Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 27, 2002, 173–182

REVERSE MARKOV INEQUALITY

Norman Levenberg and Evgeny A. Poletsky

University of Auckland, Department of Mathematics Private Bag 92019, Auckland, New Zealand Syracuse University, Department of Mathematics 215 Carnegie Hall, Syracuse, NY 13244, U.S.A.; eapolets@mailbox.syr.edu

Abstract. Let K be a compact convex set in C. For each point z_{0} ∈ ∂K and each holomorphic polynomial p = p(z) having all of its zeros in K, we prove that there exists a point z ∈ K with |z − z_{0}| ≤ 20 diam K/\sqrt{deg p} such that

|p′(z)| ≥ \frac{(deg p)^{1/2}}{20(diam K)} |p(z_{0})|;

i.e., we have a pointwise reverse Markov inequality. In particular,

∥p′∥_{K} ≥ \frac{(deg p)^{1/2}}{20(diam K)} ∥p∥_{K}.

1. Introduction

Let K be a compact set in the complex plane C and let V_{K}(z) be the extremal function of K, i.e.,

(1)  V_{K}(z) = max[0, \frac{1}{deg p} sup log |p(z)|],

where the supremum is taken over all nonconstant holomorphic polynomials p = p(z) with supremum norm ∥p∥_{K} = sup_{z∈K} |p(z)| ≤ 1. Suppose that the function V_{K} is Hölder continuous with exponent 0 < a ≤ 1. Then from the Bernstein– Walsh inequality

(2)  |p(z)| ≤ ∥p∥_{K} exp [deg pV_{K}(z)],

which follows from the definition of V_{K}(z) in (1) (cf., [R]), and the Cauchy estimates, one obtains a Markov inequality: this estimates the size of the derivative

2000 Mathematics Subject Classification: Primary 41A17, 30C10; Secondary 26C05, 26D05. E.A. Poletsky was partially supported by NSF Grant DMS-9804755.

of a polynomial p on K via the sup-norm ∥p∥_{K} of p, its degree, and the Hölder exponent a of V_{K}; namely,

(3)  ∥p′∥_{K} ≤ C(K)(deg p)^{1/a}∥p∥_{K},

where C(K) is a constant depending only on K. We outline a proof of (3); cf., [PP, Remark 3.2]. Given a polynomial p of degree n, take z_{0} ∈ K with |p′(z_{0})| = ∥p′∥_{K}. Apply the Cauchy estimates on the disk of radius r = n^{−1/a} centered at z_{0} to obtain

∥p′∥_{K} ≤ n^{1/a}∥p∥_{K_{r}}

where K_{r} := {z ∈ C : dist(z, K) ≤ r}. Since V_{K} is Hölder continuous with exponent a, (2) implies that

∥p∥_{K_{r}} ≤ ∥p∥_{K}(1 + Mr^{a})^{n}

for some constant M = M(K). The choice of r = n^{−1/a} then yields the result. For example, if K is the unit disk, then (S.N. Bernstein)

∥p′∥_{K} ≤ (deg p)∥p∥_{K}

and if K is the interval [−1, 1], then (A.A. Markov)

∥p′∥_{K} ≤ (deg p)^{2}∥p∥_{K}

(cf., [Lo]). Note that in these two examples we have precise knowledge of the constant: C(K) = 1, and this is sharp. In general, controlling C(K) is difficult.

It is natural to ask whether one can improve the Markov inequality for some natural subclasses of polynomials. Indeed, P. Lax [L] proved a conjecture of P. Erdös that if all the zeros of p lie outside the open unit disk Δ, then

∥p′∥Δ̄ ≤ \frac{1}{2}deg p∥p∥Δ̄.

An example of p(z) = z^{n} + 1 shows that this is best possible.

However, in 1939, P. Turán [T] showed that if all the zeros of p lie in Δ̄, then a reverse Markov inequality holds:

(4)  ∥p′∥Δ̄ ≥ \frac{1}{2}deg p∥p∥Δ̄.

