General Mathematics Vol. 15, No. 1 (2007), 41–50
Duffin-Schaeffer type inequalities
1
Ioan Popa
Dedicated to Professor Alexandru Lupaş
on the occasion of his 65th birthday
Abstract
We give estimation for the weighted L2 -norm of the k-th derivative of polynomial provided |pn−1 (x)| is bounded at a set of n points,
which are related in a certain way with the weight.
2000 Mathematics Subject Classification: 41A17, 41A05, 41A55,
65D30
Key words and phrases: Bouzitat quadrature, Chebyshev polynomials,
inequalities
1
Introduction
The following problem was raised by P.Turán (Varna,1970 ).
Let ϕ (x) ≥ 0 for −1 ≤ x ≤ 1 and consider the class Pn,ϕ of all polyno-
mials of degree n such that
1
Received 30 January, 2007
Accepted for publication (in revised form) 14 February, 2007
41
42
Ioan Popa
¯
¯
¯ (k) ¯
|pn (x)| ≤ ϕ (x) for −1 ≤ x ≤ 1. How large can max[−1,1] ¯pn (x)¯ be if
pn is arbitrary in Pn,ϕ ?
The aim of this paper is to consider the solution in the weighted L2 norm
for the majorant
ϕ (x) =
α−βx
√
,
(1−x) 1+x
0 ≤ β ≤ α.
Let as denote by
(1)
xi = cos
sin [(2n + 1) θ/2]
2iπ
, the zeros of Wn (x) =
,
2n + 1
sin (θ/2)
x = cos θ, the Chebyshev polynomial of the fourth kind,
(k)
xi
(2)
(k)
the zeros of Wn (x) and
Gn−1 (x) =
√
£
¤
2
′
(2nα+β) Wn′ (x)−(2n+1) βWn−1
(x)
2n(2n+1)
W,ϕ
Let Zα,β
be the class of polynomials pn−1 , of degree ≤ n − 1 such that
(3)
|pn−1 (xi )| ≤
α − βxi
√
, i = 1, 2, ..., n,
(1 − xi ) 1 + xi
where the x′i s are given by (1) and 0 ≤ β ≤ α.
Remark 1.1. From |Gn−1 (xi ) | =
α−βx
√i
(1−xi ) 1+xi
it follows:
W,ϕ
• Pn−1,ϕ ⊂ Zα,β
W,ϕ
• Gn−1 ∈ Zα,β
W,ϕ
• If pn−1 ∈ Zα,β
and i = 1, 2, ..., n,
(4)
|pn−1 (xi )| ≤ |Gn−1 (xi ) |
Duffin-Schaeffer type inequalities
2
43
Main results
W,ϕ
Theorem 2.1. If pn−1 ∈ Zα,β
(5)
Z1 r
then we have
1−x
2πn [2 (α2 + β 2 + αβ) (n + 1) − 3β 2 ]
[pn−1 (x)]2 dx ≤
1+x
3 (2n + 1)
−1
Z1
(6)
−1
√
£
¤2
(1 − x) 1 − x2 p′n−1 (x) dx ≤
3
2π (n − n)
[2(2α2 + αβ + 2β 2 )n + 8α2 + 4αβ − 7β 2 ]
15 (2n + 1)
with equality for pn−1 = Gn−1.
≤
Two cases are of special interest:
1
,
I. Case α = β = 1, ϕ (x) = √1+x
√ £
√
√
¤
′
Gn−1 = 2n2 Wn′ (x) − Wn−1
(x) = n2 Tn′ (x) = 2Un−1 (x),
Un−1 (x) = sin nθ/ sin θ, x = cos θ, the Chebyshev polynomial of the
second kind.
√
W,ϕ
Note that Pn−1,ϕ ⊂ Z1,1
,
W,ϕ
2Un−1 ∈ Z1,1
.
√
2Un−1 ∈
/ Pn−1,ϕ ,
W,ϕ
then we have
Corollary 2.1. If pn−1 ∈ Z1,1
Z1 r
(7)
1−x
[pn−1 (x)]2 dx ≤ 2πn
1+x
−1
Z1
√
(1 − x) 1 −
−1
with equality for pn−1 =
√
x2
£ ′
¤2
2π (n3 − n)
pn−1 (x) dx ≤
3
2Un−1 .
44
Ioan Popa
II. Case α = 1, β = 0, ϕ (x) =
Gn−1 =
√
2
Wn′
2n+1
1√
(1−x) 1+x
,
(x)
W,ϕ
Note that Pn−1,ϕ ⊂ Z1,0
,
√
2
Wn′
2n+1
(x) ∈ Pn−1,ϕ ,
√
2
Wn′
2n+1
W,ϕ
(x) ∈ Z1,0
.
