Real zeroes of random polynomials, I: Flip-invariance, Tur\'an's lemma, and the Newton-Hadamard polygon

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by  Ken Söze

Real zeroes of random polynomials, I
Flip-invariance, Tur´an’s lemma, and the
Newton-Hadamard polygon
Ken S¨oze
∗
6
To Ildar Ibragimov with admiration
1
0
2 Abstract

n
We showthatwith highprobabilitythe number ofrealzeroesofarandompolynomialis boundedby the
a
J number of vertices on its Newton-Hadamard polygon times the cube of the logarithm of the polynomial

9 degree. A similar estimate holds for zeroes lying on any curve in the complex plane, which is the graph
1 of a Lipschitz function in polar coordinates. The proof is based on the classical Tur´an lemma.

]
R
P 1 Introduction
.
h
t This work is motivated by the following question attributed to Larry Shepp: Let
a
m n
[ P(z) = λkzk

k=0
1 X
v bea randompolynomial of degree n> 2 with independentidentically distributedrandom coeffi-
0
cients λ . Is it true that the expected number of real zeroes of P is bounded by Clogn? Since the
5 k
8 classical work of Mark Kac [6], for many “decent” distributions of the coefficients, it has been
4
proven by Erd˝os and Offord [2], Logan and Shepp [9], Ibragimov and Maslova [4, 5], Shepp and
0
Farahmand [14] (by no means is this list complete). Here, we are interested in a bound valid for
.
1
all distributions. Several years ago, Ibragimov and Zaporozhets [3] proved that for any distribu-
0
6 tionofthecoefficients, theexpectednumberofrealzeroesiso(n)asn . Later,inworksthat
→ ∞
1 remained unpublished, this was independently improved by Kabluchko and Zaporozhets and by
:
v Krishnapur and Zeitouni to O(√n). In the opposite direction, Zaporozhets [15] constructed an
i
X example of a distribution wherein the mean number of real zeroes remains bounded as n .
→ ∞
r Inthisworkwesuggesttwoapproachestothisquestion. Thefirstone, presentedinthispart,
a
is based on tools from harmonic and complex analysis (Tura´n’s lemma and Jensen’s formula).
In the case when the coefficients of P are independent and identically distributed, it gives a
bound C log4n, which is weaker than the estimate we prove in Part II. On the other hand,
the approach of Part I needs less restrictive condition (which we call “the property (Θ)”) than
independenceand identical distribution of the coefficients. Assumingthe property (Θ), we show
that, with high probability, the number of real zeroes of P is bounded by CV(P) log3n where
V(P) is the number of vertices on the Newton-Hadamard polygon of P. It also gives the same
upper bound for the number of zeroes of P on any curve in the complex plane, which is the
graph of a Lipschitz function in polar coordinates.
The second approach, which we shall present in Part II, is based on Descartes’ rule of sign
changesandonanewanti-concentration estimate forrandompermutationsoflargeorder,which
∗290W 232nd Str,Apt 4b, Bronx, NY 10463, USA;[email protected]
1
2 Main results 2
might be of independent interest. Both approaches may be viewed as further development of
the techniques introduced in the pioneering work of Littlewood and Offord [8].
2 Main results
2.1 Key definitions
Westartwiththreedefinitionsneededtoformulateourresults. Inwhatfollows, P always stands
for a polynomial of degree n with, generally speaking, complex-valued coefficients λ .
k
2.1.1 The number of vertices on the Newton-Hadamard polygon
We denote by V(P) the number of vertices on the graph of the convex polygonal function
t max log λ +kt , t R.
k
7→ 06k6n | | ∈
(cid:0) (cid:1)
Although we will not use it, it is not difficult to see that equivalently V(P) can be defined as
the number of vertices on the Newton-Hadamard polygon, which is the the upper envelope of
convex functions ϕ such that ϕ(k) 6 log λ , 0 6 k 6 n (in other words, the lower boundary
k
− | |
of the convex hull of n+1 vertical rays (k,y): log λ 6 y < + , 0 6 k 6 n ). For more
k
− | | ∞
on this, see [12, Chapter IV, Problem 41].