Again the same example of p(z) = z^{n} + 1 shows that it is sharp. Thus for this class the Markov inequality cannot be significantly improved.

Turán also proved in [T] that for the interval I = [−1, 1] and polynomials with all their zeros in I, the reverse Markov inequality occurs in the following form:

(5)  ∥p′∥_{I} ≥ \frac{1}{6}(deg p)^{1/2}∥p∥_{I}.

In this note, we are interested in a general form of the reverse Markov inequality for a polynomial p having all of its zeros in K:

(6)  ∥p′∥_{K} ≥ C(K)(deg p)^{b}∥p∥_{K}.

Note that any polynomial p(z) = cz^{n} + 1, where |c| is sufficiently small, provides a counterexample to any form of the reverse Markov inequality for polynomials with zeros outside K.

The inequality (6) contains two parameters: C(K) and b. Since the holomorphic polynomials p with zeros in K are in one-to-one correspondence with the holomorphic polynomials p̃ with zeros in aK := {az ∈ C : z ∈ K} via p̃(z) = p(z/a), C(K) is inversely proportional to the diameter of K. We are interested in classes of compact sets K for which there is no other dependence on K for some value of b. Indeed, we show that for the class of R-circular sets, to be defined in the next section, which includes disks as well as circular arcs, the reverse Markov inequality holds with b = 1. For general convex sets we show that b = \frac{1}{2} . In both cases the value of b is best possible provided that C(K) depends only on the diameter of K.

Moreover, we prove a pointwise version of (6):

|p′(z)| ≥ C(K)(deg p)^{b}|p(z)|,

where, in the case of R-circular sets, such an inequality is valid at every point z ∈ ∂K; while, in the case of general convex sets, given any point z_{0} ∈ ∂K, we can find z ∈ K lying within a distance 20 diam K/\sqrt{deg p} from z_{0} where the above inequality holds.

Throughout the paper the following function will play a major role: given a polynomial p(z) = c(z − z_{1}) ⋯ (z − z_{n}), we define

(7)  ϕ_{p}(z) = ϕ(z) = ∑_{j=1}^{n}(z − z_{j})^{−1}.

In the polynomial p (and thus in (7)), we allow repeated roots. Note that

p′(z) = p(z)ϕ(z).

2. Reverse Markov for R-circular sets

We begin with the following result.

Proposition 2.1. Let K ⊂ C be compact. Suppose that z ∈ ∂K has the property that there exists a circle of radius R = R(z, K) passing through z such that the closed disk it bounds contains K. Then for each complex polynomial p having all of its zeros in K,

|p′(z)| ≥ \frac{deg p}{2R}|p(z)|.

Proof. Let p have degree n and zeros z_{1}, …, z_{n} in K (with repeated zeros listed as often as they occur). Since p′(z) = p(z)ϕ(z), it suffices to show that |ϕ(z)| ≥ n/(2R) for z ∈ ∂K. Rotating and translating the z -variable (which does not affect the absolute value of the derivative), we may assume that z = R and that the closed disk Δ_{R} of radius R centered at zero contains K. Then

|ϕ(z)| = |∑_{j=1}^{n}(R − z_{j})^{−1}|

and the conformal mapping w = 1/(R − z) maps Δ_{R} onto the half-plane Re w ≥ 1/(2R). Thus

|ϕ(z)| ≥ ∑_{j=1}^{n} Re \frac{1}{R − z_{j}} ≥ \frac{n}{2R}

and we have

|p′(z)| ≥ \frac{n}{2R}|p(z)|

for z ∈ ∂K.

Motivated by the proposition, we make the following definition. A compact set K ⊂ C will be called R-circular if for every point z ∈ ∂K there is a circle of radius R passing through z such that the closed disk it bounds contains K.

