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7

Inequalities for Rational Function Spaces

Overview

Precise Markov- and Bernstein-type inequalities are given for various classes of rational functions in the first section of this chapter. Extensions of the inequalities of Lax, Schur, and Russak are also presented, as are inequalities for self-reciprocal polynomials. The second section of the chapter is concerned with metric inequalities for polynomials and rational functions.

7.1 Inequalities for Rational Function Spaces

Sharp extensions of most of the polynomial inequalities of Section 5.1 are established for rational function spaces on K := ℝ (mod 2π), on the interval [−1, 1], on the unit circle of ℂ, and on the real line. The classical inequalities of Section 5.1 are then recovered as limiting cases. A sharp extension of Lax’s inequality is also given. Essentially sharp Markov- and Bernsteintype inequalities for self-reciprocal and antiself-reciprocal polynomials are presented in the exercises.

Let D := {z ∈ ℂ : |z| < 1} and ∂D := {z ∈ ℂ : |z| = 1}, as before. We study the rational function spaces:

T_{n}(a_{1}, a_{2}, …, a_{2n}; K) := \{\frac{t(θ)}{∏_{k=1}^{2n} | sin((θ − a_{k})/2)|} : t ∈ T_{n}\}

7.1 Inequalities for Rational Function Spaces 321

and

T_{n}^{c}(a_{1}, a_{2}, …, a_{2n}; K) := \{\frac{t(θ)}{∏_{k=1}^{2n} sin((θ − a_{k})/2)} : t ∈ T_{n}^{c}\}

on K with a_{1}, a_{2}, …, a_{2n} ∈ ℂ \ ℝ;

P_{n}(a_{1}, a_{2}, …, a_{n}; [−1, 1]) := \{\frac{p(x)}{∏_{k=1}^{n} |x − a_{k}|} : p ∈ P_{n}\}

and

P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; [−1, 1]) := \{\frac{p(x)}{∏_{k=1}^{n}(x − a_{k})} : p ∈ P_{n}^{c}\}

on [−1, 1] with a_{1}, a_{2}, …, a_{n} ∈ ℂ \ [−1, 1];

P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; ∂D) := \{\frac{p(z)}{∏_{k=1}^{n}(z − a_{k})} : p ∈ P_{n}^{c}\}

on ∂D with a_{1}, a_{2}, …, a_{n} ∈ ℂ \ ∂D; and

P_{n}(a_{1}, a_{2}, …, a_{n}; ℝ) := \{\frac{p(x)}{∏_{k=1}^{n} |x − a_{k}|} : p ∈ P_{n}\}

and

P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; ℝ) := \{\frac{p(z)}{∏_{k=1}^{n}(z − a_{k})} : p ∈ P_{n}^{c}\}

on ℝ with a_{1}, a_{2}, …, a_{n} ∈ ℂ \ ℝ.

The Chebyshev polynomials T̃_{n}, Ũ_{n}, and

V := (cos α)T̃_{n} + (sin α)Ũ_{n} , α ∈ K

for the rational function space T_{n}(a_{1}, a_{2}, …, a_{2n}; K) are defined in E.3 of Section 3.5, and they play a central role in this section.

322 7. Inequalities for Rational Function Spaces

Theorem 7.1.1 (Bernstein-Szegő-Type Inequality on K). Given

(a_{k})_{k=1}^{2n} ⊂ ℂ \ ℝ , Im(a_{k}) > 0 ,

let

B̃_{n}(θ) := \frac{1}{2} ∑_{k=1}^{2n} \frac{1 − |e^{ia_{k}}|^{2}}{|e^{ia_{k}} − e^{iθ}|^{2}} .

Then

f′(θ)^{2} + B̃_{n}^{2}(θ)f^{2}(θ) ≤ B̃_{n}^{2}(θ)∥f∥_{K}^{2} , θ ∈ K

for every f ∈ T_{n}(a_{1}, a_{2}, …, a_{2n}; K).

Equality holds if and only if either θ is a maximum point of |f| (that is, f(θ) = ±∥f∥_{K}) or f is a linear combination of T̃_{n} and Ũ_{n} (with real coefficients) as defined in E.3 of Section 3.5.

Corollary 7.1.2 (Bernstein-Type Inequality on K, Real Case). Given

(a_{k})_{k=1}^{2n} ⊂ ℂ \ ℝ , Im(a_{k}) > 0 ,

let the Bernstein factor B̃_{n} be defined as in Theorem 7.1.1. Then

|f′(θ)| ≤ B̃_{n}(θ)∥f∥_{K} , θ ∈ K

for every f ∈ T_{n}(a_{1}, a_{2}, …, a_{2n;}K).

Equality holds if and only if f is a linear combination of T̃_{n} and Ũ_{n} (with real coefficients) as defined in E.3 of Section 3.5, and f(θ) = 0.

Theorem 7.1.1 and Corollary 7.1.2 can be easily obtained from the extension of Theorem 3.5.3 given by E.3 of Section 3.5, which gives explicit formulas for the Chebyshev polynomials for these classes T_{n}(a_{1}, a_{2}, …, a_{2n}; K). The arguments are outlined in E.1.

The following two results can be obtained from Theorem 7.1.1 and Corollary 7.1.2 by the substitution x = cos θ; see E.2.

Corollary 7.1.3 (Bernstein-Szegő-Type Inequality on [−1, 1]). Associated with (a_{k})_{k=1}^{n} ⊂ ℂ \ [−1, 1], let the Bernstein factor B_{n} be defined by

B_{n}(x) := ∑_{k=1}^{n} Re (\frac{\sqrt{a_{k}^{2} − 1}}{a_{k} − x}) ,

where the choice of \sqrt{a_{k}^{2} − 1} is determined by |a_{k} − \sqrt{a_{k}^{2} − 1}| < 1. Then

(1 − x^{2})f′(x)^{2} + B_{n}^{2}(x)f^{2}(x) ≤ B_{n}^{2}(x)∥f∥_{[−1,1]}^{2} , x ∈ [−1, 1]

for every f ∈ P_{n}(a_{1}, a_{2}, …, a_{n}; [−1, 1]).

Equality holds if and only if either x is a maximum point of |f| (that is, f(x) = ±∥f∥_{[−1,1]}) or f = cT_{n} with c ∈ ℝ, where T_{n} is defined as in Section 3.5.

Corollary 7.1.4 (Bernstein-Type Inequality on [−1, 1], Real Case). Given (a_{k})_{k=1}^{n} ⊂ ℂ \ [−1, 1], let the Bernstein factor B_{n} be defined as in Corollary 7.1.3. Then

|f′(x)| ≤ \frac{B_{n}(x)}{\sqrt{1 − x^{2}}} ∥f∥_{[−1,1]} , x ∈ (−1, 1)

for every f ∈ P_{n}(a_{1}, a_{2}, …, a_{n}; [−1, 1]).

Equality holds if and only if f = cT_{n} with c ∈ ℝ, where T_{n} is defined as in Section 3.5, and f(x) = 0. (Note that B_{n}(x) > 0 for every x ∈ (−1, 1).)

Our next result follows from Theorem 7.1.1; see the hints to E.3.

Corollary 7.1.5 (Bernstein-Szegő-Type Inequality on ℝ). Given

(a_{k})_{k=1}^{n} ⊂ ℂ \ ℝ , Im(a_{k}) > 0 ,

let the Bernstein factor B_{n} be defined by

B_{n}(x) := ∑_{k=1}^{n} \frac{Im(a_{k})}{|x − a_{k}|^{2}} .

Then

f′(x)^{2} + B_{n}^{2}(x)f^{2}(x) ≤ B_{n}^{2}(x)∥f∥_{ℝ}^{2} , x ∈ ℝ

for every f ∈ P_{n}(a_{1}, a_{2}, …, a_{n}; ℝ).

Equality holds if and only if either x is a maximum point of |f| (that is, |f(x)| = ±∥f∥_{ℝ}) or f is a linear combination of T_{n} and U_{n} (with real coefficients) defined in E.5 of Section 3.5.

Corollary 7.1.6 (Bernstein-Type Inequality on ℝ, Real Case). Associated with (a_{k})_{k=1}^{n} ⊂ ℂ\ℝ, let the Bernstein factor B_{n} be defined as in Corollary 7.1.5. Then

|f′(x)| ≤ B_{n}(x)∥f∥_{ℝ} , x ∈ ℝ

for every f ∈ P_{n}(a_{1}, a_{2}, …, a_{n}; ℝ).

Equality holds if and only if f is a linear combination of T_{n} and U_{n} (with real coefficients) defined in E.5 of Section 3.5, and f(x) = 0.

To formulate our next theorem we introduce some notation. For a polynomial

q(z) := ∏_{k=1}^{n} (z − a_{k}) , a_{k} ∈ ℂ ,

we define

q^{∗}(z) := ∏_{k=1}^{n} (1 − \overline{a}_{k}z) =: z^{n}\overline{q}_{n}(z^{−1}) .

Then

(7.1.1)  |q(z)| = |q^{∗}(z)| ,  z ∈ ∂D .

The function

S_{n}(z) := \frac{q^{∗}(z)}{q(z)} = ∏_{k=1}^{n} \frac{1 − \overline{a}_{k}z}{z − a_{k}}

is the Blaschke product associated with (a_{k})_{k=1}^{n}.

Theorem 7.1.7 (Bernstein-Type Inequality on ∂D, Complex Case). Given (a_{k})_{k=1}^{n} ⊂ ℂ \ ∂D , let the Bernstein factor B_{n} be defined by

B_{n}(z) := max{B_{n}^{+}(z), B_{n}^{−}(z)}

with

B_{n}^{+}(z) := ∑_{{k=1 \atop |a_{k}|>1}}^{n} \frac{|a_{k}|^{2} − 1}{|a_{k} − z|^{2}}  and  B_{n}^{−}(z) := ∑_{{k=1 \atop |a_{k}|<1}}^{n} \frac{1 − |a_{k}|^{2}}{|a_{k} − z|^{2}} .

Then

|f′(z)| ≤ B_{n}(z)∥f∥_{∂D} , z ∈ ∂D

for every f ∈ P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; ∂D).

If the first sum is not less than the second sum for a fixed z ∈ ∂D, then equality holds for f = cS_{n}^{+} with c ∈ ℂ, where S_{n}^{+} is the Blaschke product associated with those a_{k} for which |a_{k}| > 1. If the second sum is not less than the first sum for a fixed z ∈ ∂D, then equality holds for f = cS_{n}^{−} with c ∈ ℂ, where S_{n}^{−} is the Blaschke product associated with those a_{k} for which |a_{k}| < 1.

