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
P
j
of modulus 1. A polynomial P of the form P (z) = n
j=0 aj z is called conjugate reciprocal
if an−j = aj , aj ∈ 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)
z∈∂D
p
4/3 m1/2 ,
for every conjugate reciprocal unimodular polynomial of degree m. We also prove that there
is an absolute constant ε > 0 such that
Mq (f ′ ) ≤ exp(ε(q − 2)/q)
p
1/3 m3/2 ,
1 ≤ q < 2,
1/3 m3/2 ,
2 < q,
and
Mq (f ′ ) ≥ exp(ε(q − 2)/q)
p
for every conjugate reciprocal unimodular polynomial of degree m, where
Mq (g) =
1
2π
Z
0
2π
1/q
|g(eit )|q dt
,
q > 0.
1. Introduction
Let Tn be the set of all real trigonometric polynomials of degree at most n. Let Pnc be the
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 K.
Let


n


X
cos(jt + γj ) , γj ∈ R
An := Q : Q(t) =


j=1
Key words and phrases. Littlewood polynomials; unimodular polynomials; conjugate reciprocal polynomials; flatness properties.
2000 Mathematics Subject Classifications. 11C08, 41A17
Typeset by AMS-TEX
1
and
Bn+1/2
We use the notation


n


X
2j + 1
:= Q : Q(t) =
cos
t + γj , γj ∈ R .


2
j=0
kQkp :=
1
2π
Z
K
|Q(t)|
p
1/p
,
p > 0,
and
kQk∞ := max |Q(t)| .
t∈K
The Bernstein–Szegő inequality (see p. 232 in [7], for instance) gives that
|Q′ (t)|2 + n2 |Q(t)|2 ≤ n2 kQk2∞ ,
Q ∈ Tn ,
t ∈ R.
Integrating the left hand side on the period and using Parseval’s formula we obtain
n(n + 1)(2n + 1) n3
+
≤ n2 kQk2∞ ,
12
2
Q ∈ An ,
and hence
(1.1)
kQk∞ ≥
p
4/3
p
n/2 ,
Q ∈ An .
One of the highlights of this paper is to improve (1.1) by showing that there is an absolute
constant ε > 0 such that
p
p
kQk∞ ≥ (1 + ε) 4/3 n/2 ,
Q ∈ An .
Let
Km


aj ∈ C , |aj | = 1 , j = 0, 1, . . . , m


n

X
aj z j ,
:= P : P (z) =

j=0
be the set of all unimodular polynomials of degree m. Associated with an algebraic polynomial P of the form
P (z) =
m
X
aj z j ,
aj ∈ C ,
j=0
let
P (z) :=
m
X
aj z j
and
j=0
2
am 6= 0 ,
P ∗ (z) := z m P (1/z) .
The polynomial P of degree m is called conjugate reciprocal if P ∗ = P . The classes An ,
Bn+1/2 , and Km and flatness properties of their elements were studied by many authors,
see [1–40], for instance. Let
it
Mq (f ) := kf (e )kq =
1
2π
Z
K
it
q
|f (e )| dt
1/q
,
q ∈ (0, ∞) ,
and
M∞ (f ) := sup |f (eit )| .
t∈K
There is a beautiful short argument to see that
(1.2)
M∞ (f ) ≥
p
4/3 m1/2
for every conjugate reciprocal unimodular polynomial f ∈ Km . Namely, Parseval’s formula
gives
1/2
m(m + 1)(2m + 1)
′
′
,
f ∈ Km .
M∞ (f ) ≥ M2 (f ) =
3
Combining this with Lax’s Bernstein-type inequality
M∞ (f ′ ) ≤
m
M∞ (f )
2
valid for all conjugate reciprocal algebraic polynomials f ∈ Pnc (see p. 438 in [7], for
instance), we obtain
2
M∞ (f ) ≥
m
m(m + 1)(2m + 1)
3
1/2
≥
p
4/3 m1/2
for all conjugate reciprocal unimodular polynomials f ∈ Km . 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)
p
4/3 m1/2 ,
for every conjugate reciprocal unimodular polynomial f ∈ Km . We also prove that there
is an absolute constant ε > 0 such that
and
p
Mq (f ′ ) ≤ exp(ε(q − 2)/q) 1/3 m3/2 ,
1 ≤ q < 2,
p
Mq (f ′ ) ≥ exp(ε(q − 2)/q) 1/3 m3/2 ,
2 < q,
for every conjugate reciprocal unimodular polynomial of degree m. See Theorem 2.7.
