Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 2, Article 67, 2006
AN INEQUALITY FOR CHEBYSHEV CONNECTION COEFFICIENTS
JAMES GUYKER
D EPARTMENT OF M ATHEMATICS
SUNY C OLLEGE AT B UFFALO
1300 E LMWOOD AVENUE
B UFFALO , N EW YORK 14222-1095
USA
guykerj@buffalostate.edu
Received 19 January, 2006; accepted 11 March, 2006
Communicated by D. Stefanescu
A BSTRACT. 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. I NTRODUCTION
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 nonnegativity 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 numerical analysis (e.g., see [9], [10], [15]), to express orthogonal polynomials in terms of Chebyshev
polynomials Tn 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 convergence of polynomial expansions of power series, and illustrate the optimality of Chebyshev
polynomials in these contexts.
For given real sequences αn , βn and nonzero γn we consider the sequence of polynomials Pn
defined by P−1 = 0, P0 = 1, and for n ≥ 0,
(1.1)
xPn = γn Pn+1 + βn Pn + αn Pn−1 .
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
021-06
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JAMES G UYKER
(α,β)
For example the Jacobi polynomials Pn
αn =
(α, β > −1) are defined by (1.1) with
β 2 − α2
2(α + n)(β + n)
,
βn =
(α + β + 2n + 1)(α + β + 2n)
(α + β + 2n + 2)(α + β + 2n)
2(n + 1)(α + β + n + 1)
.
and
γn =
(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, Corollary 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
(α,α)
{α}
n
Pn := P(α,α)
. We recall the factorial function (α)n := α(α + 1) · · · (α + n − 1) when n ≥ 1,
Pn
(1)
{− 12 }
and (α)0 := 1 for α 6= 0. Then Pn
Pn{α}
(1.2)
=
= Tn and
n
X
c(n, m)Tm
m=0
1
α 6= −
2
,
where c(n, m) = 0 if n − m is odd, and otherwise
(2 − δm0 ) α + 21 n−m α + 12 n+m n 2
2
c(n, m) =
n−m .
(2α + 1)n
2
It follows that if α > − 21 and n − m is even, then c(n, m) > 0 and the sequence hc(n, n)i is
monotonically decreasing. Moreover, we have the inversions
xn =
[ n2 ]
X
2 − δn,2k n
k=0
2n
k
{− 1 }
2
Pn−2k
([·] is the greatest integer function) and
n
x =
(1.3)
n
X
d(n, m)Pm{α}
m=0
1
α 6= −
2
,
where d(n, m) = 0 if n − m is odd, and otherwise
(2α + 1 + 2m)(2α + 1)m n−m
+ 1 n−m n 2
2
d(n, m) =
1
n+1
m
2
α + 2 n+m+2
2
{α}
is positive and hd(n, n)i is decreasing. The polynomials Pn satisfy (1.1) for sequences such
that αn > 0 and βn = 0. Furthermore, α > − 12 if and only if αn < 12 .
2. T HE C HEBYSHEV E XPANSION OF Pn
The following characterizes those polynomials that satisfy (1.1) and enjoy similar expansions
(1.2) and (1.3).
Theorem 2.1. Let Pn be defined by (1.1) for given sequences αn , βn and γn , and be normalized
by Pn (1) = 1. With n fixed, we have that
(2.1)
Pk =
k
X
a(k, m)Tm
m=0
J. Inequal. Pure and Appl. Math., 7(2) Art. 67, 2006
and
k
x =
k
X
b(k, m)Pm
(0 ≤ k ≤ n)
m=0
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A N I NEQUALITY
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C HEBYSHEV C ONNECTION C OEFFICIENTS
3
for nonnegative coefficients a(k, m) and b(k, m) such that the coefficients a(k, k) and b(k, k)
are monotonically decreasing if and only if 0 ≤ αi ≤ 21 and βi = 0 for i = 0, ..., n − 1. In this
case, the following properties hold.