W,ϕ
Corollary 2.2. If pn−1 ∈ Z1,0
then we have
Z1 r
(8)
1−x
4πn (n + 1)
[pn−1 (x)]2 dx ≤
1+x
3 (2n + 1)
−1
Z1
−1
√
£
¤2
8π (n3 − n) (n + 2)
(1 − x) 1 − x2 p′n−1 (x) dx ≤
15 (2n + 1)
with equality for pn−1 =
√
2
Wn′
2n+1
(x) .
In this second case we have a more general result:
W,ϕ
Theorem 2.2. If pn−1 ∈ Z1,0
Z1
−1
and 0 ≤ |b| ≤ a then we have
h
i2
(k)
(a+bx) (1−x)k+1/2 (1+x)k−1/2 pn−1 (x) dx ≤
≤
2π (n + k + 1)! (2ak + 2a − b)
(n − k − 1)! (2n + 1) (2k + 1) (2k + 3)
k = 1, ..., n − 2 with equality for pn−1 =
√
2
Wn′
2n+1
(x) .
Setting a = 1, b ∈ {−1, 0, 1} one obtains the following
W,ϕ
Corollary 2.3. If pn−1 ∈ Z1,0
then we have
Duffin-Schaeffer type inequalities
Z1
(9)
−1
h
i2
(k)
(1 − x)k+3/2 (1 + x)k−1/2 pn−1 (x) dx
≤
(10)
Z1
−1
45
2π (n + k + 1)!
(n − k − 1)! (2n + 1) (2k + 1)
h
i2
(k)
(1 − x)k+1/2 (1 + x)k−1/2 pn−1 (x) dx
≤
Z1
(11)
−1
4π (n + k + 1)! (k + 1)
(n − k − 1)! (2n + 1) (2k + 1) (2k + 3)
¡
1−x
≤
¢
2 k+1/2
h
(k)
pn−1
2π (n + k + 1)!
(n − k − 1)! (2n + 1) (2k + 3)
k = 1, ..., n − 2 with equality for pn−1 =
3
i2
(x) dx
√
2
Wn′
2n+1
(x) .
Auxiliary results
Here we state some lemmas which help us in proving the above theorems.
Lemma 3.1. (Duffin-Schaeffer) If q (x) = c
degree n with n distinct real zeros and if p ∈
|p′ (xi )| ≤ |q ′ (xi )|
n
Q
(x − xi ) is a polynomial of
i=1
Pn such
(i = 1, 2, ..., n) ,
then for k = 1, 2, ..., n
¯ (k+1) ¯ ¯ (k+1) ¯
¯p
(x)¯ whenever q (k) (x) = 0.
(x)¯ ≤ ¯q
that
46
Ioan Popa
Lemma 3.2. Let pn−1 be such that |pn−1 (xi )| ≤
α−βx
√i
,
(1−xi ) 1+xi
i = 1, 2, ..., n, where the x′i s are given by (1).Then we have
¯
¯ ¯
¯
¯ ′
¯ ′
(1) ¯
(1) ¯
G
(x
)
≤
p
(x
)
¯ n−1 j ¯ ¯ n−1 j ¯ , j = 1, ..., n − 1,
(12)
Proof. By the Lagrange interpolation formula based on the zeros of Wn
and using Wn′ (xi ) =
(−1)i+1 (2n+1)
(1−xi )
√
2(1+xi )
,
we can represent any algebraic polynomial pn−1 by
n
√
√
P
Wn (x)
2
pn−1 (x) = 2n+1
(−1)i+1 (1 − xi ) 1 + xi pn−1 (xi ) .
x−xi
i=1
√i
From Gn−1 (xi ) = (−1)i+1 (1−xα−βx
we have
i ) 1+xi
n
√
P Wn (x)
2
Gn−1 (x) = 2n+1
[α − βxi ] .
x−xi
i=1
Differentiating with respect to x we obtain
n
√
P
Wn′ (x)(x−xi )−Wn (x)
2
p′n−1 (x) = 2n+1
(x−xi )2
i=1
√
× (−1)i+1 (1 − xi ) 1 + xi pn−1 (xi ) .