Theorem 2.2. If K is a compact R-circular set in C, then for each complex polynomial p having all of its zeros in K, and for every point z ∈ ∂K

|p′(z)| ≥ \frac{deg p}{2R}|p(z)|.

As a corollary we obtain the first result of Turán.

Corollary 2.3. For a complex polynomial p having all of its zeros in Δ̄, and for every point z ∈ ∂Δ,

|p′(z)| ≥ \frac{1}{2}deg p|p(z)|.

In particular,

∥p′∥Δ̄ ≥ \frac{1}{2}deg p∥p∥Δ̄.

Remark. This inequality is not valid when the zeros lie very close to ∂Δ but outside Δ̄. Indeed, take p(z) = z^{n} − n^{2} . The zeros of p are the n-th roots of n^{2} and hence all have modulus n^{2/n} which tends to 1 as n → ∞. However, ∥p∥Δ̄ = n^{2} + 1 while ∥p′∥Δ̄ = n.

Remark. There exist non-convex compact sets satisfying the hypothesis of the corollary. For example, let K be an arc of a circle. But as we will see right now this estimate fails on an interval.

Reverse Markov inequality 177

3. Reverse Markov inequality for convex sets What happens for general convex compact sets K? First of all, we simplify

the proof of Turan’s second result (5) concerning the real interval I := [−1, 1].

Proposition 3.1. Let p = p(x) be a real polynomial of degree n with real zeros x_{1}, …, x_{n} ∈ I. Then

(8)  ∥p′∥_{I} ≥ \frac{\sqrt{n}}{2\sqrt{e}}∥p∥_{I}.

Proof. Order the zeros −1 ≤ x_{1} ≤ ⋯ ≤ x_{n} ≤ 1. Fix x_{0} ∈ I with |p(x_{0})| = ∥p∥_{I}. We may assume that p(x_{0}) > 0. For some m ∈ {1, …, n − 1}, x_{0} ∈ (x_{m}, x_{m+1}) or else x_{0} ∈ [0, x_{1}) ∪ (x_{n}, 1].

In this last case, Proposition 2.1 tells us that

|p′(x_{0})| ≥ \frac{1}{2}np(x_{0}).

Thus we may assume that x_{0} ∈ (x_{m}, x_{m+1}).

Since p′ = pϕ, for x ∈ (x_{m}, x_{m+1}) we have

p(x) = p(x_{0}) exp(∫_{x_{0}}^{x} ϕ(t) dt).

Hence

p′(x) = p(x_{0})ϕ(x) exp(∫_{x_{0}}^{x} ϕ(t) dt)

or, integrating by parts,

(9)  p′(x) = p(x_{0})ϕ(x) exp((x − x_{0})ϕ(x) − ∫_{x_{0}}^{x}(t − x_{0})ϕ′(t) dt).

Now since |x − x_{j}| ≤ 2,

(10)  ϕ′(x) = − ∑_{j=1}^{n} \frac{1}{(x − x_{j})^{2}} ≤ −\frac{n}{4}.

Therefore, for x_{0} ≤ x ≤ x_{m+1} we have

(11)  ϕ(x) ≤ −\frac{1}{4}n(x − x_{0})

because ϕ(x_{0}) = 0.

Since ϕ(x) approaches −∞ as x approaches x_{m+1} from the left, there exists some point x ∈ (x_{0}, x_{m+1}) at which ϕ(x) = −1/(x − x_{0}). From (11), we conclude that 1/(x − x_{0}) ≥ \frac{1}{4}n(x − x_{0}) at this point; i.e.,

(12)  x − x_{0} ≤ 2/\sqrt{n}.

Thus, at this point x, since ϕ(x) = −1/(x − x_{0}), using (9) we obtain

p′(x) = −\frac{p(x_{0})}{x − x_{0}} exp(−1 − ∫_{x_{0}}^{x}(t − x_{0})ϕ′(t) dt).

