On the maximum value of polynomials with given degree and number of roots

Sergei V. Konyagin^{1}(Moscow), Vsevolod F. Lev (Haifa)

Abstract

We show that for any polynomial P (z) = a_{n}z^{n} + ⋯ + a_{0} ∈ ℂ[z] of degree n > 0 with exactly r distinct roots holds

max_{|z|=1} |P (z)| ≥ \sqrt{|a_{0}a_{n}|} e^{n/(4r)}.

Conjecturally, e^{n/(4r)} in the right-hand side can be replaced by 2^{n/r}.

An important characteristic of a polynomial with complex coefficients is the maximum of its absolute value on the unit circle. We write 𝕌 := {z ∈ ℂ: |z| = 1} and let

M(P) := max_{z∈𝕌}|P (z)|; P ∈ ℂ[z].

Our goal in this note is to estimate M(P) in terms of the degree of P and the number of distinct roots of P, which we denote by r(P).

Conjecture 1. Let P ∈ ℂ[z] be a polynomial of degree n = deg P > 0 with r = r(P) distinct roots. Then

M(P) ≥ \sqrt{|a_{0}a_{n}|} 2^{n/r},

where a_{n} and a_{0} are the leading coefficient and the constant term of P.

Example. Suppose that r and n are positive integers such that r divides n. Fix non-zero a ∈ ℂ and θ ∈ 𝕌 and set P (z) := a(z^{r} + θ)^{n/r}. Then deg P = n, r(P) = r, the leading coefficient of P is a, the constant term of P is aθ^{n/r}, and M(P) = |a| 2^{n/r}. This shows that (if true) the assertion of Conjecture 1 is, in a sense, the strongest possible.

Our main result is

Theorem 1. Let P ∈ ℂ[z] be a polynomial of degree n = deg P > 0 with r = r(P) distinct roots. Then

M(P) > \sqrt{|a_{0}a_{n}|} e^{n/(4r)},

where a_{n} and a_{0} are the leading coefficient and the constant term of P.

1Supported by Grants 02-01-00248 and 00-15-96109 from the Russian Foundation for Basic Research.

Before we prove Theorem 1 we want to extend our discussion on Conjecture 1. We will reformulate it in several equivalent forms and indicate some particular cases when the conjecture is valid.

Given a non-zero polynomial P ∈ ℂ[z] of degree n = deg P with roots z_{1}, …, z_{n}, all distinct from zero, we define

P^{∗}(z) := ∏_{j=1}^{n}(z − z_{j}/|z_{j}|);

that is, P^{∗} is obtained by “sliding” the roots of P onto the unit circle while preserving their arguments. Also, P^{∗} is normalized so that its leading coefficient equals one.

Plainly, we have r(P^{∗}) ≤ r(P).

Lemma 1. For any polynomial P ∈ ℂ[z] with leading coefficient a_{n} and constant term a_{0} ≠ 0 and any z ∈ 𝕌 we have |P (z)| ≥ \sqrt{|a_{0}a_{n}|} |P^{∗}(z)|.

Proof. Let z_{1}, …, z_{n} be the roots of P and let θ_{j}, φ_{j} ∈ [0, 2π) and r_{j} > 0 be defined by

z_{j} = r_{j}e^{iθ_{j}}, z = e^{i(φ_{j}+θ_{j})}; j = 1, …, n.

If φ_{j} ≠ 0 for j = 1, …, n, then

\frac{|P (z)|^{2}}{|P^{∗}(z)|^{2}} = |a_{n}|^{2} ∏_{j=1}^{n} |\frac{z − z_{j}}{z − z_{j}/|z_{j}|}|^{2}

= |a_{n}|^{2} ∏_{j=1}^{n} |\frac{e^{iφ_{j}} − r_{j}}{e^{iφ_{j}} − 1} |^{2}

= |a_{n}|^{2} ∏_{j=1}^{n} \frac{(1 + r_{j}^{2}) − 2r_{j} cos φ_{j}}{2(1 − cos φ_{j})}

≥ |a_{n}|^{2} ∏_{j=1}^{n} \frac{2r_{j}(1 − cos φ_{j})}{2(1 − cos φ_{j})}

= |a_{n}|^{2} ∏_{j=1}^{n} |z_{j}|

= |a_{n}||a_{0}|.

