Marcinkiewicz inequalities based on Stieltjes zeros
Sven Ehricha;∗, Giuseppe Mastroiannib
aGSF Research Center for Environmental and Health, Institute of Biomathematics and Biometry, Ingolstadter Landstr. 1, 85764 Neuherberg, Germany
bDipartimento di Matematica, Universita degli Studi della Basilicata, 85100 Potenza, Italy Received 15 December 1997; received in revised form 30 April 1998
Abstract
The authors nd necessary and sucient conditions for GDT weighted Marcinkiewicz inequalities based at Stieltjes zeros. c1998 Elsevier Science B.V. All rights reserved.
Keywords:Marcinkiewicz inequalities; Stieltjes polynomials
1. Introduction
If P is an arbitrary polynomial of degree m−1 and u is a weight function in [−1;1]; then the following identity is well known:
Z 1
−1|
P(x)u(x)|2dx= m
X
k=1
m(u2; yk)|P(yk)|2;
where m(u2; t) is the mth Christoel function with respect to the weight function u2 and yk;
k= 1; : : : ; m; are the zeros of the mth orthogonal polynomial associated with the weight function
u2:
The above identity is generally false if we replace 2 by an arbitrary p∈(0;+∞): Thus, we can investigate on the validity of the inequality
Z 1
−1|
P(x)u(x)|pdx6C m
X
k=1
m(up; yk)|P(yk)|p; (1)
for any absolute constant C¿0 and for given −16y1¡· · ·¡ym61:
∗Corresponding author.
This problem, which is very far from having a complete solution, is strongly connected with the weighted Lp convergence of Lagrange interpolating polynomials. In particular, if P is the
Lagrange polynomial Lm(f) interpolating a continuous function f on the knots yk; k= 1; : : : ; m;
then (1) becomes
Z 1
−1|
Lm(f; x)u(x)|pdx6C m
X
k=1
m(up; yk)|f(yk)|p:
By using some imbedding theorems and some results of [8], from the above inequality we can deduce theorems on the uniform boundedness of the operator Lm in some Besov space.
However, if the points yk; k= 1; : : : ; m; are zeros of orthogonal polynomials associated with any
weight function w (i.e. yk=xm; k(w); k= 1; : : : ; m) and u and w are GDT weight functions (see
Eq. (2) in Section 1.1), then in [8] it has been proved that for 1¡p¡∞ (1) is true if and only if
u
√w’∈Lp and
√w’
u ∈L
p′
; 1 p+
1
p′ = 1; ’(x) =
√
1−x2:
Very little is known in literature when the points yk; k= 1; : : : ; m; are not zeros of orthogonal
polynomials.
In this paper we will assume that the knots yk; k= 1; : : : ; m; are the zeros of the mth Stieltjes
polynomial E
m(x) associated with an ultraspherical weight function
w(x) = (1−x2)
−1
2; ∈(0;1):
We recall that {E
m(x)}m is not a standard sequence of orthogonal polynomials (see Section 1.2) and
that the previous results in [8] cannot help us. However, in this paper we will nd necessary and sucient conditions for the validity of (1) when u is a generalized Ditzian–Totik weight function and yk; k= 1; : : : ; m+ 1; are zeros of Em+1 (x): An analogous theorem is proved if
yk; k= 1; : : : ;2m+ 1; are zeros of PmEm+1 ; where Pm is the mth ultraspherical polynomial. In the
proof of these theorems, Lemmas 2 and 3 (new bounds for E
m+1 and PmEm+1) are crucial and can
be used in several contexts. For the sake of brevity, we cannot establish here new results on the corresponding Lagrange interpolation. We will consider this topic elsewhere.
