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2

Some Special Polynomials

Overview

Chebyshev polynomials are introduced and their central role in problems in the uniform norm on [−1, 1] is explored. Sequences of orthogonal functions are then examined in some generality, although our primary interest is in orthogonal polynomials (and rational functions). The third section of this chapter is concerned with orthogonal polynomials; it introduces the most classical of these. These polynomials satisfy many extremal properties, similar to those of the Chebyshev polynomials, but with respect to (weighted) L_{2} norms. The final section of the chapter deals with polynomials with positive coefficients in various bases.

2.1 Chebyshev Polynomials

The ubiquitous Chebyshev polynomials lie at the heart of many analytic problems, particularly problems in C[a, b], the space of real-valued continuous functions equipped with the uniform (supremum) norm, ∥ · ∥_{[a,b]}. Throughout this book, for any real- or complex-valued function f defined on [a, b],

∥f∥_{[a,b]} := sup_{x∈[a,b]} |f(x)| .

30 2. Some Special Polynomials

The Chebyshev polynomials are defined by

T_{n}(x) : = cos(n arccos x) ,  x ∈ [−1, 1] ,
(2.1.1)  = \frac{1}{2}((x + \sqrt{x^{2} − 1})^{n} + (x − \sqrt{x^{2} − 1})^{n}) ,  x ∈ ℂ ,

= \frac{n}{2} ∑_{k=0}^{⌊n/2⌋}(−1)^{k} \frac{(n − k − 1)!}{k!(n − 2k)!}(2x)^{n−2k} ,  x ∈ ℂ .

These elementary equivalences are left for the reader (see E.1). The nth Chebyshev polynomial has the following equioscillation property on [−1, 1]. There exist n + 1 points ζ_{i} ∈ [−1, 1] with −1 = ζ_{n} < ζ_{n−1} < ⋯ < ζ_{0} = 1 so that

(2.1.2) T_{n}(ζ_{j}) = (−1)^{n−j}∥T_{n}∥_{[−1,1]} = (−1)^{n−j} ,  j = 0, 1, …, n .

In other words T_{n} ∈ P_{n} takes the values ±∥T_{n}∥_{[−1,1]} with alternating sign the maximum possible number of times on [−1, 1]. (These extreme points are just the points cos(kπ/n), k = 0, 1, …, n.) The Chebyshev polynomial T_{n} satisfies the following extremal property:

Theorem 2.1.1. We have

min_{p∈P_{n−1}^{c}} ∥x^{n} − p(x)∥_{[−1,1]} = ∥2^{1−n}T_{n}∥_{[−1,1]} = 2^{1−n} ,

where the minimum is uniquely attained by p(x) = x^{n} − 2^{1−n}T_{n}(x).

Proof. Observe that, while the minimum is taken over P_{n−1}^{c}, we need only consider p ∈ P_{n−1}, since taking the real part of a p ∈ P_{n−1}^{c} can only improve the estimate. From the above formulas for T_{n} we have

2^{1−n}T_{n}(x) = x^{n} + s(x) , s ∈ P_{n−1} .

Now suppose there exists q ∈ P_{n−1} with

(2.1.3)  ∥x^{n} − q(x)∥_{[−1,1]} < 2^{1−n} .

Then

2^{1−n}T_{n}(x) − (x^{n} − q(x)) = s(x) + q(x) ∈ P_{n−1}

2.1 Chebyshev Polynomials 31

changes sign between any two consecutive extrema of T_{n}, hence it has at least n zeros in (−1, 1), and thus it must vanish identically. This contradicts (2.1.3), and we are done up to proving uniqueness (this is left as E.2). ⊓ ⊔ Comments, Exercises, and Examples.

The Chebyshev polynomials T_{n} are named after the versatile Russian mathematician, P. L. Chebyshev (1821–1894). The T comes from the spelling Tchebychef (or some such variant; there are many in the literature). A wealth of information on these polynomials may be found in Rivlin [90]. Throughout later sections of this book the Chebyshev polynomials will keep recurring. The initial exercises explore elementary properties of the Chebyshev polynomials.

Erdős [39] proved that for t ∈ T_{n} with ∥t∥_{ℝ} ≤ 1, the length of the graph of t on [0, 2π] is the longest if and only if t is of the form t(θ) = cos(nθ + α) with some α ∈ ℝ (see E.6). He conjectured that for any p ∈ P_{n} with ∥p∥_{[−1,1]} ≤ 1, the maximum arc length is attained by the nth Chebyshev polynomial T_{n}. This is proved in Bojanov [82b]. Kristiansen [79] also claims a proof. In E.9 the reducibility of T_{n} is considered, and in E.11 the basic properties of the transfinite diameter are established.

E.1 Basic Properties.

a] Establish the equivalence of the three representations of T_{n} given in equation (2.1.1).

Hint: cos nθ = \frac{1}{2}[(cos θ + i sin θ)^{n} + ((cos θ − i sin θ)^{n}]. To get the third representation, use E.3 b]. ⊓ ⊔

b] The zeros of T_{n} are precisely the points

x_{k} = cos \frac{(2k−1)π}{2n} , k = 1, 2, …, n .

c] The extrema of T_{n}(x) in [−1, 1] are precisely the points

ζ_{k} = cos \frac{kπ}{n} , k = 0, 1, …, n .

d] Observe that the zeros of T_{n} and T_{n+1} interlace, as do the extrema.

E.2 Uniqueness of the Minimum in Theorem 2.1.1. Prove the uniqueness of the minimum in Theorem 2.1.1.

Hint: Assume that q ∈ P_{n−1}^{c} and

∥x^{n} − q(x)∥_{[−1,1]} ≤ 2^{1−n} .

Then

h(x) := 2^{1−n}T_{n}(x) − Re(x^{n} − q(x))

defines a polynomial from P_{n−1} on ℝ having at least n zeros (counted according to their multiplicities). Thus

2^{1−n}T_{n}(x) = Re(x^{n} − q(x)) , x ∈ ℝ ,

32 2. Some Special Polynomials

which, together with the previous inequality, implies that q(x) is real whenever T_{n}(x) = ±1. Now E.6 of Section 1.1 (Lagrange interpolation) yields that q has real coefficients. Hence

2^{1−n}T_{n}(x) = x^{n} − q(x) , x ∈ ℝ .

⊓ ⊔

E.3 Further Properties of T_{n}. a] Composition. Show that T_{nm}(x) = T_{n}(T_{m}(x)). b] Three-Term Recursion. Show that

T_{n}(x) = 2xT_{n−1}(x) − T_{n−2}(x) , n = 2, 3, ….

c] Verify that

T_{0}(x) = 1
T_{1}(x) = x
T_{2}(x) = 2x^{2} − 1
T_{3}(x) = 4x^{3} − 3x
T_{4}(x) = 8x^{4} − 8x^{2} + 1
T_{5}(x) = 16x^{5} − 20x^{3} + 5x .

Note that T_{n} is even for n even and odd for n odd. d] Another Formula for T_{n}. Show that T_{n}(x) = cosh(n cosh^{−1}(x)) for every x ∈ ℝ \ [−1, 1]. e] Differential Equation. Show that

(1 − x^{2})T′′_{n}(x) − xT′_{n}(x) + n^{2}T_{n}(x) = 0 .

f] An Identity. Show that

T_{n}(x) = \frac{T′_{n+1}(x)}{2n + 2} − \frac{T′_{n−1}(x)}{2n − 2} .

g] Orthogonality. Show that

\frac{2}{π} ∫_{−1}^{1} \frac{T_{n}(x)T_{m}(x)dx}{\sqrt{1 − x^{2}}} = δ_{n,m} := \{ \begin{matrix}0, & n ≠ m \\ 1, & n = m > 0 . \end{matrix}

h] Generating Function. Show that

\frac{1 − yx}{1 − 2yx + y^{2}} = ∑_{n=0}^{∞} T_{n}(x)y^{n} , x ∈ [−1, 1] , |y| < 1 .

Hint: Set x = cos θ and sum. ⊓ ⊔ i] Another Representation of T_{n}. Show that

T_{n}(x) = ∑_{k=0}^{⌊n/2⌋} ( \begin{matrix}n \\ 2k \end{matrix} ) x^{n−2k}(x^{2} − 1)^{k} .

j] Another Identity. Show that

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

E.4 Approximation to x^{k} on [0, 1].

a] Let T_{n}^{∗}(x) = T_{n}(2x − 1) be the nth Chebyshev polynomial shifted to the interval [0, 1]. Suppose

T_{n}^{∗}(x) = ∑_{k=0}^{n} b_{k}x^{k} .

Show that for each k = 0, 1, …, n,

min_{c_{j}∈ℝ} ∥x^{k} − ∑_{{j=0 \atop j≠k}}^{n} c_{j}x^{j}∥_{[0,1]} = ∥b_{k}^{−1}T_{n}^{∗}∥_{[0,1]} .

Hint: Proceed as in the proof of Theorem 2.1.1 and use E.6 of Section 1.3.

⊓ ⊔

b] Why does this not hold for T_{n} on [−1, 1] ?

E.5 A Composition Characterization. Suppose (p_{n})_{n=1}^{∞} is a sequence of polynomials of degree n and for all positive integers n and m

p_{n} ∘ p_{m} = p_{n·m} .

Then there exists a linear transformation w(x) = αx + β so that

w ∘ p_{n} ∘ w^{−1} = x^{n} , n = 1, 2, …

or

w ∘ p_{n} ∘ w^{−1} = T_{n} , n = 1, 2, ….

This result is due to Block and Thielman [51]. The proof outlined in this exercise follows Rivlin [90]. a] Let

q(x) := a_{0} + a_{1}x + a_{2}x^{2} , a_{2} ≠ 0 , and v(x) := \frac{x}{a_{2}} − \frac{a_{1}}{2a_{2}} .

Then

v^{−1}(q(v(x))) = x^{2} + c with c := a_{0}a_{2} + (a_{1}/2) − (a_{1}^{2}/4) .

b] Let q(x) = a_{0} + a_{1}x + a_{2}x^{2}, a_{2} ≠ 0. Then there is at most one polynomial p_{n} of degree exactly n so that

p_{n}(q(x)) = q(p_{n}(x)) .

Hint: By a] we may assume q(x) = x^{2} + c. Now suppose r, s ∈ P_{n}^{c}\P_{n−1}^{c},

r(x^{2} + c) = r^{2}(x) + c

and

s(x^{2} + c) = s^{2}(x) + c .

Then u := r − s ∈ P_{n−1}^{c} satisfies

u(x^{2} + c) = u(x)(r(x) + s(x))

from which we deduce, by comparing degrees on both sides, that n = 0. (Note that the above conditions imply r and s monic.) ⊓ ⊔

c] Finish the proof of the initial statement of this exercise.

This is a special case of a more general theorem of Ritt [23] that classifies all rational functions r and s that commute in the sense that r ∘ s = s ∘ r.

d] Another Composition Characterization. Suppose p ∈ P_{n} has the property that the closure of the set

I_{p} := {z ∈ ℂ : p^{[k]}(z) = 0 for some k = 1, 2, … }

is the interval [−1, 1], where p^{[k]} is the kth iterate of p, that is,

p^{[1]} := p and p^{[k]} := p ∘ p^{[k−1]} for k = 2, 3, ….

Then p(x) = ±T_{n}(x).

e] Let

r_{n}(x) = tan(n tan^{−1}(x)) .

Show that r_{n} is a rational function in R_{n,n}, and observe that

r_{n} ∘ r_{m} = r_{n·m} .

E.6 Trigonometric Polynomials of Longest Arc Length. Theorem 5.1.3 (Bernstein-Szegő inequality) asserts that

t′(θ)^{2} + n^{2}t^{2}(θ) ≤ n^{2}∥t∥_{ℝ}^{2}

for every t ∈ T_{n} and θ ∈ ℝ. Use this to prove the following result of Erdős. For t ∈ T_{n} with ∥t∥_{ℝ} ≤ 1, the length of the graph of t on [0, 2π] is the longest if and only if it is of the form t(θ) = cos(nθ + α) for some α ∈ ℝ.

Hint: Suppose t ∈ T_{n} with ∥t∥_{ℝ} = 1. Let s(θ) := cos nθ. If

−1 < t(θ_{1}) = s(θ_{2}) < 1

holds, then by the Bernstein-Szegő inequality (see also E.5 of Section 5.1)

|t′(θ_{1})| ≤ n(1 − t^{2}(θ_{1}))^{1/2} = n(1 − s^{2}(θ_{2}))^{1/2} = |s′(θ_{2})| ,

and if equality holds for one pair of θ_{1}, θ_{2}, then it holds for all pairs, and t(θ) ≡ cos(nθ + α) for some α ∈ ℝ. Suppose t_{n}(θ) ≢ cos(nθ + α). Let τ and σ be monotone arcs of the graphs of y = t(θ) and y = s(θ), respectively, with endpoints of each having the same ordinates y_{1} and y_{2}. Let |τ| and |σ| be the length of τ and σ, respectively, and let |τ_{x}| and |σ_{x}| be the length of the projection of τ and σ, respectively, on the x-axis. Show that

|τ| < |σ| + (|τ_{x}| − |σ_{x}|)

by approximating τ and σ by a polygonal line corresponding to a subdivision of the interval with endpoints, y_{1} and y_{2} on the y-axis. ⊓ ⊔

E.7 Monic Polynomials with Minimal Norm on an Interval.

a] The unique monic polynomial p ∈ P_{n}^{c} minimizing ∥p∥_{[a,b]} is given by

p(x) = 2 (\frac{b − a}{4})^{n} T_{n} (\frac{2x − a − b}{b − a}) .

b] Let 0 < a < b. Find all monic polynomials p ∈ P_{n}^{c} minimizing

∥p∥_{[−b,−a]∪[a,b]} .

(For two intervals of different lengths this is a much harder problem. The problem was originally due to Zolotarev and is solved in terms of elliptic functions. See Todd [88], Fischer [92], and Peherstorfer [87].)

E.8 Lower Bound for the Norm of Polynomials on the Unit Disk. Let D be the open unit disk of ℂ. Show that

∥a_{0} + a_{1}z + ⋯ + a_{n}z^{n}∥_{D} ≥ max_{0≤k≤n} |a_{k}|

for arbitrary complex numbers a_{0}, a_{1}, …, a_{n}. Thus z^{n} plays the role of the nth Chebyshev polynomial on the unit disk.

Hint: If p(z) := a_{0} + a_{1}z + ⋯ + a_{n}z^{n}, then

a_{m} = \frac{1}{2πi} ∫_{∂D} \frac{p(z)}{z^{m+1}} dz .

⊓ ⊔

The next exercise supposes some familiarity with the rudiments of reducibility over ℚ and basic properties, such as irreducibility of cyclotomic polynomials over ℚ (see Clark [71]). Details of the following observation of Schur’s are in Rivlin [90].

E.9 On the Reducibility of T_{n} over ℚ. Let n ∈ ℕ be fixed.

a] The zeros of T_{n}(x/2) are all of the form

x_{j} := e^{(2j−1)iπ/(2n)} + e^{−(2j−1)iπ/(2n)} , j = 1, 2, …, n .

b] If n ≥ 3 and ζ is a primitive nth root of unity, then ζ + ζ^{−1} is of degree φ(n)/2. (Here φ is the Euler φ function.)

c] Thus if T_{n} is irreducible over ℚ, then n must be a power of 2.

d] For a positive integer h, let

F_{h}(x) :=_{∏_{j=1}^{n} ; gcd(2j−1,2n)=h} (x − x_{j}) .

(Here gcd(m, n) denotes the greatest common divisor of m and n.) Show that if h is odd, then F_{h} is irreducible over ℚ.

e] The Factorization of T_{n}.

2T_{n}(x/2) = ∏_{h|n , h odd} F_{h}(x) .

So if n is odd, T_{n} has φ(n) factors, while if n is even, then T_{n} has φ(m) factors, where m is the largest odd divisor of n.

f] Let n ≥ 3 be odd. Then T_{n}(x)/x is irreducible over ℚ if and only if n is prime.

E.10 Chebyshev Polynomials of the Second Kind. Let the Chebyshev polynomials of the second kind be defined by

U_{n−1}(x) := \frac{1}{n}T′_{n}(x) = \frac{sin nθ}{sin θ} , x = cos θ .

a] U_{n}(x) = 2T_{n}(x) + U_{n−2}(x) .

b] T_{n}(x) = U_{n}(x) − xU_{n−1}(x) .

c] U_{n}(x) = ∑_{k=0}^{n} x^{k}T_{n−k}(x) .

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

e] Orthogonality. Show that

\frac{2}{π} ∫_{−1}^{1} U_{n}(x)U_{m}(x)\sqrt{1 − x^{2}} dx = δ_{n,m} := \{ \begin{matrix}0 , & n ≠ m \\ 1 , & n = m > 0 . \end{matrix}

f] Three-Term Recursion. Show that

U_{0}(x) = 1 , U_{1}(x) = 2x ,

U_{n}(x) = 2xU_{n−1}(x) − U_{n−2}(x) , n = 2, 3, ….

(Note that this is the same recursion as for T_{n}.)

g] The Coefficients of U_{n}. Show that

U_{n}(x) = ∑_{k=0}^{⌊n/2⌋}(−1)^{k}\binom{n − k}{k}(2x)^{n−2k} .

h] Another Form of U_{n}. Show that

U_{n}(x) = ∑_{k=0}^{⌊n/2⌋}(−1)^{k}\binom{n + 1}{2k + 1}x^{n−2k}(x^{2} − 1)^{k} .

