Fractional Generalizations of Rodrigues-Type Formulas for Laguerre Functions in Function Spaces

Article Fractional Generalizations of Rodrigues-Type Formulas for Laguerre Functions in Function Spaces

Pedro J. Miana 1,* and Natalia Romero 2

1 Departamento de Matemáticas, Facultad de Ciencias & IUMA, Universidad de Zaragoza, 50009 Zaragoza, Spain 2 Departamento de Matemáticas y Computación, Facultad de Ciencia y Tecnología, Universidad de La Rioja, 26006 Logrono,˜ Spain; [email protected] * Correspondence: [email protected]

(α) Abstract: Generalized Laguerre polynomials, Ln , verify the well-known Rodrigues’ formula. Using Weyl and Riemann–Liouville fractional calculi, we present several fractional generalizations of Rodrigues’ formula for generalized Laguerre functions and polynomials. As a consequence, we give a new addition formula and an integral representation for these polynomials. Finally, we introduce a new family of fractional Lebesgue spaces and show that some of these special functions belong to them.

Keywords: Rodrigues’ formula; Laguerre functions; function spaces; fractional calculus

1. Introduction

Citation: Miana, P.J.; Romero, N. In approximation theory, the classical orthogonal polynomials of Jacobi, Laguerre, and Fractional Generalizations of Hermite have many properties in common, namely, the Rodrigues formula, the differential Rodrigues-Type Formulas for equation, the derivative formula, and the three-term recurrence relation. Under some con- Laguerre Functions in Function ditions, these common properties are equivalent and characterize these classical orthogonal Spaces. Mathematics 2021, 9, 984. polynomials. See more details, for example, in [1] [Chapter 12] and [2] [Chapter V]. https://doi.org/10.3390/ Polynomial solutions in the differential equation math9090984 zw00(z) + (α + 1 − z)w0(z) + nw(z) = 0, Academic Editor: Ferenc Hartung (α) with n = 0, 1, 2 ... and α ∈ C are called generalized Laguerre polynomial, Ln . They verify Received: 26 March 2021 Rodrigues’ formula, Accepted: 22 April 2021 −α x n (α) x e d n+α −x Published: 27 April 2021 L (x) = (x e ), (1) n n! dxn where we have Publisher’s Note: MDPI stays neutral n m (α) m n + α x with regard to jurisdictional claims in L (x) := (−1) , n ∑ n − m m! published maps and institutional afﬁl- m=0 iations. n + α Γ(n + α + 1) with = , see, for example, [2] [p. 241] and [1] [Chapter 12]. n − m Γ(α + m + 1)(n − m)! In particular, they are

(α) Copyright: © 2021 by the authors. L0 (x) = 1, Licensee MDPI, Basel, Switzerland. ( ) L α (x) = α + 1 − x, This article is an open access article 1 distributed under the terms and (α) 1 2 L2 (x) = (α + 1)(α + 2) − 2(α + 2)x + x . conditions of the Creative Commons 2 Attribution (CC BY) license (https:// creativecommons.org/licenses/by/ 4.0/).

Mathematics 2021, 9, 984. https://doi.org/10.3390/math9090984 https://www.mdpi.com/journal/mathematics Mathematics 2021, 9, 984 2 of 15

Generalized Laguerre polynomials satisfy several recurrence equalities, see [2] [p. 241], between them, for example,

(α+1) (α) (α) xLn (x) = (n + α + 1)Ln (x) − (n + 1)Ln+1(x). (2)

Rodrigues’ formula was initially introduced for Legendre polynomials by Olinde Rodrigues in 1816. The name “Rodrigues formula” was given by Heine in 1878, after Hermite pointed out in 1865 that Rodrigues was the ﬁrst to discover it instead of Ivory and Jacobi. The term is also used to describe similar formulas for other orthogonal polynomials, mainly Laguerre and Hermite polynomials and many other sequences of orthogonal functions. These are also called the Rodrigues formula (or Rodrigues’ type formula) for that case, especially when the resulting sequence is polynomial. By means of fractional calculi, several generalizations of Rodrigues formula have appeared in the literature in the last years. We present some of them in the next lines. Several important special functions can be expressed as derivatives of complex order of elementary function, see, for example, [3]. The derivative of a complex order ν of a complex function f of a complex variable z is deﬁned by the generalized Cauchy integral,

d ν Γ(ν + 1) Z f (ν)(z) = f (z) = f (w)(w − z)−ν−1dw, dz 2πi γ

under some assumptions about ν, f and the path γ [3,4] [p. 113]. In the case of Laguerre functions, we have −µ ν (µ) z d L (z) = ez (zν+µe−z), ν Γ(1 + ν) dz which coincides with the Rodrigues’ Formula (1) for ν = n. In [5], certain Laguerre polynomials of arbitrary orders are deﬁned. The fractional Caputo derivative Dα of order α ∈ (n − 1, n] of a function f is given there by

1 Z x Dα f (x) = (x − t)n−α−1 f (n)(t)dt, x ≥ 0, Γ(n − α) 0

β see [5] [Deﬁnition 1.3]. The author deﬁnes the Laguerre polynomials Łα of order α > 0 by