Proof. For reasons of symmetry it is sufficient to prove the theorem only for z = 1. Without loss of generality we may assume that

(7.1.2)  Re (∑_{k=1}^{n} \frac{1}{1 − a_{k}} ) ≠ \frac{n}{2} ;

the remaining cases follow from this by a limiting argument. Let Q := ∂D (equipped with the usual metric topology), V := P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; ∂D) , and L(f) := f′(1) for f ∈ V. We show in this situation that n + 1 ≤ r in Theorem A.3.3 (interpolation of linear functionals). Suppose to the contrary that r ≤ n. By Theorem A.3.3, there are distinct points x_{1}, x_{2}, …, x_{r} on ∂D, and there are constants c_{1}, c_{2}, …, c_{r} ∈ ℂ such that

(7.1.3)  \frac{p′(1)q(1) − q′(1)p(1)}{q(1)^{2}} = ∑_{k=1}^{r} c_{k} \frac{p(x_{k})}{q(x_{k})} ,  p ∈ P_{n}^{c} ,

where

(7.1.4)  q(z) := ∏_{k=1}^{n} (z − a_{k}) .

We claim that x_{k} ≠ 1 for each k = 1, 2, …, r. Indeed, if there is index k such that x_{k} = 1, then Theorem A.3.3 implies that

p(z) := (z + 1)^{n−r} ∏_{k=1}^{r}(z − x_{k}) ∈ P_{n}^{c}

has a zero at 1 with multiplicity at least two, which is a contradiction. Applying (7.1.3) with the above p, we obtain

p′(1)q(1) − q′(1)p(1) = 0

and since p(1) ≠ 0 and q(1) ≠ 0, this is equivalent to

\frac{q′(1)}{q(1)} = \frac{p′(1)}{p(1)} ,

that is, in terms of the zeros of p_{n} and q_{n},

(7.1.5)  ∑_{k=1}^{n} \frac{1}{1 − a_{k}} = \frac{n − r}{2} + ∑_{k=1}^{r} \frac{1}{1 − x_{k}} .

Since x_{k} ∈ ∂D and x_{k} ≠ 1, we have

(7.1.6)  Re (\frac{1}{1 − x_{k}} ) = \frac{1}{2} ,  k = 1, 2, …, r .

It follows from (7.1.5) and (7.1.6) that

Re (∑_{k=1}^{n} \frac{1}{1 − a_{k}} ) = \frac{n}{2} ,

which contradicts assumption (7.1.2). So n + 1 ≤ r, indeed.

A compactness argument shows that there is a function f̃ ∈ V such that

∥f̃∥_{∂D} = 1  and  L(f̃) = ∥L∥ := max_{0≠f∈V} \frac{|L(f)|}{∥f∥_{∂D}} .

Theorem A.3.3 implies |f̃(x_{k})| = 1 for every k = 1, 2, …, r. Hence, if

f̃ = \frac{p̃}{q} ,  p̃ ∈ P_{n}^{c} ,  q(z) = ∏_{k=1}^{n} (z − a_{k}) ,

then

(7.1.7)  h(z) := |p̃(z)|^{2} − |q(z)|^{2} ≤ 0 ,  z ∈ ∂D

and

(7.1.8)  h(x_{k}) = 0 ,  k = 1, 2, …, r .

Note that t(θ) := h(e^{iθ}) ∈ T_{n} vanishes at each θ_{k}, where the numbers θ_{k} ∈ [−π, π) are defined by x_{k} = e^{iθ_{k}}, k = 1, 2, …, r. Because of (7.1.7), each of these zeros is of even multiplicity. Hence, n + 1 ≤ r implies that t ∈ T_{n} has at least 2n + 2 zeros and therefore t = 0. From this we can deduce that h(z) = 0 for every z ∈ ∂D, so

(7.1.9)  |p̃(z)| = |q(z)| ,  z ∈ ∂D .

We now have

z^{−n}p̃(z)p^{∗}(z) = |p̃(z)|^{2} = |q(z)|^{2} = z^{−n}q(z)q^{∗}(z) , z ∈ ∂D ,

so by the unicity theorem for analytic functions (see E.1 e] of Section 1.2)

p̃p^{∗} = qq^{∗} .

From this, it follows that there exists a constant 0 ≠ c ∈ ℂ such that

f̃(z) = \frac{p̃(z)}{q(z)} = c ∏_{k=1}^{m} \frac{z − \overline{α}_{k}^{−1}}{z − α_{k}} , z ∈ ℂ , q(z) ≠ 0

with some m ≤ n and

α_{k} := a_{j_{k}} ,  k = 1, 2, …, m ,  1 ≤ j_{1} < j_{2} < ⋯ < j_{m} ≤ n .

A straightforward calculation gives that

|f̃′(1)| = |\frac{f̃′(1)}{f̃(1)}| = |∑_{k=1}^{m} (\frac{1}{1 − \overline{α}_{k}^{−1}} − \frac{1}{1 − α_{k}} )|

= |∑_{k=1}^{m} \frac{|α_{k}|^{2} − 1}{|α_{k} − 1|^{2}} | ≤ max\{B_{n}^{+}(z), B_{n}^{−}(z)\} ,

which finishes the proof. ⊓ ⊔ Corollary 7.1.8 (Bernstein-Type Inequality on K, Complex Case). Given (a_{k})_{k=1}^{2n} ⊂ ℂ \ ℝ, let the Bernstein factor B_{n} be defined by

B̃_{n}(θ) := max{B̃_{n}^{+}(θ), B̃_{n}^{−}(θ)}

with

B̃_{n}^{+}(θ) := ∑_{{k=1 \atop Im(a_{k})<0}}^{2n} \frac{|e^{ia_{k}}|^{2} − 1}{|e^{ia_{k}} − e^{iθ}|^{2}} and B̃_{n}^{−}(θ) := ∑_{{k=1 \atop Im(a_{k})>0}}^{2n} \frac{1 − |e^{ia_{k}}|^{2}}{|e^{ia_{k}} − e^{iθ}|^{2}} .

Then

|f′(θ)| ≤ B_{n}(θ)∥f∥_{K} , θ ∈ K

for every f ∈ T_{n}^{c}(a_{1}, a_{2}, …, a_{2n}; K).

If the first sum is not less than the second sum for a fixed θ ∈ K, then equality holds for f(θ) = cS_{2n}^{+}(e^{iθ}) with c ∈ ℂ, where S_{2n}^{+} is the Blaschke product associated with those e^{ia_{k}} for which Im(a_{k}) < 0. If the second sum is not less than the first sum for a fixed θ ∈ K, then equality holds for f(θ) = cS_{2n}^{−}(e^{iθ}) with c ∈ ℂ, where S_{2n}^{−} is the Blaschke product associated with those e^{ia_{k}} for which Im(a_{k}) > 0. Note that

S_{2n}^{±}(e^{iθ}) ∈ T_{n}^{c}(a_{1}, a_{2}, …, a_{2n}; K) .

Proof. Observe that if

h(θ) := ∏_{j=1}^{2n} sin((θ − a_{j})/2) ∈ T_{n}^{c}

and t_{n} ∈ T_{n}^{c}, then there are p ∈ P_{2n}^{c} and q ∈ P_{2n}^{c} such that

\frac{t(θ)}{h(θ)} = \frac{p(e^{iθ})e^{−inθ}}{q(e^{iθ})e^{−inθ}} = \frac{p(e^{iθ})}{q(e^{iθ})} ,

where q is of the form

q(z) = c ∏_{j=1}^{2n}(z − e^{ia_{j}})

with some c ∈ ℂ. So the corollary follows from Theorem 7.1.7. ⊓ ⊔

Corollary 7.1.9 (Bernstein-Type Inequality on [−1, 1], Complex Case). Given {a_{k}}_{k=1}^{n} ⊂ ℂ \ [−1, 1], let the Bernstein factor B_{n}(x) be defined by

B_{n}(x) = max \{∑_{k=1}^{n} \frac{1 − |c_{k}|^{2}}{|c_{k} − z|^{2}} , ∑_{k=1}^{n} \frac{|c_{k}|^{−2} − 1}{|c_{k}^{−1} − z|^{2}}\} ,

where c_{k} and z are determined by

a_{k} := \frac{1}{2}(c_{k} + c_{k}^{−1}) ,  |c_{k}| < 1 ;
x := \frac{1}{2}(z + z^{−1}) ,  Im(z) > 0 .

Then

|f′(x)| ≤ \frac{B_{n}(x)}{\sqrt{1 − x^{2}}} ∥f∥_{[−1,1]} , x ∈ (−1, 1)

for every f ∈ P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; [−1, 1]). Note that

∑_{k=1}^{n} \frac{1 − |c_{k}|^{2}}{|c_{k} − z|^{2}} = Re (∑_{k=1}^{n} \frac{\sqrt{a_{k}^{2} − 1}}{a_{k} − x}) , x ∈ [−1, 1] ,

where the choice of \sqrt{a_{k}^{2} − 1} is determined by |a_{k} − \sqrt{a_{k}^{2} − 1}| < 1.

Proof. The corollary follows from Theorem 7.1.7 by the substitution x = \frac{1}{2}(z + z^{−1}). ⊓ ⊔

Bernstein’s classical polynomial inequalities discussed in Section 5.1 are contained in Theorem 7.1.7 and Corollaries 7.1.8 and 7.1.9 as limiting cases. In Theorem 7.1.7 and Corollary 7.1.9 we take

(a_{1}^{(m)}, a_{2}^{(m)}, …, a_{n}^{(m)}) ⊂ ℂ \ D

so that

lim_{m→∞} |a_{k}^{(m)}| = ∞ , k = 1, 2, …, n .

In Corollary 7.1.8 we take

(a_{1}^{(m)}, a_{2}^{(m)}, …, a_{2n}^{(m)}) ⊂ ℂ \ ℝ

so that

a_{n+k}^{(m)} = \overline{a}_{k}^{(m)} and lim_{m→∞} |Im(a_{k}^{(m)})| = ∞ , k = 1, 2, …, n .

To formulate our next result we introduce the Blaschke product

Q_{n}(z) := ∏_{k=1}^{n} \frac{z − \overline{a}_{k}}{z − a_{k}}

associated with (a_{1}, a_{2}, …, a_{n}) ⊂ ℂ \ ℝ. Obviously |Q_{n}(z)| = 1 for every z ∈ ℝ.