(i) a(k, m) = b(k, m) = 0 if k − m is odd.
(ii) If α1 > 0, then b(k, m) > 0 whenever k − m is even.
(iii) If n ≥ 3 and K is any subset of {0, ..., [ n−1
]}, then
2
X
X
(2.2)
a(n − 2, m − 2) ≤
a(n − 3, m − 1)
n−m
∈K 0
2
n−m
∈K ∗
2
where K 0 := {k ∈ K : k + 1 ∈ K} ∪ { n−2
} if n−2
is in K, and K 0 := {k ∈ K :
2
2
∗
0
k + 1 ∈ K} otherwise; and K := K ∪ {k ∈ K : k − 1 ∈ K}.
(iv) If n ≥ 2 and αi < 21 for i = 1, ..., n − 1, then a(k, m) > 0 whenever k − m is even.
(v) If αi < 21 for i = 1, ..., n − 3, then equality holds in (2.2) if and only if either K is the
whole set {0, ..., [ n−1
]} itself (i.e., (2.2) is just Pn−2 (1) = Pn−3 (1) = 1) or K 0 and K ∗
2
are the empty sets.
Proof. Suppose that P0 , ..., Pn satisfy (1.1) and (2.1) with nonnegative coefficients such that
Pk (1) = 1, and a(k, k) and b(k, k) are decreasing (0 ≤ k ≤ n). It follows that a(0, 0) =
a(1, 1) = 1 and
1
(2.3) γk a(k + 1, j) = −αk a(k − 1, j) − βk a(k, j) + (a(k, j + 1) + (1 + δ1j )a(k, j − 1))
2
for k = 0, ..., n − 1 (We use the convention that entries of vectors or matrices with any negative
indices are zero; and a(i, j) = 0 if i < j. We also assume α0 = 0.). Similarly, by substituting
(2.1) into both sides of xk+1 = xxk we have b(0, 0) = 1 and
(2.4)
b(k + 1, j) = γj−1 b(k, j − 1) + βj b(k, j) + αj+1 b(k, j + 1).
With j = k + 1 we conclude γk , a(k, k) and b(k, k) are positive; and with j = k, βk =
a(k + 1, k) = b(k + 1, k) = 0 for k = 0, ..., n − 1. Similarly property (i) follows by induction.
By the normalization we have γk + αk = 1. Thus by (2.3) and (2.4) with j = k + 1, it follows
that 21 ≤ γk ≤ 1, so 0 ≤ αk ≤ 12 (0 ≤ k ≤ n − 1), since a(k, k) and b(k, k) are decreasing.
Conversely suppose that P0 , ..., Pn are normalized and given by (1.1) for constants such that
0 ≤ αi ≤ 21 and βi = 0 (0 ≤ i ≤ n − 1). It follows that the degree of Pk is k and thus
Pk satisfies (2.1) for coefficients a(k, m) and b(k, m) generated by (2.3) and (2.4) respectively.
Since γk = 1 − αk > 0, the above argument shows that property (i) holds and b(k, m) ≥ 0; and
since 21 ≤ γk ≤ 1, a(k, k) and b(k, k) are decreasing for k = 0, ..., n.
We will show that a(k, m) ≥ 0 simultaneously with inequality (2.2) by induction
1 on n. Now
1
1
1
a(0, 0) = a(1, 1) = 1, a(2, 0) = 1−2α
,
a(2,
2)
=
,
γ
a(3,
1)
=
−α
+
+ 2 a(2, 0) and
2
2
2γ1
2γ1
2
1
a(3, 3) = 4γ1 γ2 . Hence a(k, m) ≥ 0 when n is either 2 or 3. Also (2.2) is easily checked when
n = 3.
