on an abstract segal algebra under fractional convolution

$: On an abstract Segal algebra under fractional convolution$
Montes Taurus J. Pure Appl. Math. 4 (1), 1–22, 2022

MTJPAMISSN: 2687-4814

On an abstract Segal algebra under fractional convolution

Ayse Sandıkcı ID a, Erdem Toksoy ID b

aOndokuz Mayıs University, Faculty of Art and Sciences, Departman of Mathematics, Kurupelit, Samsun, 55139, TurkeybOndokuz Mayıs University, Faculty of Art and Sciences, Departman of Mathematics, Kurupelit, Samsun, 55139, Turkey

Abstract

In this work, we find approximate identities for the spaces L1(Rd

), L1

w

(Rd

)and S α

w(Rd) under Θ convolution. Furthermore, we

determine approximate identities with compactly supported fractional Fourier transforms in the spaces L1w

(Rd

)and S α

w(Rd), wherew is weight function of regular growth. We give definitions of multipliers of these Banach algebras under Θ convolution. Also, weshow that the space S α

w(Rd) under some conditions is an abstract Segal algebra with recpect to L1w

(Rd

).

Keywords: Fractional Fourier transform, approximate identity, Segal algebras

2010 MSC: 42B10, 43A99

This work is licensed under a Creative Commons Attribution 4.0 International License.

1. Introduction

Throughout this article, we study on Rd. For any function f : Rd → C, the translation and modulation operatorsare defined as Ty f (t) = f (t − y) and Mω f (t) = eiωt f (t) for all y, ω ∈ Rd, respectively, [18]. Cc

(Rd

)denotes the

space of continuous complex function on Rd whose support is compact, [17]. Besides we write the Lebesgue space(Lp

(Rd

), ‖.‖p

), for 1 ≤ p < ∞. A weight (Beurling weight) function w on Rd is a measurable and locally bounded

function that, satisfying w (x) ≥ 1 and w (x + y) ≤ w (x) w (y) (submultiplicative, [12]) for all x, y ∈ Rd. All weights wegive throughout the article are Beurling weights that, satisfying all the conditions are given above. A weight functionw is weight function of regular growth if w

(xρ

)≤ w (x) (ρ ≥ 1) and there are constants C ≥ 1 and λ > 0 such that

w (ρx) ≤ Cρλw (x) (ρ ≥ 1) for all x ∈ Rd. If w is weight function of regular growth, then there exist constants C ≥ 1and λ > 0 such that

w (x) ≤ C1‖x‖λ (1.1)

for ‖x‖ ≥ 1, where C1 = C sup {w (x)| ‖x‖ = 1}. We define, for 1 ≤ p < ∞,

Lpw

(Rd

)=

{f | f w ∈ Lp

(Rd

)}.

†Article ID: MTJPAM-D-21-00034

Email addresses: [email protected] (Ayse Sandıkcı ID ), [email protected] (Erdem Toksoy ID )Received:19 May 2021, Accepted:18 June 2021, Published:30 June 2021?Corresponding Author: Ayse Sandıkcı

1

https://orcid.org/0000-0001-5800-5558

https://orcid.org/0000-0003-3597-6161

https://orcid.org/0000-0001-5800-5558

https://orcid.org/0000-0003-3597-6161

Sandıkcı and Toksoy / Montes Taurus J. Pure Appl. Math. 4 (1), 1–22, 2022

It is well known that Lpw

(Rd

)is a Banach space under the norm ‖ f ‖p,w = ‖ f w‖p, [16]. We define the Fourier transform

f (or F f ) of a function f ∈ L1 (R) as

f (ω) = F f (ω) =1√

2π

+∞∫−∞

f (t)e−iωtdt.

The fractional Fourier transform is a generalization of the Fourier transform through an angle parameter α andcan be considered as a rotation by an angle α in the time-frequency plane. The fractional Fourier transform with angleα of a function f ∈ L1 (R) is defined by

F α f (u) =

+∞∫−∞

Kα(u, t) f (t)dt,

where,

Kα(u, t) =

√

1−i cotα2π e

i(

u2+t22

)cotα−iut cscα

, if α , kπ, k ∈ Zδ(t − u), if α = 2kπ, k ∈ Zδ(t + u), if α = (2k + 1)π, k ∈ Z

and δ, Dirac delta function. The fractional Fourier transform with α = π2 corresponds to the Fourier transform,

[1, 2, 4, 14, 15, 23]. The fractional Fourier transform can be extended for higher dimensions as [4]:

(F α1,...,αd f

)(u1, ..., ud) =

+∞∫−∞

...

+∞∫−∞

Kα1,...,αd (u1, ..., ud; t1, ..., td) f (t1, ..., td) dt1 · · · dtd,

or shortly

F α f (u) =

+∞∫−∞

...

+∞∫−∞

Kα (u, t) f (t) dt,

whereKα (u, t) = Kα1,...,αd (u1, ..., ud; t1, ..., td) = Kα1 (u1, t1) Kα2 (u2, t2) ...Kαd (ud, td) .

Throughout this study, unless otherwise indicated, we get α = (α1, α2, ..., αd), where αi , kπ for i = 1, 2, ..., d andk ∈ Z. Let γ = (−y1 cotα1, ...,−yd cotαd) for all y = (y1, ..., yd) ∈ Rd. The Θ convolution operation is

( f Θg) (x) =

∫Rd

f (y) g (x − y) ed∑

j=1iy j(y j−x j) cotα j

dy (1.2)

=

∫Rd

f (y) TyMγg (x) dy

for all f , g ∈ L1(Rd

), [19, 20].

Let A and B be commutative Banach algebras and B ⊆ A, the space B is said to be Banach ideal of A if ‖g‖A ≤ ‖g‖Band gh ∈ B, the inequality ‖gh‖B ≤ C2 ‖g‖B ‖h‖A holds for all g ∈ B, h ∈ A, [10]. A Banach function space (B, ‖.‖B)of measurable function is said to be solid, if for every g ∈ B and any measurable function h satisfying |h (x)| ≤ |g (x)|almost everywhere, h ∈ B and ‖h‖B ≤ ‖g‖B, [9].

(B, ‖.‖B) is an abstract Segal algebra with respect to a Banach algebra (A, ‖.‖A) if it satisfies the following conditions[5]:

1. B is a dense ideal in A and B is a Banach algebra under the norm ‖.‖B.

2. There exists C1 > 0 such that ‖g‖A ≤ C1 ‖g‖B for all g ∈ B.

2


3. There exists C2 > 0 such that ‖gh‖B ≤ C2 ‖g‖A ‖h‖B for all g, h ∈ B.

Let w be a weight function on Rd. The space S w(Rd) is subalgebra of L1w

(Rd

)satisfying the following properties

[6]:

1. The space S w(Rd) is dense in L1w

(Rd

).