−β x β x e Ł (x) = Dα(e−xxα+β), x > 0, β > −1, α Γ(1 + α)

and proves β β (β) lim Łα (x) = lim Łα (x) = Ln (x). α→n+ α→n−

A wide generalization of Rodrigues’ formula is treated in [6]. The author considers the Riemann–Liouville integral to include a large numbers of special functions, in particular Laguerre polynomials and functions [6] [Section 1]. In [7], authors use a generalization of the Rodrigues’ formula to deﬁne a new special function. They study some of its properties, some recurrence relations, orthogonality property, and the continuation to the Rodrigues’ formula of the Laguerre polynomials as a limit case. In addition, the conﬂuent hypergeometric representation is given. α In this paper, we consider the Weyl and Riemann–Liouville fractional calculi, W+ and α D+ with α ∈ R in the half real line in the second section. In Section3, Theorem1, we show the following fractional Rodrigues’ formulae:

( + ) M(α, ν + 1, z) = Γ ν 1 z−νezD−α(tν−αe−t)(z), Γ(−α+ν+1) + (3) −ν z −α ν−α −t U(α, ν + 1, z) = z e W+ (t e )(z), Mathematics 2021, 9, 984 3 of 15

where M(α, ν + 1, z) and U(α, ν + 1, z) are the conﬂuent hypergeometric functions,

∞ Γ(α + j) Γ(ν) zj M(α, ν, z) := ; ∑ ( ) ( + ) j=0 Γ α Γ ν j j! Γ(1 − ν) Γ(ν − 1) U(α, ν, z) := M(α, ν, z) + z1−ν M(α + 1 − ν, 2 − ν, z). Γ(α + 1 − ν) Γ(α)

This theorem extends [8] [Theorem 3] and completes the picture given in formulae [6] [(8)–(12)], where the author only considers the Riemann–Liouville fractional calculus. In the particular case −α = n ∈ N, the conﬂuent hypergeometric functions are essentially the Laguerre polynomials and we get a second fractional Rodrigues’ formula,

n ( −n) (−1) L α (x) = exWα (tne−t)(x), x ≥ 0, n n! + in Theorem2. As a consequence, we get a new integral addition formula for Laguerre α n −λt polynomials in Corollary1. We also obtain a integral representation of W+(t e ), i.e.,

n Γ(α + k) n Z ∞ Wα (rne−λr)(t) = λαe−λttn + λα (−1)k (r − t)k−1rn−ke−λrdr, + ∑ ( ) ( ) k=1 Γ α Γ k k t

(α−n) and apply it to get a new integral representation of Ln (t), i.e.,

n n ( −n) (−1) Γ(α + k) n L α (t) = et (−1)k W−k rn−ke−r (t), t > 0. n ∑ ( ) + n! k=0 Γ α k

in Theorem3. All of these results show the deep and interesting connection between fractional calculi, in particular Weyl fractional derivation, and Laguerre polynomials. (α) µ α In the last section, we introduce new fractional Lebesgue space Tp (t + t ) which are contained in Lp(R+) with 0 ≤ µ ≤ α and p ≥ 1. Note that we understand that (0) 0 0 p + (α) µ α Tp (t + t ) = L (R ). As in the classical case, we show that the space Tp (t + t ) (α) µ α for p > 1 is module for the algebra T1 (t + t ) (Theorem4). This family of function spaces contains as a particular case some spaces which have appeared previously in the literature [9–13]. Finally, we present some special functions which belong to these fractional (α) µ α Lebesgue spaces Tp (t + t ) in Remark2.

2. Weyl and Riemann–Liouville Fractional Calculi ∞ We denote by D+ the set of test functions of compact support in [0, ∞), D+ ≡ Cc ([0, ∞)) and by S+ the Schwartz class on [0, ∞), i.e., functions which are inﬁnitely differentiable, which veriﬁes dn m ( ) < sup t n f t ∞, t≥0 dt for any m, n ∈ N ∪ {0}.

Deﬁnition 1. Given f ∈ S+, the Weyl fractional integral of f of order α > 0 is deﬁned by

Z ∞ −α 1 α−1 W+ f (u) := (t − u) f (t)dt, u ≥ 0, Γ(α) u

−α α with α > 0. This operator W+ : S+ → S+ is one to one, and its inverse, W+, is the Weyl fractional derivative of order α, and

(−1)n dn Z ∞ α ( ) = ( − )n−α−1 ( ) ≥ W+ f t n s t f s ds, t 0, Γ(n − α) dt t Mathematics 2021, 9, 984 4 of 15

holds with n = [α] + 1; see, for example, [14,15].

dn It is easy to check that, if α = n ∈ N, then Wα f = (−1)α f (α) = (−1)n and + dtn α+β α β 0 W+ f = W+(W+ f ) with α, β ∈ R, W+ = Id and f ∈ S+ and

α α α W+( fr) = r (W+ f )r, (4)

where fr(s) := f (rs) for r > 0; see more details in [14,15].

