Fractional Integration and Fractional
Differentiation for Jacobi Expansions.
Integraci´
on y Derivaci´
on Fraccionaria para desarrollos de Jacobi.
Cristina Balderrama (
cbalde@euler.ciens.ucv.ve
)
Departamento de Matem´aticas, Facultad de Ciencias, UCV. Apartado 47009, Los Chaguaramos, Caracas 1041-A Venezuela
Wilfredo Urbina (
wurbina@euler.ciens.ucv.ve
)
Departamento de Matem´aticas, Facultad de Ciencias, UCV. Apartado 47195, Los Chaguaramos, Caracas 1041-A Venezuela, and Department of Mathematics and Statistics, University of New Mexico,Albuquerque, NM 87131, USA. Abstract
In this article we study the fractional Integral and the fractional Derivative for Jacobi expansion. In order to do that we obtain an anal-ogous of P. A. Meyer’s Multipliers Theorem for Jacobi expansions. We also obtain a version of Calder´on’s reproduction formula for the Jacobi measure. Finally, as an application of the fractional differentiation, we get a characterization for Potential Spaces associated to the Jacobi mea-sure.
Key words and phrases: Fractional Integration, Fractional Differen-tiation, Jacobi expansions, Multipliers, Potential Spaces.
Resumen
En este trabajo estudiamos la integraci´on y diferenciaci´on fraccio-naria para el caso de los desarrollos de Jacobi. Para ello obtenemos un teorema an´alogo al teorema de multiplicadores de P.A. Meyer para desarrollos de Jacobi. Tambi´en obtenemos una versi´on de la f´ormula de reproducci´on de Calder´on para la medida de Jacobi. Finalmente, como una aplicacci´on de la diferenciaci´on fraccionaria, obtenemos una carac-terizaci´on de los Espacios Potenciales asociados a la medida de Jacobi.
Palabras y frases clave:Integraci´on fraccionaria, Derivaci´on fraccio-naria, Desarrollos de Jacobi, Multiplicadores, Espacios Potenciales.
1
Introduction.
Let us consider (normalized) Jacobi measure µα,β(dx) = 1
2α+β+1B(α+ 1, β+ 1)(1−x)
α(1 +x)βdx, (1)
for x ∈ [−1,1], where α, β > −1. This normalization gives a probability measure and it is not usually considered in classical orthogonal polynomial theory.
The one dimensional Jacobi operator is given by
Lα,β = (1−x2) d
2
dx2 + (β−α−(α+β+ 2)x)
d
dx. (2)
It is easy to see that this is a symmetric operator on L2([−1,1], µα,β).
Letpα,β
n be the normalized Jacobi polynomials of degreen∈N.Then the family {pα,β
n } is an orthonormal Hilbert basis of L2([−1,1], µα,β), that can be obtained by the Gram–Schmidt orthogonalization process with respect to the measure µα,β, applied to the monomials. It is well known that Jacobi polynomials are eigenfunctions of the Jacobi operator Lα,β with eigenvalue
−λn=−n(n+α+β+ 1),that is,
Lα,βpα,βn =−n(n+α+β+ 1)pα,βn . (3) Since{pα,β
n }is an orthonormal basis ofL2([−1,1], µα,β), we have the or-thogonal decomposition
L2([−1,1], µα,β) = ∞
M
n=0
Cnα,β, (4)
where, for each n, Cα,β
n is the closed subspace generated by pα,βn . This is called the Wiener–Jacobi chaos decomposition ofL2([−1,1], µα,β).
LetJα,β
n be the orthogonal projection ofL2([−1,1], µα,β) ontoCnα,β.Then, forf ∈L2([−1,1], µα,β) we have
f = ∞
X
n=0
Jnα,βf, (5)
where Jα,β
n f = ˆf(n)pα,βn with ˆ f(n) =
Z 1
−1
thenth-Jacobi–Fourier coefficient off.
