THE NUMBER OF UNIMODULAR ZEROS OF SELF-RECIPROCAL
POLYNOMIALS WITH COEFFICIENTS IN A FINITE SET
Tamás Erdélyi
February 4, 2016
Abstract. We study the number NZ(Tn ) of real zeros of trigonometric polynomials
Tn (t) =
n
X
aj,n cos(jt)
j=0
in a period [a, a + 2π), a ∈ R, and the number NZ(Pn ) of zeros of self-reciprocal algebraic
polynomials
n
X
Pn (z) =
aj,n z j
j=0
on the unit circle. One of the highlights of this paper states that limn→∞ NZ(Tn ) = ∞
whenever the set
{aj,n : j = 0, 1, . . . , n, n ∈ N} ⊂ [0, ∞)
is finite and
lim |{j ∈ {0, 1, . . . , n} : aj,n 6= 0}| = ∞ .
n→∞
This follows from a more general result stating that limn→∞ NZ(Tn ) = ∞ whenever the set
{aj,n : j = 0, 1, . . . , n, n ∈ N} ⊂ R
is finite and limn→∞ NCk (Tn ) = ∞ for every k ∈ N, where


u+k−1
X
aj,n 6=
NCk (Tn ) := u : 0 ≤ u ≤ n − k + 1,

j=u


0 .

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.
2010 Mathematics Subject Classifications. 11C08, 41A17, 26C10, 30C15
Typeset by AMS-TEX
1
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 Pnc denote the set of all algebraic polynomials of degree at most n with complex
coefficients. Let S ⊂ C. Let Pnc (S) be the set of all algebraic polynomials of degree at
most n with each of their coefficients in S. A polynomial
(1.1)
Pn (z) =
n
X
aj z j
j=0
is called conjugate-reciprocal if
aj = an−j ,
j = 0, 1, . . . , n .
A polynomial Pn of the form (1.1) is called plain-reciprocal or self-reciprocal if
aj = an−j ,
j = 0, 1, . . . , n .
If a conjugate reciprocal polynomial Pn has only real coefficients, then it is obviously
plain-reciprocal. Associated with an algebraic polynomial
Pn (z) =
n
X
aj,n z j
j=0
we introduce the numbers
NC(Pn ) := |{j ∈ {0, 1, . . . , n} : aj,n 6= 0}| .
Let NZ(Pn ) denote the number of real zeros (by counting multiplicities) of an algebraic
polynomial Pn on the unit circle. Associated with a trigonometric polynomial
Tn (t) =
n
X
aj,n cos(jt)
j=0
we introduce the numbers
NC(Tn ) := |{j ∈ {0, 1, . . . , n} : aj,n 6= 0}| .
Let NZ(Tn ) denote the number of real zeros (by counting multiplicities) of a trigonometric
polynomial Tn in a period [a, a + 2π), a ∈ R. The quotation below is from [6].
2
“Let 0 ≤ n1 < n2 < · · · < nN be integers. A cosine polynomial of the form TN (θ) =
PN
j=1 cos(nj θ) must have at least one real zero in a period [a, a + 2π), a ∈ R. This
is obvious if n1 6= 0, since then the integral of the sum on a period is 0. The above
statement is less obvious if n1 = 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 1981. See also [13, pages 285288] for a book proof. It is not difficult to prove the statement in general even in the case
n1 = 0 without using Littlewood’s Conjecture. One possible way is to use the identity
nN
X
TN ((2j − 1)π/nN ) = 0 .
j=1
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 n1 = 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
Pp−1 k k
that for an odd prime p the Fekete polynomial fp (z) = k=0 p z (the coefficients are
Legendre symbols) has ∼ κ0 p zeros on the unit circle, where 0.500813 > κ0 > 0.500668.
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 are integral and all different, what is the lower bound on the number of real zeros of
PmN
m=1 cos(nm θ)? 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
PN
paper [3] we showed that this is false. There exists a cosine polynomials m=1 cos(nm θ)
with the nm integral and all different so that the number of its real zeros in the period is
O(N 9/10 (log N )1/5 ) (here the frequencies nm = nm (N ) may vary with N ). However, there
PN
are reasons to believe that a cosine polynomial m=1 cos(nm θ) always has many zeros in
the period.”
One of the highlights of this paper is to show that the number of real zeros of the sums
PN
TN (θ) = j=1 cos(nj θ) in a period tends to ∞ whenever 0 ≤ n1 < n2 < · · · < nN 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


X
j
Ln := P : P (z) =
aj z : aj ∈ {−1, 1} .