+ −λs Example 1. Let λ ∈ C and eλ(s) := e with s ≥ 0. It is clear that eλ ∈ S+ and

−α −α −λs W+ (eλ)(s) = λ e , s ≥ 0.

α α −λs Then, W+eλ(s) = λ e for α ∈ R and s ≥ 0. In Theorem3, we give an integral expression of α n −λr W+(r e ).

Proposition 1. Take α ∈ R and f ∈ S+ then :

α α α−1 W+(s f (s))(t) = tW+ f (t) − αW+ f (t) t > 0.

Proof. If α < 0, it is shown in [14] [p. 246]; if α > 0, we have

−α α α−1 −1 −1 W+ (sW+ f (s) − αW+ f (s))(t) = t f (t) + αW+ f (t) − αW+ f (t) = t f (t)

with t > 0.

The usual convolution product ∗ on R+ is deﬁned by

Z t ( f ∗ g)(t) := f (t − s)g(s)ds, t ∈ R+, 0 for functions f , g which are “good enough”, for example, absolutely integrable functions. For functions f , g ∈ S+, the following integral equality for the convolution product holds

α−1 α ( ∗ )( ) = R s α ( ) R s (t+r−s) α ( ) W+ f g s 0 W+g r s−r Γ(α) W+ f t dt dr α−1 (5) − R ∞ α ( ) R ∞ (t+r−s) α ( ) s W+g r s Γ(α) W+ f t dt dr

for s ∈ R+ ([10,12] [Proposition 1.2]).

Deﬁnition 2. Given f ∈ S+, Riemann–Liouville fractional integral of order α > 0 is deﬁned by

Z t −α 1 α−1 D+ f (t) := (t − s) f (s)ds, t ≥ 0. Γ(α) 0

n If α = n ∈ N, the Riemann–Liouville derivative Riemann–Liouville of order n, D+, is the usual derivative and, if α ∈ R+\N, Riemann–Liouville fractional derivative of order α is deﬁned by

1 dn Z t α ( ) = ( − )n−α−1 ( ) ≥ D+ f t : n t s f s ds, t 0, Γ(n − α) dt 0

with n = [α] + 1, see, for example, [14–16]. Mathematics 2021, 9, 984 5 of 15

−α −β −(α+β) If f ∈ S+, then D+ D+ f = D+ f for α, β ≥ 0, but, in general, it is false that α β α+β D+D+ f = D+ f with α, β ∈ R. For example,

Z t −1 1 −s −s −t −t D+ D+(e )(t) = − e ds = e − 1 6= e . 0

If α > 0 and f , g ∈ S+, we apply the Fubini theorem to get that

Z ∞ Z ∞ −α −α W+ f (t)g(t)dt = f (t)D+ g(t)dt 0 0

−α α and, if we apply W+ (W+g) = g, we obtain the following “integration by parts” formula:

Z ∞ Z ∞ −α α f (t)g(t)dt = D+ f (t)W+g(t)dt. (6) 0 0 We will use both equalities in the following sections. Note that Formula (6) shows the dual behaviour between both fractional calculi.

3. Fractional Rodrigues’ Formulae for Conﬂuent Hypergeometric Functions Two linearly independent solutions of Kummer’s differential equation

zw00(z) + (ν − z)w0(z) − αw(z) = 0, α, ν ∈ R

are given by conﬂuent hypergeometric functions M(−α, 1 + ν, z) (also written by 1F1(−α, 1 + ν, z)) and U(−α, 1 + ν, z), see deﬁnitions in the Introduction, Formula (3). In the particular case of −α = n ∈ N, we have that

n ( ) Γ(ν + n + 1) (−1) L ν (z) = M(−n, 1 + ν, z) = U(−n, ν + 1, z), z ∈ C. n Γ(1 + ν)n! n!

There is a big amount of equalities which conﬂuent hypergeometric functions verify. We consider the following ones:

dn −z c−1 n −z c−n−1 dzn e z M(a, c, z) = (−1) (1 − c)ne z M(a − n, c − n, z), (7) dn −z c−a+n−1 n −z c−a−1 dzn e z U(a, c, z) = (−1) e z U(a − n, c, z). The following integral representations hold:

1−ν ( ) = Γ(ν)z R z t α−1( − )ν−α−1 < > < > M α, ν, z Γ(α)Γ(ν−α) 0 e t z t dt, ν α 0; (8) − ( ) = z1 ν R ∞ −t α−1( + )ν−α−1 < > < > U α, ν, z Γ(α) 0 e t z t dt, ν 0, z 0,

and also the recurrence relation

U(α, ν, z) − U(α, ν − 1, z) − αU(α + 1, ν, z) = 0, (9)

see all these formulae and much more in [2] [Chapter VI]. The part (i) of the next theorem includes [8] [Theorem 3] for α < 0 and ν > −1 and for α ∈ R and ν > α − 1 are presented in [6] [Section 1]. We include the proof of both parts to avoid the lack of completeness.