(cid:8) (cid:9)
2.1.2 The Lipschitz curves
By Γ we denote an arbitrary curve defined in polar coordinates by
Γ = z =reiθ: θ = θ(r),0 6 r < .
∞
If (cid:8) (cid:9)
r
θ(r ) θ(r ) 6 L log 1 ,
1 2
− r
2
(cid:12) (cid:12) (cid:12) (cid:12)
then we call Γ an L-Lipschitz curve. We denote by N(Γ;P) the number of zeroes of P on Γ
(cid:12) (cid:12) (cid:12) (cid:12)
(counted with multiplicities).
2.1.3 Flips of the coefficients
Letλ′ andλ′′ beCn+1-valuedrandomvariablesdefinedonthesameprobabilityspaceandhaving
the same distribution. For k 0,1,...,n , we put λ+ = λ′ and λ− = λ′′, and then, for any
∈ { } k k k k
(n+1)-tuple of signs σ +, n+1, let λσ = (λσ0,λσ1,...,λσn). We say that the joint law of λ′
∈ { −} 0 1 n
and λ′′ is flip-invariant if the random variables λσ are equidistributed.
σ∈{+,−}n+1
(cid:8) (cid:9)
2.1.4 The property (Θ)
Here, we introduce our assumption on the distribution of the coefficients λ Cn+1 of the
∈
polynomial P. We say that the coefficients of the random polynomial P possess property (Θ) if
there exist random variables λ′ and λ′′ equidistributed with λ whose joint law is flip-invariant
and such that, for some a C and for each k 0,1,...,n ,
∈ ∈{ }
1
λσk λ−σk > λσk a + λ−σk a a.s..
k − k 2 k − k −
(cid:12) (cid:12) (cid:2)(cid:12) (cid:12) (cid:12) (cid:12)(cid:3)
Note that for our p(cid:12)urposes, it w(cid:12) ould (cid:12)suffice to(cid:12) ha(cid:12)ve this in(cid:12)equality with any constant κ > 0
instead of 1. In the examples, which we will bring below, this condition holds with the value
2
κ= 1. To simplify our notation, we decided to fix this value of κ.
2
2 Main results 3
2.1.5 Three examples of distributions with property (Θ)
Symmetric distributions. For k 0,1,...,n , denote by τ : Cn+1 Cn+1 the map, which
k
∈ { } →
maps w w and keeps fixed the rest of coordinates of w Cn+1. Suppose that, for each
k k
7→ − ∈
k 0,1,...,n , τ λ has the same distribution as λ. Then the distribution of λ has property
k
∈ { } ◦
(Θ) with a = 0, λ′ = λ, and λ′′ = λ.
−
Note that in this example we do not assume independence of λ ’s.
k
Complex-valued independent identically distributed random variables λ , λ , ..., λ . Denote
0 1 n
by ζ the common distribution of λ ’s. We need to produce two random variables ζ± having the
k
same distribution as ζ and such that, for some a C,
∈
1
ζ+ ζ− > ζ+ a + ζ− a .
− 2 − −
(cid:12) (cid:12) (cid:2)(cid:12) (cid:12) (cid:12) (cid:12)(cid:3)
We first assume that the pro(cid:12)bability(cid:12)space(cid:12)Ω is a(cid:12)un(cid:12)ion of (cid:12)2N atoms ω having the same
i
probability 1 . Then the general case will follow by approximation1.