Henceforth let n > 3 and suppose that a(i, i − 2k) ≥ 0 (0 ≤ i ≤ n − 1; 0 ≤ k ≤ [ 2i ]) and
that (2.2) is true for all integers n0 such that 3 ≤ n0 < n. Let j be given, 0 ≤ j ≤ [ n2 ], and let K
]}. We show that a(n, n − 2j) ≥ 0 and that
be a subset of {0, ..., [ n−1
2
X
X
(2.5)
a(n − 2, n − 2k − 2) ≤
a(n − 3, n − 2k − 1)
k∈K 0
k∈K ∗
beginning with the latter.
We may assume that K is not the whole set {0, ..., [ n−1
]} since in the contrary case (2.5)
2
reduces to Pn−2 (1) = 1 ≤ Pn−3 (1) = 1. Observe that K may be written as a disjoint union
J. Inequal. Pure and Appl. Math., 7(2) Art. 67, 2006
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4
JAMES G UYKER
of sets of the form {k, ..., k + m} such that there is at least one integer between any two such
sets. Since K 0 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
δk+i, n−2 a(n − 2, n − 2(k + i) − 2) ≤
m
X
2
i=0
a(n − 3, n − 2(k + i) − 1).
i=0
By (2.3) we have
γn−3 a(n − 2, n − 2(k + i) − 2)
1
= −αn−3 a(n − 4, n − 2(k + i) − 2) + (a(n − 3, n − 2(k + i) − 1)
2
+ (1 + δ3,n−2(k+i) )a(n − 3, n − 2(k + i) − 3)),
where δ3,n−2(k+i) = 1 if and only if n is odd and k + i + 1 = n−1
= k + m (∈ K). Moreover,
2
a(n − 3, n − 2(k + i) − 3) = 0 if n is even and n−2
=
k
+
m.
Hence
the left side of (2.6) may
2
be combined as follows:
m
X
δk+i, n−2 a(n − 2, n − 2(k + i) − 2)
2
i=0
m
αn−3 X
=−
δ n−2 a(n − 4, n − 2(k + i) − 2)
γn−3 i=0 k+i, 2
+
+
1
m−1
X
γn−3
i=1
a(n − 3, n − 2(k + i) − 1)
1 a(n − 3, n − 2k − 1) + 1 + δk+m,[ n−1 ] a(n − 3, n − 2(k + m) − 1) .
2
2γn−3
Therefore (2.6) becomes
(2.7)
αn−3
−
γn−3
m
X
δk+i, n−2 a(n − 4, n − 2(k + i) − 2) −
m−1
X
2
i=0
!
a(n − 3, n − 2(k + i) − 1)
i=1
1 +
1 + δk+m,[ n−1 ] a(n − 3, n − 2(k + m) − 1)
2
2γn−3
1 − 2αn−3
≤
a(n − 3, n − 2k − 1) + a(n − 3, n − 2(k + m) − 1).
2γn−3
Let us suppose first that k + m 6= [ n−1
]. Then (2.7) may be rearranged as follows:
2
(2.8)
− αn−3
m−1
X
a(n − 4, n − 2(k + i) − 2) −
i=0
m−1
X
!
a(n − 3, n − 2(k + i) − 1)
i=1
≤
1
− αn−3 (a(n − 3, n − 2k − 1) + a(n − 3, n − 2(k + m) − 1)),
2
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A N I NEQUALITY
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which is clearly true if αn−3 = 0 so we assume αn−3 6= 0. With n0 = n − 1, replacing i by i + 1
in the second sum, we seek to show
(2.9)
m−2
X
0
0
a(n − 2, n − 2(k + i) − 2) ≤
i=0
m−1
X
a(n0 − 3, n0 − 2(k + i) − 1).
i=0
0
Since k + m < [ n−1
], we have K 0 = k, ..., k + m − 1} ⊆ {0, ..., [ n 2−1 ] . Moreover, (K 0 )0 =
2
{k, ..., k + m − 2} where the second prime is with respect to n0 . Hence (2.9) follows from the
induction hypothesis
X
(2.10)
a(n0 − 2, n0 − 2k − 2) ≤
k∈(K 0 )0
X
a(n0 − 3, n − 2k − 1).
k∈(K 0 )∗
Finally suppose that k + m = [ n−1
] so that (2.7) becomes
2
(2.11)
− αn−3
m
X
δk+i, n−2 a(n − 4, n − 2(k + i) − 2) −
m
X
2
i=0
!
a(n − 3, n − 2(k + i) − 1)
i=1
≤
1
− αn−3 a(n − 3, n − 2k − 1).
