The aim of this paper is to discuss the weak-type (1, 1) boundedness of R ∝ m + 1 and the polynomial weight w that makes the Riesz-Laguerre transforms of order greater than or equal to 2 continuous from L 1 ( w d μ α ) into L 1 , ∞ ( d μ α ) , under specific value α . Following the same notions appear in [1] [2]. The (m+ 1)th Riesz-Laguerre transform with m ∈ Z ≥ 0 d associated with the multidimensional Laguerre operator L ∝ , where ∝   = ( ∝ 1 , ⋯ , ∝ d ) is a multi-index with ∝ i   ≥ 0 , i = 1 , ⋯ , d .

The Laguerre operator L ∝ , is a self-adjoint “Laplacian” on L 2 ( d μ α ) , where μ α is the Laguerre measure of type ∝   = ( ∝ 1 , ⋯ , ∝ d ) with ∝ i   > − 1 , i = 1 , ⋯ , d ; defined on R + d = { x ∈ R d : x i > 0 ,   for   each   i = 1 , ⋯ , d } , by

d μ α ( x ) = ∏ i = 1 d x i ∝ i e − x i Γ ( ∝ i + 1 ) d x .

It is well known that the spectral resolution of L ∝ is

L ∝ = ∑ n = 0 ∞     n P n ∝ ,

where P n ∝ is the orthogonal projection on the space spanned by Laguerre polynomials of total degree n and type ∝ ind variables [3] [4]. The operator L ∝ is the infinitesimal generator of a “heat” semigroup, called the Laguerre semigroup, { e − t L ∝ : t ≥ 0 } , defined in the spectral sense as

e − t L ∝ = ∑ n = 0 ∞     e − n t P n ∝ .

For any multi-index m + 1 = ( a 1 , ⋯ , a d ) ∈ Z ≥ 0 d , the Riesz-Laguerre transforms R ∝ m + 1 of order | m + 1 | = a 1 + ⋯ + a d are defined by

R ∝ m + 1 = ∇ α m + 1 ( L α ) − | m + 1 | / 2 P 0 ∝ ⊥ ,

where ∇ ∝ is associated to L ∝ defined as ∇ ∝ = ( x 1 ∂ x 1 , ⋯ , x d ∂ x d ) , and P 0 ∝ ⊥ , denotes the orthogonal projection onto the orthogonal complement of the eigenspace corresponding to the eigenvalue 0 of L ∝ .

In order to use the well-known relationship with the Ornstein-Uhlenbeck context, but not too much exploited in the weak-type inequalities, we are going to perform a change of coordinates in R + d . If x = ( x 1 , ⋯ , x d ) is a vector R + d , then x 2 will denote the vector ( x 1 2 , ⋯ , x d 2 ) . Let Ψ : R + d → R + d be defined as Ψ ( x ) = x 2 and let d μ ˜ ∝ = d μ ∝ o Ψ − 1 be the pull-back measure from d μ ∝ . Then the modified Laguerre measure d μ ˜ ∝ is the probability measure

d μ ˜ ∝ ( x ) = 2 d ∏ i = 1 d x i 2 ∝ i + 1 e − x i 2 Γ ( ∝ i + 1 ) d x = 2 d ∏ i = 1 d x i 2 ∝ i + 1 Γ ( ∝ i + 1 ) e − | x | 2 d x , (1)

on R + d .

The map f → U Ψ f = f o Ψ is an isometry from L q ( d μ ∝ ) onto L q ( d μ ˜ ∝ ) and from L q , ∞ ( d μ α ) onto L q , ∞ ( d μ ˜ ∝ ) , for every q in [1, ∞]. So we may reduce the problem of studying the weak-type boundedness of R ∝ m + 1 to the study of the same boundedness for the modified Riesz-Laguerre transforms R ˜ ∝ m + 1 = U Ψ R ∝ m + 1 U Ψ − 1 with respect to the measure d μ ˜ ∝ .

Observe that R ˜ ∝ m + 1 coincides, up to a multiplicative constant, with ∇ m + 1 ( L ˜ ∝ ) − | m + 1 | P ˜ 0 ∝ ⊥ , being L ˜ ∝ = U Ψ L ∝ U Ψ − 1 , P ˜ 0 ∝ ⊥ = U Ψ P 0 ∝ ⊥ U Ψ − 1 andes Ñ the gradient of R d associated to the Laplacian operator [5].

For the sequel, it is convenient to express the kernel of R ˜ ∝ m + 1 with respect to the Polynomial measure ( m + 1 ) ∝ defined on R + d as

d ( m + 1 ) ∝ ( x ) = e | x | 2 d μ ˜ ∝ ( x ) . (2)

According to [6] [7], for ∝ i   > − 1 / 2 , i = 1 , ⋯ , d , the kernel of the modified Riesz-Laguerre transforms of order | m + 1 | with respect to the polynomial measure ( m + 1 ) ∝ is defined, off the diagonal, as

K m + 1 ( x , s ) = ∫ [ − 1 , 1 ] d K m + 1 ( x , s ) Π α ( s ) ( s ) d s

with

K m + 1 ( x , s ) = ∫ 0 1 ( r ) | m + 1 | − 2 ( − log r 1 − r ) | m + 1 | − 2 2 ∏ i = 1 d     H a i ( x i ( r − s i ) 1 − r ) e − q − ( r x 2 , x 2 , s ) 1 − r ( 1 − r ) | α | + d + 1 d r (3)

where H a i is the Hermite polynomial of degree a i and

q ± ( x , s ) = ∑ i = 1 d     2 x i ( 1 ± s i ) ,

Π α ( s ) = ∏ i = 1 d Γ ( α i + 1 ) Γ ( α i + 1 2 ) π ( 1 − s i 2 ) α i − 1 / 2 ,

cos θ = cos θ ( x , s ) = ∑ i = 1 d x i s i | x | 2 ,

sin θ = sin θ ( x , s ) = ( 1 − cos 2 θ ) 1 / 2 = ( 1 − ( ∑ i = 1 d x i s i | x | 2 ) 2 ) 1 / 2 .