2. The subalgebra S w(Rd) is a Banach algebra under some norm ‖.‖S wand the inequality ‖g‖1,w ≤ ‖g‖S w

holds forall g ∈ S w(Rd).

3. S w(Rd) is translation invariant and for each g ∈ S w(Rd) and all y ∈ Rd, the inequality∥∥∥Tyg

∥∥∥S w≤ w (y) ‖g‖S w

holds.

4. The mapping y→ Tyg from Rd into S w(Rd) is continuous.

Let us take Θ convolution that is given in (1.2) instead of ordinary convolution. The space S Θw(Rd) under Θ

convolution satisfies conditions of the space S w(Rd) and the norm of this space is denoted by ‖.‖S Θw, [21].

Let w be a weight function on Rd. The space S αw(Rd) consist of all f ∈ L1

w

(Rd

)such that F α f ∈S Θ

w(Rd). As thelinear space S α

w(Rd) is normed by‖ f ‖S α

w= ‖ f ‖1,w + ‖F α f ‖S Θ

w,

then S αw(Rd) is a Banach algebra under this norm. The space S α

w(Rd) is a Banach ideal on L1w

(Rd

)if the space S Θ

w(Rd)is solid. This space is also translation and modulation invariant, [21].

Let A be a Banach algebra and T is an operator from A into A. T is a multiplier of A if the equality x (Ty) = (T x) yholds for all x, y ∈ A. The set of multipliers on A is denoted by M (A). Let A be a commutative Banach algebra withoutorder (i.e xA = {0} implies x = 0). For any T ∈ M (A), the equality T (xy) = x (Ty) = (T x) y holds for all x, y ∈ A andT be a bounded linear operator, [13]. Investigating the multipliers of a Banach algebra it is important that this algebrabe without order. The Banach algebra L1

w

(Rd

)under the ordinary convolution is without order. The multiplier T on

L1w

(Rd

)is a bounded linear operator that commutes with translation operator, [3].

In this study we find approximate identities for the spaces L1(Rd

), L1

w

(Rd

)and S α

w(Rd) under Θ convolution.Furthermore, we determine approximate identities with compactly supported fractional Fourier transforms in thespaces L1

w

(Rd

)and S α

w(Rd), where w is a weight function of regular growth. Also, we give definitions of multipliersof these Banach algebras under Θ convolution. In these definitions we take a new operator corresponding to thetranslation operator for αi = π

2 with i = 1, 2, ..., d which commutes with multiplier. If we take αi = π2 with i =

1, 2, ..., d, the Θ convolution and fractional Fourier transform correspond ordinary convolution and Fourier transform,respectively. Also, we show that the space S α

w(Rd) under some conditions is an abstract Segal algebra with recpect toL1

w

(Rd

).

2. On an Abstract Segal Algebra under Fractional Convolution

First of all, we will show that the Banach algebra L1w(Rd) under Θ convolution is without order.

Theorem 2.1. Let γ = (−y1 cotα1, ...,−yd cotαd) for all y = (y1, ..., yd) ∈ Rd.

1. TyMγg ∈ L1w(Rd) for all g ∈ L1

w(Rd).

2. The mapping y→ TyMγg from Rd into L1w(Rd) is continuous.

Proof. 1. It is obvious from the invariance of translate and modulation operators of the space L1w(Rd).

3


2. It is well known that the mappings y → Tyg and y → Myg from Rd into L1w(Rd) are continuous by [11]. Let

γ = (−y1 cotα1, ...,−yd cotαd) for all y = (y1, ..., yd) ∈ Rd. Then, it is easy to see that the mapping y → MγTygfrom Rd into L1

w(Rd) is continuous by using the same method of the proof of Theorem 2.9 in [21]. Also it iswell known that

TyMγg (t) = e−iyγMγTyg (t)

for all t ∈ Rd. Thus, we haveTyMγg = e−iyγMγTyg. (2.1)

Since the mapping y→ MγTyg from Rd into L1w(Rd) is continuous, then by using (2.1), the mapping y→ TyMγg

from Rd into L1w(Rd) is also continuous.

Proposition 2.2. The algebra L1w(Rd) under Θ convolution has an approximate identity.

Proof. Let F be a finite subset of L1w

(Rd

)such that F = {g1, ..., gn}. Let ε > 0 be given. Then by Theorem 2.1 there

exist δi > 0 such that ∥∥∥TyMγgi − gi

∥∥∥1,w < ε

when ‖y‖ < δi for all i = 1, ..., n. Let δ = min {δ1, ..., δn}. Thus we have∥∥∥TyMγgi − gi

∥∥∥1,w < ε (2.2)

when ‖y‖ < δ for all i = 1, 2, ..., n. Let us take a positive function h ∈ Cc

(Rd

)such that supp h ⊂ B (0, δ) and∫

Rd

h (x)dx = 1. Therefore we write

(hΘgi) (x) − gi(x) =

∫Rd

h(y)(TyMγgi(x) − gi(x)

)dy

for all x ∈ Rd and i = 1, ..., n. Then by using (2.2), we have

‖(hΘgi) − gi‖1,w =

∥∥∥∥∥∥∥∥∥∫Rd

h(y)(TyMγgi − gi

)dy

∥∥∥∥∥∥∥∥∥1,w

≤

∫supp h

|h(y)|∥∥∥TyMγgi − gi

∥∥∥1,wdy

< ε

∫supp h

h(y)dy = ε

for all i = 1, ..., n. Hence L1w(Rd) under Θ convolution has an approximate identity by 1.3. Proposition in [8].

Since the algebra L1w(Rd) under Θ convolution has an approximate identity, then it is an algebra without order.

Now, we will give a definition of multipliers for the Banach algebra L1w(Rd) under Θ convolution.

Definition 2.3. A multiplier T is a continuous linear operator from L1w(Rd) into L1

w(Rd) which commutes with operatorTyMγ, where γ = (−y1 cotα1, ...,−yd cotαd) for all y = (y1, ..., yd) ∈ Rd. The set of multipliers on L1

w(Rd) is denotedby M

(L1

w

(Rd

)).

Proposition 2.4. Let γ = (−y1 cotα1, ...,−yd cotαd) for all y = (y1, ..., yd) ∈ Rd. Then we have

TyMγ f Θg = f ΘTyMγg (2.3)

for all f , g ∈ L1w

(Rd

).