Let us now consider{Ttα,β}t≥0 the Jacobi semigroup. This is the Markov
operator semigroup associated to the Markov probability kernel semigroup (see [2],[4])
Pt(x, dy) = ∞
X
n=0
e−λntpα,β
n (x)pα,βn (y)µα,β(dy) =pα,β(t, x, y)µα,β(dy), that is
Ttα,βf(x) =
Z 1
−1
f(y)Pt(x, dy) =
Z 1
−1
f(y)pα,β(t, x, y)µα,β(dy).
Unfortunately, there is not a reasonable explicit representation for the kernel pα,β(t, x, y).
The Jacobi semigroup {Ttα,β}t≥0 is a diffusion semigroup, conservative,
symmetric, strongly continuous on Lp([−1,1], µα,β) of positive contractions onLp, with infinitesimal generator Lα,β.
Moreover, forα, β >−12 it can be proved that{T
α,β
t }t≥0is also
hypercon-tractive, that is to say thatTtα,βis not only a contraction onLp([−1,1], µα,β), but also for any initial condition 1< q(0)<∞there exists an increasing func-tionq:R+→[q(0),∞),such that for everyf and allt≥0,
kTtα,βfkq(t)≤ kfkq(0).
The proof that we know of this fact is an indirect one, obtained by D. Bakry in [3], that is based in proving that the Jacobi operator, for the parameters α, β > −1/2, satisfies a Sobolev inequality, by checking that it satisfies a curvature-dimension inequality, and therefore a logarithmic Sobolev inequality and then use the well known equivalency due to L. Gross [8]. A detailed proof of this can be found in [2], see also [1]. From now on we will consider the Jacobi semigroup for the parameters α, β >−1
2.
On the other hand, for 0< δ <1 we define the generalized Poisson–Jacobi semigroup of orderδ,{Ptα,β,δ}, as
Ptα,β,δf(x) =
Z ∞
0
Tsα,βf(x)µδt(ds).
where {µδ
t} are the stable measures on [0,∞) of orderδ(∗). The generalized
(∗)
The stable measures on [0,∞) of orderδare Borel measures on [0,∞) such that its Laplace transform verifyR∞
0 e
−λsµδ
t(ds) =e
−λδt.
Forδfixed,{µδ
t}form a semigroup with
Poisson–Jacobi semigroup of order δ is a strongly continuous semigroup on Lp([−1,1], µα,β) with infinitesimal generator (−Lα,β)δ.
In the case δ= 1/2, we have the Poisson–Jacobi semigroup, that will be denoted asPtα,β=P
α,β,1/2
t .In this case we can explicitly computeµ
1/2
t ,
µ1t/2(ds) = t 2√πe
−t2/4s
s−3/2ds
and by Bochner’s formula we have Ptα,βf(x) =√1
π
Z ∞
0
e−u
√
uT α,β
t2/4uf(x)du. (6) Then by (3),
Ttα,βpα,βn =e−λntpα,βn , Ptα,β,δpα,βn =e−λ
γ ntpα,β
n .
Giving a function Φ :N → R the multiplier operator associated to Φ is defined as
TΦf =
∞
X
k=0
Φ(k)Jkα,βf,
forf =P∞
k=0J
α,β
k f,a polynomial.
If Φ is a bounded function, then by Parseval’s identity,TΦ is bounded on
L2([−1,1], µα,β). In the case of Hermite expansions, the P.A. Meyer’s
Multi-plier Theorem [10] gives conditions over Φ so that the multiMulti-plier TΦ can be
extended to a continuous operator on Lp for p6= 2. In the next section we will prove an analogous result for the Jacobi expansions.
In section 3 we are going to define the Fractional Integration and Differen-tiation for Jacobi expansions, as well as Bessel Potentials associated to Jacobi measure. Using Meyer’s multipliers Theorem we will see theLp continuity of the Fractional Integration and of the Bessel Potentials and we give a charac-terization of the Potential Spaces. We also study the asymptotic behavior of the Poisson-Jacobi semigroup and we give a version of Calderon’s reproducing formula.
2
P.A. Meyer’s Multiplier Theorem for Jacobi
expansions.
projectionsJnα,βcan be extended to a continuous function onLp([−1,1], µα,β).
