Consequences include sufficient conditions for the coefficients to be positive, bounds on the derivatives of the polynomials, and rates of uniform convergence for the polynomial expansions of power series

http://jipam.vu.edu.au/

Volume 7, Issue 2, Article 67, 2006

AN INEQUALITY FOR CHEBYSHEV CONNECTION COEFFICIENTS

JAMES GUYKER DEPARTMENT OFMATHEMATICS

SUNY COLLEGE ATBUFFALO

1300 ELMWOODAVENUE

BUFFALO, NEWYORK14222-1095 USA

guykerj@buffalostate.edu

Received 19 January, 2006; accepted 11 March, 2006 Communicated by D. Stefanescu

ABSTRACT. Equivalent conditions are given for the nonnegativity of the coefficients of both the Chebyshev expansions and inversions of the first n polynomials defined by a certain recursion relation. Consequences include sufficient conditions for the coefficients to be positive, bounds on the derivatives of the polynomials, and rates of uniform convergence for the polynomial expansions of power series.

Key words and phrases: Chebyshev coefficient, Ultraspherical polynomial, Polynomial expansion.

2000 Mathematics Subject Classification. 41A50, 42A05, 42A32.

1. INTRODUCTION

In the classical theory of orthogonal polynomials a standard problem is to expand one set of orthogonal polynomials as a linear combination of another with nonnegative coefficients (see [4], [6], [8], [12]). R. Askey [2] and R. Szwarc [13] give general conditions ensuring non- negativity in terms of underlying recurrence relations, while Askey [1] and W. F. Trench [14]

determine when connection coefficients are positive. It is often desirable, especially in numeri- cal analysis (e.g., see [9], [10], [15]), to express orthogonal polynomials in terms of Chebyshev polynomials T_n or cosine polynomials with nonnegative coefficients. In this note, we derive inequalities (2.2) on the connection coefficients of certain Chebyshev expansions that imply positivity. Consequently, we obtain bounds on polynomial derivatives, estimate uniform con- vergence of polynomial expansions of power series, and illustrate the optimality of Chebyshev polynomials in these contexts.

For given real sequencesα_n, β_nand nonzeroγ_nwe consider the sequence of polynomialsP_n defined byP−1 = 0, P0 = 1,and forn≥0,

(1.1) xP_n=γ_nP_n+1+β_nP_n+α_nPn−1.

ISSN (electronic): 1443-5756

c 2006 Victoria University. All rights reserved.

021-06

For example the Jacobi polynomialsPn^(α,β)(α, β >−1) are defined by (1.1) with α_n = 2(α+n)(β+n)

(α+β+ 2n+ 1)(α+β+ 2n), β_n= β²−α²

(α+β+ 2n+ 2)(α+β+ 2n) and γ_n= 2(n+ 1)(α+β+n+ 1)

(α+β+ 2n+ 1)(α+β+ 2n+ 2).

In [5] (see also [3]), Askey and G. Gasper characterize those Jacobi polynomials that are combinations of Chebyshev polynomials with nonnegative coefficients. Szwarc ([13, Corol- lary 1]) gives conditions on sequences of polynomials satisfying (1.1) which imply that their Chebyshev connection coefficients are nonnegative. Our results are based on this work and the following classical expansions of the normalized ultraspherical or Gegenbauer polynomials Pn^{α} := ^Pn(α,α)

Pn^(α,α)(1).We recall the factorial function(α)_n:=α(α+ 1)· · ·(α+n−1)whenn ≥1, and(α)₀ := 1forα6= 0. ThenP^{−

1 2}

n =T_nand

(1.2) P_n^{α} =

n

X

m=0

c(n, m)T_m

α 6=−1 2

,

wherec(n, m) = 0ifn−mis odd, and otherwise c(n, m) =

(2−δ_m0) α+ ¹₂

n−m 2

α+ ¹₂

n+m 2

(2α+ 1)_n

n

n−m 2

.

It follows that if α > −¹₂ andn−m is even, thenc(n, m) > 0and the sequence hc(n, n)iis monotonically decreasing. Moreover, we have the inversions

xⁿ = [ⁿ2] X

k=0

2−δ_n,2k 2ⁿ

n k

P^{−

1 2} n−2k

([·]is the greatest integer function) and

(1.3) xⁿ=

n

X

m=0

d(n, m)P_m^{α}

α6=−1 2

,

whered(n, m) = 0ifn−mis odd, and otherwise d(n, m) =

(2α+ 1 + 2m)(2α+ 1)_m ^n−m₂ + 1

n−m 2

2ⁿ⁺¹ α+¹₂

n+m+2 2

n m

is positive andhd(n, n)iis decreasing. The polynomialsPn^{α} satisfy (1.1) for sequences such thatαn>0andβn = 0. Furthermore,α >−¹₂ if and only ifαn< ¹₂.