j=0
Elements of Ln are often called Littlewood polynomials of degree n. Let


n


X
aj z j : aj ∈ C, |a0 | = |an | = 1, |aj | ≤ 1 ,
Kn := P : P (z) =


j=0
3
Observe that Ln ⊂ Kn . In [10] we proved that any polynomial P ∈ Kn has at least
8n1/2 log n zeros in any open disk centered at a point on the unit circle with radius
33n−1/2 log n. Thus polynomials in Kn have a few zeros near the unit circle. One may
naturally ask how many unimodular roots a polynomial in Kn can have. Mercer [34]
proved that if a Littlewood polynomial P ∈ Ln of the form (1.1) is skew reciprocal, that
is, aj = (−1)j an−j 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 aj = an−j 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
NZn := min NZ(P ) ,
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 ∈ Ln . It has
been conjectured for a long time that limn→∞ NZn = ∞. In this paper we show that
limn→∞ NZ(Pn ) = ∞ whenever Pn ∈ Ln is self-reciprocal and limn→∞ |Pn (1)| = ∞. This
follows as a consequence of a more general result in which the coefficients of the selfreciprocal polynomials Pn of degree at most n belong to a fixed finite set of real numbers.
In [6] we proved the following result.
Theorem 1.1. If the set {aj : j ∈ N} ⊂ R is finite, the set {j ∈ N : aj 6= 0} is infinite,
the sequence (aj ) is not eventually periodic, and
Tn (t) =
n
X
aj cos(jt) ,
j=0
then limn→∞ NZ(Tn ) = ∞ .
In [6] Theorem 1.1 is stated without the assumption that the sequence (aj ) is not
eventually periodic. However, as the following example shows, Lemma 3.4 in [6], dealing
with the case of eventually periodic sequences (aj ), is incorrect. Let
Tn (t) := cos t + cos((4n + 1)t) +
n−1
X
k=0
=
1 + cos((4n + 2)t)
+ cos t .
2 cos t
4
(cos((4k + 1)t) − cos((4k + 3)t))
It is easy to see that Tn (t) 6= 0 on [−π, π] \ {−π/2, π/2} and the zeros of Tn at −π/2 and
π/2 are simple. Hence Tn has only two (simple) zeros in the period. So the conclusion of
Theorem 1.1 above is false for the sequence (aj ) with a0 := 0, a1 := 2, a3 := −1, a2k := 0,
a4k+1 := 1, a4k+3 := −1 for every k = 1, 2, . . . . Nevertheless, Theorem 1.1 can be saved
even in the case of eventually periodic sequences (aj ) if we assume that aj 6= 0 for all
sufficiently large j. See Lemma 3.11. So Theorem 1 in [6] can be corrected as
Theorem 1.2. If the set {aj : j ∈ N} ⊂ R is finite, aj 6= 0 for all sufficiently large j, and
Tn (t) =
n
X
aj cos(jt) ,
j=0
then limn→∞ NZ(Tn ) = ∞ .
It was expected that the conclusion of the above theorem remains true even if the
coefficients of Tn do not come from the same sequence, that is
Tn (t) =
n
X
aj,n cos(jt) ,
j=0
where the set
{aj,n : j = 0, 1, . . . , n, n ∈ N} ⊂ R
is finite and
lim |{j : aj,n 6= 0}| = ∞ .
n→∞
The purpose of this paper is to prove such an extension of Theorem 1.1. 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.
Theorem 1.3. There is an absolute constant c > 0 such that
Z
0
2π
m
X
iλj t aj e dt ≥ cγ log m
j=1
whenever λ1 , λ2 , . . . , λn are distinct integers and a1 , a2 , . . . , am 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 285-288].
Theorem 1.4. If λ1 < λ2 < · · · < λm are integers and a1 , a2 , . . . , am are complex numbers, then
Z 2π X
m
m
1 X |aj |
iλj t aj e dt ≥
.
30 j=1 j
0
j=1
5
2. New Results
Associated with an algebraic polynomial
Pn (t) =
n
X
aj,n z j ,
aj,n ∈ C ,
j=0
let