2
As above we may assume αn−3 6= 0 and wish to show
(2.12)
m−1
X
a(n0 − 2, n0 − 2(k + i) − 2) ≤
i=0
m
X
δk+i, n−2 a(n0 − 3, n0 − 2(k + i) − 1).
2
i=0
If n is even then n0 is odd and (K 0 )0 (with respect to n0 ) 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
(K 0 )0 = {k, ..., k + m − 1} and (2.12) again reduces to Q
the hypothesis.
1
> 0 suppose further that
Next we verify a(n, n − 2j) ≥ 0. Since a(n, n) = nj=2 2γn−j+1
j ≥ 1. Let
n−1
K0 := k : k integer, 0 ≤ k ≤
, k 6= j − 1, j .
2
If n is even and j =
n−2
,
2
define K1 := K0 ; otherwise let
n−2
K1 := k : k integer, 0 ≤ k ≤
, k 6= j − 1 .
2
Note that a(n − 2, n − 2k − 2) = 0 if k = [ n−1
] > [ n−2
]. Furthermore, the sum
2
2
P
k∈K
/ 0
k+1∈K0
a(n −
2, n − 2k − 2) is zero unless j + 1 is in K0 (in which case the sum is a(n − 2, n − 2j − 2)).
Solving the equations Pn−2 (1) = 1 and Pn−1 (1) = 1 for a(n − 2, n − 2j) and a(n − 1, n − 2j +
1) + a(n − 1, n − 2j − 1) respectively, and substituting them into γn−1 a(n, n − 2j) as given by
J. Inequal. Pure and Appl. Math., 7(2) Art. 67, 2006
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6
JAMES G UYKER
(2.3), we have the following identities.
γn−1 a(n, n − 2j)
X
1
1 − 2αn−2
1
αn−1 −
1−
a(n − 2, n − 2k − 2)
= − αn−1 +
2
2
2γn−2
k∈K1
αn−2 X
+ αn−1 δn,2j+2 a(n − 2, 0) +
a(n − 3, n − 2k − 1)
2γn−2 k∈K
0
X
1 − 2αn−2
1
−
αn−2 +
a(n − 2, n − 2k − 2)
2γn−2
2
0
k∈K0
1
+ δ1,n−2j a(n − 1, n − 2j − 1)
2
X
1
=
− αn−1
a(n − 2, n − 2k − 2)
2
k∈K
/ 1
X
1
1
− αn−2
a(n − 2, n − 2k − 2)
+
2γn−2 2
k∈K
1
0
k∈K
/ 0
1
+ αn−1 δn,2j+2 a(n − 2, 0) + δ1,n−2j a(n − 1, n − 2j − 1)
2


X
X
αn−2 
+
a(n − 3, n − 2k − 1) −
a(n − 2, n − 2k − 2) .
2γn−2 k∈K
0
k∈K0
0
Each of the terms in the last expression is nonnegative, the final one a result of the induction
assumption on (2.2). Therefore a(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) = α1 γ0 b(k − 1, 0) + α1 α2 b(k − 1, 2), we have that b(k + 1, 0) > 0 for k + 1 even.
But also b(k + 1, 0) = α1 b(k, 1) so b(k, 1) > 0 for odd k. Hence (ii) is now straightforward
from (2.4).