4


Proof. Let f , g ∈ L1w

(Rd

). We may write

(TyMγ f Θg

)(x) =

∫Rd

TyMγ f (z) g (x − z)ed∑

j=1iz j(z j−x j) cotα j

dz

=

∫Rd

f (z − y) g (x − z)ed∑

j=1iy j(y j−z j) cotα j

ed∑

j=1iz j(z j−x j) cotα j

dz

for all x ∈ Rd. By substitution u = z − y, we obtain

(TyMγ f Θg

)(x) =

∫Rd

f (u) g (x − u − y)ed∑

j=1−iu jy j cotα j

ed∑

j=1i(u j+y j)(u j+y j−x j) cotα j

du

=

∫Rd

f (u) g (x − u − y)ed∑

j=1iy j(y j+u j−x j) cotα j

ed∑

j=1iu j(u j−x j) cotα j

du

=

∫Rd

f (u) TyMγg (x − u)ed∑

j=1iu j(u j−x j) cotα j

du

= f ΘTyMγg (x)

for all x ∈ Rd.

Theorem 2.5. Let T be an operator from L1w

(Rd

)into L1

w

(Rd

). Then T ∈ M

(L1

w

(Rd

))if and only if

T ( f Θg) = T f Θg = f ΘTg (2.4)

for all f , g ∈ L1w

(Rd

).

Proof. Let T be an operator from L1w

(Rd

)into L1

w

(Rd

). Firstly, we assume that equality (2.4) holds for all f , g ∈

L1w

(Rd

). Let h ∈ L1

w

(Rd

)and λ, β ∈ C. Thus by using (2.4), we have

f ΘT (λg + βh) = T f Θ (λg + βh) = λ (T f Θg) + β (T f Θh) = ( f ΘλTg) + ( f ΘβTh)

= f Θ (λTg + βTh) .

Since L1w

(Rd

)is an algebra without order, then we may write

T (λg + βh) = λTg + βTh.

Hence T is linear.Let h ∈ L1

w

(Rd

)and ( fn)n∈N is a convergent sequence such that

‖ fn − f ‖1,w → 0 and ‖T fn − g‖1,w → 0, (2.5)

where f , g ∈ L1w

(Rd

). Then we have

‖hΘg − hΘT f ‖1,w ≤ ‖hΘg − hΘT fn‖1,w + ‖hΘT fn − hΘT f ‖1,w= ‖hΘ (g − T fn)‖1,w + ‖ThΘ fn − ThΘ f ‖1,w (2.6)≤ ‖h‖1,w‖T fn − g‖1,w + ‖Th‖1,w‖ fn − f ‖1,w.

By combining (2.5) and (2.6) we obtain

hΘg − hΘT f = hΘ (g − T f ) = 0.

5


Since L1w

(Rd

)is an algebra without order, then we may write T f = g. Then by closed graph theorem T is continuous.

Finally, we will show that T commutes with operator TyMγ, where γ = (−y1 cotα1, ...,−yd cotαd) for ally = (y1, ..., yd) ∈ Rd. Let f , g ∈ L1

w

(Rd

). Then by using (2.3) and (2.4) we have

TTyMγ f Θg = T(TyMγ f Θg

)= T f ΘTyMγg = TyMγT f Θg,

where γ = (−y1 cotα1, ...,−yd cotαd) for all y = (y1, ..., yd) ∈ Rd. Since L1w

(Rd

)is an algebra without order, then we

obtainTTyMγ f = TyMγT f .

As a result, T ∈ M(L1

w

(Rd

)).

Conversely, we suppose that T ∈ M(L1

w

(Rd

)). Let φ ∈ L∞w

(Rd

). It is easy to see that the mapping g →∫

Rd

Tg(x)φ(x)dx is a linear functional on L1w

(Rd

). Also, we may write

∣∣∣∣∣∣∣∣∣∫Rd

Tg(x)φ(x)dx

∣∣∣∣∣∣∣∣∣ ≤∫Rd

|Tg(x)|w (x)|φ(x)|w (x)

dx

= ‖φ‖∞,w‖Tg‖1,w≤ ‖φ‖∞,w ‖T‖ ‖g‖1,w

such that ‖T‖ is an operator norm. Hence, this functional is bounded. Since L∞w(Rd

)is dual space of L1

w

(Rd

)(see

[16], p. 121), then there exists a function κ ∈ L∞w(Rd

)(i.e κ

w ∈ L∞(Rd

)) such that∫

Rd

Tg(x)φ(x)dx =

∫Rd

g(x)κ(x)dx (2.7)

for all g ∈ L1w

(Rd

). Let g, h ∈ L1

w

(Rd

). Thus by using (2.7), we get∫

Rd

(gΘTh) (x)φ(x)dx =

∫Rd

∫Rd

TTyMγh(x)φ(x)dxg (y) dy

=

∫Rd

∫Rd

TyMγh(x)κ(x)dxg (y) dy

=

∫Rd

∫Rd

TyMγh(x)g (y) dyκ(x)dx

=

∫Rd

(gΘh) (x)κ(x)dx

=

∫Rd

T (gΘh) (x)φ(x)dx.

Hence, by the Hahn-Banach theorem, we obtain

T (gΘh) = gΘTh.

Consequently, by commutativity of Θ convolution, we have equality (2.4).

Now, we will investigate approximate identities with compactly supported fractional Fourier transforms in thespaces L1

(Rd

)and L1

w

(Rd

), respectively. Before that, we will show that there exists a function g with compactly

supported fractional Fourier transform in the space L1(R). Throughout these examples we will set α , kπ, k ∈ Z.

6


Example 2.6. Let g be a function on R such that g (t) = 1√

2π

(sin t

2t2

)2e−

i2 t2 cotα for t , 0 and g (0) = 1

√2π

, where α , kπ,k ∈ Z. It is easy to see that this function is continuous. Also, since we have

+∞∫−∞

|g (t)|dt =

+∞∫−∞

∣∣∣∣∣∣∣ 1√

2π

(sin t

2t2

)2∣∣∣∣∣∣∣dt < ∞,

then g ∈ L1 (R). Let us take a function on R as f (t) = 1√

2π

(sin t

2t2

)2for t , 0 and f (0) = 1

√2π

. Then we write

g (t) = e−i2 t2 cotα f (t). By using Example 1.1.8 in [16] and inverse Fourier transform, we obtain

f (ω) =

{1 − |ω| , |ω| ≤ 10, |ω| > 1 . (2.8)

Hence, we get

F αg(u) = Bei2 u2 cotα

+∞∫−∞

e−iutu csc α f (t)dt = Bei2 u2 cotα f (u csc α) ,

where B =

√1−i cotα

2π . Consequently, by using (2.8), we obtain

F αg(u) =

{Be

i2 u2 cotα (1 − |u cscα|) , |u cscα| ≤ 1

0, |u cscα| > 1(2.9)

and then g has compactly supported fractional Fourier transform.