2N
Let
d = max ζ(ω ) ζ(ω )
i j
16i<j62N| − |
be the diameter of the point configuration ζ(ω ),...,ζ(ω ) in C. Pick up from this configu-
1 2N
ration a pair of points with the maximal distance. Without loss of generality, assume that they
(cid:8) (cid:9)
correspond to the atoms ω and ω , that is, d = ζ(ω ) ζ(ω ) . Then consider the
2N 2N−1 2N−1 2N
−
remaining point configuration and repeat the procedure. At the last N-th step we are left with
(cid:12) (cid:12)
two points ζ(ω ) and ζ(ω ). Thendenote by a thecente(cid:12)r of theline segmen(cid:12)t that connects these
1 2
two points, that is, a = 1(ζ(ω )+ζ(ω )).
2 1 2
By construction, for each 16 i 6 N, the point a lies at distance at most ζ(ω ) ζ(ω )
2i−1 2i
| − |
from each of the two points ζ(ω ), ζ(ω ). Hence,
2i−1 2i
1
ζ(ω ) ζ(ω ) > ζ(ω ) a + ζ(ω ) a .
2i−1 2i 2i−1 2i
− 2 − −
(cid:12) (cid:12) (cid:2)(cid:12) (cid:12) (cid:12) (cid:12)(cid:3)
It remains to let ζ+ =(cid:12) ζ, and (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
ζ−(ω ) = ζ(ω ), ζ−(ω )= ζ(ω ), 1 6 i6 N .
2i−1 2i 2i 2i−1
1 Indeed,takeasequenceofrandomvariables(ζ )thatconvergesindistributiontoζ andsuchthatζ attains
N N
2N values (not necessarily distinct ones) with probability 1 each. Let (ζ+,ζ−) be a pair of random variables
2N N N
definedon thesame probability space as ζ , equidistributed with ζ and such that, for some a ∈C,
N N N
(cid:12)(cid:12)ζN+−ζN−(cid:12)(cid:12)> 21(cid:2)(cid:12)(cid:12)ζN+−aN(cid:12)(cid:12)+(cid:12)(cid:12)ζN−−aN(cid:12)(cid:12)(cid:3). (∗)
Since ζ converge to ζ in distribution, the sequence of laws of ζ is tight. Then the sequence of joint laws of
N N
pairs(ζ+,ζ−)istightaswell,andwecanchooseasubsequence(ζ+ ,ζ− )thatconvergesindistributiontoapair
N N Nj Nj
of random variables (ζ+,ζ−) definedon thesame probability space as ζ and equidistributedwith ζ.
Furthemore,bytightnessofthesequenceoflawsof(ζ+,ζ−),wecanchoosealargepositiveconstantLsothat,
N N
for every N, P(cid:8)|ζ+−ζ−|>L(cid:9)< 1. Therefore, on an eventof probability at least 1,
N N 2 2
|a |6|ζ+−a |+|ζ−−a |+|ζ+−ζ−|63|ζ+−ζ−|63L.
N N N N N N N N N
Since both a and L are non-random, it follows that |a | 6 3L. Then, extracting from (a ) a convergent
N N Nj
subsequence,denoting bya its limit, and applying(∗) with N =N , j →∞, we get theresult.
j
2 Main results 4
Real-valued independent random variables λ , λ , ..., λ which have a common median.
0 1 n
Arguingsimilarly to the previous example, we construct the coefficients λ± equidistributed with
k
λ and satisfying λ+ +λ− = 2a, 0 6 k 6 n, where a is the common median for λ , λ , ... λ .
k k k 0 1 n
Note that in this example we have not assumed that the coefficients λ are identically
k
distributed.
2.1.6 A technical assumption
To avoid degeneracies, in what follows, we always assume that the coefficients of the random
polynomial P satisfy
P λ = 0 = P λ = 0 = 0. (1)
0 n
That is, P does not vanish at the o(cid:8)rigin an(cid:9)d the(cid:8)degree(cid:9)of P does not drop. This condition can
be dropped at the cost of a somewhat longer wording of the main result.