The concepts of transfinite diameter and capacity play a central role in potential theory, harmonic analysis, and other areas of mathematics.

E.11 Transfinite Diameter. Let E be a compact subset of ℂ. Let

Δ_{n}(E) := max_{z_{1},... ,z_{n}∈E} ∏_{{1≤i,j≤n \atop i≠j}} |z_{i} − z_{j}| .

The points z_{i} at which the above maximum are obtained are called nth Fekete points for E. If the points z_{i} are the nth Fekete points for E, then the polynomial

q_{n}(z) := ∏_{i=1}^{n} (z − z_{i})

is called an nth (monic) Fekete polynomial for E. The transfinite diameter or logarithmic capacity of E is defined by

cap(E) := lim_{n→∞}(Δ_{n}(E))^{\frac{1}{n(n−1)}} ,

where the limit exists by part c] (below).

a] Let z_{1}, z_{2}, …, z_{n} be nth Fekete points for E. Then

(Δ_{n}(E))^{1/2} = abs | \begin{matrix}1 & z_{1} & … & z_{1}^{n−1} \\ 1 & z_{2} & … & z_{2}^{n−1} \\ ⋮ & ⋮ & ... & ⋮ \\ 1 & z_{n} & … & z_{n}^{n−1} \end{matrix} | .

Hint: See E.2 b] (Vandermonde determinant) of Section 3.2. ⊓ ⊔

b] Let q_{n}(z) := ∏_{i=1}^{n}(z − z_{i}) be an nth Fekete polynomial for E. Let

m_{n} := min_{i=1,... ,n} |q′_{n}(z_{i})| and M_{n} := ∥q_{n}∥_{E} .

Then

M_{n} ≤ (\frac{Δ_{n+1}(E)}{Δ_{n}(E)})^{1/2} ≤ m_{n+1} ≤ (Δ_{n+1}(E))^{\frac{1}{n+1}} .

Outline. We have

|q_{n}(z)|^{2}Δ_{n}(E) = ∏_{i=1}^{n} |z − z_{i}|^{2} ∏_{{1≤i,j≤n \atop i≠j}} |z_{i} − z_{j}| ≤ Δ_{n+1}(E)

and

Δ_{n+1}(E) = ∏_{{1≤i,j≤n+1 \atop i≠k,j≠k,i≠j}} |z_{i} − z_{j}| ∏_{{1≤i≤n+1 \atop i≠k}} |z_{k} − z_{i}|^{2}

≤Δ_{n}(E) ∏_{{1≤i≤n+1 \atop i≠k}} |z_{k} − z_{i}|^{2} .

2.1 Chebyshev Polynomials 39

From the first line above,

M_{n}^{2}Δ_{n}(E) ≤ Δ_{n+1}(E) .

From the second line above,

Δ_{n+1}(E) ≤ Δ_{n}(E)m_{n+1}^{2} .
⊓ ⊔

c] Show that (Δ_{n}(E))^{\frac{1}{n(n−1)}} is decreasing, so the limit exists in the definition of cap(E). d] The Fekete points lie on the boundary of E. So cap(E) = cap(∂(E)). Hint: Use the maximum principle (see E.1 d] of Section 1.2). ⊓ ⊔ e] If E ⊂ F, where F ⊂ ℂ is also compact, then cap(E) ≤ cap(F). f] Chebyshev Constants. Let

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

and

M̃_{n} := {p ∈ P_{n}^{c} : p(z) = ∏_{j=1}^{n} (z − z_{j}) , z_{j} ∈ E} .

Let

μ_{n}(E) := inf{∥p∥_{E} : p ∈ M_{n}}

and

μ̃_{n}(E) := inf{∥p∥_{E} : p ∈ M̃_{n}} .

Show that the infimum in the definition of μ_{n}(E) and μ̃_{n}(E) is actually minimum. Show also that

μ_{n+m}(E) ≤ μ_{n}(E)μ_{m}(E)

and

μ̃_{n+m}(E) ≤ μ̃_{n}(E)μ̃_{m}(E)

for any two nonnegative integers n and m. Finally show that the above inequalities imply that

μ(E) := lim_{n→∞}(μ_{n}(E))^{1/n} and μ̃(E) := lim_{n→∞}(μ̃_{n}(E))^{1/n}

exist.

The numbers μ(E) and μ̃(E) are called the Chebyshev constant and modified Chebyshev constant, respectively, associated with E. Obviously μ(E) ≤ μ̃(E).

g] Transfinite Diameter and Chebyshev Constants Are the Same:

cap(E) = μ(E) = μ̃(E) .

Proof. Without loss of generality we may assume that E contains infinitely many points. Part b] yields μ̃(E) ≤ cap(E). Therefore, since μ(E) ≤ μ̃(E), it is sufficient to prove that cap(E) ≤ μ(E). Note that if p ∈ M_{n} and z_{1}, z_{2}, …, z_{n+1} ∈ E are the (n + 1)th Fekete points for E, then

(Δ_{n+1}(E))^{1/2} = abs | \begin{matrix}1 & z_{1} & … & z_{1}^{n−1} & p(z_{1}) \\ 1 & z_{2} & … & z_{2}^{n−1} & p(z_{2}) \\ ⋮ & ⋮ & ... & ⋮ & ⋮ \\ 1 & z_{n+1} & … & z_{n+1}^{n−1} & p(z_{n+1}) \end{matrix} | .

Expanding the above determinant with respect to its last column, we obtain

(Δ_{n+1}(E))^{1/2} ≤ (Δ_{n}(E))^{1/2} ∑_{j=1}^{n+1} |p(z_{j})|
≤ (n + 1)(Δ_{n}(E))^{1/2}∥p∥_{E} ,

so

(Δ_{n+1}(E))^{1/2} ≤ (n + 1)(Δ_{n}(E))^{1/2}μ_{n}(E) .

For the sake of brevity let

c_{n} := ((n + 1)^{2}(μ_{n}(E))^{2})^{1/n} and d_{n} := (Δ_{n}(E))^{\frac{2}{n(n−1)}} .

Then

d_{n+1}^{n+1} ≤ c_{n}d_{n}^{n−1} .

Since E contains infinitely many points, c_{n} > 0 and d_{n} > 0 hold for each n = 2, 3, …. Multiplying the above inequalities for n = 1, 2, …, k, we obtain after simplification that

(d_{2}d_{3} ⋯ d_{k+1})^{\frac{1}{k−1}} (d_{k+1})^{\frac{k}{k−1}} ≤ (d_{2})^{\frac{2}{k−1}} (c_{2}c_{3} ⋯ c_{k})^{\frac{1}{k−1}} .

Since lim_{k→∞} d_{k} = cap(E) and lim_{k→∞} c_{k} = (μ(E))^{2}, we conclude

(cap(E))^{2} ≤ (μ(E))^{2} ,

2.2 Orthogonal Functions 41

which finishes the proof. ⊓ ⊔ h] Show that cap([a, b]) = \frac{1}{4}(b − a).

i] Show that cap(\overline{D}_{ρ}) = ρ, where

\overline{D}_{ρ} := \{z ∈ ℂ : |z| ≤ ρ\} .

j] Show that cap(A_{α}) = sin(α/4), where A_{α} is an arc of the unit circle C of length α, 0 ≤ α ≤ 2π.

Hint: Without loss of generality we may assume that the arc A_{α} is symmetric with respect to the x-axis and 1 ∈ A_{α}. Now use part h] and the transformation x = \frac{1}{2}(z + z^{−1}). ⊓ ⊔

2.2 Orthogonal Functions

The most basic properties of orthogonal functions are explored in this section. The following section specializes the discussion to polynomials.

In this section the functions are complex-valued and the vector spaces are over the complex numbers. All the results have obvious real analogs and in many later applications we will restrict to these corresponding real cases.

An inner product on a vector space V is a function ⟨·, ·⟩ from V × V to ℂ that satisfies, for all f, g, h ∈ V and α, β ∈ ℂ,

(2.2.1)  ⟨f, f⟩ > 0 unless f = 0  (positivity)
(2.2.2)  ⟨f, g⟩ = \overline{⟨g, f⟩}  (conjugate symmetry)
(2.2.3)  ⟨αf + βg, h⟩ = α⟨f, h⟩ + β⟨g, h⟩  (linearity).

A vector space V equipped with an inner product is called an inner product space. It is a normed linear space with the norm ∥ · ∥ := ⟨·, ·⟩^{1/2}.

The canonical example for us will be the space C[a, b] of all complexvalued continuous functions on [a, b] with the inner product

(2.2.4)  ⟨f, g⟩ := ∫_{a}^{b} f(x)\overline{g(x)}w(x) dx ,

where w(x) is a nonnegative integrable function on [a, b] that is positive except possibly on a set of measure zero. It is a normed linear space with the norm

(2.2.5)  ∥f∥_{L_{2}(w)} := ⟨f, f⟩^{1/2} = (∫_{a}^{b} |f(x)|^{2}w(x) dx)^{1/2} .

More generally, if (X, μ) is a measure space (with μ nonnegative), then

(2.2.6)  ⟨f, g⟩ := ∫_{X} f(x)\overline{g(x)} dμ(x)

is an inner product on the space L_{2}(μ) of square integrable functions. More precisely, L_{2}(μ) denotes the space of equivalence classes of measurable functions for which

∥f∥_{L_{2}(μ)} := ⟨f, f⟩^{1/2} = {∫_{X} |f(x)|^{2} dμ(x)}^{1/2}

is finite. The equivalence classes are defined by the equivalence relation f ∼ g if f = g μ-almost everywhere on X.

If V is a vector space equipped with an inner product ⟨·, ·⟩, then a metric ρ can be defined on V by ρ(f, g) := ⟨f − g, f − g⟩^{1/2}. The fact that this ρ is a metric on V is an immediate consequence of (2.2.1) and Theorem 2.2.1 b]. If this metric space (V, ρ) is complete (that is, if every Cauchy sequence in (V, ρ) converges to some x ∈ V), then V is called a Hilbert space.

It can be shown that L_{2}(μ) is a Hilbert space for every measure space (X, μ) (see Rudin [87]), while C[a, b] equipped with the inner product (2.2.4), where w(x) ≡ 1, is not a Hilbert space (see E.1).

When we write L_{2}[a, b] we always mean L_{2}(μ) where μ is the Lebesgue measure on X = [a, b]. The fact that the inner product gives a norm is part of the next theorem.

Theorem 2.2.1. If (V, ⟨·, ·⟩) is an inner product space equipped with the norm ∥ · ∥ := ⟨·, ·⟩^{1/2}, then for all f, g ∈ V,

a]  |⟨f, g⟩| ≤ ∥f∥ ∥g∥  Cauchy-Schwarz inequality
b]  ∥f + g∥ ≤ ∥f∥ + ∥g∥  triangle inequality
c]  ∥f + g∥^{2} + ∥f − g∥^{2} = 2 ∥f∥^{2} + 2 ∥g∥^{2}  parallelogram law.

Proof. Let f, g ∈ V be arbitrary. To prove the Cauchy-Schwarz inequality, without loss of generality we may assume ⟨g, g⟩ = 1 and we may assume ⟨f, g⟩ is real (why?). Let ζ := ⟨f, g⟩ and note that by (2.2.1) and (2.2.3),

0 ≤ ⟨f − ζg, f − ζg⟩ = ⟨f, f⟩ − 2ζ⟨f, g⟩ + ζ^{2}⟨g, g⟩
= ∥f∥^{2} − ⟨f, g⟩^{2} ,

which finishes the proof of part a].

Using the Cauchy-Schwarz inequality, we obtain

∥f + g∥^{2} = ⟨f + g, f + g⟩ = ⟨f, f⟩ + 2Re(⟨f, g⟩) + ⟨g, g⟩
≤ ∥f∥^{2} + 2 ∥f∥ ∥g∥ + ∥g∥^{2}
≤ (∥f∥ + ∥g∥)^{2} ,

which is the triangle inequality.

The parallelogram law follows from

∥f + g∥^{2} + ∥f − g∥^{2}

= ⟨f, f⟩ + 2 Re(⟨f, g⟩) + ⟨g, g⟩ + ⟨f, f⟩ − 2 Re(⟨f, g⟩) + ⟨g, g⟩ .

⊓ ⊔

For the space L_{2}(μ) of all square integrable functions, the Cauchy-Schwarz inequality becomes

|∫_{a}^{b} f\overline{g} dμ| ≤ (∫_{a}^{b} |f|^{2} dμ(x))^{1/2} (∫_{a}^{b} |g|^{2} dμ)^{1/2} .

Applying this with f and g replaced by |f| and |g|, we obtain

(2.2.7)  ∫_{a}^{b} |fg| dμ ≤ (∫_{a}^{b} |f|^{2} dμ(x))^{1/2} (∫_{a}^{b} |g|^{2} dμ)^{1/2} .

A collection of vectors {f_{α} : α ∈ A} in an inner product space (V, ⟨·, ·⟩) is said to be orthogonal if

(2.2.8)  ⟨f_{α}, f_{β}⟩ = 0 ,  α, β ∈ A , α ≠ β .

If ⟨f_{α}, f_{β}⟩ = 0, then we write f_{α}⊥f_{β}. The collection is called orthonormal if, in addition to being orthogonal,

(2.2.9)  ⟨f_{α}, f_{α}⟩ = 1 ,  α ∈ A .

An orthogonal collection {f_{α} : α ∈ A} of nonzero vectors in an inner product space can always be orthonormalized as {∥f_{α}∥^{−1}f_{α} : α ∈ A}. The vector space over ℂ generated by {f_{α} : α ∈ A} is denoted by

span{f_{α} : α ∈ A} .

So span{f_{α} : α ∈ A} is just the set of all finite linear combinations

{∑_{i=1}^{n} c_{i}f_{α_{i}} : α_{i} ∈ A , c_{i} ∈ ℂ , n ∈ ℕ} .

Any linearly independent collection of vectors can be orthonormalized, as the next theorem shows.

Theorem 2.2.2 (Gram-Schmidt). Let (V, ⟨·, ·⟩) be an inner product space with norm ∥ · ∥ := ⟨·, ·⟩^{1/2}. Suppose {f_{i}}_{i=1}^{∞} is a linearly independent collection of vectors in V. Let

g_{1} := \frac{f_{1}}{∥f_{1}∥}

and (inductively) let

u_{n} := f_{n} − ∑_{k=1}^{n−1}⟨f_{n}, g_{k}⟩g_{k}  and  g_{n} := \frac{u_{n}}{∥u_{n}∥} .

Then {g_{n}}_{n=1}^{∞} is an orthonormal collection, and for each n,

span{g_{1}, g_{2}, …, g_{n}} = span{f_{1}, f_{2}, …, f_{n}} .

Proof. This can be proved easily by induction where the inductive step is: for m < n,

⟨u_{n}, g_{m}⟩ = ⟨f_{n}, g_{m}⟩ − ∑_{k=1}^{n−1}⟨f_{n}, g_{k}⟩⟨g_{k}, g_{m}⟩

= ⟨f_{n}, g_{m}⟩ − ∑_{k=1}^{n−1}⟨f_{n}, g_{k}⟩δ_{k,m} = 0 .

⊓ ⊔

The key approximation theoretic property orthonormal sets have is encapsulated in the following result:

Theorem 2.2.3 (Best Approximation by Linear Combinations). Let (V, ⟨·, ·⟩) be an inner product space with norm ∥ · ∥ := ⟨·, ·⟩^{1/2}. Suppose {f_{1}, …, f_{n}} is an orthonormal collection of vectors in V. Let f ∈ V. Then

min_{c_{i}∈ℂ} ∥∑_{i=1}^{n} c_{i}f_{i} − f∥

is attained if and only if

c_{i} = ⟨f, f_{i}⟩ , i = 1, 2, …, n .

In other words, the sum ∑_{i=1}^{n}⟨f, f_{i}⟩f_{i} is the best approximation to f from span{f_{1}, …, f_{n}} in the norm ⟨·, ·⟩^{1/2}.

Proof. Fix f ∈ V, and let c_{i} be as above. Let

g := ∑_{i=1}^{n} c_{i}f_{i}

and let h ∈ span{f_{1}, …, f_{n}}. Note that

(g − f)⊥f_{i} , i = 1, 2, …, n

since by orthonormality

⟨g − f, f_{i}⟩ = ∑_{j=1}^{n} c_{j}⟨f_{j}, f_{i}⟩ − c_{i} = 0 .

Thus

(g − f)⊥(h − g)

and so

∥h − f∥^{2} = ∥(h − g) + (g − f)∥^{2}
= ∥h − g∥^{2} + 2Re(⟨h − g, g − f⟩) + ∥g − f∥^{2}
= ∥h − g∥^{2} + ∥g − f∥^{2}
≥ ∥g − f∥^{2}

with strict inequality unless h = g. This finishes the proof. ⊓ ⊔

Note that the above theorem gives the following corollary:

Corollary 2.2.4. If {f_{1}, …, f_{n}} is an orthonormal collection, then every

g ∈ span{f_{1}, …, f_{n}}

can be written as

g = ∑_{i=1}^{n}⟨g, f_{i}⟩f_{i} .