Now, by using the function f which is denoted in Example 2.6, we will show that there exists a function l withcompactly supported fractional Fourier transform in the space L1

(Rd

).

Example 2.7. Let l be a function on Rd as

l (t1, ..., td) = g (t1) g (t2) · · · g (td) (2.10)

such that

g(t j

)=

1√

2π

sin t j

2t j

2

2

e−i2 t2

j cotα j

and g (0) = 1√

2π, where α j , kπ, k ∈ Z with j = 1, 2, ..., d for all t = (t1, ..., td) ∈ Rd. Let us take the function f which

is given in Example 2.6. Then we haveg(t j

)= e−

i2 t2

j cotα j f(t j

),

where α j , kπ, k ∈ Z with j = 1, 2, ..., d. Therefore by Example 2.6, g(. j)∈ L1 (R) for all j = 1, 2, ..., d. Hence by

using (2.10) and Example 1.1.12 in [16], we write l ∈ L1(Rd

). Also we may write

F αl (u1, ..., ud) =

+∞∫−∞

Kα1 (u1, t1) g (t1) dt1 · · ·

+∞∫−∞

Kαd (ud, td)g (td) dtd

= F α1 g (u1) · · · F αd g (ud) (2.11)

for all u = (u1, ..., ud) ∈ Rd. Thus by using (2.9), we obtain F αl ∈ Cc

(Rd

). Consequently, the function l has compactly

supported fractional Fourier transform.

7


Now, we will define a sequence (kn)n∈N that has integral equal to 1 and compactly supported fractional Fouriertransforms in the space L1

(Rd

).

Example 2.8. Firstly, let us take the function f which is given in Example 2.6. Let us set a sequence (kn)n∈N by

kn (t1, ..., td) = hn (t1) hn (t2) · · · hn (td) (2.12)

such thathn

(t j

)=

n

A jn

e−i2 t2

j cotα j f(nt j

),

where

A jn =

+∞∫−∞

ne−i2 t2

j cotα j f(nt j

)dt j , 0

for all n ∈ N, t = (t1, ..., td) ∈ Rd and α j , kπ, k ∈ Z with j = 1, 2, ..., d. Let us take functions

gn

(t j

)= ne−

i2 t2

j cotα j f(nt j

),

where α j , kπ, k ∈ Z with j = 1, 2, ..., d for all n ∈ N. Then we write hn

(t j

)=

gn(t j)A j

nfor all n ∈ N and j = 1, 2, ..., d.

Hence by using definition of the function f , it is easy to see that functions gn

(. j)

are continuous and

+∞∫−∞

∣∣∣∣gn

(t j

)∣∣∣∣dt j = n

+∞∫−∞

∣∣∣∣ f (nt j

)∣∣∣∣dt j = ‖ f ‖1 < ∞

for all n ∈ N and j = 1, 2, ..., d. Thus gn

(. j)∈ L1 (R) and then hn

(. j)∈ L1 (R) for all n ∈ N and j = 1, 2, ..., d.

Therefore by using (2.12) and Example 1.1.12 in [16], we have (kn)n∈N ⊂ L1(Rd

),∫Rd

kn (t) dt = 1 and

F αkn (u1, ..., ud) = F α1 hn (u1)F α2 hn (u2) · · · F αd hn (ud) (2.13)

for all u = (u1, ..., ud) ∈ Rd and n ∈ N. Also we may write

F α j hn(u j) =n

A jn

B jei2 u2

j cotα j

+∞∫−∞

e−iu jt j cscα j f(nt j

)dt j,

where B j =

√1−i cotα j

2π for all n ∈ N and j = 1, 2, ..., d. By substitution nt j = x j, we obtain

F α j hn(u j) =B j

A jn

ei2 u2

j cotα j

+∞∫−∞

e−iu jx jn cscα j f

(x j

)dx j

=B j

A jn

ei2 u2

j cotα j f(

u j

ncscα

j

),

where B j =

√1−i cotα j

2π for all n ∈ N and j = 1, 2, ..., d. Hence by using (2.8), we get

F α j hn(u j) =

B j

A jne

i2 u2

j cotα j

(1 −

∣∣∣∣ u j

n cscα j

∣∣∣∣) ,∣∣∣∣ u j

n cscα j

∣∣∣∣ ≤ 1

0,∣∣∣∣ u j

n cscα j

∣∣∣∣ > 1,

where B j =

√1−i cotα j

2π for all n ∈ N and j = 1, 2, ..., d. Thus, F α j hn ∈ Cc (R) for all n ∈ N and j = 1, 2, ..., d. Then by

using (2.13), we obtain F αkn ∈ Cc

(Rd

)for all n ∈ N.

8


Proposition 2.9. The sequence (kn)n∈N which is given in Example 2.8 is an approximate identity with compactlysupported fractional Fourier transforms in the algebra L1(Rd) under Θ convolution.

Proof. Let us take the sequence (kn)n∈N that is given in Example 2.8. Also, let g ∈ L1(Rd

)and γ = (−y1 cotα1, ...,−yd cotαd)

for all y = (y1, ..., yd) ∈ Rd. Then we may write

(knΘg) (x) − g (x) =

∫Rd

kn (y) TyMγg (x) dy − g (x)

=

∫Rd

kn (y)(TyMγg (x) − g (x)

)dy

for all x ∈ Rd and n ∈ N. Thus we get

‖knΘg − g‖1 ≤

∫Rd

∫Rd

|kn (y)|∣∣∣TyMγg (x) − g (x)

∣∣∣ dydx

=

∫Rd

|kn (y)|∥∥∥TyMγg − g

∥∥∥1dy

=

∫Rd

∣∣∣∣∣∣ nd

A1n · · · A

dn

∣∣∣∣∣∣ | f (ny1) · · · f (nyd)|∥∥∥TyMγg − g

∥∥∥1dy

for all n ∈ N. By substitution ny j = z j for all j = 1, 2, ..., d, we obtain

‖knΘg − g‖1 ≤1∣∣∣A1

n

∣∣∣ · · · ∣∣∣Adn

∣∣∣∫d

| f (z1) · · · f (zd)|∥∥∥T z

nM τ

ng − g

∥∥∥1dz, (2.14)

where z = (z1, ..., zd) ∈ Rd and τ = (−z1 cotα1, ...,−zd cotαd). It is known by definition of the sequence (kn)n∈N that