The symbol a ≲ b means a ≤ C b where C is a constant that may be different on each occurrence. And we write a ~ b whenever a ≲ b and b ≲ a .

For every multi-index ∝ we have the following result see [3].

Theorem 1: The second order Riesz-Laguerre transforms map L 1 ( d μ α ) continuously into L 1 , ∞ ( d μ α ) .

Proof: The result follows by splitting the modified Riesz-Laguerre transforms of second order into a local operator and a global one. Let us observe that for a simple covering Lemma, we may pass from estimates with respect to the measure ( m + 1 ) ∝ on the local part R 0 to estimates with respect to the modified Laguerre measure μ ˜ ∝ . Therefore the local operator is equivalent to T 0 m + 1 for | m + 1 | = 2 . The global operators bounded weak type (1, 1) and therefore so are the second order modified Riesz-Laguerre transforms.

From [8] and [9], it is known that an upper bound for | K m + 1 ( x , s ) | on G is

K ˜ m + 1 ( x , s ) = { ( | 2 x | 2 ) | m + 1 | − 2 2 e − | x | 2 , cos θ < 0 1 2 ( 4 | x | 4 sin 2 θ ) | m + 1 | − 2 4 ( 1 + cos θ 1 − cos θ ) | α | − d 2 ( 1 + ( 4 | x | 4 sin 2 θ ) 1 4 e − u 0 , cos θ ≥ 0 (4)

with

u 0 = ( q + ( x 2 , s ) q − ( x 2 , s ) ) 1 / 2 2 .

Proposition 2: For | m | = 2 ,

| K m ( x , y , s ) | ≾ e | x | 2 K * ( x , y , s ) e − | y | 2

on the global region

G = R + d × [ − 1 , 1 ] d \ R 0 = R 1 ∪ R 2 ∪ R 3 ∪ R 4 ,

with

R 0 = { ( y , s ) ∈ R + d × [ − 1 , 1 ] d : q − ( x 2 , y 2 , s ) 1 / 2 ≤ C 1 + | X | } ,

R 1 = { ( y , s ) ∉ R 0 : cos θ < 0 } ,

R 2 = { ( y , s ) ∉ R 0 : cos θ ≥ 0 , | y | ≤ | x | } ,

R 3 = { ( y , s ) ∉ R 0 : cos θ ≥ 0 , | x | ≤ | y | ≤ 2 | x | } ,

R 4 = { ( y , s ) ∉ R 0 : cos θ ≥ 0 , | y | ≤ 2 | x | } ,

and

K * ( x , y , s ) = { e − | x | 2 ( y , s ) ∈ R 1       ( 5 ) | x | 2 | ∝ | + 2 d e − c | x | q − 1 / 2 ( x 2 , y 2 , s ) ( y , s ) ∈ R 2       ( 6 ) | x | 2 | ∝ | + 2 d ( 1 ∧ e − c | x | 4 sin 2 θ | y | 2 − | x | 2 + sin θ | x | 2 ( | y | 2 − | x | 2 + sin θ | x | 2 ) 2 | ∝ | + 2 d − 1 2 ) ( y , s ) ∈ R 3       ( 7 ) ( 1 + | x | ) e − sin 2 θ | x | 2 ( y , s ) ∈ R 4       ( 8 )

Proof. In this proposition, | m | = 2 . If cos θ < 0 , it is immediate that

| K m ( x , y , s ) | ≲ e | x | 2 K * ( x , y , s ) e − | y | 2 .

Let us then assume that cos θ ≥ 0 .

1) First let us consider | x | > | y | .

Since cos θ ≥ 0 , q + 1 / 2 ≥ | x | and since | x | > | y | , then q + 1 / 2 ≤ 2 | x | . Therefore q + 1 / 2 ~ | x | . On the other hand, since q − 1 / 2 ≥ C 1 + | x | then | x | ≥ c . Thus

| K m ( x , y , s ) | ≲ [ ( q + q − ) | α | + d 2 + ( q + q − ) | α | + d 2 ( q + q − ) 1 / 4 ] e − u 0 ≲ [ | x | | α | + d ( 1 + | x | ) | α | + d + ( q + 1 / 2 ) | α | + d + 1 / 2 ( q − 1 / 2 ) | α | + d − 1 / 2 ] e − u 0 ≲ | x | 2 | α | + 2 d e − u 0 = e | x | 2 | x | 2 | α | + 2 d e − ( q + q − ) 1 / 2 2 e | y | 2 − | x | 2 2 e − | y | 2 ≲ e | x | 2 | x | 2 | α | + 2 d e − | x | q − 1 / 2 ( x 2 , y 2 , s ) 2 e − | y | 2 = e | x | 2 K * ( x , y , s ) e − | y | 2 .