Comments, Exercises, and Examples.

The theory of orthogonal functions, and in particular orthogonal polynomials, is old and far-reaching. As we will see in the next section, the names associated with the classical orthogonal polynomials including Chebyshev, Laguerre, Legendre, Hermite, Jacobi, and Stieltjes, are the “who’s who” of nineteenth century analysis. Various aspects of this beautiful body of theory are explored in the exercises of this and the next section.

Much of this material is available in G. Szegő’s [75] classical treatise “Orthogonal Polynomials.” Of course, orthogonal polynomials are intimately connected to Fourier series and parts of harmonic and functional analysis generally. The standard functional analysis in the following exercises is available in many sources. See, for example, Rudin [73, 87].

E.1 C[0, 1] Is Not a Hilbert Space. Construct a sequence of continuous functions (f_{n})_{n=1}^{∞} on [0, 1] for which

∥f_{n} − f∥_{L_{2}[0,1]} → 0

with somef ∉ C[0, 1] (in the sense that f cannot be modified on a set of measure zero to be in C[0, 1]).

So C[0, 1] equipped with the inner product (2.2.4), where [a, b] = [0, 1] and w(x) is identically 1, is not a Hilbert space. It can be shown that there is no way of putting a norm on C[0, 1] that preserves the uniform topology and makes C[0, 1] into a Hilbert space, essentially because C[0, 1] is not reflexive (see Rudin [73], Chapter 4). This, in fact, shows that C[0, 1] is not isomorphic to L_{p}[0, 1] for any p ∈ (1, ∞). For the definition of L_{p}[0, 1], see E.7.

E.2 On L_{2}(w). Consider

⟨f, g⟩ = ∫_{a}^{b} f(x)g(x)w(x) dx .

What conditions on w guarantee that ⟨f, g⟩ is an inner product on C[a, b] ?

E.3 Cauchy-Schwarz Inequality for Sequences. Show that

|∑_{i=1}^{n} α_{i}β_{i}|^{2} ≤ (∑_{i=1}^{n} |α_{i}|^{2}) (∑_{i=1}^{n} |β_{i}|^{2})

for all α_{1}, …, α_{n}, β_{1}, …, β_{n} ∈ ℂ. Equality holds if and only if there exists a γ ∈ ℂ so that either α_{i} = γ\overline{β}_{i} for each i or β_{i} = γ\overline{α}_{i} for each i.

Hint: ℂ^{n} is a Hilbert space with inner product

⟨(α_{1}, α_{2}, …, α_{n}), (β_{1}, β_{2}, …, β_{n})⟩ = ∑_{i=1}^{n} α_{i}\overline{β}_{i} .

⊓ ⊔

E.4 Bessel’s Inequality. Let (V, ⟨·, ·⟩) be an inner product space with norm ∥ · ∥ := ⟨·, ·⟩^{1/2}. Suppose {f_{i}}_{i=1}^{∞} is a countable collection of orthonormal vectors in V.

a] Show that

∑_{i=1}^{∞} |⟨f_{i}, f⟩|^{2} ≤ ∥f∥^{2} .

Hint: With h := 0 in the last expression of the proof of Theorem 2.2.3

∥f∥^{2} = ∥g∥^{2} + ∥g − f∥^{2} .

⊓ ⊔ b] Suppose

f = ∑_{i=1}^{∞}⟨f_{i}, f⟩f_{i}

in the sense that the partial sums of the right-hand side converge to f in the norm ∥ · ∥. Show that

∥f∥^{2} = ∑_{i} |⟨f_{i}, f⟩|^{2} .

E.5 The Kernel Function. Let {p_{i}}_{i=0}^{n} be a collection of orthonormal functions in L_{2}[a, b] with respect to the inner product defined by (2.2.6), where X := [a, b]. Define the kernel function by

K_{n}(x_{0}, x) := \overline{p_{0}(x_{0})}p_{0}(x) + \overline{p_{1}(x_{0})}p_{1}(x) + ⋯ + \overline{p_{n}(x_{0})}p_{n}(x) .

a] Reproducing Property. If q ∈ span{p_{0}, …, p_{n}}, then

∫_{a}^{b} K_{n}(t, x)q(t) dμ(t) = q(x) .

Hint: Expand q in terms of p_{0}, …, p_{n} as in Corollary 2.2.4. ⊓ ⊔

b] (K_{n}(x_{0}, x_{0}))^{−1/2}K_{n}(x_{0}, x) solves the following maximization problem:

max {|q(x_{0})| : q ∈ span{p_{0}, p_{1}, …, p_{n}} and ∫_{a}^{b} |q(x)|^{2} dμ(x) = 1} .

Outline. Write q = ∑_{i=0}^{n} c_{i}p_{i}. Then, as in E.4 b],

∥q∥_{L_{2}(μ)}^{2} = |c_{0}|^{2} + |c_{1}|^{2} + ⋯ + |c_{n}|^{2} = 1 .

The Cauchy-Schwarz inequality of E.3 yields that

|q(x_{0})|^{2} ≤ (∑_{i=0}^{n} |c_{i}|^{2}) (∑_{i=0}^{n} |p_{i}(x_{0})|^{2}) = K_{n}(x_{0}, x_{0}) .

However, if

c_{i} = \frac{\overline{p_{i}(x_{0})}}{(∑_{j=0}^{n} |p_{j}(x_{0})|^{2})^{1/2}} ,

so

q(x) = \frac{K_{n}(x_{0}, x)}{(K_{n}(x_{0}, x_{0}))^{1/2}} ,

then equality holds in the above inequality. ⊓ ⊔

c] Show, as in a], that if q ∈ span{p_{0}, …, p_{n}} and p_{0}, …, p_{n} are m times differentiable at x_{0}, then

|q^{(m)}(x_{0})| ≤ (∑_{k=0}^{m} |p_{k}^{(m)}(x_{0})|^{2})^{1/2} ∥q∥_{L_{2}(μ)} .

When does equality hold?

48 2. Some Special Polynomials

E.6 Completeness. Let {f_{α} : α ∈ A} be an orthonormal collection in a Hilbert space H. The collection {f_{α} : α ∈ A} is called a maximal orthonormal set in H if there is no f ≠ 0 so that ⟨f, f_{α}⟩ = 0 for every α ∈ A.

The following statements are equivalent:

(1) The set of all finite linear combinations of f_{α}, α ∈ A, is dense in H.

(2) ∥f∥^{2} = ∑_{α∈A} |⟨f_{α}, f⟩|^{2} for all f ∈ H.

(3) ⟨f, g⟩ = ∑_{α∈A}⟨f_{α}, f⟩\overline{⟨f_{α}, g⟩} for all f, g ∈ H.

(4) {f_{α} : α ∈ A} is a maximal orthonormal set in H. If any of the above holds, then the orthonormal collection is called a complete orthonormal system. (See, for example, Rudin [87].) a] Deduce (1) ⇒ (2) from Theorem 2.2.3. b] Deduce (2) ⇒ (3) from the simple identity

4⟨f, g⟩ = ∥f + g∥^{2} − ∥f − g∥^{2} + i∥f + ig∥^{2} − i∥f − ig∥^{2} .

The above identity is called polarization. c] Prove (3) ⇒ (4). d] Prove (4) ⇒ (1) by contradiction. Equality (3) is called Parseval’s identity.

The remaining exercises assumes some familiarity with measure theory.

E.7 Basic Theory of L_{p} Spaces. Let (X, μ) be a measure space (μ is nonnegative) and p ∈ (0, ∞]. The space L_{p}(μ) is defined as the collection of equivalence classes of measurable functions for which ∥f∥_{L_{p}(μ)} < ∞, where

∥f∥_{L_{p}(μ)} := (∫_{X} |f|^{p} dμ)^{1/p} , p ∈ (0, ∞)

and

∥f∥_{L_{∞}(μ)} := sup{α ∈ ℝ : μ({x ∈ X : |f(x)| > α}) > 0} < ∞ .

In any of the cases the equivalence classes are defined by the equivalence relation f ∼ g if f = g μ-almost everywhere on X. When we write L_{p}[a, b] we always mean L_{p}(μ), where μ is the Lebesgue measure on X = [a, b]. The notations L_{p}(a, b), L_{p}[a, b), and L_{p}(a, b] are also used analogously to L_{p}[a, b]. a] Hölder’s Inequality. Suppose 1 ≤ p < q ≤ ∞ and p^{−1} + q^{−1} = 1. Show that

|∫_{X} fg dμ| ≤ ∥f∥_{L_{p}(μ)}∥g∥_{L_{q}(μ)}

for every f ∈ L_{p}(μ) and g ∈ L_{q}(μ).

If 1 < p, q < ∞, then equality holds if and only if α|f|^{p} = β|g|^{q} μalmost everywhere on X for some α, β ∈ ℝ with α^{2} + β^{2} > 0, and there is a c ∈ ℂ with |c| = 1 so that cfg is nonnegative μ-almost everywhere on X.

Hölder’s inequality was proved by Rogers [1888] before Hölder [1889] proved it independently.

Hint: If the right-hand side is 0, then the inequality is obvious. If it is different from 0, then let

F := \frac{|f|}{∥f∥_{L_{p}(μ)}} and G := \frac{|g|}{∥g∥_{L_{q}(μ)}} .

If x ∈ X is such that 0 < F (x) < ∞ and 0 < G(x) < ∞, then there are real numbers s and t such that

F (x) = e^{s/p} and G(x) = e^{t/q} .

Use the convexity of the exponential function to show that

e^{s/p + t/q} ≤ p^{−1}e^{s} + q^{−1}e^{t} .

Apply this with the above choices of s and t, and integrate both sides on X with respect to μ. ⊓ ⊔

b] Minkowski’s Inequality for p ∈ [1, ∞]. Let p ∈ [1, ∞]. Show that

∥f + g∥_{L_{p}(μ)} ≤ ∥f∥_{L_{p}(μ)} + ∥g∥_{L_{p}(μ)}

for every f, g ∈ L_{p}(μ).

If 1 < p, q < ∞, then equality holds if and only if αf = βg μ-almost everywhere on X for some α, β ∈ ℝ with α^{2} + β^{2} > 0.

Hint: The cases p = 1 and p = ∞ are straightforward. Let p ∈ (1, ∞). Then

|f + g|^{p} ≤ |f| |f + g|^{p−1} + |g| |f + g|^{p−1}

and apply Hölder’s inequality (part a]) to each term separately. ⊓ ⊔

By part b], L_{p}(μ) is a vector space and ∥ · ∥_{L_{p}(μ)} is a norm on L_{p}(μ) whenever p ∈ [1, ∞]. If p ∈ (0, 1), then ∥ · ∥_{L_{p}(μ)} is still called a norm in the literature, however, for p ∈ (0, 1) the subadditive property, in general, fails. In fact, if p ∈ (0, 1), then ∥ · ∥_{L_{p}[a,b]} is superadditive for Riemann integrable functions in L_{p}[a, b]; see Pólya and Szegő [76].

c] Assume μ(X) < ∞. Show that L_{q}(μ) ⊂ L_{p}(μ) for every 0 < p < q ≤ ∞. If μ(X) ≤ 1, then prove that

∥f∥_{L_{p}(μ)} ≤ ∥f∥_{L_{q}(μ)}

for every measurable function f.

d] Assume f ∈ L_{q}(μ) for some q > 0. Show that

lim_{p→∞} ∥f∥_{L_{p}(μ)} = ∥f∥_{L_{∞}(μ)} .

e] Riesz-Fischer Theorem. Show that if 1 ≤ p ≤ ∞, then (L_{p}(μ), ρ) is a complete metric space, where

ρ(f, g) := ∥f − g∥_{L_{p}(μ)} .

Hint: Use the monotone convergence theorem and Minkowski’s inequality (part b]); see Rudin [87] for details. ⊓ ⊔

If p ∈ [1, ∞], then q ∈ [1, ∞] defined by p^{−1} + q^{−1} = 1 is called conjugate to p.

f] Bounded Linear Functionals on L_{p}(μ). Let 1 ≤ p < ∞ and g ∈ L_{q}(μ), where q is the conjugate exponent to p. Show that

Φ_{g}(f) := ∫_{X} fg dμ

is a bounded linear functional on L_{p}(μ).

Hint: Use Hölder’s inequality (part a]). ⊓ ⊔

g] Riesz Representation Theorem. Suppose 1 ≤ p < ∞, μ is (σ-) finite and Φ is a bounded linear functional on L_{p}(μ). Then there is a unique g ∈ L_{q}(μ), where q is the conjugate exponent to p, so that

Φ(f) = ∫_{X} fg dμ , f ∈ L_{p}(μ) .

Moreover, if Φ and g are related as above, then

∥Φ∥ := max{Φ(f) : f ∈ L_{p}(μ), ∥f∥_{L_{p}(μ)} = 1} = ∥g∥_{L_{q}(μ)} .

Proof. See, for example, Rudin [87] or Royden [88]. ⊓ ⊔

If X is a locally compact Hausdorff space, then the characterization of bounded linear functionals on the space C_{c}(X) of continuous functions with compact support equipped with the uniform norm is also known as the Riesz representation theorem, and its proof may be found in, for example, Rudin [87].

h] Orthogonality in L_{p}(μ). Suppose 1 ≤ p < ∞, μ is (σ-)finite, Y is a finite-dimensional subspace of L_{p}(μ). The function f ∈ L_{p}(μ) is said to be orthogonal to Y in L_{p}(μ), written f ⊥ Y, if

∥f∥_{L_{p}(μ)} ≤ ∥f + h∥_{L_{p}(μ)}

for every h ∈ Y. Show that an element f ∈ L_{p}(μ) is orthogonal to Y if and only if

∫_{X} |f|^{p−1}\overline{sign}(f)h dμ = 0

for every h ∈ Y, where

\overline{sign}(f(x)) := \{ \begin{matrix}\frac{\overline{f(x)}}{|f(x)|} & if f(x) ≠ 0 \\ 0 & if f(x) = 0 . \end{matrix}

Outline. Suppose that the integral vanishes for every h ∈ Y. Let q be the conjugate exponent to p. defined by p^{−1} + q^{−1} = 1. Observe that

g := |f|^{p−1} \overline{sign}(f) ∈ L_{q}(μ)

and

∫_{X} |g|^{q}dμ = ∫_{X} |f|^{p} dμ .

Without loss of generality we may assume that ∥f∥_{L_{p}(μ)} = 1. Then for every h ∈ Y, Hölder’s inequality yields that

∥f∥_{L_{p}(μ)} = 1 = ∫_{X} fg dμ = ∫_{X}(f + h)g dμ
≤ ∥f + h∥_{L_{p}(μ)}∥g∥_{L_{q}(μ)} = ∥f + h∥_{L_{p}(μ)} ,

proving that f ⊥ Y. (Observe that this argument is also valid for p = 1.)

Suppose now f ⊥ Y. Without loss of generality we may assume that f ∉ Y. By a standard corollary to the Hahn-Banach theorem (see, for example, Rudin [87]), there exists a linear functional M on L_{p}(μ) such that M(f) = 1, M(h) = 0 for every h ∈ Y, and ∥M∥ = ∥f∥_{L_{p}(μ)}^{−1}. This M is then representable by some element g ∈ L_{q}(μ), that is,

M(f) = ∫_{X} fg dμ, ∥g∥_{L_{p}(μ)} = ∥f∥_{L_{p}(μ)}^{−1}

52 2. Some Special Polynomials

(see part g]). Therefore

∫_{X} fg dμ = ∥f∥_{L_{p}(μ)} ∥g∥_{L_{q}(μ)} ,

and by the conditions for equality to hold in Hölder’s inequality,

g(x)f(x) ≥ 0 a.e. [μ] on X

and

|g(x)|^{q} = λ|f(x)|^{p} a.e. [μ] on X

for a suitable constant λ > 0. Hence

g(x) = λ|f(x)|^{p−1} \overline{sign}(f(x)) ,

and so M(h) = 0, h ∈ Y, implies

∫_{X} |f|^{p−1} \overline{sign}(f)h dμ = 0

for every h ∈ Y. ⊓ ⊔

The statement of part h] remains valid for closed subspaces Y instead of finite-dimensional subspaces, see, for example, Shapiro [71]. i] Minkowski’s Inequality for p ∈ (0, 1). Show that

∥f + g∥_{L_{p}(μ)} ≤ 2^{1/p−1} (∥f∥_{L_{p}(μ)} + ∥g∥_{L_{p}(μ)})

for every f, g ∈ L_{p}(μ) and p ∈ (0, 1). Hint: Verify that

∥f + g∥_{L_{p}(μ)} ≤ (∫_{X}(|f| + |g|)^{p} dμ)^{1/p}
≤ (∫_{X} |f|^{p} dμ + ∫_{X} |g|^{p} dμ)^{1/p}
≤ 2^{1/p−1} ((∫_{X} |f|^{p} dμ)^{1/p} + (∫_{X} |g|^{p} dμ)^{1/p})

whenever f, g ∈ L_{p}(μ) and p ∈ (0, 1). ⊓ ⊔ Further properties of L_{p}(μ) spaces may be found in Rudin [87].

E.8 Fourier Series.

a] Show that

\{\frac{e^{inθ}}{\sqrt{2π}} : n ∈ ℤ\}

is a maximal orthonormal collection in L_{2}[−π, π].