u+k−1

X
aj,n 6= 0 .
NCk (Pn ) := u : 0 ≤ u ≤ n − k + 1,


j=u
c
Theorem 2.1. If S ⊂ R is a finite set, P2n ∈ P2n
(S) are self-reciprocal polynomials,
Tn (t) := P2n (eit )e−int ,
and
(2.1)
lim NCk (P2n ) = ∞
n→∞
for every k ∈ N, then
(2.2)
lim NZ(Tn ) = ∞ .
n→∞
c
Corollary 2.2. If S ⊂ R is a finite set, P2n ∈ P2n
(S) are self-reciprocal polynomials,
Tn (t) := P2n (eit )e−int ,
and
(2.3)
lim |P2n (1)| = ∞ ,
n→∞
then (2.2) holds.
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, Pn ∈ Pnc (S) are self-reciprocal polynomials, and
(2.4)
lim |Pn (1)| = ∞ ,
n→∞
then
(2.5)
lim NZ(Pn ) = ∞ .
n→∞
We say that S ⊂ R has property (2.6) if (for every k ∈ N)
(2.6)
s1 + s2 + · · · + sk = 0 , s1 , s2 , . . . , sk ∈ S , implies s1 = s2 = · · · = sk = 0 ,
that is, any sum of nonzero elements of S is different from zero.
6
c
Corollary 2.4. If the finite set S ⊂ R has property (2.6), P2n ∈ P2n
(S) are self-reciprocal
polynomials,
Tn (t) := P2n (eit )e−int ,
and
(2.7)
lim NC(P2n ) = ∞ ,
n→∞
then (2.2) holds.
Our next result is slightly more general than Corollary 2.4, and it follows from Corollary
2.4 simply.
Corollary 2.5. If the finite set S ⊂ R has property (2.5), Pn ∈ Pnc (S) are self-reciprocal
polynomials, and
(2.8)
lim NC(Pn ) = ∞ ,
n→∞
then (2.5) holds.
Our next result is an obvious consequence of Corollary 2.2.
Corollary 2.6. If
Tn (t) =
n
X
aj,n cos(jt) ,
j=0
where the set
S := {aj,n : j = 0, 1, . . . , n, n ∈ N} ⊂ R
is finite and
n
X
aj,n = ∞ ,
lim n→∞
then (2.2) holds.
j=0
Our next result is an obvious consequence of Corollary 2.6.
Corollary 2.7. If
Tn (t) =
∞
X
aj,n cos(jt) ,
j=0
where the set
S := {aj,n : j = 0, 1, . . . , n, n ∈ N} ⊂ [0, ∞)
is finite, and
lim NC(Tn ) = ∞ ,
n→∞
then (2.2) holds.
7
3. Lemmas
c
In our first seven Lemmas we assume that S ⊂ C is a finite set, P2n ∈ P2n
(S), and we
use the notation
Tn (t) := P2n (eit )e−int .
c
c
Lemma 3.1. If S ⊂ C is a finite set, P2n ∈ P2n
(S), and H ∈ Pm
is a polynomial of
minimal degree m such that
(3.1)
sup NC(P2n H) < ∞ ,
n∈N
then each zero of H is a root of unity, and each zero of H is simple.
c
Proof. Let H ∈ Pm
satisfy the assumptions of the lemma and suppose to the contrary
c
that H(α) = 0, where 0 6= α ∈ C is not a root of unity. Let G ∈ Pm−1
be defined by
(3.2)
G(z) :=
H(z)
.
z−α
Let Sn∗ be the set of the coefficients of P2n G, and let
[
S ∗ :=
Sn∗ .
n∈N
As P2n ∈ P2n (S) and the set S is finite, the set S ∗ is also finite. Let
(3.3)
(P2n H)(z) =
2n+m
X
aj,n z
j
and
(P2n G)(z) =
2n+m
X
bj,n z j .
j=0
j=0
(Note that bn+m,n = 0. Due to the minimality of H we have
(3.4)
sup NC(P2n G) = ∞ .
n∈N
Observe that (3.2) implies
(3.5)
aj,n = bj−1,n − αbj,n ,
j = 1, 2, . . . , 2n + m .
Combining (3.1), (3.3), and (3.5), we can deduce that
µ := sup |j : 1 ≤ j ≤ 2n + m, bj−1,n 6= αbj,n | < ∞ .
n∈N
Let
An := {j : 1 ≤ j ≤ 2n + m, bj−1,n 6= αbj,n } = {j1,n < j2,n < · · · < jun ,n } ,
where un ≤ µ for each n ∈ N, and let j0,n := 1 and jun +1,n := 2n + m. As α ∈ C with
|α| = 1 is not a root of unity, the inequality
jl+1,n − jl,n ≥ |S ∗ |
8
for some l = 0, 1, . . . , un implies
bj,n = 0,
j = jl,n , jl,n + 1, jl,n + 2, . . . , jl+1,n − 1 .
But then bj,n 6= 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) 6= 0, the general case can easily be reduced to this.