2) Now let us assume | y | ≥ | x | and rewrite u 0 in the following way:

u 0 = | y | 2 − | x | 2 2 + ( q + ( x 2 , y 2 , s ) q − ( x 2 , y 2 , s ) ) 1 / 2 2 = | y | 2 − | x | 2 + ( q + ( x 2 , y 2 , s ) q − ( x 2 , y 2 , s ) ) 1 / 2 − ( | y | 2 − | x | 2 ) 2 = | y | 2 − | x | 2 q + q − − ( | y | 2 − | x | 2 ) 2 2 ( | y | 2 − | x | 2 + ( q + q − ) 1 / 2 ) = | y | 2 − | x | 2 + 2 sin 2 θ | x | 2 − | y | 2 | y | 2 − | x | 2 + ( q + q − ) 1 / 2 . (9)

Since

q + q − = ( | x | 2 + | y | 2 ) 2 − 4 | x | 2 | y | 2 cos 2 θ = ( | y | 2 − | x | 2 ) 2 + 4 | x | 2 | y | 2 sin 2 θ ,

and taking into account that sinθ is non-negative, we obtain that

( q + q − ) 1 / 2 ~ | y | 2 − | x | 2 + | x | | y | sin θ ≥ | y | 2 − | x | 2 + | x | 2 sin θ . (10)

Thus, from (9) together with (10) we get

u 0 ≥ | y | 2 − | x | 2 + c | x | 4 sin 2 θ | y | 2 − | x | 2 + | x | | y | sin θ . (11)

Claim 3: max ( | y | 2 − | x | 2 , | x | 2 sin θ ) ≥ 1 .

Proof. If sin θ ≥ 1 | x | 2 , the inequality is immediate.

If sin θ ≤ 1 | x | 2 , then | y | 2 ≥ | x | 2 + 1 . This inequality is immediate when | x | ≤ 1

by adjusting conveniently the constant C in the definition of the global zone and it is also immediate for d = 1 and | x | > 1 . Now let us assume that d ≥ 2 and | x | > 1 .

( C / 2 ) 2 | x | 2 ≤ C 2 ( 1 + | x | ) 2 ≤ q − ( x 2 , y 2 , s ) = | x | 2 + | y | 2 − 2 | x | | y | 1 − sin 2 θ ≤ | x | 2 + | y | 2 − 2 | x | | y | 1 − 1 | x | 4 .

Hence

| x | 2 − 2 | x | 1 − 1 | x | 4 | y | − | x | 2 − ( C / 2 ) 2 | x | 2 ≥ 0

for all | y | ≥ | x | , then

| y | ≥ | x | 1 − 1 | x | 4 + ( C / 2 ) 2 − 1 | x |

which implies that

| y | 2 ≥ | x | 2 + 2 ( C / 2 ) 2 − 1 1 − 1 | x | 4 + ( C / 2 ) 2 − 2 | x | 2 ≥ | x | 2 + 1.

Therefore by applying this claim to inequality (10) we obtain that q + q − ≥ c in this context. If | x | ≤ | y | ≤ 2 | x | , by taking into account (11), we get

u 0 ≥ | y | 2 − | x | 2 + c | x | 4 sin 2 θ | y | 2 − | x | 2 + | x | 2 sin θ , (12)

then

| K m ( x , y , s ) | ≲ ( q + q − ) | α | + d 2 ( q + q − ) 1 / 4 e − u 0 ≲ q + | α | + d 2 ( q + q − ) 2 | ∝ | + 2 d 4 ( q + q − ) 1 / 4 e − u 0 ≲ | x | 2 | α | + 2 d [ ( q + q − ) 1 / 2 ] 2 | ∝ | + 2 d − 1 2 e − u 0 ≲ e | x | 2 | x | 2 | α | + 2 d ( | y | 2 − | x | 2 + | x | 2 sin θ ) 2 | ∝ | + 2 d − 1 2 e − c | x | 4 sin 2 θ | y | 2 − | x | 2 + | x | 2 sin θ e − | y | 2 .

To get the last inequality we have used (10) and (12). On the other hand, since q + q − ≥ c it is immediate the following inequality

| K m ( x , y , s ) | ≲ | x | 2 | α | + 2 d e | x | 2 − | y | 2 .

Thus

| K m ( x , y , s ) | ≲ e | x | 2 K * ( x , y , s ) e − | y | 2 .

Now if | y | ≥ 2 | x | then ( q + q − ) 1 / 4 ≤ ( | x | 2 + | y | 2 ) 1 / 2 ≲ | y | 2 , and thus

( 1 + ( q + q − ) 1 / 4 | x | | y | | x | 2 + | y | 2 ) ≾ ( 1 + | x | ) .

Besides q − ≥ ( | y | − | x | ) 2 ≥ c | y | 2 and q + ≤ C | y | 2 therefore q + q − ≤ C . On the other

hand, from (9) together with ( q + q − ) 1 / 2 ≤ | y | 2 − | x | 2 + 2 | x | | y | sin θ and | y | 2 − | x | 2 + | x | | y | sin θ ≤ 2 | y | 2 we get

u 0 ≥ | y | 2 − | x | 2 + sin 2 θ | x | 2 | y | 2 | y | 2 − | x | 2 + | x | | y | sin θ ≥ | y | 2 − | x | 2 + sin 2 θ 2 | x | 2 .

Therefore

| K m ( x , y , s ) | ≲ e | x | 2 K * ( x , y , s ) e − | y | 2 .

Proposition 4: The operator K * defined as

K * f ( x ) = e | x | 2 ∫ R + d ∫ [ − 1 , 1 ] d χ G ( x , s ) K * ( x , s ) Π ∝ ( s ) | f ( x ) | d μ ˜ ∝ ( x ) ,

is of weak type (1, 1) with respect to the measure μ ˜ ∝ .