A jn =

+∞∫−∞

ne−i2 y2

j cotα j f(ny j

)dy j

for all n ∈ N and j = 1, 2, ..., d. By substitution ny j = z j, we obtain

A jn =

+∞∫−∞

e−i2

z2j

n2 cotα j f(z j

)dz j

for all n ∈ N and j = 1, 2, ..., d. Also we have

limn→∞

(e−

i2

z2j

n2 cotα j f(z j

))= f

(z j

)and ∣∣∣∣∣∣e− i

2

z2j

n2 cotα j f(z j

)∣∣∣∣∣∣ =∣∣∣∣ f (

z j

)∣∣∣∣ .Thus, Dominated Convergence Theorem implies that the sequence

(∣∣∣∣ 1A1

n···Adn

∣∣∣∣)n∈N

is convergent. Hence, this sequence is

bounded, that is to say, there exists M > 0 such that∣∣∣∣∣∣ 1A1

n · · · Adn

∣∣∣∣∣∣ ≤ M (2.15)

9


for all n ∈ N. Moreover, let g ∈ L1(Rd

)and γ = (−y1 cotα1, ...,−yd cotαd) for all y = (y1, ..., yd) ∈ Rd. Then the

mapping y→ TyMγg from Rd into L1(Rd

)is continuous by Theorem 2.1 (in the case of w = 1). Therefore, we get∥∥∥T z

nM τ

ng − g

∥∥∥1→ 0.

Let s (z) = f (z1) · · · f (zd) for all z = (z1, ..., zd) ∈ Rd. Then s ∈ L1(Rd

)by Example 1.1.12 in [16]. Thus, we have

|s (z)|∥∥∥T z

nM τ

ng − g

∥∥∥1→ 0

also, ∥∥∥T znM τ

ng − g

∥∥∥1≤

∥∥∥T znM τ

ng∥∥∥

1+ ‖g‖1 = 2‖g‖1 = C1

and then|s (z)|

∥∥∥T znM τ

ng − g

∥∥∥1≤ C1 |s (z)| .

Hence, Dominated Convergence Theorem implies that∫Rd

|s (z)|∥∥∥T z

nM τ

ng − g

∥∥∥1dz→ 0. (2.16)

Finally, combining (2.14), (2.15) and (2.16), we obtain

‖knΘg − g‖1 ≤1∣∣∣A1

n

∣∣∣ · · · ∣∣∣Adn

∣∣∣∫Rd

|s (z)|∥∥∥T z

nM τ

ng − g

∥∥∥1dz→ 0.

This is the desired result.

Now, we will show that there exists a function l with compactly supported fractional Fourier transform in thespace L1

w (R), where w is a weight function of regular growth on R.

Example 2.10. Let α , mπ, m ∈ Z. Let l be a continuous function as l = F −αh such that h (t) = ei2 t2 cotαg (t) for any

nonzero function g on R with compact support and possessing a continuous kth derivative (k ≥ 2). Let M =

√1+i cotα

2π .Let us take a function f on R such that

f (x) = M

+∞∫−∞

g (t) eixt cscαdt (2.17)

for all x ∈ R. Then by definition of the function l we write

l(x) = e−i2 x2 cotα f (x) (2.18)

for all x ∈ R. Also, since g has a continuous kth derivative, then integration by parts k times yields. Assume thatsupp g = [a, b]. Therefore, we may write

f (x) = M(−ix cscα)−1

b∫a

g′ (t) eixt cscαdt

for x , 0. Continuing in this way, we obtain

f (x) = M(−ix cscα)−k

+∞∫−∞

g(k) (t) eixt cscαdt (2.19)

10


for x , 0. Let us take a constant B is given by

B = max

2k |M|

+∞∫−∞

|g (t)| dt, 2k |M| |cscα|−k

+∞∫−∞

∣∣∣g(k) (t)∣∣∣ dt

. (2.20)

Also we have|x| ≤ 1⇒ 1 + |x| ≤ 2 (2.21)

and|x| ≥ 1⇒ 1 + |x| ≤ |x| + |x| = 2 |x| . (2.22)

Hence, combining (2.17), (2.20) and (2.21) we obtain

| f (x)| ≤B

(1 + |x|)k (2.23)

for all |x| ≤ 1 and also combining (2.19), (2.20) and (2.22) obtain the inequality (2.23) for all |x| ≥ 1. Since the

improper integral+∞∫−∞

B(1+|x|)k dx is convergent for k ≥ 2, then we may write

+∞∫−∞

|l(x)| dx =

+∞∫−∞

∣∣∣∣e− i2 x2 cotα

∣∣∣∣ | f (x)| dx =

+∞∫−∞

| f (x)| dx < ∞

and so l ∈ L1 (R). Since l = F −αh and l ∈ L1 (R), then we obtain F αl = h. This means that l has compactly supportedfractional Fourier transform.Now, let w be a weight function of regular growth on R. Firstly, let K = { x ∈R| |x| ≤ 1}. Since the function w is locallybounded, then there exists T > 0 such that |w (x)| ≤ T for all x ∈ K. Therefore we may write∫

K

|l(x)|w (x) dx ≤ T∫K

|l(x)| dx < ∞. (2.24)

Also, let L = { x ∈R| |x| ≥ 1}. Thus by using (1.1) and (2.23), there exist some constants B, C1 and λ > 0 such that∫L

|l(x)|w (x) dx ≤ BC1

∫L

|x|λ

(1 + |x|)k dx ≤ BC1

∫L

(1 + |x|)λ

(1 + |x|)k dx.

If the number k is taken as k ≥ [λ] + 3, we get∫L

|l(x)|w (x) dx ≤ BC1

∫L

1(1 + |x|)k−λ dx < ∞. (2.25)

By (2.24) and (2.25) we obtain

+∞∫−∞

|l(x)|w (x) dx =

∫K

|l(x)|w (x) dx +

∫L

|l(x)|w (x) dx < ∞. (2.26)

Finally, since l ∈ L1 (R) and (2.26) holds, then l ∈ L1w (R).

Now, by using the function l which is given in Example 2.10, we will show that there exists a function r withcompactly supported fractional Fourier transform in the space L1

w

(Rd

), where w is a weight function of regular growth

on Rd.

11


Example 2.11. Firstly, let us take the funcion g that is denoted in Example 2.10. Let r be a function on Rd as

r (x1, ..., xd) = l(x1)l(x2) · · · l(xd) (2.27)

such that

l(x j

)=

√1 + i cotα j

2πe−

i2 x2

j cotα j

+∞∫−∞

g(t j

)eix jt jcscα j dt

j, (2.28)

where α j , kπ, k ∈ Z with j = 1, 2, ..., d for all x = (x1, ..., xd) ∈ Rd. By using the same method in Example 2.10,it is easy to see that l

(. j)∈ L1 (R) for all j = 1, 2, ..., d. Then by using (2.27) and Example 1.1.12 in [16], we have

r ∈ L1(Rd

). Also, let w be a weight function of regular growth on Rd. If we take the maximum norm that is given by

‖x‖∞ = max {|x1| , |x2| , ..., |xd |}

for all x ∈ Rd instead of Euclid norm on Rd, then the inequality (1.1) holds for this maximum norm. Let K =

{ x ∈R| ‖ x‖∞ ≤ 1}. Since the function w is locally bounded, then there exists T > 0 such that |w (x)| ≤ T for all x ∈ K.Therefore by using l

(. j)∈ L1 (R), we may write∫

K

|r (x)|w (x) dx ≤ T∫K

|r (x)| dx (2.29)

≤ T

+∞∫−∞

|l(x1)| dx1

+∞∫−∞

|l(x2)| dx2 · · ·

+∞∫−∞

|l (xd)| dxd < ∞.