2. THECHEBYSHEVEXPANSION OFPn

The following characterizes those polynomials that satisfy (1.1) and enjoy similar expansions (1.2) and (1.3).

Theorem 2.1. LetP_nbe defined by (1.1) for given sequencesα_n, β_nandγ_n,and be normalized byP_n(1) = 1. Withnfixed, we have that

(2.1) P_k=

k

X

m=0

a(k, m)T_m and x^k =

k

X

m=0

b(k, m)P_m (0≤k ≤n)

for nonnegative coefficients a(k, m)and b(k, m) such that the coefficientsa(k, k)and b(k, k) are monotonically decreasing if and only if0≤α_i ≤ ¹₂ andβ_i = 0fori= 0, ..., n−1.In this case, the following properties hold.

(i) a(k, m) = b(k, m) = 0ifk−mis odd.

(ii) Ifα1 >0, thenb(k, m)>0wheneverk−mis even.

(iii) Ifn≥3andKis any subset of{0, ...,[ⁿ⁻¹₂ ]}, then

(2.2) X

n−m 2 ∈K⁰

a(n−2, m−2)≤ X

n−m 2 ∈K^∗

a(n−3, m−1)

where K⁰ := {k ∈ K : k + 1 ∈ K} ∪ {ⁿ⁻²₂ } if ⁿ⁻²₂ is inK, and K⁰ := {k ∈ K : k+ 1∈K}otherwise; andK^∗ :=K⁰∪ {k∈K :k−1∈K}.

(iv) Ifn≥2andα_i < ¹₂ fori= 1, ..., n−1, thena(k, m)>0wheneverk−mis even.

(v) Ifαi < ¹₂ fori= 1, ..., n−3, then equality holds in (2.2) if and only if eitherK is the whole set{0, ...,[ⁿ⁻¹₂ ]}itself (i.e., (2.2) is justPn−2(1) = Pn−3(1) = 1) orK⁰ andK^∗ are the empty sets.

Proof. Suppose that P0, ..., Pn satisfy (1.1) and (2.1) with nonnegative coefficients such that P_k(1) = 1, and a(k, k) and b(k, k) are decreasing (0 ≤ k ≤ n). It follows that a(0,0) = a(1,1) = 1and

(2.3) γ_ka(k+ 1, j) = −α_ka(k−1, j)−β_ka(k, j) + 1

2(a(k, j+ 1) + (1 +δ_1j)a(k, j−1)) fork = 0, ..., n−1(We use the convention that entries of vectors or matrices with any negative indices are zero; anda(i, j) = 0ifi < j. We also assumeα₀ = 0.). Similarly, by substituting (2.1) into both sides ofx^k+1 =xx^kwe haveb(0,0) = 1and

(2.4) b(k+ 1, j) =γj−1b(k, j−1) +β_jb(k, j) +α_j+1b(k, j+ 1).

With j = k + 1 we conclude γ_k, a(k, k) and b(k, k) are positive; and with j = k, β_k = a(k+ 1, k) =b(k+ 1, k) = 0fork = 0, ..., n−1. Similarly property (i) follows by induction.

By the normalization we haveγ_k+α_k = 1. Thus by (2.3) and (2.4) withj =k+ 1, it follows that ¹₂ ≤γ_k ≤1,so0≤α_k ≤ ¹₂ (0≤k ≤n−1), sincea(k, k)andb(k, k)are decreasing.

Conversely suppose thatP₀, ..., P_nare normalized and given by (1.1) for constants such that 0 ≤ α_i ≤ ¹₂ and β_i = 0 (0 ≤ i ≤ n −1). It follows that the degree of P_k is k and thus P_k satisfies (2.1) for coefficientsa(k, m)andb(k, m)generated by (2.3) and (2.4) respectively.

Sinceγ_k = 1−α_k >0, the above argument shows that property (i) holds andb(k, m)≥0; and since ¹₂ ≤γ_k ≤1,a(k, k)andb(k, k)are decreasing fork= 0, ..., n.

