Some operational properties of the Laguerre transform
M. M. Rodrigues, V. N. Huy, and N. M. TuanCitation: 1798, 020130 (2017); doi: 10.1063/1.4972722 View online: http://dx.doi.org/10.1063/1.4972722
View Table of Contents: http://aip.scitation.org/toc/apc/1798/1
Some operational properties of the Laguerre transform
M. M. Rodrigues,
1,a), V.N. Huy
2,b)and N.M. Tuan
3,c)1CIDMA - Center for Research and Development in Mathematics and Applications Department of Mathematics,
University of Aveiro Campus Universit´ario de Santiago, 3810-193 Aveiro, Portugal.
2Department of Mathematics, College of Science, Vietnam National University, 334 Nguyen Trai, Thanh Xuan,
Hanoi, Vietnam.
3Dept. of Mathematics, College of Education, Vietnam National University, G7 Build, 144 Xuan Thuy Rd., Cau Giay,
Dist., Hanoi, Vietnam.
a)Corresponding author: mrodrigues@ua.pt b)nhat-huy85@yahoo.com
c)nguyentuan@vnu.edu.vn
Abstract.This paper is devoted to the study of some properties of the Laguerre transform. We define new properties of the Laguerre transform in a weighted L2−space. Moreover, we present some results concerning the action of this integral transform over some
class of polynomials.
INTRODUCTION
The Laguerre polynomials appear naturally in many branches of pure and applied mathematics and mathematical physics (see e.g. [1, 2, 3, 6]). Debnath [1] introduced the Laguerre transform and derived some of its properties. He also discussed the applications in study of heat conduction [3] and to the oscillations of a very long and heavy chain with variable tension [2]. Moreover, application of the integral Laguerre transforms for forward seismic modeling can be seen in [5].
The Laguerre transform of a function f (x) is denoted by ˜fα(n) and defined by the integral
L{ f (x)} = efα(n)= ∫ ∞
0
e−xxαLαn(x) f (x)dx, n = 0, 1, 2, . . . (0.1) provided the integral exists in the sense of Lesbegue, where Lαn(x) is a generalized Laguerre polynomial of degree n with orderα > −1, and satisfies the following differential equation
d dx [ e−xxα+1 d dxL α n(x) ] + ne−xxαLα n(x)= 0. (0.2)
The sequence of Laguerre polynomial (Lαn(x))∞n=0have the following property:
∞ ∫ 0 e−xxαLα n(x)L α m(x)dx= ( n+ α n ) Γ(α + 1)δnm, (0.3)
whereδnmis Kronecker function defined by
δnm= { 1, if n = m 0, if n , m (0.4) andΓ(α + 1) = ∫ ∞ 0 xαe−xdx.
ICNPAA 2016 World Congress
AIP Conf. Proc. 1798, 020130-1–020130-10; doi: 10.1063/1.4972722 Published by AIP Publishing. 978-0-7354-1464-8/$30.00
The inverse of the Laguerre transformation is then f (x)= ∞ ∑ n=0 (δn)−1efα(n)Lαn(x) (0< x < ∞), whereδn= ( n+ α n ) Γ(α + 1).
This paper is devoted to the study of the generalized Laguerre transform and some operational properties.
MAIN RESULTS
In this section we define new properties for the Laguerre transform in a weighted L2−space. Moreover, we present
some results concerning the action of this integral transform over some class of polynomial. For 1≤ p ≤ ∞ the space Lp,αis defined via the following formula
Lp,α= { f : (0, ∞) → R : ∫ ∞ 0 | f (x)|pe−xxαdx< ∞}
with the norm
|| f ||Lp,α = ( ∫ ∞
0
| f (x)|pe−xxαdx)1/p
where the convention that
|| f ||Lp,α = (∫ ∞ 0 | f (x)|pe−xxαdx )1/p = ess supx∈R| f (x)| if p= ∞.
Clearly, if f is an arbitrary polynomial then f ∈ Lp,α. We also define
lp,α=f = (f(n))∞n=0: ∞ ∑ n=0 | f (n)|p ( n+ α n ) < ∞ with the norm
|| f ||lp,α = ∞ ∑ n=0 | f (n)|p ( n+ α n ) 1/p . We define the differential operator R via the formula
R[ f (x)] = exx−α d
dx[e−xxα+1 ddxf (x)]
= x f′′
(x)+ (α + 1 − x) f′(x) and then for n= 0, 1, ..., we have
Rn[ f (x)]= R[Rn−1[ f (x)]].