Proof. The method of proof used in [1] is an adaptation to our context of the techniques developed in [10] [11] [12] which allows us to get rid of the classical one called “forbidden regions technique”.

The kernels (5) and (6) define strong type (1, 1) operators. Indeed,

e | x | 2 ∫ R + d ∫ [ − 1 , 1 ] d χ R 1 ( x , s ) e − | x | 2 Π ∝ ( s ) | f ( x ) | d μ ˜ ∝ ( x ) ≤ C ‖ f ‖ 1 .

Moreover, for semi-integer values of the parameter α, by [6]

| x | 2 ( | ∝ | + d ) e − C | x | 2 ( 2 ( 1 − cos θ ) ) 1 / 2 (13)

is in L 1 ( d μ α ) uniformly in x and s and so the operator is of strong type with respect to μ ˜ ∝ on R 2 . Finally the result for the other values of ∝ is obtained via the multidimensional Stein’s complex interpolation Theorem. So to get the weak-type (1, 1) inequality for the operator K * it suffices to prove that the operators

S i f ( x ) = e | x | 2 ∫ R + d ∫ [ − 1 , 1 ] d χ R i + 3 ( x , s ) K * ( x , s ) Π ∝ ( s ) d s | f ( x ) | d μ ˜ ∝ ( x ) , i = 0 , 1

map L 1 ( d μ ˜ ∝ ) continuously into L 1 , ∞ ( d μ ˜ ∝ ) .

Without loss of generality, we may assume that f ≥ 0 . Fix λ > 0 and let

E i = { x ∈ R + d : S i f ( x ) > λ } ,

for i = 0 , 1 . We must prove that μ ˜ ∝ ( E i ) ≤ C ‖ f ‖ 1 λ . Let r 0 and r 1 be the positive roots of the equations

r 0 2 ( | ∝ | + d ) e r 0 2 ‖ f ‖ 1 = λ     and     r 1 e r 1 2 ‖ f ‖ 1 = λ .

We may observe that indeed, if E i ∩ { x ∈ R + d : | x | < r i } = ∅ : indeed, if | x | < r i , we have

S 0 f ( x ) ≤ | x | 2 ( | ∝ | + d ) e | x | 2 ‖ f ‖ 1 < λ ,

S 1 f ( x ) ≤ | x | e | x | 2 ‖ f ‖ 1 < λ .

On the other hand, we may take λ > K ‖ f ‖ in [1], and by choosing K large enough we may assume that both r 0 and r 1 are larger that one. Hence

μ ˜ ∝ { x ∈ R + d : | x | < 2 r i } ≤ ∫ | x | < 2 r i ∏ j = 1 d     x j 2 ∝ j + 1 e − | x | 2 d x ≲ r i 2 | α | e − 4 r i 2 ≤ C ‖ f ‖ 1 λ .

Thus we only need to estimate μ ˜ ∝ { x ∈ R + d : r i ≤ | x | ≤ 2 r i } .

We let E ′ i denote the set of x ′ ∈ S d − 1 for which there exists a ρ ∈ [ r i , 2 r i ] with ρ x ′ ∈ E . For each x ′ ∈ E ′ i we let ρ ( x ′ ) be the smallest such ρ . Observe that

sin θ ( x , s ) = sin θ ( x ′ , s ) = sin θ .

Then S i f ( ρ ( x ′ ) x ′ ) = λ , by continuity. This implies for i = 0 and x ′ ∈ E ′ 0 ,

λ = S 0 f ( ρ 0 ( x ′ ) x ′ ) = ∫ R + d ∫ [ − 1 , 1 ] d x R 3 e | x | 2 x 2 ( | α | + d ) ( 1 ∧ e − c | x | 2 sin 2 θ sin θ | x | 2 ( sin θ | x | 2 ) 2 ( | α | + d ) − 1 2 ) Π α ( s ) d s f ( x ) d μ ˜ α ( x ) ≲ e p 0 2 ( x ′ ) 2 r 0 2 ( | α | + d ) Π α ( s ) d s f ( x ) d μ ˜ α ( x )     × ∫ R + d ∫ [ − 1 , 1 ] d { | x | ≥ r 0 } x ( x ) ( 1 ∧ e − c r 0 4 sin 2 θ | x | 2 − r 0 2 + sin θ r 0 2 ( | x | 2 − r 0 2 + sin θ r 0 2 ) 2 ( | α | + d ) − 1 2 ) , (14)

and for i = 1 and x ′ ∈ E ′ 1 ,

λ = S 1 f ( p i ( x ′ ) x ′ ) = ∫ R + d ∫ [ − 1 , 1 ] d x R 5 ( x ) e | x | 2 ( 1 + | x | ) e − sin 2 θ | x | 2 Π α ( s ) d s f ( x ) d μ ˜ α ( x ) ≲ e p 0 2 ( x ′ ) 2 r 1 ∫ R + d ∫ [ − 1 , 1 ] d { | x | > r 1 } x ( x ) e − c sin 2 θ r 1 2 Π α ( s ) d s f ( x ) d μ ˜ α ( x ) (15)