1.1. Generalised Ditzian–Totik weights
We consider the so-called generalised Ditzian–Totik (GDT) weights of the form
u(x) =
M
Y
k=0
|x−tk|k!˜k(|x−tk|k); (2)
where k∈R,−1 =t0¡t1¡· · ·¡tM−1¡tM= 1,k=12 ifk∈ {0; M}andk= 1 otherwise. The
func-tion ˜!k is either equal to 1 or is a concave modulus of continuity of the rst order, i.e. ˜!k
is semi-additive, nonnegative, continuous and nondecreasing on [0;1], ˜!k(0) = 0 and 2 ˜!k((a +
b)=2)¿!˜k(a) + ˜!k(b) for all a; b∈[0;1], and for every ”¿0, ˜!k(x)=x” is a nonincreasing
func-tion on (0;1) with limx→0+!˜k(x)=x”=∞. Special cases are the generalised Jacobi weights ( ˜!k ≡1
u∈GDT being of the type (2) and m∈N, we dene
um(x) = ( √
1 +x+m−1)2 0!˜
0( √
1 +x+m−1)
M−1
Y
k=1
(|x−tk|+m−1)k
×!˜k(|x−tk|+m−1)( √
1−x+m−1)2M!˜ M(
√
1−x+m−1): (3)
We recall some results concerning GDT weights from [8]. For the convenience of the reader, we also include a short proof of the following Lemma 1 (see Section 3). Let the Hilbert transform be denoted by
H(f; t) = − Z 1
−1
f(x)
x−tdx: (4)
Lemma 1 (Mastroianni and Russo [8, Lemma 2.5]). Let 1¡p¡∞ and U∈GDT. Then for every function f such that fU∈Lp we have
kH(f)Ukp6CkfUkp; C6=C(f);
if and only if
U∈Lp; U−1 ∈Lp′
; p′= p
p−1: (5)
Lemma 2 (Mastroianni and Russo [8, Corollary 2.3]). Letw∈GDT; w∈Lp;16p6∞. Then;for
each xed 06a¡m; there exists a positive constantC; depending on a and w; such that for every
P∈Pm and E⊂[−1;1] with |arccosE|6a=m we have kPwkp6CkPwkLp([−1;1]\E):
Lemma 3 (Mastroianni and Russo [8, Lemma 2.1]). Letw∈GDT. Then there exists a polynomial
Q∈Pm; m¿1; such that for |x|61
wm(x)6Q(x)6Cwm(x); √
1−x2
m Q
′(x)6Cw
m(x);
where C6=C(m).
1.2. Stieltjes polynomials
In the sequel, C will denote a generic constant independent of the variables in the context. In dierent formulas, the same symbol C may have dierent values. Moreover, we will write A∼B;
for A; B¿0; i there exist two positive constants M1; M2 independent of A and B such that
M16
A
B
Given the ultraspherical polynomial P
m, ∈(0;1), orthogonal with respect to w(x) = (1−x2)− 1 2 on
(−1;1), the Stieltjes polynomial E
m+1 is dened, up to a multiplicative constant, by
Z 1
−1
!(x)Pm(x)E m+1(x)x
kdx= 0; k= 0;1; : : : ; m:
While this denition is possible for more general orthogonal polynomials, the zeros of the Stieltjes polynomials are not necessarily real and in (−1;1). For w and ∈(0;1) this property, as well as
the interlacing property of the zeros of P
m and Em+1 , was proved by Szego in [14]. See the surveys
[5, 10] for a more detailed overview on the history and on basic properties of Stieltjes polynomials. Zeros of Stieltjes polynomials are used as nodes for Kronrod extensions of Gaussian quadrature formulas, based on P
m, for a maximum degree of algebraic exactness. Pairs of Gauss and Gauss–
Kronrod formulas are a standard method for error estimation in automatic integration packages, see [13]. Carrying over this idea to Lagrange interpolation, extended interpolation formulas, based on the zeros of Stieltjes polynomials or other choices of nodes, have been considered in [3, 4].
In this paper, the Stieltjes polynomials E
m+1 are normalised such that we have (cf. [2])
E
m+1(cos) sin
−1= cos(m+)
−(−1)
2
+ o(1)
uniformly for ∈[”;−”], ”¿0 xed. The proofs of the results are based on the sharp bounds
sup
x∈[−1+m−2;1−m−2]|
! 2
(x)Em+1 (x)|6C; (6)
sup
x∈[−1+m−2;1−m−2]|
!(x)Pm(x)E
m+1(x)|6C; (7)
proved in [4, Theorem 2.1], where the polynomials P
m are normalised such that we have (cf. [2])
Pm(cos)Em+1 (cos) sin2−1= cos
(2m+ 2)−(2−1) 2
+ o(1)
uniformly for ∈[”;−”], ”¿0 xed. Furthermore, we shall use the asymptotic relations (see [4]) 1
|E′
m+1(k; m+1)|
∼ m1(1−(
k; m+1)
2)2; k= 1; : : : ; m+ 1; (8)
1
|K′
2m+1(yk;2m+1)|
∼ m1(1−(yk;2m+1)2); k= 1; : : : ;2m+ 1; (9)
respectively
k; m+1−k+1; m+1 ∼m+1; m+1 ∼−
1; m+1 ∼
1
m; (10)
k;2m+1−
k+1;2m+1∼
m+1; m+1 ∼−
1; m+1 ∼
1
m (11)
for the cos arguments of the zeros
1; m+1¡· · ·¡m+1; m+1 of Em+1 , respectively y1;2m+1¡· · ·¡
y
2m+1;2m+1 of K2m+1 =PmEm+1 , i.e. k; m+1= cosk; m+1 , yk;2m+1= cos k;2m+1. The above asymptotic
The following results about Stieltjes polynomials are new and of their own interest. They are, in addition to the above properties, the key to the proofs of the main results (Theorems 2 and 3 below).