Let P(x) be a polynomial. The differential operator P(R) is obtain from P(x) by substituting x → R. i.e., if P(x)=∑nk=0akxkthen P(R)[ f (x)]= n ∑ k=0 akRk[ f (x)]. Moreover, we denote by supp efα= {n ∈ Z+: efα(n), 0} the support of the Laguerre transform of f .
Theorem 0.1 Let 1 ≤ p < ∞, P(x) be a polynomial. Then for an arbitrary infinitely differentiable function f ∈ Lp,α, there exist the following limit
lim m→∞||P m(R)[ f (x)]||1/m Lp,α and lim m→∞||P m(R)[ f (x)]||1/m Lp,α = sup{|P(−n)| : n ∈ supp efα}. Proof: From the definition of differential operator R, we get
L{R[f(x)]} =∫∞ 0 e −xxαLα n(x) exx−α ddx [ e−xxα+1 d fdx]dx =∫0∞Lαn(x)dxd [e−xxα+1 d fdx]dx. Using the integral by part, we have
L{R[f(x)]} = [ e−xxα+1Lαn(x)d f dx ]∞ 0 − ∫ ∞ 0 e−xxα+1dL α n(x) dx d f dxdx = − ∫ ∞ 0 e−xxα+1dL α n(x) dx d f dxdx. (0.5) We see that − ∫ ∞ 0 e−xxα+1dL α n(x) dx d f dxdx = [ e−xxα+1dL α n(x) dx f (x) ]∞ 0 + ∫ ∞ 0 d dx [ e−xxα+1dL α n(x) dx ] f(x) dx = ∫ ∞ 0 d dx [ e−xxα+1dL α n(x) dx ] f(x) dx. (0.6)
Then, from (0.2) it follows − ∫ ∞ 0 e−xxα+1dL α n(x) dx d f dxdx= −n ∫ ∞ 0 e−xxαf(x) Lαn(x) dx. Hence L{R[f(x)]}= −nefα(n). (0.7) Similarly, we get L{Rm[f(x)]}= (−1)mnmefα(n). (0.8) Hence, from this and the definition of Pm(R) we obtain
L{Pm(R)[f (x)]}= Pm(−n)efα(n). (0.9) Now, let considerσ an arbitrary number in suppefα. Then
∫ ∞
0
e−xxαLασ(x)Pm(R)[ f (x)]dx= Rm(−σ)efα(σ). (0.10) Applying H¨older inequality, we obtain
∫0∞e−xxαLα σ(x)P m(R)[ f (x)]dx ≤ (∫0∞e−xxα|Lα σ(x)| qdx )1/q × (∫ ∞ 0 e−xxα|Pm(R)[ f (x)]|pdx )1/p . (0.11)
Therefore, it follows from (0.10) and (0.11) that |Pm(−σ)ef α(σ)| ≤ (∫ ∞ 0 e−xxα|Lασ(x)|qdx )1/q∫ ∞ 0 e−xxα|Pm(R)[ f (x)]|pdx. (0.12) That mean |Pm(−σ)ef α(σ)| ≤ (∫ ∞ 0 e−xxα|Lασ(x)|qdx )1/q ∥Pm(R)[ f (x)]∥ Lp,α. (0.13)
As Lασis a polynomial, we obtain Lασ ∈ Lq,αand then
(∫ ∞
0
e−xxα|Lασ(x)|qdx )1/q
< ∞. It follows from efα(σ) , 0 and (0.13) that
|P(−σ)| ≤ limm→∞∥P
m(R)[ f (x)]∥1/m
Lp,α. (0.14)
Since (0.14) holds for allσ ∈ suppefα, we obtain
sup{|P(−σ)| : σ ∈ suppefα} ≤ limm→∞∥Pm(R)[ f (x)]∥1L/m
p,α. (0.15)
Now, we will prove that
sup{|P(−σ)| : σ ∈ suppefα} ≥ limm→∞∥Pm(R)[ f (x)]∥1L/m
p,α. (0.16)
Indeed, if supp efαis an unbounded set then
sup{|P(−σ)| : σ ∈ suppefα} = ∞
and (0.16) holds. Now, we only need to prove (0.16) for the case that supp efα is a bounded set, that mean, f is a polynomial and M := sup{|n| : n ∈ suppefα} < ∞. Hence Pm(R)[ f (x)]= M ∑ n=−M (δn)−1Pm(−n)efα(n)Lαn(x) (0< x < ∞) and ∥Pm(R)[ f (x)]∥ Lp,α ≤ ∑M n=−M(δn)−1Pm(−σ)efα(n)∥Lαn∥Lp,α (0.17) ≤ [sup{|P(−σ)| : n ∈ suppefα}]m∑Mn=−M(δn)−1efα(n)∥Lαn∥Lp,α (0.18) which gives limm→∞∥Pm(R)[ f (x)]∥ 1/m Lp,α ≤ sup{|P(−σ)| : σ ∈ suppefα}. Therefore, (0.16) have been proved. From (0.15) and (0.16), it is easy to see
lim
m→∞||P
m(R)[ f (x)]||1/m
Lp,α = sup{|P(−n)| : n ∈ supp efα}, and then the proof is complete.