We will show thata(k, m)≥0simultaneously with inequality (2.2) by induction onn. Now a(0,0) = a(1,1) = 1, a(2,0) = ^1−2α_2γ ¹

1 , a(2,2) = _2γ¹

1, γ₂a(3,1) = −α₂+¹₂

+ ¹₂a(2,0)and a(3,3) = _4γ¹

1γ2.Hencea(k, m)≥0whenn is either 2 or 3. Also (2.2) is easily checked when n= 3.

Henceforth letn > 3and suppose thata(i, i−2k) ≥ 0(0≤ i ≤ n−1;0 ≤ k ≤ [₂ⁱ]) and that (2.2) is true for all integersn⁰such that3≤n⁰ < n. Letjbe given,0≤j ≤[ⁿ₂], and letK be a subset of{0, ...,[ⁿ⁻¹₂ ]}. We show thata(n, n−2j)≥0and that

(2.5) X

k∈K⁰

a(n−2, n−2k−2)≤ X

k∈K^∗

a(n−3, n−2k−1) beginning with the latter.

We may assume that K is not the whole set {0, ...,[ⁿ⁻¹₂ ]} since in the contrary case (2.5) reduces to Pn−2(1) = 1 ≤ Pn−3(1) = 1. Observe that K may be written as a disjoint union

of sets of the form{k, ..., k+m}such that there is at least one integer between any two such sets. Since K⁰ and K^∗ are then corresponding disjoint unions of subsets of these sets, and both sides of (2.5) may be separated into sums over these subsets accordingly, we may assume K ={k, ..., k+m}.

Thus we wish to show (2.6)

m

X

i=0

δ_k+i,ⁿ⁻²

2 a(n−2, n−2(k+i)−2)≤

m

X

i=0

a(n−3, n−2(k+i)−1).

By (2.3) we have

γn−3a(n−2, n−2(k+i)−2)

=−α_n−3a(n−4, n−2(k+i)−2) + 1

2(a(n−3, n−2(k+i)−1)

+ (1 +δ3,n−2(k+i))a(n−3, n−2(k+i)−3)), whereδ3,n−2(k+i) = 1if and only ifnis odd andk+i+ 1 = ⁿ⁻¹₂ =k+m(∈K). Moreover, a(n−3, n−2(k+i)−3) = 0ifnis even and ⁿ⁻²₂ =k+m. Hence the left side of (2.6) may be combined as follows:

m

X

i=0

δ_k+i,ⁿ⁻²

2 a(n−2, n−2(k+i)−2)

=−αn−3

γn−3 m

X

i=0

δ_k+i,ⁿ⁻²

2 a(n−4, n−2(k+i)−2) + 1

γn−3 m−1

X

i=1

a(n−3, n−2(k+i)−1)

+ 1

2γn−3

a(n−3, n−2k−1) +

1 +δ_k+m,[ⁿ⁻¹

2 ]

a(n−3, n−2(k+m)−1)

.

Therefore (2.6) becomes

(2.7) − αn−3

γn−3 m

X

i=0

δ_k+i,ⁿ⁻²

2 a(n−4, n−2(k+i)−2)−

m−1

X

i=1

a(n−3, n−2(k+i)−1)

!

+ 1

2γn−3

1 +δ_k+m,[ⁿ⁻¹

2 ]

a(n−3, n−2(k+m)−1)

≤ 1−2αn−3

2γn−3

a(n−3, n−2k−1) +a(n−3, n−2(k+m)−1).

Let us suppose first thatk+m 6= [ⁿ⁻¹₂ ]. Then (2.7) may be rearranged as follows:

(2.8) −α_n−3

m−1

X

i=0

a(n−4, n−2(k+i)−2)−

m−1

X

i=1

a(n−3, n−2(k+i)−1)

!

≤ 1

2 −αn−3

(a(n−3, n−2k−1) +a(n−3, n−2(k+m)−1)),

which is clearly true ifαn−3 = 0so we assumeαn−3 6= 0. Withn⁰ =n−1, replacingibyi+ 1 in the second sum, we seek to show

(2.9)

m−2

X

i=0

a(n⁰−2, n⁰ −2(k+i)−2)≤

m−1

X

i=0

a(n⁰−3, n⁰−2(k+i)−1).

