,π] λ j j=1 Also, for every c R the function R m c has at most K zeros in the period [,π), otherwise Rolle s Theorem implies that Q m = (R m c) has at least K zeros in the period [, π). Hence π π Q m (t) dt = R m (t) dt = V π (R m) 2K max R m(t) t [,π] m 1 2KA π + 2KA(5+logm), λ j=1 j where V π (R m ) is the total variation of R m on the interval [,π], and the lemma is proved. The lemma below is needed only in the proof of Lemma Lemma Suppose k N. Let be the kth roots of unity. Suppose ( ) 2πji z j := exp, j = 0,1,...,k 1, k {b 0,b 1,...,b k 1 } R, b 0 0, 17

18 and k 1 Q(z) := b j z j. Then there is a value of j {0,1,... k 1} for which Re(Q(z j )) 0. Proof. If the statement of the lemma were false, then k 2 z k 1 (Q(z)+Q(1/z)) = (z k 1) α ν z ν withsomeα ν R, ν = 0,1,...,k 2. Observethatthecoefficientofz k 1 ontherighthand side is 0, while the coefficient of z k 1 on the left hand side is 2b 0 0, a contradiction. Our final lemma has already been used in Section 1 of this paper, where Theorem 1 in [6] has been corrected Theorem 1.2. Lemma If 0 / {b 0,b 1,...,b k 1 } R, {a 0,a 1,...,a m 1 } R, where m = uk with some integer u 0, ν=0 a m+lk+j = b j, l = 0,1,..., j = 0,1,...,k 1, and n = m+lk +r with integers m 0, l 0, k 1, and 0 r k 1, then there is a constant c 3 > 0 depending only on the sequence (a j ) but independent of n such that n T n (t) := Re a j e ijt has at least c 3 n zeros in [,π). Proof. Note that n a j z j = m 1 k 1 a j z j +z m b j z j z(l+1)k 1 z k 1 r +z m+lk b j z j = P 1 (z)+p 2 (z), where and with P 1 (z) := k 1 P 2 (z) := z uk m 1 r a j z j +z m+lk b j z j b j z jz(l+1)k 1 z k 1 k 1 Q(z) := b j z j. 18 = Q(z)z ukz(l+1)k 1 z k 1

19 By Lemma 3.10 there is a kth root of unity ξ = e iτ such that Re(Q(ξ)) 0. Then, for every K > 0 there is a δ (0,2π/k) such that Re(P 2 (e it )) oscillates between K and K at least c 4 (l+1)kδ times on the interval [τ δ,τ +δ], where c 4 > 0 is a constant independent of n. Now we choose δ (0,2π/k) for K := 1+ m 1 k 1 a j + b j. Then n T n (t) := Re a j e ijt = Re(P 1 (e it ))+Re(P 2 (e it )) has at least one zero on each interval on which Re(P 2 (e it )) oscillates between K and K, and hence it has at least c 10 (l +1)kδ > c 3 n zeros on [,π), where c 3 > 0 is a constant independent of n. Proof of the Theorems We denote the set of all real trigonometric polynomials of degree at most k by T k. Proof of Theorems 2.1. Suppose the theorem is false. Then there are k N, a strictly increasingsequence(n ν ) ν=1 ofpositiveintegers, andeventrigonometricpolynomialsq nν T k with maximum norm 1 such that T nν has a sign change on the period [,π) exactly at t 1,nν < t 2,nν < < t mν,n ν, where m ν are nonnegative even integers and m ν k for each ν, and hence the even trigonometric polynomials Q nν T k defined by m ν Q nν (t) := h nν j=1 sin t t j,n ν 2 with an appropriate choice of h nν R have maximum norm 1 on the period [,π) and (4.1) T nν (t)q nν (t) 0, t R. Picking a subsequence of (n ν ) ν=1 if necessary, without loss of generality we may assume that Q nν converges to a Q T k uniformly on the period [,π). That is, (4.2) lim ν ε ν = 0 with ε ν := max t [,π] Q(t) Q n ν (t). We introduce the notation n ν T nν (t) = a j,ν cos(jt), 19

20 n ν K ν (4.3) T nν (t)q(t) 3 = a j,ν cos(jt) Q(t) 3 = b j,ν cos(β j,ν t), b jν 0, j = 0,1,...,K ν, and n ν L ν (4.4) T nν (t)q(t) 4 = a j,ν cos(jt) Q(t) 4 = d j,ν cos(δ j,ν t), d j,ν 0, j = 0,1,...,L ν, where β 0,ν < β 1,ν < < β Kν,ν and δ 0,ν < δ 1,ν < < δ Lν,ν are nonnegative integers. Since the set S := {a j,ν : j {0,1,...,n ν }, ν N} R is finite, the sets {b j,ν : j {0,1,...,K ν }, ν N} R and {d j,ν : j {0,1,...,L ν }, ν N} R are finite as well. Hence there are ρ,m (0, ) such that (4.5) a j,ν M, j = 0,1,...,n ν, ν N, (4.6) ρ b j,ν M, j = 0,1,...,K ν, ν N, and (4.7) ρ d j,ν M, j = 0,1,...,L ν, ν N. As n ν T nν (t) = a j,ν cos(jt), and the set S := {a j,ν : j {0,1,...,n ν }, ν N} R is finite, orthogonality and Q nν T k imply that (4.8) π T nν (t)q nν (t)dt 2πM(k+1) max t [,π] Q n,ν(t) = 2πM(k+1), where M := max{ z : z S } 2max{ z : z S} depends only on the finite set S. Observe that our indirect assumption together with Lemma 3.7 implies that (4.9) lim ν K ν = and lim ν L ν =. Indeed, if lim ν K ν <, 20