3
2. New Results
Theorem 2.1. Let Q ∈ An and P = (Q′ )2 + n2 Q2 . There is an absolute constant δ > 0
such that
kP k1/2 ≤ (1 − δ)kP k1 .
Theorem 2.1*. Let Q ∈ Bn+1/2 and P = (Q′ )2 + (n + 1/2)2 Q2 . There is an absolute
constant δ > 0 such that
kP k1/2 ≤ (1 − δ)kP k1 .
Theorem 2.2. Let Q ∈ An and P = (Q′ )2 + n2 Q2 . There is an absolute constant δ > 0
such that
kP k∞ ≥ (1 + δ)kP k1 .
Theorem 2.2*. Let Q ∈ Bn+1/2 and P = (Q′ )2 + (n + 1/2)2 Q2 . There is an absolute
constant δ > 0 such that
kP k∞ ≥ (1 + δ)kP k1 .
Theorem 2.3. There is an absolute constant δ > 0 such that
p
p
kQk∞ ≥ (1 + δ) 4/3 n/2
for every Q ∈ An .
Theorem 2.3*. There is an absolute constant δ > 0 such that
p
p
kQk∞ ≥ (1 + δ) 4/3 n/2
for every Q ∈ Bn+1/2 .
Let ∂D be the unit circle of the complex plane. Let f be a continuous function on ∂D
and let
Z
1
q
|f (eit )|q dt
Iq (f ) := Mq (f ) =
2π K
and
M∞ (f ) := max |f (z)| .
z∈∂D
Theorem 2.4. There is an absolute constant ε > 0 such that
p
M1 (f ′ ) ≤ (1 − ε) 1/3 m3/2
for every conjugate reciprocal unimodular polynomial f ∈ Km .
Theorem 2.5. There is an absolute constant ε > 0 such that
p
M∞ (f ′ ) ≥ (1 + ε) 1/3 m3/2
for every conjugate reciprocal unimodular polynomial f ∈ Km .
4
Theorem 2.6. There is an absolute constant ε > 0 such that
M∞ (f ) ≥ (1 + ε)
p
4/3 m1/2
for every conjugate reciprocal unimodular polynomial f ∈ Km .
Theorem 2.7. There is an absolute constant ε > 0 such that
and
p
Mq (f ′ ) ≤ exp(ε(q − 2)/q) 1/3 m3/2 ,
1 ≤ q < 2,
p
Mq (f ′ ) ≥ exp(ε(q − 2)/q) 1/3 m3/2 ,
2 < q,
for every conjugate reciprocal unimodular polynomial f ∈ Km .
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.
p
√
Remark 2.2. The factor (1 + δ) 4/3 in Theorem 2.3 cannot be replaced by (1 + δ) 2.
p
p
Remark 2.3. The factor (1 + ε) 1/3 in Theorem 2.5 cannot be replaced by (1 + ε) 1/2.
p
√
Remark 2.4. The factor (1 + ε) 4/3 in Theorem 2.6 cannot be replaced by (1 + ε) 2.
c
c
A polynomial f ∈ Pm
is called skew-reciprocal if f ∗ (z) = f (−z). A polynomial f ∈ Pm
m
∗
is called plain-reciprocal if f = f , that is, f (z) = z 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 |f ′ (z)| − min |f ′ (z)| ≥ εm3/2
z∈∂D
z∈∂D
for all conjugate reciprocal, plain-reciprocal, and skew-reciprocal unimodular polynomials
f ∈ Km .
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)
p
1/3 m3/2
holds for all plain-reciprocal and skew-reciprocal unimodular polynomials f ∈ Km ?