2.2 The main theorem
At last, we are ready to state the main result of this note:
Theorem 1. Let P be a random polynomial of degree n > 2 with coefficients having the property
(Θ) and satisfying the non-degeneracy condition (1). Let L > 0 and A > 0. Then, with
probability at least 1 n−A, we have
−
sup N(Γ;P): Γ is L Lipschitz 6 C(A,L)V(P) log3n.
−
Here, C(A,L) is a posi(cid:8)tive value that depends only(cid:9)on the parameters A and L.
Note thatthereis nohopefor asimilarly strongnon-randomestimate: a construction, which
goes back to Bloch and Po´lya [1], allows one to construct a polynomial P of any degree n > 2
with V(P) = 2 and with at least n/logn positive zeroes.
p
2.3 A corollary for the case of i.i.d. coefficients
As an almost immediate corollary, we obtain
Corollary 2. Suppose that the coefficients of P are independent identically distributed random
variables satisfying the non-degeneracy condition (1). Then
sup E[N(Γ;P)]: Γ is L Lipschitz 6 C(L) log4n, n > 2.
−
In particular, E[N(R(cid:8);P)] 6 C log4n with a positiv(cid:9)e numerical constant C. As we have already
mentioned, the latter estimate will be improved in Part II by a different technique.
Proof. We use Theorem 1 with A = 1. Since the total number of zeroes of P on Γ cannot
exceed n, a set of probability n−1 can contribute to the expectation E[N(Γ;P)] by at most 1.
Therefore,
sup E[N(Γ;P)]: Γ is L Lipschitz 6 C(L)log3n E[V(P)].
− ·
To estimate the mea(cid:8)n E[V(P)], we note that V(P)(cid:9)equals the cardinality of the set of indices
ν 0,1,...,n such that, for some r (0, ),
∈ { } ∈ ∞
ν is the largest index satisfying λ rν = max λ rk, (2)
ν k
| | 06k6n| |
If (2) holds for some r (0,1), then
∈
λ > λ for each k 0,1,...,ν 1 .
ν k
| | | | ∈ { − }
2 Main results 5
By symmetry, the probability of this event does not exceed 1 . Similarly, if (2) holds for some
ν+1
r [1, ), then
∈ ∞
λ > λ for each k ν +1,ν +2,...,n ,
ν k
| | | | ∈ { }
and the probability of this event is 6 1 . Thus,
n−ν+1
E[V(P)] 6 2 1+ 1 + ... + 1 6 Clogn, n > 2,
2 n
proving the corollary. (cid:0) (cid:1)
2.4 A probabilistic lower bound for a random polynomial on an arc
The following result is the main tool needed for the proof of Theorem 1. Likely, it may be of
independent interest. Put
n
S(r,P) = λ rk.
k
| |
k=0
X
Theorem 3. Let P be a random polynomial of degree n > 2 with coefficients having the property
(Θ). Let m N, let r > 0, and let I R be an interval of length at most 2π. Then, for some
∈ ⊂
positive numerical constant c,
P max P(reiθ) 6 n−2 c I 6mS(r,P) 6 2−m.
θ∈I | | | |
n o
(cid:0) (cid:1)
The proof of this theorem will be given in Section 3.
2.5 The reduction principle
Our starting point is the following claim:
Lemma 4. Suppose that the coefficients of the random polynomial P possess the property (Θ).
Then, for any Borel set Λ Cn+1, we have
⊂
P λ Λ 6sup P υ Λ ,
∈ ∈
where the supremum is taken over(cid:8)all ra(cid:9)ndom var(cid:8)iables υ(cid:9): +, n+1 Cn+1 of the form
{ −} →
υσ = (υσ0,υσ1, ...,υσn) such that the random variables υ are independent, take the values υ±
0 1 n k k
with probability 1 and, for some a C,
2 ∈
1
υ+ υ− > υ+ a + υ− a , k 0,1,...,n . (3)
k − k 2 k − k − ∈ { }
(cid:12) (cid:12) (cid:2)(cid:12) (cid:12) (cid:12) (cid:12)(cid:3)
It is worth not(cid:12)ing that(cid:12)for ind(cid:12)epende(cid:12)nt (cid:12)real-valu(cid:12)ed random variables, this reduction was
used already by Kolmogorov in [7], where he proved a slightly weaker version of what is called
nowadays the Kolmogorov-Rogozin concentration inequality.