Let P(x)= x. Then, according Theorem 0.1 we obtain
Corollary 0.2 Let 1 ≤ p ≤ ∞, P(x) be a polynomial. Then for an arbitrary arbitrary infinitely differentiable function f ∈ Lp,α, there exist the following limit
lim m→∞||R m[ f (x)]||1/m Lp,α and lim m→∞||R m[ f (x)]||1/m Lp,α = deg( f ).
Now, let consider an infinitely differentiable function f : (0, ∞) → R. Assume that we can define the sequence of function (Sm[ f (x)])∞m=1satisfying the condition
R(Sm)[ f (x)]= Sm−1[ f (x)], ∀m ∈ N.
We also see that, for an arbitrary polynomial f we get efα(0)= 0 if and only if the sequence of function (Sm[ f (x)])∞m=1 is well defined and
Sm[ f (x)]= ∞ ∑ n=1 (δn)−1 ( −1 n )m efα(n) Lαn(x). Next, we state the following theorem
Theorem 0.3 Let consider P(x) a polynomial and 1 ≤ p ≤ ∞. Then for an arbitrary infinitely differentiable function f ∈ Lp,αsatisfying efα(0)= 0, we have
lim m→∞||P m (S )[ f (x)]||1L/m 2,α ≥ sup { P(−1n ) : n ∈ suppefα } . Moreover, if f is a polynomial then
lim m→∞||P m(S )[ f (x)]||1/m L2,α = sup { P(−1n ) : n ∈ suppefα } . Proof: We have known that
L{ f (x)} = L{R(S )[ f (x)]} = nL{S [ f (x)]}. (0.19) We deduce that L{S[f(x)]}= −1 n efα(n). (0.20) Similarly, we get L{Sm[f(x)]}= ( −1 n )m efα(n). (0.21)
From this and the definition of Pm(S ), we obtain
L{Pm(S )[f(x)]}= Pm ( −1 n ) efα(n). (0.22)
Now, we considerσ an arbitrary number in suppefα. Then ∫ ∞ 0 e−xxαLασ(x)Pm(S )[ f (x)]dx= Rm ( −1 σ ) efα(σ). (0.23)
Applying H¨older inequality we obtain ∫0∞e−xxαLα σ(x)P m(S )[ f (x)]dx ≤ (∫0∞e−xxα|Lα σ(x)| qdx )1/q × (∫ ∞ 0 e−xxα|Pm(S )[ f (x)]|pdx )1/p . (0.24)
Therefore, it follows from (0.23) and (0.24) that |Pm ( −1 σ ) efα(σ)| ≤ (∫ ∞ 0 e−xxα|Lασ(x)|qdx )1/q∫ ∞ 0 e−xxα|Pm(S )[ f (x)]|pdx. That mean |Pm ( −1 σ ) efα(σ)| ≤ (∫ ∞ 0 e−xxα|Lασ(x)|qdx )1/q ∥Pm(S )[ f (x)]∥ Lp,α. (0.25)
As Lασis a polynomial, we obtain Lασ ∈ Lq,αand then
(∫ ∞ 0 e−xxα|Lα σ(x)| qdx )1/q < ∞. Then it follows from efα(σ) , 0 and (0.25) that
|P ( −1 σ ) | ≤ limm→∞∥P m(S )[ f (x)]∥1/m Lp,α. (0.26)
Since (0.26) holds for allσ ∈ suppefα, we obtain
sup{|P ( −1 σ ) | : σ ∈ suppefα} ≤ limm→∞∥P m(S )[ f (x)]∥1/m Lp,α. (0.27)
Now, we will prove
sup{|P ( −1 σ ) | : σ ∈ suppefα} ≥ limm→∞∥Pm(S )[ f (x)]∥1L/m p,α (0.28)
in the case where f is a polynomial. Indeed, put M := deg( f ). Hence,
Pm(S )[ f (x)]= M ∑ n=−M (δn)−1Pm ( −1 n ) efα(n)Lαn(x) (0< x < ∞) and ∥Pm(S )[ f (x)]∥ Lp,α ≤ ∑M n=−M(δn)−1Pm (−1 σ ) efα(n)∥Lαn∥Lp,α ≤ [sup{|P(−1 σ)| : n ∈ suppefα}]m ∑M n=−M(δn)−1efα(n)∥Lαn∥Lp,α which gives limm→∞∥Pm(S )[ f (x)]∥1L/mp,α ≤ sup{|P ( −1 σ ) | : σ ∈ suppefα}.