The result follows.

Assume for a moment that Conjecture 1 is established for monic polynomials with all roots on the unit circle. Then for any P (z) = a_{n}z^{n} + ⋯ + a_{0} ∈ ℂ[z] with a_{n}, a_{0} ≠ 0 using Lemma 1 we get

M(P) ≥ \sqrt{|a_{0}a_{n}|} M(P^{∗}) ≥ \sqrt{|a_{0}a_{n}|} 2^{n/r(P^{∗})} ≥ \sqrt{|a_{0}a_{n}|} 2^{n/r(P)}.

This shows that Conjecture 1 can be equivalently reformulated as follows.

Conjecture 1′. Let P ∈ ℂ[z] be a monic polynomial of degree n = deg P > 0 with r = r(P) distinct roots, all situated on the unit circle. Then M(P) ≥ 2^{n/r}.

Suppose that P (z) ∈ ℂ[z] satisfies the assumptions of Conjecture 1′; that is, P (z) = ∏_{j=1}^{r}(z − z_{j})^{n_{j}} with pairwise distinct z_{1}, …, z_{r} ∈ 𝕌 and positive integers n_{1}, …, n_{r} such that n_{1} + ⋯ + n_{r} = n. For j = 1, …, r set p_{j} := rn_{j}/n so that the condition |P (z)| ≥ 2^{n/r} can be rewritten as

∏_{j=1}^{r} |z − z_{j}|^{p_{j}} ≥ 2.  (1)

Observe that p_{1}, …, p_{r} are positive rational numbers satisfying p_{1} + ⋯ + p_{r} = r. Conversely, for any rational p_{1}, …, p_{r} > 0 with p_{1}+⋯+p_{r} = r one can find integers n_{1}, …, n_{r} > 0 so that p_{j} = rn_{j}/n (j = 1, …, r) and n_{1} + ⋯ + n_{r} = n. This shows that Conjecture 1′ is equivalent to the assertion that for any rational p_{1}, …, p_{r} > 0 satisfying p_{1} + ⋯ + p_{r} = r and any pairwise distinct z_{1}, …, z_{r} ∈ 𝕌 there exists z ∈ 𝕌 such that ∏_{j=1}^{r} |z − z_{j}|^{p_{j}} ≥ 2. In fact, from continuity considerations the requirements that p_{j} are rational and z_{j} are pairwise distinct can be dropped, and p_{j} > 0 can be relaxed to p_{j} ≥ 0. This gives yet another reformulation of our conjecture.

Conjecture 1′′. Let r be a positive integer. Then for any z_{1}, …, z_{r} ∈ 𝕌 and any p_{1}, …, p_{r} ≥ 0 such that p_{1} + ⋯ + p_{r} = r there exists z ∈ 𝕌 satisfying (1).

We note that Conjectures 1–1′′ hold true in the following cases.

(i) All roots of P are pairwise distinct: r = n. Indeed, define θ ∈ 𝕌 so that the complex numbers a_{0} and θ^{n}a_{n} have the same argument. Then

M(P) ≥ |\frac{1}{n} ∑_{j=0}^{n−1} P (θe^{2πij/n})| = |a_{0}| + |a_{n}| ≥ 2\sqrt{|a_{0}a_{n}|} = \sqrt{|a_{0}a_{n}|} 2^{n/r}.

(ii) More generally, all roots of P have the same multiplicity. For, if m is the common multiplicity of the roots, then we can write P (z) = (Q(z))^{m} for some Q(z) ∈ ℂ[z] such that all roots of Q are pairwise distinct. Now the claim follows from (i) in view of M(P) = (M(Q))^{m}, deg(P) = m deg(Q), and r(P) = r(Q).