Lemma 1. If 1 < p < ∞ then for every n ∈ N, Jnα,β can be extended to a continuous operator to Lp([−1,1], µα,β), that will also be denoted as Jα,β
n ,
that is, there exists Cn,p∈R+ such that
kJnα,βfkp≤Cn,pkfkp,
forf ∈Lp([−1,1], µα,β).
Proof. Let us consider p > 2 and for the initial condition q(0) = 2, let t0 be a positive number such that q(t0) = p. Taking g = Jnα,βf, by the hypercontractive property, Parseval identity and H¨older inequality we obtain,
kTtα,β0 gkp=kT α,β t0 J
α,β
n fkp≤ kJnα,βfk2≤ kfk2≤ kfkp.
Now, asTt0J α,β
n f =e−t0λnJnf we get
kJnα,βfkp≤Cn,pkfkp,
withCn,p=et0λn. For 1< p <2 the result follows by duality. ¤
We also need the following technical result,
Lemma 2. Let1< p <∞.Then, for eachm∈Nthere exists a constantCm such that
kTtα,β(I−J0α,β−J
α,β
1 − · · · −J
α,β
m−1)fkp≤Cme−tmkfkp.
Proof. Letp >2 and for the initial conditionq(0) = 2, let t0 be a positive
number such thatq(t0) =p.
Ift≤t0, since Ttα,β is a contraction, by theLp- continuity of the projec-tions
kTtα,β(I−J0α,β− · · · −J
α,β
m−1)fkp ≤ k(I−J0α,β− · · · −J
α,β m−1)fkp
≤ kfkp+ m−1
X
k=0
kJkα,βfkp
≤ (1 + m−1
X
k=0
But since et0λk ≤et0λm for all 0 ≤k ≤m−1 and λm ≥m for all m > 1,
k f, by the hypercontractive property, kTtα,β Now, by the Minkowski integral inequality, we have an analogous result for the generalized Poisson–Jacobi semigroup.
From the generalized Poisson–Jacobi semigroup let us define a new family of operators {Pk,γ,mα,β }k∈Nby the formula
Pk,γ,mα,β f = 1 (k−1)!
Z ∞
0
tk−1Pα,β,γ t (I−J
α,β
0 −J
α,β
1 − · · · −J
α,β m−1)f dt.
By the preceding lemma and the Minkowski integral inequality we have,
Proposition 4. If 1< p <∞, then for every m∈N there is a constantCm such that
kPk,γ,mα,β fkp≤ Cm mγkkfkp.
Observe that in particular iff =pα,β
n , n≥m
Pk,γ,mα,β pα,β n =
1 λγkn
pα,β
n . (7)
Now we are ready to present P.A. Meyer’s Multipliers Theorem for Jacobi expansions.
Theorem 5. If for some n0∈Nand0< γ <1
Φ(k) =h
µ 1
λγk
¶
, k≥n0,
with han analytic function in a neighborhood of zero, thenTΦ, the multiplier
operator associated to Φ, admits a continuous extension to Lp([−1,1], µα,β).
Proof. Let
TΦf =Tφ1f+TΦ2f =
n0−1
X
k=0
Φ(k)Jkα,βf+ ∞
X
k=n0
Φ(k)Jkα,βf.
By the lemma 1 we have that
kTΦ1fkp≤ n0−1
X
k=0
|Φ(k)|kJkα,βfkp≤
Ãn0−1
X
k=0
|Φ(k)|Ck
!
kfkp,
that is, T1
Φ is Lp continuous. It remains to be seen that TΦ2 is also Lp
By the hypothesis let us assume that h can be written as h(x) =
pcontinuous, by proposition 4, we obtain,
kT2
3
Fractional Integration and Differentiation.
As in the classical case, forγ >0 we define the Fractional Integral of orderγ, Iα,β
γ , with respect to Jacobi measure, as
Iγα,β= (−Lα,β)−γ/2. (8) Iα,β
γ is also called Riesz Potential of orderγ.
Observe that, since zero is an eigenvalue of Lα,β, then Iα,β
γ is not de-fined over all L2([−1,1], µα,β). Let Π
0 = I−J0α,β and denote also by Iγα,β the operator (−Lα,β)−γ/2Π
0. Then, this operator is well defined over all
Thus, for f a polynomial in L2([−1,1], µα,β) with Jacobi expansion
P∞
k=0J
α,β
k f, we have Iα,β
γ f = ∞
X
k=1
1 λγ/k 2J
α,β k f.