Hint: The orthonormality is obvious. In order to show the maximality, first note that L_{2}[−π, π] is a Hilbert space by the Riesz-Fischer theorem (E.7 e]). Hence, by E.6, it is sufficient to show that the set T^{c} of all complex trigonometric polynomials is dense in L_{2}[−π, π]. By the Stone-Weierstrass theorem (E.2 of Section 4.1) T^{c} is dense in C^{∗}[−π, π], where C^{∗}[−π, π] is the space of all complex-valued 2π-periodic continuous functions on ℝ equipped with the uniform norm on ℝ. Finally, it is a standard measure theoretic argument to show that C^{∗}[−π, π] is dense in L_{2}[−π, π]; see, for example, Rudin [87]. ⊓ ⊔

The kth Fourier coefficient f̂(k) of a function f ∈ L_{1}[−π, π] is defined by

f̂(k) := \frac{1}{2π} ∫_{−π}^{π} f(θ)e^{−ikθ} dθ .

The (formal) Fourier series of a function f ∈ L_{1}[−π, π] is defined by

f ∼ ∑_{k=−∞}^{∞} f̂(k)e^{ikθ} .

The functions

S_{n}(θ) := ∑_{k=−n}^{n} f̂(k)e^{ikθ}

are called the nth partial sums of the Fourier series of f.

b] Show that if f ∈ L_{2}[−π, π], then

∑_{k=−∞}^{∞} |f̂(k)|^{2} = ∥f∥_{L_{2}[−π,π]}^{2} < ∞ .

Hint: Use part a], E.6, and E.7 e]. ⊓ ⊔

c] Show that if f ∈ L_{2}[−π, π], then

lim_{n→∞} ∥f − S_{n}∥_{L_{2}[−π,π]} = 0 ,

so f is the L_{2}[−π, π] limit of the partial sums of its formal Fourier series.

Hint: Use part b]. ⊓ ⊔

Carleson, in 1966, solved Luzin’s problem by showing that S_{n} → f almost everywhere on [−π, π] for every f ∈ L_{2}[−π, π]. Earlier, Kolmogorov showed that there is a function f ∈ L_{1}[−π, π] so that S_{n} diverges everywhere on [−π, π].

d] Isometry of L_{2}[−π, π] and ℓ_{2}^{c}. Let

ℓ_{2}^{c} := {x = (x_{k})_{k=−∞}^{∞} : x_{k} ∈ ℂ , ∑_{k=−∞}^{∞}|x_{k}|^{2} < ∞}

and

∥x∥_{ℓ_{2}^{c}} := ( ∑_{k=−∞}^{∞} |x_{k}|^{2})^{1/2} , x = (x_{k})_{−∞}^{∞} , x_{k} ∈ ℂ .

Show that the function I : L_{2}[−π, π] → ℓ_{2}^{c} defined by

I(f) := f̂:= (f̂(k))_{−∞}^{∞}

is one-to-one and onto, and

∥I(f)∥_{ℓ_{2}^{c}} = ∥f∥_{L_{2}[−π,π]} .

Hint: Use part a] to show that I is one-to-one. Use the Riesz-Fischer theorem (E.7 e]) to show that I is onto. The norm-preserving property is the content of part b]. ⊓ ⊔

Part d] shows that the structure of L_{2}[−π, π] is the same as that of ℓ_{2}^{c}. Hence L_{2}[−π, π] is a separable Hilbert space, that is, it has a countable dense subset. So if ϵ > 0 is fixed, then any collection {f_{α} : α ∈ A} from L_{2}[−π, π] for which

∥f_{α} − f_{β}∥_{L_{2}[−π,π]} ≥ ϵ , α, β ∈ A , α ≠ β

must be countable.

e] The Riemann-Lebesgue Lemma. If f ∈ L_{1}[−π, π], then f̂(k) → 0 as k → ∞.

Hint: First prove it for step functions, then extend the result to every f ∈ L_{1}[−π, π] by using the fact that step functions form a dense set in L_{1}[−π, π]. ⊓ ⊔

f] Show that

∑_{n=1}^{∞} \frac{1}{n^{2}} = \frac{π^{2}}{6} ,  ∑_{n=1}^{∞} \frac{1}{n^{4}} = \frac{π^{4}}{90} , and ∑_{n=1}^{∞} \frac{1}{n^{2k}} = r_{k}π^{2k}

for every k = 3, 4, ⋯, where r_{k} is a rational number.

Hint: Let f be the 2π periodic function defined by

f(θ) := (\frac{π − θ}{2} )^{k} , θ ∈ [0, 2π) .

Apply part b]. ⊓ ⊔

2.2 Orthogonal Functions 55

E.9 Denseness of Polynomials in L_{2}(μ) on ℝ. a] Let μ be a finite Borel measure on [a, b] and f ∈ L_{1}(μ). Show that if

∫_{a}^{b} f(x)e^{itx} dμ(x) = 0 , t = \frac{2πk}{b − a} , k = 0, ±1, ±2, …,

then f(x) = 0 a.e. [μ] on [a, b]. Outline. Use the fact that the set T^{c} of all complex trigonometric polynomials is dense in C^{∗}[−π, π] (see the hint given for E.8 a]) and standard measure theoretic arguments to show that the assumption of part a] implies

∫_{a}^{b} f(x)g(x) dμ(x) = 0

for every bounded measurable function g defined on [a, b]. Now, choosing

g(x) := \overline{sign}(f(x)) := \{ \begin{matrix}\frac{\overline{f(x)}}{|f(x)|} & if f(x) ≠ 0 \\ 0 & if f(x) = 0 , \end{matrix}

we obtain

∫_{−π}^{π} |f(x)| dμ(x) = 0 ,

and the result follows. ⊓ ⊔ b] Let μ be a finite Borel measure on ℝ and f ∈ L_{1}(μ). Show that if

∫_{ℝ} f(x)e^{itx} dμ(x) = 0 , t ∈ ℝ ,

then f(x) = 0 a.e. [μ] on ℝ. Hint: Use part a] to show that the assumption of part b] implies

∫_{ℝ} f(x)g(x) dμ(x) = 0

for every bounded measurable function defined on ℝ (first assume that g has compact support, then eliminate this assumption). Finish the proof as in part a]. ⊓ ⊔ c] Let μ be a Borel measure on ℝ satisfying

∫_{ℝ} e^{r|x|} dμ(x) < ∞

with some r > 0. Show that the set P^{c} of all complex algebraic polynomials is dense in L_{2}(μ).

Outline. First observe that the assumption on μ implies P^{c} ⊂ L_{2}(μ). The fact that L_{2}(μ) is a Hilbert space (see E.7 e]), Theorem 2.2.2 (Gram-Schmidt), and E.6 imply that it is sufficient to prove that if f ∈ L_{2}(μ) and

∫_{ℝ} f(x)x^{k} dμ(x) = 0 , k = 0, 1, 2, …,

then f(x) = 0 a.e. [μ] on ℝ. Assume that f ∈ L_{2}(μ) satisfies the above orthogonality relation. Use Theorem 2.2.1 a] (Cauchy-Schwarz inequality) to show that

F (t) := ∫_{ℝ} f(x)e^{−itx} dμ(x)

is well-defined on ℝ. For every t_{0} ∈ ℝ, we have

f(x)e^{−itx} = f(x)e^{−it_{0}x}e^{−i(t−t_{0})x}

= ∑_{k=0}^{∞} (−i)^{k} \frac{(t − t_{0})^{k}}{k!} f(x)e^{−it_{0}x}x^{k} .

Note that if |t − t_{0}| ≤ r/2, then the integral of the right-hand side with respect to μ(x) on ℝ can be calculated by integrating term by term since

∑_{k=0}^{∞} ∫_{ℝ} |(−i)^{k} \frac{(t − t_{0})^{k}}{k!} f(x)e^{−it_{0}x}x^{k}| dμ(x)

≤ ∑_{k=0}^{∞} ∫_{ℝ} \frac{|t − t_{0}|^{k}}{k!} |f(x)||x|^{k} dμ(x)

= ∫_{ℝ} |f(x)|e^{|t−t_{0}||x|} dμ(x)

≤ |∫_{ℝ} |f(x)|^{2} dμ(x) ∫_{ℝ} e^{2|t−t_{0}||x|} dμ(x)|^{1/2} < ∞ .

Therefore, if |t − t_{0}| ≤ r/2, then

F (t) = ∑_{k=0}^{∞}(−i)^{k} \frac{(t − t_{0})^{k}}{k!} ∫_{ℝ} f(x)e^{−it_{0}x}x^{k} dμ(x) .

This means that F has a Taylor series expansion about every t_{0} ∈ ℝ with radius of convergence at least r/2. Also, with the choice t_{0} = 0, by the assumed orthogonality relations, we have F (t) = 0 whenever |t| ≤ r/2. We can now deduce that F (t) = 0 for every t ∈ ℝ. Hence it follows from part b] that f(x) = 0 a.e. [μ] on ℝ. ⊓ ⊔

2.3 Orthogonal Polynomials 57

2.3 Orthogonal Polynomials

The classical orthogonal polynomials arise on orthogonalizing the sequence

(1, x, x^{2}, …)

with respect to various particularly nice weights, w(x), on an interval, which, after a linear transformation, may be taken to be one of [−1, 1], [0, ∞), or (−∞, ∞). The main examples we consider are the Jacobi polynomials

(2.3.1) P_{n}^{(α,β)}(x) , where w(x) := (1−x)^{α}(1+x)^{β} on [−1, 1] , α, β > −1 .

When α = β = −1/2 the Jacobi polynomials are the Chebyshev polynomials of the first kind,

(2.3.2)  T_{n}(x) , where w(x) := (1 − x^{2})^{−1/2} on [−1, 1] .

When α = β = 1/2 they are the Chebyshev polynomials of the second kind,

(2.3.3)  U_{n}(x) , where w(x) := (1 − x^{2})^{1/2} on [−1, 1] .

Another special case of importance is α = β = 0, which gives the Legendre polynomials,

(2.3.4)  P_{n}(x) , where w(x) = 1 on [−1, 1] .

The Laguerre polynomials are

(2.3.5)  L_{n}(x) , where w(x) := e^{−x} on [0, ∞) .

The Hermite polynomials are

(2.3.6)  H_{n}(x) , where w(x) := e^{−x^{2}} on (−∞, ∞) .

The above notation is traditionally used to denote orthogonal polynomials with a standard normalization; see the exercises. It is not usually the case that this normalization gives orthonormality. All of these much studied polynomials arise naturally, as do all the special functions, in the study of differential equations. We catalog some of the special properties of these classical orthogonal polynomials in the exercises.

In general, a nondecreasing bounded function α (typically the distribution function of a finite measure) defined on ℝ is called an m-distribution if it takes infinitely many distinct values, and its moments, that is, the improper Stieltjes integrals

∫_{−∞}^{∞} x^{n} dα(x) = lim_{{ω_{1}→−∞ \atop ω_{2}→+∞}} ∫_{ω_{1}}^{ω_{2}}x^{n} dα(x) ,

exist and are finite for n = 0, 1, ….

Theorem 2.3.1 (Existence and Uniqueness of Orthonormal Polynomials). For every m-distribution α there is a unique sequence of polynomials (p_{n})_{n=0}^{∞} with the following properties:

(i)  p_{n}(x) = γ_{n}x^{n} + r_{n−1}(x) ,  γ_{n} > 0 , r_{n−1} ∈ P_{n−1} ,

(ii)  ∫_{ℝ} p_{n}(x)p_{m}(x) dα(x) = δ_{n,m} = \{ \begin{matrix}1 & for n = m \\ 0 & for n ≠ m . \end{matrix}

Proof. The result follows from Theorem 2.2.2 (Gram-Schmidt). Note that the defining property of an m-distribution α ensures that

⟨p, q⟩ := ∫_{ℝ} pq dα

is an inner product on P_{n}^{c}. ⊓ ⊔

The sequence (p_{n})_{n=0}^{∞} defined by Theorem 2.3.1 is called the sequence of orthonormal polynomials associated with an m-distribution α. The sequence (q_{n})_{n=0}^{∞} is called a sequence of orthogonal polynomials associated with an m-distribution α if

q_{n} = c_{n}p_{n} , 0 ≠ c_{n} ∈ ℂ , n = 0, 1, …,

where (p_{n})_{n=0}^{∞} is the sequence of orthonormal polynomials associated with α. The support supp(α) of an m-distribution α is defined as the closure of the set

{x ∈ ℝ : α is increasing at x} .

If α is absolutely continuous on ℝ, then

dα(x) = w(x) dx with some 0 ≤ w ∈ L_{1}(∞, ∞)

in which case α may be identified as a nonnegative weight function w ∈ L_{1}(−∞, ∞) whose integral takes infinitely many distinct values. If (a, b) is an interval and w ∈ L_{1}[a, b] has an integral that takes infinitely many distinct values, then the sequence of orthogonal (orthonormal) polynomials associated with

w̃(x) = \{ \begin{matrix}w(x) & if x ∈ (a, b) \\ 0 & if x ∉ (a, b) \end{matrix}

is said to be orthogonal (orthonormal) with respect to the weight w.

One thing distinguishing orthogonal polynomials from general orthogonal systems is the existence of a three-term recursion.

Theorem 2.3.2 (Three-Term Recursion). Suppose (p_{n})_{n=0}^{∞} is a sequence of orthonormal polynomials with respect to an m-distribution α. Then

xp_{n}(x) = a_{n}p_{n+1}(x) + b_{n}p_{n}(x) + a_{n−1}p_{n−1}(x) , n = 0, 1, …,

where

p_{−1} := 0 , a_{−1} = 0 , a_{n} = \frac{γ_{n}}{γ_{n+1}} > 0 , b_{n} ∈ ℝ , n = 0, 1, …

(γ_{n} is the leading coefficient of p_{n}).

This theorem has a converse due to Favard [35]; see E.12.

Proof. Since xp_{n}(x) ∈ P_{n+1}, we may write

(2.3.7)  xp_{n}(x) = ∑_{k=0}^{n+1} d_{k}p_{k}(x) ,  d_{k} ∈ ℝ .

For notational convenience, let

⟨p, q⟩ := ∫_{ℝ} p(x)q(x) dα(x)

for any two polynomials p and q. Since ⟨p_{n}, q⟩ = 0 for every q ∈ P_{n−1}, we have

⟨xp_{n}(x), q(x)⟩ = ⟨p_{n}(x), xq(x)⟩ = 0

for every q ∈ P_{n−2}. In particular,

⟨xp_{n}(x), p_{k}(x)⟩ = 0 , k = 0, 1, …, n − 2 .

On the other hand, using (2.3.7) and the orthonormality of (p_{n})_{n=0}^{∞}, we obtain

⟨xp_{n}(x), p_{k}(x)⟩ = d_{k}⟨p_{k}, p_{k}⟩ = d_{k} .

Hence d_{k} = 0 for each k = 0, 1, …, n − 2 and

(2.3.8)  xp_{n}(x) = d_{n+1}p_{n+1}(x) + d_{n}p_{n}(x) + d_{n−1}p_{n−1}(x) .

Here the lead coefficient of the left-hand side polynomial is γ_{n}, while the lead coefficient of the right-hand side polynomial is d_{n+1}γ_{n+1}, so

a_{n} := d_{n+1} = γ_{n}/γ_{n+1} .

In order to show that a_{n−1} := d_{n−1} = γ_{n−1}/γ_{n}, note that (2.3.8) and the orthonormality of (p_{n})_{n=0}^{∞} imply

0 = ⟨p_{n+1}, p_{n−1}⟩

= \frac{1}{d_{n+1}} ⟨xp_{n}(x), p_{n−1}(x)⟩ − \frac{d_{n}}{d_{n+1}} ⟨p_{n}, p_{n−1}⟩ − \frac{d_{n−1}}{d_{n+1}} ⟨p_{n−1}, p_{n−1}⟩

= \frac{1}{d_{n+1}} ⟨p_{n}(x), γ_{n−1}x^{n}⟩ − \frac{d_{n−1}}{d_{n+1}}

= \frac{1}{d_{n+1}} \frac{γ_{n−1}}{γ_{n}} − \frac{d_{n−1}}{d_{n+1}} .

Hence

a_{n−1} := d_{n−1} = \frac{γ_{n−1}}{γ_{n}} .

⊓ ⊔

Theorem 2.3.3 (Christoffel-Darboux Formula). With the notation of the previous theorem,

∑_{k=0}^{n} p_{k}(x)p_{k}(y) = \frac{γ_{n}}{γ_{n+1}} (\frac{p_{n+1}(x)p_{n}(y) − p_{n}(x)p_{n+1}(y)}{x − y})

for all x ≠ y ∈ ℂ.

Proof. Theorem 2.3.2 (three-term recursion) yields that

Δ_{k} : = p_{k+1}(x)p_{k}(y) − p_{k}(x)p_{k+1}(y)

= \frac{1}{a_{k}} (x − y)p_{k}(x)p_{k}(y) + \frac{a_{k−1}}{a_{k}} (p_{k}(x)p_{k−1}(y) − p_{k−1}(x)p_{k}(y)) .