Clearly, since r₀ and r₁ are greater than one, we have

μ ˜ α { x ∈ E i : r i ≤ | x | ≤ 2 r i } ≤ ∫ E ′ i d σ ( x ′ ) ∫ ρ i ( x ′ ) 2 r i     e − ρ 2 ρ 2 ( | ∝ | + d ) − 1 d ρ ≲ ∫ E ′ i e − ρ i ( x ′ ) 2 r i 2 ( | ∝ | + d − 1 ) d σ ( x ′ )

combining this estimate for i = 0 with (14), we get

μ ˜ α { x ∈ E i : r i ≤ | x | ≤ 2 r i } ≤ C λ ∫ E ′ 0 r 0 2 ( ( 2 | ∝ | + 2 d ) − 1 ) d σ ( x ′ ) ( I 0 + I I 0 ) , (16)

with

I 0 = ∫ [ − 1 , 1 ] d ∫ { sin θ r 0 2 ≤ c } e − c r 0 4 sin 2 θ | x | 2 − r 0 2 + sin θ r 0 2 ( | x | 2 − r 0 2 + sin θ r 0 2 ) 2 ( | ∝ | + d ) − 1 2 f ( x ) d μ ˜ α ( x ) Π α ( s ) d s ,

and

I I 0 = ∫ [ − 1 , 1 ] d ∫ { | x | ≥ r 1 sin θ r 0 2 ≤ c } f ( x ) d μ ˜ α ( x ) Π α ( s ) d s .

Similarly for i = 1 with (15), we obtain

μ ˜ α { x ∈ E 1 : r 1 ≤ | x | ≤ 2 r 1 } ≤ C λ ∫ E ′ 1 r 1 2 ( | ∝ | + d ) − 1 d σ ( x ¯ ) ( I 1 + I I 1 ) , (17)

with

I 1 = ∫ [ − 1 , 1 ] d ∫ { | x | ≥ r 1 sin θ r 1 2 ≥ c } e − c sin 2 θ r 1 2 f ( x ) d μ ˜ α ( x ) Π α ( s ) d s ,

and

I I 1 = ∫ [ − 1 , 1 ] d ∫ { sin θ r 1 2 ≤ c } f ( x ) d μ ˜ α ( x ) Π α ( s ) d s .

It is immediate to verify that

r 0 2 ( 2 ( | ∝ | + d ) − 1 ) ∫ [ − 1 , 1 ] d ∫ { x ′ : sin θ r 0 2 ≤ c } σ ( x ′ ) Π α ( s ) d s ≤ C

and

r 1 2 ( | ∝ | + d ) − 1 ∫ [ − 1 , 1 ] d ∫ { x ′ : sin θ r 0 2 ≤ c } d σ ( x ′ ) Π α ( s ) d s ≤ C .

Which give, after changing the order of integration in (16) and (17), the desired estimate for the terms involving II₀ and II₁, respectively as in [5]. Now let us prove that for | x | ≥ r 0

r 0 2 ( 2 ( | ∝ | + d ) − 1 ) ∫ [ − 1 , 1 ] d ∫ { x ′ : sin θ r 0 2 ≤ c } e − c r 0 4 sin 2 θ | x | 2 − r 0 2 + sin θ r 0 2 ( | x | 2 − r 0 2 + sin θ r 0 2 ) 2 ( | ∝ | + d ) − 1 2 d σ ( x ′ ) Π α ( s ) d s ≤ C

and for | x | > r 1

r 1 2 ( | ∝ | + d ) − 1 ∫ [ − 1 , 1 ] d ∫ { x ′ : sin θ r 1 2 ≥ c } e − c sin 2 θ r 1 2 d σ ( x ′ ) Π α ( s ) d s ≤ C .

Firstly, one considers the case where ∝   = ( n 1 2 , − 1 , ⋯ , n d 2 − 1 ) with n i ∈ N and

n i > 1 for each i = 1 , ⋯ , d . In this case the inner integrals can be interpreted as integrals over S | n | − 1 with respect to the Lebesgue measure, expressed in polyradial

coordinates in [11]. The same estimates are obtained also for ∝   ∈ N d 2 − 1 + i R d .

Finally the result for the other values of ∝ are obtained via the multidimensional Stein’s complex interpolation Theorem. Indeed, let F : C d → C the function defined by

F ( ξ ) = r 0 2 ( 2 ξ + 2 d − 1 ) ∫ { sin θ r 0 2 ≤ c } e − c r 0 4 sin 2 θ | x | 2 − r 0 2 + sin θ r 0 2 ( | x | 2 − r 0 2 + sin θ r 0 2 ) 2 ( | ξ | + d ) − 1 2 Π ξ ( s ) d s .

We have seen that | F ( n 2 − 1 ) | ≤ C and it is easy to prove that | F ( n 2 − 1 + i ζ ) | ≤ | F ( n 2 − 1 ) | , whenever n is a integer vector and ζ ∈ R d .

Now we introduce the possible roots of the equations mentioned in the following Remark see [13]

Remark 5: 1) a) if r 0 = r 1 then we have

r 0 2 | α | + 2 d − 1 = 1 ,

and

2 | α | + 2 d − 1 = 0 ,

which implies that

| α | = 1 2 ( 1 − 2 d ) ,

b) if r 0 ≠ r 1 we have the quadratic equation

( 2 | α | + 2 d ) ln r 0 = ln r 1 + r 1 2 − r 0 2

we assume, for simplicity, that r 0 = e n and r 1 = e 2 n we can find

( 2 | α | + 2 d ) ln e n = ln e 2 n + e 4 n − e 2 n

( e 2 n ) 2 − e 2 n + 2 n ( 1 − | α | − d ) = 0

so that

e 2 n = 1 ± 1 − 8 n ( 1 − | α | − d ) 2 ,

where n ≥ 1 , we can easily find r 0 .

2) S 0 and S 1 aremonotone.

3) Since | x | 2 ( | ∝ | + d ) | x | < 1 , then | x | 2 ( | ∝ | + d ) ≤ | x | < C .