Proof. Let Ω bethe underlyingprobability space of λ′ and λ′′ in the definition of flip-invariance,
letΩ = Ω +, n+1 betheproductspacewiththeuniformdistributionover allsignsequences
×{ −}
σ = (σ ,σ ,...,σ ), and let λσ d=ef (λσ0,λσ1,...,λσn). Then λσ: Ω Cn+1 and, for each
σ e +,0 n1+1, thne random variables λσ0and1λ are enquidistributed. The→refore,
∈ { −}
e
PΩ λ Λ =PΩ×{+,−}n+1 λσ Λ 6 esssupP{+,−}n+1 λσ(ω) Λ .
{ ∈ } { ∈ } { ∈ }
ω∈Ω
It remains to observe that, for a.e. ω Ω, the random variable υσ = λσ(ω) satisfies (3) with
∈
the same value a as in the condition (Θ). Hence, the essential supremum on the RHS does not
exceed the supremum in the conclusion of the lemma.
3 Proof of Theorem 3 6
Thus,itsufficestoproveTheorems1and3foraspecialclass ofrandompolynomials. Hence,
in what follows, we assume that:
(a) the underlying probability space is +, n+1 with the uniform distribution over sign
{ −}
sequences, and, as above, we denote the elements of this space by σ = (σ ,σ ,...,σ );
0 1 n
(b) (λ±) are 2n+2 complex numbers, a is a complex number, and for each k 0,1,...,n ,
k ∈{ }
1
λ+ λ− > λ+ a + λ− a ;
k − k 2 k − k −
(cid:12) (cid:12) (cid:2)(cid:12) (cid:12) (cid:12) (cid:12)(cid:3)
(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)
(c) the random variables λ are independent and λ takes the values λ± with probability 1
k k k 2
each.
3 Proof of Theorem 3
The main ingredient of the proof of Theorem 3 is Tura´n’s lemma [10, Chapter 5, Lemma 1] (see
also [11, Chapter 1]):
Lemma 5. Let
m
p(t) = a eiℓkt, a C, ℓ Z,ℓ = ℓ for k =j.
k k k k j
∈ ∈ 6 6
k=1
X
Then for every interval I R of length at most 2π,
⊂
m
max p > b I m−1 a
k
I | | | | | |
k=1
(cid:0) (cid:1) X
with a positive numerical constant b.
Note that the conclusion of this lemma is usually stated in the form
max p > b I m−1 max p .
I | | | | [−π,π]| |
(cid:0) (cid:1)
Since
m m 1/2 1 π 1/2
a 6 √m a 2 = √m p(t)2dt 6 √m max p ,
k k
Xk=1| | (cid:16)Xk=1| | (cid:17) (cid:16)2π Z−π| | (cid:17) [−π,π]| |
the version we will be using readily follows from the usual one.
3.1 The case of few large coefficients
Given m N and r > 0, we assume that, for some a C,
∈ ∈
# k: λ a rk > δS(r;P) 6 2m (4)
k
| − |
and show that for every interval(cid:8)I R of length at most 2(cid:9)π,
⊂
max P(reiθ) > cn−1 b I 4m+1S(r;P) (5)
θ∈I | |
(cid:12) (cid:12) (cid:0) (cid:1)
provided that δ = c n−2 b I 4m(cid:12)+1 with(cid:12)a sufficiently small constant c .
1 1
| |
(cid:0) (cid:1)
3 Proof of Theorem 3 7
3.1.1 The polynomial P¯
Put P¯(z) = (1 z)P(z). We need this polynomial to get rid of the dependence on the value of
−
a. Note that when a = 0 this polynomial is not needed.