Theorem 1. Given α ∈ R and 0, the following equalities hold: Γ(ν + 1) (i) M(α, ν + 1, z) = z−νezD−α(tν−αe−t)(z), for ν > α − 1. Γ(−α + ν + 1) + −ν z −α ν−α −t (ii) U(α, ν + 1, z) = z e W+ (t e )(z), for ν ∈ R. Mathematics 2021, 9, 984 6 of 15

Proof. (i) Let ν > α − 1. For α = 0, M(0, ν + 1, z) = 1 and the equality holds. Take α > 0 and

Γ(ν + 1)z−ν Z z M(α, ν + 1, z) = ettα−1(z − t)ν−αdt Γ(α)Γ(ν + 1 − α) 0 Γ(ν + 1)z−ν Z z = ez e−t(z − t)α−1tν−αdt Γ(α)Γ(ν + 1 − α) 0 Γ(ν + 1) = z−νezD−α(tν−αe−t)(z), Γ(−α + ν + 1) +

for z > 0 and by holomorphy 0. Finally, take α < 0. We write n = [−α] + 1

dn D−α(tν−αe−t)(z) = (D−n−α(tν−αe−t))(z), + dzn + and we apply the above equality for α > 0 to get

Γ(ν + 1 − α) D−n−α(tν−αe−t)(z) = e−zzν+n M(n + α, ν + n + 1, z), + Γ(ν + n + 1)

for z > 0. We apply the ﬁrst formula in (7) to get

Γ(ν + 1 − α) dn D−α(tν−αe−t)(z) = e−zzν+n M(n + α, ν + n + 1, z)(z) + Γ(ν + n + 1) dzn Γ(ν + 1 − α) −z ν n = e z (−1) (−ν − n)n M(α, ν + 1, z) Γ(ν + n + 1) Γ(ν + 1 − α) = e−zzν M(α, ν + 1, z), Γ(ν + 1)

and we get the equality for z > 0 and then 0. (ii) Let ν ∈ R. For α = 0, U(0, ν + 1, z) = 1 and the equality holds. Take α > 0. We apply the second formula in (8) to get

z−ν Z ∞ z−νez Z ∞ U(α, ν + 1, z) = e−ttα−1(z + t)ν−αdt = e−t(t − z)α−1tν−αdt Γ(α) 0 Γ(α) z −ν z −α ν−α −t = z e W+ (t e )(z),

for z > 0 and by holomorphy 0. Finally, take α < 0 and n = [−α] + 1. Firstly, we consider n = 1, i.e., −1 < α < 0. By the second formula in (7) for n = 1 and the above equality for α > 0, we get that

d U(α, ν + 1, z) = (−1)ezz−(ν−α−1) e−zzν−αU(1 + α, ν + 1, z) dz d −( + ) = (−1)ezz−(ν−α−1) z−αW 1 α (tν−α−1e−t)(z) dz + z −ν −(1+α) ν−(1+α) −z z −(ν−1) −α) ν−1−α) −z = αe z W+ (t e ) + e z W+ (t e ) z −(ν−1) −α ν−1−α −z = αU(1 + α, ν + 1, z) + e z W+ (t e ),

for z > 0. By the recurrence relation (9), we have that

U(α, ν + 1, z) = αU(1 + α, ν + 1, z) + U(α + 1, ν, z),

z −(ν−1) −α ν−1−α −z and we conclude that U(α, ν + 1, z) = e z W+ (t e ), for z > 0 and 0. Now, we suppose that the equality holds for n + 1 > −α > n, and we claim that it also is Mathematics 2021, 9, 984 7 of 15

true for n + 1 > −α > n. Again, by the second formula in (7) for n = 1 and our hypothesis, we get that

d U(α, ν + 1, z) = (−1)ezz−(ν−α−1) e−zzν−αU(1 + α, ν + 1, z) dz z −ν −(1+α) ν−(1+α) −z z −(ν−1) −α ν−1−α −z = αe z W+ (t e ) + e z W+ (t e ) z −(ν−1) −α ν−1−α −z = αU(1 + α, ν + 1, z) + e z W+ (t e ),

for z > 0. Again, by the recurrence relation (9), we conclude that

z −(ν−1) −α ν−1−α −z U(α, ν + 1, z) = e z W+ (t e ),

for z > 0 and 0, and the proof is ﬁnished.

4. Fractional Rodrigues’ Formulae for Laguerre Polynomials n (α−n) (−1) x α n −t Now, ﬁrst we check Ln (x) = n! e W+(t e )(x) with α ∈ R and x ≥ 0. In the case α = 0, the equality holds directly (see [2] [p. 240]) since

n n (−n) (−x) (n) (−x) L (x) = L (x) = , n n! 0 n! and then n n (−1) (−x) (−n) exW0 (tne−t)(x) = = L (x). n! + n! n (α) Analogously, we have L0 (x) = 1.

Theorem 2. Let α ∈ R, n ∈ N. Then,

n ( −n) (−1) L α (x) = exWα (tne−t)(x), x ≥ 0. n n! +

Proof. We give a proof by induction. Take α > 0; for n = 1, we apply Proposition1 to get

α −t α −t α−1 −t −x (α−1) W+(te )(x) = xW+(e )(x) − αW+ (e )(x) = e (x − α) = −L1 (x).

