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3

Chebyshev and Descartes Systems

Overview

A Chebyshev space is a finite-dimensional subspace of C(A) of dimension n + 1 that has the property that any element that vanishes at n + 1 points vanishes identically. Such spaces, whose prototype is the space P_{n} of real algebraic polynomials of degree at most n, share with the polynomials many basic properties. The first section is an introduction to these Chebyshev spaces. A basis for a Chebyshev space is called a Chebyshev system. Two special families of Chebyshev systems, namely, Markov systems and Descartes systems, are examined in the second section. The third section examines the Chebyshev “polynomials” associated with Chebyshev spaces. These associated Chebyshev polynomials, which equioscillate like the usual Chebyshev polynomials, are extremal for various problems in the supremum norm. The fourth section studies particular Descartes systems

(x^{λ_{0}}, x^{λ_{1}}, …) , λ_{0} < λ_{1} < …

on (0, ∞) in detail. These systems, which we call Müntz systems, can be very explicitly orthonormalized on [0, 1], and this orthogonalization is also examined. The final section constructs Chebyshev “polynomials” associated with the Chebyshev spaces

span \{1 , \frac{1}{x − a_{1}} , …, \frac{1}{x − a_{n}} \} , a_{i} ∈ ℝ \ [−1, 1]

on [−1, 1] and explores their various properties.

92 3. Chebyshev and Descartes Systems

3.1 Chebyshev Systems

From an approximation theoretic point of view an essential property that polynomials of degree at most n have is that they can uniquely interpolate at n + 1 points. This is equivalent to the fact that a polynomial of degree at most n that vanishes at n + 1 points vanishes identically. Any (n + 1)- dimensional vector space of continuous functions with this property is called a Chebyshev space or sometimes a Haar space. Many basic approximation properties extend to these spaces. The precise definition is the following.

Definition 3.1.1 (Chebyshev System). Let A be a Hausdorff space. The sequence (f_{0}, …, f_{n}) is called a real (or complex) Chebyshev system or Haar system of dimension n + 1 on A if f_{0}, …, f_{n} are real- (or complex-) valued continuous functions on A, span{f_{0}, …, f_{n}} over ℝ (or ℂ) is an (n+1)-dimensional subspace of C(A), and any element of span{f_{0}, …, f_{n}} that has n + 1 distinct zeros in A is identically zero.

If (f_{0}, …, f_{n}) is a Chebyshev system on A, then span{f_{0}, …, f_{n}} is called a Chebyshev space or Haar space on A.

Chebyshev systems and spaces will be assumed to be real, unless we explicitly specify otherwise. If A ⊂ ℝ, then the topology on A is always meant to be the usual metric topology.

Implicit in the definition is that A contains at least n + 1 points. Being a Chebyshev system is a property of the space spanned by the elements of the system, so every basis of a Chebyshev space is a Chebyshev system.

A point x_{0} ∈ (a, b) is called a double zero of an f ∈ C[a, b] if f(x_{0}) = 0 and f(x_{0} − ϵ)f(x_{0} + ϵ) > 0 for all sufficiently small ϵ > 0 (in other words, if f vanishes without changing sign at x_{0}). It is easy to see that if (f_{0}, …, f_{n}) is a Chebyshev system on [a, b] ⊂ ℝ, then every 0 ≠ f ∈ span{f_{0}, …, f_{n}} has at most n zeros even if each double zero is counted twice; see E.10. Chebyshev spaces are defined via zero counting, and many of the theorems in the theory of Chebyshev spaces are proved by zero counting arguments. So it is important to make the agreement that, unless it is stated explicitly otherwise, we count the zeros of an element f from a Chebyshev space on [a, b] so that each double zero of f is counted twice.

The following simple equivalences hold:

Proposition 3.1.2 (Equivalences). Let f_{0}, …, f_{n} be real- (or complex-) valued continuous functions on a Hausdorff space A (containing at least n + 1 points). Then the following are equivalent:

3.1 Chebyshev Systems 93

a] Every 0 ≠ p ∈ span{f_{0}, …, f_{n}} has at most n distinct zeros in A. b] If x_{0}, …, x_{n} are distinct elements of A and y_{0}, …, y_{n} are real (or complex) numbers, then there exists a unique p ∈ span{f_{0}, …, f_{n}} such that

p(x_{i}) = y_{i} , i = 1, 2, …, n .

c] If x_{0}, …, x_{n} are distinct points of A, then

D(x_{0}, …, x_{n}) := | \begin{matrix}f_{0}(x_{0}) & … & f_{n}(x_{0}) \\ ⋮ & ... & ⋮ \\ f_{0}(x_{n}) & … & f_{n}(x_{n}) \end{matrix} | ≠ 0 .

Proof. These equivalences are all elementary facts in linear algebra. ⊓ ⊔

On an interval there is a sign regularity to the determinants in c].

Proposition 3.1.3. Suppose (f_{0}, …, f_{n}) is a (real) Chebyshev system on [a, b] ⊂ ℝ. Then there exists a δ := −1 or δ := 1 such that

δ | \begin{matrix}f_{0}(x_{0}) & … & f_{n}(x_{0}) \\ ⋮ & ... & ⋮ \\ f_{0}(x_{n}) & … & f_{n}(x_{n}) \end{matrix} | > 0

for any a ≤ x_{0} < x_{1} < ⋯ < x_{n} ≤ b.

Proof. This follows immediately from part c] of the previous proposition and continuity considerations. That is, if D(x_{0}, …, x_{n}) < 0 while D(y_{0}, …, y_{n}) > 0, then for some λ ∈ (0, 1)

D(λx_{0} + (1 − λ)y_{0}, …, λx_{n} + (1 − λ)y_{n}) = 0 ,

which is impossible. ⊓ ⊔

The intimate relationship between Chebyshev systems and best approximation in the uniform norm is indicated by the next result. In order to state it we need to introduce the notion of an alternation sequence.

Definition 3.1.4 (Alternation Sequence). Let A ⊂ ℝ and let

x_{0} < x_{1} < ⋯ < x_{n}

be n + 1 points of A. Then (x_{0}, x_{1}, …, x_{n}) is said to be an alternation sequence of length n + 1 for a real valued f ∈ C(A) if

|f(x_{i})| = ∥f∥_{A} , i = 0, 1, …, n

and

sign(f(x_{i+1})) = −sign(f(x_{i})) , i = 0, 1, …, n − 1 .

Definition 3.1.5 (Best Approximation). Suppose that U is a (finitedimensional) subspace of a normed space (V, ∥ · ∥). If g ∈ V and p ∈ U satisfy

∥g − p∥ = inf_{h∈U} ∥g − h∥ ,

then p is said to be a best approximation to g from U.

As a result of the finite dimensionality of the subspace U, at least one best approximation to any g ∈ V from U exists. This is straightforward since

T := {p ∈ U : ∥p∥ ≤ ∥g∥ + 1}

is a compact subset of U, so any sequence (p_{j}) of approximations to g from U satisfying

∥g − p_{j}∥ ≤ j^{−1} + inf_{h∈U} ∥g − h∥

has a convergent subsequence with limit in U. This limit is then a best approximation to g from U.

Theorem 3.1.6 (Alternation of Best Approximations). Suppose (f_{0}, …, f_{n}) is a Chebyshev system on [a, b] ⊂ ℝ. Let A be a closed subset of [a, b] containing at least n + 2 distinct points. Then p ∈ H_{n} := span{f_{0}, …, f_{n}} is a best approximation to g ∈ C(A) from H_{n} in the uniform norm on A if and only if there exists an alternation sequence of length n + 2 for g − p on A.

Proof. The proof of the only if part of the theorem is mostly an example of a standard type of perturbation argument that will recur later.

The perturbation argument goes as follows. Suppose p is a best approximation of required type and suppose a alternation sequence of maximal length for g − p is

(x_{0} < x_{1}, < ⋯ < x_{m})

where x_{i} ∈ A and where m is strictly less than n + 1. Suppose, without loss of generality, that

g(x_{0}) − p(x_{0}) > 0

(otherwise multiply by −1). Now let

Y := {x ∈ A : |g(x) − p(x)| = ∥g − p∥_{A}} .

Note that Y is compact. Since (x_{0} < x_{1} < … < x_{m}) is an alternation sequence of maximal length, we can divide Y into m + 1 disjoint compact subsets Y_{0}, Y_{1}, …, Y_{m} with

x_{0} ∈ Y_{0} , x_{1} ∈ Y_{1} , …, x_{m} ∈ Y_{m}

so that

sign(g(x) − p(x)) = −sign(g(y) − p(y)) ≠ 0 , x ∈ Y_{i} , y ∈ Y_{i+1} .

Now choose a p^{∗} ∈ span{f_{0}, …, f_{n}} satisfying

sign_{x∈Y_{i}}(p^{∗}(x)) = (−1)^{i} , i = 0, 1, …, m .

This can be done by choosing points z_{i} with

max Y_{i−1} < z_{i} < min Y_{i} , i = 1, 2, …, m

and then applying E.11. We now claim that, for δ > 0 sufficiently small,

(3.1.1)  ∥g − (p + δp^{∗})∥_{A} < ∥g − p∥_{A} ,

which contradicts the fact that p is a best approximation, and so there must exist an alternation set of length n + 2 for g − p on A. To verify (3.1.1) we proceed as follows:

For each i = 0, 1, …, m choose an open set O_{i} ⊂ [a, b] (in the usual metric topology relative to [a, b]) containing Y_{i} so that for every x ∈ \overline{O}_{i},

(3.1.2)  sign(g(x) − p(x)) = sign(p^{∗}(x))

and

(3.1.3)  |g(x) − p(x)| ≥ \frac{1}{2}∥g − p∥_{A} .

Now pick a δ_{1} > 0 such that for every x ∈ B := A \ ⋃_{i=0}^{m} O_{i} and δ ∈ (0, δ_{1}),

|g(x) − (p(x) + δp^{∗}(x))| < ∥g − p∥_{A} ,

which can be done since B is compact and by construction

∥g − p∥_{B} < ∥g − p∥_{A} .

Note that (3.1.2) and (3.1.3) allow us to pick a δ_{2} > 0 such that for x ∈ ⋃_{i=0}^{m} \overline{O}_{i} and δ ∈ (0, δ_{2}),

|g(x) − (p(x) + δp^{∗}(x))| < |g(x) − p(x)| .

This verifies (3.1.1) and finishes the direct half of the theorem.

The proof of the converse is simple. Suppose there is an alternation sequence of length n + 2 for g − p on A, and suppose there exists a p^{∗} with

∥g − p^{∗}∥_{A} < ∥g − p∥_{A}.

Then p^{∗}−p has at least n+1 zeros on [a, b], one between any two consecutive alternation points for g − p on A, and hence it vanishes identically. This contradiction finishes the proof. ⊓ ⊔

In the setting of Theorem 3.1.6 the best approximation is unique; see E.5.

Comments, Exercises, and Examples.

The terminology is not entirely standard in the literature with Chebyshev systems often referred to as Haar systems on intervals of ℝ. There are various proofs of Theorem 3.1.6, of which ours is by no means the most elegant (see Cheney [66], for example). The point of this proof is that it easily modifies to deal with characterizations of extremal functions for various other extremal problems. Many good books cover this standard material. See, for example, Cheney [66], Lorentz [86a], or Pinkus [89]; see also Appendix 3. An extensive treatment of Chebyshev systems is available in Karlin and Studden [66] or Nürnberger [89], where E.3 can be found. E.4 shows that real Chebyshev systems are intrinsically one-dimensional.

E.1 Examples of Chebyshev systems.

a] Suppose 0 = λ_{0} < λ_{1} < ⋯ < λ_{n}. Show that

(x^{λ_{0}}, x^{λ_{1}}, …, x^{λ_{n}})

is a Chebyshev system on [0, ∞).

b] Suppose λ_{0} < λ_{1} < ⋯ < λ_{n}. Show that

(x^{λ_{0}}, x^{λ_{1}}, …, x^{λ_{n}})

is a Chebyshev system on (0, ∞).

c] Suppose λ_{0} < λ_{1} < ⋯ < λ_{n}. Show that

(x^{λ_{0}}, x^{λ_{0}} log x , x^{λ_{1}}, x^{λ_{1}} log x , …, x^{λ_{n}}, x^{λ_{n}} log x)

is a Chebyshev system on (0, ∞).

d] Suppose λ_{0} < λ_{1} < ⋯ < λ_{n}. Show that

(\frac{1}{x − λ_{0}} , \frac{1}{x − λ_{1}} , …, \frac{1}{x − λ_{n}} )

is a Chebyshev system on (−∞, ∞) \ {λ_{0}, λ_{1}, …, λ_{n}}.

e] Suppose λ_{0} < λ_{1} < ⋯ < λ_{n}. Show that

(e^{λ_{0}x}, e^{λ_{1}x}, …, e^{λ_{n}x})

is a Chebyshev system on (−∞, ∞).

f] Show that

(1, cos θ, sin θ, cos 2θ, sin 2θ, …, cos nθ, sin nθ)

is a Chebyshev system on [0, 2π).

g] Show that

(1, cos θ, cos 2θ, …, cos nθ)

is a Chebyshev system on [0, π).

E.2 More Examples.

a] If (f_{0}, …, f_{n}) is a Chebyshev system on A, then it is also a Chebyshev system on any subset B of A containing at least n + 1 points.

b] If (f_{0}, …, f_{n}) is a Chebyshev system on A and g ∈ C(A) is strictly positive on A, then (gf_{0}, …, gf_{n}) is also a Chebyshev system on A.

c] If (f_{0}, …, f_{n}) is a Chebyshev system on [0, 1], then

(1 , ∫_{0}^{x} f_{0}(t) dt , …, ∫_{0}^{x} f_{n}(t) dt)

is also a Chebyshev system on [0, 1].

See E.8 of Section 3.2, which treats the effect of differentiation on a Chebyshev system.

E.3 Extended Complete Chebyshev Systems. Let (g_{0}, …, g_{n}) be a sequence of functions in C^{n}[a, b]. Define the Wronskian determinant

W (g_{0}, …, g_{m})(t) := | \begin{matrix}g_{0}(t) & g_{1}(t) & … & g_{m}(t) \\ g′_{0}(t) & g′_{1}(t) & … & g′_{m}(t) \\ ⋮ & ⋮ & ... & ⋮ \\ g_{0}^{(m)}(t) & g_{1}^{(m)}(t) & … & g_{m}^{(m)}(t) \end{matrix} | .

We say that (g_{0}, …, g_{n}) is an extended complete Chebyshev system (ECT system) on [a, b] if

W (g_{0}, …, g_{m})(t) > 0 , m = 0, 1, …, n , t ∈ [a, b] .

a] Let span{g_{0}, …, g_{n}} be an (n + 1)-dimensional subspace of C^{n}[a, b]. Show that the following statements are equivalent:

(i) For every m = 0, 1, …, n, 0 ≠ f ∈ span{g_{0}, …, g_{m}} has at most m zeros in [a, b] counting multiplicities (x_{0} ∈ [a, b] is a zero of f with multiplicity k if f(x_{0}) = f′(x_{0}) = ⋯ f^{(k−1)}(x_{0}) = 0 and f^{(k)}(x_{0}) ≠ 0).

(ii) For each i = 0, 1, …, n, there exists a choice of δ_{i} := 1 or δ_{i} := −1 such that

(δ_{0}g_{0}, δ_{1}g_{1}, …, δ_{n}g_{n})

is an ECT system on [a, b].

In particular, every ECT system on [a, b] is a Chebyshev system on [a, b].

Proof. For details see Karlin and Studden [66]. ⊓ ⊔ b] Characterization Theorem. The following statements are equivalent:

(i) (g_{0}, …, g_{n}) is an ECT system on [a, b].

(ii) There exist w_{i} ∈ C^{n−i}[a, b], i = 0, 1, …, n, with w_{i} strictly positive such that

g_{0}(t) = w_{0}(t) ,
g_{1}(t) = w_{0}(t) ∫_{a}^{t} w_{1}(t_{1}) dt_{1} ,
⋮
g_{n}(t) = w_{0}(t) ∫_{a}^{t} w_{1}(t_{1}) ∫_{a}^{t_{1}} w_{2}(t_{2}) ⋯ ∫_{a}^{t_{n−1}} w_{n}(t_{n}) dt_{n} ⋯ dt_{2} dt_{1} .

Proof. This is proved by induction on n. See Karlin and Studden [66]. ⊓ ⊔

c] Suppose λ_{0} < λ_{1} < ⋯ < λ_{n}. Show that (x^{λ_{0}}, …, x^{λ_{n}}) is an ECT system on [a, b] provided a > 0.

E.4 Railway Track Theorem. Real Chebyshev systems exist only on very special subsets of ℝ^{m}. Indeed, real Chebyshev systems intrinsically live on one-dimensional subsets.

a] Suppose A ⊂ ℝ^{m} contains three distinct arcs that join at a point x_{0}. Then, for n ≥ 2, there exists no real Chebyshev system (f_{0}, …, f_{n}) on A.

Proof. Suppose there exists a real Chebyshev system (f_{0}, …, f_{n}) on such a set A. Let

V (x, y) := D(x, y, x_{2}, x_{3}, …, x_{n})

(D is defined in Proposition 3.1.2) which is never zero for distinct points x, y, x_{2}, x_{3}, …, x_{n}. Choose distinct points x_{0}, x_{1}, …, x_{n} on one of the three distinct arcs so that x_{0} is adjacent to x_{1}. Pick the points z_{1} ≠ x_{0} and z_{2} ≠ x_{0} so that z_{1}, z_{2}, and x_{1} are on different arcs. Now consider interchanging x := x_{0} and y := x_{1} by moving x from x_{0} to z_{1}, y from x_{1} to z_{2}, x from z_{1} to x_{1}, and y from z_{2} to x_{0}. Since x, y, x_{2}, x_{3}, …, x_{n} remain distinct, V (x, y) does not vanish in this process. This contradicts the fact that V (x, y) is continuous and V (x_{0}, x_{1}) = −V (x_{1}, x_{0}). ⊓ ⊔

The following more general result of Mairhuber [56] also holds:

b] Mairhuber’s Theorem. If (f_{0}, …, f_{n}) is a real Chebyshev space on A, then A is homeomorphic to a subset of the unit circle.

E.5 Uniqueness of Best Approximations. Prove that a best approximation from a Chebyshev space satisfying the conditions of Theorem 3.1.6 is unique.

Hint: Suppose f has two best approximations p_{1} ∈ H_{n} and p_{2} ∈ H_{n}. Then, by the alternation characterization, p_{1} − p_{2} ∈ H_{n} has at least n + 1 zeros on [a, b] (we count each internal zero without sign change twice). Now E.10 implies that p_{1} − p_{2} = 0. ⊓ ⊔

E.6 De la Vallée Poussin Theorem. Suppose H_{n} is a Chebyshev space of dimension (n + 1) on [a, b]. If p ∈ H_{n} and there exist n + 2 points

a ≤ x_{0} < x_{1} < ⋯ < x_{n+1} ≤ b

so that

sign(f(x_{i}) − p(x_{i})) = −sign(f(x_{i+1}) − p(x_{i+1})) , i = 0, 1, …, n ,

then

inf_{p∈H_{n}} ∥f − p∥_{[a,b]} ≥ min_{i=0,... ,n+1} |f(x_{i}) − p(x_{i})| .

E.7 Haar’s Characterization of Chebyshev Spaces. The following pretty theorem is due to Haar (for a proof, see E.3 of Appendix 3):

Theorem. Let f_{0}, …, f_{n} ∈ C(A) where A is a compact Hausdorff space containing at least n + 1 points. Then (f_{0}, …, f_{n}) is a Chebyshev system on A if and only if every g ∈ C(A) has a unique best approximation from span{f_{0}, …, f_{n}} in the uniform norm on A.