5
Problem 2.2. Is there an absolute constant ε > 0 such that
M∞ (f ′ ) ≥ (1 + ε)
or at least
p
1/3 m3/2
max |f ′ (z)| − min |f ′ (z)| ≥ εm3/2
z∈∂D
z∈∂D
holds for all unimodular polynomials f ∈ Km ?
Our method to prove Theorem 2.5 does not seem to work for all unimodular polynomials
f ∈ Km . In an e-mail 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 Lm 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
M∞ (f ′ ) ≥ (1 + ε)
or at least
p
1/3 m3/2
max |f ′ (z)| − min |f ′ (z)| ≥ εm3/2
z∈∂D
z∈∂D
holds for all Littlewood polynomials f ∈ Lm ?
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
M∞ (f ) ≥ (1 + ε)m1/2
or at least
max |f (z)| − min |f (z)| ≥ εm1/2
z∈∂D
z∈∂D
holds for all Littlewood polynomials f ∈ Lm ?
The same problem may be raised only for all skew reciprocal Littlewood polynomials
f ∈ Lm 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 ∈ Tn be of the form
R(t) = Rn (t) =
n
X
aj cos(jt + γj )
j=1
6
aj , γj ∈ R ,
j = 1, 2, . . . , n .
Let sm :=
Pm
j=1
a2j , m = 1, 2, . . . , n, and let µ := kRk2 , that is, µ2 = sn . Suppose
kRk1 ≥ cµ ,
where c > 0 is a constant (necessarily not greater than 1). Suppose also that the coefficients
of R satisfy
X
s[n/h] /sn = µ−2
a2j ≤ 2−9 c6
1≤j≤n/h
for some constant h > 0. Let V = 2−5 c3 . Then there exists a constant B > 0 depending
only on c and h such that
m({t ∈ K : v1 µ ≤ |R(t)| ≤ v2 µ}) ≥ B(v2 − v1 )2
for every v1 and v2 such that −V ≤ v1 < v2 ≤ V .
Lemma 3.2. Associated with Q ∈ Tn we define the sets
Eδ := {t ∈ K : |Q′ (t)| < δn3/2 }
and
Fδ := {t ∈ K : |Q′′ (t)| ≤ δ 1/2 n5/2 } .
We have
m(Eδ \ Fδ ) ≤ 8δ 1/2 .
Proof of Lemma 3.2. Observe that Eδ \Fδ is the union of at most 4n pairwise disjoint open
subintervals of the period. Let these intervals be (xj , yj ), j = 1, 2, . . . , µ, where µ ≤ 4n.
By the Mean Value Theorem we can deduce that there are ξj ∈ (xj , yj ) such that
2δn3/2 ≥ |Q′ (yj ) − Q′ (xj )| = |Q′′ (ξj )|(yj − xj ) ≥ δ 1/2 n5/2 (yj − xj ) ,
and hence
yj − xj ≤ 2δ 1/2 n−1 ,
Hence
m(Eδ \ Fδ ) =
µ
X
j=1
j = 1, 2, . . . , µ .
(yj − xj ) ≤ µ(2δ 1/2 n−1 ) ≤ 8δ 1/2 .
Lemma 3.3. Let Q ∈ An , P := (Q′ )2 + n2 Q2 , δ ∈ (0, 1), and
Suppose
(3.1)
1/2 1/2
Gδ := {t ∈ K : P (t)|1/2 − kP k1 ≤ δ 1/4 kP k1 } .
kP k1/2 ≥ (1 − δ)kP k1 .
7
Then
m(K \ Gδ ) ≤ 4πδ 1/2 .
Proof of Lemma 3.3. Using (3.1) we have
1/2
1/2
kP k1/2 ≥ (1 − δ)1/2 kP k1
(3.2)
1/2
≥ (1 − δ)kP k1
.
Hence
Z
K
|P (t)|1/2 − kP k1/2 2 dt = 4π kP k1/2 kP k1/2 − kP k1/2
1
1
1
1/2
1/2
1/2
≤ 4πkP k1 δkP k1
= 4πδkP k1 .