Taking the case n + 1 and again by Proposition1, we obtain

α n+1 −t α n −t α−1 n −t W+(t e )(x) = xW+(t e )(x) − αW+ (t e )(x), x > 0

and, using induction hypothesis,

α n+1 −t −x n (α−n) (α−1−n) W+(t e )(x) = e n!(−1) xLn (x) − αLn+1 (x) , x > 0.

We apply the recurrence Formula (2) and

α n+1 −t −x n (α−n−1) W+(t e )(x) = e n!(−1) (−1)(n + 1)Ln+1 (x) −x n+1 (α−(n+1)) = e (−1) (n + 1)!Ln+1 (x)

and we get the equality. In the case α < 0, we work in a similar way.

Remark 1. In the case α = m ∈ N, we obtain an equivalent formula to

−β x m (β) x e d L (x) = (xm+βe−x). m m! dxm

(−k) (n − k)! (k) To check this, we use L (x) = (−x)k L (x), 1 ≤ k ≤ n, ([2] p. 240). n n! n−k Mathematics 2021, 9, 984 8 of 15

If λ > 0, we also get

α n −λt α−n −λx n (α−n) W+(t e )(x) = λ e (−1) n!Ln (λx) ! n α n! = λαe−λx λm−n(−1)n−m xm . ∑ − m=0 n m m!

The addition formula for Laguerre polynomials states that

n (α+β+1) (α) (β) R Ln (t + r) = ∑ Lj (t)Ln−j(r), t, r ∈ . j=0

([2] [p. 249]). The following corollary shows a new integral addition formula for Laguerre polynomials.

Corollary 1. Let α ∈ R and n, m ∈ N. Then,

Z ∞ Z ∞ α−1 −s (α−(n+m+1)) −r (α−n) (t + r − s) −t (α−m) e Ln+m+1 (s) = e Ln (r) e Lm (t)dt dr s s Γ(α) Z s Z s α−1 −r (α−n) (t + r − s) −t (α−m) − e Ln (r) e Lm (t)dt dr 0 s−r Γ(α)

for s ∈ R+.

n s −s + Proof. We write p (s) = e for s ∈ R . Note that p ∗ p = p + + for n, m ∈ N. By n n! n m n m 1 α n −s (α−n) Theorem2, W (pn)(s) = (−1) e Ln (s) for α, s ∈ R and n ∈ N. Finally, we apply Formula (5) to conclude the equality.

α n −λt Now, we want to give another representation, an integral representation of W+(t e ) (α−n) and Ln . To do this, we check the following Lemma about the Pochhammer symbol (α)j, where

Γ(α + j) (α) := α(α + 1) ... (α + j − 1) = , j ∈ N, α ∈ C. j Γ(α)

Lemma 1. Let l ∈ N and α ∈ C. Then,

l (α) l (−α) (−1)k k = l . (10) ∑ ( − ) ( − ) k=1 k 1 ! k l 1 !

Proof. Since (−α) α(α − 1) ... (α − l + 1) l = (−1)l , (l − 1)! (l − 1)! l it is enough to prove that Pl is a polynomial in α of degree l, the leading coefﬁcient (−1) /l! and roots {0, 1, . . . , l − 1} where

l (α) l l α(α + 1) ... (α + k − 1) l P (α) := (−1)k k = (−1)k . l ∑ ( − ) ∑ ( − ) k=1 k 1 ! k k=1 k 1 ! k

Let j ∈ {1, . . . , l − 1} and we consider the polynomial:

l l xj−1(1 − x)l = (−1)l ∑ (−1)kxk+j−1. k=0 k Mathematics 2021, 9, 984 9 of 15

We derive j times and evaluate in x = 1 to obtain

l l ∑ (−1)k(k + j − 1) ... k = 0, k=0 k

and then

l j(j + 1) ... (j + k − 1) l P (j) = (−1)k l ∑ ( − ) k=1 k 1 ! k 1 l l = (−1)kk(k + 1) ... (j + k − 1) = 0, ( − ) ∑ j 1 ! k=0 k

and we conclude the proof.

Theorem 3. If n ∈ N and λ, α > 0, we have that

n Γ(α + k) n Wα (rne−λr)(t) = λα (−1)k W−k rn−ke−λr (t), t > 0, + ∑ ( ) + k=0 Γ α k

and n n ( −n) (−1) Γ(α + k) n L α (t) = et (−1)k W−k rn−ke−r (t), t > 0. n ∑ ( ) + n! k=0 Γ α k