For the Fractional Integral of order γ >0 we have the following integral representation,
Iα,β γ f =
1 Γ(γ)
Z ∞
0
tγ−1Pα,β
t f dt, (10)
for f polynomial, where Ptα,β is the Poisson–Jacobi semigroup. In order to prove that observe that for the Jacobi polynomials, we have, by the change of variabless=λ1k/2t,
1 Γ(γ)
Z ∞
0
tγ−1Pα,β
t pα,βk dt = 1 Γ(γ)
Z ∞
0
tγ−1e−λ1k/2tdt pα,β
k
= 1
Γ(γ) 1 λγ/k 2
Z ∞
0
sγ−1e−sds pα,β k =
1 λγ/k 2p
α,β k .
The Meyer’s multiplier theorem allows us to extend Iα,β
γ as a bounded operator onLp([−1,1], µα,β), as next theorem shows.
Theorem 6. The the Fractional Integral of orderγ,Iα,β
γ admits a continuous
extension, that it will also be denoted as denote Iα,β
γ , toLp([−1,1], µα,β).
Proof. Ifγ/2<1, thenIα,β
γ is a multiplier with associated function
Φ(k) = 1 λγ/k 2 =h
Ã
1 λγ/k 2
!
whereh(z) =z,which is analytic in a neighborhood of zero. Then the results follows immediately by Meyer’s theorem.
Now, if γ/2 ≥ 1, let us consider β ∈ R, 0 < β < 1 and δ = γ
2β. Then δβ= γ2. Leth(z) =zδ, which is analytic in a neighborhood of zero. Then we have
h
Ã
1 λβk
!
= 1
λδβk = 1
Again the results follows applying Meyer’s theorem. ¤ Now the Bessel Potential of orderγ > 0, Jα,β
γ , associated to the Jacobi measure is defined as
Jα,β
γ = (I− Lα,β)
−γ/2. (11)
Observe that for the Jacobi polynomials we have
Jγα,βp α,β k =
1 (1 +λk)γ/2p
α,β k ,
and, therefore iff ∈L2([−1,1], µα,β) polynomial with expansionP∞
k=0J
α,β k f
Jγα,βf = ∞
X
k=0
1 (1 +λk)γ/2J
α,β
k f. (12)
Again Meyer’s theorem allows us to extend Bessel Potentials to a continuous operator onLp([−1,1], µα,β),
Theorem 7. The operator Jα,β
γ admits a continuous extension, that it will
also be denoted as Jα,β
γ , toLp([−1,1], µα,β).
Proof. Bessel Potential of orderγ is a multiplier associated to the function Φ(k) =³ 1
1+λk
´γ/2
. Letβ ∈R,β >1 and h(z) =³ zβ zβ+1
´γ/2
.Then his an analytic function on a neighborhood of zero and
h
Ã
1 λ1k/β
!
=
µ
1 1 +λk
¶γ/2
= Φ(k).
The results follows applying Meyer’s theorem. ¤
Finally as in the classical case, we define the Fractional Derivative of order γ >0,Dα,β
γ , with respect to Jacobi measure as
Dγα,β= (−Lα,β)γ/2. (13) Observe that, for the Jacobi polynomials we have,
Dα,βγ p α,β k =λ
γ/2
k p α,β
k , (14)
and therefore, by the density of the polynomials in Lp([−1,1], µα,β), 1< p <
∞, Dα,β
For the Fractional Derivative of order 0< γ <1 we also have a integral representation,
Dγα,βf = 1 cγ
Z ∞
0
t−γ−1(Ptα,βf−f)dt, (15)
for f polynomial, where cγ = R∞
0 s
−γ−1(e−s−1)ds, since, for the Jacobi polynomials, we have, by the change of variabless=λ1k/2t,
Z ∞
0
t−γ−1(Pα,β t p
α,β k −p
α,β k )dt =
Z ∞
0
t−γ−1(e−λ1/2
k t−1)dt pα,β
k
= λγ/k 2
Z ∞
0
s−γ−1(e−s
−1)ds pα,βk = λγ/k 2cγpα,βk .