Letting a := m(K \ Gδ ) we have
aδ 1/2 kP k1 ≤ 4πδkP k1 ,
and the lemma follows.
31
kP k1 . Then R :=
Lemma 3.4. Let Q ∈ An , P := (Q′ )2 + n2 Q2 , and kP k1/2 ≥ 32
2 2
′′
n Q + Q satisfies the assumptions of Lemma 3.1 with c := 1/32 and h = 29 326 for all
sufficiently large n.
Proof of Lemma 3.4. Let n ≥ 3. Observe that
R(t) =
n
X
aj := n2 − j 2 ,
aj cos(jt + γj ) ,
j=1
Then
j = 1, 2, . . . , n .
n
1 X 2
(n − j 2 )2 ,
2 j=1
sn = µ2 = kRk22 =
hence
γj ∈ R ,
n5
n5
≤ s n = µ2 ≤
.
6
2
Using Parseval’s formula, we have
n
(3.3)
(3.4)
2n3
1X 2
(j + n2 ) ≥
,
kP k1 =
2 j=1
3

kQ′′ k1 ≤ kQ′′ k2 ≤ 
n
1X
2
j=1
1/2
j4
8
n5/2
≤ (1 + o(1)) √ ,
10
and
(3.5)
and hence
(3.6)

kQ′ k2 ≤ 
knQk1 = (2π)
−1
≥ (2π)
−1
n
1X
2
j=1
Z
ZK
K
1/2
j2
n3/2
≤ (1 + o(1)) √ ,
6
|P (t) − Q′ (t)2 |1/2 dt
|P (t)|
1/2
dt − (2π)
1/2
−1
Z
K
|Q′ (t)| dt
31
1/2
1/2
kP k1 − kQ′ k2
= kP k1/2 − kQ′ k1 ≥
32
r
r
1 3/2
31 2 3/2
n − (1 + o(1))
n .
≥
32 3
6
Combining (3.5) and (3.4) we conclude
kRk1 = kn2 Q + Q′′ k1 ≥ kn2 Qk1 − kQ′′ k1
r
r
n5/2
31 2 5/2
1 5/2
n − (1 + o(1))
n − (1 + o(1)) √
≥
32 3
6
10
1 5/2
≥
n
32
for all sufficiently large n. Also, s[n/h] ≤ (n/h)n4 = n5 /h, hence s[n/h] /sn ≤ 1/h. Therefore
the assumptions of Lemma 3.1 are satisfied with c := 1/32 and h = 29 326 . 4. Proof of the Theorems
Proof of Theorem 2.1. Let Q ∈ An , P = (Q′ )2 +n2 Q2 , and R := n2 Q2 +Q′′ . Let δ ∈ (0, 1).
Suppose kP k1/2 ≥ (1 − δ)kP k1 . 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 = 29 326 . Now let
Eδ := {t ∈ K : |Q′ (t)| < δn3/2 } ,
Fδ := {t ∈ K : |Q′′ (t)| ≤ δn5/2 } ,
1/2
1/2
Gδ := {t ∈ K : |P (t)1/2 − kP k1 | ≤ δ 1/4 kP k1 } ,
and
Hγ := {t ∈ K : γn5/2 ≤ |R(t)| < 2γn5/2 } .
Recall that by the Parseval’formula we have
(4.1)
kP k1 =
n(n + 1)(2n + 1)
n3
+
.
2
12
9
Hence, if t ∈ Gδ ∩ Eδ ∩ Fδ and the absolute constant δ > 0 is sufficiently small, then
|R(t)| ≥n2 |Q(t)| − |Q′′ (t)| = n(P (t) − Q′ (t)2 )1/2 − |Q′′ (t)|
3
1/2
n
n(n + 1)(2n + 1)
1/4 2
2 3
≥n (1 − δ )
−δ n
− δn5/2
+
2
12
1
≥ n5/2 ,
2
that is,
(4.2)
|R(t)| ≥
1 5/2
n ,
2
t ∈ Gδ ∩ Eδ ∩ Fδ .
By Lemma 3.2 we have
m(Eδ \ Fδ ) ≤ 8δ 1/2 .
