Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2007, Article ID 93815, 10 pages
doi:10.1155/2007/93815
Research Article
A Cohen-Type Inequality for Jacobi-Sobolev Expansions
Bujar Xh. Fejzullahu
Received 21 August 2007; Revised 20 November 2007; Accepted 11 December 2007
Recommended by Wing-Sum Cheung
Let μ be the Jacobi measure supported on the interval [−1, 1]. Let us introduce the
1
Sobolev-type inner product f ,g = −1 f (x)g(x)dμ(x) + M f (1)g(1) + N f (1)g (1),
where M,N ≥ 0. In this paper we prove a Cohen-type inequality for the Fourier expansion in terms of the orthonormal polynomials associated with the above Sobolev inner
product. We follow Dreseler and Soardi (1982) and Markett (1983) papers, where such
inequalities were proved for classical orthogonal expansions.
Copyright © 2007 Bujar Xh. Fejzullahu. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1. Introduction and main result
Let dμ(x) = (1 − x)α (1 + x)β dx, α > −1, β > −1, be the Jacobi measure supported on the
interval [−1,1]. We will say that f (x) ∈ L p (dμ) if f (x) is measurable on [−1,1] and
f L p (dμ) < ∞, where
f L p (dμ) =
⎧ 1
1/ p
⎪
⎪
f (x) p dμ(x)
⎪
⎨
if 1 ≤ p ≤ ∞,
⎪
⎪
⎪
⎩ess sup f (x)
if p = ∞.
−1
(1.1)
−1<x<1
Now let us introduce the Sobolev-type spaces
p
p
S p = f : f S p = f L p (dμ) + M | f (1) + N f (1) < ∞ ,
p
p
S∞ = f : f S∞ = f L∞ (dμ) < ∞ ,
1 ≤ p < ∞,
(1.2)
p = ∞.
2
Journal of Inequalities and Applications
Let f and g function in S2 . We can introduce the discrete Sobolev-type inner product
f ,g =
1
−1
f (x)g(x)dμ(x) + M f (1)g(1) + N f (1)g (1),
(1.3)
(α,β)
where M ≥ 0, N ≥ 0. We denote by {qn }n≥0 the sequence of orthonormal polynomials
with respect to the inner product (1.3) (see [1, 2]). These polynomials are known in
the literature as Jacobi-Sobolev-type polynomials. For M = N = 0, the classical Jacobi
(α,β)
orthonormal polynomials appear. We will denote them by { pn }n≥0 .
For f ∈ S1 , the Fourier expansion in terms of Jacobi-Sobolev-type polynomials is
∞
f(k)qk
(α,β)
(x),
(1.4)
k =0
where
f(k) = f , qk
(α,β)
.
(1.5)
The Cesàro means of order δ of the Fourier expansion (1.4) are defined by (see [3,
pages 76-77])
σ δn f (x) =
n
Aδ
n −k
k =0
Aδn
f(k)qk
(α,β)
(x),
(1.6)
where Aδk = ( k+δ
k ).
For a function f ∈ S p and a given sequence {ck,n }nk=0 ,n ∈ N ∪ {0}, of complex numα,β,M,N
bers with |cn,n | > 0, we define the operators Tn
by
α,β,M,N
Tn
(f)=
n
ck,n f(k)qk
(α,β)
.
(1.7)
k =0
Let us denote p0 = (4β + 4)/(2β + 3) and its conjugate q0 = (4β + 4)/(2β + 1). Here is
the main result.
Theorem 1.1. Let β ≥ α ≥ −1/2, β > −1/2, and 1 ≤ p ≤ ∞. There exists a positive constant
c, independent of n, such that
α,β,M,N Tn
⎧
(2β+2)/ p−(2β+3)/2
⎪
⎪
⎪n
⎨
(2β+1)/(4β+4)
[S p ]
≥ c|cn,n | (log n)
⎪
⎪
⎪
⎩n(2β+1)/2−(2β+2)/ p
if 1 ≤ p < p0 ,
if p = p0 , p = q0 ,
if q0 ≤ p < ∞,
(1.8)
where by [S p ] one denotes the space of all bounded, linear operators from the space S p into
itself, with the usual operator norm ·[S p ] .
Bujar Xh. Fejzullahu 3
Corollary 1.2. Let α, β, and p be as in Theorem 1.1. For ck,n = 1, k = 0,...,n, and for p
outside the Pollard interval (p0 , q0 ),
Sn [S p ] −→ ∞,
n −→ ∞,
(1.9)
where Sn denotes the nth partial sum of the expansion (1.4).