E.8 Best Approximation to x^{n}. Reprove Theorem 2.1.1 by using the alternation characterization of best approximations.

E.9 Best Rational Approximations. Let f ∈ C[a, b]. Then p/q ∈ R_{n,m} is a best approximation to f from R_{n,m} in C[a, b] if and only if f − p/q has an alternation set of length at least

2 + max{n + deg(q), m + deg(p)}

[a, b]. (Here we must assume p/q is written in a reduced form.)

The proof of this is a fairly complicated variant of the proof of Theorem 3.1.6 (see, for example, Cheney [66]).

E.10 Zeros of Functions in Chebyshev Spaces. As before, we call the point x_{0} ∈ (a, b) a double zero of f ∈ C[a, b] if f(x_{0}) = 0 and

f(x_{0} − ϵ)f(x_{0} + ϵ) > 0

for all sufficiently small ϵ > 0 (in other words, if f vanishes without changing sign at x_{0}). Let (f_{0}, …, f_{n}) be a Chebyshev system on [a, b] ⊂ ℝ. Show that every 0 ≠ f ∈ span{f_{0}, …, f_{n}} has at most n zeros even if each double zero is counted twice.

Hint: Use Proposition 3.1.2 b]. ⊓ ⊔ E.11 Functions in a Chebyshev Space with Prescribed Sign Changes. Let (f_{0}, …, f_{n}) be a Chebyshev system on [a, b], and let

a < z_{1} < z_{2} < ⋯ < z_{m} < b , 0 ≤ m ≤ n .

Show that there is a function p^{∗} ∈ span{f_{0}, …, f_{n}} such that

(i) p^{∗}(x) = 0 if and only if x = z_{i} for some i = 1, 2, …, m,

(ii) p^{∗}(x) changes sign at each z_{i}, i = 1, 2, …, m.

Hint: If m = n, then use Proposition 3.1.3 and a continuity argument to show that

p^{∗}(x) = | \begin{matrix}f_{0}(x) & f_{1}(x) & … & f_{n}(x) \\ f_{0}(z_{1}) & f_{1}(z_{1}) & … & f_{n}(z_{1}) \\ ⋮ & ⋮ & ... & ⋮ \\ f_{0}(z_{n}) & f_{1}(z_{n}) & … & f_{n}(z_{n}) \end{matrix} |

satisfies the requirements.

If m < n, then use the already proved case, a limiting argument, and E.10 to show that there are p_{j} ∈ span{f_{0}, …, f_{n}}, j = 1, 2, such that

(1) p_{j}(x) changes sign at x if and only if x = z_{i}, i = 1, 2, …, m,

(2) p_{1}(x) ≠ 0 for every x ∈ [a, z_{m}] \ {z_{1}, z_{2}, …, z_{m}},

(3) p_{2}(x) ≠ 0 for every x ∈ [z_{1}, b] \ {z_{1}, z_{2}, …, z_{m}}.

Now show that either p^{∗} := p_{1} + p_{2} or p^{∗} := p_{1} − p_{2} satisfies the requirements. ⊓ ⊔

E.12 The Dimension of a Chebyshev Space on a Circle. Let (f_{0}, …, f_{n}) be a Chebyshev system on a circle C. Show that n must be even. Observe that such Chebyshev systems exist.

Hint: Show that for every set of n distinct points x_{1}, x_{2}, …, x_{n} on the circle there is a p ∈ span{f_{0}, …, f_{n}} such that p(x) = 0 if and only if x ∈ {x_{1}, x_{2}, …, x_{n}} and p(x) changes sign at each x_{i}. ⊓ ⊔

3.2 Descartes Systems

Chebyshev systems capture some of the essential properties of polynomials. There are two additional types of systems that capture some additional properties.

Definition 3.2.1 (Markov System). We say that (f_{0}, …, f_{n}) is a Markov system on a Hausdorff space A if each f_{i} ∈ C(A), and {f_{0}, …, f_{m}} is a Chebyshev system for each m = 0, 1, …, n. (We allow n to tend to +∞, in which case we call the system an infinite Markov system on A.)

3.2 Descartes Systems 101

A Markov system is just a Chebyshev system with each initial segment also a Chebyshev system. Being a Chebyshev system is a property of the space not of the basis. However, the Markov system depends on the basis. For example,

(x^{λ_{0}}, x^{λ_{1}}, …) , λ_{0} < λ_{1} < ⋯

is a Markov system on any A ⊂ (0, ∞) containing infinitely many points, but not every basis of span{x^{λ_{0}}, x^{λ_{1}}, … } is a Markov system on A (see E.1).

Proposition 3.2.2. (f_{0}, …, f_{n}) is a Markov system on an interval [a, b] if and only if for each i = 0, 1, …, n, there exists a choice of δ_{i} := 1 or δ_{i} := −1 such that with g_{i} := δ_{i}f_{i},

D ( \begin{matrix}g_{0} & g_{1} & … & g_{m} \\ x_{0} & x_{1} & … & x_{m} \end{matrix} ) := | \begin{matrix}g_{0}(x_{0}) & … & g_{m}(x_{0}) \\ ⋮ & ... & ⋮ \\ g_{0}(x_{m}) & … & g_{m}(x_{m}) \end{matrix} | > 0

for every a ≤ x_{0} < x_{1} < ⋯ < x_{m} ≤ b and m = 0, 1, …, n.

Proof. This is an easy consequence of Proposition 3.1.3 by induction on n. ⊓ ⊔

A stronger property that a system on an interval can have is the following:

Definition 3.2.3 (Descartes System). The system (f_{0}, …, f_{n}) is said to be a Descartes system (or order complete Chebyshev system) on an interval I if each f_{i} ∈ C(I) and

D ( \begin{matrix}f_{i_{0}} & f_{i_{1}} & … & f_{i_{m}} \\ x_{0} & x_{1} & … & x_{m} \end{matrix} ) > 0

for any 0 ≤ i_{0} < i_{1} < ⋯ < i_{m} ≤ n and x_{0} < x_{1} < ⋯ < x_{m} from I. (Once again we allow n to tend to ∞.)

This again is a property of the basis. It implies that any finitedimensional subspace generated by some basis elements is a Chebyshev space on I. The canonical example of a Descartes system on [a, b], a > 0, is

(x^{λ_{0}}, x^{λ_{1}}, …) , λ_{0} < λ_{1} < ⋯

(see E.2). A Descartes system on I is obviously a Descartes system on any subinterval of I.

The following version of Descartes’ rule of signs holds for Descartes systems.

Theorem 3.2.4 (Descartes’ Rule of Signs). If (f_{0}, …, f_{n}) is a Descartes system on [a, b], then the number of distinct zeros of any

0 ≠ f = ∑_{i=0}^{n} a_{i}f_{i} , a_{i} ∈ ℝ

is not greater than the number of sign changes in (a_{0}, …, a_{n}).

A sign change occurs between a_{i} and a_{i+k} exactly when a_{i}a_{i+k} < 0 and a_{i+1} = a_{i+2} = ⋯ = a_{i+k−1} = 0.

Proof. Suppose (a_{0}, …, a_{n}) has p sign changes. Then we can partition {a_{0}, …, a_{n}} into exactly p + 1 blocks so that each block is of the form

a_{n_{m}+1}, a_{n_{m}+2}, …, a_{n_{m+1}}, m = 0, 1, …, p

(n_{0} := −1, n_{p+1} := n), where all of the coefficients in each of the blocks are of the same sign and not all the coefficients in a block vanish. Now let

g_{m} := ∑_{i=n_{m}+1}^{n_{m+1}} |a_{i}|f_{i} , m = 0, 1, …, p .

Then, for a ≤ x_{0} < x_{1} < ⋯ < x_{p} ≤ b,

D ( \begin{matrix}g_{0} & g_{1} & … & g_{p} \\ x_{0} & x_{1} & … & x_{p} \end{matrix} )

= ∑_{i_{0}=n_{0}+1}^{n_{1}} ⋯ ∑_{i_{p}=n_{p}+1}^{n_{p+1}} |a_{i_{0}}| ⋯ |a_{i_{p}}|D ( \begin{matrix}f_{i_{0}} & f_{i_{1}} & … & f_{i_{p}} \\ x_{0} & x_{1} & … & x_{p} \end{matrix} ) > 0

since each of the determinants in the sum is positive. Thus {g_{0}, ….g_{p}} is a (p + 1)-dimensional Chebyshev system on [a, b], and hence

f := ∑_{i=0}^{p} δ_{i}g_{i} , δ_{i} = ±1

has at most p zeros. This finishes the proof. ⊓ ⊔

A refined version of Descartes’ rule of signs for ordinary polynomials is presented in the exercises. The following comparison theorem due to Pinkus and, independently, Smith [78] will be of use later.

Theorem 3.2.5. Suppose (f_{0}, …, f_{n}) is a Descartes system on [a, b]. Suppose

p = f_{α} + ∑_{i=1}^{k} a_{i}f_{λ_{i}} ,  and  q = f_{α} + ∑_{i=1}^{k} b_{i}f_{γ_{i}} ,

where 0 ≤ λ_{1} < λ_{2} < ⋯ < λ_{k} ≤ n, 0 ≤ γ_{1} < γ_{2} < ⋯ < γ_{k} ≤ n,

0 ≤ γ_{i} ≤ λ_{i} < α , i = 1, 2, …, m ,

and

α < λ_{i} ≤ γ_{i} ≤ n , i = m + 1, m + 2, …, k

with strict inequality for at least one index i = 1, 2, …, k. If

p(x_{i}) = q(x_{i}) = 0 , i = 1, 2, …, k ,

where x_{i} ∈ [a, b] are distinct, then

|p(x)| ≤ |q(x)|

for all x ∈ [a, b] with strict inequality for x ≠ x_{i}.

The proof is left as a guided exercise (see E.4) with some interesting consequences presented in E.5.

Comments, Exercises, and Examples.

Theorem 3.2.4 characterizes Descartes systems; see Karlin and Studden [66, p. 25]. Some caution must be exercised since, as in the previous section, definitions are not entirely standard. We will explore two particular Descartes systems in greater detail later; see E.2 and E.3. For further material, the reader is referred to Karlin and Studden [66], Karlin [68], and Nürnberger [89].

E.1 Distinctions.

a] Given λ_{0} < λ_{1} < ⋯ and A ⊂ (0, ∞) with infinitely many points, show that (x^{λ_{0}}, x^{λ_{1}}, …) is a Markov system on A, but there is a basis for span{x^{λ_{0}}, x^{λ_{1}}, … }, which is not a Markov system on A.

b] Find a Markov system that is not a Descartes system.

E.2 Examples of Descartes Systems.

a] Suppose λ_{0} < λ_{1} < · · · . Show that the Müntz system

(x^{λ_{0}}, x^{λ_{1}}, …)

is a Descartes system on (0, ∞).

Hint: For every 0 ≤ i_{0} < i_{1} < ⋯ < i_{m}, the determinant

D ( \begin{matrix}x^{λ_{i_{0}}} & x^{λ_{i_{1}}} & … & x^{λ_{im}} \\ x_{0} & x_{1} & … & x_{m} \end{matrix} ) = | \begin{matrix}x_{0}^{λ_{i_{0}}} & … & x_{0}^{λ_{im}} \\ ⋮ & ... & ⋮ \\ x_{m}^{λ_{i_{0}}} & … & x_{m}^{λ_{im}} \end{matrix} |

is nonzero for any 0 < x_{0} < x_{1} < ⋯ < x_{m} < ∞ by Proposition 3.1.2 and E.1 a] of Section 3.1. It only remains to prove that it is positive whenever 0 < x_{0} < x_{1} < ⋯ < x_{m} < ∞. Observe that the exponents λ_{i_{j}} can be varied continuously (for fixed x_{i}) without changing the sign of the determinant provided no two ever become equal. Now perturb (λ_{i_{0}}, λ_{i_{1}}, …, λ_{i_{m}}) into (0, 1, …, m) and observe that the determinant becomes a Vandermonde determinant, as in part b], which in this case is positive. ⊓ ⊔

b] Vandermonde Determinant. Show that

| \begin{matrix}1 & x_{0} & … & x_{0}^{m} \\ 1 & x_{1} & … & x_{1}^{m} \\ ⋮ & ⋮ & ... & ⋮ \\ 1 & x_{m} & … & x_{m}^{m} \end{matrix} | = ∏_{0≤i<j≤m}(x_{j} − x_{i}) .

Hint: The determinant is a polynomial in x_{0}, x_{1}, …, x_{m} of degree m in each variable that vanishes whenever x_{i} = x_{j}. ⊓ ⊔

c] Suppose λ_{0} < λ_{1} < · · ·. Show that the exponential system

(e^{λ_{0}t}, e^{λ_{1}t}, …)

is a Descartes system on (−∞, ∞).

Hint: Use part a]. ⊓ ⊔

d] Suppose 0 < λ_{0} < λ_{1} < · · ·. Show that

(sinh λ_{0}t, sinh λ_{1}t, …)

is a Descartes system on (0, ∞).

Outline. Let 0 ≤ i_{0} < i_{1} < ⋯ < i_{m} be fixed integers. First we show that

(sinh λ_{i_{0}}t, sinh λ_{i_{1}}t, …, sinh λ_{i_{m}}t)

is a Chebyshev system on (0, ∞). Indeed, let

0 ≠ f ∈ span{sinh λ_{i_{0}}t, sinh λ_{i_{1}}t, …, sinh λ_{i_{m}}t} .

Then

0 ≠ f ∈ span{e^{±λ_{i_{0}} t}, e^{±λ_{i_{1}} t}, …, e^{±λ_{im} t}}

and by E.1 e] of Section 3.1, f has at most 2m zeros in (−∞, ∞). Since f is odd, it has at most m zeros in (0, ∞).

Since for every 0 ≤ i_{0} < i_{1} < ⋯ < i_{m}, (sinh λ_{i_{0}}t, …, sinh λ_{i_{m}}t) is a Chebyshev system on (0, ∞), the determinant

D ( \begin{matrix}sinh λ_{i_{0}}t & sinh λ_{i_{1}}t & … & sinh λ_{i_{m}}t \\ x_{0} & x_{1} & … & x_{m} \end{matrix} )

= | \begin{matrix}sinh λ_{i_{0}}x_{0} & sinh λ_{i_{1}}x_{0} & … & sinh λ_{i_{m}}x_{0} \\ sinh λ_{i_{0}}x_{1} & sinh λ_{i_{1}}x_{1} & … & sinh λ_{i_{m}}x_{1} \\ ⋮ & ⋮ & ... & ⋮ \\ sinh λ_{i_{0}}x_{m} & sinh λ_{i_{1}}x_{m} & … & sinh λ_{i_{m}}x_{m} \end{matrix} |

is nonzero for any 0 < x_{0} < x_{1} < ⋯ < x_{m} < ∞ Proposition 3.1.2. So it only remains to prove that it is positive whenever 0 < x_{0} < x_{1} < ⋯ < x_{m} < ∞. Now let

D(α) := D ( \begin{matrix}sinh λ_{i_{0}}t & sinh λ_{i_{1}}t & … & sinh λ_{i_{m}}t \\ αx_{0} & αx_{1} & … & αx_{m} \end{matrix} )

= | \begin{matrix}sinh λ_{i_{0}}αx_{0} & sinh λ_{i_{1}}αx_{0} & … & sinh λ_{i_{m}}αx_{0} \\ sinh λ_{i_{0}}αx_{1} & sinh λ_{i_{1}}αx_{1} & … & sinh λ_{i_{m}}αx_{1} \\ ⋮ & ⋮ & ... & ⋮ \\ sinh λ_{i_{0}}αx_{m} & sinh λ_{i_{1}}αx_{m} & … & sinh λ_{i_{m}}αx_{m} \end{matrix} |

and

D^{∗}(α) := D ( \begin{matrix}\frac{1}{2}e^{λ_{i_{0}} t} & \frac{1}{2}e^{λ_{i_{1}} t} & … & \frac{1}{2}e^{λ_{im} t} \\ αx_{0} & αx_{1} & … & αx_{m} \end{matrix} )

= | \begin{matrix}\frac{1}{2}e^{λ_{i_{0}} αx_{0}} & \frac{1}{2}e^{λ_{i_{1}} αx_{0}} & … & \frac{1}{2}e^{λ_{im} αx_{0}} \\ \frac{1}{2}e^{λ_{i_{0}} αx_{1}} & \frac{1}{2}e^{λ_{i_{1}} αx_{1}} & … & \frac{1}{2}e^{λ_{im} αx_{1}} \\ ⋮ & ⋮ & ... & ⋮ \\ \frac{1}{2}e^{λ_{i_{0}} αx_{m}} & \frac{1}{2}e^{λ_{i_{1}} αx_{m}} & … & \frac{1}{2}e^{λ_{im} αx_{m}} \end{matrix} |

where 0 < x_{0} < x_{1} < ⋯ < x_{m} < ∞ are fixed. Since

(sinh λ_{i_{0}}t, sinh λ_{i_{1}}t, …, sinh λ_{i_{m}}t)

and

(e^{λ_{i_{0}} t}, e^{λ_{i_{1}} t}, …, e^{λ_{im} t})

are Chebyshev systems on (0, ∞), D(α) and D^{∗}(α) are continuous nonvanishing functions of α on (0, ∞). Now, observe that

lim_{α→∞} |D(α)| = lim_{α→∞} |D^{∗}(α)| = ∞  and  lim_{α→∞} \frac{D(α)}{D^{∗}(α)} = 1 .

By part c],

(e^{λ_{i_{0}} t}, e^{λ_{i_{1}} t}, …, e^{λ_{im} t})

106 3. Chebyshev and Descartes Systems

is a Descartes system on (−∞, ∞), hence D^{∗}(α) > 0 for every α > 0. So the above limit relations imply that D(α) > 0 for every large enough α, hence for every α > 0. In particular,

D(1) = D ( \begin{matrix}sinh λ_{i_{0}}t & sinh λ_{i_{1}}t & … & sinh λ_{i_{m}}t \\ x_{0} & x_{1} & … & x_{m} \end{matrix} ) > 0 ,

which finishes the proof. ⊓ ⊔ e] Suppose 0 < λ_{0} < λ_{1} < · · ·. Show that

(cosh λ_{0}t, cosh λ_{1}t, …)

is a Descartes system on (0, ∞). Hint: Proceed as in the outline for part d]. ⊓ ⊔

E.3 Rational Systems. a] Cauchy Determinants. Show that

| \begin{matrix}\frac{1}{α_{1}+β_{1}} & … & \frac{1}{α_{1}+β_{m}} \\ ⋮ & ... & ⋮ \\ \frac{1}{α_{m}+β_{1}} & … & \frac{1}{α_{m}+β_{m}} \end{matrix} | = \frac{∏_{1≤i<j≤m}(α_{j} − α_{i})(β_{j} − β_{i})}{∏_{1≤i,j≤m}(α_{i} + β_{j})} .

Hint: Multiply both sides above by ∏_{1≤i,j≤m}(α_{i} + β_{j}) and observe that both sides are polynomials of the same degree, m−1, in each variable α_{i}, β_{i}. Also both sides vanish exactly when α_{i} = α_{j} or β_{i} = β_{j}. So up to a constant both sides are the same. Now show that the constant is 1. ⊓ ⊔ b] Let α_{1} > α_{2} > ⋯ > b. Show that

(\frac{1}{α_{1} − x} , \frac{1}{α_{2} − x} , …)

is a Descartes system on [a, b].

Let α_{1} < α_{2} < ⋯ < a. Show that

(\frac{1}{x − α_{1}} , \frac{1}{x − α_{2}} , …)

is a Descartes system on [a, b] (see also E.6 e]).