Sincek+m < [ⁿ⁻¹₂ ], we haveK⁰ =

k, ..., k+m−1} ⊆ {0, ...,[ⁿ⁰₂⁻¹] .Moreover,(K⁰)⁰ = {k, ..., k+m−2}where the second prime is with respect ton⁰. Hence (2.9) follows from the induction hypothesis

(2.10) X

k∈(K⁰)⁰

a(n⁰−2, n⁰ −2k−2)≤ X

k∈(K⁰)^∗

a(n⁰ −3, n−2k−1).

Finally suppose thatk+m = [ⁿ⁻¹₂ ]so that (2.7) becomes

(2.11) −αn−3 m

X

i=0

δ_k+i,ⁿ⁻²

2 a(n−4, n−2(k+i)−2)−

m

X

i=1

a(n−3, n−2(k+i)−1)

!

≤ 1

2 −α_n−3

a(n−3, n−2k−1).

As above we may assumeαn−3 6= 0and wish to show

(2.12)

m−1

X

i=0

a(n⁰−2, n⁰−2(k+i)−2)≤

m

X

i=0

δ_k+i,ⁿ⁻²

2 a(n⁰ −3, n⁰−2(k+i)−1).

If n is even then n⁰ is odd and (K⁰)⁰ (with respect ton⁰) is {k, ..., k +m −1}. Thus (2.12) is a consequence of the induction hypothesis as above. On the other hand, if n is odd, then (K⁰)⁰ ={k, ..., k+m−1}and (2.12) again reduces to the hypothesis.

Next we verify a(n, n−2j) ≥ 0. Sincea(n, n) = Qn j=2

1

2γn−j+1 > 0 suppose further that j ≥1. Let

K₀ :=

k :kinteger,0≤k≤

n−1 2

, k 6=j−1, j

.

Ifnis even andj = ⁿ⁻²₂ , defineK₁ :=K₀; otherwise let K1 :=

k :kinteger,0≤k≤

n−2 2

, k 6=j−1

.

Note thata(n−2, n−2k−2) = 0ifk = [ⁿ⁻¹₂ ]>[ⁿ⁻²₂ ]. Furthermore, the sumP

k /∈K0 k+1∈K0

a(n− 2, n−2k−2)is zero unlessj + 1is inK₀ (in which case the sum isa(n −2, n−2j −2)).

Solving the equationsPn−2(1) = 1andPn−1(1) = 1fora(n−2, n−2j)anda(n−1, n−2j+ 1) +a(n−1, n−2j−1)respectively, and substituting them intoγn−1a(n, n−2j)as given by

(2.3), we have the following identities.

γn−1a(n, n−2j)

= 1

2−αn−1+ X

k∈K1

αn−1− 1 2

1− 1−2αn−2

2γn−2

a(n−2, n−2k−2) +αn−1δ_n,2j+2a(n−2,0) + αn−2

2γ_n−2 X

k∈K0

a(n−3, n−2k−1)

− 1 2γn−2

αn−2+ 1−2α_n−2 2

X

k∈K₀⁰

a(n−2, n−2k−2)

+ 1

2δ_1,n−2ja(n−1, n−2j−1)

= 1

2 −αn−1

X

k /∈K1

a(n−2, n−2k−2)

+ 1

2γn−2

1

2 −αn−2

X

k∈K1 k /∈K0 0

a(n−2, n−2k−2)

+αn−1δn,2j+2a(n−2,0) + 1

2δ1,n−2ja(n−1, n−2j−1) + αn−2

2γn−2



 X

k∈K0

a(n−3, n−2k−1)− X

k∈K⁰₀

a(n−2, n−2k−2)



.

Each of the terms in the last expression is nonnegative, the final one a result of the induction assumption on (2.2). Thereforea(n, n−2j)≥0.

Property (i) has already been shown so for (ii), assume α₁ > 0. Since b(0,0) = 1 and b(k+ 1,0) =α₁γ₀b(k−1,0) +α₁α₂b(k−1,2),we have thatb(k+ 1,0)>0fork+ 1even.

But also b(k + 1,0) = α1b(k,1)so b(k,1) > 0 for oddk. Hence (ii) is now straightforward from (2.4).

For property (iv), supposeα_i < ¹₂ (i = 1, ..., n−1).By the induction argument above, the first term of the last identity forγn−1a(n, n−2j)is positive sincek =j −1 ∈/ K1. Since the other terms are nonnegative,a(n, n−2j)>0.