Claim 6.
1+r
S(r;P¯)> S(r;P), 0 < r < . (6)
2(n+1) ∞
Proof. First, assume that 0 < r 6 1. Then
n
S(r;P¯) = λ + λ λ rk + λ rn+1
0 k k−1 n
| | | − | | |
k=1
X
n
1
> (n+1)λ + (n+1 k)λ λ rk
0 k k−1
n+1 | | − | − |
h Xk=1 i
n n
1
> (n+1)λ + (n+1 k)λ rk (n+1 k)λ rk
0 k k−1
n+1 | | − | | − − | |
h Xk=1 Xk=1 i
n−1
1
= (n+1)λ + (n+1 k) (n k)r λ rk + λ rn n λ r
0 k n 0
n+1 | | − − − | | | | − | |
h Xk=1(cid:0) (cid:1) i
n−1
1
= (n+1 k) (n k)r λ rk + λ rn .
k n
n+1 − − − | | | |
hXk=0(cid:0) (cid:1) i
For 0< r 6 1, we have (n+1 k) (n k)r > 1. Thus, the RHS of the previous estimate is
− − −
n
1 0<r61 1+r
> λ rk > S(r;P).
k
n+1 | | 2(n+1)
k=0
X
Now, let 1 6 r < . Then
∞
n
S(r;P¯) = λ + λ λ rk + λ rn+1
0 k k−1 n
| | | − | | |
k=1
X
n
1
> k λ rk λ rk +(n+1)λ rn+1
k−1 k n
n+1 | | −| | | |
hXk=1 (cid:0) (cid:1) i
n
1
= (k+1)r k λ rk.
k
n+1 − | |
k=0
X(cid:0) (cid:1)
Since r > 1, we have (k+1)r k > r, and therefore, the RHS of the previous estimate is
−
n
r r>1 1+r
> λ rk > S(r;P),
k
n+1 | | 2(n+1)
k=0
X
proving the claim.
3.1.2 Proof of the lower bound (5) assuming (4)
First, we observe that
# k: λ¯ rk > 2δ(1+r)S(r;P) 6 4m+2, (7)
k
| |
(cid:8) (cid:9)
3 Proof of Theorem 3 8
where λ¯ are coefficients of the polynomial P¯. Indeed,
k
n
P¯(z) = λ + (λ λ )zk λ zn+1.
0 k k−1 n
− −
k=1
X
Suppose that for some k 1, ...,n and δ > 0,
∈
(cid:8) λ (cid:9)λ rk > 2δ(1+r)S(r;P).
k k−1
| − |
Then
λ a rk + λ a rk > 2δ(1+r)S(r;P).
k k−1
| − | | − |
That is, at least one of the following estimates holds: either λ a rk > δ(1 +r)S(r;P), or
k
| − |
λ a rk > δ(1+r)S(r;P), proving (7).
k−1
| − |
Now, we split the polynomial P¯ into large and small parts. The small part P¯sm will consists
of the terms λ¯ rk with
k
λ¯ rk 6 2δ(1+r)S(r;P).
k
| |
The rest goes to the large part P¯la, which is a sum of at most 4m+2 terms. Using Tura´n’s
lemma, we get
max P(reiθ) > (1+r)−1max P¯(reiθ)
θ∈I θ∈I
(cid:12)(cid:12) (cid:12)(cid:12) > (1+r)−1 max(cid:12)(cid:12) P¯la(rei(cid:12)(cid:12)θ) max P¯sm(reiθ)
θ∈I − θ∈I
> (1+r)−1h(b I(cid:12)(cid:12))4m+1S(r(cid:12)(cid:12);P¯la) (cid:12)(cid:12)S(r;P¯sm)(cid:12)(cid:12)i
| | −
h i
> (1+r)−1 (b I )4m+1 S(r;P¯) (n+1)2δ(1+r)S(r;P)
| | −
h
(cid:0) (n+1)2δ(1+r)S(r;P) (cid:1)
−
(6) 1 i
> (b I )4m+1 2(n+1)δ 2(n+1)δ S(r;P)
| | 2(n+1) − −
h (cid:16) (cid:17) i
1
> (b I )4m+1 4nδ 4nδ S(r;P).