Proposition 6: For all m, the operator

T 0 m + 1 f ( x ) = p . v . ∫ R + d ∫ [ − 1 , 1 ] d χ R 0 ( x , s ) K m + 1 ( x , s ) Π ∝ ( s ) d s f ( x ) d ( m + 1 ) ∝ ( x ) ,

which is the modified Riesz-Laguerre transform restricted to the local regionR₀, is of weak type (1, 1) with respect to the measure μ ˜ ∝

Proof. The proof of this result follows the same steps like the proof of the weak-type boundedness on the local zone of the first order Riesz-Laguerre transforms done in [4] [9]. For the former we have the Calderon–Zygmund-type estimates for the kernel K m + 1 .

Lemma 7: There exists a constant C such that

| K m + 1 ( x , s ) | φ ( x , s ) ≤ C ( 2 | x | 2 ( 1 − cos θ ) ) − ( | ∝ | + d ) ,

| ∇ ( x , x ) ( K m + 1 ( x , s ) φ ( x , s ) ) | ≤ C ( 2 | x | 2 ( 1 − cos θ ) ) − ( | ∝ | + d + 1 / 2 )

being φ ( x , s ) acut-off function defined in [3] and ( x , s ) ∈ R 0 .

Proof. Since | x i ( r − s i ) | ≤ q − 1 / 2 ( r x 2 , x 2 , s ) , then

| ∏ i = 1 d     H a i ( x i ( r − s i ) 1 − r ) | e − q − ( r x 2 , x 2 , s ) 1 − r ≲ ∑ k = 0 | m + 1 | ( q − 1 / 2 ( r x 2 , x 2 , s ) 1 − r ) k / 2 e − q − ( r x 2 , x 2 , s ) 1 − r

≲ e − q − ( r x 2 , x 2 , s ) 2 ( 1 − r ) ≲ e − c q − ( r x 2 , x 2 , s ) 1 − r ,

where last inequality follows from this one:

q − ( r x 2 , x 2 , s ) ≥ ( 2 | x | 2 ( 1 − cos θ ) ) 1 / 2 − 2 C ( 1 − r 1 / 2 )

when ( x , s ) ∈ R 0 in [6]. Thus on R 0

| K m + 1 ( x , s ) | φ ( x , s ) ≲ ∫ 1 / 2 1 ( r ) | m + 1 | − 1 ( − log r 1 − r ) | m + 1 | − 2 2 e − c q − ( r x 2 , x 2 , s ) 1 − r ( 1 − r ) | ∝ | + d + 1 d r ≲ ∫ 0 1 / 2 ( r ) | m + 1 | − 2 ( − log r ) | m + 1 | − 2 2 d r + ∫ 1 / 2 1 e − c q − ( r x 2 , x 2 , s ) 1 − r ( 1 − r ) | ∝ | + d + 1 d r ≲ 1 + ( 2 | x | 2 ( 1 − cos θ ) ) − ( | ∝ | + d ) .

In computing the gradient to f the kernel with respect to x we are going to have integrals such as K m + 1 ( x , s ) ∂ x j ( x , s ) ,

∫ 0 1 ( r ) | m + 1 | − 1 ( − log r 1 − r ) | m + 1 | − 2 2 ∏ i ≠ j     H a i ( x i ( r − s i ) 1 − r )   × H a i − 1 ( x i ( r − s i ) 1 − r ) e − q − ( r x 2 , x 2 , s ) 1 − r ( 1 − r ) | ∝ | + d + 3 / 2 d r ,

with H − 1 ≡ 0 and

∫ 0 1 ( r ) | m + 1 | − 2 ( − log r 1 − r ) | m + 1 | − 2 2 H a i ( x i ( r − s i ) 1 − r ) × x j ( r − s j ) 1 − r e − q − ( r x 2 , x 2 , s ) 1 − r ( 1 − r ) | ∝ | + d + 3 / 2 d r .

In order to estimate these three integrals we use the same estimates described at the beginning of this proof. For the first one we have to use

2 | ∇ x φ ( x , s ) | ≲ 1 ( 2 | x | 2 ( 1 − cos θ ) ) 1 / 2

The gradient with respect to x is treated similarly.

For the latter we have the following Theorem regarding the L p d μ ˜ ∝ -boundedness for 1 < p < ∞ of the modified Riesz-Laguerre transform of any order on G.

Theorem8: The operator

R g m + 1 f ( x ) = ∫ R + d ∫ [ − 1 , 1 ] d χ G ( x , s ) K m + 1 ( x , s ) Π ∝ ( s ) d s f ( x ) d μ ∝ ( x )

is strong-type ( p , p ) for 1 < p < ∞ with respect to the measure μ ˜ ∝ .

Proof.The proof of this result in [1] is an adaptation to our context of the same result for the higher order Riesz–Gauss transform s done in [6]. Taking into account that on G, q + ( x 2 , s ) q − ( x 2 , s ) ≥ c when cos θ ≥ 0 , an upperbound for | K m + 1 ( x , s ) | is

K ˜ ˜ m + 1 ( x , s ) = { ( 2 | x | 2 ) | m + 1 | − 2 2 e − | x | 2     if   cos θ < 0 , ( 2 | x | 2 ( 1 + cos θ ) ) | α | + d ( 2 | x | 2 sin θ ) | m + 1 | − 1 2 e − | x | 2 sin θ     if   cos θ ≥ 0.