Next letα_i < ¹₂ (i = 1, ..., n−3)and suppose that equality holds in (2.2) for some subset K of{0, ...,[ⁿ⁻¹₂ ]}. As above, it follows that equality holds in (2.2) over each subset of K of the form{k, ..., k+m}. Hence assumeK = {k, ..., k+m}. Ifm = 0andk 6= ⁿ⁻²₂ thenK⁰ and K^∗ are empty. If K = {ⁿ⁻²₂ }, then by equality in (2.11) we have−αn−3a(n −4,0) =

1

2 −αn−3

a(n−3,1)which is impossible since the left side is nonpositive and the right side is positive by property (iv).

Therefore, letm≥ 1. Ifk+m 6= [ⁿ⁻¹₂ ]then equality holds in (2.8), and by (2.9) and (2.10) we have

−αn−3



 X

k∈(K⁰)^∗

a(n⁰−3, n⁰ −2k−1)− X

k∈(K⁰)⁰

a(n⁰−2, n⁰−2k−2)





= 1

2 −αn−3

(a(n−3, n−2k−1) +a(n−3, n−2(k+m)−1)).

This is impossible since both sides must be zero which impliesa(n−3, n−2(k+m)−1) = 0.

Finally, ifk+m = [ⁿ⁻¹₂ ]then equality holds in (2.11) and a similar argument showsa(n− 3, n−2k−1) = 0 which by property (iv) implies k = 0and thus K must be the whole set

{0, ...,[ⁿ⁻¹₂ ]}.

3. BOUNDS ONPOLYNOMIAL DERIVATIVES

It is well known [9] thatT_nis bounded by one and for1≤k ≤n T_n^(k)(x)

≤T_n^(k)(1) = n²(n²−1)(n²−4)· · ·(n²−(k−1)²) 1·3·5· · ·(2k−1)

when −1 ≤ x ≤ 1; and if 1 ≤ k < n and

Tn^(k)(x)

= Tn^(k)(1), then x = ±1. More generally, it follows from a result of R.J. Duffin and A.C. Schaeffer ([7], [9, Thm. 2.24]) that if P_n is any polynomial of degree n that is bounded by one in [−1,1] and 1 ≤ k < n, then

Pn^(k)(x)

≤ Tn^(k)(1) with equality holding only when P_n = ±T_n and x = ±1. For the polynomials in Theorem 2.1, we may be more precise.

Corollary 3.1. LetP₀, ..., P_nbe defined by (1.1) with0≤α_i ≤ ¹₂, β_i = 0,andγ_i = 1−α_ifor i= 0, ..., n−1.Then

(a) |P_n(x)| ≤ 1 = P_n(1) = T_n(1); and if n ≥ 1, |P_n(x)| = 1, and α_i < ¹₂ for i = 1, ..., n−1, thenx=±1.

(b) If1≤k ≤n,then

P_n^(k)(x)

≤P_n^(k)(1)≤T_n^(k)(1) for allxin[−1,1]. Moreover in this case:

(i) Ifk < nand

Pn^(k)(x)

=Pn^(k)(1),thenx=±1.

(ii) If Pn^(k)(1) = Tn^(k)(1), then P_n = T_n. In particular, if n ≥ 2 and α_i < ¹₂ (i = 1, ..., n−1), thenPn^(k)(1) < Tn^(k)(1).

Proof. By Theorem 2.1,P_n=Pn

m=0a(n, m)T_m wherea(n, m)≥0. Therefore

|P_n(x)| ≤

n

X

m=0

a(n, m)|T_m(x)| ≤P_n(1) = 1.

Suppose that n ≥ 1 and|P_n(x)| = 1. Ifn = 1, thenx = ±1,so assume n ≥ 2andα_i < ¹₂ (i= 1, ..., n−1). ThenP_n(x) =±P_n(1)and hencea(n, m)(1±T_m(x)) = 0for allm. Since T₀ = 1, T₁ =xandT₂ = 2x²−1, property (iv) of Theorem 2.1 implies thatx=±1.

Next assumek ≥1. SinceD

Tm^(k)(1)E

m

is increasing, P_n^(k)(x)

=

n

X

m=k

a(n, m)T_m^(k)(x)

≤

n

X

m=k

a(n, m)

T_m^(k)(x)

≤

n

X

m=k

a(n, m)T_m^(k)(1) =P_n^(k)(1)

≤T_n^(k)(1)

n

X

m=k

a(n, m)≤T_n^(k)(1).