(4.3)
By Lemma 3.3 we have
(4.4)
m(K \ Gδ ) ≤ 4πδ 1/2 .
Observe that if 0 < γ < 1/4 then (4.2) implies that Hγ ⊂ K \ (Gδ ∩ Eδ ∩ Fδ ), hence
Hγ ∩ Eδ ⊂ (Eδ \ Gδ ) ∪ (Eδ \ Fδ ) .
Therefore, by (4.3) and (4.4) we can deduce that
(4.5)
m(Hγ ∩ Eδ ) ≤ m(Eδ \ Gδ ) + m(Eδ \ Fδ )
≤ 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
m(Hγ ) ≥ Bγ 2 .
(4.6)
It follows from (4.5) and (4.6) that
(4.7)
m(Hγ \ Eδ ) ≥
1
Bγ 2
2
for all sufficiently small absolute constants δ > 0. Observe that
(4.8)
|2Q′ (t)R(t)| ≥ 2δn3/2 γn5/2 = 2γδn4 ,
and
(4.9)
P ′ (t) = 2Q′ (t)R(t) .
10
t ∈ Hγ \ E δ ,
Combining (4.7), (4.8), and (4.9), we obtain
m({t ∈ K : |P ′ (t)| ≥ 2γδn4 }) ≥
1
Bγ 2 ,
2
and hence
Z
(4.10)
K
1
Bγ 2 (2γδn4 ) = Bγ 3 δn4 .
2
|P ′ (t)| dt ≥
Now let Pe := P − 2πkP k1 ∈ T2n . Then (4.10) can be rewritten as
Z
|Pe ′ (t)| dt ≥ Bγ 3 δn4 ,
K
and by Bernstein’s inequality in L1 (see p. 390 of [7], for instance), we have
Z
1
e
(4.11)
2πkP k1 =
|Pe (t)| dt ≥ Bγ 3 δn3 .
2
K
Observe that
(4.12)
2πkPek1 =
=
Z
ZK
K
≤
Z
|Pe(t)| dt
P (t) − kP k1 dt ≤
K
=2π
Z
K
P (t)1/2 − kP k1/2 P (t)1/2 + kP k1/2 dt
1
1
1/2 Z
1/2
(P (t)1/2 − kP k1/2 2 dt
(P (t)1/2 + kP k1/2 2 dt
1
1
1/2
1/2
2kP k1 (kP k1
1/2
≤4πkP k1
1/2 1/2
kP k1/2
−
kP k1 − kP k1/2
1/2
K
1/2
2kP k1
= 4πn3/2
kP k1 − kP k1/2 ≥
!2
≥ δ ∗ kP k1
≥
1/2 1/2
+ kP k1/2
1/2
.
kP k1 − kP k1/2
Combining (4.11), (4.12), and (4.1), we conclude
2πkPek1
4πn3/2
1/2
kP k1
Bγ 3 δn3/2
8π
2
≥ δ ∗ n3
with an absolute constant δ ∗ > 0. Proof of Theorem 2.2. By Theorem 2.1 there is an absolute constant δ > 0 such that
Z
Z
Z
1/2
1/2
kP k1 = 2π
|P (t)| dt = 2π
|P (t)| |P (t)| dt ≤ 2π
|P (t)|1/2 kP k1/2
∞ dt
≤
K
1/2
kP k1/2 kP k1/2
∞
K
≤ (1 − δ)
K
1/2
1/2
kP k1 kP k1/2
∞
11
.
Hence
1/2
kP k1
≤ (1 − δ)1/2 kP k1/2
∞ ,
and the result follows. Proof of Theorem 2.3. The Bernstein–Szeg̋o inequality (see p. 232 of [7],for instance) yields
Q′ (t)2 + n2 Q(t)2 ≤ n2 kQk2∞ ,
t ∈ R , Q ∈ An ⊂ Tn ,
hence if P = (Q′ )2 + n2 Q2 , then
kP k∞ ≤ n2 kQk∞ .