Lemma 4. Let ∈(0;1), r∈N; Im:= [m−1;−m−1]. Then
lim
m→∞
Z
Im
(Em+1 (cos) sin−1)r−cosr
(m+)−(−1) 2
2
d= 0;
lim
m→∞
Z
Im
(K2m+1 (cos) sin2−1)r−cosr
(2m+ 2)−(2−1) 2
2
d= 0:
Lemma 5. Let ∈(0;1); 1¡p¡∞; u∈Lp; r∈N. Then there exists C¿0 such that
lim inf
m→∞ k(E
m+1)
ru
kp¿C
u
(w 2
)r
p
¿0; (12)
lim inf
m→∞ k(K
2m+1)
ru
kp¿C
u
(w)r
p
¿0: (13)
2. Marcinkiewicz inequalities
For a nonnegative weight w with 0¡kwk1¡∞, the Christoel function is dened by
m(w; t) = m+1
X
k=1
p2k(w; t) !−1
;
where pk(w) are the orthonormal polynomials with respect to w. In [1] and [7, Theorem 5] it is
proved that, for w∈GDT; we have
m(w; x)∼ √
1−x2
m +
1
m2
!
wm(x); (14)
where wm is like um in (3) and |t|61: The following result is proved in [8, Theorem. 2.6].
Theorem 1. Letu∈GDT; u∈Lp;16p¡∞. Let z
k= cosk; k∈[0;];−1¡z1¡· · ·¡zm¡1;and
k+1−k ∼m∼−1 ∼m−1. Then for every polynomial P∈Plm and l xed integer; we have
m
X
k=1
m(up; zk)|P(zk)|p
!1=p
6CkPukp;
Remark. The statement is slightly dierent from [8, Theorem 2.6] (with respect to the endpoints), but the proof can be carried over completely.
The following theorems are the main results of this paper.
Theorem 2. Let ∈(0;1); 1¡p¡∞; u∈GDT; u∈Lp. The following assertions are equivalent.
(1) For all p∈Pm;
kPukp6C
m+1
X
k=1
m+1(up; k; m+1)|P(k; m+1)|p
!1=p
; (15)
where C is independent of m and P:
(2)
w 2
u ∈L
p′
; (16)
where (1=p) + (1=p′) = 1.
Theorem 3. Let ∈(0;1); 1¡p¡∞; u∈GDT; u∈Lp. The following assertions are equivalent.
(1) For all p∈P2m;
kPukp6C
2m+1
X
k=1
m(up; yk;2m+1)|P(yk;2m+1)|p
!1=p
; (17)
where C is independent of m and P:
(2) u
w ∈
Lp and w
u ∈L
p′
; (18)
where (1=p) + (1=p′) = 1.
Remark. In view of Lemma 1, a third equivalent statement is the boundedness of the weighted Hilbert transform in Lp,
H(f) u
w 2
p
6C
f u w
2
p
;
respectively,
H(f) u
w
p
6C
f u w
p
:
3. Proofs
Proof of Lemma 1. We have
if and only if U∈Ap, i.e. for each interval I⊂(−1;1) there holds
kUkLp(I)kU−1kLp′
(I)6C|I|; p
′= p
p−1; (19)
with C being independent of I, and where |I| is the measure of I (cf. [6, 11]). If (19) holds, then obviously (5) follows. Suppose (5) holds. Since U and U−1 are bounded functions in each
subinterval of [−1;1] not including the possible singularities tk, k= 0;1; : : : ; M, it is sucient to
consider U(x) =|x|aw˜(|x|), −1=p¡a¡1=p′, I= (0; d), 0¡d¡1, where ˜w is either equal to 1 or is a concave modulus of continuity of the rst order that satises the additional assumptions in Section 1.1. Hence, ˜w(|x|) is a nondecreasing function, and we have
A:= Z d
0
xapw˜p(x) dx
!1=p
6w˜(d)d
a+1=p
(1 +ap)1=p =
U(d)d1=p
(1 +ap)1=p:
Since U∈GDT, for all ”¿0 ˜w(|x|)=x” is a nonincreasing function. For every xed ”¡(1=p′)−a,
B := Z d
0
dx xap′
˜
wp′(|x|) !p1′
= Z d
0
x−ap′−”p′
x”
˜
w(|x|) p
′
dx
!1=p
′
6 d
”
˜
w(d)
d1=p′
da+”(1−ap′−”p′)1=p′ =
d1=p′
U(d)
1
(1−ap′−”p′)1=p′:
Thus it follows that A·B6C(a; p)d, i.e., (19).