E.4 Proof of Theorem 3.2.5. Assume the notation of this Theorem 3.2.5. a] Let 0 ≤ δ_{0} < δ_{1} < ⋯ < δ_{μ} ≤ n and a < x_{1} < x_{2} < ⋯ < x_{μ} < b. Show that there exists a unique p = f_{δ_{μ}} + ∑_{i=0}^{μ−1} a_{i}f_{δ_{i}} such that (1) p(x_{i}) = 0 , i = 1, 2, …, μ .

Show also that the above p has the following properties:

(2) p(x) changes sign at each x_{i} ,

(3) p(x) ≠ 0 if x ∉ {x_{1}, x_{2}, …, x_{μ}} ,

(4) a_{i}a_{i+1} < 0 , i = 0, 1, …, μ − 1 , a_{μ} := 1 ,

(5) p(x) > 0 , x ∈ (x_{μ}, b] .

Hint: Since (f_{δ_{0}}, …, f_{δ_{μ}}) and (f_{δ_{0}}, …, f_{δ_{μ−1}}) are Chebyshev systems, E.11 of Section 3.1 shows that there exists a p of the desired form satisfying (1). Since (f_{δ_{0}}, …, f_{δ_{μ−1}}) is a Chebyshev system, this p is unique. Now E.11 of Section 3.1 yields that p satisfies (2) and (3). By Theorem 3.2.4, p satisfies (4). The fact that (5) holds for p follows from expanding the determinant

D ( \begin{matrix}f_{δ_{0}} & f_{δ_{1}} & … & f_{δ_{μ−1}} & f_{δ_{μ}} \\ x_{1} & x_{2} & … & x_{μ} & x \end{matrix} )

by Cramer’s rule. This determinant is just cp(x) with some c > 0 since it vanishes at each x_{i}, and the coefficient of f_{δ_{μ}} is positive; see Definition 3.2.3. Also, the above determinant is positive for all x ∈ (x_{μ}, b]; see Definition 3.2.3 again. ⊓ ⊔

b] Prove Theorem 3.2.5.

Outline. For notational simplicity assume that α = n (hence m = k); the general case is analogous. Further, we may assume that there is an index j such that

γ_{j} < λ_{j} and γ_{i} = λ_{i} whenever i ≠ j

since the result follows from this by a finite number of pairwise comparisons. So we assume

p = f_{n} + a_{j}f_{λ_{j}} + ∑_{{i=1 \atop i≠j}}^{k} a_{i}f_{λ_{i}}

and

q = f_{n} + b_{j}f_{γ_{j}} + ∑_{{i=1 \atop i≠j}}^{k} b_{i}f_{λ_{i}} ,

where 0 ≤ λ_{1} < λ_{2} < ⋯ < λ_{k} < n and 0 ≤ λ_{j−1} < γ_{j} < λ_{j} for some 1 ≤ j ≤ k (of course, the inequality λ_{j−1} < γ_{j} holds only if λ_{j−1} is defined, that is, only if j ≥ 2). Then

p − q = a_{j}f_{λ_{j}} − b_{j}f_{γ_{j}} + ∑_{{i=1 \atop i≠j}}^{k} (a_{i} − b_{i})f_{λ_{i}}

has exactly k zeros on [a, b] at x_{1}, x_{2}, …, x_{k} because p − q is in a (k + 1)- dimensional Chebyshev subspace.

108 3. Chebyshev and Descartes Systems

By property (5) in part a] applied to p and q, respectively,

p(x) > 0 and q(x) > 0 , x ∈ (x_{k}, b] .

Now by property (4) in part a] applied to c(p−q), where c is chosen so that the lead coefficient of c(p − q) is 1, and by the fact that p and p − q have the same coefficient for f_{λ_{j}}, the lead coefficient of p − q (a_{k} − b_{k} provided λ_{k} > λ_{j}) is negative. So property (5) in part a] implies that

p(x) − q(x) < 0, x ∈ (x_{k}, b] .

Hence 0 < p(x) < q(x), x ∈ (x_{k}, b].

Now use property (3) in part a] and the fact that all of p, q, and p − q change sign only at x_{i} to finish the proof. ⊓ ⊔

The following extension of part a] will be used later: c] Suppose

0 ≤ δ_{0} < δ_{1} < ⋯ < δ_{μ} ≤ n , a ≤ x_{1} ≤ x_{2} ≤ ⋯ ≤ x_{μ} ≤ b , a < x_{2}, x_{μ−1} < b, and x_{i} < x_{i+2}, i = 1, 2, …, μ − 2. Show that there exists a unique p = f_{δ_{μ}} + ∑_{i=0}^{μ−1} a_{i}f_{δ_{i}} (with a_{i} ∈ ℝ) such that

(1) p(x_{i}) = 0, i = 1, 2, …, μ ,

(2) p(x) changes sign at x_{i} if and only if x_{i} ∉ {a, b, x_{i−1}, x_{i+1}} . Show also that

(3) p(x) ≠ 0 if x ∉ {x_{1}, x_{2}, …, x_{μ}} ,

(4) a_{i}a_{i+1} ≤ 0 , i = 0, 1, …, μ − 1 , a_{μ} := 1 ,

(5) p(x) > 0 , x ∈ (x_{μ}, b) ,

(6) (−1)^{μ}p(x) > 0 , x ∈ (a, x_{1}) ,

(7) (−1)^{μ−i}p(x) > 0 , x ∈ (x_{i}, x_{i+1}) , i = 1, 2, …, μ − 1 . Hint: Use part a] and a limiting argument. The uniqueness follows from E.10 of Section 3.1. ⊓ ⊔

The next exercise provides a solution to a problem of Lorentz, which is settled in Borosh, Chui, and Smith [77]. E.5 A Problem of Lorentz on Best Approximation to x^{λ}. Suppose that [a, b] ⊂ [0, ∞), n ∈ ℕ, and p ∈ (0, ∞] are fixed. Let μ be a finite Borel measure on [a, b]. a] Suppose λ_{1}, λ_{2}, …, λ_{n} are arbitrary fixed real numbers if a > 0, or fixed real numbers greater than −1/p if a = 0. Let f ∈ L_{p}(μ) be fixed. Show that

E_{p,μ}(λ_{1}, λ_{2}, …, λ_{n}; f) := min_{a_{i}∈ℝ} ∥f(x) − ∑_{i=1}^{n} a_{i}x^{λ_{i}} ∥_{L_{p}(μ)}

exists and is finite.

Outline. Use a standard compactness argument. ⊓ ⊔

b] Suppose 1 < p < ∞, the support of μ contains at least n + 1 points, λ_{1}, λ_{2}, …, λ_{n}, λ are arbitrary fixed distinct real numbers if a > 0, or fixed real numbers greater than −1/p if a = 0. Show that if (ã_{i})_{i=1}^{n} ⊂ ℝ satisfies

E_{p,μ}(λ_{1}, λ_{2}, …, λ_{n}; x^{λ}) = ∥x^{λ} − ∑_{i=1}^{n} ã_{i}x^{λ_{i}} ∥_{L_{p}(μ)} ,

then

f(x) := x^{λ} − ∑_{i=1}^{n} ã_{i}x^{λ_{i}}

has exactly n sign changes on (a, b).

Hint: Since (x^{λ_{1}}, …, x^{λ_{n}}, x^{λ}) is a Chebyshev system, it is sufficient to prove that f has at least n sign changes on (a, b). Suppose f has at most n − 1 sign changes on [a, b]. Then, since (x^{λ_{1}}, …, x^{λ_{n}}) is a Chebyshev system, by E.11 of Section 3.1 there exists an element

h ∈ span{x^{λ_{1}}, …, x^{λ_{n}}}

such that

|f(x)|^{p−1} sign(f(x))h(x) ≥ 0

on [a, b] with strict inequality at all but n points (at every point where f does not vanish). Using that the support of μ contains at least n+1 distinct points, this implies

∫_{a}^{b} |f|^{p−1} sign(f)h dμ > 0 ,

which contradicts E.7 h] of Section 2.2. ⊓ ⊔

c] Suppose p = ∞, supp(μ) = [a, b], λ_{1}, λ_{2}, …, λ_{n}, λ are arbitrary fixed distinct real numbers if a > 0, or fixed distinct nonnegative real numbers if a = 0. Show that if (ã_{i})_{i=1}^{n} ⊂ ℝ satisfies

E_{∞,μ}(λ_{1}, λ_{2}, …, λ_{n}; x^{λ}) = ∥x^{λ} − ∑_{i=1}^{n} ã_{i}x^{λ_{i}} ∥_{L_{∞}(μ)} ,

then

f(x) := x^{λ} − ∑_{i=1}^{n} ã_{i}x^{λ_{i}}

has exactly n sign changes on (a, b).

Hint: Use Theorem 3.1.6. ⊓ ⊔ d] Best Approximation to x^{λ} from Certain Classes of Müntz Polynomials. Let 1 < p ≤ ∞. Suppose the support of μ contains at least n + 1 distinct points if 1 < p < ∞ or supp(μ) = [a, b] if p = ∞. Let λ > γ_{1} > γ_{2} > ⋯ be arbitrary fixed real numbers if a > 0, or fixed nonnegative real numbers if a = 0. Suppose we wish to minimize

E_{p,μ}(λ_{1}, λ_{2}, …, λ_{n}; x^{λ})

for all sets of n distinct real numbers λ_{1} > λ_{2} > ⋯ > λ_{n} satisfying

{λ_{1}, λ_{2}, …, λ_{n}} ⊂ {γ_{1}, γ_{2}, … } .

Show that the minimum occurs if and only if

{λ_{1}, λ_{2}, …, λ_{n}} = {γ_{1}, γ_{2}, …, γ_{n}} .

Hint: Let

E_{p,μ}(λ_{1}, λ_{2}, …, λ_{n}; x^{λ}) = ∥x^{λ} − ∑_{i=1}^{n} ã_{i}x^{λ_{i}} ∥_{L_{p}(μ)} ,

where {λ_{1}, λ_{2}, …, λ_{n}} is a set of n distinct real numbers for which the minimum is taken; see part a]. By parts b] and c]

f(x) := x^{λ} − ∑_{i=1}^{n} ã_{i}x^{λ_{i}}

has exactly n sign changes x_{1}, x_{2}, …, x_{n} on (a, b). Let

g ∈ span{x^{γ_{1}}, x^{γ_{2}}, …, x^{γ_{n}}}

interpolate x^{λ} at the points x_{1}, x_{2}, …, x_{n}. Now use Theorem 3.2.5 to finish the proof. ⊓ ⊔

E.6 Strictly Totally Positive Kernels (Karlin [68]). A (continuous) function K(s, t) is an STP kernel on [a, b] × [c, d] if

| \begin{matrix}K(s_{0}, t_{0}) & … & K(s_{0}, t_{n}) \\ ⋮ & ... & ⋮ \\ K(s_{n}, t_{0}) & … & K(s_{n}, t_{n}) \end{matrix} | > 0

for all a ≤ s_{0} < ⋯ < s_{n} ≤ b, c ≤ t_{0} < ⋯ < t_{n} ≤ d, and for all n > 0. a] Observe that E.3 b] implies that

K(s, t) = \frac{1}{s + t} is STP on [a, b] × [a, b] , a > 0 .

Observe also that E.2 b] implies that

K(s, t) = e^{st} is STP on (−∞, ∞) × (−∞, ∞) .

b] Suppose K is STP on [a, b] × [c, d], and (f_{0}, …, f_{n}) is a Chebyshev system on [a, b]. Show that if

v_{i}(x) = ∫_{a}^{b} K(t, x)f_{i}(t) dt , i = 0, 1, …, n ,

then (v_{0}, …, v_{n}) is a Chebyshev system on [c, d].

c] Variation Diminishing Property. Suppose K is STP on [a, b] × [c, d] and suppose f ∈ C[a, b]. Let

g(x) := ∫_{a}^{b} K(t, x)f(t) dt .

Then g has no more sign changes on [c, d] than f has on [a, b].

d] The Laplace transform of a function f ∈ C[0, ∞) ∩ L_{1}[0, ∞)

L(f)(x) := ∫_{0}^{∞} f(t)e^{−tx} dt

has no more sign changes on [0, ∞) than f does.

Proof. This follows from parts a] and c]. It may also be proved directly by induction as follows. Suppose f has exactly n sign changes on [0, ∞), one at x_{0}. Then g(x) := (x_{0} − x)f(x) has exactly n − 1 sign changes on [0, ∞). Now observe that

e^{−x_{0}x} \frac{d}{dx}(e^{x_{0}x}L(f)(x)) = L(g)(x) ,

so L(f) has at most one more sign change on [0, ∞) than L(g) does. ⊓ ⊔

e] Use part d] and E.2 a] to reprove that

(\frac{1}{x + α_{1}} , \frac{1}{x + α_{2}} , …) , −a < α_{1} < α_{2} < ⋯

satisfies Descartes’ rule of signs on [a, b]. ⊓ ⊔

Proof. Observe that

∫_{0}^{∞}e^{−α_{i}t}e^{−tx} dt = \frac{1}{x + α_{i}}

for every x ∈ (−α_{i}, ∞). ⊓ ⊔ E.7 Descartes’ Rule of Signs for Polynomials.

a] Prove by induction that ∑_{k=0}^{n} a_{k}x^{k} ∈ P_{n} has no more zeros in (0, ∞) (repeated zeros are counted according to their multiplicities) than the number of sign changes in (a_{0}, a_{1}, …, a_{n}).

b] Let α > 0. Let p(x) = ∑_{k=0}^{n} a_{k}x^{k} and q(x) := (x−α)p(x) = ∑_{k=0}^{n+1} b_{k}x^{k}. Show that if the number of sign changes in {a_{0}, a_{1}, …, a_{n}} is m, then the number of sign changes in {b_{0}, b_{1}, …, b_{n+1}} is at least m + 1.

c] Give another proof of a] based on b].

d] In part a] the number of sign changes in (a_{0}, a_{1}, …, a_{n}) exceeds the number of positive zeros by an even integer.

Hint: See Pólya and Szegő [76]. ⊓ ⊔

Refinements of the above exercise are presented in E.6 of Appendix 1, where Cauchy indices are discussed.

The first part of the following exercise is a version of a result from Zielke [79]:

E.8 The Effect of Differentiation on Weak Markov Systems. The system (f_{0}, …, f_{n}) is called a weak Chebyshev system on [a, b] if f_{i} ∈ C[a, b] for each i and every f ∈ span{f_{0}, …, f_{n}} has at most n sign changes on [a, b] (so the only difference between a Chebyshev system and a weak Chebyshev system is that in the definition of the latter, zeros without sign change are not counted).

Analogously, the system (g_{0}, …, g_{n}) is called a weak Markov system on [a, b] if g_{i} ∈ C[a, b] for each i and (g_{0}, …, g_{m}) is a weak Chebyshev system on [a, b] for every m = 0, 1, …, n (so the only difference between a Markov system and a weak Markov system is that in the definition of the latter, zeros without sign change are not counted).

a] Suppose (1, f_{1}, …, f_{n}) is a weak Markov system of C^{1} functions on [a, b]. Show that (f′_{1}, …, f′_{n}) is a weak Markov system on [a, b].

Outline. Proceed by induction on n. If n = 0, then the statement is obvious. Suppose that the statement is true for n − 1. By the inductive hypothesis (f′_{1}, …, f′_{n−1}) is a weak Markov system on [a, b], hence Rolle’s theorem implies that (1, f_{1}, …, f_{n−1}) is a Markov system on [a, b].

Suppose that g ∈ span{f′_{1}, …, f′_{n−1}} is of the form

g := ∑_{i=1}^{n} a_{i}f′_{i} , a_{i} ∈ ℝ

and g has at least n sign changes on [a, b]. Then there exist n + 2 distinct points

a < x_{1} < x_{2} < ⋯ < x_{n+2} < b

and ϵ = ±1 such that

F := ϵ ∑_{i=1}^{n} a_{i}f_{i}

satisfies

(−1)^{i}(F (x_{i+1}) − F (x_{i})) > 0 , i = 1, 2, …, n + 1 .

Since (1, f_{1}, …, f_{n−1}) is a Markov system on [a, b], by Proposition 3.1.2, there exist functions

G_{δ} ∈ span{1, f_{1}, …, f_{n−1}}

such that

G_{δ}(x_{i}) = F (x_{i}) + δ(−1)^{i} , i = 2, 3, …, n + 1

for every δ > 0. Then by the inductive hypothesis, G′_{δ} has at most n − 2 sign changes on [a, b]. It follows that if δ > 0 is sufficiently small, then

(−1)(G_{δ}(x_{2}) − G_{δ}(x_{1})) < 0

and

(−1)^{n+1}(G_{δ}(x_{n+2}) − G_{δ}(x_{n+1})) < 0 ,

otherwise G′_{δ} would have at least n sign changes on [a, b]. Now show that for sufficiently small δ > 0

F − G_{δ} ∈ span{1, f_{1}, …, f_{n}}

has at least n + 1 sign changes, which is a contradiction. ⊓ ⊔

b] Suppose that (1, f_{1}, …, f_{n}) is an ECT system on [a, b] and suppose that each f_{i} ∈ C^{n}[a, b] (see E.3 of Section 3.1). Show that (f′_{1}, …, f′_{n}) is also an ECT system on [a, b] with each f′_{i} ∈ C^{n−1}[a, b].

Hint: Use the definition given in E.3 of Section 3.1. ⊓ ⊔

c] Suppose (1, f_{1}, …, f_{n}) is a weak Markov system on [a, b] with each f_{i} ∈ C^{1}[a, b]. Show that (1, f_{1}, …, f_{n}) is a Markov system on [a, b].

Hint: Use Rolle’s theorem and part a]. ⊓ ⊔

3.3 Chebyshev Polynomials in Chebyshev Spaces

Suppose

H_{n} := span{f_{0}, f_{1}, …, f_{n}}

is a Chebyshev space on [a, b], and A is a compact subset of [a, b] with at least n + 1 points. We can define the generalized Chebyshev polynomial

T_{n} := T_{n}{f_{0}, f_{1}, …, f_{n}; A}

for H_{n} on A by the following three properties:

(3.3.1)  T_{n} ∈ span{f_{0}, f_{1}, …, f_{n}}

there exists an alternation sequence (x_{0} < x_{1} < ⋯ < x_{n}) for T_{n} on A, that is,

(3.3.2)  sign(T_{n}(x_{i+1})) = −sign(T_{n}(x_{i})) = ±∥T_{n}∥_{A}

for i = 0, 1, …, n − 1, and

(3.3.3)  ∥T_{n}∥_{A} = 1 with T_{n}(max A) > 0 .

Of course the existence and uniqueness of such a T_{n} has to be proved. Note that if together with span{f_{0}, …, f_{n}}, span{f_{0}, …, f_{n−1}} is also a Chebyshev space, then Theorem 3.1.6 implies that

T_{n} = c (f_{n} − ∑_{k=0}^{n−1} a_{k}f_{k}) ,

where the numbers a_{0}, a_{1}, …, a_{n−1} ∈ ℝ are chosen to minimize

(3.3.4)  ∥f_{n} − ∑_{k=0}^{n−1} a_{k}f_{k}∥_{A} ,

satisfies properties (3.3.1) and (3.3.2), and the normalization constant c ∈ ℝ can be chosen so that T_{n} satisfies property (3.3.3) as well. In E.1 we outline the proof of the existence and uniqueness of a T_{n} satisfying properties (3.3.1) to (3.3.3) without assuming that span{f_{0}, …, f_{n−1}} is a Chebyshev space.