| | 4n − −
h (cid:16) (cid:17) i
Choosing δ = c n−2 b I 4m+1 with a sufficiently small constant c , we see that the RHS of the
1 1
| |
previous estimate is > c n−1 b I 4m+1S(r;P), proving (5). ✷
(cid:0) 2(cid:1)
| |
(cid:0) (cid:1)
3.2 The dangerous configurations are rare
Fix an interval I R of length at most 2π and fix δ as above. Taking into account what we
⊂
have just proven, we see that in order to prove Theorem 3, we need to estimate the number of
sign sequences σ +, n+1 such that there exist at least 2m (6 n) indices k satisfying
∈ { −}
λσk a rk >δS(r;P) (8)
k −
(cid:12) (cid:12)
but still (cid:12) (cid:12)
max P(reiθ) 6 δ S(r;P) (9)
1
θ∈I
(cid:12) (cid:12)
with some positive δ1 δ to be chosen(cid:12) momen(cid:12)tarily. We call the corresponding sequence of
≪
signs σ dangerous and aim to show that the number of dangerous sequences does not exceed
2n+1−m.
3 Proof of Theorem 3 9
Takeanydangeroussignsequenceσ andanm-elementsubsetofthesetof“largecoefficients”
that appear in condition (8), and flip all the signs σ corresponding to this m-element subset.
k
Runningoverallpossiblem-elementssubsetsofthesetof“largecoefficients”ofagivendangerous
sign sequence σ, we obtain at least 2m > 2m different sign sequences. Claim 7 (given few lines
m
below) will yield that all new sign sequences obtained from all dangerous sign sequences σ are
(cid:0) (cid:1)
different, provided that the parameter δ is chosen as
1
δ = 1δ b I 2m−1. (10)
1 4 | |
(cid:0) (cid:1)
Therefore, with the choice of the parameters as in (10), the total number of all dangerous sign
sequences multiplied by 2m cannot exceed 2n+1. At the same time, for any non-dangerous σ,
m
we automatically have
(cid:0) (cid:1)
c
max P(reiθ) > 1 n−2 b I 4m+1 b I 2m−1S(r;P) > n−2 c I 6mS(r;P).
θ∈I 4 | | · | | | |
(cid:12) (cid:12) (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
Therefore, Th(cid:12)eorem 3(cid:12) follows if we prove the following claim:
Claim 7. Let σ +, n+1 be any sign sequence. Suppose that there exist two different m-
∈ { −}
element subsets U ,U 0,1,2,...n , U = U , so that the sets of flips corresponding to U
1 2 1 2 1
⊂ { } 6
and U turn σ into a dangerous sign sequence with all coefficients corresponding to flipped signs
2
becoming “large” as in condition (8). Then the parameter δ cannot be as small as in (10).
1
Proof. Once again, we will rely on Tura´n’s lemma. We fix the sign sequence σ and denote by
σ1, σ2 the flipped sign sequences, i.e.,
σ for k U ,
j k j
σ = − ∈
k (σk for k / Uj.
∈
ByP ,j = 1,2,wedenotethecorrespondingpolynomials. Choosingk U U andk U U ,
j 1 1 2 2 2 1
∈ \ ∈ \
we have
(9) (8) δ σj
max P (reiθ) 6 δ S(r;P ) 6 1 λ kj a rkj , j = 1,2.