Corollary 7.1.10 (Bernstein-Type Inequality on ℝ, Complex Case). Given (a_{k})_{k=1}^{n} ⊂ ℂ \ ℝ, let the Bernstein factor B_{n}(x) be defined by

B_{n}(x) := max{B_{n}^{+}(x), B_{n}^{−}(x)}

with

B_{n}^{+}(x) := ∑_{{k=1 \atop Im(a_{k})>0}}^{n} \frac{2 |Im(a_{k})|}{|x − a_{k}|^{2}} and B_{n}^{−}(x) := ∑_{{k=1 \atop Im(a_{k})<0}}^{n} \frac{2 |Im(a_{k})|}{|x − a_{k}|^{2}}

for every x ∈ ℝ. Then

|f′(x)| ≤ B_{n}(x)∥f∥_{ℝ} , x ∈ ℝ

for every f ∈ P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; ℝ).

If the first sum is not less than the second sum for a fixed x ∈ ℝ, then equality holds for f = cQ_{n}^{+} with c ∈ ℂ, where Q_{n}^{+} is the Blaschke product associated with the poles a_{k} lying in the open upper half-plane

H^{+} := {z ∈ ℂ : Im(z) > 0} .

If the second sum is not less than the first sum for a fixed x ∈ ℝ, then equality holds for f = cQ_{n}^{−} with c ∈ ℂ, where Q_{n}^{−} is the Blaschke product associated with the poles a_{k} lying in the open lower half-plane

H^{−} := {z ∈ ℂ : Im(z) < 0} .

Corollary 7.1.10 follows from Theorem 7.1.7; see E.4.

The next theorem improves the Bernstein-type inequality of Theorem 7.1.7 in the case when \{a_{k}\}_{k=1}^{n} ⊂ ℂ \ \overline{D} and f has all its zeros in ℂ \ D. It extends Lax [44].

Theorem 7.1.11 (Lax-Type Inequality). Given (a_{k})_{k=1}^{n} ⊂ ℂ \ \overline{D}, let the Bernstein factor B_{n} be, as in Theorem 7.1.7, defined by

B_{n}(z) := ∑_{k=1}^{n} \frac{|a_{k}|^{2} − 1}{|a_{k} − z|^{2}} .

Then

|h′(z)| ≤ \frac{1}{2}B_{n}(z)∥h∥_{∂D} , z ∈ ∂D

for every h ∈ P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; ∂D) having all its zeros in ℂ \ D.

Equality holds for h = c(S_{n} + 1) with c ∈ ℂ, where S_{n} is the Blaschke product associated with (a_{k})_{k=1}^{n}.

Note that B_{n}(z) = |S′_{n}(z)|. Note also that
h := c(S_{n} + 1) ∈ P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; ∂D) , c ≠ 0

has all its zeros on ∂D.

Proof. First assume that each zero of h is on ∂D, the general case can be reduced to this (see E.5). Thus let h := p/q, where p ∈ P_{n}^{c} has all its zeros on ∂D and where

q(z) := ∏_{k=1}^{n} (z − a_{k}) , |a_{k}| > 1 .

Let

q^{∗}(z) := ∏_{k=1}^{n} (1 − \overline{a}_{k}z) .

We study

u(θ) := \frac{p(e^{2iθ})e^{−inθ}}{|q(e^{2iθ})|} = \frac{p(e^{2iθ})}{\sqrt{q(e^{2iθ)}}\sqrt{q^{∗}(e^{2iθ})}}

for θ ∈ ℝ, where the square roots are taken so that \sqrt{q} is analytic in a neighborhood of the closed unit disk, and \sqrt{q^{∗}} is analytic in a neighborhood of the complement of the open unit disk. Since p ∈ P_{n}^{c} has all its zeros on ∂D, there exists a 0 ≠ β ∈ ℂ such that

t(θ) := βp(e^{2iθ})e^{−inθ}

is a real trigonometric polynomial of degree at most n (see E.5 a]). Also

|q(e^{2iθ)}| = |q^{∗}(e^{2iθ})| = ∏_{k=1}^{n} |1 − \overline{a}_{k}e^{2iθ}|

= γ ∏_{k=1}^{2n} | sin((θ − c_{k})/2)| ,

where γ > 0,

e^{ic_{k}} = \overline{a}_{k}^{−1/2} , Im(c_{k}) > 0 , k = 1, 2, …, n

and

e^{ic_{k}} = −\overline{a}_{k}^{−1/2} , Im(c_{k}) > 0 , k = n + 1, n + 2, …, 2n .

Applying Theorem 7.1.1 to

β^{−1}u ∈ T_{n}(c_{1}, c_{2}, …, c_{2n}; K) ,

7.1 Inequalities for Rational Function Spaces 331

we obtain

(7.1.10) |u′(θ) − iB̃_{n}(θ)u(θ)| ≤ B̃_{n}(θ)∥u∥_{K} , θ ∈ K ,

where

(7.1.11)  B̃_{n}(θ) = \frac{1}{2} ∑_{k=1}^{n} (\frac{1 − |a_{k}|^{−1}}{|\overline{a}_{k}^{−1/2} − e^{iθ}|^{2}} +\frac{1 − |a_{k}|^{−1}}{| − \overline{a}_{k}^{−1/2} − e^{iθ}|^{2}} )

= ∑_{k=1}^{n} \frac{1 − |a_{k}|^{−2}}{|\overline{a}_{k}^{−1} − e^{2iθ}|^{2}} = ∑_{k=1}^{n} \frac{|a_{k}|^{2} − 1}{|a_{k} − e^{2iθ}|^{2}} .

Observe that

(7.1.12)  u(θ) = \frac{p(e^{2iθ})}{q(e^{2iθ})} \begin{matrix}√ & \overline{q(e^{2iθ})} \\ √ & \overline{q^{∗}(e^{2iθ})} \end{matrix} = \frac{p(e^{2iθ})}{q(e^{2iθ})}f_{n}(e^{iθ}) ,

where

(7.1.13)  f_{n}(z) := \begin{matrix}√ & \overline{q(z^{2})} \\ √ & \overline{q^{∗}(z^{2})} \end{matrix} .

A simple calculation (see E.4. of Section 3.5) shows that

(7.1.14) B̃_{n}(θ) = e^{iθ} \frac{f′_{n}(e^{iθ})}{f_{n}(e^{iθ})} , θ ∈ K .

Also, since |f_{n}(e^{iθ})| = 1 for every θ ∈ K, we have

(7.1.15)  ∥u∥_{K} = ∥\frac{p}{q}∥_{∂D} = ∥h∥_{∂D} .

Now (7.1.10) to (7.1.15) yield

|\frac{d}{dθ} (\frac{p(e^{2iθ})}{q(e^{2iθ})}f_{n}(e^{iθ})) − ie^{iθ} \frac{f′_{n}(e^{iθ})}{f_{n}(e^{iθ})} \frac{p(e^{2iθ})}{q(e^{2iθ})}f_{n}(e^{iθ})| ≤ B̃_{n}(θ)∥h∥_{∂D} .

So

|2ie^{2iθ}h′(e^{2iθ})f_{n}(e^{2iθ}) + ie^{iθ}f′_{n}(e^{iθ})h(e^{2iθ})
− ie^{iθ}f′_{n}(e^{iθ})h(e^{2iθ})| ≤ B̃_{n}(θ)∥h∥_{∂D} .

Thus

2 |h′(e^{2iθ})| ≤ B̃_{n}(θ)∥h∥_{∂D} ,

which, together with (7.1.11), finishes the proof. ⊓ ⊔ Comments, Exercises, and Examples.

Most of the results in this section have been proved in Borwein and Erdélyi [to appear 4] and in Borwein, Erdélyi, and Zhang [94a]. A weaker version of Corollary 7.1.9 has been obtained by Russak (see Petrushev and Popov [87]). Theorem 7.1.11 contains, as a limiting case, an inequality of Lax [44] conjectured by Erdős. Lax’s inequality establishes the sharp Bernstein-type inequality on the unit disk for polynomials p ∈ P_{n}^{c} having no zeros in the open unit disk. That is,

∥p′∥_{D} ≤ \frac{n}{2} ∥p∥_{D}

for such polynomials. Various extensions of this inequality are given by Ankeny and Rivlin [55], Govil [73], Malik [69], and others. We discuss some of these in E.16 of Appendix 5.

E.1 Proof of Theorem 7.1.1 and Corollary 7.1.2. Given (a_{k})_{k=1}^{2n} ⊂ ℂ \ ℝ, let

T_{n,a} := T_{n}(a_{1}, a_{2}, …, a_{2n}; K) .

a] Show that T_{n,a} is a Hermite interpolation space. That is, if the points x_{1}, x_{2}, …, x_{k} ∈ K are distinct, and m_{1}, m_{2}, …, m_{k} are positive integers with ∑_{i=1}^{k} m_{i} ≤ 2n + 1, then for any choice of real numbers y_{i,j}, there is a function f ∈ T_{n,a} such that

f^{(j)}(x_{i}) = y_{i,j} , i = 1, 2, …, k , j = 0, 1, …, m_{i} − 1 .

Hint: See the hint to E.7 of Section 1.1. ⊓ ⊔

b] Show that for every fixed θ ∈ K, the value

max_{0≠f∈T_{n,a}} \frac{f′(θ)^{2} + B_{n}^{2}(θ)f^{2}(θ)}{∥f∥_{K}^{2}}

is attained by an f̃ ∈ T_{n,a}.

Hint: Use a compactness argument. ⊓ ⊔

c] Show that f̃ = cV, where c ∈ ℝ and V is one of the Chebyshev polynomials for T_{n,a} defined in Theorem 3.5.3 and E.3 of Section 3.5.

Hint: Use a variational method with the help of part a]. ⊓ ⊔

d] Prove Theorem 7.1.1.

Hint: Use part c] and E.4 of Section 3.5. ⊓ ⊔

e] Prove Corollary 7.1.2.

E.2 Proof of Corollaries 7.1.3 and 7.1.4.

a] Prove the inequality of Corollary 7.1.3.

7.1 Inequalities for Rational Function Spaces 333

Hint: Use Theorem 7.1.1 with the substitution x = cos θ. Note that f ∈ P_{n}(a_{1}, a_{2}, …, a_{n}; [−1, 1]) implies

g(θ) := f(cos θ) = T_{n}(c_{1}, c_{1}, c_{2}, c_{2}, …, c_{n}, c_{n}; K) ,

where the numbers c_{k} ∈ ℂ are defined by

e^{ic_{k}} = a_{k} − \sqrt{a_{k}^{2} − 1} , a_{k} = \frac{1}{2}(e^{ic_{k}} + e^{−ic_{k}}), Im(c_{k}) > 0 .