Hence, using Theorem 2.1 and Parseval’s formula we can deduce that
kQk2∞ ≥n−2 kP k∞ ≥ n−2 (1 + δ)kP k1 = n−2 (1 + δ) kQ′ k22 + n2 kQk22
n(n + 1)(2n + 1) n3
−2
=n (1 + δ)
≥ (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 =P2n is even and f ∈ Km is a conjugate
m
reciprocal unimodular polynomial. Let f (z) = j=0 aj z j , where aj ∈ C and |aj | = 1 for
each j = 0, 1, . . . m. As f is conjugate reciprocal, we have
am−j = aj ,
j = 0, 1, . . . , m ,
and an ∈ {−1, 1}, in particular. Let Q ∈ An be defined by 2Q(t) = e−int f (eit ) − an . Then
ieit f ′ (eit ) = eint (2Q′ (t) + in(2Q(t) + an )) ,
hence the triangle inequality implies that
|f ′ (eit )| ≤ 2|eint (Q′ (t) + inQ(t))| + |eint inan | = 2|Q′ (t) + inQ(t)| + n
= 2|P (t)|1/2 + n ,
where P := (Q′ )2 + n2 Q2 is the same as in Theorem 2.1, and the theorem follows from
Theorem 2.1 as
1/2
1/2
M1 (f ′ ) ≤ 2kP k1/2 + n ≤ 2(1 − δ)1/2 kP k1 + n
1/2
n(n + 1)(2n + 1) n3
1/2
+
+n
≤ 2(1 − δ)
12
2
1/2
2(n + 1)3
1/2
+n
≤ 2(1 − δ)
3
p
≤ (1 − δ)1/2 1/3 m3/2 + o(m3/2 ) .
12
Now assume that m = 2n + 1 is odd and f ∈ Km is a conjugate reciprocal unimodular
polynomial. Let Q ∈ Bm/2 be defined by 2Q(t) = e−imt/2 f (eit ). Then
ieit f ′ (eit ) = 2eimt/2 (Q′ (t) + (im/2)Q(t))
implies that
|f ′ (eit )| = 2|eimt/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 Q2 is the same as in Theorem 2.1*, and the theorem follows
from Theorem 2.1* as
1/2
1/2
M1 (f ′ ) ≤ 2kP k1/2 ≤ 2(1 − δ)1/2 kP k1
1/2
(n + 1)(n + 2)(2n + 3) (n + 1)3
1/2
≤ 2(1 − δ)
+
12
2
1/2
2(n + 1)3
≤ 2(1 − δ)1/2
≤
3
p
≤ (1 − δ)1/2 1/3 m3/2 + o(m3/2 ) .
Proof of Theorem 2.5. Let f ∈ Km be a conjugate reciprocal unimodular polynomial. By
Theorem 2.4 there is an absolute constant ε > 0 such that
Z
Z
m(m + 1)(2m + 1)
′ 2
′ it 2
= M2 (f ) = 2π
|f (e )| dt = 2π
|f ′ (eit )||f ′ (eit )| dt
6
K
K
Z
≤ 2π
|f ′ (eit )| max |f ′ (eiτ )| dt
K
′
Hence
τ ∈K
≤ M1 (f )M∞ (f ′ )
p
≤ (1 − ε) 1/3 m3/2 M∞ (f ′ ) .
and the theorem follows. p
1/3 m3/2 ≤ (1 − ε)M∞ (f ′ ) ,
Proof 1 of Theorem 2.6. First assume that m P
= 2n is even and f ∈ Km is a conjugate
m
reciprocal unimodular polynomial. Let f (z) = j=0 aj z j , where aj ∈ C and |aj | = 1 for
each j = 0, 1, . . . m. As f is conjugate reciprocal, we have
am−j = aj ,
j = 0, 1, . . . , m ,
13
and an ∈ {−1, 1}, in particular. Let Q ∈ An be defined by 2Q(t) = e−int f (eit ) − an .
Observe that
max |f (z)| − k2Qk∞ ≤ 1 ,
z∈∂D
hence the theorem follows from Theorem 2.3. Now assume that m = 2n + 1 is odd and
f ∈ Km is a conjugate reciprocal unimodular polynomial. Let Q ∈ Bn/2 be defined by
2Q(t) = e−imt/2 f (eit ). Observe that
max |f (z)| = k2Qk∞ ,
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 kf ′ k∞ = (m/2) kf k∞ .