So

a_{k} \frac{Δ_{k}}{x − y} = p_{k}(x)p_{k}(y) + a_{k−1} \frac{Δ_{k−1}}{x − y} ,

and we sum the above from 0 to n to get the desired formula. ⊓ ⊔

Corollary 2.3.4. In the notation of Theorem 2.3.2

∑_{k=0}^{n} p_{k}^{2}(x) = \frac{γ_{n}}{γ_{n+1}} (p′_{n+1}(x)p_{n}(x) − p′_{n}(x)p_{n+1}(x)) .

Proof. Let y → x in Theorem 2.3.3. ⊓ ⊔

We can deduce quite easily from this that orthogonal polynomials associated with an m-distribution α have real interlacing zeros lying in the interior of the smallest interval containing supp(α); see E.1 and E.2.

Comments, Exercises, and Examples.

Askey, in comments following an outline of the history of orthogonal polynomials by Szegő [82, vol. III], writes:

“The classical orthogonal polynomials are mostly attributed to someone other than the person who introduced them. Szegő refers to Abel and Lagrange and Tschebyscheff in [75, chapter 5] for work on the Laguerre polynomials L_{n}^{0}(x). Abel’s work was published posthumously in 1881. Probably the first published work on these polynomials that uses their orthonormality was by Murphy (1833). Hermite polynomials were studied extensively by Laplace in connection with work on probability theory. Hermite’s real contribution to these polynomials was to introduce Hermite polynomials in several variables. Lagrange came across the recurrence relation for Legendre polynomials.”

Perhaps this is not very surprising given the many diverse ways in which these polynomials can arise.

There are many sources for the basic properties of orthogonal polynomials. In particular, Askey and Ismail [84], Chihara [78], Erdélyi et al. [53], Freud [71], Nevai [79b], [86], Szegő [75], and, in tabular form, Abramowitz and Stegun [65] are such sources. Exercises include a treatment of the elementary properties of the most familiar orthogonal polynomials. The connections linking orthogonal polynomials, the moment problem, and Favard’s converse theorem to the three-term recursion are also examined in the exercises.

E.1 Simple Real Zeros. Let (p_{n})_{n=0}^{∞} be the sequence of orthonormal polynomials associated with an m-distribution α. Show that each p_{n} has exactly n simple real zeros lying in the interior of the smallest interval containing supp(α).

Hint: Suppose the statement is false. Then p_{n} has at most n−1 sign changes on [a, b], hence there exists 0 ≠ q ∈ P_{n−1} so that

p_{n}(x)q(x) ≥ 0 , x ∈ [a, b] .

Show that this contradicts the orthogonality relation

0 = ∫_{ℝ} p_{n}(x)q(x) dα(x) = ∫_{a}^{b} p_{n}(x)q(x) dα(x) .
⊓ ⊔

E.2 Interlacing of Zeros. Let (p_{n})_{n=0}^{∞} be the sequence of orthonormal polynomials associated with an m-distribution α. Then the zeros of p_{n} and p_{n+1} strictly interlace. That is, there is exactly one zero of p_{n} strictly between any two consecutive zeros of p_{n+1}.

Hint: From Corollary 2.3.4,

p′_{n+1}(x)p_{n}(x) − p′_{n}(x)p_{n+1}(x)

is positive on ℝ. Since p_{n+1} has n + 1 simple real zeros (see E.1), we see that if γ and δ are two consecutive zeros of p_{n+1}, then

sign(p′_{n+1}(γ)) = −sign(p′_{n+1}(δ)) ,

and hence

sign(p_{n}(γ)) = −sign(p_{n}(δ)) .
⊓ ⊔

E.3 Orthogonality of (K_{n}(x_{0}, x))_{n=0}^{∞}. Let (p_{n})_{n=0}^{∞} be the sequence of orthonormal polynomials associated with an m-distribution α. Let

x_{0} < min supp(α) or x_{0} > max supp(α) .

Let (K_{n}(x_{0}, x))_{n=0}^{∞} be the sequence of associated kernel functions (as in E.5 of Section 2.2). Show that

∫_{ℝ} K_{n}(x_{0}, x)K_{m}(x_{0}, x)|x − x_{0}| dα(x) = 0 ,

for any two nonnegative integers n ≠ m.

E.4 Hypergeometric Functions. We introduce the following standard notation: the rising factorial (or Pochammer symbol)

(a)_{n} := a(a + 1) ⋯ (a + n − 1) , (a)_{0} := 1

for a ∈ ℂ and n = 1, 2, …; the binomial coefficient

\binom{a}{n} := \frac{a(a − 1) ⋯ (a − n + 1)}{n!} , \binom{a}{0} := 1

for a ∈ ℂ and n = 1, 2, …; and the Gaussian hypergeometric series

2F_{1}(a, b; c; z) := F (a, b; c; z) := ∑_{n=0}^{∞} \frac{(a)_{n}(b)_{n}}{(c)_{n}} \frac{z^{n}}{n!}

for a, b, c ∈ ℂ.

a] For Re(c) > Re(b) > 0 ,

F (a, b; c; z) = \frac{Γ (c)}{Γ (b)Γ (c − b)} ∫_{0}^{1} t^{b−1}(1 − t)^{c−b−1}(1 − tz)^{−a} dt ,

2.3 Orthogonal Polynomials 63

where Γ is, as usual, the gamma function defined by

Γ (z) := ∫_{0}^{∞} t^{z−1}e^{−t} dt , Re(z) > 0 .

Proof. See, for example, Szegő [75]. ⊓ ⊔ b] Hypergeometric Differential Equation. The function y = F (a, b; c; z) satisfies

z(1 − z) \frac{d^{2}y}{dz^{2}} + [c − (a + b + 1)z] \frac{dy}{dz} − aby = 0 .

Proof. See, for example, Szegő [75]. ⊓ ⊔

In E.5, E.6, and E.7 we catalog some of the basic properties of some of the classical orthogonal polynomials. Proofs are available in Szegő [75], for example.

E.5 Jacobi Polynomials. a] Rodrigues’ Formula. Let

P_{n}^{(α,β)}(x) := (−1)^{n} \frac{2^{−n}}{n!} (1−x)^{−α}(1+x)^{−β} \frac{d^{n}}{dx^{n}} [(1 − x)^{α}(1 + x)^{β}(1 − x^{2})^{n}] .

Then (P_{n}^{(α,β)})_{n=0}^{∞} is a sequence of orthogonal polynomials on [−1, 1] associated with the weight function

w(x) := (1 − x)^{α}(1 + x)^{β} , −1 < α, β < ∞ .

That is,

P_{n}^{(α,β)} ∈ P_{n} and ∫_{−1}^{1} P_{n}^{(α,β)}P_{m}^{(α,β)}(1 − x)^{α}(1 + x)^{β} dx = 0

for any two nonnegative integers n ≠ m. In the rest of the exercise, the polynomials P_{n}^{(α,β)} are as in part a]. b] Normalization. We have

P_{n}^{(α,β)}(1) = \binom{n + α}{n} = \frac{(α + 1)_{n}}{n!}

and

∫_{−1}^{1}(P_{n}^{(α,β)}(x))^{2}(1 − x)^{α}(1 + x)^{β} dx
= \frac{2^{α+β+1}}{2n + α + β + 1} \frac{Γ (n + α + 1)Γ (n + β + 1)}{Γ (n + 1)Γ (n + α + β + 1)} .

64 2. Some Special Polynomials

c] Explicit Form.

P_{n}^{(α,β)}(x) = \frac{1}{2^{n}} ∑_{m=0}^{n} \binom{n + α}{m}\binom{n + β}{n − m}(x − 1)^{n−m}(x + 1)^{m}

= ∑_{m=0}^{n} \binom{α + n}{n − m}\binom{α + β + n + m}{m} (\frac{x − 1}{2})^{m}

= \binom{n + α}{n}_{2}F_{1} (−n, n + α + β + 1 ; α + 1 ; \frac{1 − x}{2}) .

d] Differential Equation. The function y = P_{n}^{(α,β)}(x) satisfies

(1 − x^{2}) \frac{d^{2}y}{dx^{2}} + [β − α − (α + β + 2)x] \frac{dy}{dx} + n(n + α + β + 1)y = 0 .

e] Recurrence Relation. The sequence (P_{n}^{(α,β)}(x))_{n=0}^{∞} satisfies

D_{n}P_{n+1}^{(α,β)}(x) = (A_{n} + B_{n}x)P_{n}^{(α,β)}(x) − C_{n}P_{n−1}^{(α,β)}(x) ,

where

P_{0}^{(α,β)} = 1 and P_{1}^{(α,β)}(x) = \frac{1}{2}[α − β + (α + β + 2)x]

and

D_{n} = 2(n + 1)(n + α + β + 1)(2n + α + β)
A_{n} = (2n + α + β + 1)(α^{2} − β^{2})
B_{n} = (2n + α + β + 2)(2n + α + β + 1)(2n + α + β)
C_{n} = 2(n + α)(n + β)(2n + α + β + 2) .

f] Generating Function.

∑_{n=0}^{∞} P_{n}^{(α,β)}(x)z^{n} = \frac{2^{α+β}}{R(1 − z + R)^{α}(1 + z + R)^{β}} ,

where R = \sqrt{1 − 2xz + z^{2}}.

There are various special cases, some of which we have previously defined. The Legendre polynomials P_{n} are defined by

P_{n} := P_{n}^{(0,0)} , n = 0, 1, ….

2.3 Orthogonal Polynomials 65

The Chebyshev polynomials T_{n} defined in Section 2.1 satisfy

T_{n} = \frac{4^{n}}{\binom{2n}{n}}P_{n}^{(−1/2,−1/2)} , n = 0, 1, ….

The ultraspherical (or Gegenbauer) polynomials C_{n}^{(α)} are defined by

C_{n}^{(α)} := \frac{Γ (2α + n)Γ (α + \frac{1}{2})}{Γ (2α)Γ (α + n + \frac{1}{2})}P_{n}^{(α−1/2 ,α−1/2)} , n = 0, 1, ….

In terms of C_{n}^{(α)}, the Chebyshev polynomials of the first and second kind are given by

T_{n} = \frac{n}{2}C_{n}^{(0)} and U_{n} = C_{n}^{(1)} , n = 0, 1, ….

E.6 Hermite Polynomials. a] Rodrigues’ Formula. Let

H_{n}(x) := \frac{(−1)^{n}}{exp(−x^{2})} \frac{d^{n}}{dx^{n}} exp(−x^{2}) .

Then (H_{n})_{n=0}^{∞} is a sequence of orthogonal polynomials on (−∞, ∞) associated with the weight function

w(x) := exp(−x^{2}) .

That is,

H_{n} ∈ P_{n} and ∫_{ℝ} H_{n}(x)H_{m}(x) exp(−x^{2}) dx = 0

for any two nonnegative integers n ≠ m. In the rest of the exercise, the polynomials H_{n} are as in part a]. b] Normalization. We have

∫_{−∞}^{∞}(H_{n}(x))^{2} exp(−x^{2}) dx = \sqrt{π} 2^{n}n!

and

H_{2n+1}(0) = 0 , H_{2n}(0) = (−1)^{n} \frac{(2n)!}{n!} .

c] Explicit Form.

H_{n}(x) = n! ∑_{m=0}^{⌊n/2⌋} \frac{(−1)^{m}(2x)^{n−2m}}{m!(n − 2m)!} .

66 2. Some Special Polynomials

d] Differential Equation. The function y = H_{n}(x) satisfies

\frac{d^{2}y}{dx^{2}} − 2x \frac{dy}{dx} + 2ny = 0 .

e] Recurrence Relation. The sequence (H_{n}(x))_{n=0}^{∞} satisfies

H_{n+1}(x) = 2xH_{n}(x) − 2nH_{n−1}(x)

with

H_{0}(x) = 1  and  H_{1}(x) = 2x .

f] Generating Function.

∑_{n=0}^{∞} H_{n}(x)\frac{z^{n}}{n!} = exp(2xz − z^{2}) .

E.7 Laguerre Polynomials. Let α ∈ (−1, ∞). a] Rodrigues’ Formula. Let

L_{n}^{(α)}(x) := \frac{1}{n!e^{−x}x^{α}} \frac{d^{n}}{dx^{n}} (e^{−x}x^{α+n}) .

Then (L_{n}^{(α)})_{n=0}^{∞} is a sequence of orthogonal polynomials on [0, ∞) associated with the weight function

w(x) := x^{α} exp(−x) .

That is,

L_{n}^{(α)} ∈ P_{n} and ∫_{0}^{∞} L_{n}^{(α)}(x)L_{m}^{(α)}(x)x^{α} exp(−x) dx = 0

for any two nonnegative integers n ≠ m. In the rest of the exercise, the polynomials L_{n}^{(α)} are defined as in part a]. b] Normalization. We have

∫_{0}^{∞}(L_{n}^{(α)}(x))^{2}x^{α}e^{−x} dx = \frac{Γ (α + n + 1)}{n!}

and

L_{n}^{(α)}(0) = ( \begin{matrix}n + α \\ n \end{matrix} ) .

c] Explicit Form.

L_{n}^{(α)}(x) = ∑_{m=0}^{n} \frac{(−1)^{m}}{m!} ( \begin{matrix}n + α \\ n − m \end{matrix} ) x^{m} .

d] Differential Equation. The function y = L_{n}^{(α)}(x) satisfies

x \frac{d^{2}y}{dx^{2}} + (α + 1 − x) \frac{dy}{dx} + ny = 0 .

e] Recurrence Relation. The sequence (L_{n}^{(α)}(x))_{n=0}^{∞} satisfies

(n + 1)L_{n+1}^{(α)}(x) = [(2n + α + 1) − x]L_{n}^{(α)}(x) − (n + α)L_{n−1}^{(α)}(x)

with

L_{0}^{(α)} = 1  and  L_{1}^{(α)}(x) = −x + α + 1 .

f] Generating Function.

∑_{n=0}^{∞} L_{n}^{(α)}(x)z^{n} = exp (\frac{xz}{z − 1}) (1 − z)^{−α−1} .

E.8 Christoffel Numbers and Gauss-Jacobi Quadrature. Let (p_{n})_{n=0}^{∞} be the sequence of orthonormal polynomials associated with an m-distribution α. Let x_{ν,n}, ν = 1, 2, …, n, denote the zeros of p_{n}. Let

λ_{ν,n} := \frac{1}{p′_{n}(x_{ν,n})} ∫_{ℝ} \frac{p_{n}(x)}{x − x_{ν,n}} dα(x) , ν = 1, 2, …, n .

The numbers λ_{ν,n} are called the Christoffel or Cotes numbers.

a] Show that, for any q ∈ P_{2n−1},

∫_{ℝ} q(x)dα(x) = ∑_{ν=1}^{n} λ_{ν,n}q(x_{ν,n}) .

Hint: First show the equality for every q ∈ P_{n−1} by using the Lagrange interpolation formula (E.6 of Section 1.1). If q ∈ P_{2n−1}, then q = sp_{n} + r with some s, r ∈ P_{n−1}, where s is orthogonal to p_{n}. ⊓ ⊔

b] Show that λ_{ν,n} > 0 for every ν = 1, 2, …, n.

Hint: Use part a] to show that

λ_{ν,n} = \frac{1}{(p′_{n}(x_{ν,n}))^{2}} ∫_{ℝ} (\frac{p_{n}(x)}{x − x_{ν,n}} )^{2} dα(x) .

⊓ ⊔

68 2. Some Special Polynomials

c] Suppose [a, b] is a finite interval containing supp(α). Let f ∈ C[a, b]. Show that

|∫_{a}^{b} f(x) dα(x) − ∑_{ν=1}^{n} λ_{ν,n}f(x_{ν,n})|

≤ 2(α(b+) − α(a−)) min_{p∈P_{2n−1}} ∥f − p∥_{[a,b]} .

Hint: Use parts a] and b] together with the observation

∑_{ν=1}^{n} λ_{ν,n} = ∫_{a}^{b} dα(x) = α(b+) − α(a−) .

⊓ ⊔

d] Suppose supp(α) ⊂ [a, b], where a, b ∈ ℝ. Show that

∑_{ν=1}^{n} λ_{ν,n}f(x_{ν,n}) ⟶_{n→∞} ∫_{a}^{b} f(x)dα(x)

for every Riemann-Stieltjes integrable function on [a, b] with respect to α. Hint: First show that f is Riemann-Stieltjes integrable on [a, b] with respect to α if and only if for every ϵ > 0 there are g_{1}, g_{2} ∈ C[a, b] so that

g_{1}(x) ≤ f(x) ≤ g_{2}(x) , x ∈ [a, b]

and

∫_{a}^{b} (g_{2}(x) − g_{1}(x)) dα(x) < ϵ .

Finish the proof by part c] and the Weierstrass approximation theorem (see E.1 of Section 4.1). ⊓ ⊔ e] Suppose supp(α) is compact. Let

Z := {x_{ν,n} : ν = 1, 2, …, n, n = 1, 2, … } .

Show that supp(α) ⊂ \overline{Z}, where \overline{Z} denotes the closure of Z. Hint: Use part d]. ⊓ ⊔ f] Show by an example that supp(α) ≠ \overline{Z} is possible.

E.9 Characterization of Compact Support. Using the notation of Theorem 2.3.2 and E.8, show that the following statements are equivalent:

(1) supp(α) is compact.

(2) sup_{n∈ℕ}{|a_{n}| + |b_{n}|} < ∞.

(3) The set Z := {x_{ν,n} : ν = 1, 2, …, n, n = 1, 2, … } is bounded.