For property (iv), suppose αi < 21 (i = 1, ..., n − 1). By the induction argument above, the
first term of the last identity for γn−1 a(n, n − 2j) is positive since k = j − 1 ∈
/ K1 . Since the
other terms are nonnegative, a(n, n − 2j) > 0.
Next let αi < 21 (i = 1, ..., n − 3) and suppose that equality holds in (2.2) for some subset
K of {0, ..., [ n−1
]}. As above, it follows that equality holds in (2.2) over each subset of K of
2
then K 0
the form {k, ..., k + m}. Hence assume K = {k, ..., k + m}. If m = 0 and k 6= n−2
2
and K ∗ are empty. If K = { n−2
}, then by equality in (2.11) we have −αn−3 a(n − 4, 0) =
2
1
− αn−3 a(n − 3, 1) which is impossible since the left side is nonpositive and the right side
2
is positive by property (iv).
Therefore, let m ≥ 1. If k + m 6= [ n−1
] then equality holds in (2.8), and by (2.9) and (2.10)
2
we have


X
X
− αn−3 
a(n0 − 3, n0 − 2k − 1) −
a(n0 − 2, n0 − 2k − 2)
k∈(K 0 )∗
k∈(K 0 )0
=
1
− αn−3 (a(n − 3, n − 2k − 1) + a(n − 3, n − 2(k + m) − 1)).
2
This is impossible since both sides must be zero which implies a(n − 3, n − 2(k + m) − 1) = 0.
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A N I NEQUALITY
FOR
C HEBYSHEV C ONNECTION C OEFFICIENTS
7
Finally, if k + m = [ n−1
] then equality holds in (2.11) and a similar argument shows a(n −
2
3, n − 2k − 1) = 0 which by property (iv) implies k = 0 and thus K must be the whole set
{0, ..., [ n−1
]}.
2
3. B OUNDS ON P OLYNOMIAL D ERIVATIVES
It is well known [9] that Tn is bounded by one and for 1 ≤ k ≤ n
2
2
2
2
2
(k) Tn (x) ≤ Tn(k) (1) = n (n − 1)(n − 4) · · · (n − (k − 1) )
1 · 3 · 5 · · · (2k − 1)
(k) (k)
when −1 ≤ x ≤ 1; and if 1 ≤ k < n and Tn (x) = Tn (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 Pn is any
polynomial of degree n that is bounded by one in [−1, 1] and 1 ≤ k < n,
(k) (k)
then Pn (x) ≤ Tn (1) with equality holding only when Pn = ±Tn and x = ±1. For the
polynomials in Theorem 2.1, we may be more precise.
Corollary 3.1. Let P0 , ..., Pn be defined by (1.1) with 0 ≤ αi ≤ 12 , βi = 0, and γi = 1 − αi for
i = 0, ..., n − 1. Then
(a) |Pn (x)| ≤ 1 = Pn (1) = Tn (1); and if n ≥ 1, |Pn (x)| = 1, and αi < 12 for i =
1, ..., n − 1, then x = ±1.
(b) If 1 ≤ k ≤ n, then
(k) Pn (x) ≤ Pn(k) (1) ≤ Tn(k) (1)
for all x in [−1, 1]. Moreover
in this case:
(k) (k)
(i) If k < n and Pn (x) = Pn (1), then x = ±1.
(k)
(k)
(ii) If Pn (1) = Tn (1), then Pn = Tn . In particular, if n ≥ 2 and αi <
(k)
(k)
1, ..., n − 1), then Pn (1) < Tn (1).
P
Proof. By Theorem 2.1, Pn = nm=0 a(n, m)Tm where a(n, m) ≥ 0. Therefore
|Pn (x)| ≤
n
X
1
2
(i =
a(n, m) |Tm (x)| ≤ Pn (1) = 1.
m=0
Suppose that n ≥ 1 and |Pn (x)| = 1. If n = 1, then x = ±1, so assume n ≥ 2 and αi < 21
(i = 1, ..., n − 1). Then Pn (x) = ±Pn (1) and hence a(n, m)(1 ± Tm (x)) = 0 for all m. Since
T0 = 1, T1 = x and T2 = 2x2D− 1, property
(iv) of Theorem 2.1 implies that x = ±1.