Thus

∫ R + d | ∫ G ∩ { cos θ < 0 } K m + 1 ( x , s ) Π ∝ ( s ) d s f ( x ) d μ ∝ ( x ) | p d μ ˜ ∝ ≲ ∫ R + d ( ∫ G ∩ { cos θ < 0 } K ˜ ˜ m + 1 ( x , s ) Π ∝ ( s ) d s | f ( x ) | d μ ∝ ( x ) ) p d μ ˜ ∝ ≲ ∫ R + d ( ∫ R + d ( 2 | x | 2 ) p ′ ( | m + 1 | − 2 ) 2 d μ ˜ ∝ ) p − 1 d μ ˜ ∝ ( x ) ‖ f ‖ L p ( d μ ˜ ∝ ) p .

For the region G ∩ { cos θ ≥ 0 } we are going to use the following estimates:

2 x 2 ≤ q + ≤ | 2 x | 2 , q − ≥ 0 , ( q + q − ) 1 2 ≥ 0 ,

0 ≤ | 1 p − 1 2 | | x | 2 sin θ < | x | 2 sin θ ,     since   p > 1 ,

∫ R + d | ∫ G ∩ { cos θ < 0 } K m + 1 ( x , s ) Π ∝ ( s ) d s f ( x ) d μ ∝ ( x ) | p d μ ˜ ∝ ≲ ∫ R + d ( ∫ G ∩ { cos θ < 0 } K ˜ ˜ m + 1 ( x , s ) Π ∝ ( s ) d s | f ( x ) | d μ ∝ ( x ) ) p d μ ˜ ∝ ≲ ∫ R + d ( ∫ G ∩ { cos θ < 0 } | 2 x | 2 ( | ∝ | + d ) ( 2 | x | 2 sin θ ) ( m + 1 ) − 1 2 e − 2 | x | 2 sin θ 2 Π ∝ ( s ) d s × | f ( x ) | e − | x | 2 p d μ ∝ ( x ) ) p d μ ∝ ( x ) ≲ ∫ R + d ( ∫ G ∩ { cos θ < 0 } | 2 x | 2 ( | ∝ | + d ) ( 2 | x | 2 sin θ ) ( m + 1 ) − 1 2 e − ( 1 2 − | 1 p − 1 2 | ) | x | 2 sin θ Π ∝ ( s ) d s × | f ( x ) | e − | x | 2 p d μ ∝ ( x ) ) p d μ ∝ ( x ) ≲ ∫ R + d ( ∫ G ∩ { cos θ < 0 } | 2 x | 2 ( | ∝ | + d ) ( 2 | x | 2 sin θ ) ( m + 1 ) − 1 2 e − c | x | 2 sin θ Π ∝ ( s ) d s × | f ( x ) | e − | x | 2 p d μ ∝ ( x ) ) p d μ ∝ ( x ) .

To finish the proof we just need to check that the kernel

H ( x , s ) : = | 2 x | 2 ( | ∝ | + d ) e − c p 2 | x | 2 sin θ χ G ∩ { cos θ ≥ 0 } ,

for G = { ( x , s ) : q − 1 2 ( x 2 , s ) ≥ c 1 + 2 | x | } is in L 1 ( d ( m + 1 ) ∝ ( x ) )

and independently of the remaining variables. Due to the symmetry of the kernel we are going to check only the first Claim given in [1].

∫ R + d H ( x , s ) d ( m + 1 ) ∝ ( x ) ≲ ∫ 0 ≤ 1 | x | 2 ( | ∝ | + d ) e − c p | x | ( 2 | x | 2 ( 1 − cos θ ) ) 1 / 2 d ( m + 1 ) ∝ ( x )       + ∫ 0 > 1 2 | x | 2 ( | ∝ | + d ) e − c ˜ p ( 2 | x | ) d ( m + 1 ) ∝ ( x ) .

It is clear that the second integral is bounded independentl y of x and s, for the first one see (13) for any x.

It is known that the first order Riesz-Laguerre transforms are weak-type (1, 1). Furthermore, we also know from that the the Riesz-Laguerre transforms of order higher than 2 need not be weak-type (1, 1) with respect to μ α . However, we can prove the following result that has to do with certain kind of weights we can add on the domain of these transforms to make them satisfy a weak-type inequality.

Let us mention that in the Gaussian context something quite similar occur with the higher order Riesz-Gauss transforms. Perez proved that for | m + 1 | > 2 , the Riesz-Gauss transforms of order | m + 1 | associated to the Ornstein-Uhlenbeck semigroup, map L 1 ( ( 1 + | x | | m + 1 | − 2 ) d γ ) continuously into L 1 , ∞ ( d γ ) , with d γ ( x ) = e − | x | 2 d x . Regarding the weights for the Riesz-Laguerre transforms of order higher than 2, then [1] proved the following

Theorem 9: The Riesz-Laguerre transforms order | m + 1 | with | m + 1 | > 2 , map L 1 ( w d μ α ) continuously into L 1 , ∞ ( d μ α ) . Where

w ( x ) = ( 1 + | x | ) | m + 1 | − 2

Proof. As we mention in the preliminaries to prove this theorem is equivalent to prove that the modified Riesz-Laguerre transforms of order higher than 2 map L 1 ( w ˜ ε d μ ˜ ∝ ) continuously into L 1 , ∞ ( d μ ˜ ∝ ) , with w ˜ ( x ) = ( 1 + | x | ) | m + 1 | − 2 . For each x ∈ R + d . Let us write

R + d × [ − 1 , 1 ] d = ∪ i = 0 4 R i .