Outline. (1) ⇒ (2). Note that the orthogonality of {p_{n}}_{n=0}^{∞} implies

a_{n} = ∫_{ℝ} xp_{n−1}(x)p_{n}(x) dα(x)  and  b_{n} = ∫_{ℝ} xp_{n}^{2}(x) dα(x) .

So supp(α) ⊂ [−K, K], the Cauchy-Schwarz inequality, and the orthonormality of (p_{n})_{n=0}^{∞} yield

|a_{n}| ≤ K ∫_{−K}^{K} |p_{n−1}(x)p_{n}(x)| dα(x)

≤ K (∫_{−K}^{K} p_{n−1}^{2}(x) dα(x))^{1/2} (∫_{−K}^{K} p_{n}^{2}(x) dα(x))^{1/2} ≤ K

and

|b_{n}| ≤ K ∫_{−K}^{K} p_{n}^{2}(x)dα(x) = K .

(2) ⇒ (3). Use Theorem 2.3.2 (three-term recursion) to show that

x_{ν,n} ∑_{k=0}^{n−1} p_{k}^{2}(x_{ν,n}) ≤ 2 ∑_{k=0}^{n−1} a_{k+1}p_{k}(x_{ν,n})p_{k+1}(x_{ν,n}) + ∑_{k=0}^{n−1} b_{k}p_{k}^{2}(x_{ν,n}) .

Hence

|x_{ν,n}| ∑_{k=0}^{n−1} p_{k}^{2}(x_{ν,n}) ≤ (2 max_{0≤k≤n−1} |a_{k}| + max_{0≤k≤n−1} |b_{k}|) ∑_{k=0}^{n−1} p_{k}^{2}(x_{ν,n}) .

(3) ⇒ (1). If Z ⊂ [−K, K], then by E.8 a]

∫_{ℝ} x^{2n−2} dα(x) = ∑_{ν=1}^{n} λ_{ν,n}x_{ν,n}^{2n−2} ≤ K^{2n−2} ∫_{ℝ} dα(x) ,

which implies supp(α) ⊂ [−K, K]. ⊓ ⊔

E.10 A Condition for supp(α) ⊂ [0, ∞). Let (p_{n})_{n=0}^{∞} be the sequence of orthonormal polynomials associated with an m-distribution α. Suppose supp(α) is compact and

p_{n}(0)p_{n+1}(0) < 0 , n = 0, 1, ….

Show that supp(α) ⊂ [0, ∞).

Hint: Use the interlacing property of the zeros of p_{n} (E.2) to show that

Z := {x_{ν,n} : ν = 1, 2, …, n, n = 1, 2, … } ⊂ [0, ∞).

Now use E.8 e] to obtain supp(α) ⊂ [0, ∞). ⊓ ⊔ E.11 The Solvability of the Moment Problem. Let (μ_{n})_{n=0}^{∞} be a sequence of real numbers. We would like to characterize those sequences (μ_{n})_{n=0}^{∞} for which there exists an m-distribution α so that

∫_{ℝ} x^{n} dα(x) = μ_{n} , n = 0, 1, ….

Let

μ(p) := ∑_{k=0}^{n} a_{k}μ_{k}

for every p ∈ P_{n} of the form p(x) = ∑_{k=0}^{n} a_{k}x^{k}.

A polynomial p is called nonnegative if it takes nonnegative values on the real line. The sequence (μ_{n})_{n=0}^{∞} is called positive definite if

μ(p) := ∑_{k=0}^{n} a_{k}μ_{k} > 0 , n = 0, 1, …

holds for every nonnegative polynomial p ∈ P_{n} of the form

p(x) = ∑_{k=0}^{n} a_{k}x^{k} .

The aim of this exercise is to outline the proof of Hamburger’s characterization of the solvability of the moment problem by the positive definiteness of the sequence of moments. See part o].

a] Show that if there exists an m-distribution α so that

∫_{ℝ} x^{n} dα(x) = μ_{n} , n = 0, 1, …,

then (μ_{n})_{n=0}^{∞} is positive definite.

Hint: An m-distribution α is increasing at infinitely many points. ⊓ ⊔

b] Show that (μ_{n})_{n=0}^{∞} is positive definite if and only if μ(p^{2}) > 0 holds for every 0 ≠ p ∈ P_{n}, n = 0, 1, ….

Hint: Use E.3 of Section 2.4. ⊓ ⊔

c] Show that (μ_{n})_{n=0}^{∞} is positive definite if and only if

| \begin{matrix}μ_{0} & μ_{1} & … & μ_{n} \\ μ_{1} & μ_{2} & … & μ_{n+1} \\ ⋮ & ⋮ & ... & ⋮ \\ μ_{n} & μ_{n+1} & … & μ_{2n} \end{matrix} | > 0 , n = 0, 1, ….

Hint: Use part b] and the law of inertia of Sylvester. See, for example, van der Waerden [50]. ⊓ ⊔

d] Helly’s Selection Theorem. Suppose the functions f_{n}, n = 1, 2, …, are nondecreasing on ℝ, and

sup_{n∈ℕ} ∥f_{n}∥_{ℝ} < ∞ .

Then there exists a subsequence of (f_{n})_{n=1}^{∞} that converges for every x ∈ ℝ. That is, we can select a pointwise convergent subsequence.

Hint: See, for example, Freud [71]. ⊓ ⊔

e] Helly’s Convergence Theorem. Let [a, b] be a finite interval. Suppose the functions α_{n}, n = 1, 2, …, are nondecreasing on [a, b] and

sup_{n∈ℕ} ∥α_{n}∥_{[a,b]} < ∞ .

Suppose also that (α_{n}(x))_{n=1}^{∞} converges to α(x) for every x ∈ [a, b]. Then

lim_{n→∞} ∫_{a}^{b} f(x) dα_{n}(x) = ∫_{a}^{b} f(x) dα(x)

for every f ∈ C[a, b].

Hint: See, for example, Freud [71]. ⊓ ⊔

In the rest of the exercise (except for the last part) we assume that (μ_{n})_{n=0}^{∞} is positive definite. Our goal is to prove the converse of part a]. Let

p_{n}^{∗}(x) := | \begin{matrix}μ_{0} & μ_{1} & … & μ_{n−1} & 1 \\ μ_{1} & μ_{2} & … & μ_{n} & x \\ ⋮ & ⋮ & ... & ⋮ & ⋮ \\ μ_{n} & μ_{n+1} & … & μ_{2n−1} & x^{n} \end{matrix} | .

f] Show that

μ(p_{n}^{∗}q) = 0 , q ∈ P_{n−1} .

g] Show that each p_{n}^{∗} has n simple real zeros.

Hint: Use part f]. ⊓ ⊔

Let x_{1,n} > x_{2,n} > ⋯ > x_{n,n} be the zeros of p_{n}^{∗}. Let

l_{ν,n}(x) := \frac{p_{n}^{∗}(x)}{p_{n}^{∗′}(x_{ν,n})(x − x_{ν,n})} , ν = 1, 2, …, n , n = 1, 2, …

(see E.6 of Section 1.1), and let

λ_{ν,n} := μ(l_{ν,n}) .

72 2. Some Special Polynomials

h] Show that

μ(q) = ∑_{ν=1}^{n} λ_{ν,n}q(x_{ν,n})

for every q ∈ P_{2n−1}. Hint: Use part f]. ⊓ ⊔ i] Show that

λ_{ν,n} = μ(l_{ν,n}^{2}) > 0 , ν = 1, 2, …, n , n = 1, 2, ….

Hint: Use part h]. ⊓ ⊔ For x ∈ ℝ, let

α_{n}(x) := ∑_{{ν: x_{ν,n}≤x}} λ_{ν,n} , n = 1, 2, ….

j] Show that 0 ≤ α_{n}(x) ≤ μ_{0} on ℝ for each n, and there is a subsequence of (α_{n})_{n=1}^{∞} that converges pointwise to a nondecreasing real-valued function α on ℝ. Hint: Use parts h], i], and d]. ⊓ ⊔ k] Show that for every finite interval [a, b],

lim_{k→∞} ∫_{a}^{b} x^{m} dα_{n_{k}}(x) = ∫_{a}^{b} x^{m} dα(x) , m = 0, 1, 2, …,

where α is defined in part j]. Hint: Use part e]. ⊓ ⊔ l] Let m be a fixed nonnegative integer and let r := ⌊m/2⌋ + 1. Show that if n_{k} ≥ r + 1, a ≤ −1 and b ≥ 1, then

|∫_{−∞}^{a} x^{m} dα_{n_{k}}(x) + ∫_{b}^{∞} x^{m} dα_{n_{k}}(x)|

≤ (\frac{1}{|a|} + \frac{1}{|b|}) ∫_{ℝ} x^{2r} dα_{n_{k}}(x) = (\frac{1}{|a|} + \frac{1}{|b|}) μ_{2r} .

Hint: Use part h]. ⊓ ⊔ m] Show that

∫_{ℝ} x^{m} dα(x) = μ_{m} , m = 0, 1, 2, …

where α is defined in part j]. Hint: Use parts k] and l]. ⊓ ⊔

2.3 Orthogonal Polynomials 73

n] Show that α defined in part j] is an m-distribution. o] There exists an m-distribution α so that

μ_{n} = ∫_{ℝ} x^{n} dα(x)

if and only if (μ_{n})_{n=0}^{∞} is positive definite, that is, if and only if

| \begin{matrix}μ_{0} & μ_{1} & … & μ_{n} \\ μ_{1} & μ_{2} & … & μ_{n+1} \\ ⋮ & ⋮ & ... & ⋮ \\ μ_{n} & μ_{n+1} & … & μ_{2n} \end{matrix} | > 0 , n = 0, 1, ….

Hint: Combine parts a], c], m], and n]. ⊓ ⊔

Necessary and sufficient conditions for the uniqueness of the solution of the moment problem are given in Freud [71], for example.

E.12 Favard’s Theorem. Given (a_{n})_{n=0}^{∞} ⊂ (0, ∞) and (b_{n})_{n=0}^{∞} ⊂ ℝ, the polynomials p_{n} ∈ P_{n} are defined by

xp_{n}(x) = a_{n}p_{n−1}(x) + b_{n}p_{n}(x) + a_{n+1}p_{n+1}(x) ,
p_{−1} = 0 , p_{0} = γ_{0} > 0 .

Then there exists an m-distribution α such that

∫_{ℝ} p_{n}(x)p_{m}(x) dα(x) = 0

for any two distinct nonnegative integers n and m. In other words, the converse of Theorem 2.3.2 is true.

In order to prove Favard’s theorem, proceed as follows: a] Show that the polynomials p_{n} are of the form

p_{n}(x) = γ_{n}x^{n} + r(x) , γ_{n} > 0 , r ∈ P_{n−1} .

b] Let p̃_{n} := γ_{n}^{−1}p_{n}, n = 0, 1, . . .. The sequence (μ_{n})_{n=0}^{∞} is defined as follows. Let

μ_{0} := 1 , μ(q) := c if q = c , c ∈ ℝ .

If μ_{0}, μ_{1}, …, μ_{n} have already been defined, then let

μ(q) := ∑_{k=0}^{n} c_{k}μ_{k} whenever q(x) = ∑_{k=0}^{n} c_{k}x^{k} , c_{k} ∈ ℝ

74 2. Some Special Polynomials

and let

μ_{n+1} := μ(x^{n+1} − p̃_{n+1}(x)) .

Show that

μ(p̃_{n}p̃_{m}) = 0 , m = 0, 1, …, n − 1 , n = 0, 1, …

and

μ(p_{n}^{2}) > 0 , n = 0, 1, ….

Hint: It is sufficient to prove that

μ(p̃_{n}(x)x^{m}) = 0 , m = 0, 1, …, n − 1 , n = 0, 1, … and

μ(p̃_{n}(x)x^{n}) > 0 , n = 0, 1, ….

These can be obtained from the definition of μ and from Theorem 2.3.2 (three-term recursion) by induction on n. ⊓ ⊔ c] Show that every q ∈ P_{n} is of the form

q = ∑_{k=0}^{n} c_{k}p̃_{k} , c_{k} ∈ ℝ

and if q ≠ 0, then

μ(q^{2}) = ∑_{k=0}^{n} c_{k}^{2}μ(p_{k}^{2}) > 0 .

d] Show that (μ_{n})_{n=0}^{∞} is positive definite in the sense of E.11. Hint: Use the previous part and E.11 b]. ⊓ ⊔ e] Prove Favard’s theorem. Hint: Use part o] of E.11, parts d] and b] of this exercise, and the definition of μ. ⊓ ⊔

E.13 Christoffel Function. Let α be an m-distribution. For a fixed n ∈ ℕ, the function

λ_{n}(z) = inf {∫_{ℝ} q^{2}(x) dα(x) : q ∈ P_{n−1} , |q(z)| = 1} , z ∈ ℂ

is called the nth Christoffel function associated with α. a] Show that

λ_{n}(z) := (∑_{k=0}^{n−1} |p_{k}(z)|^{2})^{−1} ,

where (p_{n})_{n=0}^{∞} is the sequence of orthonormal polynomials associated with α.

Show also that the infimum in the definition of λ_{n}(z) is actually a minimum, and it is attained if and only if

q(x) = \frac{∑_{k=0}^{n−1} \overline{p_{k}(z)}p_{k}(x)}{∑_{k=0}^{n−1} |p_{k}(z)|^{2}} .

Hint: Write

q = ∑_{k=0}^{n−1} c_{k}p_{k} , c_{k} ∈ ℂ

and observe that the orthonormality of (p_{n})_{n=0}^{∞} implies

∫_{ℝ} q^{2}(x) dα(x) = ∑_{k=0}^{n−1} |c_{k}|^{2} .

Now use the Cauchy-Schwarz inequality (E.3 of Section 2.2) to find the maximum of |q(z)| for polynomials q ∈ P_{n−1}^{c} satisfying

∫_{ℝ} q^{2}(x) dα(x) ≤ 1

where z ∈ ℂ is fixed.

b] Let λ_{ν,n}, ν = 1, 2, …, n, be the Christoffel numbers associated with an m-distribution α, that is, the coefficients in the Gauss-Jacobi quadrature formula, as in E.8. Show that

λ_{ν,n} = λ_{n}(x_{ν,n}) , ν = 1, 2, …, n ,

that is, the Christoffel numbers are the values of the Christoffel function at the zeros of the nth orthonormal polynomial p_{n}.

Hint: Use parts E.8 a] and E.8 b]. ⊓ ⊔

c] Let x ∈ ℝ be fixed. Show that

∑_{n=0}^{∞} p_{n}^{2}(x) < ∞

if and only if x is a mass point of α, that is, α(x−) < α(x+), in which case

∑_{n=0}^{∞} p_{n}^{2}(x) = (α(x+) − α(x−))^{−1} .

Hint: Use part a] and the Weierstrass approximation theorem. See E.1 of Section 4.1. ⊓ ⊔ E.14 The Markov-Stieltjes Inequality. Let α be an m-distribution with associated orthonormal polynomials (p_{n})_{n=0}^{∞}. Let x_{1,n} > x_{2,n} > ⋯ > x_{n,n} denote the zeros of p_{n}. Let x_{0,n} := ∞ and x_{n+1,n} := −∞. As in E.8 let λ_{ν,n}, ν = 1, 2, …, n, be the Christoffel numbers associated with α . Show that

λ_{ν,n} ≤ ∫_{x_{ν+1,n}}^{x_{ν−1,n}} dα(x) , ν = 1, 2, …, n

and

∫_{x_{ν,n}}^{x_{ν−1,n}} dα(x) ≤ λ_{ν,n} + λ_{ν−1,n} , ν = 2, 3, …, n .

Hint: Let 1 ≤ k ≤ n be fixed. Use E.7 of Section 1.1 (Hermite interpolation) to find polynomials P ∈ P_{2n−1} and Q ∈ P_{2n−1} with the following properties:

(1) P (x_{j,n}) = Q(x_{j,n}) = 1 , j = 1, 2, …, k − 1 ,

(2) P (x_{k,n}) = 0 , Q(x_{k,n}) = 1 ,

(3) P (x_{j,n}) = Q(x_{j,n}) = 0 , j = k + 1, k + 2, …, n ,

(4) P (x) ≤ χ_{(−∞, x_{k,n}]}(x) ≤ Q(x) , x ∈ ℝ ,

where

χ_{(−∞, x_{k,n}]}(x) := \{ \begin{matrix}1 & if − ∞ < x ≤ x_{k,n} \\ 0 & if x_{k,n} < x < ∞ . \end{matrix}

Now apply E.8 (Gauss-Jacobi quadrature formula) to P and Q. ⊓ ⊔

E.15 Orthonormal Polynomials as Determinants. Suppose α is an mdistribution with moments

μ_{n} = ∫_{ℝ} x^{n} dα(x) , n = 0, 1, ….