ρ
ρ
For ck,n = An−k /An , 0 ≤ k ≤ n, Theorem 1.1 yields the following.
Corollary 1.3. Let α, β, p, and δ be given numbers such that β > −1/2,
1
≤ α ≤ β,
2
1 ≤ p ≤ ∞,
−
2β + 2 2β + 3
−
if 1 ≤ p < p0 ,
p
2
2β + 1 2β + 2
0≤δ<
−
if q0 < p ≤ ∞.
2
p
(1.10)
0≤δ<
Then, for p ∈ [p0 , q0 ],
δ
σ n [S p ]
−→ ∞,
n −→ ∞.
(1.11)
2. Preliminaries
We summarize some properties of Jacobi-Sobolev-type polynomials that we will need in
the sequel (cf. [1]). Throughout this paper, positive constants are denoted by c,c1 ,... and
they may vary at every occurrence. The notation un ∼ vn means c1 ≤ un /vn ≤ c2 for n large
enough, and by un ∼
= vn , we mean that the sequence un /vn converges to 1.
(α,β)
The representation of the polynomials qn in terms of the Jacobi orthonormal poly(α,β)
nomials pn is
(α,β)
qn
(α,β)
(x) = An pn
(α+2,β)
(x) + Bn (x − 1)pn−1
(α+4,β)
(x) + Cn (x − 1)2 pn−2
(x),
(2.1)
where
(a) if M > 0 and N > 0, then An ∼
= − cn−2α−2 , Bn ∼
= cn−2α−2 , Cn ∼
= 1,
∼
∼
−
1/(α
+
2),
B
1,
C
1/(α
+ 2),
(b) if M = 0 and N > 0, then An ∼
=
n=
n=
−2α−2
∼
∼
∼
, Bn = 1, Cn = 0.
(c) if M > 0 and N = 0, then An = cn
(α,β)
The maximum of qn on [−1,1] is
max qn
x∈[−1,1]
(α,β)
(x) ∼ nβ+1/2
1
if β ≥ α ≥ − .
2
(2.2)
4
Journal of Inequalities and Applications
(α,β)
The polynomials qn
satisfy the estimate
⎧ ⎪
O θ −α−1/2 (π − θ)−β−1/2
⎪
⎪
⎪
⎨ c
c
≤θ≤π− ,
n
n
c
if 0 ≤ θ ≤ ,
n
c
if π − ≤ θ ≤ π,
n
if
(α,β)
qn (cosθ) = O nα+1/2
⎪
⎪
⎪
⎪ β+1/2 ⎩
O n
(2.3)
for α ≥ −1/2, β ≥ −1/2, and n ≥ 1.
The Mehler-Heine-type formula for Jacobi orthonormal polynomials is (see [4, Theorem 8.1.1] and [4, Formula (4.3.4)])
(α,β)
lim (−1)n n−β−1/2 pn
cos π −
n→∞
z
n
−(α+β)/2 =2
z
2
−β
Jβ (z),
(2.4)
where α, β are real numbers, and Jβ (z) is the Bessel function. This formula holds uniformly for |z| ≤ R, for R a given positive real number.
From (2.4),
(α,β)
lim (−1)n n−β−1/2 pn
n→∞
cos π −
z
n+ j
−(α+β)/2 =2
z
2
−β
Jβ (z)
(2.5)
holds uniformly for |z| ≤ R, R > 0 fixed, and uniformly on j ∈ N ∪ {0}.
Lemma 2.1. Let α,β > −1 and M,N ≥ 0. There exists a positive constant c such that
(α,β)
lim (−1)n n−β−1/2 qn
n→∞
cos π −
z
n
−β
z
2
=c
Jβ (z),
(2.6)
uniformly for |z| ≤ R, R > 0 fixed.
Proof. Here we will only analyze the case when M = 0 and N > 0. The proof of the other
cases can be done in a similar way. From (2.1), we have
(α,β)
(−1)n n−β−1/2 qn
cos π −
z
n+ j
(α,β)
= An (−1) n−β−1/2 pn
n
− Bn cos π −
cos π −
z
n+ j
z
n −1
− 1 (−1)
n+ j
z
(α+2,β)
×n−β−1/2 pn−1
cos π −
n+ j
2
z
n −2
− 1 (−1)
n+ j
z
(α+4,β)
×n−β−1/2 pn−2
cos π −
,
n+ j
+ Cn cos π −
where j ∈ N ∪ {0}.