c
c
Assume to the contrary that H(1) = 0 and H ′ (1) 6= 0. Let G1 ∈ Pm−1
and G2 ∈ Pm−2
be
defined by
H(z)
H(z)
G1 (z)
(3.6)
G1 (z) :=
and
G2 (z) :=
=
,
z−1
(z − 1)2
z−1
respectively. Let
(3.7)
(P2n H)(z) =
2n+m
X
aj,n z j ,
j=0
(P2n G1 )(z) =
2n+m
X
bj,n z
j
and
(P2n G2 )(z) =
j=0
cj,n z j .
j=0
Due to the minimality of the degree of H we have
(3.8)
2n+m
X
sup NC(P2n G1 ) = ∞ .
n∈N
Observe that (3.6) implies
(3.9)
aj,n = bj−1,n − bj,n ,
j = 1, 2, . . . , 2n + m .
bj,n = cj−1,n − cj,n
j = 1, 2, . . . , 2n + m .
and
(3.10)
Combining (3.1), (3.7), and (3.9), we can deduce that
(3.11)
µ := sup |j : 1 ≤ j ≤ 2n + m, bj−1,n 6= bj,n | < ∞ .
n∈N
By using (3.8) and (3.11), for every n ∈ N and N ∈ N there is an L ∈ N such that
0 6= b := bL,n = bL+1,n = · · · = bL+N,n .
Combining this with (3.10), we get
cj−1,n = cj,n − b ,
j = L + 1, L + 2, . . . , L + N .
Hence
(3.12)
sup max{|cj,n | : j = 0, 1, . . . , 2n + m} = ∞ .
n∈N
c
On the other hand P2n ∈ P2n
(S) together with the fact that the set S is finite implies that
the set
{|cj,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. 9
c
Lemma 3.2. If S ⊂ C is a finite set, P2n ∈ P2n
(S), H(z) := z k − 1,
(3.13)
µ := sup NC(P2n H) < ∞ ,
n∈N
then there are constants c1 > 0 and c2 > 0 depending only on µ, k, and S and independent
of n and δ such that
Z δ
|P2n (eit )| dt > c1 log NCk (P2n − c2 δ −1
−δ
for every δ ∈ (0, π), and hence assumption (2.1) implies
Z δ
|P2n (eit )| dt = ∞
lim
n→∞
−δ
for every δ ∈ (0, π).
Proof. We define
G(z) :=
k−1
X
zj
j=0
so that H(z) = G(z)(z − 1). Let
Sn∗
be the set of the coefficients of P2n G. We define
∞
[
∗
S :=
Sn∗ .
n=1
and the set S is finite, the set S ∗ is also finite. So by Theorem 1.3 there
As P2n ∈
is an absolute constant c > 0 such that
Z 2π
(3.14)
|(P2n G)(eit )| dt ≥ cγ log(NC(P2n G)) ≥ cγ log(NCk (P2n )
n∈N
c
(S)
P2n
0
with
γ :=
min
z∈S ∗ \{0}
|z| .
Observe that
|(P2n G)(eit )| =
Hence
(3.15)
Z
1
µ
µ
πµ
it
|(P
H)(e
)|
≤
=
≤
,
2n
|eit − 1|
|eit − 1|
2 sin(t/2)
2t
2π−δ
it
|(P2n G)(e )| dt =
δ
Z
π−δ
|(P2n G)(eit )| dt ≤ 2π
−π+δ
t ∈ (−π, π) .
π2µ
πµ
=
2δ
δ
Now (3.14) and (3.15) give
Z δ
Z
1 δ
it
|P2n (e )| dt ≥
|(P2n G)(eit )| dt
k
−δ
−δ
!
Z 2π
Z 2π−δ
1
it
it
=
|(P2n G)(e )| dt −
|(P2n G)(e )| dt
k
0
δ
≥
π2µ
1
cγ log(NCk (P2n ) −
.
k
δ
10
c
Lemma 3.3. If S ⊂ R is a finite set, P2n ∈ P2n
(S) are self-reciprocal, H(z) := z k − 1,
(3.13) holds,
Z x
it −int
Tn (t) := P2n (e )e
and
Rn (x) :=
Tn (t) dt ,
0
and 0 < δ ≤ (2k)−1 , then
sup max |Rn (x)| < ∞ .
n∈N x∈[−δ,δ]
Proof. Let
Tn (t) = a0,n +
n
X
2aj,n cos(jt) ,
aj,n ∈ S .
j=1
Observe that (3.13) implies that
(3.16)
sup |{j : k ≤ j ≤ n, aj−k,n 6= aj,n }| < ∞ .
n∈N
We have
Rn (x) = a0,n x +
n
X
2aj,n
j=1
Now (3.16) implies that
Rn (x) = a0,n x +
j
un
X
sin(jx) .
Fm,n (x) ,
m=1
where
Fm,n (x) :=
nm
X
2Am sin((jm + jk)x)
jm + jk
j=0
with some Am ∈ S, m = 1, 2, . . . , un , and jm ∈ N, where
µ := sup un < ∞ .
n∈N
Hence it is sufficient to prove that
(3.17)
sup max |Fm,n (x)| < ∞ ,
n∈N x∈[−δ,δ]
m = 1, 2, . . . , un .
Let x ∈ (0, δ], where 0 < δ ≤ (2k)−1 . We break the sum as
(3.18)
Fm,n = 2Am (Rm,n + Sm,n ) ,
where
Rm,n (x) :=
nm
X
j=0
jm +jk≤x−1
11
sin((jm + jk)x)
jm + jk
and
nm