Proof of Lemma 4. Let ”¿0. For the rst statement, we split the integral,
Z
Im
=
Z ”
1 m
+ Z −1
m
−”
+ Z −”
”
=:I1+I2:
Now, using (6),
|I1|6C
Z ”
1 m
+ Z −1
m
−”
d6C”;˜
where ˜C is independent of ” and n. Furthermore,
sup
∈[”;−”]
Em+1 (cos) sin−1−cos
(m+)−(−1) 2
= o(1)
for m→ ∞ (cf. [2]). For A=E
m+1(cos) sin
−1 and B= cos
{(m+)−(−1)
2} we have
|Ar−Br|6|A−B|
r−1
X
i=0
|A|r−1−i|B|i
Proof of Lemma 5. For orthogonal polynomials and r= 1 the assertion was proved in [12, Theorem 32]. For the present case, the proof is based on Lemma 4 and follows analogously as in [12, Proof of Theorem 3.2]. Therefore we omit the details.
Proof of Theorem 2. Assume that condition (16) holds. We consider the Lagrange interpolation operator
with Q being a yet unspecied positive polynomial of degree6m, is a polynomial of degree 62m. Using (8), we estimate
6C
Using (14), we obtain
q
where the constant C is independent of m and k, and hence
Z 1
and the last inequality follows from Theorem 1, (10) and (14). From Lemma 2 we obtain
and H denotes the Hilbert transform as in (4). We have
using (6). For x∈Am, we have um∼u, and from Lemma 3 we obtain that there exists a polynomial
Therefore, the condition (16) is sucient for the inequality (15).
We now show the necessity of (16). We rst estimate the fundamental Lagrange polynomials, using (8),
and hence, for u being a GDT weight,
where the second inequality comes from an application of (15) and (14). Assume
w
Since u∈GDT, i.e. ˜!j in the denition of u are of subalgebraic growth, at least one of the following
assertions must be true:
Using Lemma 5, we have
and since we already proved that uw−1
2 Similarly as in (21) we obtain
sup
Using Lemma 5, we have
lim sup
where A denotes the characteristic function of A. Using this inequality, we have
i.e., there exists a constant D, independent of , such that
logt1+ 1
2 6D for all 0¡¡ t1+ 1
4 ;
which is a contradiction. Hence, (16) must be valid.
Proof of Theorem 3. We obtain the suciency of (18) as well as the necessity of the second condition in (18) in a completely analogous way as in the previous proof. To prove the necessity of the rst condition in (18), assume that for every polynomial P∈P2m we have
kPukpp6C
2m+1
X
k=1
2m+1(up; yk;2m+1)|P(y k;2m+1)|
p:
Then for every continuous function f we have
kL
2m+1(f)ukp 6C 2m+1
X
k=1
2m+1(up; yk;2m+1)|f(y k;2m+1)|
6Ckfk∞
2m+1
X
k=1
2m+1(up; yk;2m+1)6Ckfk∞kukp6Ckfk∞;
using Theorem 1 and since u∈Lp. Therefore,
kL
2m+1ukp= sup
kfk∞=1
kL
2m+1(f)ukp6C:
Applying Theorem 2.2 of [9, p. 433], we get
kK2m+1 ukp6CkK2m+1 wk1kL
2m+1ukp:
Using (7), we have
kK2m+1 wk16C;
C independent of m. Hence
kK2m+1 ukp¡CkL
2m+1ukp¡C¡˜ ∞;
where ˜C is independent of m. By Lemma 5 with r= 1, this implies the rst condition in (18).
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