(2.7)
Bujar Xh. Fejzullahu 5
Finally, if n→∞ and using (2.1) and (2.5), we get
(α,β)
lim (−1)n n−β−1/2 qn
cos π −
n→∞
z
n+ j
1
1 −(α+β)/2
+ 2·2−(α+β+2)/2 +
2
4·2−(α+β+4)/2
= −
α+2
α+2
−β
−(α+β)/2 z
=2
Jβ (z).
2
−β
z
2
Jβ (z)
(2.8)
We also need to know the S p norms for Jacobi-Sobolev-type polynomials
1 (α,β) p
(α,β) p
(α,β) p
(α,β) p
qn =
qn (x) dμ(x) + M qn (1) + M qn
(1) ,
Sp
−1
(2.9)
(α,β)
where 1 ≤ p < ∞. Hence, it is sufficient to estimate the L p (dμ) norms for qn . For M =
N = 0, the calculation of these norms is given in [4, page 391, Exercise 91] (see also [5,
Formula (2.2)]).
Lemma 2.2. Let M, N ≥ 0 and γ > −1/ p. For β ≥ −1/2,
0
−1
(α,β)
(1 + x)γ qn
⎧
⎪
⎪
c
⎪
⎪
⎪
⎪
⎨
p
p
if 2γ > pβ − 2 + ,
2
p
if 2γ = pβ − 2 + ,
2
p
if 2γ < pβ − 2 + .
2
(x) dx ∼ ⎪log n
⎪
⎪
⎪
⎪
⎪
⎩n pβ+p/2−2γ−2
(2.10)
Proof. From (2.3), for pβ + p/2 − 2γ − 2 = 0, we have
0
−1
(α,β)
(1 + x)γ qn
p
(x) dx = O(1)
= O(1)
π
π/2
π −1/n
π/2
π
p
(cosθ) dθ
(π − θ)2γ+1 (π − θ)− pβ− p/2 dθ
(2.11)
(π − θ)2γ+1 n pβ+p/2 dθ
+ O(1)
=O n
(α,β)
(π − θ)2γ+1 qn
π −1/n
pβ+p/2−2γ−2
+ O(1);
and for (pβ + p/2 − 2γ − 2) = 0, we have
0
−1
(α,β)
(1 + x)γ qn
p
(x) dx = O (logn).
(2.12)
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Journal of Inequalities and Applications
On the other hand, according to Lemma 2.1, we have
π
π/2
(α,β)
(π − θ)2γ+1 qn
p
(cosθ) dθ >
π
π −1/n
1
(π − θ)2γ+1 qn
(α,β)
p
(cosθ) dθ
2γ+1 p
(α,β)
z qn
n−1 dz
cos
π
−
n 0
−β
p
1 2γ+1
−1
z
pβ+p/2 z
n dz
∼
c
n
J
(z)
=
β
n
2
=
z
n
∼n
0
pβ+p/2−2γ−2
(2.13)
.
Using a similar argument as above, for 2γ = pβ − 2 + p/2, we have
π
π/2
(α,β)
(π − θ)2γ+1 qn
p
(cosθ) dx >
π
∼
=c
π −n−1/2
n1/2
−β
z
z2γ+1 0
(α,β)
(π − θ)2γ+1 qn
2
p
(cosθ) dx
p
γ+1
Jβ (z)
≥ c log n.
dz ∼ n
(2.14)
Finally, from [1, Theorem 5], we get
π
π/2
(α,β)
(π − θ)2γ+1 qn
p
(cosθ) dθ >
3π/4
π/2
(α,β)
(π − θ)2γ+1 qn
p
(cosθ) dθ ∼ c.
(2.15)
Notice that some of the above results appear in [6].
3. Proof of Theorem 1.1
For the proof of Theorem 1.1, we will use the test functions
α,β, j
gn
j
(α+ j,β+ j)
(x) = 1 − x2 pn
(x),
(3.1)
α,β,M,N
where β ≥ α ≥ −1/2, β > −1/2, and j ∈ N \ {1}. By applying the operators Tn
α,β, j
test functions gn , for some j > β + 1/2 − (2β + 2)/ p, we get
α,β,M,N
Tn
α,β, j
gn
=
n
α,β, j
ck,n gn
(α,β)
(k)qk
,
to the
(3.2)
k =0
where
α,β, j
gn
α,β, j (α,β) ,
(k) = gn , qk
k = 0,1,...,n.