X
Sm,n (x) :=
j=0
x−1 <jm +jk
Here
(3.19)
|Rm,n (x)| ≤
X
j=0
jm +jk≤x−1
sin((jm + jk)x)
.
jm + jk
sin((jm + jk)x) ≤ |x−1 ||x| ≤ 1 .
jm + jk
Further, using Abel rearrangement, we have
Bm,v,k (x) Bm,u,k (x)
+
jm + vk
jm + n m k
n
m
X
1
1
+
Bm,j,k (x)
.
−
jm + jk jm + (j + 1)k
j=0
Sm,n (x) = −
x−1 ≤jm +jk
with
Bm,j,k (x) :=
j
X
sin((jm + αk)x)
α=0
−1
and with some u, v ∈ N0 for which x < jm + (u + 1)k and x−1 < jm + (v + 1)k. Hence,
Bm,v,k (x) Bm,u,k (x) +
(3.20)
|Sm,n (x)| ≤ jm + vk jm + uk nm
X
1
1
|Bm,j,k (x)|
+
.
−
jm + jk jm + (j + 1)k
j=0
x−1 <jm +jk
Observe that
x−1 < jm + (w + 1)k < 2(jm + wk)
if w ≥ 1 ,
and
2k ≤ δ −1 ≤ x−1 < jm + k
if w = 0 ,
hence,
(3.21)
1
≤ 2x ,
jm + wk
w ∈ N0
Observe that x ∈ (0, δ] and 0 < δ ≤ (2k)−1 imply that 0 < x < πk −1 . Hence, with z = eix
we have
(3.22)
j
! j
j
1
1 X
1 X
X
|Bm,j,k (x)| = Im
z jm +αk = z αk z jm +αk ≤ 2
2
2
α=0
α=0
α=0
1 1 − z (j+1)k 1
1
1
≤ |1 − z (j+1)k |
≤
=
k
k
2 1−z
2
|1 − z |
|1 − z k |
1
π
≤
≤
.
sin(kx/2)
kx
12
Combining (3.20), (3.21), and (3.22), we conclude
(3.23)
π
π
6π
π
2x +
2x +
2x ≤
.
kx
kx
kx
k
|Sm,n (x)| ≤
As Fn,m is odd, (3.18), (3.19), and (3.23) give (3.17). 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,
Z
δ
|R′ (x)| dx = L
and
max |R(x)| = M ,
x∈[−δ,δ]
−δ
then there is an η ∈ [−M, M ] such that R − η has at least L(2M )−1 zeros in [−δ, δ].
c
Lemma 3.5. If S ⊂ R is a finite set, P2n ∈ P2n
(S) are self-reciprocal,
Tn (t) := P2n (eit )e−int ,
H(z) := z k − 1, and (3.13) and (2.1) hold, then (2.2) also holds.
Proof. Let 0 < δ ≤ (2k)−1 . Let Rn be defined by
Rn (x) :=
Z
x
Tn (t) dt .
0
Observe that |Tn (x)| = |P2n (eix )| for all x ∈ R. By Lemmas 3.2 and 3.3 we have
lim
n→∞
Z
δ
−δ
|Rn′ (x)| dx
= lim
n→∞
Z
δ
|Tn (x)| dx = lim
−δ
n→∞
Z
δ
|P2n (eix )| dx = ∞
−δ
and
sup max |Rn (x)| < ∞ .
n∈N [−δ,δ]
Therefore, by Lemma 3.4 there are cn ∈ R such that
lim NZ(Rn − cn ) = ∞ .
n→∞
However, Tn (x) = (Rn − cn )′ (x) for all x ∈ R, and hence
lim NZ(Tn ) = ∞ .
n→∞
Our next lemma follows immediately from Lemmas 3.1 and 3.5.
13
c
Lemma 3.6. If S ⊂ R is a finite set, P2n ∈ P2n
(S) are self-reciprocal,
Tn (t) := P2n (eit )e−int ,
(2.1) holds, and there is a polynomial H ∈ Pm such that (3.13) holds, then (2.2) also
holds.
Moreover, we have the following two observation.
Lemma 3.7. Let (nν ) be a strictly increasing sequence of positive integers. If S ⊂ R is a
c
(S) are self-reciprocal, Tnν (t) := P2nν (eit )e−inν t ,
finite set, P2nν ∈ P2n
ν
lim NCk (P2nµ ) = ∞
µ→∞
for every k ∈ N, and there is a polynomial H ∈ Pm such that
sup NC(P2nν H) < ∞ ,
ν∈N
then
lim NZ(Tnν ) = ∞ .
ν→∞
Proof. We define
P2n := P2nν ,
nν ≤ n < nν+1 ,
and apply Lemma 3.6. The next lemma is straightforward consequences of Theorem 1.4.
Lemma 3.8. Let λ0 < λ1 < · · · < λm be nonnegative integers and let
Qm (t) =
m
X
Aj cos(λj t) ,
Aj ∈ R , j = 0, 1, . . . , m .
j=0
Then
Z
m
1 X |Am−j |
|Qm (t)| dt ≥
.
60 j=0 j + 1
−π
π
We will also need the lemma below in the proof of Theorem 2.1..
Lemma 3.9. Let λ0 < λ1 < · · · < λm be nonnegative integers and let
Qm (t) =
m
X
Aj cos(λj t) ,
Aj ∈ R , j = 0, 1, . . . , m .
j=0
14
Let A := max{|Aj | : j = 0, 1, . . . , m} . Suppose Qm has at most K − 1 zeros in the period
[−π, π). Then