Proof. By Lemma1, we have that

n! l Γ(α + k) l n! α (−1)k = (−1)l , ∑ ( ) ( ) j!l k=1 Γ α Γ k k j! l

for l ≥ 1 and 0 ≤ j ≤ n. By the second remark in Theorem2 and taking in the last expression l = n − j, we have that

n−1 α n! λ−αeλtWα (rne−λr)(t) = tn + λj−n(−1)n−j tj + ∑ − j=0 n j j! − n−1 n! n j Γ(α + k) n − j = tn + λj−ntj (−1)k ∑ ( − ) ∑ ( ) ( ) j=0 j! n j k=1 Γ α Γ k k n Γ(α + k) n n−k n − k = tn + (−1)k λj−ntj (n − 1 − j)! ∑ ( ) ( ) ∑ k=1 Γ α Γ k k j=0 j

n! n − j nn − k due to = (n − j)! . j! k k j In the other hand, we apply Newton’s formula, to get the equality:

Z ∞ n−k n − k Z ∞ (r − t)k−1rn−ke−λrdr = ∑ tj (r − t)n−j−1e−λrdr t j=0 j t n−k n − k = e−λt ∑ tjλj−n(n − 1 − j)! j=0 j

with t ≥ 0. From here, we have that

n Γ(α + k) n Z ∞ Wα (rne−λr)(t) = λαe−λttn + λα (−1)k (r − t)k−1rn−ke−λrdr, + ∑ ( ) ( ) k=1 Γ α Γ k k t Mathematics 2021, 9, 984 10 of 15

and we obtain the equality. As a consequence, we apply Theorem2 and the ﬁrst equality to obtain the second one.

As a corollary of this theorem, we show an equality considered in [14] [p. 315]. However, ﬁrst, we need to comment some aspects about Laplace transform. Given f ∈ S+, the Laplace transform of f , L( f ), is given by

Z ∞ L( f )(z) := f (t)e−ztdt,

see, for example [17]. If f is a function in two variables f = f (t, s), L( f ; t) and L( f ; s) are Laplace transforms, if there exist, in each parameter. If we apply equality (6) in the integral representation of the Laplace transform, we get that, for α > 0 Z ∞ α L( f )(z) = W+ f (t)Sα(t, z)dt,

Z t −α 1 α−1 −zs Sα(t, z) := D+ (ez)(t) = (t − s) e ds

In fact, function Sα(t, z) may be deﬁned for any z ∈ C and

Z t 1 β−1 Sα+β(t, z) = (t − s) Sα(s, z)ds, z ∈ C, t > 0. (12) Γ(β) 0

Corollary 2. Let a, α > 0, λ > a and n ∈ N ∪ {0}. Then, we have that

1 λα n n Γ(α + k) = (−1)k L(S (r, −a)rn−k)(λ). ( − )n+1 ∑ ( ) α+k λ a n! k=0 k Γ α

Proof. By the equality (6) and Theorem3, we get that

1 Z ∞ tn 1 Z ∞ = at −λt = α ( n −λr)( ) −α( ar)( ) n+ e e dt W+ r e t D+ e t dt (λ − a) 1 0 n! n! 0 λα n Γ(α + k) n Z ∞ Z ∞ = (−1)k S (t, −a) (r − t)k−1rn−ke−λrdrdt. ∑ ( ) ( ) α n! k=0 Γ α Γ k k 0 t

We apply the Fubini theorem to get

1 λα n Γ(α + k) n Z ∞ Z r = (−1)k rn−ke−λr (r − t)k−1S (t, −a)dtdr ( − )n+1 ∑ ( ) ( ) α λ a n! k=0 Γ α Γ k k 0 0 λα n Γ(α + k) n Z ∞ = (−1)k rn−ke−λrS (r, −a)dr, ∑ ( ) α+k n! k=0 Γ α k 0

and we conclude the equality.

5. Fractional Lebesgue Spaces The well-known Hardy inequality states that

1 !p Z ∞ Γ( ) Z ∞ | −α ( )|p ≤ p αp| ( )|p W+ f t dt 1 t f t dt, (13) 0 Γ(α + p ) 0

for p ≥ 1 and α ≥ 0, see, for example, [18] [pp. 244–245]. Mathematics 2021, 9, 984 11 of 15

(α) µ α For 0 ≤ µ ≤ α, we introduce a family of subspaces Tp (t + t ) that are contained in Lp(R+).

(α) µ α Deﬁnition 3. For α > 0, let the Banach space Tp (t + t ) be deﬁned as the completion of the Schwartz class S+ in the norm

1 Z ∞ p 1 α p µ α p || f ||(µ,α),p := |W+ f (t)| (t + t ) dt . Γ(α + 1) 0

(0) 0 0 p + We understand that Tp (t + t ) = L (R ) and || ||(0,0),p = || ||p. The case p = 1, µ = 0 and α ∈ N were introduced in [9], and, for p = 1, 0 ≤ µ ≤ α and α > 0 in [10] [Section 1]. Finally, the case p > 1, µ = α and α > 0 were considered in [11] [Deﬁnition 2.1] and [13] [Section 1.2]. In [19], other families of function spaces are studied connected with a kernel function k on [0, ∞). (α) µ α In the next proposition, we present some results for spaces Tp (t + t ) which extends [11] [Proposition 2.2].