Now, iff is a polynomial, by (9) and (14) we have,
Iγα,β(Dα,βγ f) =Dα,βγ (Iγα,βf) = Π0f. (16)
In [5] H. Bavinck has defined Fractional Integration and Differentiation for Jacobi expansions. Nevertheless the motivation, the methods and techniques use in his paper are totally different from ours.
Now we are going to give an alternative representation ofDα,β
γ andIγα,β that are very useful in what follows. Before that, we need the following technical result of the asymptotic behavior of{Ptα,β}t≥0 at infinity.
Lemma 8. IfR1
−1f(y)µα,β(dy) = 0andf has continuos derivatives up to the
second order, then
¯ ¯ ¯ ¯
∂ ∂tP
α,β t f(x)
¯ ¯ ¯ ¯≤
Cf,α,β(1 +|x|)e−(α+β+2)1/2
t. (17)
As a consequence we have that the Poisson-Jacobi semigroup {Ptα,β}t≥0, has
exponential decay on (C0α,β)⊥ =
L∞
n=1Cnα,β. More precisely, if we have
R1
−1f(y)µα,β(dy) = 0,
|Ptα,βf(x)| ≤Cf,α,β(1 +|x|)e
−(α+β+2)1/2
Proof. First, let us prove that ¯¯
and asfhas continue derivatives up to the second order, there exist a constant Cf such that¯¯
By hypothesis, since we are assuming that R1
−1f(y)µα,β(dy) = 0,
lim t→∞P
α,β
Then Remember that, since{Ptα,β}t≥0 is an strongly continuos semigroup, we
have
lim t→0+P
α,β
t f(x) =f(x) (20) Now we are ready to give the alternate representation ofDα,β
γ andIγα,β,
Proof. Let us start proving (21). Integrating by parts in (15) we get
Dγα,βf(x) =
since, by (19), (20) and the previous lemma, we have
Let us prove now (22). Again, by integrating by parts, we have since, by the previous result
lim Observe that the previous proposition is also true for the Jacobi polyno-mials of order n >0,and therefore is true for any nonconstant polynomialf such that R1
−1f(y)µα,β(dy) = 0.
Now let us write Ptα,βf(x) = and define the operator Qα,βt as
Qα,βt f(x) =−t
Corollary 10. Supposef differentiable with continuos derivatives up to the second order such that R1
−1f(y)µα,β(dy) = 0, then we have An interesting use of the family{Qα,βt }is that it allows to give a version of Calder´on’s reproduction formula for the Jacobi measure,
Theorem 11. i) Suppose f ∈L1([−1,1], µα,β)such thatR1
ii) Suppose f a polynomial such that R1
−1f(y)µα,β(dy) = 0, then we have Formula (28) is the version of Calder´on’s reproduction formula for the Jacobi measure.
Proof. i) Using (20) and (19) we have,
Z ∞
Let us prove (28), given f a polynomial such that R1
−1f(y)µα,β(dy) = 0, by
Substituting in (30)
f =Dα,βγ
¡
Iγα,βf
¢
=Cγ
Z ∞
0
Z ∞
0
t−γ−1sγ−1Qα,β t
¡
Qα,βs f
¢
dsdt,
withCγ =− 1
γ2c
γΓ(γ).
To show (29), let us integrate by parts, and by Lemma 8 we have
Z ∞
0
u∂
2
∂u2P
α,β
u f(x)du = u ∂ ∂uP
α,β u f(x)
¯ ¯ ¯ ¯
∞
0
−
Z ∞
0
∂ ∂uP
α,β u f(x)du
= −
Z ∞
0
∂ ∂uP
α,β
u f(x)du=−Puα,βf(x)
¯ ¯
∞
0
= P0α,βf(x) =f(x).
¤ Now let us consider the Jacobi Potential Spaces. The Jacobi Potential space of order γ > 0, Lp
γ([−1,1], µα,β), for 1 < p < ∞, is defined as the completion of the polynomials with respect to the norm
kfkp,γ:=k(I− Lα,β)γ/2fkp. That is to say f ∈ Lp
γ([−1,1], µα,β) if, and only if, there is a sequence of polynomials {fn} such that limn→∞kfn−fkp,γ= 0.