Z π
m
X
1
|Qm (t)| dt ≤ 2KA π +
≤ 2KA(5 + log m) .
λ
−π
j=1 j
Proof. We may assume that λ0 = 0, the case λ0 > 0 can be handled similarly. Associated
with Qm in the lemma let
m
X
Aj
Rm (t) := A0 t +
sin(λj t) .
λ
j
j=0
Clearly

max |Rm (t)| ≤ A π +
t∈[−π,π]
m
X
j=1

1
.
λj
Also, for every c ∈ R the function Rm − c has at most K zeros in the period [−π, π),
otherwise Rolle’s Theorem implies that Qm = (Rm − c)′ has at least K zeros in the period
[−π, π). Hence
Z π
Z π
′
π
|Qm (t)| dt =
|Rm
(t)| dt = V−π
(Rm ) ≤ 2K max |Rm (t)|
t∈[−π,π]
−π
−π


m
X
1

≤ 2KA(5 + log m) ,
≤ 2KA π +
λ
j=1 j
and the lemma is proved.
Lemma 3.10. Suppose k ∈ N. Let
2πji
zj := exp
,
k
j = 0, 1, . . . , k − 1 ,
be the kth roots of unity. Suppose
0∈
/ {b0 , b1 , . . . , bk−1 } ⊂ R
and
Q(z) :=
k−1
X
bj z j .
j=0
Then there is a value of j ∈ {0, 1, . . . k − 1} for which Re(Q(zj )) 6= 0 .
Proof. If the statement of the lemma were false, then
z
k−1
k
(Q(z) + Q(1/z)) = (z − 1)
k−2
X
αν z ν .
ν=0
k−1
Observe that the coefficient of z
on the right hand side is 0, while the coefficient of
k−1
z
on the left hand side is b1 =
6 0, a contradiction. 15
Lemma 3.11. If 0 ∈
/ {b0 , b1 , . . . , bk−1 } ⊂ R, {a0 , a1 , . . . , am−1 } ⊂ R, where m = uk with
some integer u ≥ 0,
am+lk+j = bj ,
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 c9 > 0 independent of n such that


n
X
Tn (t) := Re 
aj eijt 
j=0
has at least c9 n zeros in [−π, π).
Proof. Note that
n
X
j=0
where
j
aj z =
m−1
X
j=0
j
m