That mean (0.28) have been proved. From (0.27) and (0.28), it is easy to see
lim m→∞||P m (S )[ f (x)]||1L/m p,α = sup{|P ( −1 n ) | : n ∈ supp efα} and the proof is complete.
Let P(x)= x, then we have the following result
Corollary 0.4 Let 1 ≤ p ≤ ∞ and let f an arbitrary polynomial satisfying efα(0) = 0. Then, there exist the following limit lim m→∞||S m[ f (x)]||1/m Lp,α and lim m→∞||S m[ f (x)]||1/m Lp,α = 1 deg1( f ), where deg1( f )= inf{n ∈ Z+: efα(n), 0}.
We denote by Tmthe set of the algebra polynomials of degree≤ m, and the error of approximation Em( f ) of f by
elements from Tmis
Em( f )= inf P∈Tm
∥ f − P∥L2,α.
Take it in mind, we have the following theorem
Theorem 0.5 Assume that f (x)∈ L2,α. Then L{ f (x)} ∈ l2,αand
|| f ||L2,α = √ Γ(α + 1)||L{ f (x)}||l2,α. Moreover, Em( f )= || f ||L2,α− m ∑ n=0 ( efα(n))2Γ(α + 1) ( n+ α n ) . Proof: We see that
|| f (x) −∑m n=0(δn)−1c(n)Lαn(x)||L2,α = ∫∞ 0 e −xxα(f (x)−∑m n=0(δn)−1c(n)Lαn(x) )2 dx (0.29) =∫0∞e−xxα( f (x))2dx− 2∫∞ 0 e−xxαf (x) ∑m n=0(δn)−1c(n)Lαn(x)dx (0.30) +∫0∞e−xxα(∑nm=0(δn)−1c(n)Lαn(x) )2 dx. (0.31) Note that ∫ ∞ 0 e−xxαf (x) m ∑ n=0 (δn)−1c(n)Lαn(x)dx = ∑m n=0 ∫∞ 0 e−xxαf (x)(δn)−1c(n)Lnα(x)dx (0.32) =∑m n=0(δn)−1c(n) efα(n). (0.33) Using (0.3) we obtain ∫ ∞ 0 e−xxα (∑m n=0 (δn)−1c(n)Lαn(x) )2 dx =∑1≤i, j≤m∫∞ 0 e −xxα((δ n)−2(c(n))2Lαi(x)Lαj(x) ) dx =∑1≤i= j≤m ∫∞ 0 e −xxα((δ n)−2(c(n))2Lαi(x)Lαj(x) ) dx =∑m n=0(δn)−1(c(n))2. (0.34)
Using (0.29), (0.32) and (0.34) we get || f (x) − m ∑ n=0 (δn)−1c(n)Lαn(x)||L2,α = || f ||2L2,α − 2 ∑m n=0(δn)−1c(n) efα(n)+ ∑m n=0(δn)−1(c(n))2 = || f ||2 L2,α− ∑m n=0(δn)−1( efα(n))2+ ∑m n=0(δn)−1( efα(n)− c(n))2 (0.35) which gives inf (c0,...,cm)∈Rm || f (x) − m ∑ n=0 (δn)−1c(n)Lαn(x)||L2,α = || f ||2L2,α− m ∑ n=0 (δn)−1( efα(n))2.
Then, it follows from∑mn=0(δn)−1c(n)Lαn(x) which is the algebra polynomial of degree m Em( f )= || f ||2L2,α− m ∑ n=0 (δn)−1( efα(n))2= || f ||L2,α − m ∑ n=0 ( efα(n))2Γ(α + 1) ( n+ α n ) . Let c(n)= efα(n) for all n= 0, 1, . . . , m, we obtain
|| f (x) − m ∑ n=0 (δn)−1efα(n)Lnα(x)||L2,α = || f ||2L2,α− m ∑ n=0 (δn)−1( efα(n))2.
Therefore, since lim
m→∞|| f (x) − m ∑ n=0 (δn)−1efα(n)Lαn(x)||L2,α = 0, we have L{ f (x)} ∈ l2,αand || f ||2 L2,α = m ∑ n=0 (δn)−1( efα(n))2, and then √ Γ(α + 1)||L{ f (x)}||l2,α = || f ||L2,α. The proof is complete.