Hence the theorem follows from a combination of this and Theorem 2.5. Proof of Theorem 2.7. Let f ∈ Km be conjugate reciprocal. Observe that by Parseval’s
formula we have
1/2
m3/2
m(m + 1)(2m + 1)
′
= (1 + o(1)) √ .
(4.13)
M2 (f ) =
6
3
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) := q log(Mq (f )) on (0, ∞). Let f be a
continuous function on ∂D and let
Z
1
q
|f (eit )|q dt .
Iq (f ) := Mq (f ) =
2π K
Then h(q) := log(Iq (f )) = q log(Mq (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
r−q
p−r
Ir (f ) ≤ Ip (f ) p−q Iq (f ) p−q ,
that is,
(4.14)
1
2π
Z
p−q r−q p−r
Z
Z
1
1
it p
it q
|f (e )| dt
≤
|f (e )| dt
|f (e )| dt
.
2π K
2π K
K
it
r
To see this let
α :=
p−q
,
r−q
β :=
p−q
,
p−r
γ :=
p
,
α
δ :=
hence 1/α + 1/β = 1 and γ + δ = r. Let
F (t) := |f (eit )|γ = |f (eit )|
14
p(r−q)
p−q
,
q
,
β
and
G(t) := |f (eit )|δ = |f (eit )|
q(p−r)
p−q
.
Then by Hölder’s inequality we conclude
Z
K
F (t)G(t) dt ≤
Z
K
α
F (t) dt
1/α Z
K
1/β
G(t) dt
,
β
and (4.14) follows.
Let q ∈ [1, 2). Then, using the convexity property of the function h(q) := q log(Mq (f ))
on (0, ∞), we obtain
2 log M2 − log M1
2 log M2 − q log Mq
≥
.
2−q
2−1
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) :=
q log(Mq (f )) on (0, ∞), we obtain
q log Mq − 2 log M2
2 log M2 − log M1
≥
.
q−2
2−1
Combining this with the Theorem 2.4 and (4.13) gives theorem.
Proof of Remark 2.1. Let (fn )√be an ultraflat sequence of unimodular polynomials fn ∈ Kn
satisfying M∞ (fn ) ≤ (1 + εn ) n with a sequence (εn ) of numbers εn > 0 converging to 0.
It is shown in [7] that such a sequence (fn ) exists. Let gn (z) = zfn−1 (z). Let Qn ∈ An be
defined by Qn (t) := Re(gn (eit )). Then the Bernstein–Szegő inequality (see p. 232 in [7],
for instance) gives that Pn := (Q′n )2 + n2 Q2n satisfy
kPn k∞ ≤ n2 kQn k2∞ ≤ (1 + εn )2 n3 ,
while by Parseval’s formula we have
kPn k1 =
n(n + 1)(2n + 1)
2n3
n3
+
≥
.
2
12
3
Proof of Remark 2.2. Let Qn ∈ An be the same as in the proof of Remark 2.1. Then
3/2
kQn k∞ ≤ kPn k1/2
.
∞ ≤ (1 + εn )n
Proof of Remark 2.4. Let fn ∈ Kn and gn (z) = zfn−1 (z) be the same as in the proof of
Remark 2.1. For m = 2n we define hm ∈ Km by
hm (z) := z n (gn (z) + g n (1/z) + 1 .
15
We have
√ √
√
M∞ (hm ) ≤ 2(1 + εn ) n + 1 ≤ (1 + εn ) 2 m + 1 .
Proof of Remark 2.3. For m = 2n let hm ∈ Km be the same as in the proof of Remark 2.4.
Then using the well-known Bernstein-type inequality for conjugate reciprocal polynomials
(see p. 438 in [7], for instance), we have
M∞ (h′m ) ≤
√ √
1
m
(1 + εn ) 2 m ≤ √ (1 + εn )m3/2 .
2
2
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E-mail address: terdelyi@math.tamu.edu
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