By (10) and (12) the integral

∫_{x_{0}}^{x}(t − x_{0})ϕ′(t) dt ≤ −\frac{n}{4} ∫_{x_{0}}^{x}(t − x_{0}) dt = −\frac{n(x − x_{0})^{2}}{8} .

Thus

|p′(x)| ≥ \frac{p(x_{0})}{x − x_{0}} exp(−1 + \frac{n(x − x_{0})^{2}}{8} ).

The function of x in the right side of the inequality above is decreasing when 0 < x − x_{0} ≤ 2/\sqrt{n} and attains its minimum when x − x_{0} = 2/\sqrt{n}. Therefore

|p′(x)| ≥ \frac{p(x_{0})\sqrt{n}}{2\sqrt{e}}.

The polynomial p(x) = (x^{2} − 1)^{n/2} (n even) shows that the \frac{1}{2} power of n is the correct exponent in this inequality. However, the constant is not sharp. Indeed, Erod [E] proved a sharp result for each degree.

We proceed to show that a reverse Markov inequality with exponent \frac{1}{2} is valid for general convex compact sets.

Theorem 3.2. For any convex compact set K in C, any polynomial p = p(z) of degree n with all of its zeros in K, and any point z_{0} ∈ ∂K, there is a point z ∈ K with |z − z_{0}| ≤ 20(diam K)/\sqrt{n} such that

|p′(z)| ≥ \frac{\sqrt{n}}{20(diam K)}|p(z_{0})|.

In particular,

∥p′∥_{K} ≥ \frac{\sqrt{n}}{20(diam K)}∥p∥_{K}.

Proof. For simplicity, we assume the diameter of K is 1. Fix p = p(z) of degree n with all of its zeros z_{1}, …, z_{n} in K and let z_{0} ∈ ∂K. We may assume that z_{0} = 0. We want to prove that there is a constant C > 0—we will see that C = 1/20 will work—such that

|p′(z)| ≥ C\sqrt{n} |p(0)|

for some point z ∈ K with |z| ≤ r_{n} := 1/(C\sqrt{n} ). We may assume that the set K lies in the upper half plane H. Then the points w_{k} := −1/z_{k} lie in H as well. We fix the angle α = π/12; note that

\frac{1}{4} < sin α < \frac{1}{3}.

We divide H into the sector S_{1} := {w ∈ H : α < arg w < π −α} and S_{2} = H \S_{1} . As before, we set

ϕ(z) := ∑_{k=1}^{n} \frac{1}{z − z_{k}} = ϕ_{1}(z) + ϕ_{2}(z),

where

ϕ_{1}(z) = ∑′ \frac{1}{z − z_{k}} and ϕ_{2}(z) = ∑′′ \frac{1}{z − z_{k}} .

Here ∑′ denotes the sum over k with w_{k} ∈ S_{1} and ∑′′ denotes the sum over k with w_{k} ∈ S_{2} .

Let n_{1} be the number of of points w_{k} in S_{1} and let n_{2} be the number of points w_{k} in S_{2} . Since Im w_{k} ≥ 0, Im w_{k} ≥ |w_{k}| sin α when w_{k} ∈ S_{1} , and since the diameter of K is 1 implies |w_{k}| ≥ 1, we see that

(13)  |ϕ(0)| ≥ ∑_{k=1}^{n} Im w_{k} ≥ sin α∑′|w_{k}| ≥ \frac{1}{4}a ≥ \frac{1}{4}n_{1},

where a = ∑′|w_{k}|.

We let d denote the distance from z_{0} = 0 to the nearest zero of p; i.e., d = min{|z_{1}|, …, |z_{n}|}. Let t = d sin α. We want to estimate the values of

ϕ′(z) = ϕ′_{1}(z) + ϕ′_{2}(z) = −∑′ \frac{1}{(z − z_{k})^{2}} − ∑′′ \frac{1}{(z − z_{k})^{2}}

when |z| ≤ t.