Note that if (f_{0}, …, f_{n}) is a Descartes system on [a, b], then the normalization constant (that is, the lead coefficient) c in T_{n} is positive. This follows from E.4 of Section 3.2.

3.3 Chebyshev Polynomials in Chebyshev Spaces 115

On intervals, with f_{i}(x) := x^{i}, the definition (3.3.1) to (3.3.3) gives the usual Chebyshev polynomials; see E.7 of Section 2.1.

The Chebyshev polynomials T_{n} for H_{n} on A encode much of the information of how the space H_{n} behaves with respect to the uniform norm on A. Many extremal problems are solved by the Chebyshev polynomials.

When (f_{0}, f_{1}, …) is a Markov system on [a, b] we can introduce the sequence (T_{n})_{n=0}^{∞} of associated Chebyshev polynomials

T_{n} := T_{n}{f_{0}, f_{1}, …, f_{n}; [a, b]}

for H_{n} on [a, b]. Then (T_{0}, T_{1}, …) is a Markov system on [a, b] again with the same span. (One reason for not always choosing this as a canonical basis is that it is never a Descartes system.)

The denseness of Markov spaces in C[a, b] is intimately tied to the location of the zeros of the associated Chebyshev polynomials; see Section 4.1.

An example of an extremal problem solved by the Chebyshev polynomials is the following:

Theorem 3.3.1. Suppose H_{n} := span{f_{0}, …, f_{n}} is a Chebyshev space on [a, b] with associated Chebyshev polynomial

T_{n} := T_{n}{f_{0}, f_{1}, …, f_{n}; [a, b]}

and each f_{i} is differentiable at b. Then

max{|p′(b)| : p ∈ H_{n} , ∥p∥_{[a,b]} ≤ 1 , p(b) = T_{n}(b)}

is attained by T_{n}.

Proof. Suppose p ∈ H_{n}, ∥p∥_{[a,b]} ≤ 1, and p(b) = T_{n}(b). We need to show that |p′(b)| ≤ |T′_{n}(b)|. Let a ≤ ζ_{0} < ζ_{1} < ⋯ < ζ_{n} ≤ b be the points of alternation for T_{n}, that is,

T_{n}(ζ_{i}) = ±(−1)^{i} , i = 0, 1, …, n .

Note that T_{n} − p has at least n zeros in [ζ_{0}, ζ_{n}], one in each [ζ_{i−1}, ζ_{i}], i = 1, 2, …, n (we count each internal zero without sign change twice, as in E.10 of Section 3.1). So if b ≠ ζ_{n}, then T_{n} − p has n + 1 zeros on [a, b] including the zero at b, hence p = T_{n}, and the proof is finished. We may thus assume that T_{n}(b) = 1. Assume that |p′(b)| > |T′_{n}(b)|. Since T′_{n}(b) ≥ 0, without loss of generality we may assume that p′(b) > T′_{n}(b), otherwise we study −p. Then T_{n} −p has two zeros on [ζ_{n}, b], and hence has n+1 zeros in [a, b] (again, we count each internal zero without sign change twice). Thus by E.10 of Section 3.1 we have p = T_{n}, which contradicts the assumption p′(b) > T′_{n}(b). ⊓ ⊔

An extension of the above theorem to interior points is considered in E.3.

Theorem 3.3.2. Suppose (f_{0}, …, f_{n−1}, g) and (f_{0}, …, f_{n−1}, h) are both Chebyshev systems on [a, b] with associated Chebyshev polynomials

T_{n} := T_{n}{f_{0}, f_{1}, …, f_{n−1}, g; [a, b]}

and

S_{n} := T_{n}{f_{0}, f_{1}, …, f_{n−1}, h; [a, b]} ,

respectively. Suppose (f_{0}, f_{1}, …, f_{n−1}, g, h) is also a Chebyshev system. Then the zeros of T_{n} and S_{n} interlace (there is exactly one zero of S_{n} between any two consecutive zeros of T_{n}).

Proof. Since (f_{0}, …, f_{n−1}, g, h) is a Chebyshev system on [a, b], T_{n} ± S_{n} has at most n + 1 zeros. However, between any two consecutive alternation points of T_{n}, of which there are n+1, there is a zero of T_{n} ±S_{n} (which may be at an internal alternation point of T_{n} only if it is a zero without sign change, which is then counted twice). Likewise, there is a zero of T_{n} ± S_{n} between any two consecutive alternation points of S_{n}. Thus between any three successive alternation points of say T_{n} there can be at most three zeros of T_{n} ± S_{n}. However, if S_{n} had two zeros between two consecutive zeros of T_{n}, then there would be three consecutive alternation points of either T_{n} or S_{n} with at least four zeros of either T_{n} + S_{n} or T_{n} − S_{n} between them, which is impossible. ⊓ ⊔

Theorem 3.3.3. Suppose (f_{0}, f_{1}, …) is a Markov system on [a, b] with associated Chebyshev polynomials

T_{n} := T_{n}{f_{0}, f_{1}, …, f_{n}; [a, b]} .

Then the zeros of T_{n} and T_{n−1} strictly interlace (there is exactly one zero of T_{n−1} strictly between any two consecutive zeros of T_{n}).

Proof. The proof is analogous to that of Theorem 3.3.2. ⊓ ⊔

Theorem 3.3.4 (Lexicographic Property). Let (f_{0}, f_{1}, …) be a Descartes system on [a, b]. Suppose λ_{0} < λ_{1} < ⋯ < λ_{n} and γ_{0} < γ_{1} < ⋯ < γ_{n} are nonnegative integers satisfying

λ_{i} ≤ γ_{i} , i = 0, 1, …, n .

Let

T_{n} := T_{n}{f_{λ_{0}}, f_{λ_{1}}, …, f_{λ_{n}}; [a, b]}

and

S_{n} := T_{n}{f_{γ_{0}}, f_{γ_{0}}, …, f_{γ_{n}}; [a, b]}

denote the associated Chebyshev polynomials.

Let

α_{1} < α_{2} < ⋯ < α_{n} and β_{1} < β_{2} < ⋯ < β_{n}

denote the zeros of T_{n} and S_{n}, respectively. Then

α_{i} ≤ β_{i} , i = 1, 2, …, n

with strict inequality if λ_{i} ≠ γ_{i} for at least one index i. (In other words, the zeros of T_{n} lie to the left of the zeros of S_{n}.)

Proof. It is clearly sufficient to prove the theorem in the case that λ_{i} = γ_{i} for each i ≠ m, and λ_{m} < γ_{m} < λ_{m+1} for a fixed index m, and then to proceed by a sequence of pairwise comparisons (if m = n, then λ_{m+1} is meant to be replaced by ∞). So suppose

T_{n} := T_{n}{f_{λ_{0}}, …, f_{λ_{m}}, …, f_{λ_{n}}; [a, b]} = ∑_{i=0}^{n} c_{i}f_{λ_{i}}

and

S_{n} := T_{n}\{f_{λ_{0}}, …, f_{γ_{m}}, …, f_{λ_{n}}; [a, b]\} = d_{m}f_{γ_{m}} + ∑_{{i=0 \atop i≠m}}^{n} d_{i}f_{λ_{i}}

with λ_{m} < γ_{m} < λ_{m+1}. Then by Theorem 3.3.2 the zeros of S_{n} and T_{n} interlace and all that remains to prove is that the largest zero of S_{n} is larger than the largest zero of T_{n}. That is, we must show that α_{n} < β_{n}. For this we argue as follows. It follows from Theorem 3.2.5 that the lead coefficient of T_{n} is less than the lead coefficient of S_{n} (c_{n} < d_{n} provided m < n). Since both T_{n} and S_{n} have an alternation sequence of length n + 1 on [a, b], and since

∥T_{n}∥_{[a,b]} = ∥S_{n}∥_{[a,b]} = 1 , T_{n}(b) > 0 , S_{n}(b) > 0 ,

it follows from E.1 b] that

S_{n} − T_{n} ∈ span{f_{λ_{0}}, f_{λ_{1}} …, f_{λ_{n}}, f_{γ_{m}}}

has n+1 zeros x_{1} ≤ x_{2} ≤ ⋯ ≤ x_{n+1} on [a, b]. Therefore, it follows from E.4 c] of Section 3.2 that (S_{n} −T_{n})(x) > 0 on (x_{n+1}, b) and (S_{n} −T_{n})(x) < 0 on (x_{n}, x_{n+1}). Hence the assumption β_{n} ≤ α_{n} would imply that S_{n} −T_{n} has at least n + 2 zeros on [a, b] (counting each internal zero without sign change twice), which is a contradiction. (Draw a picture and use the alternation characterization of the Chebyshev polynomials T_{n} and S_{n} to make the proof of the above statement transparent.) So β_{n} > α_{n}, indeed, and the proof is finished. ⊓ ⊔ Comments, Exercises, and Examples.

If H_{n} := span(f_{0}, …, f_{n}) is a Chebyshev space on [a, b], A is a compact subset of [a, b], and p ∈ (0, ∞), then

T_{n} = c (f_{n} − ∑_{k=0}^{n−1} a_{k}f_{k})

with a_{0}, a_{1}, …, a_{n−1} ∈ ℝ minimizing

∫_{A} |f_{n} − ∑_{k=0}^{n−1} a_{k}f_{k}|^{p}

is called an L_{p} Chebyshev polynomial for H_{n} on A. When A = [a, b] and p ∈ (1, ∞], the properties of the zeros of these L_{p} Chebyshev polynomials are explored in Pinkus and Ziegler [79], where much of the material of this section may be found. For example, an L_{p} analog of Theorem 3.3.2 still holds.

E.1 Existence and Uniqueness of Chebyshev Polynomials. Let A ⊂ [a, b] be a compact set containing at least n + 1 points. Let (f_{0}, …, f_{n}) be a Chebyshev system on [a, b].

a] Existence of Chebyshev Polynomials. Show that there exists a T_{n} satisfying properties (3.3.1) to (3.3.3).

Hint: If A contains exactly n + 1 points, then the existence of T_{n} is just the interpolation property of a Chebyshev space formulated in Proposition 3.1.2 b]. So assume that A contains at least n + 2 points. Then there is a δ > 0 so that A ∩ [a, c] contains at least n + 1 points for every c ∈ (b − δ, b). Show that for every c ∈ (b−δ, b), there is a g_{c} ∈ span{f_{0}, …, f_{n}} for which

sup{f(b) : f ∈ span{f_{0}, f_{1}, …, f_{n}} , ∥f∥_{[a,c]} = 1}

is attained. Use a variational method to show that g_{c} satisfies properties (3.3.1) to (3.3.3) with A replaced by A ∩ [a, c].

Now let (c_{k})_{k=1}^{∞} be a sequence of numbers from (b−δ, b) that converges to b. Let g_{c_{k}} ∈ span{f_{0}, …, f_{n}} satisfy properties (3.3.1) to (3.3.3) with A replaced by A ∩ [a, c_{k}]. Show that there is a subsequence of (g_{c_{k}})_{k=1}^{∞} that converges to a g ∈ span{f_{0}, …, f_{n}} uniformly on [a, b]. Show that T_{n} := g satisfies properties (3.3.1) to (3.3.3). ⊓ ⊔

b] A Lemma for Part c]. Suppose f, g ∈ C[a, b] with ∥f∥_{A} = ∥g∥_{A} > 0 and there are alternation sequences

(x_{1} < x_{2} < ⋯ < x_{n+1}) and (y_{1} < y_{2} < ⋯ < y_{n+1})

for f and g, respectively, on A.

Suppose also that

sign(f(x_{1})) = sign(g(y_{1})).

Show that f − g has at least n + 1 zeros on [a, b].

c] Uniqueness of Chebyshev Polynomials. Show that the Chebyshev polynomials

T_{n}{f_{0}, f_{1}, …, f_{n}; A}

satisfying properties (3.3.1) to (3.3.3) are unique.

Hint: Use part a] and E.10 of Section 3.1. ⊓ ⊔

E.2 More on Chebyshev Polynomials. Let H_{n} := span{f_{0}, …, f_{n}} be a Chebyshev space on [a, b] with associated Chebyshev polynomial denoted by T_{n} := T_{n}{f_{0}, …, f_{n}; [a, b]}. Show the following statements.

a] If 1 ∈ H_{n}, then |T_{n}(a)| = |T_{n}(b)| = 1.

b] If 1 ∈ H_{n}, then T_{n} is monotone between two successive points of its alternation sequence.

Note that the conclusions of parts a] and b] do not necessarily hold in general.

c] If T_{n} =: ∑_{i=0}^{n} a_{i}f_{i}, a_{i} ∈ ℝ, then the coefficient sequence of T_{n}/a_{m} solves

min_{b_{i}∈ℝ} ∥f_{m} + ∑_{{i=0 \atop i≠m}}^{n} b_{i}f_{i}∥_{[a,b]}

uniquely, provided that {f_{0}, … f_{m−1}, f_{m+1}, … f_{n}} is also a Chebyshev system on [a, b]. (So this applies to ordinary polynomials on [0, 1] but not on [−1, 1].)

d] Suppose (f_{0}, …, f_{n}) is a Descartes system on [a, b] with associated Chebyshev polynomial T_{n} := T_{n}{f_{0}, …, f_{n}; [a, b]} =: ∑_{i=0}^{n} a_{i}f_{i}, a_{i} ∈ ℝ. Show that a_{n} > 0 and a_{i}a_{i+1} < 0 for each i = 0, 1, …, n − 1.

Hint: Use E.4 a] of Section 3.2. ⊓ ⊔

E.3 Extension of Theorem 3.3.1. Let H_{n} := span{f_{0}, …, f_{n}} be a Chebyshev space on [a, b] with associated Chebyshev polynomial

T_{n} := T_{n}{f_{0}, f_{1}, …, f_{n}; [a, b]}

and suppose each f_{i} is differentiable at x_{0} ∈ [a, b].

a] If T′_{n}(x_{0}) > 0, then

max{p′(x_{0}) : p ∈ H_{n} , ∥p∥_{[a,b]} ≤ 1 , p(x_{0}) = T_{n}(x_{0})}

is attained only by T_{n}.

120 3. Chebyshev and Descartes Systems

b] If T′_{n}(x_{0}) < 0, then

min{p′(x_{0}) : p ∈ H_{n} , ∥p∥_{[a,b]} ≤ 1 , p(x_{0}) = T_{n}(x_{0})}

is attained and only by T_{n}. Hint: Consider the number of zeros of T_{n} − p. ⊓ ⊔

E.4 More Lexicographic Properties of Müntz Spaces. Let [a, b] ⊂ [0, ∞). Suppose

λ_{0} < λ_{1} < ⋯ < λ_{n} and γ_{0} < γ_{1} < ⋯ < γ_{n}

are arbitrary real numbers if a > 0, or arbitrary nonnegative numbers if a = 0. Suppose λ_{i} ≤ γ_{i} for each i with strict inequality for at least one index i. Let

H_{n} := span{x^{λ_{0}}, x^{λ_{1}}, …, x^{λ_{n}}} and G_{n} := span{x^{γ_{0}}, x^{γ_{1}}, …, x^{γ_{n}}} .

Denote the associated Chebyshev polynomials for H_{n} and G_{n} on [a, b] by

T_{n,λ} := T_{n}{x^{λ_{0}}, x^{λ_{1}}, …, x^{λ_{n}}; [a, b]}

and

T_{n,γ} := T_{n}{x^{γ_{0}}, x^{γ_{1}}, …, x^{γ_{n}}; [a, b]} ,

respectively. a] Show that λ_{n} ≥ 0 implies T_{n,λ}(b) = 1. Hint: T_{n,λ}(b) ≠ 1 would imply that T′_{n,λ} has at least n + 1 distinct zeros in [a, ∞) if λ_{0} > 0 and at least n distinct zeros if λ_{0} = 0. ⊓ ⊔ b] Let x_{0} = a or x_{0} = b. If x_{0} = a = 0, then assume that λ_{0} = 0 and λ_{1} = 1. Show that

max{|p′(x_{0})| : p ∈ H_{n} , ∥p∥_{[a,b]} ≤ 1}

is attained uniquely by ±T_{n,λ}. c] Let x_{0} ∈ [0, ∞) \ [a, b]. If x_{0} = 0, then assume that λ_{0} = 0. Show that

max{|p(x_{0})| : p ∈ H_{n} , ∥p∥_{[a,b]} ≤ 1}

is attained uniquely by ±T_{n,λ}. Hint for b] and c]: First prove that an extremal p^{∗} ∈ H_{n} exists. Then show, by a variational method, that p^{∗} equioscillates n + 1 times between ±1 on [a, b]. ⊓ ⊔ d] Let λ_{n} ≥ 0, γ_{n} ≥ 0, and a > 0. Show that

|T′_{n,λ}(b)| < |T′_{n,γ}(b)| .

Show also that if a > 0 and there exists an index k, 0 ≤ k ≤ n, such that λ_{k} = γ_{k} = 0, then

|T′_{n,λ}(a)| > |T′_{n,γ}(a)| .

e] Let λ_{n} ≥ 0, γ_{n} ≥ 0, and a > 0. Show that

|T_{n,λ}(x_{0})| < |T_{n,γ}(x_{0})| , x_{0} ∈ (b, ∞) .

Show also that if λ_{0} ≥ 0, γ_{0} ≥ 0, and a > 0, then

|T_{n,λ}(x_{0})| > |T_{n,γ}(x_{0})| , x_{0} ∈ [0, a)

(when x_{0} = 0, we need the assumption λ_{0} = γ_{0} = 0).

Hint for d] and e]: Suppose to the contrary that one of the inequalities of parts d] and e] fails. Assume, without loss of generality, that there is an index m such that λ_{i} = γ_{i} whenever i ≠ m, and λ_{m} < γ_{m}. Note that by a], λ_{n} ≥ 0 and γ_{n} ≥ 0 imply T_{n,λ}(b) = T_{n,γ}(b) = 1. Also, by E.1 a], λ_{k} = γ_{k} = 0 implies T_{n,λ}(a) = T_{n,γ}(a) = (−1)^{n}. Now use Theorem 3.3.4 to show that

T_{n,λ} − T_{n,γ} ∈ span{x^{λ_{0}}, x^{λ_{1}}, …, x^{λ_{n}}, x^{γ_{m}}}

has at least n + 2 zeros in (0, ∞) (in the cases when λ_{n} ≥ 0 and γ_{n} ≥ 0 are assumed) or in [0, ∞) (in the cases when λ_{k} = γ_{k} = 0 is assumed). This contradiction finishes the proof. ⊓ ⊔

f] Let 0 < a < b. Show that if λ_{n} ≥ 0 and γ_{n} ≥ 0, then

max_{p∈H_{n}} \frac{|p′(b)|}{∥p∥_{[a,b]}} < max_{q∈G_{n}} \frac{|q′(b)|}{∥q∥_{[a,b]}} .

Show also that if there exists an index k, 0 ≤ k ≤ n, such that λ_{k} = γ_{k} = 0, then

max_{p∈H_{n}} \frac{|p′(a)|}{∥p∥_{[a,b]}} > max_{q∈G_{n}} \frac{|q′(a)|}{∥q∥_{[a,b]}} .

Hint: Combine parts b] and d]. ⊓ ⊔

g] Let 0 < a < b. Let λ_{n} ≥ 0 and γ_{n} ≥ 0. Show that

max_{p∈H_{n}} \frac{|p(x_{0})|}{∥p∥_{[a,b]}} < max_{q∈G_{n}} \frac{|q(x_{0})|}{∥q∥_{[a,b]}} , x_{0} ∈ (b, ∞) .

Show also that

max_{p∈H_{n}} \frac{|p(x_{0})|}{∥p∥_{[a,b]}} > max_{q∈G_{n}} \frac{|q(x_{0})|}{∥q∥_{[a,b]}} , x_{0} ∈ [0, a)

(when x_{0} = 0, we need the assumption λ_{0} = γ_{0} = 0).