Verify that if B̃_{n} is the Bernstein factor given in Theorem 7.1.1 associated with

(c_{1}, c_{1}, c_{2}, c_{2}, …, c_{n}, c_{n}) ,

then

B̃_{n}(θ) = ∑_{k=1}^{n} Re (\frac{\sqrt{a_{k}^{2} − 1}}{a_{k} − cos θ} ) , θ ∈ K ,

where the choice of \sqrt{a_{k}^{2} − 1} is determined by |a_{k} − \sqrt{a_{k}^{2} − 1}| < 1. ⊓ ⊔ b] Given x ∈ [−1, 1], prove that equality holds in the inequality of Corollary 7.1.3 if and only if either x is a maximum point of |f| (that is, f(x) = ±∥f∥_{[−1,1]}) or f = cT_{n} with c ∈ ℝ, where T_{n} is defined in Section 3.5. Hint: Observe that

V = (cos α)T̃_{n} + (sin α)Ũ_{n}

is even if and only if V = ±T̃_{n} and use Theorem 7.1.1. ⊓ ⊔ c] Prove Corollary 7.1.4.

E.3 Proof of Corollaries 7.1.5 and 7.1.6. a] Prove Corollary 7.1.5.

Hint: Use Theorem 7.1.1 and the substitution x = i\frac{e^{iθ} + 1}{e^{iθ} − 1} , which maps K onto ℝ ∪ {∞}. ⊓ ⊔ b] Prove Corollary 7.1.6.

E.4 Proof of Corollary 7.1.10. Prove Corollary 7.1.10. Hint: Use Theorem 7.1.7 and the substitution x = i\frac{z + 1}{z − 1} , which maps ∂D onto ℝ ∪ {∞} . ⊓ ⊔

E.5 Completion of the Proof of Theorem 7.1.11. a] Show that if p ∈ P_{n}^{c} has all its zeros on the unit circle, then there is a 0 ≠ β ∈ ℂ such that g(θ) := βe^{−inθ}p(e^{2iθ}) is a real trigonometric polynomial of degree at most n.

334 7. Inequalities for Rational Function Spaces

For f = p/q with

q(z) := ∏_{k=1}^{n} (z − a_{k}) , a_{k} ∈ ℂ

and

p(z) := γ ∏_{k=1}^{m} (z − b_{k}) , b_{k} ∈ ℂ , γ ∈ ℂ , m ≤ n ,

we define f^{∗} := p^{∗}/q, where

p^{∗}(z) := \overline{γ} ∏_{k=1}^{m} (1 − z\overline{b}_{k}) .

Let D := {z ∈ ℂ : |z| < 1}. b] Show that |f^{∗}(z)| = |f(z)| for every z ∈ ∂D.

In each of the remaining parts of the exercise suppose that |a_{k}| > 1 for each k. c] Show that if ϵ ∈ ∂D and |b_{k}| ≥ 1 for each k, then f + ϵf^{∗} has all its zeros on the unit circle. d] Show that if |b_{k}| ≥ 1 for each k, then

|f′(z)| ≤ |f^{∗′}(z)| , z ∈ ∂D .

Hint: First observe that it is sufficient to study the case z = 1. We have

|Re (\frac{f′(1)}{f(1)})| = |Re (∑_{k=1}^{n} \frac{1}{1 − b_{k}} ) − Re (∑_{k=1}^{n} \frac{1}{1 − a_{k}} )|

≤ |Re (∑_{k=1}^{n} \frac{1}{1 − b_{k}} ) − \frac{n}{2}| + |\frac{n}{2} − Re (∑_{k=1}^{n} \frac{1}{1 − a_{k}} )|

= \frac{n}{2} − Re (∑_{k=1}^{n} \frac{1}{1 − b_{k}} ) + \frac{n}{2} − Re (∑_{k=1}^{n} \frac{1}{1 − a_{k}} )

= n − Re (∑_{k=1}^{n} \frac{1}{1 − b_{k}} ) − Re (∑_{k=1}^{n} \frac{1}{1 − a_{k}} )

= Re (∑_{k=1}^{n} \frac{1}{1 − b_{k}^{−1}} ) − Re (∑_{k=1}^{n} \frac{1}{1 − a_{k}} )

= Re (∑_{k=1}^{n} \frac{1}{1 − \overline{b}_{k}^{−1}} ) − Re (∑_{k=1}^{n} \frac{1}{1 − a_{k}} )

= |Re (\frac{f^{∗′}(1)}{f^{∗}(1)})|

and

|Im (\frac{f′(1)}{f(1)})| = |Im (∑_{k=1}^{n} \frac{1}{1 − b_{k}} ) − Im (∑_{k=1}^{n} \frac{1}{1 − a_{k}} )|

= |Im (n − ∑_{k=1}^{n} \frac{1}{1 − b_{k}} ) − Im (∑_{k=1}^{n} \frac{1}{1 − a_{k}} )|

= |−Im (∑_{k=1}^{n} \frac{1}{1 − \overline{b}_{k}^{−1}} ) + Im (∑_{k=1}^{n} \frac{1}{1 − a_{k}} )|

= |Im (\frac{f^{∗′}(1)}{f^{∗}(1)})| .

The result now follows from the combination of part b] and the above two inequalities. ⊓ ⊔

e] Prove that if |b_{k}| ≥ 1 for each k, then

2|f′(z)| ≤ |f′(z)| + |f^{∗′}(z)| ≤ B_{n}(z)∥f∥_{∂D} , z ∈ ∂D ,

where B_{n}(z) is the Bernstein factor defined in Theorem 7.1.11.

Hint: Use parts c] and d] and the already proved part of Theorem 7.1.11 (when f has all its zeros on the unit circle). ⊓ ⊔

f] Show that if |b_{k}| ≥ 1 for each k, then

|f′(z)| ≤ \frac{1}{2}B_{n}(z) (max_{z∈∂D} |f(z)| − min_{z∈∂D} |f(z)|) , z ∈ ∂D ,

where B_{n}(z) is the Bernstein factor defined in Theorem 7.1.7. This extends a result of Aziz and Dawood [88].

Hint: Assume that ∥f∥_{∂D} = 1. Let m := min_{z∈∂D} |f(z)|. Let α be a constant of modulus less than 1. Let g(z) := f(z) − αm. Observe that the argument of α can be chosen so that

|g^{∗′}(z)| = |f^{∗′}(z)| − |α|mB_{n}(z) .

By Rouché’s theorem, g has no zeros in D. So parts d] and e] imply that

2 |f^{∗′}(z)| − 2 |α| mB_{n}(z) = 2 |g^{∗′}(z)| ≤ |g′(z)| + |g^{∗′}(z)|
= |f′(z)| + |f^{∗′}(z)| − |α|mB_{n}(z)
≤ B_{n}(z) − |α|mB_{n}(z) .

Since |α| can be chosen arbitrarily close to 1, the result follows. ⊓ ⊔

336 7. Inequalities for Rational Function Spaces

E.6 Extensions of Russak’s Inequalities. a] Given (a_{k})_{k=1}^{2n} ⊂ ℂ \ ℝ, show that

∥f′∥_{L_{1}(K)} ≤ 2πn ∥f∥_{K}

for every f ∈ T_{n}(a_{1}, a_{2}, …, a_{2n}; K) and

∥f′∥_{L_{1}(K)} ≤ 4πn ∥f∥_{K}

for every f ∈ T_{n}^{c}(a_{1}, a_{2}, …, a_{2n}; K). b] Given (a_{k})_{k=1}^{n} ⊂ ℂ \ ℝ, show that

∥f′∥_{L_{1}(ℝ)} ≤ πn ∥f∥_{ℝ}

for every f ∈ P_{n}(a_{1}, a_{2}, …, a_{n}; ℝ,) and

∥f′∥_{L_{1}(ℝ)} ≤ 2πn ∥f∥_{ℝ}

for every f ∈ P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; ℝ). Hint: Use Corollaries 7.1.2, 7.1.8, 7.1.6, and 7.1.10. Write the Bernstein factors in a form so that the integral (of each term in the maximum if the Bernstein factor is defined by a maximum) can be evaluated by the residue theorem (in part a]) and by finding the antiderivative (in part b]). ⊓ ⊔ c] Are any of the inequalities of parts a] and b] sharp? If so, in which cases?

E.7 Markov-Type Inequality. Given (a_{k})_{k=1}^{n} ⊂ ℝ \ [−1, 1], show that

∥f′∥_{[−1,1]} ≤ \frac{n}{n − 1} (∑_{k=1}^{n} \frac{1 + |c_{k}|}{1 − |c_{k}|})^{2} ∥f∥_{[−1,1]}

for every f ∈ P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; [−1, 1]), where the numbers c_{k} are defined by

c_{k} := a_{k} − \sqrt{a_{k}^{2} − 1} , a_{k} = \frac{1}{2}(c_{k} + c_{k}^{−1}) , |c_{k}| < 1 .

Proceed as follows: a] Given (a_{k})_{k=1}^{n} ⊂ ℝ \ [−1, 1], let

a_{k}(y) := \{ \begin{matrix}\frac{2a_{k}}{1 + y} + \frac{1 − y}{1 + y} & if & 0 ≤ y ≤ 1 \\ \frac{2a_{k}}{1 − y} + \frac{1 + y}{1 − y} & if & −1 ≤ y ≤ 0 \end{matrix}

and let c_{k}(y) be defined by

a_{k}(y) =: \frac{1}{2}(c_{k}(y) + c_{k}(y)^{−1}) , |c_{k}(y)| < 1 .

Show that

|f′(y)| ≤ \frac{2}{1 + |y|} (∑_{k=1}^{n} \frac{1 + c_{k}(y)}{1 − c_{k}(y)})^{2} ∥f∥_{[−1,1]}

for every f ∈ P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; [−1, 1]).

Hint: Show by a variational method that

max_{f} \frac{|f′(±1)|}{∥f∥_{[−1,1]}} = |T′_{n}(±1)| ,

where the maximum is taken for all 0 ≠ f ∈ P_{n}(a_{1}, a_{2}, …, a_{n}; [−1, 1]), and T_{n} is the Chebyshev polynomial for P_{n}(a_{1}, a_{2}, …, a_{n}; [−1, 1]) defined in Section 3.5. Now the result follows from E.1 c] of Section 3.5 by a linear shift from [−1, 1] to [−1, y] if 0 ≤ y ≤ 1, or to [y, 1] if −1 ≤ y ≤ 0. ⊓ ⊔

b] Given (a_{k})_{k=1}^{n} ⊂ ℝ \ [−1, 1], show that

|f′(y)| ≤ \frac{n}{1 − |y|}∥f∥_{[−1,1]} , y ∈ (−1, 1)

for every f ∈ P_{n}(a_{1}, a_{2}, …, a_{n}; [−1, 1]).