On the roots of Wn′ (x) and
using
(3) we find
¯
´¯
³
¯
³
´¯
n ¯¯Wn x(1) ¯¯
√
P
¯ ′
j
(1) ¯
2
³
´ 2 [α − βxi ]
¯pn−1 xj ¯ ≤ 2n+1
(1)
i=1
=
³
´¯
√ ¯¯
(1) ¯ n
2¯Wn xj ¯ P
2n+1
i=1
α−βxi
³
(1)
xj −xi
xj −xi
´2
¯
³
´¯
¯ ′
(1) ¯
= ¯Gn−1 xj ¯ .
Lemma 3.3. Let pn−1 be such that |pn−1 (xi )| ≤
1√
,
(1−xi ) 1+xi
i = 1, 2, ..., n,where the x′i s are given by (1).Then we have
(13)
(k)
¯
¯
¯ (k) (k) ¯
¯pn−1 (xj )¯ ≤
(k)
√
2 ¯¯ (k+1) (k) ¯¯
(xj )¯ ,
¯Wn
2n + 1
whenever Wn (xj ) = 0, k = 0, 1, ..., n − 1.
√
2
Wn′
Proof. For α = 1, β = 0, Gn−1 = 2n+1
¯
¯
¯
¯
√
¯
(1) ¯
2 ¯
′′ (1) ¯
W
(x
)
and (12) give ¯p′n−1 (xj )¯ ≤ 2n+1
¯ n j ¯.
Duffin-Schaeffer type inequalities
47
Now the proof is concluded by applying Duffin-Schaeffer Lemma.
We need the following quadrature formulae:
(k+1)
Lemma 3.4. For any given n and k, 0 ≤ k ≤ n − 1, let xi
i = 1, ..., n − k − 1, be the zeros of
(14)
Z1
−1
¡
1−x
¢
2 k
r
(k+1)
.
Wn
,
Then the quadrature formula
1−x
f (x) dx = Af (−1) + Bf (1) +
1+x
+
n−k−1
X
i=1
³
´
(k+1)
s i f xi
,
2k
A=
2 (2n + 1) Γ (k + 1/2) Γ (k + 3/2) (n − k − 1)!
,
(n + k + 1)
B=
22k+2 Γ (k + 3/2) Γ (k + 5/2) (n − k − 1)!
,
(2n + 1) (n + k + 1)!
si > 0 have algebric degree of precision 2n − 2k − 1.
Proof. The quadrature formula (14) is the Bouzitat formula of the second
(k+ 23 ,k+ 21 )
(k+1)
kind [3, formula (4.8.1)], for the zeros of Wn
= cPn−k−1
. Setting
α = k + 1/2, β = k − 1/2, m = n − k − 1 in [3, formula (4.8.5)] we find A,
B and si > 0 (cf. [3, formula (4.8.4)]).
4
Proof of the Theorems
Because Wn (x) =
(15)
( ,
(2n)!!
P 2
(2n−1)!! n
1 −1
2
)
(x) we recall the formulae:
α + β + m + 1 (α+1,β+1)
d (α,β)
Pm (x) =
Pm−1
(x) ,
dx
2
Γ (m + α + 1)
Pm(α,β) (1) =
,
Γ (α + 1) Γ (m + 1)
48
Ioan Popa
(−1)m Γ (m + β + 1)
Γ (β + 1) Γ (m + 1)
Pm(α,β) (−1) =
Setting α = k + 1/2, β = k − 1/2, m = n − k in [3, formula (4.8.5)] on
(k+ 1 ,k− 1 )
(k)
obtains the Gauss formula for the zeros of Wn = cPn−k 2 2
Z1
(16)
−1
¡
1−x
¢
2 k
r
n−k
X (k) ³ (k) ´
1−x
,
H i f xi
f (x) dx =
1+x
i=1
with algebric degree of precision 2n − 2k − 1 and
(k)
(17)
Hi
> 0 (cf.[3, formula(4.8.4)]).
Proof of Theorem 2.1
For k = 0 in (16) we obtain
n
R1 q 1−x
P
(0)
f
(x)
dx
=
Hi f (xi ) of degree 2n−1.
1+x
i=1
−1
According to this quadrature formula and using (4) and (17) we have
n
R1 q 1−x
P
(0)
2
Hi (pn−1 (xi ))2
[p
(x)]
dx
=
n−1
1+x
−1
≤
n
P
i=1
i=1
R1
(0)
Hi (Gn−1 (xi ))2 =
−1
q
1−x
1+x
[Gn−1 (x)]2 dx.