Let ‖x‖∞ =∣∣∣x j0

∣∣∣ for all x ∈ Rd. Also, let us take L = { x ∈R| ‖ x‖∞ ≥ 1}. Then by using (1.1), there exist someconstants C1 and λ > 0 such that∫

L

|r (x)|w (x) dx ≤ C1

∫L

|r (x)| ‖x‖λ∞ dx (2.30)

≤ C1

+∞∫−∞

|l(x1)| dx1 · · ·

+∞∫−∞

∣∣∣l(x j0 )∣∣∣ ∣∣∣x j0

∣∣∣λdx j0 · · ·

+∞∫−∞

|l (xd)| dxd.

Also similar to Example 2.10, let us take a function f such that

f(x j

)=

√1 + i cotα j

2π

+∞∫−∞

g(t j

)eix jt jcscα j dt

j,

where α j , kπ, k ∈ Z with j = 1, 2, ..., d for all x j ∈ R. Thus by using (2.28) we write

l(x j

)= e−

i2 x2

j cotα j f(x j

)(2.31)

for all x j ∈ R. Let M j =

√1+i cotα j

2π . By using the same method in Example 2.10, we get∣∣∣∣ f (x j

)∣∣∣∣ ≤ B j(1 +

∣∣∣x j

∣∣∣)k , (2.32)

where

B j = max

2k∣∣∣M j

∣∣∣ +∞∫−∞

∣∣∣∣g (t j

)∣∣∣∣ dt j, 2k∣∣∣M j

∣∣∣ ∣∣∣cscα j

∣∣∣−k+∞∫−∞

∣∣∣∣g(k)(t j

)∣∣∣∣ dt j

12


for all j = 1, 2, ..., d. Letd∏

j=1

(+∞∫−∞

∣∣∣∣l (x j

)∣∣∣∣ dx j

)= C2 for j , j0. Hence by using (2.30) and (2.32), we get

∫L

|r (x)|w (x) dx ≤ C1C2

+∞∫−∞

∣∣∣l(x j0 )∣∣∣ ∣∣∣x j0

∣∣∣λdx j0

≤ B j0C1C2

+∞∫−∞

(1 +

∣∣∣x j0

∣∣∣)λ(1 +

∣∣∣x j0

∣∣∣)k dx j0 .

If the number k is taken as k ≥ [λ] + 3, then we have∫L

|r (x)|w (x) dx ≤ B j0C1C2

+∞∫−∞

1(1 +

∣∣∣x j0

∣∣∣)k−λ dx j0 < ∞. (2.33)

Therefore combining (2.29) and (2.33), we obtain∫Rd

|r(x)|w (x) dx =

∫K

|r(x)|w (x) dx +

∫L

|r(x)|w (x) dx < ∞. (2.34)

Since r ∈ L1(Rd

)and (2.34) holds, then r ∈ L1

w

(Rd

). Furthermore, by using (2.27) we may write

F αr (u1, ..., ud) = F α1 l (u1)F α2 l (u2) · · · F αd l (ud) . (2.35)

for all u = (u1, ..., ud) ∈ Rd. Let us take a function h such that

h(t j

)= e−

i2 x2

j cotα j g(t j

),

where α j , kπ, k ∈ Z with j = 1, 2, ..., d for all x j ∈ R. Then by using (2.28), we have F −α j h = l for all j = 1, 2, ..., d.Since l

(. j)∈ L1

(Rd

)for all j = 1, 2, ..., d, then F α j l = h ∈ Cc (R). Hence by using (2.35) we obtain F αr ∈ Cc

(Rd

).

Now, we will set a sequence (pn)n∈N that has integral equal to 1 and compactly supported fractional Fouriertransforms in the space L1

w

(Rd

), where w be a weight function of regular growth on Rd.

Example 2.12. Firstly, let us take the functions l and f are denoted in Example 2.11. Let us define a sequence (pn)n∈Nby

pn (t1, ..., td) = kn (t1) kn (t2) · · · kn (td) (2.36)

such thatkn

(t j

)=

n

A jn

e−i2 t2

j cotα j f(nt j

),

where

A jn =

+∞∫−∞

ne−i2 t2

j cotα j f(nt j

)dt j , 0

for all n ∈ N, t = (t1, ..., td) ∈ Rd and α j , kπ, k ∈ Z with j = 1, 2, ..., d. Let us take functions

gn

(t j

)= ne−

i2 t2

j cotα j f(nt j

),

where α j , kπ, k ∈ Z with j = 1, 2, ..., d for all n ∈ N. Then we write kn

(t j

)=

gn(t j)A j

nfor all n ∈ N and j = 1, 2, ..., d.

Also by using continuity of the function l and (2.31), we say that the function f is continuous. Then functions gn

(. j)

are continuous and+∞∫−∞

∣∣∣∣gn

(t j

)∣∣∣∣dt j = n

+∞∫−∞

∣∣∣∣ f (nt j

)∣∣∣∣dt j = ‖ f ‖1 < ∞

13


for all n ∈ N and j = 1, 2, ..., d. Thus gn

(. j)∈ L1 (R) and then kn

(. j)∈ L1 (R) for all n ∈ N and j = 1, 2, ..., d. Hence

by using (2.36) and Example 1.1.12 in [16], we have (pn)n∈N ⊂ L1(Rd

)and

∫Rd

pn (t) dt = 1 for all n ∈ N. Also, using

the same method in the proof of Proposition 2.9, there exists C > 0 such that∣∣∣∣∣∣ 1A1

n · · · Adn

∣∣∣∣∣∣ ≤ C (2.37)

for all n ∈ N. Let w be a weight function of regular growth on Rd. First of all, let K ={

x ∈Rd∣∣∣ ‖x‖∞ ≤ 1

}. Since the

function w is locally bounded, then there exists T > 0 such that |w (x)| ≤ T for all x ∈ K. Then, by using (2.37) andsubstitution z j = nt j for all j = 1, 2, ..., d, we may write∫

K

|pn (t)|w (t) dt ≤ T

∣∣∣∣∣∣ nd

A1n · · · A

dn

∣∣∣∣∣∣∫Rd

|kn(t1)| |kn(t2)| · · · |kn (td)| dt

≤ TCnd

+∞∫−∞

| f (nt1)| dt1 · · ·

+∞∫−∞

| f (ntd)| dtd (2.38)

= TC

+∞∫−∞

| f (z1)| dz1 · · ·

+∞∫−∞

| f (zd)| dzd

= TC ‖ f ‖d1 < ∞.