We are now able to present the following result
Theorem 0.6 Let P(x) be the polynomial. Then for an arbitrary polynomial f , we always have ||Pm(R)[ f (x)]|| L2,α≤ sup{|P(−n)| : n ∈ suppefα}||Pm−1(R)[ f (x)]||L2,α, (0.36) ||Pm(R)[ f (x)]|| L2,α ≥ inf { |P(−n)| : n ∈ suppefα } ||Pm−1(R)[ f (x)]|| L2,α (0.37) and lim m→∞||P m(R)[ f (x)]||1/m L2,α = deg( f ). (0.38)
Proof: We have known that
L{Pm(R)[f(x)]}= Pm(−n) efα(n). Then it follows from Theorem 0.5 that
||Pm(R)[ f (x)]|| L2,α = √ Γ(α + 1)||L{Pm(R)[f(x)]}|| l2,α = √Γ(α + 1) ( ∑∞ n=0 ( Pm(−n)ef α(n))2 ( n+ α n ))1/2 . Hence ||Pm(R)[f(x)]|| L2,α = √ Γ(α + 1) ∑ n∈suppefα ( Pm(−n)efα(n))2 ( n+ α n ) 1/2 ≤ √Γ(α + 1) sup{|P(−n)| : n ∈ suppefα } × ∑ n∈suppefα ( Pm−1(−n)efα(n))2 ( n+ α n ) 1/2 = sup{|P(−n)| : n ∈ suppefα } ||Pm−1(R)[f (x)]|| L2,α
and (0.36) just been proved.
Now, we will prove (0.37). Indeed, we have ||Pm(R)[f (x)]|| L2,α= √ Γ(α + 1) ∞ ∑ n=0 (Pm(−n)efα(n))2 ( n+ α n ) 1/2 which gives ||Pm(R)[f(x)]|| L2,α = √ Γ(α + 1) ∑ n∈suppefα (Pm(−n)efα(n))2 ( n+ α n ) 1/2 . It follows that ||Pm(R)[f(x)]|| L2,α ≥ √ Γ(α + 1) inf{|P(−n)| : n ∈ suppefα} × ∑ n∈suppefα (Pm−1(−n)efα(n))2 ( n+ α n ) 1/2 = inf{|P(−n)| : n ∈ suppefα}||Pm−1(R)[f(x)]||L2,α, and (0.37) just been proved. Note that, we check (0.37) from Theorem 0.1. The proof is complete.
Since for all a polynomial f and n > deg( f ) then efα(n) = 0, we obtain suppefα ⊂ {0, 1, . . . deg( f )}. Then, by applying Theorem 0.6 we have the following corollary
Corollary 0.7 Let P(x) be the polynomial. Then for an arbitrary polynomial f , we always have ||Pm(R)[ f (x)]|| L2,α ≤ sup{|P(−n)| : 0 ≤ n ≤ deg( f )}||Pm−1(R)[ f (x)]||L2,α, ||Pm (R)[ f (x)]||L2,α ≥ inf { |P(−n)| : 0 ≤ n ≤ deg( f )}||Pm−1 (R)[ f (x)]||L2,α and lim m→∞||P m(R)[ f (x)]||1/m L2,α = deg( f ).
Let P(x)= x, then it follows from Theorem 0.6
Corollary 0.8 Let P(x) be the polynomial. Then for an arbitrary polynomial f , we always have ||Rm[ f (x)]|| L2,α ≤ deg( f )||P m−1(R)[ f (x)]|| L2,α, ||Rm[ f (x)]|| L2,α ≥ deg1( f )||P m−1(R)[ f (x)]|| L2,α,
where deg1( f )= inf{n ∈ Z+: efα(n), 0} and lim
m→∞||R
m[ f (x)]||1/m
L2,α = deg( f ).
Remark.Since f is the polynomial with degree d, f (x) can express via f (x)= d ∑ n=0 anLnα(x). Then {n ∈ Z+: efα(n), 0} = {n ∈ N : an, 0} and C= inf{|P(−n)| : n ∈ Z+, efα(n), 0} = inf{|P(−n)| : n ∈ Z+, an, 0}.
ACKNOWLEDGMENTS
The work of the first author was supported by Portuguese funds through the CIDMA - Center for Research and Development in Mathematics and Applications, and the Portuguese Foundation for Science and Technology (“FCT– Fundac¸˜ao para a Ciˆencia e a Tecnologia”), within project UID/MAT/ 0416/2013. The work of V.N.Huy was supported by Vietnam National University under grant number QG.16.08.
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