Hint: When y = 0, this follows from Corollary 7.1.3. When y ∈ (−1, 1) is arbitrary, use a linear shift from [−1, 1] to [2y − 1, 1] if 0 ≤ y ≤ 1, or to [−1, 2y + 1] if −1 < y ≤ 0. ⊓ ⊔

c] Prove the Markov-type inequality of the exercise.

Hint: Combine parts a] and b]. Note that

|a_{k}(y)| ≥ |a_{k}|  and  |c_{k}(y)| ≤ |c_{k}| < 1 ,  k = 1, 2, …, n

holds for every y ∈ [−1, 1]. ⊓ ⊔

E.8 Schur-Type Inequality. Given {a_{k}}_{k=1}^{n} ⊂ ℝ \ [−1, 1], show that

∥f∥_{[−1,1]} ≤ max\{|U_{n}(1)|, |U_{n}(−1)|\} · ∥f(x)\sqrt{1 − x^{2}}∥_{[−1,1]}

for every f ∈ P_{n}(a_{1}, a_{2}, …, a_{n}; [−1, 1]), where U_{n} is the Chebyshev polynomial (of the second kind) for P_{n}(a_{1}, a_{2}, …, a_{n}; [−1, 1]) defined in Section 3.5, and

|U_{n}(±1)| = |∑_{k=1}^{n} \frac{\sqrt{a_{k}^{2} − 1}}{a_{k} ∓ 1}|

with the choice of \sqrt{a_{k}^{2} − 1} determined by |a_{k} − \sqrt{a_{k}^{2} − 1}| < 1. Show that equality holds if and only if f = cU_{n}, c ∈ ℝ.

Hint: First show that if f̃ ∈ P_{n}(a_{1}, a_{2}, …, a_{n}; [−1, 1]) is extremal for

max_{f} \frac{|f(±1)|}{∥f(x)\sqrt{1 − x^{2}}∥_{[−1,1]}} ,

where the maximum is taken for all 0 ≠ f ∈ P_{n}(a_{1}, a_{2}, …, a_{n}; [−1, 1]), then f = cU_{n} with some c ∈ ℝ. Observe that U_{n}(±1) can be evaluated by L’Hospital’s rule since

U_{n}(x)^{2} = \frac{1 − T_{n}(x)^{2}}{1 − x^{2}} ;

hence |U_{n}(±1)|^{2} = |T′_{n}(±1)| = |B_{n}(±1)| · |U_{n}(±1)| with the notation of Section 3.5. Thus |U_{n}(±1)| = |B_{n}(±1)| .

If y ∈ [−1, 1] is arbitrary, then use a linear shift from [−1, 1] to [−y, y] (some caution must be exercised about the change of poles). ⊓ ⊔

E.9 Extension of Lax’s Inequality on the Half-Plane. Associated with (a_{k})_{k=1}^{n} ⊂ H^{+} := {z ∈ ℂ : Im(z) > 0}, let the Bernstein factor B_{n} be, as in Corollary 7.1.10, defined by

B_{n}(x) := ∑_{k=1}^{n} \frac{2 Im(a_{k})}{|x − a_{k}|^{2}} .

Show that

|h′(x)| ≤ \frac{1}{2}B_{n}(x)∥h∥_{ℝ} , x ∈ ℝ

for every h ∈ P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; ℝ) having all its zeros in H^{+}.

Equality holds for h = c(S̃_{n} + 1) with c ∈ ℂ, where S̃_{n} is the Blaschke product associated with (a_{k})_{k=1}^{n}. Note that B_{n}(z) = |S̃′_{n}(z)|. Note also that

h = c(S̃ + 1) ∈ P_{n}^{c}(a_{1}, a_{2}, …, a_{n}; ℝ) , c ≠ 0

has all its zeros on ℝ.

Hint: Use Theorem 7.1.11 with the substitution x = i\frac{z + 1}{z − 1} . ⊓ ⊔

E.10 Remarks on Theorem 7.1.7 and Corollary 7.1.10.

a] Given (a_{k})_{k=1}^{n} ⊂ D and z ∈ ∂D, show that equality holds in the inequality of Theorem 7.1.7 if and only if f = cS_{n} with c ∈ ℂ, where S_{n} is the Blaschke product associated with (a_{k})_{k=1}^{n}.

Hint: Analyze the proof of Theorem 7.1.7. ⊓ ⊔

b] Given (a_{k})_{k=1}^{n} ⊂ H^{+} = {z ∈ ℂ : Im(z) > 0} and x ∈ ℝ, show that equality holds in the inequality of Corollary 7.1.10 if and only if f = cQ_{n} with c ∈ ℂ, where Q_{n} is the Blaschke product associated with (a_{k})_{k=1}^{n}.

Note that the only if parts of E.10 a] and E.10 b] above are not claimed in the general case of Theorem 7.1.7 and Corollary 7.1.10 (why?).

E.11 Markov-Bernstein-Type Inequality for SR_{n}^{c} and ASR_{n}^{c}. Let SR_{n}^{c} denote the set of all self-reciprocal polynomials p ∈ P_{n}^{c} satisfying

p(z) ≡ z^{n}p(z^{−1}) .

Let SR_{n} denote the set of all real self-reciprocal polynomials of degree at most n, that is, SR_{n} := SR_{n}^{c} ∩ P_{n}. For a polynomial p ∈ P_{n}^{c} of the form

(7.1.16)  p(z) = ∑_{j=0}^{n} c_{j}z^{j} ,  c_{j} ∈ ℂ ,

p ∈ SR_{n}^{c} if and only if

c_{j} = c_{n−j} , j = 0, 1, …, n .

Let ASR_{n}^{c} denote the set of all antiself-reciprocal polynomials p ∈ P_{n}^{c} satisfying

p(z) ≡ −z^{n}p(z^{−1}) .

Let ASR_{n} denote the set of all real antiself-reciprocal polynomials of degree at most n, that is, ASR_{n} := ASR_{n}^{c} ∩ P_{n}. Let ASR_{n} := ASR_{n}^{c} ∩ P_{n}. For a polynomial p ∈ P_{n}^{c} of the form (7.1.16) p ∈ ASR_{n}^{c} if and only if

c_{j} = −c_{n−j} , j = 0, 1, …, n .

a] There exists an absolute constant c such that

|p′(x)| ≤ cn min \{(1 + log n), log (\frac{e}{1 − x^{2}} )\} ∥p∥_{[−1,1]}

holds for every x ∈ [−1, 1] and for every p ∈ P_{n}^{c} satisfying

(7.1.17)  |p(x)| ≤ (1 + |x|^{n})∥p∥_{[−1,1]} ,  x ∈ ℝ ,

in particular, for every p ∈ SR_{n}^{c} and for every p ∈ ASR_{n}^{c}.

The inequality

|p′(x)| ≤ cn(1 + log n) ∥p∥_{[−1,1]}

for all p ∈ SR_{n} and for all p ∈ ASR_{n} was first obtained by Kroó and Szabados [94a]. They also showed that up to the constant c > 0 the above inequality is sharp for both SR_{n} and ASR_{n}. Here we present a distinct proof. The sharpness is studied in E.12 f].

Outline. Suppose p ∈ P_{n}^{c} satisfies (7.1.17). Then

(7.1.18)  f(x) := \frac{p(x)}{1 + x^{2n}}

satisfies

(7.1.19)  ∥f∥_{ℝ} ≤ 2 ∥p∥_{[−1,1]} .

Let

(7.1.20)  a_{k} := e^{i(2k−1)π/(2n)} ,  k = 1, 2, …, 2n

be the zeros of the equation z^{2n} + 1 = 0. Now Corollary 7.1.10, together with (7.1.18) to (7.1.20), yields that

(7.1.21)  |f′(x)| ≤ 2 (∑_{k=1}^{n} \frac{Im(a_{k})}{|x − a_{k}|^{2}} ) ∥f∥_{ℝ}

≤ 4 (∑_{k=1}^{n} \frac{Im(a_{k})}{|x − a_{k}|^{2}} ) ∥p∥_{[−1,1]} , x ∈ ℝ .

Show that if n ∈ ℕ and x ∈ [−1, 1], then

(7.1.22)  4 ∑_{k=1}^{n} \frac{Im(a_{k})}{|x − a_{k}|^{2}} ∼ n min \{(1 + log n), log (\frac{e}{1 − x^{2}} )\} ,

where here the ∼ symbol means that there are absolute constants c_{1} > 0 and c_{2} > 0 (independent of n ∈ ℕ and x ∈ [−1, 1]) such that the left-hand side is between c_{1} times the right-hand side and c_{2} times the right-hand side for every n ∈ ℕ and x ∈ [−1, 1]. Combining (7.1.18), (7.1.21), and (7.1.22), we conclude that there is an absolute constant c_{2} such that

|\frac{p′(x)}{1 + x^{2n}} − \frac{2nx^{2n−1}}{1 + x^{2n}} \frac{p(x)}{1 + x^{2n}} |

≤ c_{2}n min \{(1 + log n), log (\frac{e}{1 − x^{2}} )\} ∥p∥_{[−1,1]} , x ∈ ℝ .

So if x ∈ [−1, 1], then

|p′(x)| ≤ (2c_{2}n min \{(1 + log n), log (\frac{e}{1 − x^{2}} )\} + 2n) ∥p∥_{[−1,1]}

and the proof is finished. ⊓ ⊔

b] There exists an absolute constant c > 0 such that

∥\frac{p′(x)}{1 + x^{2n}} ∥_{ℝ} ≤ cn(1 + log n) ∥\frac{p(x)}{1 + x^{2n}} ∥_{ℝ}

for every p ∈ P_{2n}^{c}.

Proof. Associated with p ∈ P_{2n}^{c}, let

f(x) := \frac{p(x)}{1 + x^{2n}} .

Let

a_{k} := e^{i(2k−1)π/(2n)} , k = 1, 2, …, 2n

be the zeros of the equation z^{2n} + 1 = 0. Using Corollary 7.1.10, we can deduce that there are absolute constants c_{1} and c_{2} such that

∥f′∥_{ℝ} ≤ (∑_{k=1}^{n} 2 |Im(a_{k})|^{−1}) ∥f∥_{ℝ} ≤ c_{1} (∑_{k=1}^{n} (\frac{k}{n})^{−1}) ∥f∥_{ℝ}
≤ c_{2}n(1 + log n)∥f∥_{ℝ} .