122 3. Chebyshev and Descartes Systems

Hint: Combine parts c] and e]. ⊓ ⊔ h] Extend the validity, with < and > replaced by ≤ and ≥, respectively, of the inequalities of parts f] and g] to the case when the interval [a, b] ⊂ (0, ∞) is replaced by [0, b]. i] Suppose there exists an index k, 0 ≤ k ≤ n, such that λ_{k} = γ_{k} = 0. Let x_{0} ∈ (0, a). Show that the second inequalities of parts e] and g] hold true. Hint: Modify the arguments given in the hints to parts d], e], f], and g]. Note that 1 ∈ H_{n} ∩ G_{n} ensures, as in E.1 a], that

T_{n,λ}(b) = T_{n,γ}(b) = 1  and  T_{n,λ}(a) = T_{n,γ}(a) = (−1)^{n} .
⊓ ⊔

E.5 Lexicographic Properties of (sinh λ_{0}t, …, sinh λ_{n}t). Let

0 < λ_{0} < λ_{1} < ⋯ < λ_{n} and 0 < γ_{0} < γ_{1} < ⋯ < γ_{n} .

Suppose λ_{i} ≤ γ_{i} for each i. Let

H_{n} := span{sinh λ_{0}t, sinh λ_{1}t, …, sinh λ_{n}t}

and

G_{n} := span{sinh γ_{0}t, sinh γ_{1}t, …, sinh γ_{n}t} .

Denote the associated Chebyshev polynomials for H_{n} and G_{n} on [0, 1] by

T_{n,λ} := T_{n}{sinh λ_{0}t, sinh λ_{1}t, …, sinh λ_{n}t; [0, 1]}

and

T_{n,γ} := T_{n}{sinh γ_{0}t, sinh γ_{1}t, …, sinh γ_{n}t; [0, 1]} ,

respectively. a] Let

α_{1} < α_{2} < … < α_{n} and β_{1} < β_{2} < ⋯ < β_{n}

denote the zeros of T_{n,λ} and T_{n,γ}, respectively. Show that

α_{i} ≤ β_{i}, i = 1, 2, …, n

(in other words, the zeros of T_{n,λ} lie to the left of the zeros of T_{n,γ}). Outline. By E.2 d] of Section 3.2, (sinh λ_{0}t, …, sinh λ_{n}t) is a Descartes system on (0, ∞). Hence, by Theorem 3.3.3, the zeros of

T_{n,λ,δ} := T_{n}{sinh λ_{0}t, sinh λ_{1}t, …, sinh λ_{n}t; [δ, 1]}

on [δ, 1] lie to the left of the zeros of

T_{n,γ,δ} := T_{n}{sinh γ_{0}t, sinh γ_{1}t, …, sinh γ_{n}t; [δ, 1]}

on [δ, 1] for every δ ∈ (0, 1). Show that

lim_{δ→0} ∥T_{n,λ} − T_{n,λ,δ}∥ = lim_{δ→0} ∥T_{n,γ} − T_{n,γ,δ}∥ = 0 ;

hence the desired result follows by a continuity argument. ⊓ ⊔

b] Show that

max{|p′(0)| : p ∈ H_{n} , ∥p∥_{[0,1]} ≤ 1}

is attained uniquely by ±T_{n,λ}.

c] Show that

T_{n,λ}(1) = T_{n,γ}(1) = 1 .

Hint: T_{n,λ}(1) ≠ 1 would imply that

T′_{n,λ} ∈ span{cosh λ_{0}t, cosh λ_{1}t, …, cosh λ_{n}t}

has at least n + 1 distinct zeros in (0, ∞). This is impossible, since by E.2 e] of Section 3.2, (cosh λ_{0}t, …, cosh λ_{n}t) is a Descartes (hence Chebyshev) system on (0, ∞). ⊓ ⊔

d] Show that

|T′_{n,λ}(0)| ≥ |T′_{n,γ}(0)| .

Hint: Suppose to the contrary that the above inequality fails to hold. Assume, without loss of generality, that there is an index m such that λ_{i} = γ_{i} whenever i ≠ m, and λ_{m} < γ_{m}. Obviously

T_{n,λ}(0) = T_{n,γ}(0) = 0 .

Part c] implies that T_{n,λ}(1) = T_{n,γ}(1) = 1 . Now use part a] and the above observation to show that

T_{n,λ} − T_{n,γ} ∈ span{sinh λ_{0}t, sinh λ_{1}t, …, sinh λ_{n}t, sinh γ_{m}t}

has at least n + 2 zeros in (0, ∞). This contradicts E.2 d] of Section 3.2. ⊓ ⊔

e] Show that

max_{0≠p∈H_{n}} \frac{|p′(0)|}{∥p∥_{[0,1]}} ≥ max_{0≠q∈G_{n}} \frac{|q′(0)|}{∥q∥_{[0,1]}} .

Hint: Combine parts b] and d]. ⊓ ⊔

The result of the following exercise has been observed independently by Lubinsky and Ziegler [90] and Kroó and Szabados [94]. Various coefficient estimates for polynomials are discussed in Milovanović, Mitrinović, and Rassias [94]. An estimate for the coefficients of polynomials having a given number of terms is obtained in Baishanski and Bojanic [80]. Approximation by such polynomials is studied in Baishanski [83].

124 3. Chebyshev and Descartes Systems

E.6 Coefficient Bounds for Polynomials in a Special Basis. Show that

|c_{m}| ≤ 2^{−n}\binom{2n}{2m}∥p∥_{[−1,1]} , m = 0, 1, …, n

for every polynomial p of the form

p(x) = ∑_{m=0}^{n} c_{m}(1 − x)^{m}(x + 1)^{n−m} , c_{m} ∈ ℝ .

Outline. By E.5 a] and b] of Section 2.3

T_{n}(x) = ∑_{m=0}^{n} d_{m,n}(1 − x)^{m}(x + 1)^{n−m} ,

where

d_{m,n} = (−1)^{m}2^{−n} \begin{matrix}\binom{n−1/2}{m} & \binom{n−1/2}{n−m} \\ ( & {n−1/2 \atop n} & ) \end{matrix} = 2^{−n}\binom{2n}{2m} .

If |c_{m}| > d_{m,n}∥p∥_{[−1,1]} for some index m, then the polynomial

q(x) = \frac{T_{n}(x)}{d_{m,n}} − \frac{p(x)}{c_{m}}

has at least n distinct zeros in (−1, 1). However,

q(x) = ∑_{{j=0 \atop j≠m}}^{n} a_{j}(1 − x)^{j}(x + 1)^{n−j} = (1 − x)^{n} ∑_{{j=0 \atop j≠m}}^{n} a_{j} (\frac{1 + x}{1 − x})^{n−j}

can have at most n − 1 distinct zeros in (−1, 1) since

(u^{0}, u^{1}, … u^{n−m−1}, u^{n−m+1}, u^{n−m+2}, …, u^{n})

is a Chebyshev system on (0, ∞) by E.1 a] of Section 3.1. ⊓ ⊔ E.7 On the Zeros of the Chebyshev Polynomials for Müntz Spaces. Let 0 =: λ_{0} < λ_{1} < ⋯ < λ_{n}, and let

H_{n} := span{x^{λ_{0}}, x^{λ_{1}}, …, x^{λ_{n}}} .

3.4 Müntz-Legendre Polynomials 125

Denote the associated Chebyshev polynomials for H_{n} on [0, 1] by T_{n} := T_{n}{x^{λ_{0}}, x^{λ_{1}}, …, x^{λ_{n}}; [0, 1]} . a] Let ϵ ∈ (0, \frac{1}{2}). Suppose α < β are two consecutive zeros of T_{n} lying in [ϵ, 1 − ϵ]. Show that

\frac{ϵ^{2}}{n} ≤ β − α .

Hint: It is clear that Q_{n}(x) := x(1 − x)T_{n}(x) has two consecutive zeros γ < δ in [α, 1] such that

δ − γ ≥ \frac{ϵ}{n} .

Show that if

β − α < \frac{ϵ^{2}}{n} ,

then

R_{n}(x) := T_{n}(x) − T_{n} (\frac{γx}{α}) ∈ H_{n}

has at least n + 1 zeros on [0, α/γ], which is a contradiction. ⊓ ⊔

b] Denote the zeros of T_{n} in (0, 1) by x_{1} < x_{2} < ⋯ < x_{n}. Show that

log x_{k+1} − log x_{k} ≤ log x_{k} − log x_{k−1} , k = 2, 3, …, n − 1 .

Hint: Use a zero counting argument, as in the hint to part a]. ⊓ ⊔

3.4 Müntz-Legendre Polynomials

We examine in some detail the system

(x^{λ_{0}}, x^{λ_{1}}, …)

on [0, 1] which we call a Müntz system. In particular, we explicitly construct orthogonal “polynomials” for this system. This allows us to derive various extremal properties of these systems and leads to a very simple proof of the classical Müntz-Szász theorem in Section 4.2.

We adopt the following definition for x^{λ}:

(3.4.1)  x^{λ} = e^{λ log x} , x ∈ (0, ∞) , λ ∈ ℂ

with value at 0 defined to be the limit of x^{λ} as x → 0 from (0, ∞) whenever the limit exists.

Given a sequence Λ := (λ_{i})_{i=0}^{∞} of complex numbers, an element of span{x^{λ_{0}}, x^{λ_{1}}, …, x^{λ_{n}}} is called a Müntz polynomial or a Λ-polynomial. We denote the set of all such polynomials by M_{n}(Λ), that is,

(3.4.2)  M_{n}(Λ) := span{x^{λ_{0}}, x^{λ_{1}}, …, x^{λ_{n}}} ,

where the linear span is over ℝ or ℂ according to context. Let

(3.4.3)  M(Λ) := ⋃_{n=0}^{∞} M_{n}(Λ) = span{x^{λ_{0}}, x^{λ_{1}}, … } .

For the L_{2}[0, 1] theory of Müntz systems, we consider

(3.4.4)  Λ = (λ_{i})_{i=0}^{∞} , Re(λ_{i}) > −1/2 , and λ_{i} ≠ λ_{j} , i ≠ j ,

where Re(λ) denotes the real part of λ. This ensures that the Λ-polynomials

p(x) = ∑_{k=0}^{n} a_{k}x^{λ_{k}} , a_{k} ∈ ℂ

are in L_{2}[0, 1]. We can then define the orthogonal Λ-polynomials with respect to Lebesgue measure. We call these Müntz-Legendre polynomials. Although we often assume (3.4.4), the following definition requires neither the distinctness of the numbers λ_{i} nor the assumption Re(λ_{i}) > −1/2.

Definition 3.4.1 (Müntz-Legendre Polynomials). Let Λ := (λ_{i})_{i=0}^{∞} be a sequence of complex numbers. We define the nth Müntz-Legendre polynomial on (0, ∞) by

(3.4.5)  L_{n}(x) := L_{n}{λ_{0}, …, λ_{n}}(x)

:= \frac{1}{2πi} ∫_{Γ} ∏_{k=0}^{n−1} \frac{t + \overline{λ}_{k} + 1}{t − λ_{k}} \frac{x^{t} dt}{t − λ_{n}} , n = 0, 1, …,

where the positively oriented, simple closed contour Γ surrounds the zeros of the denominator in the integrand, and \overline{λ}_{k} denotes the conjugate of λ_{k}.

The orthogonality of the above functions with respect to the Lebesgue measure is proved in Theorem 3.4.3. However, first we give a simple explicit representation of L_{n} in the case that the numbers λ_{i} are distinct. This is deduced immediately from evaluating the above integral by the residue theorem.

Proposition 3.4.2. Let Λ := (λ_{i})_{i=0}^{∞} be a sequence of distinct complex numbers. Then

(3.4.6)  L_{n}{λ_{0}, …, λ_{n}}(x) = ∑_{k=0}^{n} c_{k,n}x^{λ_{k}} ,  x ∈ (0, ∞)

with

c_{k,n} := \frac{∏_{j=0}^{n−1}(λ_{k} + \overline{λ}_{j} + 1)}{∏_{j=0,j≠k}^{n}(λ_{k} − λ_{j})} ,

where L_{n}{λ_{0}, …, λ_{n}}(x) is defined by (3.4.5).

So L_{n}{λ_{0}, …, λ_{n}} is indeed a Λ polynomial provided the numbers λ_{i} are distinct. Its value at x = 0 is defined if for all i either Re(λ_{i}) > 0 or λ_{i} = 0.

From either Definition 3.4.1 or the above proposition it is obvious that the order of λ_{0}, …, λ_{n−1} in L_{n}{λ_{0}, …, λ_{n}} does not make any difference, as long as λ_{n} is kept last. For example,

L_{2}{λ_{0}, λ_{1}, λ_{2}} = L_{2}{λ_{1}, λ_{0}, λ_{2}}

while both, in general, are different from L_{2}{λ_{0}, λ_{2}, λ_{1}}. For a fixed sequence Λ, we let L_{n}(Λ), or simply L_{n}, denote the nth Müntz-Legendre polynomial L_{n}{λ_{0}, …, λ_{n}}, whenever there is no ambiguity.

An analog of Proposition 3.4.2 can be established even if the numbers λ_{i} are not distinct, however, in the nondistinct case, L_{n}(Λ) does not belong to the space M_{n}(Λ); see E.7 b]. In the very special case that all the indices are the same we recover the Laguerre polynomials; see E.1.

The orthogonality of {L_{n}}_{n=0}^{∞} is the content of the main theorem of this section.

Theorem 3.4.3 (Orthogonality). Let Λ = (λ_{i})_{i=0}^{∞} be a sequence of complex numbers with Re(λ_{i}) > −1/2 for i = 0, 1, . . .. The functions L_{n} defined by (3.4.5) satisfy

(3.4.7)  ∫_{0}^{1} L_{n}(x)\overline{L_{m}(x)} dx = \frac{δ_{n,m}}{1 + λ_{n} + \overline{λ}_{n}}

for all nonnegative integers n and m. (Here δ_{n,m} is the Kronecker symbol.)

Proof. We may assume that the numbers λ_{i} are distinct. Note that

L_{n}{λ_{0}, λ_{1}, …, λ_{n}}(x)

is uniformly continuous in λ_{0}, …, λ_{n} for x in closed subintervals of (0, 1], and the nondistinct λ_{i} case can be handled by a limiting argument. We may further suppose that m ≤ n. Since Re(λ_{i}) > −1/2, we can pick a simple closed contour Γ such that Γ lies completely to the right of the vertical line Re(t) = −1/2 and Γ surrounds all zeros of the denominator of the integrand in (3.4.5). When t ∈ Γ, we have Re(t + \overline{λ}_{m}) > −1, and

∫_{0}^{1} x^{t+\overline{λ}_{m}} dx = \frac{1}{t + \overline{λ}_{m} + 1}

for every m = 0, 1, …. Hence Fubini’s theorem yields

∫_{0}^{1} L_{n}(x)x^{\overline{λ}^{m}} dx = \frac{1}{2πi} ∫_{Γ} ∏_{k=0}^{n−1} \frac{t + \overline{λ}_{k} + 1}{t − λ_{k}} \frac{dt}{(t − λ_{n})(t + \overline{λ}_{m} + 1)} .

Notice that for m < n, the new factor, t + \overline{λ}_{m} + 1, in the denominator can be cancelled, and for m = n the new pole −(\overline{λ}_{n} + 1) is outside Γ since Re(−\overline{λ}_{n} − 1) < −1/2. Changing the contour from Γ to |t| = R with R > max{λ_{0}| + 1, …, |λ_{n}| + 1}, gives

∫_{0}^{1} L_{n}(x)x^{\overline{λ}^{m}} dx = \frac{1}{2πi} ∫_{|t|=R} ∏_{k=0}^{n−1} \frac{t + \overline{λ}_{k} + 1}{t − λ_{k}} \frac{dt}{(t − λ_{n})(t + \overline{λ}_{m} + 1)}

− \frac{δ_{m,n}}{−\overline{λ}_{n} − 1 − λ_{n}} ∏_{k=0}^{n−1} \frac{−\overline{λ}_{n} + \overline{λ}_{k}}{−\overline{λ}_{n} − 1 − λ_{k}} .

On letting R → ∞, we see that the integral on the right-hand side is actually 0, which gives

∫_{0}^{1} L_{n}(x)x^{\overline{λ}^{m}} dx = \frac{δ_{n,m}}{\overline{λ}_{n} + λ_{n} + 1} ∏_{k=0}^{n−1} \frac{\overline{λ}_{n} − \overline{λ}_{k}}{\overline{λ}_{n} + λ_{k} + 1} .

Therefore Proposition 3.4.2 and m ≤ n yield

∫_{0}^{1} L_{n}(x)\overline{L_{m}(x)} dx = ∫_{0}^{1} L_{n}(x) ∑_{k=0}^{m} \overline{c}_{k,m}x^{λ_{k}} dx

= \overline{c}_{m,m} ∫_{0}^{1} L_{n}(x)x^{\overline{λ}^{m}} dx = \frac{δ_{m,n}}{λ_{n} + \overline{λ}_{n} + 1} ,

and the proof is finished. ⊓ ⊔

An alternative proof of orthogonality is suggested in E.3. If we let

(3.4.8)  L_{n}^{∗} := (1 + λ_{n} + \overline{λ}_{n})^{1/2}L_{n} ,

then we get an orthonormal system, that is,

∫_{0}^{1} L_{n}^{∗}(x)\overline{L_{m}^{∗}(x)} dx = δ_{m,n} , m, n = 0, 1, ….

We call these L_{n}^{∗} the orthonormal Müntz-Legendre polynomials.

There is also a Rodrigues-type formula for the Müntz-Legendre polynomials (see E.2). Let

p_{n}(x) = ∑_{k=0}^{n} \frac{x^{λ_{k}}}{∏_{j=0,j≠k}^{n}(λ_{k} − λ_{j})} .

Then

L_{n}(x) = (D_{λ_{0}}D_{λ_{1}} ⋯ D_{λ_{n−1}})(p_{n})(x) ,

where the differential operators D_{λ} are defined by

D_{λ}(f)(x) := x^{−\overline{λ}} \frac{d}{dx}(x^{1+\overline{λ}}f(x)) .

The following is a differential recurrence formula for (L_{n})_{n=0}^{∞}:

Theorem 3.4.4. For a fixed sequence Λ := (λ_{i})_{i=0}^{∞} of complex numbers, let L_{n} be defined by (3.4.5). The identity

(3.4.9)  xL′_{n}(x) − xL′_{n−1}(x) = λ_{n}L_{n}(x) + (1 + \overline{λ}_{n−1})L_{n−1}(x)

holds for every x ∈ (0, ∞) and n = 1, 2, ….

Proof. From (3.4.5) we get

(x^{−λ_{n}}L_{n}(x))′ = \frac{1}{2πi} ∫_{Γ} \frac{∏_{k=0}^{n−2}(t + \overline{λ}_{k} + 1)}{∏_{k=0}^{n−1}(t − λ_{k})}(t + \overline{λ}_{n−1} + 1)x^{t−λ_{n}−1} dt .

On multiplying both sides by x^{λ_{n}+\overline{λ}_{n−1}+1}, we obtain

x^{λ_{n}+\overline{λ}_{n−1}+1}(x^{−λ_{n}}L_{n}(x))′
=\frac{1}{2πi} ∫_{Γ} \frac{∏_{k=0}^{n−2}(t + \overline{λ}_{k} + 1)}{∏_{k=0}^{n−1}(t − λ_{k})}(t + \overline{λ}_{n−1} + 1)x^{t+\overline{λ}_{n−1}} dt ,

and again by the definition of L_{n−1},

x^{λ_{n}+\overline{λ}_{n−1}+1}(x^{−λ_{n}}L_{n}(x))′ = (x^{\overline{λ}_{n−1}+1}L_{n−1}(x))′ .