Let ‖x‖∞ =∣∣∣x j0

∣∣∣ for all x ∈ Rd. Also, let us take L = { x ∈R| ‖x‖∞ ≥ 1}. Then by using (1.1) and (2.37), there existsome constants C1 and λ > 0 such that∫

L

|pn (t)|w (t) dt ≤ C1

∫L

|pn (t)| ‖t‖λ∞ dt

≤ C1

∫Rd

|pn (t)|∣∣∣t j0

∣∣∣λdt (2.39)

≤ C1Cnd

+∞∫−∞

| f (nt1)| dt1 · · ·

+∞∫−∞

∣∣∣∣ f (nt j0

)∣∣∣∣ ∣∣∣t j0

∣∣∣λdt j0 · · ·

+∞∫−∞

| f (ntd)| dtd

for all n ∈ N. By substitution z j = nt j for all j = 1, 2, ..., d, we get∫L

|pn (t)|w (t) dt ≤ CC1

+∞∫−∞

| f (z1)| dz1 · · ·

+∞∫−∞

∣∣∣∣ f (z j0

)∣∣∣∣ ∣∣∣∣∣ z j0

n

∣∣∣∣∣λdz j0 · · ·

+∞∫−∞

| f (zd)| dzd

≤ CC1

+∞∫−∞

| f (z1)| dz1 · · ·

+∞∫−∞

∣∣∣∣ f (z j0

)∣∣∣∣ ∣∣∣z j0

∣∣∣λdz j0 · · ·

+∞∫−∞

| f (zd)| dzd

for all n ∈ N. Letd∏

j=1

(+∞∫−∞

∣∣∣∣ f (z j

)∣∣∣∣ dz j

)= C2 for j , j0. Thus, combining (2.39) and (2.32), we obtain

∫L

|pn (t)|w (t) dt ≤ CC1C2

+∞∫−∞

∣∣∣∣ f (z j0

)∣∣∣∣ ∣∣∣z j0

∣∣∣λdz j0 (2.40)

≤ CC1C2B j0

+∞∫−∞

(1 +

∣∣∣z j0

∣∣∣)λ(1 +

∣∣∣z j0

∣∣∣)k dz j0

14


for all n ∈ N. If the number k is taken as k ≥ [λ] + 3 and

C3 = CC1C2B j0

+∞∫−∞

1(1 +

∣∣∣z j0

∣∣∣)k−λ dz j0 ,

then we have ∫L

|pn (t)|w (t) dt ≤ C3 < ∞ (2.41)

for all n ∈ N. By combining (2.38) and (2.41) we obtain∫Rd

|pn(x)|w (x) dx =

∫K

|pn(x)|w (x) dx +

∫L

|pn(x)|w (x) dx < ∞. (2.42)

for all n ∈ N. Since (pn)n∈N ⊂ L1(Rd

)and (2.42) holds, then (pn)n∈N ⊂ L1

w

(Rd

). Let C4 = TC ‖ f ‖d1 + C3. Then,

combining (2.38) and (2.41), we have

‖pn‖1,w =

∫Rd

|pn(x)|w (x) dx ≤ C4

for all n ∈ N. This means that the sequence (pn)n∈N is bounded in the space L1w

(Rd

). Besides by using (2.36), we

haveF αpn (u1, ..., ud) = F α1 kn (u1)F α2 kn (u2) · · · F αd kn (ud) (2.43)

for all u = (u1, ..., ud) ∈ Rd and n ∈ N. Also we write

F α j kn(u j) =n

A jn

N jei2 u2

j cotα j

+∞∫−∞

e−iu jt j cscα j f(nt j

)dt j,

where N j =

√1−i cotα j

2π for all n ∈ N and j = 1, 2, ..., d. By substitution nt j = y j, we have

F α j kn(u j) =N j

A jn

ei2 u2

j cotα j

+∞∫−∞

e−iu jn y j cscα j f

(y j

)dy j, (2.44)

where N j =

√1−i cotα j

2π for all n ∈ N and j = 1, 2, ..., d. Let us take the functions h and g are denoted in Example 2.11.Then F α j l = h for all j = 1, 2, ..., d. By using the definition of fractional Fourier transform, we get

g(u j

)= N j

+∞∫−∞

f(t j

)e−iu jt jcscα j

dt j,

where N j =

√1−i cotα j

2π for all j = 1, 2, ..., d. Therefore by using (2.44), we write

F α j kn(u j) =1

A jn

ei2 u2

j cotα j g(u j

n

)for all n ∈ N and j = 1, 2, ..., d. Since the function g ∈ Cc (R), then F α j kn ∈ Cc (R) for all n ∈ N and j = 1, 2, ..., d.Thus, by using (2.43), we obtain F αpn ∈ Cc

(Rd

)for all n ∈ N.

15


Proposition 2.13. Let w be a weight function of regular growth on Rd. Then the sequence (pn)n∈N which is givenin Example 2.12 is a bounded approximate identity with compactly supported fractional Fourier transforms in thealgebra L1

w(Rd) under Θ convolution.

Proof. Let us take the sequence (pn)n∈N that is given in Example 2.12. Also, let g ∈ L1w

(Rd

)and γ = (−y1 cotα1, ...,−yd cotαd)

for all y = (y1, ..., yd) ∈ Rd. Then we may write

(pnΘg) (x) − g (x) =

∫Rd

pn (y) TyMγg (x) dy − g (x)

=

∫Rd

pn (y)(TyMγg (x) − g (x)

)dy

for all x ∈ Rd and n ∈ N. Thus we get

‖pnΘg − g‖1,w ≤

∫Rd

∫Rd

|pn (y)|∣∣∣TyMγg (x) − g (x)