Therefore

|\frac{p′(x)}{1 + x^{2n}} − \frac{2nx^{2n−1}}{1 + x^{2n}} \frac{p(x)}{1 + x^{2n}} | ≤ c_{2}n(1 + log n) ∥\frac{p(x)}{1 + x^{2n}} ∥_{ℝ}

for every x ∈ ℝ, which implies

∥\frac{p′(x)}{1 + x^{2n}} ∥_{ℝ} ≤ (c_{2}n(1 + log n) + 2n) ∥\frac{p(x)}{1 + x^{2n}} ∥_{ℝ} ,

and the result follows. ⊓ ⊔

c] For every m ∈ ℕ, there exists a constant c(m) depending only on m so that

∥p^{(m)}∥_{[−1,1]} ≤ c(m)(n(1 + log n))^{m}∥p∥_{[−1,1]}

for every p ∈ P_{n}^{c} satisfying

(7.1.23)  |p(x)| ≤ (1 + |x|^{n})∥p∥_{[−1,1]} .

Proof. Using part b] and induction on m, we see that there exists a constant c_{1}(m) depending only on m such that

∥\frac{p^{(m)}(x)}{1 + x^{2n}} ∥_{ℝ} ≤ c_{1}(m)(n(1 + log n))^{m} ∥\frac{p(x)}{1 + x^{2n}} ∥_{ℝ}

for every p ∈ P_{2n}^{c}. Note that if p ∈ P_{n}^{c} satisfies (7.1.23), then

∥\frac{p(x)}{1 + x^{2n}} ∥_{ℝ} ≤ 2 ∥\frac{p(x)}{1 + |x|^{n}} ∥_{ℝ} ≤ 2∥p∥_{[−1,1]} ,

and the result follows. ⊓ ⊔ E.12 Quasi-Chebyshev Polynomials for SR_{2n} and ASR_{2n}. Let

a_{k} := e^{i(2k−1)π/(2n)} , k = 1, 2, … n

be the zeros of the equation z^{2n} + 1 = 0 in the upper half-plane. Let

M_{2n}(z) := ∏_{k=1}^{n} (z − a_{k})^{2} ,  z ∈ ℂ ,
P_{2n}(x) :=Re(M_{2n}(x)) ,  x ∈ ℝ ,

and

Q_{2n}(x) :=Im(M_{2n}(x)) ,  x ∈ ℝ .

a] Show that if n is even, then P_{2n} ∈ SR_{2n} and Q_{2n} ∈ ASR_{2n}, while if n is odd, then Q_{2n} ∈ SR_{2n} and P_{2n} ∈ ASR_{2n}.

b] Show that

|M_{2n}(x)| = 1 + x^{2n} , x ∈ ℝ

and

P_{2n}(x)^{2} + Q_{2n}(x)^{2} = (1 + x^{2n})^{2} , x ∈ ℝ ,

in particular

∥P_{2n}∥_{[−1,1]} ≤ 2  and  ∥Q_{2n}∥_{[−1,1]} ≤ 2 .

Proof. Note that

(7.1.24)  \frac{M_{2n}(x)}{1 + x^{2n}} = ∏_{k=1}^{n} \frac{x − a_{k}}{x − \overline{a}_{k}} ,

which implies the first equality. The rest is straightforward from the definitions. ⊓ ⊔

c] Show that there are extended real numbers

∞ = z_{0} > z_{1} > ⋯ > z_{2n} = −∞

such that

\frac{M_{2n}(z_{j})}{1 + z_{j}^{2n}} = (−1)^{j} , j = 0, 1, …, 2n

(the value of the left-hand side at ±∞ is defined by taking the limit when x → ±∞).

Hint: Use (7.1.24) and the argument principle (see, for example, Ash [71]). ⊓ ⊔

7.1 Inequalities for Rational Function Spaces 343

d] Let n ∈ ℕ be even. Show that there exist

1 = x_{0} > y_{1} > x_{1} > y_{2} > ⋯ > x_{n−1} > y_{n} > x_{n} = −1

such that

\frac{P_{2n}(x_{j})}{1 + x_{j}^{2n}} = (−1)^{j+n/2} , Q_{2n}(x_{j}) = 0 , j = 0, 1, …, n

and

\frac{Q_{2n}(y_{j})}{1 + y_{j}^{2n}} = (−1)^{j+1+n/2}, P_{2n}(y_{j}) = 0 , j = 1, 2, … n .

Formulate the analog statement when n ∈ ℕ is odd. e] Show that there exists an absolute constant c > 0 such that

|P′_{2n}(x)| + |Q′_{2n}(x)| ≥ cn min \{(1 + log n), log (\frac{e}{1 − x^{2}} )\}

for every n ∈ ℕ and x ∈ [−1, 1].

Proof. If x ∈ [−1, 1], then

|P′_{2n}(x)| + |Q′_{2n}(x)| ≥ |M′_{2n}(x)| = |M′_{2n}(x)| |\frac{1 + x^{2n}}{M_{2n}(x)}|

≥ |\frac{M′_{2n}(x)}{M_{2n}(x)}| = |∑_{k=1}^{n} \frac{2}{x − a_{k}} | ≥ 2 ∑_{k=1}^{n} \frac{Im(a_{k})}{|x − a_{k}|^{2}}

≥ cn min \{(1 + log n), log (\frac{e}{1 − x^{2}} )\}

with an absolute constant c > 0, where the last inequality follows from (7.1.22). ⊓ ⊔

The next part shows the sharpness of the inequality of E.11 a]. f] Let c > 0 be the same absolute constant as in part e]. For the sake of brevity let

δ_{n}(x) := (cn min \{(1 + log n), log (\frac{e}{1 − x^{2}} )\})^{−1} .

Show that for every interval

I_{n,x} := [x, x + 8δ_{n}(x)] ⊂ [0, 1] ,

there exist y_{1} ∈ I_{n,x} and y_{2} ∈ I_{n,x} such that

|P′_{2n}(y_{1})| ≥ \frac{1}{2}δ_{n}(x)^{−1}  and  |Q′_{2n}(y_{2})| ≥ \frac{1}{2}δ_{n}(x)^{−1} .

Proof. Suppose that

|P′_{2n}(y)| < \frac{1}{2}δ_{n}(x)^{−1}

for every y ∈ I_{n,x}. Then, from part e] we can deduce that

|Q′_{2n}(y)| > \frac{1}{2}δ_{n}(x)^{−1}

for every y ∈ I_{n,x}. Therefore

∥Q_{2n}∥_{[−1,1]} ≥ \frac{1}{2} ∫_{I_{n,x}} |Q′_{2n}(y)| dy > \frac{1}{2}8δ_{n}(x)\frac{1}{2}δ_{n}(x)^{−1} = 2 ,

which contradicts the last inequality of part b]. This finishes the proof of the first inequality. The second inequality can be proven in the same way.

⊓ ⊔

g] Show that

p′(1) = \frac{1}{2}n p(1)

for every p ∈ SR_{n}^{c}.

By using the quasi Chebyshev polynomials for SR_{2n} and ASR_{2n}, it can be shown that the inequality of E.11 c] is essentially sharp for the classes SR_{n} and ASR_{2n} for every m. This has been pointed out to us by Szabados. The argument requires some more technical details than the proof in the m = 1 case discussed in the above exercise.

7.2 Inequalities for Logarithmic Derivatives

We derive a series of metric inequalities of the form

m (\{x ∈ ℝ : \frac{r′(x)}{r(x)} ≥ α\}) ≤ \frac{cn}{α} , α > 0 ,

where r is a rational function of type (n, n) and c is a constant independent of n. Here m is the Lebesgue measure, although, since the sets in question are usually just finite unions of intervals, this is mostly a notational convenience. One of the interesting features of these inequalities is their easy extension from the polynomial case to the rational case.

The basic inequality is due to Loomis [46]. Note the invariance of the measure of the set in this case.

7.2 Inequalities for Logarithmic Derivatives 345

Theorem 7.2.1. If p ∈ P_{n} has n real roots, then

m (\{x ∈ ℝ : \frac{p′(x)}{p(x)} ≥ α\}) = \frac{n}{α} , α > 0 .

Proof. By considering α^{−1}p instead of p, it is sufficient to prove the theorem for α = 1. We first consider the case where p has distinct roots, which are denoted by α_{1} < α_{2} < ⋯ < α_{n}. Then

\frac{p′(x)}{p(x)} = ∑_{k=1}^{n} \frac{1}{x − α_{i}} .

Let β_{1} < β_{2} < ⋯ < β_{n} be the roots of p − p′, which must all be real. Note that these are the points where p′(x)/p(x) = 1. It is now easy to see from the graph that

\{x ∈ ℝ : \frac{p′(x)}{p(x)} ≥ 1\} = [α_{1}, β_{1}] ∪ [α_{2}, β_{2}] ∪ ⋯ ∪ [α_{n}, β_{n}]

and

m (\{x ∈ ℝ : \frac{p′(x)}{p(x)} ≥ 1\}) = ∑_{i=1}^{n} β_{i} − ∑_{i=1}^{n} α_{i} .

However, if p(x) := x^{n} + a_{n−1}x^{n−1} + ⋯ + a_{0}, then

∑_{i=1}^{n} α_{i} = −a_{n−1} ,

while

∑_{i=1}^{n} β_{i} = −(a_{n−1} − n)

is −1 times the second coefficient of p − p′.

This gives the result for distinct roots. The case when some of the roots of p are repeated can be handled by an easy limiting argument. ⊓ ⊔

Corollary 7.2.2. If α_{i} ∈ ℝ, c_{i} > 0, i = 1, 2, …, n, and ∑_{i=1}^{n} c_{i} = 1, then

m (\{x ∈ ℝ : ∑_{i=1}^{n} \frac{c_{i}}{x − α_{i}} ≥ α\}) = \frac{1}{α} , α > 0 .

Proof. For c_{i} rational this follows immediately from Theorem 7.2.1 on clearing the denominators of c_{i} by multiplying by an integer. The real case is an obvious limiting argument. ⊓ ⊔

In order to extend Theorem 7.2.1 to arbitrary polynomials we need the following generalization of E.3 of Section 2.4 due to Videnskii [51]. The proof is indicated in the exercises.

Lemma 7.2.3. If p ∈ P_{n} is positive on [a, b], then there exists q, s ∈ P_{n} nonnegative on [a, b] with all real roots (in [a, b]) so that

p(x) = q(x) + s(x) .

We now prove the unrestricted case of Theorem 7.2.1.

Theorem 7.2.4. Let p ∈ P_{n}. Then

m (\{x ∈ ℝ : \frac{p′(x)}{p(x)} ≥ α\}) ≤ \frac{2n}{α} , α > 0 .