21 then Lemma 3.7 with H P m defined by H(e it ) := e imt/2 Q(t) 3, m := 6deg(Q), while if lim L ν <, ν then Lemma 3.7 with H P m defined by H(e it ) := e imt/2 Q(t) 4, m := 8deg(Q), gives (2.2) lim n NZ(T n) =. which is the conclusion of the theorem contradicting our indirect assumption that the theorem is false. We claim that (4.10) K ν c 5 L ν with some c 5 > 0 independent of ν N. Indeed, using Parseval s formula (4.2), (4.3), and (4.7) we deduce (4.11) 1 π π T nν (t) 2 Q(t) 4 Q nν (t) 2 dt = 1 π π (T nν (t)q(t) 2 Q nν (t)) 2 dt 1 2 ρ2 K ν for every sufficiently large ν N. Also, (4.1) (4.8) imply (4.12) 1 π nν (t) π T 2 Q(t) 4 Q nν (t) 2 dt = 1 π (T nν (t)q nν (t))(t nν (t)q(t) 4 )Q nν (t)dt π 1 ( π )( )( ) T nν (t)q nν (t)dt max π T n ν (t)q(t) 4 max Q n ν (t) t [,π] t [,π] 1 ( π ) ( ) T nν (t)q nν (t)dt L ν M max π Q n ν (t) t [,π] c 6 L ν with a constant c 6 > 0 independent of ν for every ν N. Now (4.10) follows from (4.11) and (4.12). From Lemma 3.8 we deduce (4.13) π T nν (t)q(t) 4 dt c 7 ρlogl ν with some constant c 7 > 0 independent of ν N. On the other hand, using (4.1), Lemma 21

22 3.9, (4.2), (4.4), (4.8), and (4.10), we obtain (4.14) π π π T nν (t)q(t) 4 dt T nν (t)q(t) 3 Q nν (t) dt+ π π T nν (t)q(t) 3 Q(t) Q nν (t) dt = T nν (t)q nν (t) Q(t) 3 dt+ T nν (t)q(t) 3 Q(t) Q nν (t) dt ( π )( ) T nν (t)q nν (t) dt max t [,π] Q(t) 3 ( π )( ) + T nν (t)q(t) 3 dt max Q(t) Q n ν (t) t [,π] ( π )( ) ( π ) = T nν (t)q nν (t)dt max t [,π] Q(t) 3 + T nν (t)q(t) 3 dt c 8 +c 9 (logk ν )ε ν c 8 +c 9 (log(c 5 L ν ))ε ν c 10 +c 9 (logl ν )ε ν, ε ν where c 8,c 9, and c 10 are constants independent of ν N and ε ν 0 as ν. Combining (4.13) and (4.14), we obtain c 7 ρlogl ν c 10 +c 9 (logl ν )ε ν, which contradicts(4.9). Hence our indirect assumption is false, and the theorem is true. Proof of Corollary 2.2. Observe that assumption (2.3) implies assumption (2.1), and hence follows from Theorem 2.1 Proof of Corollary 2.3. Corollary 2.2 implies (4.15) lim k NZ(P 2k) =. and (4.16) lim k NZ(P 2k+1) =. Note that (4.15) is an obvious consequence of Theorem 2.1. To see (4.16) observe that if P 2k+1 P c 2k+1 (S) are self-reciprocal then P 2k+2 defined by (4.17) P2k+2 (z) := (z +1)P 2k+1 (z) P c 2k+2 ( S) are also self-reciprocal, where the fact that S is finite implies that the set (4.18) S := {s1 +s 2 : s 1,s 2 S {0}} R 22

23 is also finite. Also, (2.4) lim n P n(1) = implies lim P 2k+2 (1) = lim 2 P 2k+1(1) =. k k Hence the polynomials P 2k+2 P c 2k+2 ( S) satisfy the assumptions of Corollary 2.2, and it follows that (4.19) lim k NZ( P 2k+2 ) =. Combining this with (4.20) NZ(P 2k+1 ) = NZ( P 2k+2 ) 1 we obtain (4.16). Combining (4.15) and (4.16), we obtain the conclusion of the corollary: (2.5) lim n NZ(P n) =. Proof of Corollary 2.4. If (2.7) holds and the finite set S R has property (2.6), then assumption (2.1) is satisfied and (2.2) lim n NZ(P 2n) = lim n NZ(T n) = follows from Theorem 2.1. Proof of Corollary 2.5. Corollary 2.4 implies (4.15) and (4.16). Note that (4.15) is an obvious consequence of Corollary 2.4. To see (4.16) observe that if P 2k+1 P2k+1 c (S) are self-reciprocal, then P 2k+2 defined by (4.17) are also self-reciprocal, where the fact that S is finite implies that the set S R defined by (4.18) is also finite. It is easy to see that the fact that S satisfies (2.6) implies that S also satisfies (2.6), that is, (for every k N) s 1 +s 2 + +s k = 0, s 1,s 2,...,s k S, implies s 1 = s 2 = = s k = 0, that is, any sum of nonzero elements of S is different from 0. Similarly, (2.6) implies that NC( P 2k+2 ) NC(P 2k+1 ). Combining this with (2.8) it follows that lim NC( P 2k+2 ) =. k Hence the polynomials P 2k+2 P c 2k+2 ( S) defined by (4.17) satisfy the assumptions of Corollary 2.4, and (4.19) follows. Combining this with (4.20) we obtain (4.16). Combining (4.15) and (4.16), we obtain (2.5), the conclusion of the corollary. Proof of Corollary 2.6. This is an obvious consequence of Corollary 2.2. Proof of Corollary 2.7. This is an obvious consequence of Corollary

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