E
(k)
Next assume k ≥ 1. Since Tm (1)
is increasing,
m
n
(k) X
Pn (x) = a(n, m)Tm(k) (x)
m=k
≤
≤
n
X
m=k
n
X
a(n, m) Tm(k) (x)
a(n, m)Tm(k) (1) = Pn(k) (1)
m=k
≤
Tn(k) (1)
n
X
a(n, m) ≤ Tn(k) (1).
m=k
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8
JAMES G UYKER
(k) (k)
Assume 1 ≤ k < n and Pn (x) = Pn (1). Then
0=
n
X
a(n, m) Tm(k) (1) − Tm(k) (x) ,
m=k
(k) (k)
where each term is nonnegative. Since a(n, n) > 0 we have that Tn (x) = Tn (1) so x = ±1.
P
(k)
(k)
Finally suppose that k ≥ 1 and Pn (1) = Tn (1). Then nm=k a(n, m) = 1 so a(n, m) = 0
(k)
(k)
for m < k. Also a(n, m)Tm (1) = a(n, m)Tn (1) so a(n, m) = 0 for m = k, ..., n − 1.
Therefore Pn = a(n, n)Tn and thus a(n, n) = 1.
However the previous case is impossible if n ≥ 2 and αi < 12 (i = 1, ..., n − 1) since by
property (iv) of Theorem 2.1 we would have a(n, 0) > 0 or a(n, 1) > 0.
(k)
Remark 3.2. For fixed k the sequence Pn (1) of bounds is increasing. In fact by (1.1),
(k)
(1)
(k)
Pn (1) may be generated recursively as follows: Initially we have P1 (1) = 1, Pk (1) =
(k−1)
(k)
k
k
P
(1) (k ≥ 2) and set ekk := k+α
P (1). Then for n ≥ 1,
1−αk−1 k−1
1−αk k
(k)
Pn+1 (1) = Pn(k) (1) + enk ≥ Pn(k) (1) ≥ 0,
where the differences enk are defined by
enk :=
k
αn
en−1,k +
P (k−1) (1).
1 − αn
1 − αn n
{α}
Ultraspherical polynomials y = Pn
with α ≥ − 21 satisfy the differential equation
(1 − x2 )y 00 − 2(α + 1)xy 0 + n(n + 2α + 1)y = 0
(k)
and thus a closed form for Pn (1) is possible in this case since
Pn(k) (1) =
n(n + 2α + 1) − (k − 1)(2α + 1) − (k − 1)2 (k−1)
Pn
(1).
2(α + k)
This extends known Chebyshev and Legendre identities ([9, p. 33], [11, p. 251]).
4. P OLYNOMIAL E XPANSIONS OF P OWER S ERIES
A standard application of the theory of orthogonal polynomials is the least squares or uniform 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
for expansions of power series in terms of the polynomials that satisfy (1.1):
P coefficients
i
Let ai x be a convergent
P power series on (−1, 1), and for every n let Pn be a polynomial of
n
degree n. Then xP
= nm=0 b(n, m)Pm for some numbers b(n, m); and we define the Fourier
coefficient cj of ai xi with respect to the sequence hPn i by
X
cj :=
ai b(i, j)
i
P
whenever this sum converges.PNote that cnj := ni=0 ai b(i, j) is then the jth coefficient in the
expansion of the partial sum ni=0 ai xi : since b(i, j) = 0 for j > i,
n
X
i
ai x =
i=0
J. Inequal. Pure and Appl. Math., 7(2) Art. 67, 2006
n
X
i=0
ai
n
X
j=0
b(i, j)Pj =
n
X
cnj Pj .