Proof. Let α > 0 and let p_{n} ∈ P_{n}. Choose a and b such that

\{x ∈ ℝ : \frac{p′(x)}{p(x)} ≥ α\} ⊂ [a, b] .

By Lemma 7.2.3 we can find polynomials q ∈ P_{2n} and s ∈ P_{2n} such that

p^{2}(x) = q(x) + s(x)

where, for x ∈ [a, b],

0 ≤ q(x) ≤ p^{2}(x)  and  0 ≤ s(x) ≤ p^{2}(x)

and both q and s have only real roots. Now

\{x ∈ ℝ : \frac{p′(x)}{p(x)} ≥ α\} = \{x ∈ ℝ : \frac{(p^{2})′(x)}{p^{2}(x)} ≥ 2α\} .

Also,

(p^{2})′(x) ≥ 2α p^{2}(x)

holds exactly when

q′(x) + s′(x) ≥ 2α(q(x) + s(x)) .

By Theorem 7.2.1

m (\{x ∈ ℝ : \frac{q′(x)}{q(x)} ≥ 2α\}) = m (\{x ∈ ℝ : \frac{s′(x)}{s(x)} ≥ 2α\}) = \frac{n}{α} .

Since q and s are nonnegative on [a, b], it follows that

m(\{x ∈ [a, b] : q′(x) + s′(x) ≥ 2α(q(x) + s(x))\}) ≤ \frac{2n}{α} ,

and the proof is finished. ⊓ ⊔

It can be shown that this inequality is asymptotically sharp to the extent that the constant 2 cannot be replaced by any smaller constant for large n; see Kristiansen [82].

Theorem 7.2.4 extends easily to rational functions.

Theorem 7.2.5. If r = p/q, where p, q ∈ P_{n}, then

m (\{x ∈ ℝ : \frac{r′(x)}{r(x)} ≥ α\}) ≤ \frac{8n}{α} , α > 0 .

Proof. We have

\frac{r′}{r} = \frac{p′}{p} − \frac{q′}{q}

and for α > 0,

\{x ∈ ℝ : \frac{r′(x)}{r(x)} ≥ α\} ⊂ \{x ∈ ℝ : \frac{p′(x)}{p(x)} ≥ \frac{α}{2}\} ∪ \{x ∈ ℝ : \frac{q′(x)}{q(x)} ≤ −\frac{α}{2}\} .

By Theorem 7.2.4

m (\{x ∈ ℝ : \frac{p′(x)}{p(x)} ≥ \frac{α}{2}\}) ≤ \frac{4n}{α} ,

and with s(x) := q(−x),

m (\{x ∈ ℝ : \frac{q′(x)}{q(x)} ≤ −\frac{α}{2}\}) = m (\{x ∈ ℝ : \frac{s′(x)}{s(x)} ≥ \frac{α}{2}\}) ≤ \frac{4n}{α} .

It follows that

m (\{x ∈ ℝ : \frac{r′(x)}{r(x)} ≥ α\}) ≤ \frac{8n}{α}

for every α > 0. ⊓ ⊔

This inequality probably does not have the exact constant. It can be shown (see Borwein, Rakhmanov, and Saff [to appear]) that the constant 8 cannot be replaced by any constant less than or equal to 2π.

Comments, Exercises, and Examples.

Many variants on the inequalities of this section are presented in E.2, E.3, E.4, and E.5. Some of these are in Borwein [82].

E.5 explores some metric properties of the lemniscate

E(p) := {z ∈ ℂ : |p(z)| ≤ 1}

of a monic polynomial p ∈ P_{n}^{c} of the form

p(z) = ∏_{i=1}^{n} (z − z_{i}) , z_{i} ∈ ℂ .

The reader is referred to Erdős, Herzog, and Piranian [58] for many results and open problems concerning the lemniscate of monic polynomials. One in particular, which is still open, conjectures that for monic polynomials p ∈ P_{n}^{c}, the length of the boundary of E(p) is maximal for p(z) := z^{n} − 1 and so is O(n). Pommerenke [61] has shown that this length is O(n^{2}). Borwein [95] improves this to O(n); see E.7. E.9 solves another problem of Erdős, namely, the diameter of E(p) for a monic polynomial p ∈ P_{n}^{c} is always at least 2.

Erdős [76] contains several other related open problems.

E.1 Polynomials as Sums of Polynomials with Real Roots.

a] Suppose p ∈ P_{2n} \ P_{2n−1}, and suppose that p > 0 on [a, b]. Then

p(x) = (x − a)(b − x)u^{2}(x) + v^{2}(x)

for some u ∈ P_{n−1} and v ∈ P_{n}, which have all their zeros in [a, b].

b] Suppose p ∈ P_{2n+1} \ P_{2n} and suppose that p > 0 on [a, b]. Then

p(x) = (b − x)u^{2}(x) + (x − a)v^{2}(x)

for some u, v ∈ P_{n}, which have all their zeros in [a, b].

Hint: Let p be a polynomial of degree 2n that is strictly positive on [a, b]. Let T_{n} be the Chebyshev polynomial for the Chebyshev system

\{\frac{1}{\sqrt{p(x)}} , \frac{x}{\sqrt{p(x)}} , …, \frac{x^{n}}{\sqrt{p(x)}}\} .

Then T_{n}(x) = v(x)/\sqrt{p(x)} with some v ∈ P_{n}. Show that, for part a], v and u defined by

(x − a)(b − x)u^{2}(x) = p(x) − v^{2}(x)

are the required polynomials. Use a similar construction for part b]. ⊓ ⊔

E.2 Various Specializations. For the next exercises we use the notation P_{n}^{+} to denote the polynomial of degree at most n with nonnegative coefficients, and P_{n}^{↑} to denote those elements of P_{n} that are nondecreasing on [0, ∞).

a] If r = p/q, where p, q ∈ P_{n} and both p and q have only real roots, then

m (\{x ∈ ℝ : \frac{r′(x)}{r(x)} ≥ α\}) ≤ \frac{4n}{α} , α > 0 .

b] Let r(x) := x^{n}/(4n − x)^{n}. Then

m (\{x ∈ ℝ : \frac{r′(x)}{r(x)} ≥ 1\}) = 4n .

c] If r = p/q, where p ∈ P_{n} and q ∈ P_{n}^{↑}, then

m (\{x ≥ 0 : \frac{r′(x)}{r(x)} ≥ α\}) ≤ \frac{2n}{α} , α > 0 .

d] If r = p/q, where p ∈ P_{n}^{+} and q ∈ P_{n}^{↑}, then

m (\{x ≥ 0 : \frac{r′(x)}{r(x)} ≥ α\}) ≤ \frac{n}{α} , α > 0 .

e] Let r(x) := x^{n}. Then

m (\{x ≥ 0 : \frac{r′(x)}{r(x)} ≥ α\}) = \frac{n}{α} , α > 0 .

E.3 Another Metric Inequality. If p ∈ P_{n} has n real roots lying in the interval (a, b), then

m (\{x ∈ ℝ : |\frac{p′(x)}{p(x)}| ≤ \frac{α}{|(x − a)(b − x)|}\}) = \frac{2n}{α} , α > 0 .

Outline. Prove that

m (\{x ∈ ℝ : 0 ≤ \frac{(x − a)(b − x)p′(x)}{p(x)} ≤ α\}) = \frac{α}{n}

and

m (\{x ∈ ℝ : 0 ≥ \frac{(x − a)(b − x)p′(x)}{p(x)} ≥ −α\}) = \frac{α}{n} .

First consider the case when p has distinct zeros. Let y_{0} < y_{1} < ⋯ < y_{n} denote the n + 1 roots of (x − a)(b − x)p′(x), and let x_{0} < x_{1} < ⋯ < x_{n−1} denote the n roots of p(x). Then

y_{0} < x_{0} ≤ y_{1} < ⋯ x_{n−1} < y_{n} < x_{n} := ∞ .

Since

lim_{x→x_{i}−}(x − a)(b − x)\frac{p′(x)}{p(x)} = −∞ ,

we can deduce that for each interval (y_{i}, x_{i}) there exists a point δ_{i} ∈ (y_{i}, x_{i}) such that

(δ_{i} − a)(b − δ_{i})p′(δ_{i}) = −αp(δ_{i}) .

Since the above equation can have at most n + 1 solutions, we have

m (\{x ∈ ℝ : 0 ≥ \frac{(x − a)(b − x)p′(x)}{p(x)} ≥ −α\}) = ∑_{i=0}^{n}(δ_{i} − y_{i}) .

Now proceed as in the proof of Theorem 7.2.1. ⊓ ⊔ E.4 Extensions to P_{n}^{c}.

a] If p ∈ P_{n}^{c} , then

m (\{x ∈ ℝ : |\frac{p′(x)}{p(x)}| ≥ α\}) ≤ \frac{8\sqrt{2} n}{α} , α > 0 .

Hint: Write p as the sum of its real and imaginary parts. ⊓ ⊔

b] If r = p/q with p, q ∈ P_{n}^{c}, then

m (\{x ∈ ℝ : |\frac{r′(x)}{r(x)}| ≥ α\}) ≤ \frac{32\sqrt{2} n}{α} , α > 0 .

E.5 On the Lemniscate E(p). Let

p(z) := ∏_{i=1}^{n} (z − z_{i}) , z_{i} ∈ ℂ

and let

E(p) := {z ∈ ℂ : |p(z)| ≤ 1} .

a] Show that

m(E(p) ∩ ℝ)) ≤ 4 · 2^{−1/n}

with equality only for the Chebyshev polynomial of degree n normalized to have lead coefficient 1; see Pólya [28].

Hint: Analogously to the proof of the Remez inequality of Section 5.1, show that the Chebyshev polynomial transformed to an interval of length 4 is extremal for this problem. ⊓ ⊔

b] Let m_{2}(·) denote the planar Lebesgue measure. Show that

m_{2}(E(p)) ≤ 4π .

(In fact, m_{2}(E(p)) ≤ π, which is exact for z^{n}; this is due to Pólya [28].)

E.6 Cartan’s Lemma. Let

p(z) := ∏_{j=1}^{n} (z − z_{j}) , z_{j} ∈ ℂ

and β > 0 be fixed. Then there exist at most n open disks, the sum of whose radii is at most 2β, so that if z ∈ ℂ is outside the union of these open disks then

|p(z)| > (\frac{β}{e} )^{n} .

Prove Cartan’s Lemma as follows:

a] Let α_{1}, α_{2}, …, α_{ν} be fixed complex numbers. Let β > 0. Show that there exists a positive integer μ less than or equal to ν for which there exists an open disk with radius μβ/n containing α_{j} for exactly μ distinct values of j = 1, 2, …, ν.