k−1
X
aj z + z 
bj z
j=0
P1 (z) :=

j
m−1
X
r
X
z (l+1)k − 1
m+lk
+z
bj z j = P1 (z) + P2 (z) ,
k
z −1
j=0
j
aj z + z
m+lk
P2 (z) := z
uk
k−1
X
bj z j
j=0
with
bj z j
j=0
j=0
and
r
X
(l+1)k
z (l+1)k − 1
−1
uk z
=
Q(z)z
k
k
z −1
z −1
Q(z) :=
k−1
X
bj z j .
j=0
By Lemma 3.10 there is a kth root of unity ξ = eiτ such that Re(Q(ξ)) 6= 0. Then, for
every K > 0 there is a δ ∈ (0, 2π/k) such that Re(P2 (eit )) oscillates between −K and K at
least c10 (l + 1)kδ times on an interval [τ − δ, τ + δ], where c10 > 0 is a constant independent
of n. Now we choose δ ∈ (0, 2π/k) for
K := 1 +
m−1
X
j=0
Then
|aj | +
k−1
X
|bj | .
j=0


n
X
Tn (t) := Re 
aj eijt  = Re(P1 (eit )) + Re(P2 (eit ))
j=0
has at least one zero on each interval on which Re(P2 (eit )) oscillates between −K and K,
and hence it has at least c10 (l + 1)kδ > c9 n zeros on [−π, π), where c9 > 0 is a constant
independent of n. 16
Proof of the Theorems
We denote the set of all real trigonometric polynomials of degree at most k by Tk .
Proof of Theorems 2.1. Suppose the theorem is false. Then there are k ∈ N, a strictly
increasing sequence (nν )∞
ν=1 of positive integers and even trigonometric polynomials Qnν ∈
Tk with maximum norm 1 on the period such that
(4.1)
Tnν (t)Qnν (t) ≥ 0 ,
t ∈ R.
We can pick a subsequence of (nν )∞
ν=1 (without loss of generality we may assume that it is
the sequence (nν )∞
itself)
that
converges
to a Q ∈ Tk uniformly on the period [−π, π).
ν=1
That is,
(4.2)
lim εν = 0
ν→∞
with
εν := max |Q(t) − Qnν (t)| .
t∈[−π,π]
We introduce the notation
(4.3)


Kν
nν
X
X
3
3


bj,ν cos(βj,ν t) ,
aj,ν cos(jt) Q(t) =
Tnν (t)Q(t) =
j=0
j=0
bjν 6= 0,
j = 0, 1, . . . , Kν ,
and
(4.4)