j=0
http://jipam.vu.edu.au/
A N I NEQUALITY
FOR
C HEBYSHEV C ONNECTION C OEFFICIENTS
9
We have the following estimate where kf k denotes 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. Let hPn i be given
P by (1.1) with hαn i, hβn i and hγn i nonnegative, and
P suppose
that Pn (1) = 1 for all n. If
ai converges absolutely, then the coefficient cj of
ai xi with
respect to hPn i exists for every j and
n
X
X
ai b(i, j) ≤
|ai | .
cj −
i>n
i=0
Moreover, if
P
i−j even
iai converges absolutely, then
n
X
X
X
i
ai x −
cj Pj (x) ≤
(|ai | + |iai |)
j=0
i>n
for all x in [−1, 1]. In this case, if αi ≤ 12 and
P βi i= 0 for i = 0, ..., n − 1, and if ai ≥ 0 for all i
and dk is the kth Chebyshev coefficient of ai x , then
n
n
X
X
X
X
i
i
(4.1)
ai x −
c j Pj ≥ ai x −
dk Tk .
j=0
k=0
In addition, if αi < 12 (i = 1, ..., n − 1), then equality holds in (4.1) if and only if
polynomial of degree at most n.
P
ai xi is a
Proof. Asume that hαn i, hβn i and hγn i arePnonnegative, and Pn (1) = 1 for all n. By (1.1) the
degree of Pn is n for all n. Thus xn = nm=0 b(n, m)P
Pn m where b(n, m) ≥ 0 by (2.4), and
n
b(n, m) ≤ 1 by the
normalization
since
we
have
1
=
m=0 b(n, m).
P
P
Suppose that
|ai | converges. Then i ai b(i, j) converges absolutely by the comparison
test so cj exists and
X
X
a
b(i,
j)
|ai | .
≤
i
i>n
i>n
Next assume
we have
P
|iai | < ∞. Then
P
i−j even
|ai | < ∞ so cj exists for every j. Thus by Corollary 3.1
n
n
X
X
X
X
ai x i −
cj Pj (x) ≤ ai xi + (cj − cnj )Pj (x)
j=0
i>n
≤
X
i>n
|ai | +
j=0
n
X
|cj − cnj | ,
j=0
where
n X
n
X
X
X 1
n+1X
|iai | ≤
|iai | .
|cj − cnj | =
ai b(i, j) ≤
i
n + 1 i>n
j=0
j=0 i>n
j=0 i≥n+1
n
X
Suppose further that αi ≤ 21 , βi = 0 (i = 0, ..., n − 1) and ai is nonnegative for all i. Then
P
cj ≥ 0 for all j and by Theorem 2.1, Pj = jk=0 a(j, k)Tk where a(j, k) ≥ 0. Since we also
J. Inequal. Pure and Appl. Math., 7(2) Art. 67, 2006
http://jipam.vu.edu.au/
10
have
JAMES G UYKER
P
ai x i =
P
cj Pj uniformly on [−1, 1] in this case, it follows that
n
X
X
X
X
ai x i −
c j Pj = c j Pj =
cj
j=0
j>n
j>n
and similarly
n
X
X
X
i
ai x −
dk Tk =
dk .
k=0
k>n
But
X
dk Tk =
X
ai xi
=
X
c j Pj
=
X
cj
j
∞
X
a(j, k)Tk
k=0
!
X X
=
k
cj a(j, k) Tk
j≥k
since
X
X
X
X
cj ≤
cj Pj (1) =
ai < ∞.
cj a(j, k) ≤
j≥k
j≥k
Since
the coefficients in a uniformly convergent Chebyshev expansion are unique, dk =
P
c
j≥k j a(j, k). Therefore
X X
X
a(j, k)
dk =
cj
j>n
k>n
=
X
k>n
j
X
cj
j>n
=
X
cj −
j>n
≤
X
a(j, k) −
k=0
n X
X
n
X
!
a(j, k)
k=0
cj a(j, k)
k=0 j>n
cj .
j>n
Finally, assume that αi < 21 (i = 1, ..., n − 1). By property (iv) of Theorem 2.1, a(j, 0) +
a(j,
P 1)i > 0Pfor j > n so if equality holds in the last inequality then cj = 0 for all j > n. Thus
ai x = cj Pj is a polynomial of degree at most n.