Therefore, in order to get the result, it will be enough to prove that each of the following operators

T i m + 1 f ( x ) = ∫ R + d ∫ [ − 1 , 1 ] d χ R i ( x , s ) K m + 1 ( x , s ) Π ∝ ( s ) d s f ( x ) d μ ∝ ( x ) ,

for i = 0 , ⋯ , 4 maps L 1 ( w ˜ ε d μ ˜ ∝ ) continuously into L 1 , ∞ ( d μ ˜ ∝ ) .

0

| K m + 1 ( x , s ) | ≲ { ( 2 | x | 2 ) | m + 1 | − 2 2 K * ( x , s ) ,                     if   cos θ < 0 , ( 2 | x | 2 sin θ ) | m + 1 | − 2 2 K * ( x , s ) ,       cos θ ≥ 0

If ( x , s ) ∈ R i , | k m + 1 ( x , s ) | is controlled by C ( 1 + { | x | } ) | m + 1 | − 2 e − | x | 2 and there for it is immediate to prove that T 1 m + 1 maps L 1 ( w ˜ ε d μ ˜ ∝ ) into L 1 ( d μ ˜ ∝ ) .

Now if ( x , s ) ∈ R i , with i = 2 , 3 , 4 , weclaim that

| K m + 1 ( x , s ) | ≲ w ˜ ( x ) K * ( x , s )

If ( x , s ) ∈ R 2 since

q + ≤ ( 2 | x | ) 2 ≲ | x | 2 ,

then

| K m + 1 ( x , s ) | ≲ ( 2 | x | 2 sin θ ) | m + 1 | − 2 2 e − C ( | x | 4 ( 1 − cos θ ) ) 1 / 2 ≲ w ˜ ( x ) e − C ( | x | 4 ( 1 − cos θ ) ) 1 / 2 .

Also

q + q − = 4 | x | 4 sin 2 θ .

Thus

| K m + 1 ( x , s ) | ≲ ( 2 | x | 2 sin θ ) | m + 1 | − 2 2 K * ( x , s ) ≲ w ˜ ( x ) K * ( x , s ) .

And this concludes the proof of the Theorem.

It should be noted that there is another proof of Theorem 9 for multi-indices of half-integer type by taking f w as the function f in [3] [7].

Now we introduce a sharp estimate forw.

Corollary 10: The Riesz-Laguerre transforms of order | m + 1 | with | m + 1 | > 2 , map L 1 ( w d μ α ) continuously into L 1 , ∞ ( d μ α ) . Where

| ∝ | = 8 n ( 1 − d ) − 1 − ( 2 e 2 n − 1 ) 2 8 n

and

w ( x ) ≤ ( 1 + | C | ) | m + 1 | − 2 = K | m + 1 | − 2

Proof. From Theorem 9 and Remark 5: We can directly see that

w ( x ) = ( 1 + | x | ) | m + 1 | − 2 ≤ ( 1 + | C | ) | m + 1 | − 2 ≤ K | m + 1 | − 2 ,

where m ≥ 2 .

Theorem 11: The weight w is the optimal polynomial weight needed to get the weak type (1, 1) inequality for the Riesz-Laguerre transforms of order | m + 1 | .

Proof. This proof follows essentially in [9]. With the notation of that Theorem 1 One takes η ∈ ℝ + d with | η | sufficiently large, away from the axis and obtains the following lower bound for K m + 1 ( x , η )

K m + 1 ( x , η ) = C ∫ [ − 1 , 1 ] d K m + 1 ( x , η , s ) Π ∝ ( s ) d s ≥ C | η | | m + 1 | − 2 | ∝ | − d − 1 e ξ 2 − | η | 2 (18)

for x ∈ J = { ξ η | η | + v : v ⊥ η , | v | < 1 , 1 2 | η | < ξ < 3 2 | η | } .

Now if we assume that the Riesz-Laguerre transforms of order | m + 1 | > 2 map L 1 ( w ˜ ε d μ ˜ ∝ ) continuously into L 1 , ∞ ( d μ ˜ ∝ ) with w ˜ ε = ( 1 + | x | ) ε and

0 < ε < | m + 1 | − 2 then by taking f ≥ 0 in L 1 ( w ˜ ε d μ ˜ ∝ ) close to an approxima tion of a point mass at η, with ‖ f ‖ L 1 ( w ˜ ε d μ ˜ ∝ ) = 1 we have that R ∝ m + 1 f ( x ) is close

to e | η | 2 K m + 1 ( x , η ) | η | − ε and by applying in equality (18) we get that

e | n | k m + 1 ( x , η ) | η | − ϵ ≥ | η | | m + 1 | − 2 | α | − d − 1 − ϵ e − ( | n | 2 ) 2 . Therefore setting

λ = | η | | m + 1 | − 2 | α | − d − 1 − ϵ e − ( | n | 2 ) 2

we obtain

e − ( | η | 2 ) 2 | η | 2 | α | + d − 1 ≲ μ ˜ α ( J ) ≤ μ ˜ α { x ∈ R + d : R α m + 1 f ( x ) > λ } ≲ 1 λ = C | η | 2 | α | + d − | m + 1 | + 1 + ϵ e − ( | η | 2 ) 2 .

Hence | η | | m + 1 | − 2 − ϵ must be bounded which is a contradiction. Therefore the conclusion of Theorem 11 holds.

The authors declare no conflicts of interest regarding the publication of this paper.

Elobied, A., Siddig, A. and Widatalla, S. (2022) Inequalities for Higher Order Riesz-Laguerre Transforms. Advances in Pure Mathematics, 12, 332-347. https://doi.org/10.4236/apm.2022.124025