(iii) The distinct roots of P^{∗} are equally spaced on the unit circle. Set ρ := r(P^{∗}) and ω := e^{2πi/ρ} and for j ∈ [1, ρ] define P_{j}(z) := P (ω^{j}z). Then P_{j}^{∗}(z) = ω^{−jn}P^{∗}(ω^{j}z) whence M(P_{j}^{∗}) = M(P^{∗}) and therefore

(M(P^{∗}))^{ρ} = M(P_{1}^{∗}) ⋯ M(P_{ρ}^{∗}) ≥ M(P_{1}^{∗} ⋯ P_{ρ}^{∗}).

However, the polynomial P_{1}^{∗} ⋯ P_{ρ}^{∗} has the same ρ distinct roots as P^{∗}, and the multiplicity of each root is n. By (ii) we conclude that M(P^{∗}) ≥ 2^{n/ρ} ≥ 2^{n/r}, hence M(P) ≥ \sqrt{|a_{0}a_{n}|} 2^{n/r} by Lemma 1.

(iv) All roots of P are situated on a half-line starting at the origin; that is, there exists φ ∈ [0, 2π) such that arg z_{0} = φ for any z_{0} satisfying P (z_{0}) = 0. Indeed, we have P^{∗}(z) = (z − e^{2πiφ})^{n} which yields M(P^{∗}) = 2^{n}, and application of Lemma 1 completes the argument.

(v) All roots of P are situated on two half-lines starting at the origin; that is, there exist φ_{1}, φ_{2} ∈ [0, 2π) such that φ_{1} ≠ φ_{2} and arg z_{0} ∈ {φ_{1}, φ_{2}} for any z_{0} satisfying P (z_{0}) = 0. In this case there are integers n_{1}, n_{2} > 0 such that n_{1} + n_{2} = n and P^{∗}(z) = (z − e^{2πiφ_{1}})^{n_{1}}(z − e^{2πiφ_{2}})^{n_{2}}. Suppose that φ ∈ [0, 2π) is so chosen that |e^{2πiφ} − e^{2πiφ_{j}}| ≥ \sqrt{2} (j = 1, 2). Then

M(P^{∗}) ≥ |P^{∗}(e^{2πiφ})| ≥ 2^{(n_{1}+n_{2})/2} ≥ 2^{n/r}

and it remains to apply Lemma 1.

In particular, these observations establish Conjectures 1–1′′ if P has at most two distinct roots, and also if all roots of P are real.

We conclude our note with the proof of Theorem 1. We derive the theorem from

Proposition 1. Suppose that P (z) = ∑_{j=0}^{n} a_{j}z^{j} ∈ ℂ[z] is a monic polynomial of degree n = deg P > 0 with r = r(P) distinct roots, all situated on the unit circle. Then

( ∑_{j=0}^{n} |a_{j}|^{2})^{1/2} ≥ e^{n/(4r)−1}.

To see how Proposition 1 implies Theorem 1, suppose that P ∈ ℂ[z] is a polynomial of degree n = deg P > 0 with r = r(P) distinct roots, constant term a_{0}, and leading coefficient a_{n}. Fix a positive integer m and let F (z) := (P (z))^{m} so that r(F) = r, the constant term of F is a_{0}^{m}, and the leading coefficient of F is a_{n}^{m}. Write F^{∗}(z) = ∑_{j=0}^{mn} b_{j}z^{j} (b_{0}, …, b_{nm} ∈ ℂ). By Lemma 1 and Proposition 1 as applied to the polynomial F^{∗} we get then

(M(P))^{m} = M(F) ≥ \sqrt{|a_{0}^{m}a_{n}^{m}|} M(F^{∗})

≥ (\sqrt{|a_{0}a_{n}|} )^{m} (∫_{0}^{1} |F^{∗}(e^{2πiφ})|^{2} dφ)^{1/2} = (\sqrt{|a_{0}a_{n}|} )^{m} (∑_{j=0}^{mn} |b_{j}|^{2})^{1/2}

≥ (\sqrt{|a_{0}a_{n}|} )^{m}e^{(mn)/(4r)−1}

and the assertion of the theorem follows. It remains to prove the proposition. We need yet another lemma.