We finish the proof by simplifying by the product rule and dividing both sides by x^{\overline{λ}^{n−1}}. ⊓ ⊔

Corollary 3.4.5. For a fixed sequence Λ := (λ_{i})_{i=0}^{∞} of complex numbers, let L_{n} and L_{n}^{∗} be defined by (3.4.5) and (3.4.8), respectively. Then for every x ∈ (0, ∞) and for every n = 0, 1, …,

a] xL′_{n}(x) = λ_{n}L_{n}(x) + ∑_{k=0}^{n−1} (λ_{k} + \overline{λ}_{k} + 1)L_{k}(x) ,

b] xL_{n}^{∗′}(x) = λ_{n}L_{n}^{∗}(x) + \sqrt{λ_{n} + \overline{λ}_{n} + 1} ∑_{k=0}^{n−1} \sqrt{λ_{k} + \overline{λ}_{k} + 1} L_{k}^{∗}(x) ,

and

c] xL′′_{n}(x) = (λ_{n} − 1)L′_{n}(x) + ∑_{k=0}^{n−1} (λ_{k} + \overline{λ}_{k} + 1)L′_{k}(x) .

Proof. The first identity follows from Theorem 3.4.4 on expressing

xL′_{n}(x) − xL′_{0}(x)

as a telescoping sum. From this and the relation

L_{k}^{∗} = (λ_{k} + \overline{λ}_{k} + 1)^{1/2}L_{k}

we get part b]. Differentiating the identity of part a] gives part c]. ⊓ ⊔

The values and derivative values of the Müntz-Legendre polynomials at 1 can now all be calculated.

Corollary 3.4.6. For a fixed sequence Λ := (λ_{i})_{i=0}^{∞} of complex numbers, let L_{n} be defined by (3.4.5). Then

a] L_{n}(1) = 1 ,

b] L′_{n}(1) = λ_{n} + ∑_{k=0}^{n−1} (λ_{k} + \overline{λ}_{k} + 1) ,

and

c] L′′_{n}(1) = (λ_{n} − 1)L′_{n}(1) + ∑_{k=0}^{n−1} (λ_{k} + \overline{λ}_{k} + 1)L′_{k}(1) .

Proof. It suffices to show that L_{n}(1) = 1; the rest follows from Corollary 3.4.5. Notice that

L_{n}(1) = \frac{1}{2πi} ∫_{Γ} ∏_{k=0}^{n−1} \frac{t + \overline{λ}_{k} + 1}{t − λ_{k}} \frac{dt}{t − λ_{n}} .

Now, since Γ surrounds all zeros of the denominator, and the degree of the denominator is one higher than that of the numerator, we can evaluate the integral on circles of radius R → ∞ to get the result. ⊓ ⊔

Comments, Exercises, and Examples.

Müntz polynomials are just exponential polynomials ∑ a_{k}e^{−λ_{k}t} under the change of variables x = e^{−t} and have received considerable scrutiny (Schwartz [59] is a monograph on this topic). The orthogonalizations of Müntz systems exist in the Russian literature (see, for example, Badalyan [55] and Taslakyan [84]; it has been further explored in McCarthy, Sayre, and Shawyer [93]). Borwein, Erdélyi, and Zhang [94b] contains most of the content of this section.

Various properties of the Müntz-Legendre polynomials are examined in the exercises. Note that if λ_{0} < λ_{1} < λ_{2} < ⋯, then the Müntz system (x^{λ_{0}}, x^{λ_{1}}, …) is a Descartes system on [a, b], a > 0, and so we can apply Theorem 3.3.4 to the associated Chebyshev polynomials on [a, b] to deduce how the zeros shift when the exponents are varied lexicographically. Similar results are given for the Müntz-Legendre polynomials in E.7.

E.1 Laguerre Polynomials.

a] Let L_{n}{λ_{0}, …, λ_{n}}(x) be defined by (3.4.5). If λ_{0} = ⋯ = λ_{n} = λ, then

L_{n}\{λ_{0}, λ_{1}, …, λ_{n}\}(x) = x^{λ}L_{n}(−(1 + λ + \overline{λ}) log x) ,

where L_{n} is the nth Laguerre polynomial orthonormal with respect to the weight e^{−x} on [0, ∞) with L_{n}(0) = 1 (see E.7 of Section 2.3, where L_{n} is denoted by L_{n} as is standard).

Proof. Since λ_{k} = λ, (3.4.5) yields

L_{n}\{λ_{0}, λ_{1}, …, λ_{n}\}(x) = \frac{1}{2πi} ∫_{Γ} \frac{x^{t}(t + \overline{λ} + 1)^{n}}{(t − λ)^{n+1}} dt ,

where the contour Γ can be taken to be any circle centered at λ. By the residue theorem,

L_{n}\{λ_{0}, λ_{1}, …, λ_{n}\}(x) = \frac{1}{n!} [\frac{d^{n}}{dt^{n}} (x^{t}(t + \overline{λ} + 1)^{n}]_{t=λ}

= \frac{1}{n!} ∑_{k=0}^{n} \binom{n}{k}x^{λ}(log x)^{k}[n(n − 1) ⋯ (k + 1)](λ + \overline{λ} + 1)^{k}

= x^{λ} ∑_{k=0}^{n} \frac{1}{k!}\binom{n}{k}(1 + λ + \overline{λ})^{k} log^{k} x .

See also part b]. ⊓ ⊔

b] Let

L_{n}(x) := ∑_{k=0}^{n} \frac{1}{k!}\binom{n}{k}(−x)^{k}.

Then (L_{n})_{n=0}^{∞} is an orthonormal sequence of polynomials on [0, ∞) with respect to the inner product

⟨f, g⟩ = ∫_{0}^{∞} f(x)g(x)e^{−x} dx .

Deduce the orthonormality from a] and Theorem 3.4.3 by substituting

y = −(1 + λ + \overline{λ}) log x .

E.2 Rodrigues-Type Formula. Let Λ = (λ_{i})_{i=0}^{∞} be a sequence of distinct complex numbers. Let L_{n} be defined by (3.4.5).

a] Let

p_{n}(x) := ∑_{k=0}^{n} \frac{x^{λ_{k}}}{∏_{j=0,j≠k}^{n}(λ_{k} − λ_{j})} .

Show that

p_{n}(x) = \frac{1}{2πi} ∫_{Γ} \frac{x^{t} dt}{∏_{j=0}^{n}(t − λ_{j})} ,

where Γ is any contour surrounding λ_{0}, λ_{1}, …, λ_{n}. Use this to show that

p_{n}^{(k)}(1) = 0 , k = 0, 1, …, n − 1 .

b] Show that

L_{n}(x) = (D_{λ_{0}}D_{λ_{1}} ⋯ D_{λ_{n−1}})(p_{n})(x) ,

where

D_{λ}(f)(x) := x^{−\overline{λ}}(x^{1+\overline{λ}}f(x))′ .

c] If 0 = λ_{0} < λ_{1} < ⋯, then

p_{n}(0) = (−1)^{n} ∏_{j=1}^{n} λ_{j}^{−1}

and (−1)^{n}p_{n} is strictly decreasing on [0, 1].

E.3 Another Proof of Orthogonality. Deduce the orthogonality of the sequence (L_{n})_{n=0}^{∞} on [0, 1] from Theorem 3.4.4 by using integration by parts and induction.

E.4 Integral Recursion. For a given sequence Λ := (λ_{i})_{i=0}^{∞} of complex numbers satisfying (3.4.4), let L_{n} be defined by (3.4.5). Show that

L_{n}(x) = L_{n−1}(x) − (λ_{n} + \overline{λ}_{n−1} + 1)x^{λ_{n}} ∫_{x}^{1} t^{−λ_{n}−1}L_{n−1}(t) dt , x ∈ (0, 1] .

Hint: Use Theorem 3.4.4. ⊓ ⊔

E.5 On the Maximum of L_{n} on [0, 1]. If Λ = (λ_{i})_{i=0}^{∞} is a sequence of nonnegative numbers satisfying

(3.4.10)  λ_{n} ≥ ∑_{k=0}^{n−1} (1 + 2λ_{k}) ,  n = 1, 2, …

and L_{n} is defined by (3.4.5), then

|L_{n}(x)| < L_{n}(1) = 1 , x ∈ [0, 1) , n = 2, 3, ….

Hint: Use Theorem 3.4.4. ⊓ ⊔

If λ_{k} = ρ^{k}, then (3.4.10) holds if and only if ρ ≥ 2 + \sqrt{3}.

E.6 The Reproducing Kernel. Let Λ = (λ_{i})_{i=0}^{∞} be as in (3.4.4), and let L_{n} and L_{n}^{∗} be defined by (3.4.5) and (3.4.8), respectively. Then for every p ∈ M_{n}(Λ), we have

p(x) = ∫_{0}^{1} K_{n}(x, t)p(t) dt ,

where

K_{n}(x, t) := ∑_{k=0}^{n} L_{k}^{∗}(x)\overline{L_{k}^{∗}(t)}

is the nth reproducing kernel (see also E.5 of Section 2.2).

E.7 On the Zeros of Müntz-Legendre Polynomials. Assume that

(λ_{0}, λ_{1}, …, λ_{n}) ⊂ (−\frac{1}{2}, ∞) .

a] For a function f ∈ C(0, 1) let S^{−}(f) and Z(f) denote the number of sign changes and the number of zeros, respectively, of f in (0, 1) (we count each zero without sign change twice). Let Φ and Ψ ∈ C(0, 1). Show that if

n ≤ S^{−}(αΦ + βΨ) ≤ Z(αΦ + βΦ) ≤ n + 1

for every real α and β with α^{2} + β^{2} > 0, then the zeros of Φ and Ψ strictly interlace.

Proof: This result is due to Pinkus and Ziegler [79]. ⊓ ⊔

b] Assume that

{λ_{0}, λ_{1}, …, λ_{n}} = {λ̃_{0}, λ̃_{1}, …, λ̃_{m}} ,

where the numbers λ̃_{0}, λ̃_{1}, …, λ̃_{m} are distinct, and let m_{j}, j = 0, 1, …, m, be the number of indices i = 0, 1, …, n for which λ_{i} = λ̃_{j}. Show that L_{n}{λ_{0}, …, λ_{n}} is in the space

H_{n} := span{x^{λ_{j}}(log x)^{i} : j = 0, 1, …, m, i = 0, 1, …, m_{j} − 1} ,

which is a Chebyshev space on (0, ∞).

Hint: Use the definition and the residue theorem. ⊓ ⊔

c] Show that {L_{k}{λ_{0}, …, λ_{k}}}_{k=0}^{n} is a basis for the Chebyshev space H_{n} defined in part b].

Hint: Use Theorem 3.4.3 (orthogonality). ⊓ ⊔

d] Show that L_{n} := L_{n}{λ_{0}, …, λ_{n}} has exactly n distinct zeros in (0, 1) and L_{n} changes sign at each of these zeros.

Hint: Assume to the contrary that the number of sign changes of L_{n} is less than n. Use part c] to find a function p ∈ span{L_{k}}_{k=0}^{n−1} that changes sign exactly at those points in (0, 1) where L_{n} changes sign. Then ∫_{0}^{1} L_{n}p ≠ 0, which contradicts Theorem 3.4.3. ⊓ ⊔

e] Suppose λ_{n} < λ_{n}^{∗}. Show that the zeros of

Φ := L_{n}{λ_{0}, λ_{1}, …, λ_{n−1}, λ_{n}}

and

Ψ := L_{n}{λ_{0}, λ_{1}, …, λ_{n−1}, λ_{n}^{∗}}

in (0, 1) strictly interlace.

Hint: Note that Theorem 3.4.3 (orthogonality) implies

∫_{0}^{1} (αΦ + βΨ)p = 0

for every p ∈ H_{n−1}, where H_{n−1} is defined by part b] with respect to the sequence (λ_{0}, λ_{1}, …, λ_{n−1}). Use the hint given to part d] to show that αΦ + βΨ has at least n sign changes in (0, 1) whenever α and β are real with α^{2} + β^{2} > 0. Use part b] to obtain that αΦ + βΨ cannot have more than n + 1 zeros in (0, 1) whenever α and β are real with α^{2} + β^{2} > 0. Finish the proof by part a]. ⊓ ⊔

f] Let λ_{0}, …, λ_{k−1}, λ_{k+1}, …, λ_{n} be fixed distinct numbers. Suppose

(λ_{k,i})_{i=1}^{∞} ⊂ (−1/2, ∞)

is a sequence with lim_{i→∞} λ_{k,i} = ∞. Show that the largest zero of

L_{n,k,i} := L_{n}{λ_{0}, …, λ_{k−1}, λ_{k,i}, λ_{k+1}, …, λ_{n}}

in (0, 1) tends to 1.

Outline. Assume, without loss, that λ_{k,i} is greater than each of the numbers λ_{j}, j = 0, 1, …, n, j ≠ k. Let

g_{i}(x) := λ_{k,i}(L_{n,k,i}(x) − c_{k,n}^{(i)} x^{λ_{k,i}}) ,

where

c_{k,n}^{(i)} = \frac{∏_{j=0}^{n−1}(λ_{k,i} + λ_{j} + 1)}{∏_{j=0,j≠k}^{n}(λ_{k,i} − λ_{j})}

is the coefficients of x^{λ_{k,i}} in L_{n,k,i}. Use (3.4.6) to show that the functions g_{i} converge uniformly on [δ, 1], δ ∈ (0, 1), to a function

0 ≠ g ∈ H_{n−1} := span{x^{λ_{0}}, …, x^{λ_{k−1}}, x^{λ_{k+1}}, …, x^{λ_{n}}} .

Use L_{n,k,i}(1) = 1 (see Corollary 3.4.6) and the explicit formula for c_{k,n}^{(i)} to show that g(1) ≤ 0 and that the functions

L_{n,k,i}(x) = (L_{n,n,i}(x) − c_{k,n}^{(i)} x^{λ_{k,i}}) + c_{k,n}^{(i)} x^{λ_{k,i}}

converge to g(x), as i → ∞, for every x ∈ (0, 1).

Now assume that the statement of part f] is false. Then there is an ϵ ∈ (0, 1) and a subsequence (λ_{k,i_{j}})_{j=1}^{∞} of (λ_{k,i})_{i=1}^{∞} such that the Müntz-Legendre polynomials L_{n,k,i_{j}} have no zeros in [1−ϵ, 1]. Deduce from this and L′_{n,k,i_{j}}(1) > 0, λ_{n,i_{j}} > 0 (see Corollary 3.4.6 a]), that g_{i_{j}} is nondecreasing on [1 − ϵ, 1] whenever λ_{n,i_{j}} > 0.

Therefore g is nondecreasing on [1 − ϵ, 1], which, together with 0 ≠ g ∈ H_{n−1} and g(1) ≤ 0, implies that g(1 − ϵ) < 0. Hence L_{n,k,i}(1 − ϵ) < 0 if i is large enough. Since L_{n,k,i}(1) = 1 (see Corollary 3.4.6), each L_{n,k,i} has a zero in (1 − ϵ, 1), provided i is large enough, which contradicts our assumption. ⊓ ⊔

g] Let Φ and Ψ be as in part e]. Let

x_{1} < x_{2} < ⋯ < x_{n} and x_{1}^{∗} < x_{2}^{∗} < ⋯ < x_{n}^{∗}

be the zeros of Φ and Ψ, respectively, in (0, 1). Show that λ_{n} < λ_{n}^{∗} implies that

x_{j} < x_{j}^{∗} , j = 1, 2, …, n .

Hint: Combine parts e] and f]. ⊓ ⊔

h] Let λ_{k} ≠ λ_{n}. Show that the zeros of

Φ := L_{n}{λ_{0}, …, λ_{k−1}, λ_{k}, λ_{k+1} …, λ_{n−1}, λ_{n}}

and

Ψ := L_{n}{λ_{0}, …, λ_{k−1}, λ_{n}, λ_{k+1} …, λ_{n−1}, λ_{k}}

in (0, 1) strictly interlace.

Hint: Use part a] and arguments similar to those given in the hints to part e]. Note that

Φ(1) = Ψ(1) = 1  and  Ψ′(1) − Φ′(1) = λ_{n} − λ_{k} ≠ 0

(see Corollary 3.4.6 a]) imply that αΦ + βΨ is not identically 0 whenever α and β are real with α^{2} + β^{2} > 0. ⊓ ⊔

i] Let Φ and Ψ be as in part h]. Let

x_{1} < x_{2} < ⋯ < x_{n} and x_{1}^{∗} < x_{2}^{∗} < ⋯ < x_{n}^{∗}

be the zeros of Φ and Ψ, respectively, in (0, 1). Show that λ_{k} < λ_{n} implies that

x_{j} < x_{j}^{∗} , j = 1, 2, …, n .

Hint: By part h] it is sufficient to prove that x_{n} ≤ x_{n}^{∗}. Let H_{n} be the Chebyshev space defined in part b]. Corollary 3.4.6 implies that

Φ(1) = Ψ(1) = 1  and  Ψ′(1) − Ψ′(1) = λ_{n} − λ_{k} > 0 .

Deduce from this and part i] that x_{n}^{∗} < x_{n} would imply that 0 ≠ Ψ −Φ ∈ H_{n} has at least n + 1 distinct zeros in (0, 1], which is a contradiction. ⊓ ⊔

136 3. Chebyshev and Descartes Systems

j] Lexicographic Property. Suppose

max_{0≤i≤n} λ_{i} ≤ min_{0≤j≤n} μ_{j}

and λ_{i} < μ_{j} for some indices i and j. Let

x_{1} < x_{2} < ⋯ < x_{n} and x_{1}^{∗} < x_{2}^{∗} < ⋯ < x_{n}^{∗}

be the zeros of

L_{n}{λ_{0}, λ_{1}, …, λ_{n}} and L_{n}{μ_{0}, μ_{1}, …, μ_{n}} ,

respectively, in (0, 1). Show that

x_{j} < x_{j}^{∗} , j = 1, 2, …, n .

Hint: Repeated applications of parts g] and i] give the desired result. ⊓ ⊔ k] Let λ_{0} < λ_{n}. Let

x_{1} < x_{2} < ⋯ < x_{n} and x_{1}^{∗} < x_{2}^{∗} < ⋯ < x_{n}^{∗}

be the zeros of

L_{n}{λ_{0}, λ_{1}, …, λ_{n}} and L_{n}{λ_{n}, λ_{n−1}, …, λ_{0}} ,

respectively, in (0, 1). Show that (x_{j})_{j=1}^{n} and (x_{j}^{∗})_{j=1}^{n} strictly interlace and

x_{j} < x_{j}^{∗} , j = 1, 2, …, n .

Hint: Use parts h] and i] and the comment given after Proposition 3.4.2. ⊓ ⊔ l] Show that the zeros of

Φ := L_{n−1}{λ_{0}, …, λ_{n−1}} and Ψ := L_{n}{λ_{0}, …, λ_{n}}

in (0, 1) strictly interlace. Hint: Use part a] and arguments similar to those given in the hints for part e]. ⊓ ⊔

E.8 A Global Estimate for the Zeros. Let (λ_{i})_{i=1}^{n} ⊂ (−1/2, ∞). Assume that x_{1} < x_{2} < ⋯ < x_{n} are the zeros of L_{n}{λ_{0}, …, λ_{n}} in (0, 1). Then

exp (−\frac{4n + 2}{1 + 2λ_{∗}} ) < x_{1} < x_{2} < ⋯ < x_{n} < exp (\frac{−j_{1}^{2}}{(1 + 2λ^{∗})(4n + 2)}) ,

where λ_{∗} := min{λ_{0}, …, λ_{n}}, λ^{∗} := max{λ_{0}, …, λ_{n}} and j_{1} > 3π/4 is the smallest positive zero of the Bessel function

J_{0}(z) := ∑_{k=0}^{∞} \frac{(−z^{2})^{k}}{(k! 2^{k})^{2}} .