As in the classical case, the Jacobi Potential space Lp
γ([−1,1], µα,β) can also be defined as the image of Lp([−1,1], µα,β) under the Bessel Potential
Jα,β
γ , that is,
Lγp([−1,1], µα,β) =Jγα,βLp([−1,1], µα,β). For the details of the proof of this equivalence, we refer to [2].
Now, let us consider some inclusion properties among Jacobi Potential Spaces,
Proposition 12. i) If p < q, then Lq
γ([−1,1], µα,β) ⊆ Lpγ([−1,1], µα,β)
for eachγ >0.
ii) If0< γ < δ,thenLpδ([−1,1], µα,β)⊆Lp
γ([−1,1], µα,β)for each0< p <
∞.
ii)Letf be a polynomial and consider
φ= (I− Lα,β)δ/2f = ∞
X
k=0
(1 +λk)δ/2Jkα,βf,
which is also a polynomial. Then φ∈ Lpδ([−1,1], µα,β), kφkp =kfkp,δ and
J(α,βγ−δ)φ= (I−Lα,β)(γ
−δ)/2φ= (I−Lα,β)γ/2f.By theLpcontinuity of Bessel Potentials,
kfkp,γ=k(I− Lα,β)γ/2fkp=kJ(γ−δ)φkp≤Cpkfkp,δ.
Now letf ∈Lpδ([−1,1], µα,β). Then there existsg∈Lp([−1,1], µα,β) such that f =Jδα,βg and a sequence of polynomials{gn}in Lp([−1,1], µα,β) such that limn→∞kgn−gkp= 0.Setfn =Jδα,βgn.Then limn→∞kfn−fkp,δ= 0, and
kfn−fkp,γ=k(I− Lα,β)γ/2(fn−f)kp=k(I− Lα,β)γ/2(I− Lα,β)
−δ/2
(gn−g)kp
=k(I− Lα,β)(γ−δ)/2
(gn−g)kp=kJ(γ−δ)(gn−g)kp.
By the Lp continuity of Bessel Potentials lim
n→∞kfn−fkp,γ= 0. Therefore,
kfkp,γ ≤ kfn−fkp,γ+kfnkp,γ
≤ kfn−fkp,γ+kfnkp,δ,
taking limit as ngoes to infinity, we obtain the result. ¤ Let us consider the space
Lγ([−1,1], µα,β) = [ p>1
Lpγ([−1,1], µα,β).
Lγ([−1,1], µα,β) is the natural domain ofDα,β
γ . We define it in this space as follows.
Let f ∈ Lγ([−1,1], µα,β), then there is p > 1 such that f ∈
Lp
γ([−1,1], µα,β) and a sequence{fn} polynomials such that limn→∞fn =f in Lpγ([−1,1], µα,β).We define forf ∈Lγ([−1,1], µα,β)
Dα,βγ f = lim n→∞D
α,β γ fn
in Lp([−1,1], µα,β). The next theorem shows that Dα,β
Theorem 13. Let γ >0 and1< p, q <∞.
i) If {fn} is a sequence of polynomials such that limn→∞fn = f in Lp
γ([−1,1], µα,β), thenlimn→∞Dγα,βfn∈Lp([−1,1], µα,β) and the limit
does not depend on the choice of the sequence{fn}.
If f ∈ Lp
γ([−1,1], µα,β)
T
Lq
γ([−1,1], µα,β), then the limit does not
de-pend on the choice ofporq. Thus Dα,β
γ is well defined onLγ([−1,1], µα,β).
ii) f ∈ Lp
γ([−1,1], µα,β) if, and only if, Dγα,βf ∈ Lp([−1,1], µα,β).