Proposition 2. Take p ≥ 1 and β > α ≥ µ > 0. (α) µ α p + (i) The operator Dµ,α : Tp (t + t ) → L (R ) deﬁned by

1 ( ) D f (t) := (tµ + tα)Wα f (t), t ∈ R+, f ∈ T α (tµ + tα), µ,α Γ(α + 1) + p

−1 p + (α) µ α is an isometry whose inverse operator (Dµ,α) : L (R ) → Tp (t + t ) is given by

−1 −α µ α −1 + (α) µ α (Dµ,α) f (t) = W+ ((t + t ) f )(t), t ∈ R f ∈ Tp (t + t ).

(β) µ+β−α β (α) µ α p + (ii) Tp (t + t ) ,→ Tp (t + t ) ,→ L (R ). 0 1 1 (α) µ α (α) µ α (iii) If p > 1 and p satisﬁes p + p0 = 1, then the dual of Tp (t + t ) is Tp0 (t + t ), where the duality is given by

Z ∞ 1 α α µ α 2 h f , giµ,α = W+ f (t)W+g(t)(t + t ) dt = hDµ,α f , Dµ,αgi0 Γ(α + 1)2 0

(α) µ α (α) µ α for f ∈ Tp (t + t ), g ∈ Tp0 (t + t ).

Proof. (i) By deﬁnition, we have

Z ∞ 1/p 1 µ α α p kDµ,α f kp = ((t + t )|W+ f (t)|) dt = k f k(µ,α),p. Γ(α + 1) 0

(ii) Let f ∈ S+ and 0 < µ ≤ α < β. For p = 1, see [10] [Section 1, p. 16]. Let 1 < p < ∞, then

1 Z ∞ k kp = ( µ + α)p|( −(β−α)( β ))|p( ) f (µ α) p p t t W W f t dt , , (Γ(α + 1)) 0 1 !p Γ( ) Z ∞ ≤ p (β−α)p( µ + α)p| β |p( ) 1 t t t W f t dt Γ(α + 1)Γ(β − α + p ) 0 p Γ( 1 )Γ(β + 1) ! ≤ p k kp 1 f (µ+β−α,β),p Γ(α + 1)Γ(β − α + p ) Mathematics 2021, 9, 984 12 of 15

where we have used Hardy’s inequality (13). The part (iii) is a straightforward consequence of (i) and the duality of Lp(R+).

Note that, in fact,

k f k(µ,α),p = kDµ,α f kp, h f , giν,α = hDµ,α f , Dµ,αgi0, (14)

(α) µ α (α) µ α 1 1 for f ∈ Tp (t + t ), g ∈ Tp0 (t + t ) with p + p0 = 1.

(α) α µ Example 2. In this example, we consider some functions which belong (or not) to Tp (t + t ) for p ≥ 1 and 0 ≤ µ ≤ α.

β (α) α µ β p + (i) Note that t 6∈ Tp (t + t ) for β ∈ C due to t does not belong to L (R ). < < > −γ( + )−δ = Γ(δ−γ) ( + )γ−δ (ii) For 0 γ δ and a 0, it is well known that W+ a t Γ(δ) t a ; see, for example, [20] [p. 201]. With this formula, it is easy to check that

Γ(α + β) Wα (a + t)−β = (t + a)−(α+β). + Γ(β)

Write f (t) := (a + t)−β and then

Z ∞ (tα + tµ)p Z ∞ 1 || ||p = ≤ < f (µ,α),p Cα,p ( + ) dt Cα,p p dt ∞, 0 |(t + a) α β p| 0 (t + a) β

−β (α) α µ and we conclude functions (a + t) ∈ Tp (t + t ) for β > 1/p and a > 0. c−1 (1 − t) −α (iii) We deﬁne functions jc(t) := χ (t) for t ≥ 0. It is easy to check that W (jc) = Γ(c) (0,1) + (α) α µ 1 jα+c for α > 0. Then, jc ∈ Tp (t + t ) if and only if c > α + 1 − p .

(α) α µ + Note that, for f ∈ Tp (t + t ) for p, α ≥ 1, then f ∈ C(R ), limt→∞ f (t) = 0 and

p (α) α µ sup t | f (t)| ≤ Cα,pk f k(µ,α),p, f ∈ Tp (t + t ), t>0

where Cα,p is independent of f , compare with [11] [Proposition 2.4]. The following theorem extends the case β = α, which was proved in [13] [Proposition 4.1.9] and p = 1 in [10] [Proposition 1.4].

Theorem 4. Take p ≥ 1 and α ≥ µ > 0. Then,

(α) µ α (α) µ α (α) µ α Tp (t + t ) ∗ T1 (t + t ) ,→ Tp (t + t ),

(α) µ α (α) µ α i.e., k f ∗ gk(µ,α),p ≤ Cα,pk f k(µ,α),pkgk(µ,α),1 for f ∈ Tp (t + t ), and g ∈ T1 (t + t ), where Cα,p is a constant independent of f and g.