Note that for such z,

|z_{k}|(1 − sin α) ≤ |z − z_{k}| ≤ |z_{k}|(1 + sin α).

Hence

(14) \begin{aligned}|ϕ′_{1}(z)| & = |∑′ \frac{1}{(z − z_{k})^{2}} | ≤ ∑′ \frac{1}{|z − z_{k}|^{2}} \\ & ≤ \frac{1}{(1 − sin α)^{2}} (∑′ \frac{1}{|z_{k}|^{2}} ) ≤ \frac{1}{(1 − sin α)^{2}} (∑′ \frac{1}{|z_{k}|})^{2} ≤ \frac{9a^{2}}{4} . \end{aligned}

Next we consider

− Re ϕ′_{2}(z) = Re ∑′′ \frac{1}{(z − z_{k})^{2}} .

If |z| ≤ t and w_{k} ∈ S_{2} , then | arg 1/z_{k}^{2}| does not exceed 2α. Since |zw_{k}| ≤ sin α, we see that | arg(1 + zw_{k})| ≤ α and

β_{k}(z) := |arg \frac{1}{(z − z_{k})^{2}} | = |arg \frac{1}{z_{k}^{2}(1 + zw_{k})^{2}} | ≤ 4α.

Now

Re \frac{1}{(z_{k} − z)^{2}} = \frac{1}{|z_{k} − z|^{2}} cos β_{k}(z) ≥ \frac{1}{|z_{k}|^{2}(1 + sin α)^{2}} cos 4α.

Thus

(15)  Re ∑′′ \frac{1}{(z − z_{k})^{2}} ≥ \frac{cos 4α}{(1 + sin α)^{2}} ∑′′ \frac{1}{|z_{k}|^{2}} ≥ \frac{9}{32}∑′′ \frac{1}{|z_{k}|^{2}} .

Recall again that the diameter of K is 1 implies |w_{k}|^{2} ≥ 1, so that

(16)  ∑′′ \frac{1}{|z_{k}|^{2}} ≥ n_{2}.

Plugging (16) into (15) we obtain

(17)  − Re ϕ′_{2}(z) = Re ∑′′ \frac{1}{(z − z_{k})^{2}} ≥ \frac{9}{32}n_{2}.

Fixing a constant C > 0 which we will specify later, recall that our goal is to prove that |p′(z)| ≥ C\sqrt{n} |p(0)| for some point z ∈ K with |z| ≤ r_{n} = 1/(C\sqrt{n} ) (the conclusion of the theorem is that we can take C = 1/20). First of all, we note that it suffices to consider the case when d > r_{n}. For if d ≤ r_{n}, then |z_{k}| ≤ 1/(C\sqrt{n} ) for some k and |p′(z)| ≥ C\sqrt{n} |p(0)| for some point z ∈ [0, z_{k}]; this interval lies in K by convexity of K.

If |ϕ(0)| > C\sqrt{n}, then |p′(0)| = |p(0)ϕ(0)| > C\sqrt{n} |p(0)| and the desired inequality is true. Thus we may assume that |ϕ(0)| ≤ C\sqrt{n}, and, by (13), that both \frac{1}{4}a and \frac{1}{4}n_{1} are less than C\sqrt{n} .

Let t_{n} = r_{n} sin α. For the sake of obtaining a contradiction, we suppose that

(18)  |p′(z)| < C\sqrt{n} |p(0)| for all z ∈ K satisfying |z| < r_{n}.

Then for |z| < t_{n} we find that

|p(z)| ≥ |p(0)| − |∫_{0}^{z} p′(ζ) dζ| > |p(0)|(1 − sin α) > \frac{2}{3}|p(0)|.