Assume1≤k < nand

Pn^(k)(x)

=Pn^(k)(1). Then 0 =

n

X

m=k

a(n, m) T_m^(k)(1)−

T_m^(k)(x)

,

where each term is nonnegative. Sincea(n, n)>0we have that

Tn^(k)(x)

=Tn^(k)(1)sox=±1.

Finally suppose thatk ≥1andPn^(k)(1) =Tn^(k)(1). ThenPn

m=ka(n, m) = 1soa(n, m) = 0 for m < k. Also a(n, m)Tm^(k)(1) = a(n, m)Tn^(k)(1) so a(n, m) = 0 for m = k, ..., n− 1.

ThereforeP_n =a(n, n)T_nand thusa(n, n) = 1.

However the previous case is impossible if n ≥ 2and α_i < ¹₂ (i = 1, ..., n−1) since by property (iv) of Theorem 2.1 we would havea(n,0)>0ora(n,1)>0.

Remark 3.2. For fixed k the sequence Pn^(k)(1) of bounds is increasing. In fact by (1.1), Pn^(k)(1) may be generated recursively as follows: Initially we have P₁⁽¹⁾(1) = 1, P_k^(k)(1) =

k

1−αk−1P_k−1^(k−1)(1) (k ≥2)and sete_kk:= ^k+α_1−α^k

kP_k^(k)(1).Then forn ≥1, P_n+1^(k)(1) =P_n^(k)(1) +e_nk ≥P_n^(k)(1) ≥0, where the differencese_nkare defined by

e_nk := α_n 1−αn

en−1,k+ k 1−αn

P_n^(k−1)(1).

Ultraspherical polynomialsy=Pn^{α} withα ≥ −¹₂ satisfy the differential equation (1−x²)y⁰⁰−2(α+ 1)xy⁰+n(n+ 2α+ 1)y= 0

and thus a closed form forPn^(k)(1)is possible in this case since

P_n^(k)(1) = n(n+ 2α+ 1)−(k−1)(2α+ 1)−(k−1)²

2(α+k) P_n^(k−1)(1).

This extends known Chebyshev and Legendre identities ([9, p. 33], [11, p. 251]).

4. POLYNOMIAL EXPANSIONS OFPOWER SERIES

A standard application of the theory of orthogonal polynomials is the least squares or uni- form approximation of functions by partial sums of generalized Fourier expansions in terms of orthogonal polynomials, especially Chebyshev polynomials. The coefficients of the expansion are given by an inner product used in generating the polynomials. In our case, we may define Fourier coefficients for expansions of power series in terms of the polynomials that satisfy (1.1):

LetP

aixⁱ be a convergent power series on(−1,1), and for everynletPn be a polynomial of degreen. Thenxⁿ = Pn

m=0b(n, m)P_m for some numbersb(n, m); and we define the Fourier coefficientcj ofP

aixⁱ with respect to the sequencehPniby cj :=X

i

aib(i, j)

whenever this sum converges. Note thatc_nj := Pn

i=0a_ib(i, j)is then thejth coefficient in the expansion of the partial sumPn

i=0a_ixⁱ: sinceb(i, j) = 0forj > i,

n

X

i=0

a_ixⁱ =

n

X

i=0

a_i

n

X

j=0

b(i, j)P_j =

n

X

j=0

c_njP_j.

We have the following estimate where kfkdenotes the uniform norm max{|f(x)| : −1 ≤ x ≤ 1}. The optimal property of Chebyshev expansions extends a result of T.J. Rivlin and M.W. Wilson ([10], [9, Thm. 3.17]).

Corollary 4.1. LethP_nibe given by (1.1) with hα_ni, hβ_niandhγ_ninonnegative, and suppose thatP_n(1) = 1 for all n. If P

a_i converges absolutely, then the coefficientc_j of P

a_ixⁱ with respect tohPniexists for everyjand

c_j −

n

X

i=0

a_ib(i, j)

≤ X

i>n i−jeven

|a_i|.

Moreover, ifP

ia_iconverges absolutely, then

Xa_ixⁱ−

n

X

j=0

c_jP_j(x)

≤X

i>n

(|a_i|+|ia_i|)

for allxin[−1,1]. In this case, ifα_i ≤ ¹₂ andβ_i = 0fori= 0, ..., n−1, and ifa_i ≥0for alli anddkis thekth Chebyshev coefficient ofP

aixⁱ, then

(4.1)

Xa_ixⁱ−

n

X

j=0

c_jP_j

≥

Xa_ixⁱ−

n

X

k=0

d_kT_k .