Let

Δ_{n} := | \begin{matrix}μ_{0} & μ_{1} & … & μ_{n} \\ μ_{1} & μ_{2} & … & μ_{n+1} \\ ⋮ & ⋮ & ... & ⋮ \\ μ_{n} & μ_{n+1} & … & μ_{2n} \end{matrix} | , n = 0, 1, ….

a] Show that Δ_{n} > 0, n = 0, 1, 2, ….

b] Show that the orthonormal polynomials p_{n} associated with α are of the form

p_{n}(x) = (Δ_{n}Δ_{n−1})^{−1/2} | \begin{matrix}μ_{0} & μ_{1} & … & μ_{n−1} & 1 \\ μ_{1} & μ_{2} & … & μ_{n} & x \\ ⋮ & ⋮ & ... & ⋮ & ⋮ \\ μ_{n} & μ_{n+1} & … & μ_{2n−1} & x^{n} \end{matrix} | .

c] Let (a_{n})_{n=0}^{∞} ⊂ (0, ∞) and (b_{n})_{n=0}^{∞} ⊂ ℝ be the coefficients in the threeterm recursion for the sequence (p_{n})_{n=0}^{∞} of orthonormal polynomials associated with α as in Theorem 2.3.2. Show that the monic orthogonal polynomials p̃_{n} := γ_{n}^{−1}p_{n} are of the form

p̃_{n}(x) = det (xI_{n} − J_{n})

where J_{n} is the tridiagonal n by n Jacobi matrix

J_{n} := ( \begin{matrix}b_{0} & a_{1} \\ a_{1} & b_{1} & a_{2} \\ & a_{2} & b_{2} & a_{3} \\ & & ... & ... & ... \\ & & & a_{n} & b_{n} \end{matrix} )

and I_{n} is the n by n unit matrix.

E.16 The Support of α. Let (a_{n})_{n=0}^{∞} ⊂ (0, ∞) and (b_{n})_{n=0}^{∞} ⊂ ℝ be the coefficient sequences in the three-term recursion for the sequence of (p_{n})_{n=0}^{∞} of orthonormal polynomials associated with an m-distribution α as in Theorem 2.3.2.

a] Show that if supp(α) ⊂ [b − a, b + a] with some a > 0 and b ∈ ℝ, then

a_{n} ≤ a and |b_{n} − b| ≤ a , n = 0, 1, ….

Hint: Use the orthonormality of (p_{n})_{n=0}^{∞} to show that

a_{n} = ∫_{b−a}^{b+a} (x − b)p_{n−1}(x)p_{n}(x) dα(x)

and

b_{n} − b = ∫_{b−a}^{b+a} (x − b)p_{n}^{2}(x) dα(x) .

Now apply the Cauchy-Schwarz inequality, and use the orthonormality of (p_{n})_{n=0}^{∞} again. ⊓ ⊔

b] Show that

supp(α) ⊂ [−K, K]

where

K := 2 sup{a_{n} : n ∈ ℕ} + sup{|b_{n}| : n ∈ ℕ}

(the suprema are taken over all nonnegative integers).

Hint: Suppose K < ∞; otherwise there is nothing to prove. Combine E.9, the inequality in the hint to the direction (2) ⇒ (3) of E.9, and E.8 e]. ⊓ ⊔ c] Blumenthal’s Theorem. Assume that

lim_{n→∞} a_{n} =: \frac{a}{2} > 0  and  lim_{n→∞} b_{n} =: b ∈ ℝ .

Then

supp(α) = [b − a, b + a] ∪ F

where F ⊂ ℝ \ [b − a, b + a] is a countable bounded set for which

F \ (b − a − ϵ, b + a + ϵ)

is finite for every ϵ > 0.

Proof. See Nevai [79b] or Máté, Nevai, and Van Assche [91]. ⊓ ⊔

d] Rakhmanov’s Theorem. Suppose supp(α) ⊂ [b − a, b + a] with some a > 0 and b ∈ ℝ. Suppose also that α′(x) > 0 a.e. in [b − a, b + a]. Then

lim_{n→∞} a_{n} = \frac{a}{2} and lim_{n→∞} b_{n} = b .

Proof. See, for example, Máté, Nevai, and Totik [85], or Nevai [91]. ⊓ ⊔

There is an analogous theory of orthogonal polynomials on the unit circle initiated by Geronimus, Shohat, and Achiezer. An important contribution, called Szegő theory, may be found in Freud [71].

E.17 A Theorem of Stieltjes [14]. Let w be a positive continuous weight function on [−a, a]. Denote the nth moment by

μ_{n} := ∫_{−a}^{a} x^{n}w(x) dx .

Let p̃_{n} ∈ P_{n} denote the nth monic orthogonal polynomial on [−a, a] associated with the weight w. Then (p̃_{n})_{n=0}^{∞} satisfies a three-term recursion

p̃_{n}(x) = (x − A_{n})p̃_{n−1}(x) − B_{n}p̃_{n−2}(x)

with p̃_{0}(x) = 1 and p̃_{1}(x) := x − A_{1}; see Theorem 2.3.2.

Suppose the sequence of polynomials (q_{n})_{n=0}^{∞} satisfies the same recursion commencing with q_{0}(x) := 0 and q_{1}(x) := B_{1}. Stieltjes’ theorem (see, for example, Cheney [66]) states the following.

Theorem. For any x ∉ [−a, a],

∫_{−a}^{a} \frac{w(t) dt}{x − t} = \frac{μ_{0}}{x} + \frac{μ_{1}}{x^{2}} + ⋯

= \frac{B_{1}}{x − A_{1} − \frac{B_{2}}{x − A_{2} − \frac{B_{3}}{x − A_{3} − · · · .}}}

2.4 Polynomials with Nonnegative Coefficients 79

Furthermore, the nth convergent q_{n}/p̃_{n} satisfies

\frac{q_{n}(x)}{p̃_{n}(x)} = \frac{B_{1}}{x − A_{1} − \frac{B_{2}}{... \frac{B_{n}}{x − A_{n}} .}}

E.18 Completeness of Orthogonal Polynomials. Let (p_{n})_{n=0}^{∞} be the sequence of orthonormal polynomials associated with an m-distribution α. If supp(α) ⊂ [a, b], where [a, b] is a finite interval, then (p_{n})_{n=0}^{∞} is a maximal orthogonal collection in L_{2}[a, b].

Hint: Use the Weierstrass approximation theorem (E.1 of Section 4.1) on [a, b]. ⊓ ⊔

E.19 Bounds for Jacobi Polynomials. For all Jacobi weight functions w(x) = (1 − x)^{α}(1 + x)^{β} with α ≥ −1/2 and β ≥ −1/2, the inequalities

max_{x∈[−1,1]} \frac{p_{n}^{2}(x)}{∑_{k=0}^{n} p_{k}^{2}(x)} ≤ \frac{4 (2 + \sqrt{α^{2} + β^{2}})}{2n + α + β + 2}

and

max_{x∈[−1,1]} \sqrt{1 − x^{2}}w(x)p_{n}^{2}(x) ≤ \frac{2e (2 + \sqrt{α^{2} + β^{2}})}{π}

hold, where (p_{n})_{n=0}^{∞} is the sequence of orthonormal Jacobi polynomials associated with the weight function w.

Proof. See Erdélyi, Magnus, and Nevai [94]. ⊓ ⊔

2.4 Polynomials with Nonnegative Coefficients

A quadratic polynomial x^{2} + αx + β with real coefficients has both roots in the halfplane {z ∈ ℂ : Re(z) ≤ 0} if and only if β ≥ 0 and α ≥ 0. This easy consequence of the quadratic formula gives the following lemma:

Lemma 2.4.1. If p ∈ P_{n} has all its zeros in {z ∈ ℂ : Re(z) ≤ 0}, then either p or −p has all nonnegative coefficients.

The converse of this is far from true. Indeed, the following result of Meissner holds (see Pólya and Szegő [76]). We denote by P_{n}^{+} the set of polynomials in P_{n}, that have all nonnegative coefficients.

Theorem 2.4.2. If p ∈ P_{n}^{c} and p(x) > 0 for x > 0, then p = s/t, where s and t are both polynomials with all nonnegative coefficients.

Since a polynomial p that is real-valued on the positive real axis has real coefficients, and since p(x) > 0 for all x > 0 implies that the leading coefficient of p is positive, Theorem 2.4.2 will follow immediately from the next lemma.

Lemma 2.4.3. Suppose α, β ∈ ℝ and suppose x^{2} − αx + β has no nonnegative root. Then x^{2} − αx + β = p(x)/q(x), where p, q ∈ P_{m} both have all nonnegative coefficients, and where

m ≤ 10 (4 − \frac{α^{2}}{β} )^{−1/2} .

Proof. The quadratic polynomial x^{2} − αx + β has no positive root if and only if α^{2} < 4β. We set c := α^{2}/β and note that c < 4. Consider

(x^{2} − αx + β)(x^{2} + αx + β) = x^{4} + (2β − α^{2})x^{2} + β^{2}
= x^{4} + β(2 − c)x^{2} + β^{2} .

If c ≤ 2 we have the desired factorization. If c > 2, consider

(x^{4} + β(2 − c)x^{2} + β^{2})(x^{4} − β(2 − c)x^{2} + β^{2})
= x^{8} + β^{2}(2 − (2 − c)^{2})x^{4} + β^{4} .

If 2 − (2 − c)^{2} > 0 we are finished. In general, we proceed as follows:

Let

P_{n}(x) : = x^{2^{n+1}} + β^{2^{n−1}}(2 − (2 − (2 − ⋯ 2 − (2 − c)^{2})^{2} ⋯)^{2})x^{2^{n}} + β^{2^{n}}
= x^{2^{n+1}} + β^{2^{n−1}}c_{n}x^{2^{n}} + β^{2^{n}} ,

where c_{n} has n nested terms. Let

Q_{n}(x) := x^{2^{n+1}} − β^{2^{n−1}}c_{n}x^{2^{n}} + β^{2^{n}} .

Note that, since c_{n+1} = 2 − c_{n}^{2}

P_{n}(x)Q_{n}(x) = x^{2^{(n+1)+1}} − β^{2^{n}}c_{n}^{2}x^{2^{n+1}} + 2β^{2^{n}}x^{2^{n+1}} + β^{2^{n+1}}
= x^{2^{n+2}} + β^{2^{n}}c_{n+1}x^{2^{n+1}} + β^{2^{n+1}}
= P_{n+1}(x) .

Consider the smallest n (if it exists) such that c_{n} is nonnegative. Then

(x^{2} − αx + β)(x^{2} + αx + β) = P_{1}(x)

and

P_{1}Q_{1}Q_{2} ⋯ Q_{n−1} = P_{n} ,

where Q_{1}Q_{2} ⋯ Q_{n−1} ∈ P_{2^{n+1}−4}^{+} since each c_{k} < 0 for k < n, and where P_{n} ∈ P_{2^{n+1}}^{+} since c_{n} ≥ 0. Thus, we have the desired representation

x^{2} − αx + β = \frac{P_{n}(x)}{(x^{2} + αx + β)(Q_{1}Q_{2} ⋯ Q_{n−1})(x)} ,

where n is the smallest integer such that c_{n} > 0.

Now suppose c_{1}, …, c_{n−1}, c_{n} are all nonpositive. Then
c_{k} = −\sqrt{2 − c_{k+1}} , k = 1, 2, …, n − 1

and c_{1} = 2 − c imply

(2.4.1)  c > 2 + (2 + ⋯ (2 + (2 + 2^{1/2})^{1/2})^{1/2} ⋯)^{1/2} =: δ_{n} ,

where the above formula contains n iterations. Since, by assumption, c < 4, and since δ_{n} → 4 as n → ∞, it is clear that (2.4.1) is not satisfied for some n, and eventually some c_{n} is greater than zero.

The estimate on the degree requires analyzing the rate of convergence of (δ_{n})_{n=0}^{∞}. Since δ_{n} = 2 + \sqrt{δ_{n−1}}, we have

4 − δ_{n} = 2 − \sqrt{δ_{n−1}} = \frac{4 − δ_{n−1}}{2 + \sqrt{δ_{n−1}}} ≤ \frac{4 − δ_{n−1}}{2} .

By repeated applications of the above,

4 − δ_{n} ≤ \frac{4 − δ_{0}}{2^{n}} = \frac{1}{2^{n−1}} .

Now we can improve the above estimate as follows: We have

4 − δ_{n} =\frac{4 − δ_{n−1}}{2 + \sqrt{δ_{n−1}}} = \frac{4 − δ_{n−2}}{(2 + \sqrt{δ_{n−1}})(2 + \sqrt{δ_{n−2}})}
=\frac{4 − δ_{0}}{(2 + \sqrt{δ_{n−1}})(2 + \sqrt{δ_{n−2}}) ⋯ (2 + \sqrt{δ_{0}})}
≤\frac{2}{(2 + \sqrt{4 − 2^{2−n}})(2 + \sqrt{4 − 2^{3−n}}) ⋯ (2 + \sqrt{4 − 2^{(n+1)−n}})}
≤\frac{2}{(2 + 2 − 2^{1−n})(2 + 2 − 2^{2−n}) ⋯ (2 + 2 − 2^{n−n})}

≤2 · 4^{−n} ∏_{j=2}^{n+1} \frac{1}{1 − 2^{−j}} ≤ 2 · 4^{−n} ∏_{j=2}^{n+1} (1 + 2 · 2^{−j}) ≤ 2e4^{−n} .

So if

m := 2^{n+1} ≥ \frac{2\sqrt{2e}}{\sqrt{4 − c}} ,

then

4 − δ_{n} ≤ 2e4^{−n} ≤ 4 − c ,

that is, δ_{n} ≥ c. ⊓ ⊔

We note that in the above proof a little additional effort yields a slightly better constant than 10.

Let k ∈ ℕ and ϵ ∈ (0, π). It follows easily from Lemma 2.3.4 that if p ∈ P_{k} has no zeros in the cone

{z ∈ ℂ : | arg(z)| < ϵ} ,

then there are s, t ∈ P_{m}^{+} with m ≤ \frac{5}{4}πkϵ^{−1} so that p = s/t; see E.1 d]. The essential sharpness of this upper bound is shown by E.1 e]. An easier proof of Theorem 2.4.2 that gives a weaker bound for the degree of the numerator and denominator in the representation is given by E.1 f].

A similar sort of representation theorem due to Bernstein [15] is the following:

Theorem 2.4.4. If p ∈ P_{n} and p(x) > 0 for x ∈ (−1, 1), then there is a representation

p(x) = ∑_{j=0}^{d} a_{j}(1 − x)^{j}(1 + x)^{d−j}

with each a_{j} ≥ 0. (The smallest d := d(p) for which such a representation exists is called the Lorentz degree of p.)

It suffices to prove this result for quadratic polynomials; this is left as an exercise; see E.1 f].

The proof of the following interesting result of Barnard et al. [91] is surprisingly complicated, and we do not reproduce it here.

Theorem 2.4.5. Suppose that p ∈ P_{n} has all nonnegative coefficients. Suppose that the zeros of p are z_{1}, z_{1}, …, z_{n} ∈ ℂ. For τ ≥ 0, let

p_{τ} (z) = ∏_{{j=1 \atop |arg(z_{j})|>τ}}^{n} (1 − \frac{z}{z_{j}} ) ,

where arg(z) is defined so that arg(z) ∈ [−π, π). Then p_{τ} (z) has all nonnegative coefficients.

It follows from this result that if p ∈ P_{n} has all nonnegative coefficients and if q(x) = x^{2} + αx + β is a quadratic polynomial with zeros forming a pair of conjugate zeros of p that have least angular distance from the positive x-axis, then p/q also has all nonnegative coefficients.

Comments, Exercises, and Examples.

Polynomials with all nonnegative coefficients have a number of distinguishing properties that are explored in the exercises. For example, only analytic functions with all nonnegative coefficients can be approximated uniformly on [0, 1] by such polynomials; see E.2. So a Weierstrass-type theorem does not hold for these polynomials. This is quite different from approximation by polynomials of the form

(2.4.2)  ∑ a_{i,j}(x + 1)^{i}(1 − x)^{j} ,  a_{i,j} ≥ 0 .

Since every polynomial that is strictly positive on (−1, 1) has such a representation (E.1 b]), it follows from the Weierstrass approximation theorem that all nonnegative functions from C[−1, 1] are in the uniform closure.

It follows from Theorem 2.4.2 and the Weierstrass approximation theorem (see E.1 of Section 4.1) that fractions of polynomials with all nonnegative coefficients form a dense set in the uniform norm on [0, 1] in the set of nonnegative continuous functions on a finite closed interval [0, 1]. Hence they have a much larger uniform closure on [0, 1] than that of the polynomials with all nonnegative coefficients.

Various inequalities for polynomials of the form (2.4.2) are considered in Appendix 5.

E.1 Remarks on Theorem 2.4.2.

a] Suppose α, β ∈ ℝ, ϵ ∈ (0, π), and suppose x^{2} + αx + β has no zeros in the cone

{z ∈ ℂ : | arg(z)| < ϵ} .

Show that there are p, q ∈ P_{m}^{+} with m ≤ \frac{5}{2}πϵ^{−1} such that

x^{2} + αx + β = \frac{p(x)}{q(x)} .

Hint: This is a reformulation of Theorem 2.4.3 by introducing the angle between the positive x-axis and the zero of the quadratic polynomial. ⊓ ⊔

b] Let n ∈ ℕ. Show that if p ∈ P_{n}^{+}, then p has no zeros in the cone

{z ∈ ℂ : |arg(z)| < π/(2n)} .

This is sharp, as the example p(x) := x^{n} + 1 shows.

c] Show that the result of part a] is sharp up to the constant \frac{5}{2}.