More-over, there exists positive constantsAp,γ andBp,γ such that
Bp,γkfkp,γ ≤ kDα,βγ fkp≤Ap,γkfkp,γ. (31)
Proof.
ii)First, let us note that forp=P∞
n=0Jnα,βppolynomial,
Dα,βγ Jγα,βp= ∞
X
n=0
µ
λn 1 +λn
¶γ/2
Jnα,βp,
that is, the operator Dα,β
γ Jγα,β is a multiplier with associated function Φ(n) = ³ λn
1+λn
´γ/2
= h( 1
λn) where h(z) =
³
1
z+1
´γ/2
, and therefore by Meyer’s theorem it is Lp-continuos.
Letf be a polynomial and let φ be a polynomial such thatf =Jα,β γ φ. We have that kfkp,γ=kφkp and by the continuity of the operatorDγα,βJγα,β
kDα,βγ fkp=kDα,βγ Jγα,βφkp≤Ap,γkφkp=Ap,γkfkp,γ.
To prove the converse, let us suppose that f polynomial, thenDα,β γ f is also a polynomial, and thereforeDα,β
γ f ∈Lp([−1,1], µα,β).Consider
φ= (I− Lα,β)γ/2f = ∞
X
k=0
(1 +λk)γ/2Jkα,βf = ∞
X
k=0
µ
1 +λk λk
¶γ/2
Jkα,β(Dα,βγ f).
The mapping
g= ∞
X
k=0
Jkα,βg7→
∞
X
k=0
µ
1 +λk λk
¶γ/2
is a multiplier with associated function Φ(k) = ³1+λk
λk
´γ/2
= h(λ1k) where h(z) = (z+ 1)γ/2,so by Meyer’s theorem, takingg=Dα,β
γ f we have
kfkp,γ=kφkp≤Bp,γkDγα,βfkp.
Thus we get (31) for polynomials. For the general case consider f ∈
Lp
γ([−1,1], µα,β), then there existsg∈Lp([−1,1], µα,β) such thatf =Jγα,βg and a sequence {gn} of polynomials such that limn→∞kgn −gkp = 0. Let fn =Jα,β
γ gn,then limn→∞kfn−fkp,γ = 0. Then, by the continuity of the operator Dα,β
γ Jγα,β and as limn→∞kgn−gkp= 0, lim
n→∞kD α,β
γ (fn−f)kp= lim n→∞kD
α,β
γ Jγα,β(gn−g)kp= 0. Then, as
Bp,γkfnkp,γ ≤ kDα,βγ fnkp≤Ap,γkfnkp,γ,
the results follows by taking the limit asngoes to infinity in this inequality. i) Let {fn} be a sequence of polynomials such that limn→∞fn = f in Lp
γ([−1,1], µα,β). Then, by (31) lim
n→∞kD α,β
γ fnkp≤Bp,γ lim
n→∞kfnkp,γ =Bp,γkfkp,γ, hence, limn→∞Dα,βγ fn∈Lp([−1,1], µα,β).
Now suppose that {qn} is another sequence of polynomials such that limn→∞qn =f in Lpγ([−1,1], µα,β). By (31)
Bp,γkfnkp,γ≤ kDα,βγ fnkp≤Ap,γkfnkp,γ
and
Bp,γkqnkp,γ≤ kDα,βγ qnkp≤Ap,γkqnkp,γ. Taking the limit asngoes to infinity
Bp,γkfkp,γ≤ lim n→∞kD
α,β
γ fnkp≤Ap,γkfkp,γ
Bp,γkfkp,γ≤ lim n→∞kD
α,β
γ qnkp≤Ap,γkfkp,γ,
therefore limn→∞Dγα,βfn = limn→∞Dγα,βqn inLp([−1,1], µα,β) and the limit does not depends on the choice of the sequence.
Finally, let us suppose that f ∈ Lp
γ([−1,1], µα,β)
T
Lq
Lqγ([−1,1], µα,β) ⊆ Lpγ([−1,1], µα,β) and therefore f ∈ Lqγ([−1,1], µα,β). Then, if {fn} is a sequence of polynomials such that limn→∞fn = f in Lq
γ([−1,1], µα,β) (hence inLpγ([−1,1], µα,β)), we have lim
n→∞D α,β
γ fn∈Lq([−1,1], µα,β) =Lp([−1,1], µα,β)
\
Lq([−1,1], µα,β). Therefore the limit does not depends on the choice of porq. ¤
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