Hint: Suppose to the contrary that there is no such positive integer μ. Show that this would imply the existence of an open disk with radius νβ/n containing α_{j} for at least ν + 1 distinct values of j = 1, 2, …, ν, which is a contradiction. ⊓ ⊔

b] Show that there exist open disks D_{1}, D_{2}, …, D_{k} and positive integers m_{1}, m_{2}, …, m_{k} with the following properties:

(1) ∑_{j=1}^{k} m_{j} = n , m_{1} ≥ m_{2} ≥ ⋯ ≥ m_{k};

(2) D_{j} has radius \frac{m_{j}β}{n} ;

(3) D_{j} ∩ E_{j} contains exactly m_{j} zeros of p, where

E_{j} := ℂ \ (D_{1} ∪ D_{2} ∪ … ∪ D_{j−1}) ;

(4) for every integer m > m_{j}, no open disk of radius \frac{mβ}{n} contains exactly m zeros of p in E_{j}.

Hint: Use part a]. ⊓ ⊔

c] Let D_{1}, D_{2}, …, D_{k} be the open disks specified in part b]. For j = 1, 2, …, k, let D_{j}^{∗} be the disk with the same center as D_{j} and with twice its radius. Show that for every

z ∈ E^{∗} := ℂ \ (D_{1}^{∗} ∪ D_{2}^{∗} ∪ ⋯ ∪ D_{k}^{∗})

there is a permutation z̃_{1}, z̃_{2}, …, z̃_{n} of the zeros z_{1}, z_{2}, …, z_{n} such that

|z − z̃_{j}| ≥ \frac{jβ}{n} , j = 1, 2, …, n .

Hint: Let z ∈ E^{∗} be fixed. Show, by induction on i, that for every i = 0, 1, …, n − 1, there are at least i + 1 zeros of p outside the open disk with center z and radius \frac{(n−i)β}{n} . Distinguish the cases

(1) n − i > m_{1} ;

(2) m_{j} ≥ n − i > m_{j+1}, j = 1, 2, …, k − 1 ;

(3) m_{k} ≥ n − i > 0 . ⊓ ⊔

d] Finish the proof of Cartan’s lemma.

E.7 The Length of the Boundary of E(p). Let

p(z) := ∏_{j=1}^{n} (z − z_{j}) , z_{j} ∈ ℂ

and let

E := E(p) := {z ∈ ℂ : |p(z)| ≤ 1} .

Show that the boundary ∂E of E is of length at most 4eπn. (In fact, with E.10 a], the estimate can be improved to (5.2) πn.)

Outline. Proceed as follows:

a] Let L be an arbitrary line in the complex plane. Show that the set ∂E ∩ L contains at most 2n distinct points.

Hint: By performing a translation and a rotation, if necessary, we may assume that L = ℝ. Now observe that there is a polynomial P (x, y) of degree at most 2n, in two real variables x and y, with complex coefficients, such that

∂E = {z ∈ ℂ : |p(z)|^{2} = 1}
= \{z ∈ ℂ : p(z)\overline{p(z)} = 1\}
= {z = x + iy : x, y ∈ ℝ , P (x, y) = 1} .

Hence

E ∩ ℝ = {x ∈ ℝ : P (x, 0) = 1}

and since P (x, 0) ∈ P_{2n}^{c}, the result follows. ⊓ ⊔

b] For a ∈ ℂ and r > 0, let Q be the square

Q := {z ∈ ℂ : |Re(z − a)| < r , |Im(z − a)| < r} .

Show that Q ∩ ∂E is of length at most 8rn.

Hint: Divide Q ∩ ∂E into subcurves C_{1}, C_{2}, …, C_{m} so that every vertical and horizontal line contains at most one point of each C_{j}. Let l_{x}^{(j)} and l_{y}^{(j)} denote the length of the interval obtained by projecting C_{j} to the x axes and y axes, respectively. Let

l_{x} := ∑_{j=1}^{m} l_{x}^{(j)}  and  l_{y} := ∑_{j=1}^{m} l_{y}^{(j)} .

Use part a] to show that l_{x} ≤ 4nr and l_{y} ≤ 4rn. Hence, if l denotes the length of Q ∩ ∂E_{n}, then

l ≤ l_{x} + l_{y} ≤ 8rn .
⊓ ⊔

c] Show that the boundary ∂E of E is of length at most 16en.

Hint: Combine Cartan’s lemma (see E.7) and part b]. ⊓ ⊔

The rest of the exercise is about improving 16en to 4πen.

d] Let B be the open disk with center a ∈ ℂ and radius r > 0. Show that B ∩ ∂E is of length at most 2πrn.

Outline. Let Q be the square

Q := {z ∈ ℂ : |Re(z)| < r , |Im(z)| < r} .

For α ∈ [0, 2π) let Q_{α} be the square

Q_{α} := {a + e^{iα}(z − a) : z ∈ Q} .

Let l(α) denote the length of Q_{α} ∩ ∂E. Let l_{x}(α) and l_{y}(α) denote the total length of the intervals obtained by projecting Q_{α} ∩ ∂E to the lines

{z = re^{iα} : r ∈ ℝ}  and  {z = re^{i(α+π/2)} : r ∈ ℝ} ,

respectively (counting multiplicities). The precise definition of l_{x}(α) and l_{y}(α) can be formulated in the same way as in the hint to part b], which is left to the reader. Use part a] to show that

l_{x}(α) + l_{y}(α) ≤ 8rn , α ∈ [0, 2π) .

Hence

\frac{4}{π} l(0) = \frac{1}{2π} ∫_{0}^{2π} l(0)(| sin α| + | cos α|) dα

= \frac{1}{2π} ∫_{0}^{2π} (l_{x}(α) + l_{y}(α)) dα ≤ 8rn ,

and l(0) ≤ 2πrn follows. ⊓ ⊔

e] Prove the initial statement of the exercise.

Hint: Combine Cartan’s lemma (see E.6) and part d]. ⊓ ⊔

E.8 On the Length of Another Lemniscate. Suppose p ∈ P_{n}^{c}. Show that the length of the lemniscate

F = F (p) := \{z ∈ ℂ : |\frac{p′(z)}{p(z)}| = n\}

is at most 16n(1 + log n) (actually at most 4πn(1 + log n)).

Hint: The arguments are very similar to those given in the outline to E.7. First show that if L is an arbitrary line in the complex plane, then F ∩ L contains at most 2n distinct points. Next prove that if D_{r} is an open disk of radius r in the complex plane, then D_{r} ∩ F is of length at most 8rn (actually at most 2πrn). Now use E.6 c] with β = 1 + log n. ⊓ ⊔

The proof of the following exercise requires some familiarity with harmonic functions.

E.9 On the Diameter of E(p). The diameter diam(A) of a nonempty set A ⊂ ℂ is defined by

diam(A) := sup{|z_{1} − z_{2}| : z_{1}, z_{2} ∈ A} .

Let p ∈ P_{n}^{c} be an arbitrary monic polynomial of degree n, that is,

p(z) := ∏_{j=1}^{n} (z − z_{j}) , z_{j} ∈ ℂ .

Then the diameter of the set

E(p) := {z ∈ ℂ : |p(z)| ≤ 1}

is at least 2.

Proceed as follows:

a] Let

ℂ̂ := ℂ ∪ {∞}  and  Δ := {z ∈ ℂ̂ : |z| > 1} .

Let p ∈ P_{n}^{c} be monic. Let E := co(E(p)), that is, the convex hull of E(p). Use the Riemann mapping theorem (see Ahlfors [53]) to show that there exists a function g of the form

g(z) = bz + ∑_{j=0}^{∞} b_{j}z^{−j} , z ∈ Δ , b, b_{j} ∈ ℂ

such that g is analytic and one-to-one on Δ, and

g(Δ) = ℂ̂ \ E .

Hint: Note that ℂ̂ \ E is simply connected. ⊓ ⊔

b] Let b be the same as in part a]. Show that |b| ≥ 1.

Outline. Because of the definition of E, for every ε > 0 there exists a δ > 0 such that

−ε ≤ log |z^{−n}p(g(z))| , |z| = 1 + δ .

Since G(z) := log |z^{−n}p(g(z))| is harmonic on Δ, we have

G(∞) = \frac{1}{2π(1 + δ)} ∫_{0}^{2π} G((1 + δ)e^{iθ}) dθ ≥ \frac{−ε}{2π(1 + δ)}

for every ε > 0, so G(∞) ≥ 0. On the other hand, since p ∈ P_{n}^{c} is a monic polynomial of degree n,

G(∞) = log |b|^{n} = n log |b| ,

from which |b| ≥ 1 follows. ⊓ ⊔

c] Show that diam(E) ≥ 2 .

Outline. Assume to the contrary that diam(E) < 2. Then there exists a δ > 0 such that

|z^{−1}(g(z) − g(−z))| < 2 , |z| = 1 + δ .

Since

F (z) := z^{−1}(g(z) − g(−z)) = 2b + 2b_{1}z^{−2} + 2b_{3}z^{−4} + ⋯

is analytic on Δ, the maximum principle (E.1 d] of Section 1.2) yields

2|b| = |F (∞)| ≤ max_{|z|=1+δ} |F (z)| = max_{|z|=1+δ} |z^{−1}(g(z) − g(−z))| < 2 ,

that is, |b| < 1, which contradicts part b]. ⊓ ⊔

d] Note that diam(A) = diam(co(A)) for every nonempty A ⊂ ℂ, in particular, diam(E(p)) = diam(E).

The more general result that diam(A) ≥ 2 cap(A) for every nonempty A ⊂ ℂ is observed in Pommerenke [75].

E.10 More on E(p). Suppose p is a monic polynomial with complex coefficients. As before, let

E := E(p) := {z ∈ ℂ : |p(z)| ≤ 1} .

a] It follows from E.6 (Cartan’s lemma) that the set E(p) can be covered by disks the sum of whose radii is at most 2e. It is conjectured in Erdős, Herzog, and Piranian [58] that the correct value in this problem is 2. (The current best constant is less than 2.6.)

b] If E(p) is connected, then it is contained in a disk with radius 2 centered at \frac{1}{n} ∑_{k=1}^{n} z_{k}, where z_{1}, z_{2}, …, z_{n} are the zeros of p.

Proof. This is conjectured in Erdős, Herzog, and Piranian [58] and proved in Pommerenke [59b]. ⊓ ⊔

c] If E(p) is connected, then its circumference is at least 2π.

Proof. This is also conjectured in Erdős, Herzog, and Piranian [58], and proved in Pommerenke [59b]. ⊓ ⊔