P
Remark 4.2. If hPn i is defined as in Corollary 4.1 and, more generally, (iai )2 converges,
then by the Schwarz inequality it follows that
! 21
X
|ai | ≤
i>n
J. Inequal. Pure and Appl. Math., 7(2) Art. 67, 2006
X
i>n
(iai )2
X1
i2
i>n
! 12
http://jipam.vu.edu.au/
A N I NEQUALITY FOR C HEBYSHEV C ONNECTION C OEFFICIENTS
so
P
11
|ai | < ∞ and cj exists for every j. Moreover in this case we have
n
X
X
i
ai x −
cj Pj (x)
j=0
n X
X
X
≤
ai x i +
ai b(i, j)
i>n
j=0 i>n
!1
!1
! 21
! 21
n
X
X b(i, j) 2 2
X x i 2 2 X
X
+
(iai )2
≤
(iai )2
i
i
j=0
i>n
i>n
i>n
i>n
!1
! 12
! 12
X
X1 2 X
n
+
2
(iai )2
≤ √
(iai )2
≤ (n + 2)
2
i
n
i>n
i>n
i>n
where the last inequality follows from the proof of the integral test.
R EFERENCES
[1] R. ASKEY, Orthogonal expansions with positive coefficients, Proc. Amer. Math. Soc., 16 (1965),
1191–1194.
[2] R. ASKEY, Orthogonal expansions with positive coefficients. II, SIAM J. Math. Anal., 2 (1971),
340–346.
[3] R. ASKEY, Jacobi polynomial expansions with positive coefficients and imbeddings of projective
spaces, Bull. Amer. Math. Soc., 74 (1968), 301–304.
[4] R. ASKEY, Orthogonal Polynomials and Special Functions, Regional Conference Series in Applied Mathematics, 21, SIAM, Philadelphia, PA, 1975.
[5] R. ASKEY AND G. GASPER, Jacobi polynomial expansions of Jacobi polynomials with nonnegative coefficients, Proc. Camb. Phil. Soc., 70 (1971), 243–255.
[6] H. BATEMAN, Higher Transcendental Functions, V. 2, Bateman Manuscript Project, McGrawHill, New York, 1953.
[7] R.J. DUFFIN AND A.C. SCHAEFFER, A refinement of an inequality of the brothers Markoff,
Trans. Amer. Math. Soc., 50 (1941), 517–528.
[8] E.D. RAINVILLE, Special Functions, Macmillan, New York, 1960.
[9] T.J. RIVLIN, The Chebyshev Polynomials, Wiley, New York, 1974.
[10] T.J. RIVLIN AND M.W. WILSON, An optimal property of Chebyshev expansions, J. Approx. Theory, 2 (1969), 312–317.
[11] G. SANSONE, Orthogonal Functions, Interscience, New York, 1959.
[12] G. SZEGÖ, Orthogonal Polynomials, Amer. Math. Soc. Colloq. Publ., 23, Amer. Math. Soc., Providence, RI, 1939.
[13] R. SZWARC, Connection coefficients of orthogonal polynomials, Canad. Math. Bull., 35 (1992),
548–556.
[14] W.F. TRENCH, Proof of a conjecture of Askey on orthogonal expansions with positive coefficients,
Bull. Amer. Math. Soc., 81 (1975), 954–956.
[15] M.W. WILSON, Nonnegative expansions of polynomials, Proc. Amer. Math. Soc., 24 (1970), 100–
102.
J. Inequal. Pure and Appl. Math., 7(2) Art. 67, 2006
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