In addition, ifα_i < ¹₂ (i = 1, ..., n−1), then equality holds in (4.1) if and only ifP

a_ixⁱ is a polynomial of degree at mostn.

Proof. Asume thathα_ni, hβ_niandhγ_niare nonnegative, andP_n(1) = 1for alln. By (1.1) the degree of P_n is n for all n. Thusxⁿ = Pn

m=0b(n, m)P_m where b(n, m) ≥ 0 by (2.4), and b(n, m)≤1by the normalization since we have 1ⁿ =Pn

m=0b(n, m).

Suppose thatP

|a_i|converges. Then P

ia_ib(i, j) converges absolutely by the comparison test soc_j exists and

X

i>n

a_ib(i, j)

≤ X

i>n i−jeven

|a_i|.

Next assumeP|ia_i| < ∞.Then P|a_i| < ∞soc_j exists for everyj. Thus by Corollary 3.1 we have

Xa_ixⁱ−

n

X

j=0

c_jP_j(x)

≤

X

i>n

a_ixⁱ

+

n

X

j=0

(c_j −c_nj)P_j(x)

≤X

i>n

|a_i|+

n

X

j=0

|c_j −c_nj|,

where

n

X

j=0

|c_j−c_nj|=

n

X

j=0

X

i>n

a_ib(i, j)

≤

n

X

j=0

X

i≥n+1

1

i |ia_i| ≤ n+ 1 n+ 1

X

i>n

|ia_i|.

Suppose further that α_i ≤ ¹₂, β_i = 0(i = 0, ..., n−1) anda_i is nonnegative for all i. Then c_j ≥ 0for all j and by Theorem 2.1, P_j = Pj

k=0a(j, k)T_k wherea(j, k) ≥ 0.Since we also

haveP

a_ixⁱ =P

c_jP_j uniformly on[−1,1]in this case, it follows that

Xa_ixⁱ−

n

X

j=0

c_jP_j

=

X

j>n

c_jP_j

=X

j>n

c_j

and similarly

Xa_ixⁱ−

n

X

k=0

d_kT_k

=X

k>n

d_k.

But

Xd_kT_k =X a_ixⁱ

=X

cjPj

=X

j

cj

∞

X

k=0

a(j, k)Tk

=X

k

X

j≥k

cja(j, k)

! Tk

since

X

j≥k

c_ja(j, k)

≤X

j≥k

c_j ≤X

c_jP_j(1) =X

a_i <∞.

Since the coefficients in a uniformly convergent Chebyshev expansion are unique, d_k = P

j≥kcja(j, k).Therefore X

k>n

dk=X

j>n

cj

X

k>n

a(j, k)

=X

j>n

c_j

j

X

k=0

a(j, k)−

n

X

k=0

a(j, k)

!

=X

j>n

c_j −

n

X

k=0

X

j>n

c_ja(j, k)

≤X

j>n

c_j.

Finally, assume that α_i < ¹₂ (i = 1, ..., n−1). By property (iv) of Theorem 2.1, a(j,0) + a(j,1)> 0forj > nso if equality holds in the last inequality thencj = 0for allj > n. Thus Paixⁱ =P

cjPj is a polynomial of degree at mostn.

Remark 4.2. If hP_ni is defined as in Corollary 4.1 and, more generally, P

(ia_i)² converges, then by the Schwarz inequality it follows that

X

i>n

|a_i| ≤ X

i>n

(ia_i)²

!¹₂ X

i>n

1 i²

!¹₂

soP|a_i|<∞andc_j exists for everyj. Moreover in this case we have

Xa_ixⁱ−

n

X

j=0

c_jP_j(x)

≤

X

i>n

a_ixⁱ

+

n

X

j=0

X

i>n

a_ib(i, j)

≤ X

i>n

(ia_i)²

!¹₂ X

i>n

xⁱ i

2!¹₂ +

n

X

j=0

X

i>n

(ia_i)²

!¹₂ X

i>n

b(i, j) i

2!¹₂

≤(n+ 2) X

i>n

1 i²

!¹₂ X

i>n

(ia_i)²

!¹₂

≤ n+ 2

√n

X

i>n

(ia_i)²

!¹₂

where the last inequality follows from the proof of the integral test.

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