(3.3)
Bujar Xh. Fejzullahu 7
From (2.1), we have
α,β, j
gn
(k) =
1
= Ak
j
(α+ j,β+ j)
1 − x 2 pn
−1
1
1
−1
k,n
(α,β)
j
(α+ j,β+ j)
j
(α+ j,β+ j)
1 − x 2 pn
k,n
(x)dμ(x)
(x)pk
1 − x 2 pn
−1
1
+ Ck
(α+ j,β+ j)
1 − x 2 pn
−1
+ Bk
j
(α,β)
(x)qk
(x)dμ(x)
(α+2,β)
(x)(x − 1)pk−1
(x)dμ(x)
(α+4,β)
(x)(x − 1)2 pk−2
(3.4)
(x)dμ(x)
k,n
= I1 + I2 + I3 ,
(γ,ρ)
where 0 ≤ k ≤ n, and it is assumed that pi (x) = 0 for i = −1, −2.
According to [5, Formula (2.8)] and [4, Formula (4.3.4)], we get
j
(α+ j,β+ j)
1 − x 2 pn
2j
α+ j,β+ j −1/2 (x) = hn
α,β
bm, j (α,β,n) hn+m
1/2
(α,β)
pn+m (x).
(3.5)
m=0
Taking into account (3.5)
2j
α+ j,β+ j −1/2 I1k,n = Ak hn
α,β
bm, j (α,β,n) hn+m
1/2
m=0
×
1
−1
(α,β)
(α,β)
pn+m (x)pk
(x)dμ(x).
(3.6)
Thus
I1k,n = 0,
α+ j,β+ j −1/2
I1n,n = An hn
0 ≤ k ≤ n − 1,
α,β 1/2
hn
b0, j (α,β,n),
n ≥ 0, m = 0.
(3.7)
Again, according to [5, Formula (2.8)] and [4, Formula (4.3.4)],
α+2,β −1/2
α+ j,β+ j −1/2
h k −1
I2k,n = Bk hn
×
1
−1
j
(α+ j,β+ j)
1 − x 2 Pn
(α+2,β)
(x)(x − 1)Pk−1
(x)dμ(x)
2j
α+2,β −1/2 α+ j,β+ j −1/2
= Bk h n
h k −1
bm, j (α,β,n)
m=0
1
(α,β)
(α+2,β)
×
Pn+m (x)(x − 1)Pk−1 (x)dμ(x).
−1
(3.8)
Since (see [4, Formula (4.5.4)])
(α+2,β)
(x − 1)Pk−1
(x) =
2k
2(k + α + 1) (α+1,β)
(α+1,β)
Pk
P
(x) −
(x),
2k + α + β + 1
2k + α + β + 1 k−1
(3.9)
8
Journal of Inequalities and Applications
(α+1,β)
and deg Pk−1
≤ n − 1, we have
I2k,n =
α+ j,β+ j −1/2 α+2,β −1/2
2k Bk
hn
h k −1
2k + α + β + 1
2j
×
1
bm, j (α,β,n)
m=0
−1
(3.10)
(α,β)
(α+1,β)
Pn+m (x)Pk
(x)dμ(x).
Formula 16.4 (11) in [7, page 285] shows that
1
(α,β)
−1
Pn
(α+1,β)
(x)Pn
(x)dμ(x) =
2α+β+1 Γ(α + n + 1)Γ(β + n + 1) 2n + α + β + 1 α,β
=
hn .
Γ(n + 1)Γ(α + β + n + 2)
n+α+β+1
(3.11)
This formula can also be proved by using [4, page 257, Identity (9.4.3)].
Thus
I2k,n = 0, 0 ≤ k ≤ n − 1,
α+ j,β+ j −1/2 α+2,β −1/2
2nBn
hn
h n −1
I2n,n =
n+α+β+1
α,β
× hn b0, j (α,β,n),
(3.12)
n ≥ 1, m = 0.
In a similar way,
2j
α+4,β −1/2 α+ j,β+ j −1/2
h k −2
bm, j (α,β,n)
m=0
I3k,n = Ck hn
×
1
−1
(3.13)
(α,β)
(α+4,β)
Pn+m (x)(x − 1)2 Pk−2 (x)dμ(x).
Again, as applications of [4, Formula (4.5.4)] and [4, Formula (9.4.3)], we point out the
following formulas:
(α+4,β)
(x − 1)2 Pk−2
(x) =
4k(k − 1)
(α+2,β)
+ Qk−1 (x),
P
(2k + α + β + 1)(2k + α + β + 2) k
(3.14)
where deg Qk−1 ≤ n − 1, and
1
−1
(α,β)
Pn
(α+2,β)
(x)Pn
(x)dμ(x) =
(2n + α + β + 1)(2n + α + β + 2) α,β
hn .