Using the following formula ( k = 0 in (14))
R1 q 1−x
3π
f (x) dx = π(2n+1)
f (−1) + 2n(n+1)(2n+1)
f (1)
1+x
2n(n+1)
−1
+
n−1
P
i=1
we find
R1 q 1−x
1+x
³
´
(1)
s i f xi
[Gn−1 (x)]2 =
2πn[2(α2 +β 2 +αβ )(n+1)−3β 2 ]
.
3(2n+1)
−1
Setting k = 1 in (16) we get
n−1
√
R1
P (1) ³ (1) ´
2
(1−x) 1−x f (x) dx =
H i f xi
of degree 2n−3.
−1
i=1
According to this formula and using (12) and (17) we have
³
´´2
n−1
√
£
¤2
R1
P (1) ³
(1)
Hi
pn−1 xi
(1−x) 1−x2 p′n−1 (x) dx =
−1
i=1
Duffin-Schaeffer type inequalities
≤
n−1
P
i=1
(1)
Hi
49
³
´´2 R1
³
√
£
¤2
(1)
= (1−x) 1−x2 G′n−1 (x) dx.
Gn−1 xi
−1
From (14) with k = 1 we find
√
£
¤2
R1
(1−x) 1−x2 G′n−1 (x) dx
−1
=
2π (n3 −n)
[2(2α2
15(2n+1)
+ αβ + 2β 2 )n + 8α2 + 4αβ − 7β 2 ].
Proof of Theorem 2.2
If we replace f (x) with (a + bx) f (x), 0 ≤ |b| ≤ a in (16) we get
q
R1
k
1−x
f (x) dx
(a + bx) (1−x2 )
1+x
−1
=
n−k
P³
i=1
(k)
a + bxi
´
³
´
(k)
(k)
of degree 2n − 2k − 2
H i f xi
According to this formula and using (13) and (17) we have
i2
q h
R1
(k)
1−x
2 k
(a + bx) (1−x )
pn−1 (x) dx
1+x
−1
n−k
P³
´i2
³
(k)
xi
=
a+
i=1
´
h
³
´i2
n−k
P³
(k)
(k)
(k+1)
(k)
2
a
+
bx
H
W
x
≤ (2n+1)
n
2
i
i
i
(k)
bxi
´
(k)
Hi
h
(k)
pn−1
i=1
=
2
(2n+1)2
R1
k
(a + bx) (1−x2 )
−1
q
1−x
1+x
h
(k+1)
Wn
i2
(x) dx
If we replace f (x) with (a + bx) f (x), 0 ≤ |b| ≤ a in (14) we get
q
R1
k
1−x
(a + bx) (1−x2 )
f (x) dx = A (a−b) f (−1) + B (a + b) f (1)
1+x
−1
+
n−k−1
P ³
i=1
(k)
a + bxi
´
³
´
(k+1)
s i f xi
of degree 2n − 2k − 2.
In order to complete the proof we apply this formula to
h
i2
(k+1)
2
.
f = (2n+1)
W
(x)
n
2
³
´
(k+1)
(k+1)
xi
= 0 and the following relations deduced
Having in mind Wn
from (15)
(k+1)
Wn
(−1) =
(−1)n−k−1 (n+k+1)!
(n−k−1)!(2k+1)!!
³
´2
(k+1)
, Wn
(1) =
(2n+1)2
(2k+3)2
³
(k+1)
Wn
´2
(1) ,
50
Ioan Popa
we find
h
i2
R1
(k+1)
2
W
(x)
dx
(a + bx) (1−x)k+1/2 (1+x)k−1/2 (2n+1)
n
2
−1
h
h
i2
i2
(k+1)
(k+1)
2
2
= (2n+1)2 A (a−b) Wn
(−1) + (2n+1)2 B (a + b) Wn
(1)
=
2π(n+k+1)!(2ak+2a−b)
.
(n−k−1)!(2n+1)(2k+1)(2k+3)
References
[1] D.K. Dimitrov, Markov inequalities for weight functions of Chebyshev
type, J.Approx.Theory 83 (1995), 175–181.
[2] R.J. Duffin and A.C. Schaeffer, A refinement of an inequality of the
brothers Markoff, Trans.Amer.Math.Soc. 50 (1941), 517–528.
[3] A. Ghizzetti and A. Ossicini, Quadrature formulae, Akademie-Verlag,
Berlin, 1970.
[4] A. Lupaş, Numerical Methods, Constant Verlag, Sibiu, 2001.
Department of Mathematics,
Edmond Nicolau College, Ciortea 9,
Cluj-Napoca, Romania
E-mail:ioanpopa.cluj@personal.ro