Proof. Let L_{n} be the nth Laguerre polynomial with respect to the weight e^{−x} on [0, ∞), and let the zeros of L_{n} be z_{1} < z_{2} < ⋯ < z_{n}. Then by [Szegő [75], pp. 127–131])

\frac{j_{1}^{2}}{4n + 2} < z_{1} < z_{2} < ⋯ < z_{n} < 4n + 2 ,

where the upper estimate is asymptotically sharp, and the lower estimate is sharp up to a multiplicative constant (not exceeding 4^{4}/(9π^{2})). Now use E.1 and E.7 j]. ⊓ ⊔

E.9 The Order of the Zero at 1 of Certain Polynomials. This exercise, due in part to Kós, gives precise estimates on the maximum order of the zero at 1 of a polynomial whose coefficients are bounded in modulus by the leading coefficient.

a] Suppose a_{0}, a_{1}, …, a_{n−1} are complex numbers with modulus at most 1, and suppose a_{n} = 1. Then the multiplicity of the zero of

p(x) = a_{0} + a_{1}x + a_{2}x^{2} + ⋯ + a_{n}x^{n}

at 1 is at most 5\sqrt{n}.

Proof. If p has a zero at 1 of multiplicity m, then for every polynomial f of degree less than m, we have

(3.4.11)  a_{0}f(0) + a_{1}f(1) + ⋯ + a_{n}f(n) = 0 .

We construct a polynomial f of degree at most 5\sqrt{n}, for which

f(n) > |f(0)| + |f(1)| + ⋯ + |f(n − 1)| .

Equality (3.4.11) cannot hold with this f, so the multiplicity of the zero of p at 1 is at most the degree of f.

Let T_{ν} be the νth Chebyshev polynomial defined by (2.1.1). Let k ∈ ℕ, and let

g := T_{0} + T_{1} + ⋯ + T_{k} ∈ P_{k} .

Note that g(1) = k + 1. Also, for 0 < y ≤ π,

g(cos y) = 1 + cos y + cos 2y + ⋯ + cos ky = \frac{sin(k + \frac{1}{2})y + sin \frac{1}{2}y}{2 sin \frac{1}{2}y} .

Hence, for −1 ≤ x < 1,

|g(x)| ≤ \begin{matrix}√ & \overline{2} \\ √ & \overline{1 − x} \end{matrix} .

Let f(x) := g^{4}(\frac{2x}{n} − 1). Then f(n) = g^{4}(1) = (k + 1)^{4} and

|f(0)| + |f(1)| + ⋯ + |f(n − 1)| ≤ ∑_{j=1}^{n} \frac{4}{(\frac{2j}{n} )2} < n^{2} ∑_{j=1}^{∞} \frac{1}{j^{2}} = \frac{π^{2}n^{2}}{6} .

If k := ⌊(π^{2}/6)^{1/4}\sqrt{n}⌋, then

f(n) > |f(0)| + |f(1)| + ⋯ + |f(n − 1)| .

In this case the degree of f is 4k ≤ 5\sqrt{n}. ⊓ ⊔

The result of part a] is essentially sharp.

b] For every n ∈ ℕ, there exists a polynomial

p_{n}(x) = a_{0} + a_{1}x + ⋯ + a_{2n^{2}}x^{2n^{2}}

such that a_{2n^{2}} = 1, |a_{0}|, |a_{1}|, …, |a_{2n^{2}−1}| are real numbers with modulus at most 1, and p_{n} has a zero at 1 with multiplicity at least n.

Proof. Define

L_{n}(x) := \frac{(n!)^{2}}{2πi} ∫_{Γ} \frac{x^{t} dt}{∏_{k=0}^{n} (t − k^{2})} , n = 0, 1, …,

where the simple closed contour Γ surrounds the zeros of the denominator of the integrand. Then L_{n} is a polynomial of degree n^{2} with a zero of multiplicity at least n at 1. (This can easily be seen by repeated differentiation and then evaluation of the above contour integral by expanding the contour to infinity.)

Also, by the residue theorem,

L_{n}(x) = 1 + ∑_{k=1}^{n} c_{k,n}x^{k^{2}}

where

c_{k,n} = \frac{(−1)^{n}(n!)^{2}}{∏_{j=0,j≠k}^{n}(k^{2} − j^{2})} = \frac{(−1)^{k}2(n!)^{2}}{(n − k)!(n + k)!} .

It follows that

|c_{k,n}| ≤ 2 , k = 1, 2, …, n .

Hence,

P_{n}(x) := \frac{L_{n}(x) + L_{n}(x^{2})}{2}

3.5 Chebyshev Polynomials in Rational Spaces 139

is a polynomial of degree 2n^{2} with a zero of order n at 1. Also P_{n} has constant coefficient 1 and each of its other coefficients is a real number of modulus less than 1. Now let p_{n}(x) := x^{2n^{2}}P_{n}(1/x). ⊓ ⊔ c] For every n ∈ ℕ, there exists a polynomial

p_{n}(x) = a_{0} + a_{1}x + ⋯ + a_{n}x^{n}

such that a_{n} = 1, a_{0}, a_{1}, …, a_{n−1} are real numbers of modulus less than 1, and p has a zero at 1 with multiplicity at least ⌊\sqrt{n/2}⌋.

3.5 Chebyshev Polynomials in Rational Spaces

There are very few situations where Chebyshev polynomials can be explicitly computed. Indeed, only the classical case of Section 2.1 is well known.

However, the explicit formulas for the Chebyshev polynomials for the trigonometric rational system

(3.5.1) (1 , \frac{1 ± sin θ}{cos θ − a_{1}} , \frac{1 ± sin θ}{cos θ − a_{2}} , …, \frac{1 ± sin θ}{cos θ − a_{n}} ) , θ ∈ [0, 2π)

and therefore also for the rational system

(3.5.2)  (1 , \frac{1}{x − a_{1}} , \frac{1}{x − a_{2}} , …, \frac{1}{x − a_{n}} ) , x ∈ [−1, 1]

with distinct real poles outside [−1, 1] are implicitly contained in Achiezer [56].

The case (3.5.1) does not perfectly fit our discussion of Section 3.3 because of the periodicity or because [0, 2π) is not a compact subset of ℝ. This leads to nonuniqueness of the Chebyshev polynomials. Note that ordinary polynomials arise as a limiting case of the span of system (3.5.2) on letting all the poles tend to ±∞.

We are primarily interested in the linear span of (3.5.2) and its trigonometric counterpart obtained with the substitution x = cos θ. Let

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

and

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

where (a_{k})_{k=1}^{n} ⊂ ℂ\[−1, 1] is a fixed sequence of poles.

When the poles a_{1}, a_{2}, …, a_{n} are distinct and real, (3.5.3) and (3.5.4) are simply the real spans of the systems

(3.5.5)  (1 , \frac{1}{x − a_{1}} , \frac{1}{x − a_{2}} , …, \frac{1}{x − a_{n}} )  on [−1, 1]

and

(3.5.6) (1 , \frac{1 ± sin θ}{cos θ − a_{1}} , \frac{1 ± sin θ}{cos θ − a_{2}} , …, , \frac{1 ± sin θ}{cos θ − a_{n}} )  on [0, 2π) ,

respectively.

We can construct Chebyshev polynomials of the first and second kinds, which are analogous to T_{n} and U_{n} of Section 2.1, for the spaces P_{n}(a_{1}, a_{2}, …, a_{n}) and T_{n}(a_{1}, a_{2}, …, a_{n}) as follows. Given a sequence (a_{k})_{k=1}^{n} ⊂ ℂ\[−1, 1], we define the sequence (c_{k})_{k=1}^{n} by

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

that is,

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

Note that

(a_{k} + \sqrt{a_{k}^{2} − 1}) (a_{k} − \sqrt{a_{k}^{2} − 1}) = 1 .

In what follows, \sqrt{a_{k}^{2} − 1} is always defined by (3.5.8) (this specifies the choice of root). Let D := {z ∈ ℂ : |z| < 1}, let

(3.5.9)  M_{n}(z) := (∏_{k=1}^{n}(z − c_{k})(z − \overline{c}_{k}))^{1/2} ,

where the square root is defined so that M_{n}^{∗}(z) := z^{n}M_{n}(z^{−1}) is an analytic function in a neighborhood of the closed unit disk \overline{D}, and let

(3.5.10)  f_{n}(z) := \frac{M_{n}(z)}{z^{n}M_{n}(z^{−1})} .

Note that f_{n}^{2} is actually a finite Blaschke product (see E.12 of Section 4.2). Also, f_{n}(z^{−1}) = f_{n}(z)^{−1} whenever |z| = 1.

The Chebyshev polynomials of the first kind for the spaces

P_{n}(a_{1}, a_{2}, …, a_{n}) and T_{n}(a_{1}, a_{2}, …, a_{n})

are now defined by

(3.5.11) T_{n}(x) := \frac{1}{2}(f_{n}(z) + f_{n}(z)^{−1}) , where x := \frac{1}{2}(z + z^{−1}) , |z| = 1

and

(3.5.12) T̃_{n}(θ) := T_{n}(cos θ) , θ ∈ ℝ ,

respectively.

The Chebyshev polynomials of the second kind for these two spaces are defined by

(3.5.13) U_{n}(x) := \frac{f_{n}(z) − f_{n}(z)^{−1}}{z − z^{−1}} , where x := \frac{1}{2}(z + z^{−1}) , |z| = 1

and

(3.5.14) Ũ_{n}(θ) := U_{n}(cos θ) sin θ , θ ∈ ℝ ,

respectively.

As we will see, these Chebyshev polynomials preserve many of the elementary properties of the classical trigonometric and algebraic Chebyshev polynomials. This is the content of the next three results.

Theorem 3.5.1 (Chebyshev Polynomials of the First and Second Kinds in Trigonometric Rational Spaces). Given (a_{k})_{k=1}^{n} ⊂ ℂ\[−1, 1], let T̃_{n} and Ũ_{n} be defined by (3.5.12) and (3.5.14), respectively. Then the following statements hold:

a] T̃_{n} ∈ T_{n}(a_{1}, a_{2}, …, a_{n}) and Ũ_{n} ∈ T_{n}(a_{1}, a_{2}, …, a_{n}) .

b] ∥T̃_{n}∥_{ℝ} = 1 and ∥Ũ_{n}∥_{ℝ} = 1 .

c] There exist 0 = θ_{0} < θ_{1} < ⋯ < θ_{n} = π such that

T̃_{n}(θ_{j}) = T̃_{n}(−θ_{j}) = (−1)^{j} , j = 0, 1, …, n .

d] There exist 0 < τ_{1} < τ_{2} < ⋯ < τ_{n} < π such that

Ũ_{n}(τ_{j}) = −Ũ_{n}(−τ_{j}) = (−1)^{j−1} , j = 1, 2, …, n .

e] For every θ ∈ ℝ,

T̃_{n}^{2}(θ) + Ũ_{n}^{2}(θ) = 1 .

Proof. Observe that there are polynomials p_{1} ∈ P_{n} and p_{2} ∈ P_{n−1} such that

(3.5.15)  T̃_{n}(θ) = T_{n}(cos θ) = \frac{e^{−inθ}M_{n}^{2}(e^{iθ}) + e^{inθ}M_{n}^{2}(e^{−iθ})}{2M_{n}(e^{iθ})M_{n}(e^{−iθ})}

= \frac{p_{1}(cos θ)}{∏_{k=1}^{n} |cos θ − a_{k}|}

and

(3.5.16) Ũ_{n}(θ) = U_{n}(cos θ) sin θ = \frac{e^{−inθ}M_{n}^{2}(e^{iθ}) − e^{inθ}M_{n}^{2}(e^{−iθ})}{2iM_{n}(e^{iθ})M_{n}(e^{−iθ})}

= \frac{p_{2}(cos θ) sin θ}{∏_{k=1}^{n} |cos θ − a_{k}|} .

Thus a] is proved.

Since |c_{k}| < 1 and f_{n}^{2} is a finite Blaschke product, we have

(3.5.17)  |f_{n}(z)| = 1 whenever |z| = 1 .

Now b] follows immediately from (3.5.10) to (3.5.14).

Note that T̃_{n}(θ) is the real part, and Ũ_{n}(θ) is the imaginary part of f_{n}(e^{iθ}), that is,

f_{n}(e^{iθ}) = T̃_{n}(θ) + iŨ_{n}(θ) , θ ∈ ℝ ,

which together with (3.5.17) implies e].

To prove parts c] and d], we note that T̃_{n}(θ) = ±1 if and only if f_{n}(e^{iθ}) = ±1, and Ũ_{n}(θ) = ±1 if and only if f_{n}(e^{iθ}) = ±i. Since |c_{k}| < 1 for k = 1, 2, …, n, f_{n}^{2} has exactly 2n zeros in the open unit disk D. Since f_{n}^{2} is analytic in a region containing the closed unit disk \overline{D}, c] and d] follow by the argument principle (see, for example, Ash [71]). ⊓ ⊔

With the transformation x = cos θ = \frac{1}{2}(z + z^{−1}) and z = e^{iθ}, Theorem 3.5.1 can be reformulated as follows:

Theorem 3.5.2 (Chebyshev Polynomials in Algebraic Rational Spaces). Given (a_{k})_{k=1}^{n} ⊂ ℂ \ [−1, 1], let T_{n} and U_{n} be defined by (3.5.11) and (3.5.13), respectively. Then

a] T_{n} ∈ P_{n}(a_{1}, a_{2}, …, a_{n}) and U_{n} ∈ P_{n}(a_{1}, a_{2}, …, a_{n}) .

b] ∥T_{n}∥_{[−1,1]} = 1 and ∥\sqrt{1 − x^{2}} U_{n}(x)∥_{[−1,1]} = 1 .

c] There exist 1 = x_{0} > x_{1} > ⋯ > x_{n} = −1 such that

T_{n}(x_{j}) = (−1)^{j} , j = 0, 1, 2, …, n .

d] There exist 1 > y_{1} > y_{2} > ⋯ > y_{n} > −1 such that

\sqrt{1 − y_{j}^{2}} U_{n}(y_{j}) = (−1)^{j−1} , j = 1, 2, …, n .

e] For every x ∈ [−1, 1],

(T_{n}(x))^{2} + (\sqrt{1 − x^{2}} U_{n}(x))^{2} = 1 .

Parts c] and d] of Theorems 3.5.1 and 3.5.2 establish the equioscillation property of the Chebyshev polynomials, which also extends to certain linear combinations of Chebyshev polynomials. In the trigonometric polynomial case this is the fact that cos α cos nθ+sin α sin nθ = cos(nθ−α) equioscillates 2n times on the unit circle [0, 2π]. Our next theorem characterizes the Chebyshev polynomials for T_{n}(a_{1}, a_{2}, …, a_{n}) and records a monotonicity property that we require later.

Theorem 3.5.3 (Chebyshev Polynomials in Trigonometric Rational Spaces). Let (a_{k})_{k=1}^{n} ⊂ ℂ \ [−1, 1]. Then (i) and (ii) below are equivalent:

(i) There is an α ∈ ℝ such that

V = (cos α) T̃_{n} + (sin α) Ũ_{n} ,

where T̃_{n} and Ũ_{n} are defined by (3.5.12) and (3.5.14).

(ii) V ∈ T_{n}(a_{1}, a_{2}, …, a_{n}) has uniform norm 1 on ℝ, and it equioscillates 2n times on ℝ (mod 2π). That is, there exist

0 ≤ θ_{0} < θ_{1} < ⋯ < θ_{2n−1} < 2π

so that

V (θ_{j}) = ±(−1)^{j} , j = 0, 1, …, 2n − 1 .

Furthermore, if V is of the form in (i) (or characterized by (ii)), then

V′ = (cos α) T̃′_{n} + (sin α) Ũ′_{n}

does not vanish between any two consecutive alternation points of V (that is, between θ_{j−1} and θ_{j} for j = 1, 2, …, 2n − 1 and between θ_{2n−1} and 2π + θ_{0}).

Proof. (i) ⇒ (ii). By Theorem 3.5.1 e] and Cauchy’s inequality, we have

(3.5.18) |(cos α) T̃_{n} + (sin α) Ũ_{n}|^{2} ≤ (cos^{2} α + sin^{2} α)(T̃_{n}^{2} + Ũ_{n}^{2}) = 1

on the real line. From Theorem 3.5.1 c], d], and e], we obtain that T̃_{n}/Ũ_{n} oscillates between +∞ and −∞ exactly 2n times on ℝ (mod 2π), and hence it takes the value cot α exactly 2n times. At each such point, (3.5.18) becomes an equality, namely, (cos α) T̃_{n} + (sin α) Ũ_{n} = ±1 with different signs for every two consecutive such points.

(ii) ⇒ (i). Let V be as specified in part (ii) of the theorem. Let θ_{0} be a point where V achieves its maximum on ℝ, so V (θ_{0}) = 1. We want to show that V is equal to p := T̃_{n}(θ_{0})T̃_{n} +Ũ_{n}(θ_{0})Ũ_{n}. Since V (θ_{0}) = p(θ_{0}) = 1 and V′(θ_{0}) = p′(θ_{0}) = 0, V − p has a zero at θ_{0} with multiplicity at least 2. There are at least 2n − 1 more zeros (we count multiplicities) of V − p in ℝ (mod 2π), with one between any two consecutive alternation points of p if the first zero of p to the right of θ_{0} is greater than the first zero of V to the right of θ_{0}. If the first zero of V to the right of θ_{0} is greater than or equal to the first zero of p to the right of θ_{0}, then there is one zero of p − V between any two consecutive alternation points of V. In any case V − p has at least 2n + 1 zeros in ℝ (mod 2π). Hence V − p is identically 0.

To prove the final part of the theorem let V ∈ T_{n}(a_{1}, a_{2}, …, a_{n}) be such that ∥V ∥_{ℝ} = 1 and V equioscillates 2n times on ℝ (mod 2π) between ±1. Assume there is a θ_{0} ∈ [0, 2π) such that |V (θ_{0})| < 1 and V′(θ_{0}) = 0. Then V (θ_{0}) ≠ 0; otherwise the numerator of V would have at least 2n + 1 zeros in ℝ (mod 2π), which is a contradiction. Observe that there is a trigonometric polynomial t ∈ T_{2n} such that

V^{2}(θ) − V^{2}(θ_{0}) = \frac{t(θ)}{∏_{k=1}^{n}(cos θ − a_{k})(cos θ − \overline{a}_{k})} .

This t has at least 4n + 1 zeros in ℝ (mod 2π), which is a contradiction again. Therefore V′(θ) ≠ 0 if |V (θ)| < 1, which means that V is strictly monotone between any two of its consecutive alternation points. ⊓ ⊔

Under some assumptions on (a_{k})_{k=1}^{n} it is easy to write down the explicit partial fraction decompositions for T_{n} and U_{n}.