Proof. Take p > 1 and f , g ∈ S+. Then, we apply Formula (5) to get

Z s Z s p α p α α−1 α |W ( f ∗ g)(s)| ≤ Cα,p |W g(r)| (t + r − s) |W f (t)| dt dr 0 s−r Z ∞ Z ∞ p α α−1 α +Cα,p |W g(r)| (t + r − s) |W f (t)| dt dr s s

for s ≥ 0. From now, we write all constants by Cα,p, which may be different in each line. Mathematics 2021, 9, 984 13 of 15

We start with the second summand. By the Minkowski inequality ([21] [p. 200]), we get that

1 1 In the inner integral, we apply the Holder¨ integral with + = 1 to obtain: p q

Z ∞ (t + r − s)α−1 |Wα f (t)| dt s 1 ! q 1 Z ∞ (α−1)q Z ∞ p (t + r − s) α µ p α p ≤ µ q dt (t + t ) |W f (t)| dt s (tα + t ) s ! 1 Z ∞ (t + r − s)(α−1)q q ≤ k f k(µ,α),p αq dt . s t

We change the variable t − s = u to have that

Z ∞ (t + r − s)(α−1)q Z ∞ u + r αq 1 q Z ∞ rαq 1 rαq dt = du ≤ du = C , αq + + αq ( + )q q q s t 0 u s u r 0 s u r sαq r p

where we have applied that (u + r)/(u + s)−1 ≤ r/s for u > 0 and 0 ≤ s ≤ r. Finally, as 0 ≤ µ ≤ α, we get that

1 Z r Z ∞ p p α µ p α−1 α α (s + s ) (t + r − s) |W f (t)| dt ds ≤ Cα,pk f k(µ,α),p r . 0 s

Now, we consider the ﬁrst integral in the bound of |Wα( f ∗ g)(s)|p. Again, by the Minkowski inequality, we obtain that

We split the interval (r, ∞) in two parts (r, 2r) ∪ [2r, ∞). In the ﬁrst summand, as s ≤ 2r, then (sα + sµ)p ≤ 2αp (rα + rµ)p, and

1 Z 2r Z s p p (sα + sµ)p (t + r − s)α−1 |Wα f (t)| dt ds r s−r 1 Z 2rZ s p p ≤ 2α(rα + rµ) (t + r − s)α−1 |Wα f (t)| dt ds . r s−r

We change the variable s − r = x, and by the Hardy inequality (13), we get that

1 Z ∞Z ∞ p p α α µ α−1 α α µ ≤ 2 (r + r ) (t − x) |W f (t)| dt dx ≤ Cα,p(r + r )k f k(µ,α),p. 0 x Mathematics 2021, 9, 984 14 of 15

In the second summand, we change the variable x = s − r ≥ r, and then

1 Z ∞ Z r+x p p ((x + r)α + (x + r)β)p (t − x)α−1 |Wα f (t)| dt dx r x 1 Z ∞ Z r+x p p ≤ 2α (xα + xβ)p (t − x)α−1 |Wα f (t)| dt dx . r x

We change the variable t − x = u, in the inner integral, and we apply the Minkowski inequality to get

Finally, we join every summand to get that

k f ∗ gk(µ,α),p ≤ k f k(µ,α),pkgk(µ,α),1,

and the proof is ﬁnished.

Remark 2. (a) We consider the conﬂuent hypergeometric functions U(α, ν, z) treated in Section3, and we deﬁned ν −z uα,ν(z) := z e U(α, ν, z), z > 0.

(α) µ α 1 By Theorem1 (ii), functions uα,ν ∈ Tp (t + t ) for ν > α − µ − p with α ≥ µ ≥ 0 and p ≥ 1. n t −t −t (b) For n ≥ 0, we write functions q (t) := e . Note that q (t) = e and q = (q − ∗ q ) n n! 0 n n 1 0 (α) µ α for n ≥ 1. It is straightforward to check that q0 ∈ T1 (t + t ),

Γ(µ + 1) ||q || = 1 + , 0 (µ,α),1 Γ(α + 1)

(α) µ α q0 ∈ Tp (t + t ) and

1 2 2Γ(αp + 1) p ||q || ≤ , 0 (µ,α),p Γ(α + 1)pµ p

for p > 1. By Theorems2 and4, we conclude that

1 1 Z ∞ p p n α µ p −tp (α−n) p n 2 2Γ(αp + 1) Γ(µ + 1) (t + t ) e |Ln (t)| dt ≤ Cα,p µ 1 + , 0 p p Γ(α + 1)

for p ≥ 1, n ∈ N and α ≥ µ ≥ 0.

Author Contributions: This article is a joint research between both authors. First, second and forth sections are written together; results in third section are mainly obtained by N.R. and results in the Mathematics 2021, 9, 984 15 of 15

fifth section by P.J.M. Both authors are collaborated to improve final version of this paper. Both authors have read and agreed to the published version of the manuscript. Funding: This research received no external funding. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Not applicable. Acknowledgments: The first author has been partially supported by Project MTM ID2019-105979GB- I00 of the MCEI and Project E-64, D.G. Aragón, Universidad de Zaragoza, Spain. The second author has been supported by the Ministerio de Economía y Competitividad under Grant MTM2018-095896- B-C21. Conflicts of Interest: The authors declare no conflict of interest.

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