nν
Lν
X
X
4
4


Tnν (t)Q(t) =
aj,ν cos(jt) Q(t) =
dj,ν cos(δj,ν t) ,
j=0
dj,ν 6= 0 ,
j=0
j = 0, 1, . . . , Lν ,
where β0,ν < β1,ν < · · · < βKν ,ν and δ0,ν < δ1,ν < · · · < δLν ,ν are nonnegative integers.
Since the set {aj,ν : j = 0, 1, . . . , nν , ν ∈ N} ⊂ R is finite, the sets
{bj,ν : j = 0, 1, . . . , Kν , ν ∈ N} ⊂ R
and
{dj,ν : j = 0, 1, . . . , Lν , ν ∈ N} ⊂ R
are finite as well. Hence there are ρ, M ∈ (0, ∞) such that
(4.5)
|aj,ν | ≤ M ,
(4.6)
ρ ≤ |bj,ν | ,
j = 0, 1, . . . , nν , ν ∈ N ,
j = 0, 1, . . . , Kν , ν ∈ N ,
and
(4.7)
|dj,ν | ≤ M ,
j = 0, 1, . . . , Lν , ν ∈ N .
17
Observe that our indirect assumption together with Lemma 3.7 implies that
(4.8)
lim Kν = ∞
and
ν→∞
lim Lν = ∞ .
ν→∞
We claim that
(4.9)
Kν ≤ c3 Lν
with some c3 > 0 independent of ν ∈ N. Indeed, using Parseval’s formula (4.2), (4.3), and
(4.6) we deduce
Z π
Z π
1
1
1
2
4
2
Tnν (t) Q(t) Qnν (t) dt =
(Tnν (t)Q(t)2 Qnν (t))2 dt ≥ ρ2 Kν
(4.10)
π −π
π −π
2
for every sufficiently large ν ∈ N . Also, (4.1) – (4.8) imply
(4.11)
Z π
Z π
1
1
2
4
2
Tn (t) Q(t) Qnν (t) dt =
(Tnν (t)Qnν (t))(Tnν (t)Q(t)4 )Qnν (t) dt
π −π ν
π −π
Z π
1
4
≤
Tnν (t)Qnν (t) dt
max |Tnν (t)Q(t) |
max |Qnν (t)|
t∈[−π,π]
t∈[−π,π]
π
−π
Z π
1
Tnν (t)Qnν (t) dt Lν M
max |Qnν (t)|
≤
π
t∈[−π,π]
−π
≤ c4 Lν
with a constant c4 > 0 independent of ν for every ν ∈ N. Now (4.9) follows from (4.10)
and (4.11). From Lemma 3.8 we deduce
Z π
(4.12)
|Tnν (t)Q(t)4 | dt ≥ c5 ρ log Lν
−π
with some constant c5 > 0 independent of ν ∈ N . On the other hand, using (4.1), Lemma
3.9, (4.2), (4.4), (4.9), and (4.8), we obtain
(4.13)
Z π
|Tnν (t)Q(t)4 | dt
−π
Z π
Z π
3
≤
|Tnν (t)Q(t) ||Qnν (t)| dt +
|Tnν (t)Q(t)3 | |Q(t) − Qnν (t)| dt
−π
−π
Z π
Z π
3
≤
|Tnν (t)Qnν (t)||Q(t) | dt +
|Tnν (t)Q(t)3 | |Q(t) − Qnν (t)| dt
−π
−π
Z π
3
|Tnν (t)Qnν (t)| dt
max |Q(t)|
≤
t∈[−π,π]
−π
Z π
3
|Tnν (t)Q(t) | dt
max |Q(t) − Qnν (t)|
+
t∈[−π,π]
−π
Z π
Z π
3
3
|Tnν (t)Q(t) | dt εν
≤
Tnν (t)Qnν (t) dt
max |Q(t)| +
−π
t∈[−π,π]
−π
≤c6 + c7 (log Kν )εν ≤ c6 + c7 (log(c3 Lν ))εν
≤c8 + c7 (log Lν )εν = o(log Lν ) ,
18
where c6 , c7 , and c8 are constants independent of ν ∈ N. Since (4.13) contradicts (4.12),
the proof of the theorem is finished. Proof of Corollary 2.2. Observe that assumption (2.3) implies assumption (2.1). Proof of Corollary 2.3. Corollary 2.2 implies
(4.14)
lim NZ(P2k ) = ∞ .
k→∞
and
(4.15)
lim NZ(P2k+1 ) = ∞ .
k→∞
Note that (4.14) is an obvious consequence of Theorem 2.1. To see (4.15) observe that if
c
P2k+1 ∈ P2k+1
(S) are self-reciprocal then Pe2k+2 defined by
c
e
Pe2k+2 (z) := (z + 1)P2k+1 (z) ∈ P2k+2
(S)
e Also,
are also self-reciprocal, where the finiteness of S implies the finiteness of S.
lim |Pn (1)| = ∞
n→∞
implies
lim |Pe2k+2 (1)| = lim 2|P2k+1 (1)| = ∞ .
k→∞
k→∞
c
e satisfy the assumptions of Corollary 2.2. Hence the polynomials Pe2k+2 ∈ P2k+2
(S)
Proof of Corollary 2.4. If the finite set S ⊂ R has property (2.6), then assumption (2.1) is
satisfied. Proof of Corollary 2.5. Corollary 2.4 implies (4.14) and (4.15). Note that (4.14) is an
c
(S) are
obvious consequence of Corollary 2.4. To see (4.15) observe that if P2k+1 ∈ P2k+1
self-reciprocal, then Pe2k+2 defined by
c
e
Pe2k+2 (z) := (z + 1)P2k+1 (z) ∈ P2k+2
(S)
are also self-reciprocal, where the finiteness of S implies the finiteness of
Se := {s1 + s2 : s1 , s2 ∈ S ∪ {0}} .
Also, it is easy to see that if
s1 + s2 + · · · + sk = 0
with some s1 , s2 , . . . , sk ∈ Se and k ∈ N, then s1 = s2 = · · · = sk = 0. Hence the
c
e satisfy the assumptions of Corollary 2.4. polynomials Pe2k+2 ∈ P2k+2
(S)
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 2.6. 19
14. Acknowledgements
The author wishes to thank Stephen Choi and Jonas Jankauskas for their reading earlier
versions of my paper carefully, pointing out many misprints, and their suggestions to make
the paper more readable.
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Department of Mathematics, Texas A&M University, College Station, Texas 77843, College Station, Texas 77843 (T. Erdélyi)
22