Lemma 2. Let m be a positive integer, and suppose that z_{1}, …, z_{m} ∈ ℂ are the roots of the polynomial Q(z) = ∑_{j=0}^{m} b_{j}z^{j} ∈ ℂ[z]. Then

( ∑_{j=0}^{m} |b_{j}|^{2})^{1/2} ≥ |b_{m}| ∏_{j=1}^{m} max{1, |z_{j}|}.

Proof. This is a well-known property of Mahler’s measure.

Proof of Proposition 1. Set K := ( ∑_{j=0}^{n} |a_{j}|^{2})^{1/2} and let z_{1}, …, z_{n} ∈ ℂ be the roots of P. For any integer k ∈ [0, n] and real λ > 0 we have

\frac{1}{k!} P^{(k)}(λz) = ∑_{j=k}^{n} \binom{j}{k}a_{j}λ^{j−k}z^{j−k}.  (2)

We choose here λ := 1 − (k/n) so that

\binom{k}{k}λ^{k−k} ≤ \binom{k + 1}{k}λ^{(k+1)−k} ≤ ⋯ ≤ \binom{n}{k}λ^{n−k}  (3)

in view of

\begin{matrix}( & {j+1 \atop k} & ) \\ ( & {j \atop k} & ) \end{matrix} = \frac{j + 1}{j + 1 − k} = 1 + \frac{k}{j + 1 − k} ≥ 1 + \frac{k}{n − k} = λ^{−1}; j = k, …, n − 1.

Let y_{1}, …, y_{n−k} ∈ ℂ be the roots of P^{(k)} so that λ^{−1}y_{1}, …, λ^{−1}y_{n−k} are the roots of Q(z) := (1/k!)P^{(k)}(λz). Applying Lemma 2 and taking into account (2) and (3) we get

\binom{n}{k}λ^{n−k}K ≥ ( ∑_{j=k}^{n} (\binom{j}{k}|a_{j}| λ^{j−k})^{2})^{1/2} = \binom{n}{k}λ^{n−k} ∏_{j=1}^{n−k} max\{1, λ^{−1}|y_{j}|\},

whence

K ≥ ∏_{j=1}^{n−k} max{1, λ^{−1}|y_{j}|}.  (4)

We claim now that the sequences {z_{1}, …, z_{n}} and {y_{1}, …, y_{n−k}} have at least n − rk common terms. For, if n_{1}, …, n_{r} are the multiplicities of the distinct roots of P, then the number of common terms of the two sequences is at least

∑_{j=1}^{r} max{n_{j} − k, 0} ≥ ∑_{j=1}^{r}(n_{j} − k) = n − kr.

It follows that there are at least n − kr indices j ∈ [1, n − k] such that |y_{j}| = 1, and then from (4) we obtain

K ≥ λ^{−(n−kr)} > e^{\frac{k}{n} (n−kr)}.  (5)

To optimize we choose k := ⌊n/(2r)⌋ and notice that then

\frac{k}{n} (n − kr) = k − \frac{k^{2}r}{n} > \frac{n}{2r} − 1 − \frac{n}{4r} = \frac{n}{4r} − 1.  (6)

The result now follows from (5) and (6).

Remark 1. Taking λ := 1 − (αk/n) with a suitable α > 1 in the proof of Proposition 1 one can improve the estimate of Theorem 1 to

M(P) ≥ \sqrt{|a_{0}a_{n}|} c^{n/r}

with some c > e^{1/4}.

Remark 2. The method we used in the proof of Proposition 1 can be applied to show that for any polynomial P (z) = ∑_{j=0}^{n} a_{j}z^{j} ∈ ℂ[z] of degree n = deg P > 0 with r = r(P) distinct roots holds

( ∑_{j=0}^{n} |a_{j}|^{2})^{1/2} ≥ \sqrt{|a_{0}a_{n}|} e^{n/(8r)−1/2}.

That is, the assumptions that P is monic and has all roots on the unit circle can be dropped if the exponent n/(4r) − 1 is replaced by n/(8r) − 1/2.

Moscow Lomonosov State University, Department of Mechanics and Mathematics Haifa University at Oranim, Department of Mathematics