We now take any point z′ ∈ K with |z′| = sin α/(C\sqrt{n} ). By the convexity of K and the assumption d > r_{n} > t_{n} = |z′|, such points exist. Since n_{2} = n − n_{1} ≥ n − 4C\sqrt{n} , it follows from (17) and the inequality sin α > \frac{1}{4} that

(19)  |∫_{0}^{z′} ϕ′_{2}(z) dz| ≥ \frac{9}{32}(n − 4C\sqrt{n} )\frac{sin α}{C\sqrt{n}} = \frac{9}{32}(\frac{\sqrt{n}}{4C} − 1).

Equation (14) and the inequalities a < 4C\sqrt{n} , |z′| < 1/[3C\sqrt{n}] give

(20)  |∫_{0}^{z′} ϕ′_{1}(z) dz| ≤ \frac{9a^{2}}{4} |z′| < 12C\sqrt{n} .

Now

|ϕ(z′)| ≥ |∫_{0}^{z′} ϕ′_{2}(z) dz| − |∫_{0}^{z′} ϕ′_{1}(z) dz| − |ϕ(0)|.

Plugging (19) and (20) and the assumption |ϕ(0)| < C\sqrt{n} (from (18) into the latter inequality we obtain

|ϕ(z′)| > \frac{9}{32}(\frac{\sqrt{n}}{4C} − 1) − 12C\sqrt{n} − C\sqrt{n} = \sqrt{n}(\frac{9}{128C} − 13C) − \frac{9}{32}

and

|p′(z′)| = |p(z′)| · |ϕ(z′)| > \frac{2}{3}[\sqrt{n}(\frac{9}{128C} − 13C) − \frac{9}{32}]|p(0)|.

If C = 1/20 then

|p′(z′)| > .05\sqrt{n} |p(0)| = C\sqrt{n} |p(0)|

and this contradicts our assumption (18) since

|z′| = sin α/(C\sqrt{n} ) < 1/[3C\sqrt{n} ] < 1/[C\sqrt{n} ] = r_{n}.

We end this section (and the paper) with a family of examples to show that for a star-shaped compact set the exponent b in a reverse Markov inequality can be arbitrarily small.

Proposition 3.3. Given ε > 0, there exists a compact, star-shaped set K ⊂ C such that there is no constant C > 0 with

∥p′∥_{K} ≥ C(deg p)^{ε}∥p∥_{K}

for all polynomials p.

182 Norman Levenberg and Evgeny A. Poletsky

Proof. Fix a positive integer m which will be chosen appropriately later and let z_{1}, …, z_{m} be the m-th roots of unity. Consider the set consisting of m “spokes of a wheel”

K_{m} := {tz_{j} : 0 ≤ t ≤ 1, j = 1, …, m}.

For a positive integer n, let p(z) = (z^{m} − 1)^{n}. Then deg p = nm; ∥p∥_{K_{m}} = 1; and a calculation shows that

∥p′∥_{K_{m}} = mn(\frac{m − 1}{mn − 1})^{(m−1)/m}[\frac{m − 1}{mn − 1} − 1]^{n}.

For n large,

∥p′∥_{K_{m}} ≍ m^{2−1/m}e^{−1+1/m}n^{1/m}.

Thus, taking m > 1/ε proves the proposition.

References

[E] Erod, J.: Bizonyus polinomok maximumanak. - Mat. Fiz. Lapok. 46, 1939, 58–82.

[L] Lax, P.D.: Proof of a conjecture of P. Erdös on the derivative of a polynomial. - Bull. Amer. Math. Soc. 50, 1944, 509–513.

[Lo] Lorentz, G.G.: Approximation of Functions. - Holt, Rinehart and Winston, New York, 1966.

[PP] Pawlucki, W., and W. Pleśniak: Markov’s inequality and C^{∞} functions on sets with polynomial cusps. - Math. Ann. 275, 1986, 467–480.

[R] Ransford, T.: Potential Theory in the Complex Plane. - Cambridge University Press, Cambridge, 1995.

[T] Turán, P.: Über die Ableitung von Polynomen. - Comp. Math. 7, 1939, 89–95.

Received 21 November 2000