(n + α + β + 1)(n + α + β + 2)
(3.15)
Thus
I3k,n = 0,
I3n,n =
0 ≤ k ≤ n − 1,
α+ j,β+ j −1/2 α+4,β −1/2
4n(n − 1)Cn
hn
h n −2
(n + α + β + 1)(n + α + β + 2)
α,β
×hn b0, j (α,β,n),
n ≥ 2, m = 0.
(3.16)
Bujar Xh. Fejzullahu 9
α,β, j
In order to estimate (gn ) (k), we will distinguish the following three cases.
(1) M > 0, N > 0, then
∼ − 2 j cn−2α−2 ,
I1n,n =
−2α−2
∼ j
I n,n
,
2 = 2 c1 n
I3n,n ∼
= 2j.
(3.17)
Thus
α,β, j
gn
n,n
n,n
n,n
(n) = I1 + I2 + I3
∼
= 2j.
(3.18)
(2) M = 0, N > 0, then
−2 j
I1n,n ∼
=
α+2
I3n,n ∼
=
I2n,n ∼
= 2j,
,
−2 j
α+2
.
(3.19)
Thus
α,β, j
gn
(n) ∼
= 2j.
(3.20)
(3) M > 0, N = 0, then
∼ − 2 j cn−2α−2 ,
I1n,n =
I2n,n ∼
= 2j,
I3n,n = 0.
(3.21)
Thus
α,β, j
gn
(n) ∼
= 2j.
(3.22)
As a conclusion,
α,β, j
gn
α,β, j
gn
(k) = 0,
0 ≤ k ≤ n − 1,
(3.23)
∼ 2j.
(n) =
On the other hand, for 1 ≤ p < ∞,
α,β, j p
α,β, j p
gn = gn p
Sp
=
1
−1
≤ c1
L (dμ)
(α+ j,β+ j)
(1 − x) j p+α (1 + x) j p+β pn
0
−1
0
+ c2
(β+ j,α+ j)
(1 + x) j p+α pn
−1
p
(3.24)
(x) dx
(α+ j,β+ j)
(1 + x) j p+β pn
p
(x) dx
p
(x) dx.
Taking M = N = 0 in lemma, we have
α,β, j p
g n ≤ c
Sp
for j > β + 1/2 − (2β + 2)/ p > α + 1/2 − (2α + 2)/ p and q0 ≤ p < ∞.
(3.25)
10
Journal of Inequalities and Applications
It is well known (see, e.g., [8, Theorem 1]) that
(α+ j,β+ j) pn
(x) ≤ c(1 − x)− j/2−α/2−1/4 (1 + x)− j/2−β/2−1/4
(3.26)
for α,β ≥ −1/2, and x ∈ (−1,1). Therefore,
α,β, j g n S∞
α,β, j = gn 1/2( j −α−1/2)
L∞ (dμ)
≤ c(1 − x)
(1 + x)1/2( j −β−1/2) ≤ c,
(3.27)
for j > β + 1/2 ≥ α + 1/2.
Now, we will prove our main result.
Proof of Theorem 1.1. Let β ≥ α ≥ −1/2 and β > −1/2. By duality, it is enough to assume
that q0 ≤ p ≤ ∞. From (3.2), (3.23), (3.25), and (3.27), we have
α,β,M,N Tn
[S p ]
α,β, j −1 α,β,M,N α,β, j Tn
≥ g n gn
Sp
Sp
(α,β) ≥ ccn,n qn .
Sp
(3.28)
On the other hand, from (2.9) [1, Theorem 2], and lemma, we have
(α,β) qn Sp
⎧
⎨ (logn)1/ p
≥ c (2β+1)/2−(2β+2)/ p
⎩n
if p = q0 ,
if q0 < p < ∞.
(3.29)
From this expression, taking into account (2.2) and (3.28), the statement of the theorem
follows.
References
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[2] I. A. Rocha, F. Marcellán, and L. Salto, “Relative asymptotics and Fourier series of orthogonal
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Bujar Xh. Fejzullahu: Faculty of Mathematics and Sciences, University of Prishtina,
Mother Teresa 5, 10000 Prishtina, Kosovo, Serbia
Email address: bujarfej@uni-pr.edu