Theorem 3.5.4. Let (a_{k})_{k=1}^{n} ⊂ ℂ \ [−1, 1] be a sequence of distinct numbers such that its nonreal elements are paired by complex conjugation. Let T_{n} and U_{n} be the Chebyshev polynomials of the first and second kinds defined by (3.5.11) and (3.5.13), respectively. Then

(3.5.19)  T_{n}(x) = A_{0,n} + \frac{A_{1,n}}{x − a_{1}} + ⋯ + \frac{A_{n,n}}{x − a_{n}}

and

(3.5.20)  U_{n}(x) = \frac{B_{1,n}}{x − a_{1}} + \frac{B_{2,n}}{x − a_{2}} + ⋯ + \frac{B_{n,n}}{x − a_{n}} ,

where

A_{0,n} = \frac{(−1)^{n}}{2} (c_{1}^{−1}c_{2}^{−1} ⋯ c_{n}^{−1} + c_{1}c_{2} ⋯ c_{n}) ,

A_{k,n} = (\frac{c_{k} − c_{k}^{−1}}{2} )^{2} ∏_{{j=1 \atop j≠k}}^{n} \frac{1 − c_{k}c_{j}}{c_{k} − c_{j}} , k = 1, 2, …, n ,

and

B_{k,n} = \frac{c_{k} − c_{k}^{−1}}{2} ∏_{{j=1 \atop j≠k}}^{n} \frac{1 − c_{k}c_{j}}{c_{k} − c_{j}} , k = 1, 2, …, n .

Proof. It follows from Theorems 3.5.1 a] and 3.5.2 a] that T_{n} and U_{n} can be written as the partial fraction forms above. Now it is quite easy to calculate the coefficients A_{k,n} and B_{k,n}. For example,

A_{0,n} = lim_{x→∞} T_{n}(x) = lim_{z→0} \frac{1}{2} (\frac{M_{n}(z)}{z^{n}M_{n}(z^{−1})} + \frac{z^{n}M_{n}(z^{−1})}{M_{n}(z)})

= \frac{(−1)^{n}}{2} (c_{1}^{−1}c_{2}^{−1} ⋯ c_{n}^{−1} + c_{1}c_{2} ⋯ c_{n})

and for k = 1, 2, …, n ,

A_{k,n} = lim_{x→a_{k}}(x − a_{k})T_{n}(x)

= lim_{z→c_{k}} \frac{1}{4}(z − c_{k})(1 − c_{k}^{−1}z^{−1}) (\frac{M_{n}(z)}{z^{n}M_{n}(z^{−1})} + \frac{z^{n}M_{n}(z^{−1})}{M_{n}(z)})

= (\frac{c_{k} − c_{k}^{−1}}{2} )^{2} ∏_{{j=1 \atop j≠k}}^{n} \frac{1 − c_{k}c_{j}}{c_{k} − c_{j}} , k = 1, 2, …, n .

The coefficients B_{k,n} can be calculated in the same fashion. ⊓ ⊔

Comments, Exercises, and Examples.

The explicit formulas of this section are tremendously useful. They allow, for example, derivation of sharp Bernstein-type inequalities for rational functions; see Section 7.1. Various further properties of these Chebyshev polynomials for rational function spaces are explored in the exercises, which follow, Borwein, Erdélyi, and Zhang [94b]. In particular, the orthogonalization of such rational systems on [−1, 1] with respect to the weight w(x) = (1 − x^{2})^{−1/2} can be made explicit in terms of the Chebyshev polynomials. Various other aspects of these orthogonalizations may be found in Achiezer [56], Bultheel et al. [91], and Van Assche and Vanherwegen [92].

E.1 Further Properties of T̃_{n} and T_{n}. Given (a_{k})_{k=1}^{n} ⊂ ℂ \ [−1, 1], let (c_{k})_{k=1}^{n} be defined by

c_{k} := a_{k} − \sqrt{a_{k}^{2} − 1} , |c_{k}| < 1 ,

as before. We introduce the Bernstein factors

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

and

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

where the choice of \sqrt{a_{k}^{2} − 1} is determined by the restriction |c_{k}| < 1. Note that for x ∈ [−1, 1], we have

Re (\frac{\sqrt{a_{k}^{2} − 1}}{a_{k} − x}) = Re ( \frac{c_{k}^{−1} − c_{k}}{\frac{1}{2}c_{k}^{−1} − x}) ≥ \frac{(1 − |c_{k}|^{2})(1 − |c_{k}|)^{2}}{|1 − 2c_{k}x|^{2}} > 0 .

The following result generalizes the trigonometric identities

(cos nt)′ = −n sin nt , (sin nt)′ = n cos nt ,

and

((cos nt)′)^{2} + ((sin nt)′)^{2} = n^{2} ,

which are limiting cases (if n ∈ ℕ and t ∈ ℝ are fixed, then lim B̃_{n}(t) = n as all a_{k} → ±∞).

a] Show that, on the real line,

T̃′_{n} = −B̃_{n}Ũ_{n} ,  Ũ′_{n} = B̃_{n}T̃_{n} ,

and

(T̃′_{n})^{2} + (Ũ′_{n})^{2} = B̃_{n}^{2} .

Hint: For example,

T̃′_{n}(θ) = \frac{1}{2} (f′_{n}(e^{iθ}) − \frac{f′_{n}(e^{iθ})}{f_{n}^{2}(e^{iθ})}) ie^{iθ}

= \frac{−e^{iθ}f′_{n}(e^{iθ})}{f_{n}(e^{iθ})} \frac{f_{n}(e^{iθ}) − f_{n}(e^{iθ})^{−1}}{2i} = −B̃_{n}(θ)Ũ_{n}(θ) .

⊓ ⊔

b] If V := (cos α) T̃_{n} + (sin α) Ũ_{n} for some α ∈ ℝ, then

(V′)^{2} + B̃_{n}^{2}V^{2} = B̃_{n}^{2}

holds on the real line.

c] The Derivative of T_{n} at ±1. Let T_{n} be defined by (3.5.11). Then

T′_{n}(1) = (∑_{k=1}^{n} Re (\frac{1 + c_{k}}{1 − c_{k}} ))^{2}

and

T′_{n}(−1) = (−1)^{n} (∑_{k=1}^{n} Re (\frac{1 − c_{k}}{1 + c_{k}} ))^{2} .

3.5 Chebyshev Polynomials in Rational Spaces 147

d] Contour Integral for T_{n}. Show that

T_{n}(x) = \frac{1}{2πi} ∫_{γ} (∏_{j=1}^{n} \frac{(t − c_{j})(t − \overline{c}_{j})}{(1 − c_{j}t)(1 − \overline{c}_{j}t)})^{1/2} \frac{t − x}{t^{2} − 2tx + 1} dt

for every x ∈ [−1, 1], where γ is a circle centered at the origin with radius 1 < r < min{|c_{j}^{−1}| : 1 ≤ j ≤ n}, and the square root in the integrand is an analytic function of t in a neighborhood of γ. Hint: Cauchy’s integral formula and the map x = \frac{1}{2}(z + z^{−1}) give

T_{n}(x) = \frac{1}{2} (\frac{M_{n}(z)}{z^{n}M_{n}(z^{−1})} + \frac{z^{n}M_{n}(z^{−1})}{M_{n}(z)})

= \frac{1}{2πi} ∫_{γ} \frac{1}{2} \frac{M_{n}(t)}{t^{n}M_{n}(t^{−1})} (\frac{1}{t − z} + \frac{1}{t − z^{−1}} ) dt

= \frac{1}{2πi} ∫_{γ} \frac{M_{n}(t)}{t^{n}M_{n}(t^{−1})} \frac{t − x}{t^{2} − 2tx + 1} dt ,

where M_{n} is defined by (3.5.9). ⊓ ⊔

E.2 Orthogonality. Given (a_{k})_{k=1}^{∞} ⊂ ℝ\[−1, 1], let (c_{k})_{k=1}^{∞} be defined by

a_{k} = \frac{1}{2}(c_{k} + c_{k}^{−1}) , c_{k} = a_{k} − \sqrt{a_{k}^{2} − 1} , c_{k} ∈ (−1, 1)

and let (T_{n})_{n=0}^{∞} be defined by (3.5.11). a] Show that

∫_{−1}^{1} T_{n}(x)T_{m}(x)\frac{dx}{\sqrt{1 − x^{2}}} = \frac{π}{2} (−1)^{n+m}(1 + c_{1}^{2} ⋯ c_{m}^{2})c_{m+1} ⋯ c_{n}

for all integers 0 ≤ m ≤ n. (The empty product is understood to be 1.) b] Given a ∈ ℝ \ [−1, 1], let c ∈ (−1, 1) be defined by

a = \frac{1}{2}(c + c^{−1}) , c = a − \sqrt{a^{2} − 1} , c ∈ (−1, 1) .

Show that

∫_{−1}^{1} T_{n}(x)\frac{1}{x − a} \frac{dx}{\sqrt{1 − x^{2}}} = \frac{2π}{c − c^{−1}} ∏_{j=1}^{n} \frac{c − c_{j}}{1 − cc_{j}} .

c] Show that

∫_{−1}^{1} T_{n}(x)\frac{dx}{\sqrt{1 − x^{2}}} = (−1)^{n}πc_{1}c_{2} ⋯ c_{n}

and

∫_{−1}^{1} T_{n}(x)\frac{1}{x − a_{k}} \frac{dx}{\sqrt{1 − x^{2}}} = 0 , k = 1, 2, …, n .

Given a sequence (a_{k})_{k=1}^{∞} ⊂ ℝ\[−1, 1], we define
R_{0} := 1 , R_{n} := T_{n} + c_{n}T_{n−1}

and

R_{0}^{∗} := \frac{1}{\sqrt{π}} , R_{n}^{∗} := \sqrt{\frac{2}{π(1 − c_{n}^{2})}} (T_{n} + c_{n}T_{n−1}) .

The following part of this exercise indicates that these simple linear combinations of T_{n} and T_{n−1} give the orthogonalization of the rational system

(1 , \frac{1}{x − a_{1}} , \frac{1}{x − a_{2}} , …)

whenever (a_{k})_{k=1}^{∞} ⊂ ℝ \ [−1, 1] is a sequence of distinct real numbers.

d] Show that, for all nonnegative integers n and m,

∫_{−1}^{1} R_{n}^{∗}(x)R_{m}^{∗}(x)\frac{dx}{\sqrt{1 − x^{2}}} = δ_{m,n} ,

where δ_{m,n} is the Kronecker symbol.

Proof. Let m ≤ n. By part c],

∫_{−1}^{1} R_{n}(x)\frac{1}{x − a_{k}} \frac{dx}{\sqrt{1 − x^{2}}} = 0

holds for k = 1, 2, …, n − 1. Also

∫_{−1}^{1} R_{n}(x)\frac{dx}{\sqrt{1 − x^{2}}} = ∫_{−1}^{1} (T_{n}(x) + c_{n}T_{n−1}(x))\frac{dx}{\sqrt{1 − x^{2}}}
= (−1)^{n}(c_{1}c_{2} ⋯ c_{n}) + c_{n}(−1)^{n−1}(c_{1}c_{2} ⋯ c_{n−1}) = 0 .

This implies that

∫_{−1}^{1} R_{n}(x)R_{m}(x)\frac{dx}{\sqrt{1 − x^{2}}} = 0 , m = 0, 1, …, n − 1 .

Finally, it follows from part a] that

∫_{−1}^{1} R_{n}^{∗}(x)^{2} \frac{dx}{\sqrt{1 − x^{2}}} = 1 .

⊓ ⊔

e] Assume (a_{k})_{k=1}^{∞} ⊂ ℝ \ [−1, 1]. Then T_{n} and R_{n} have exactly n zeros in [−1, 1], and the zeros of T_{n−1} and T_{n} strictly interlace.

E.3 Extension of Theorems 3.5.1 and 3.5.3. Given (a_{k})_{k=1}^{2n} ⊂ ℂ \ ℝ, let

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

Without loss of generality we may assume that

Im(a_{k}) > 0 , k = 1, 2, …, 2n .

a] Show that there is a polynomial q_{2n} ∈ P_{2n}^{c} of the form

q_{2n}(z) = γ ∏_{k=1}^{2n} (z − c_{k}) , |c_{k}| < 1 , γ ∈ ℂ

such that

|q_{2n}(e^{iθ})| = ∏_{k=1}^{2n} |sin((θ − a_{k})/2)| , θ ∈ ℝ .

Hint: Use the fact that |z − c| = |1 − \overline{c}z| whenever |z| = 1 and c ∈ ℂ. ⊓ ⊔

Associated with q_{2n} ∈ P_{2n}^{c} defined in part a], let
M_{n}(z) := \sqrt{q_{2n}(z)}

and

M_{n}^{∗}(z) := (\overline{γ} ∏_{k=1}^{2n}(1 − \overline{c}z))^{1/2} ,

where the square roots are defined so that M_{n}^{∗} is analytic in a neighborhood of the closed unit disk, and M_{n} is analytic in a neighborhood of the complement of the open unit disk. Let

f_{n}(z) := \frac{M_{n}(z)}{M_{n}^{∗}(z)} .

For θ ∈ ℝ, we define

T̃_{n}(θ) := Re(f_{n}(e^{iθ})) = \frac{1}{2} (\frac{M_{n}(e^{iθ})}{M_{n}^{∗}(e^{iθ})} + \frac{M_{n}^{∗}(e^{iθ})}{M_{n}(e^{iθ})})

and

Ũ_{n}(θ) := Im(f_{n}(e^{iθ})) = \frac{1}{2i} (\frac{M_{n}(e^{iθ})}{M_{n}^{∗}(e^{iθ})} − \frac{M_{n}^{∗}(e^{iθ})}{M_{n}(e^{iθ})}) .

Using the new (extended) definitions, show the following:

150 3. Chebyshev and Descartes Systems

b] T̃_{n} ∈ T_{n}(a_{1}, a_{2}, …, a_{2n}) and Ũ_{n} ∈ T_{n}(a_{1}, a_{2}, …, a_{2n}) . c] ∥T̃_{n}∥_{ℝ} = 1 and ∥Ũ_{n}∥_{ℝ} = 1 . d] There are numbers θ_{1} < θ_{2} < ⋯ < θ_{2n} in [−π, π) such that

T̃(θ_{j}) = ±(−1)^{j} , j = 1, 2, …, 2n .

e] There are numbers τ_{1} < τ_{2} < ⋯ < τ_{2n} in [−π, π) such that

Ũ(τ_{j}) = ±(−1)^{j} , j = 1, 2, …, 2n .

f] T̃(θ)^{2} + Ũ(θ)^{2} = 1 for every θ ∈ ℝ . g] Both T̃_{n} and Ũ_{n} have exactly 2n simple zeros in the period [−π, π), and the zeros of T̃_{n} and Ũ_{n} strictly interlace. h] The statements of Theorem 3.5.3 remain valid.

E.4 Extension of the Bernstein Factor B̃_{n}. Let

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

With the notation of the previous exercise we define

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

a] Show that for every θ ∈ ℝ,

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

b] Show that, on the real line,

T̃′_{n} = −B̃_{n}Ũ_{n} ,  Ũ′_{n} = B̃_{n}T̃_{n} ,

and

(T̃′_{n})^{2} + (Ũ′_{n})^{2} = B̃_{n}^{2} .

c] Show that

(V′)^{2} + B̃_{n}^{2}V^{2} = B̃_{n}^{2}

holds on the real line for every V of the form

V = (cos α) T̃_{n} + (sin α) Ũ_{n} , α ∈ ℝ .

3.5 Chebyshev Polynomials in Rational Spaces 151

E.5 Chebyshev Polynomials for P_{n}(a_{1}, a_{2}, …, a_{n}) on ℝ. Let

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

Let

M_{n}(z) := (∏_{k=1}^{n}(z − a_{k}))^{1/2}

and

M_{n}^{∗}(z) := (∏_{k=1}^{n}(z − \overline{a}_{k}))^{1/2} ,

where the square roots are defined so that M_{n}^{∗} is analytic in a neighborhood of the closed upper half-plane, and M_{n} is analytic in a neighborhood of the closed lower half-plane. Let

f_{n}(z) := \frac{M_{n}(z)}{M_{n}^{∗}(z)} .

For x ∈ ℝ, we define

T_{n}(x) := Re(f_{n}(x)) = \frac{1}{2} (\frac{M_{n}(x)}{M_{n}^{∗}(x)} + \frac{M_{n}^{∗}(x)}{M_{n}(x)})

and

U_{n}(x) := Im(f_{n}(x)) = \frac{1}{2i} (\frac{M_{n}(x)}{M_{n}^{∗}(x)} − \frac{M_{n}^{∗}(x)}{M_{n}(x)}) .

Show the following: a] T_{n} ∈ P_{n}(a_{1}, a_{2}, …, a_{n}) and U_{n} ∈ P_{n}(a_{1}, a_{2}, …, a_{n}) . b] ∥T_{n}∥_{ℝ} = 1 and ∥U_{n}∥_{ℝ} = 1 . c] There are real numbers x_{1} > x_{2} > ⋯ > x_{n−1} such that

T_{n}(x_{j}) = (−1)^{j} , lim_{x→∞} T_{n}(x) = 1 , and lim_{x→−∞} T_{n}(x) = (−1)^{n} .

d] There are real numbers y_{1} > y_{2} > ⋯ > y_{n−1} such that

U_{n}(y_{j}) = (−1)^{j+1} and lim_{x→±∞} U_{n}(x) = 0 .

e] T_{n}(x)^{2} + U_{n}(x)^{2} = 1 for every x ∈ ℝ . f] Both T_{n} and U_{n} have exactly n simple zeros on ℝ, and the zeros of T_{n} and U_{n} strictly interlace. g] The following statements are equivalent:

(i) There exists an α ∈ ℝ such that

V = (cos α) T_{n} + (sin α) U_{n} .

(ii) V ∈ P_{n}(a_{1}, a_{2}, ⋯, a_{n}) has uniform norm 1 on ℝ, and it equioscillates n times on the extended real line. That is, there are extended real numbers ∞ ≥ z_{1} > z_{2} > ⋯ > z_{n} > −∞ such that

V (z_{j}) = ±(−1)^{j} , j = 1, 2, …, n ,

where

V (∞) := lim_{x→∞} V (x) .

h] With the notation of part g], V is strictly monotone on each of the intervals

(z_{1}, ∞), (z_{2}, z_{1}), …, (z_{n}, z_{n−1}), (−∞, z_{n}) .

E.6 Bernstein Factor on ℝ. Let (a_{k})_{k=1}^{n} ⊂ ℂ \ ℝ with

Im(a_{k}) > 0 , k = 1, 2, …, n .

With the notation of E.5 let

B_{n}(x) := \frac{f′_{n}(x)}{f_{n}(x)} , x ∈ ℝ .

a] Show that

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

b] Show that, on the real line,

T′_{n} = −B_{n}U_{n} , U′_{n} = B_{n}T_{n} ,

and

(T′_{n})^{2} + (U′_{n})^{2} = B_{n}^{2} .

c] Show that

(V′)^{2} + B_{n}^{2}V^{2} = B_{n}^{2}

holds on the real line for every V of the form

V = (cos α) T_{n} + (sin α) U_{n} .

3.5 Chebyshev Polynomials in Rational Spaces 153

E.7 Coefficient Bounds in Nondense Rational Function Spaces. Suppose (a_{k})_{k=1}^{∞} ⊂ ℝ \ [−1, 1] is a sequence of distinct numbers satisfying

∑_{j=1}^{∞} \sqrt{1 − |a_{j}|^{−2}} < ∞ .

Show that there are numbers K_{j} > 0 such that

|D_{j,n}| ≤ K_{j}∥p∥_{[−1,1]} , j = 0, 1, …, n , n ∈ ℕ

for every p ∈ P_{n}(a_{1}, a_{2}, …, a_{n}) of the form

p(x) = D_{0,n} + \frac{D_{1,n}}{x − a_{1}} + ⋯ + \frac{D_{n,n}}{x − a_{n}} , D_{j,n} ∈ ℝ .

Hint: Use E.2 c] of Section 3.3 and Theorem 3.5.4. ⊓ ⊔