θ∈I j 1 j δ kj −
(cid:12) (cid:12) (cid:12) (cid:12)
Therefore,
(cid:12) (cid:12) (cid:12) (cid:12)
max (P P )(reiθ) 6 δ1 λσk11 a rk1 + λσk22 a rk2 . (11)
θ∈I 1 − 2 δ k1 − k2 −
(cid:12) (cid:12) h(cid:12) (cid:12) (cid:12) (cid:12) i
On the other hand, the difference P P has at most 2m terms:
(cid:12) (cid:12)1 2(cid:12) (cid:12) (cid:12) (cid:12)
−
P P (z) = λσk1 λσk2 zk
1− 2 k − k
(cid:0) (cid:1) k∈UX1△U2(cid:0) (cid:1)
where, as usual, denotes the symmetric difference. For k U U , we have σ2 = σ1.
△ ∈ 1 △ 2 k − k
Then, by assumption (b) in Section 2.5,
σ1 σ2 1 σ1 σ2
λ k λ k > λ k a + λ k a , k U U .
k − k 2 k − k − ∈ 1△ 2
(cid:12) (cid:12) h(cid:12) (cid:12) (cid:12) (cid:12)i
In particular, this h(cid:12)olds for k(cid:12)= k ,k(cid:12) . There(cid:12)fo(cid:12)re, the R(cid:12)HS of (11) is
1 2
6 2δ1 λσk11 λσk21 rk1 + λσk12 λσk22 rk2 6 2δ1 S(r;P P ).
δ k1 − k1 k2 − k2 δ 1− 2
h(cid:12) (cid:12) (cid:12) (cid:12) i
If δ is as small as in ((cid:12)10), this con(cid:12)tradict(cid:12)s Tura´n’s lem(cid:12) ma applied to P P . This proves the
1 1 2
−
claim and finishes off the proof of Theorem 3.
4 Proof of Theorem 1 10
4 Proof of Theorem 1
4.1 Preliminaries
4.1.1
First, we observe that it suffices to prove Theorem 1 only for zeroes of P lying in the closed
unit disk z 6 1 . To get the result for the rest of the zeroes, all one needs is to consider the
{| | }
polynomial P∗(z) = znP(z−1).
4.1.2
Itwillbeconvenienttomaketheexponentialchangeofvariablez = e−2πw,w = t+is,0 6 t < ,
∞
and to deal with the exponential polynomial
n
Q(w) = P(z) = λ e−2πkw.
k
k=0
X
4.1.3
Put
h(t) = max log λ 2πkt , H(t) = eh(t).
k
06k6n | |−
(cid:0) (cid:1)
By ν(t) we denote the central index, that is, the largest of the indices ν, for which
log λ 2πνt > log λ 2πkt, k 0,1,2, ...n .
ν k
| |− | |− ∈{ }
Obviously, H(t) 6 S(e−2πt;P) 6 (n+1)H(t). This will allow us to replace S by H in our
estimates. The advantage of H over S is that the former has sharper transitions at the points
where the central index changes its value.
4.1.4
In the new notation, Theorem 3 says that given t > 0, given an interval I = [s′,s′′] of length
less than 1, and given a positive integer parameter m, there exists an event
Σ(t,I,m) +, n+1, with P Σ(t,I,m) 6 2−m
⊂ { −}
such that for every σ +, n+1 Σ(t,I,m), (cid:0) (cid:1)
∈ { −} \
max Q(t+is) > n−2(c I )6mH(t). (12)
s∈I | |
(cid:12) (cid:12)
This estimate is complemented by(cid:12)the obvio(cid:12)us upper bound
max Q(t+is) 6 (n+1)H(t). (13)
s
(cid:12) (cid:12)
(cid:12) (cid:12)
4.2 The test sets and exceptional sign sequences
n+1
Our exceptional event Σ +, will be a union of the events Σ(t,I,m) taken over a
⊂ −
certain finite sets of “test points” t and “test intervals” I. So we start by defining these sets.
(cid:8) (cid:9)