Hint: Let n ∈ ℕ. Consider

x^{2} + αx + β = (x − exp (\frac{2πi}{2n})) (x − exp (\frac{−2πi}{2n})) .

84 2. Some Special Polynomials

Show that if there are p, q ∈ P_{m}^{+} so that x^{2} + αx + β = p(x)/q(x), then m ≥ n − 1. ⊓ ⊔ d] Let ϵ ∈ (0, π) and k ∈ ℕ. Suppose p ∈ P_{k} has no zeros in the cone

{z ∈ ℂ : | arg(z)| < ϵ} .

Show that there are s, t ∈ P_{m}^{+} with m ≤ \frac{5}{4}πkϵ^{−1} so that p = s/t . e] Let ϵ ∈ (0, π) and k ∈ ℕ. Let

f_{k}(x) := ((x − z_{0})(x − \overline{z}_{0}))^{k} ,

where arg(z_{0}) = ϵ. Assume that f_{k} = s/t, where s, t ∈ P_{m}^{+}. Show that

m ≥ (log 2)kϵ^{−1} .

Hint: First observe that

s(y) ≤ s(y + yδm^{−1}) ≤ e^{δ}s(y)

for every s ∈ P_{m}^{+}, y ∈ (0, ∞), and δ ∈ (0, ∞). Therefore

f_{k}(y + yδm^{−1}) ≤ e^{δ}f_{k}(y)

for every y ∈ (0, ∞) and δ ∈ (0, ∞). Now let y ≥ |z_{0}| be chosen so that |y − z_{0}| = ϵ. Applying the above inequality with this y and δ := mα, we obtain

f_{k}(y + ϵy) ≤ e^{mϵ}f_{k}(y) ,

hence 2^{k} ≤ e^{mϵ}, that is, k log 2 ≤ mϵ. ⊓ ⊔ f] Prove that if r ∈ P_{n} and r(x) > 0 for all x > 0, then there is an integer d ≥ n such that

r(x) = \frac{q(x)}{(1 + x)^{d−n}} ,

where q ∈ P_{d}^{+}. Hint: Let α, β ∈ ℝ and α^{2} < 4β. Consider

(x^{2} − αx + β)(1 + x)^{d} = ∑_{j=0}^{d+2} c_{j}x^{j}

and compute c_{j} explicitly. ⊓ ⊔ g] If p ∈ P_{n} and p(x) > 0 for all x ∈ (−1, 1), then it is of the form

p(x) = ∑_{j=0}^{d} a_{j}(1 − x)^{j}(1 + x)^{d−j} , a_{j} ≥ 0

for some d ≥ n. Hint: Apply a] to

r(u) := (1 + u)^{n}p (\frac{1 − u}{1 + u}) , u := \frac{1 − x}{1 + x} .

⊓ ⊔

E.2 Polynomials with Nonnegative Coefficients.

a] If p ∈ P_{n}^{+}, then for x > 0

0 ≤ p′(x) < \frac{n}{x} p(x) .

b] If (p_{n})_{n=1}^{∞} is a sequence of polynomials with p_{n} ∈ P^{+} := ∪_{k=0}^{∞}P_{k}^{+} and (p_{n})_{n=1}^{∞} converges to f uniformly on [0, 1], then f is the restriction to [0, 1] of a function analytic in D := {z ∈ ℂ : |z| < 1} of the form

f(z) = ∑_{n=0}^{∞} a_{n}z^{n} , a_{n} ≥ 0 .

Hint: Since (p_{n}(1))_{n=1}^{∞} converges and each p_{n} has nonnegative coefficients, there is a constant C such that

∥p_{n}∥_{D} ≤ p_{n}(1) ≤ C , n = 1, 2, ….

Now Montel’s theorem (see, for example, Ash [71]) implies that (p_{n})_{n=1}^{∞} has a locally uniformly convergent subsequence on D. Deduce that this subsequence converges to an f with nonnegative coefficients. ⊓ ⊔

E.3 Nonnegative-Valued Polynomials and Sums of Squares.

a] Suppose p ∈ P_{2n} is nonnegative on ℝ. Then there exist s, t ∈ P_{n} such that

p(x) = s^{2}(x) + t^{2}(x) .

Hint: If p ∈ P_{2} and p is nonnegative, then for some real numbers α and β,

p(x) = (x − α)^{2} + β^{2} .

Now use the identity

(a^{2} + b^{2})(c^{2} + d^{2}) = (ac + bd)^{2} + (ad − bc)^{2} .
⊓ ⊔

b] If p ∈ P_{2n} is nonnegative for x ≥ 0, then there exist s, t, u, v ∈ P_{n} such that

p(x) = s^{2}(x) + t^{2}(x) + xu^{2}(x) + xv^{2}(x) .

c] Suppose t ∈ T_{n} is nonnegative on ℝ. Show that there exists a q ∈ P_{n}^{c} such that

t(θ) = |q(e^{iθ})|^{2} , θ ∈ ℝ .

Show also that if, in addition, t ∈ T_{n} is even, then there exists a q ∈ P_{n} such that the above holds. d] If p ∈ P_{n} is nonnegative on [−1, 1], then there exist s, t ∈ P_{n} such that

p(x) = s^{2}(x) + (1 − x^{2})t^{2}(x) .

Hint: Write, by c],

p(cos θ) = |s(cos θ) + it(cos θ) sin θ|^{2} .
⊓ ⊔

The above exercise follows Pólya and Szegő [76]; see also E.1 of Section 7.2 where this result is extended.

The following two exercises discuss results proved in Erdélyi and Szabados [88], [89b], and Erdélyi [91c].

E.4 Lorentz Degree of Polynomials. Given a polynomial p ∈ P_{n}, let d = d(p) be the minimal nonnegative integer for which the polynomial p is of the form

p(x) = ± ∑_{j=0}^{d} a_{j}(1 − x)^{j}(x + 1)^{d−j} , a_{j} ≥ 0 .

If there is no such d, then let d(p) := ∞. We call d := d(p) the Lorentz degree of the polynomial p.

a] Let p ∈ P_{n} \ P_{n−1} be of the form

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

For m ≥ n, let the numbers b_{j,m}, j = 0, 1, …, m, be defined by

p(x) = (∑_{j=0}^{n} b_{j}(1 − x)^{j}(x + 1)^{n−j}) (\frac{1 − x}{2} + \frac{x + 1}{2})^{m−n}

= ∑_{j=0}^{m} b_{j,m}(1 − x)^{j}(x + 1)^{m−j} .

Show that if d(p) is finite, then it is the smallest value of m for which each b_{j,m} is nonnegative or each b_{j,m} is nonpositive.

b] Show that if p ∈ P_{1} \ P_{0} has no zeros in (−1, 1), then d(p) = 1.

c] Show that if p ∈ P_{2}\P_{1} has no zeros in the open unit disk, then d(p) = 2.

d] Show that if the zeros of a polynomial p ∈ P_{2} \ P_{1} lie on the ellipse

B_{ϵ} := {z = x + iy : y^{2} = ϵ^{2}(1 − x^{2}), x ∈ (−1, 1)}

with ϵ ∈ (0, 1], then

ϵ^{−2} ≤ d(p) < 2ϵ^{−2} + 1.

2.4 Polynomials with Nonnegative Coefficients 87

e] Let ϵ ∈ (0, 1) be such that ϵ^{−2} is an integer. Let

p_{1}(x) := x^{2} + 2\frac{3ϵ^{2} − 1}{1 − ϵ^{2}}x + \frac{8ϵ^{4} − 5ϵ^{2} + 1}{1 − ϵ^{2}} .

Show that p_{1} has its zeros on B_{ϵ} defined in d], and d(p_{1}) = ϵ^{−2}. f] Let ϵ ∈ (0, 1) be such that 2ϵ^{−2} is an integer. Let

p_{2}(x) := x^{2} − 2\frac{2 − 3ϵ^{2}}{(1 − ϵ^{2})(2 − ϵ^{2})}x + \frac{−ϵ^{8} − 5ϵ^{6} + ϵ^{4} − 8ϵ^{2} + 4}{(1 − ϵ^{2})(2 − ϵ^{2})^{2}} .

Show that p_{2} has its zeros on B_{ϵ} defined in part d], and d(p_{2}) = 2ϵ^{−2}. g] Show that d(pq) ≤ d(p) + d(q) for any two polynomials p and q. h] Let ϵ ∈ (0, 1]. Show that if p ∈ P_{n} has no zeros in

D_{ϵ} := {z = x + iy : y^{2} < ϵ^{2}(1 − x^{2}), x ∈ (−1, 1), y ∈ ℝ} ,

then

d(p) < 2nϵ^{−2} + n < 3nϵ^{−2} .

i] Let p be a polynomial. Show that d(p) < ∞ if and only if p = 0 or p has no zeros in (−1, 1). j] Show that

|p (y + \sqrt[\frac{i}{2}]{\frac{(1−y^{2})n}{d}})| ≤ 2^{n} |p (y − \frac{n}{4d})|

for every p ∈ B_{d}(−1, 1), 1 ≤ n ≤ d, and y ∈ [0, 1) (i is the imaginary unit). Hint: Modify the proof of Lemma A.5.4. ⊓ ⊔ k] Let b ∈ [0, 1]. Show that

p′(b) ≤ d p(b)

for every p ∈ B_{d}(−1, 1), positive in (−1, 1). Hint: If q_{j,d}(x) := (1 − x)^{j}(x + 1)^{d−j} , then

q′_{j,d}(b) = q_{j,d}(b) (\frac{d − j}{1 + b} − \frac{j}{1 − b}) ≤ d q_{j,d}(b) , j = 0, 1, …, d

for every b ∈ [0, 1]. ⊓ ⊔ l] Show that d(p) ≥ \frac{1}{17}nϵ^{−2} whenever

p(x) = ((x − z_{0})(x − z̄_{0}))^{n} , z_{0} ∈ B_{ϵ} , ϵ ∈ (0, 1] ,

where B_{ϵ} is defined in part d].

88 2. Some Special Polynomials

Proof. Let z_{0} = y + iϵ(1 − y^{2})^{1/2}, y ∈ (−1, 1). Without loss of generality it may be assumed that 0 ≤ y < 1. Distinguish two cases. Case 1: 1 − 2ϵ^{2} ≤ y < 1. By part k],

d(p) ≥ \frac{p′(1)}{p(1)} = \frac{2n(1 − y)}{(1 − y)^{2} + ϵ^{2}(1 − y^{2})}

= \frac{2n}{(1 − y) + ϵ^{2}(1 + y)} > \frac{n}{2ϵ^{2}} .

Case 2: 0 ≤ y < 1 − 2ϵ^{2}. Applying part k] with

b := y + ϵ(1 − y^{2})^{1/2} ∈ [0, 1]

deduce that

(2.4.3)  d(p) ≥ \frac{p′(b)}{p(b)} = \frac{2n(b − y)}{(b − y)^{2} + ϵ^{2}(1 − y^{2})} = \frac{n}{ϵ(1 − y^{2})^{1/2}} .

Use part j] to obtain

(2.4.4) |(1 − y^{2})ϵ^{2} − \frac{(1 − y^{2})n}{4d}|^{n} ≤ 2^{n} ((1 − y^{2})ϵ^{2} + \frac{n^{2}}{16d^{2}} )^{n} ,

where d := d(p) and n = \frac{1}{2} deg(p). If

(1 − y^{2})ϵ^{2} ≥ \frac{(1 − y^{2})n}{8d} ,

then there is nothing to prove. Therefore assume that

(2.4.5)  (1 − y^{2})ϵ^{2} ≤ \frac{(1 − y^{2})n}{8d} .

Now (2.4.3) to (2.4.5) yield

(\frac{(1 − y^{2})n}{8d})^{n} ≤ 2^{n} ((1 − y^{2})ϵ^{2} + \frac{n^{2}}{16d^{2}} )^{n}

and so

\frac{(1 − y^{2})n}{8d} ≤ 2 ((1 − y^{2})ϵ^{2} + \frac{n^{2}}{16d^{2}} ) .

Since, by (2.4.3), n^{2}d^{−2} ≤ (1 − y^{2})ϵ^{2}, the above inequality implies

\frac{(1 − y^{2})n}{8d} ≤ 2 \frac{17}{16}(1 − y^{2})ϵ^{2}

and so d ≥ \frac{1}{17}nϵ^{−2}. ⊓ ⊔ m] Show that p ∈ P_{n} \ P_{n−1} and d(p) = n imply that the zeros z_{1}, z_{2}, …, z_{n} of p satisfy |z_{1}z_{2} ⋯ z_{n}| ≥ 1.

n] Show that d(pq) < max{d(p), d(q)} can happen.

Hint: Let

p(x) := (1 − x)^{2} − 2(1 − x^{2}) + 4(x + 1)^{2} and q(x) := (x + 1) + \frac{1}{2}(1 − x) .

Show that d(p) = 4, d(q) = 1, and d(pq) = 3. ⊓ ⊔

o] Show that if p ∈ P_{k} \ P_{k−1} has no zeros in (−1, 1), z ∈ ℂ, |z| > 1, and

p(x) = ((x − z)(x − \overline{z}))^{m}q(x) ,

then d(p) = deg(p) = k + 2m if m is sufficiently large. This shows that polynomials p with the property d(p) = deg(p) can have arbitrary many prescribed zeros in ℂ \ (−1, 1).

E.5 Lorentz Degree of Trigonometric Polynomials. Given ω ∈ (0, π] and a real trigonometric polynomial t ∈ T_{n}, let d = d_{ω}(t) be the minimal nonnegative integer for which t is of the form

t(θ) = ± ∑_{j=0}^{2d} a_{j} sin^{j} \frac{ω − θ}{2} sin^{2d−j} \frac{θ + ω}{2} , a_{j} ≥ 0 .

If there is no such d, then let d_{ω}(t) := ∞. We call d = d_{ω}(t) the Lorentz degree of t.

a] Let t ∈ T_{n} \ T_{n−1} be of the form

t(θ) = ∑_{j=0}^{2n} b_{j} sin^{j} \frac{ω − θ}{2} sin^{2n−j} \frac{θ + ω}{2} .

For m ≥ n let the numbers b_{j,m}, j = 0, 1, …, 2m, be defined by

t(θ) = (∑_{j=0}^{2n} b_{j} sin^{j} \frac{ω − θ}{2} sin^{2n−j} \frac{θ + ω}{2} )

× (\frac{1}{sin^{2} ω} (sin^{2} \frac{ω − θ}{2} + 2 cos ω sin \frac{ω − θ}{2} sin \frac{θ + ω}{2} + sin^{2} \frac{θ + ω}{2} ))^{m−n}

= ∑_{j=0}^{2m} b_{j,m} sin^{j} \frac{ω − θ}{2} sin^{2m−j} \frac{θ + ω}{2} .

Show that if d_{ω}(t) is finite, then it is the smallest value of m for which each b_{j,m} is nonnegative or each b_{j,m} is nonpositive.

90 2. Some Special Polynomials

Hint: The second factor in the representation of t is identically 1. ⊓ ⊔ We introduce the notation

G := {z = x + iy : −π ≤ x < π, y ∈ ℝ}

and

G_{ω} := {z = x + iy : cos ω cosh y ≥ cos x, −π ≤ x < π, y ∈ ℝ} .

b] Let t ∈ T_{1} \ T_{0}. Show that d_{ω}(t) = 1 if and only if t has its zeros in G_{ω}. c] Assume 0 < ω < π/2, t ∈ T_{1}, and t(z) = 0 for some

z := x + iy ∈ G \ (G_{ω} ∪ (−ω, ω)) .

Show that

d_{ω}(t) < max \frac{4 sin(ω ± x)(sin ω cosh y ∓ sin x)}{cos ω sinh^{2} y} − 1 ,

where the maximum is to be taken over both sets of signs. d] Suppose π/2 ≤ ω ≤ π, and t ∈ T_{1}, t(z) = 0 for some z ∈ G \ G_{ω}. Show that d_{ω}(t) = ∞. e] Show that d_{ω}(t_{1}t_{2}) ≤ d_{ω}(t_{1}) + d_{ω}(t_{2}) for any two trigonometric polynomials t_{1} and t_{2}. f] Let 0 < ω < π/2 and 0 < ϵ < ∞. Show that if t ∈ T_{n} has no zeros in

E_{ω,ϵ} := {z = x + iy : y^{2} < ϵ^{2}(ω^{2} − x^{2}) , x ∈ (−ω, ω) , y ∈ ℝ} ,

then

d_{ω}(t) ≤ n (\frac{4}{cos ω} ϵ^{−2} + 2 tan ω + 1) .

g] Let p be a trigonometric polynomial and 0 < ω < π/2. Show that d_{ω}(t) < ∞ if and only if t = 0 or t has no zeros in (−ω, ω). (Note that part d] shows that this conclusion fails to hold when π/2 ≤ ω ≤ π.) h] Show that there is an absolute constant c > 0 (independent of n, ω and z_{0}) so that d_{ω}(t) ≥ cnϵ^{−2} whenever

t(θ) = (sin \frac{θ − z_{0}}{2} sin \frac{θ − \overline{z}_{0}}{2} )^{n}

with z_{0} ∈ ∂E_{ω,ϵ} \ {−ω, ω}, 0 < ϵ < ∞, where ∂E_{ω,ϵ} denotes the boundary of E_{ω,ϵ} defined in part f].