∣∣∣ dyw (x) dx

=

∫Rd

|pn (y)|∥∥∥TyMγg − g

∥∥∥1,wdy

=

∫Rd

∣∣∣∣∣∣ nd

A1n · · · A

dn

∣∣∣∣∣∣ | f (ny1) · · · f (nyd)|∥∥∥TyMγg − g

∥∥∥1,wdy

for all n ∈ N. By substitution ny j = z j for j = 1, 2, ..., d, we obtain

‖pnΘg − g‖1,w ≤1∣∣∣A1

n

∣∣∣ · · · ∣∣∣Adn

∣∣∣∫Rd

| f (z1) · · · f (zd)|∥∥∥T z

nM τ

ng − g

∥∥∥1,w

dz, (2.45)

where z = (z1, ..., zd) ∈ Rd and τ = (−z1 cotα1, ...,−zd cotαd). Also, using the same method in the proof of theProposition 2.9, there exists C > 0 such that ∣∣∣∣∣∣ 1

A1n · · · A

dn

∣∣∣∣∣∣ ≤ C (2.46)

for all n ∈ N. Moreover, let g ∈ L1w

(Rd

)and γ = (−y1 cotα1, ...,−yd cotαd) for all y = (y1, ..., yd) ∈ Rd. Then the

mapping y→ TyMγg from Rd into L1w

(Rd

)is continuous by Theorem 2.1. Therefore, we get∥∥∥T z

nM τ

ng − g

∥∥∥1,w→ 0.

Let s (z) = f (z1) · · · f (zd) for all z = (z1, ..., zd) ∈ Rd. Hence, it is known that s ∈ L1w

(Rd

)by Example 2.11. Thus, we

have|s (z)|

∥∥∥T znM τ

ng − g

∥∥∥1,w→ 0.

Since w is a weight function of regular growth on Rd, then we write∥∥∥T znM τ

ng − g

∥∥∥1,w

≤ w( zn

)‖g‖1,w + ‖g‖1,w

≤ w( zn

)‖g‖1,w + w

( zn

)‖g‖1,w

≤ 2‖g‖1,ww (z) .

Let C1 = 2‖g‖1,w. Thus we get|s (z)|

∥∥∥T znM τ

ng − g

∥∥∥1,w≤ C1 |s (z)|w (z) .

16


Since sw ∈ L1(Rd

), then Dominated Convergence Theorem implies that∫

Rd

|s (z)|∥∥∥T z

nM τ

ng − g

∥∥∥1,w

dz→ 0. (2.47)

Consequently, combining (2.45), (2.46) and (2.47), we obtain

‖pnΘg − g‖1,w ≤1∣∣∣A1

n

∣∣∣ · · · ∣∣∣Adn

∣∣∣∫Rd

|s (z)|∥∥∥T z

nM τ

ng − g

∥∥∥1,w

dz→ 0.

This is the desired result.

Corollary 2.14. Let w be a weight function of regular growth on Rd. Then the set Fα0,w

(Rd

)=

{f ∈ L1

w

(Rd

)∣∣∣∣ F α f ∈ Cc

(Rd

)}is dense in L1

w

(Rd

).

Proof. Let w be a weight function of regular growth on Rd. Also, let us take the sequence (pn)n∈N is given in Example2.12 and g ∈ L1

w

(Rd

). It is known that the sequence (pn)n∈N is an approximate identity with compactly supported

fractional Fourier transforms by Proposition 2.13. Let ε > 0 be given. Then there exists n0 ∈ N such that∥∥∥(pn0Θg)− g

∥∥∥1,w < ε (2.48)

for all n ≥ n0. Therefore we may write

F α(pn0Θg

)(u) = Me

d∑j=1− i

2 u2j cotα j

Fαpn0 (u) Fαg (u) , (2.49)

where M =

[d∏

j=1

√2π

1−i cotα j

]for all u ∈ Rd by Theorem 7 in [20]. Since F αpn0 ∈ Cc

(Rd

), then by using (2.49), we

have F α(pn0Θg

)∈ Cc

(Rd

). This means the function F α

(pn0Θg

)belongs to Fα

0,w

(Rd

). Hence, the set

Fα0,w

(Rd

)=

{f ∈ L1

w

(Rd

)∣∣∣∣ F α f ∈ Cc

(Rd

)}is dense in L1

w

(Rd

).

Proposition 2.15. Let w be a weight function of regular growth on Rd. If Cc

(Rd

)⊂ S Θ

w(Rd), then S αw(Rd) is dense in

L1w

(Rd

).

Proof. Let w be a weight function of regular growth on Rd and Cc

(Rd

)⊂ S Θ

w(Rd). We have

Fα0,w

(Rd

)⊂ S α

w(Rd) ⊂ L1w

(Rd

)(2.50)

by the definition of the space S αw(Rd). Also, it is known that the set

Fα0,w

(Rd

)=

{f ∈ L1

w

(Rd

)∣∣∣∣ F α f ∈ Cc

(Rd

)}is dense in L1

w

(Rd

)by Corollary 2.14. Let ε > 0 be given. Then there exists a function h ∈ Fα

0,w

(Rd

)such that

‖g − h‖1,w < ε

for all g ∈ L1w

(Rd

). Therefore by using (2.50), the function h also belongs to S α

w(Rd). This is the desired result.

17


Proposition 2.16. Let w be a weight function of regular growth on Rd. If S Θw(Rd) is a solid space and Cc

(Rd

)⊂

S Θw(Rd), then S α

w(Rd) is an abstract Segal algebra with respect to L1w

(Rd

).

Proof. Let w be a weight function of regular growth on Rd. Then, it is known that S αw(Rd) is a Banach algebra and is

a Banach ideal on L1w

(Rd

), also inequalities ‖g‖1,w ≤ ‖g‖S α

wand ‖gΘh‖S α

w≤ ‖g‖S α

w‖h‖1,w holds for all g, h ∈ S α

w(Rd) by

[21]. Furthermore, S αw(Rd) is dense in L1

w

(Rd

)by Proposition 2.15. Hence, S α

w(Rd) is an abstract Segal algebra with

respect to L1w

(Rd

).

Theorem 2.17. Let S Θw(Rd) be a solid space and γ = (−y1 cotα1, ...,−yd cotαd) for all y = (y1, ..., yd) ∈ Rd.

1. TyMγg ∈ S αw(Rd) and ∥∥∥TyMγg

∥∥∥S α

w≤ w (y) ‖g‖S α

w(2.51)

for all g ∈ S αw(Rd).

2. Let Cc(Rd) ∩ S Θw(Rd) is dense in S Θ

w(Rd). Then the mapping y→ TyMγg from Rd into S αw(Rd) is continuous.

Proof. 1. Let g ∈ S αw(Rd). Then g ∈ L1

w

(Rd

)and F αg ∈ S Θ

w(Rd). It is easy to see that∥∥∥Mγg

∥∥∥1,w = ‖g‖1,w

and Mγg ∈ L1w(Rd). Also it is well known that the space L1

w(Rd) is translation invariant and holds∥∥∥Tyg

∥∥∥1,w ≤

w (y) ‖g‖1,w for all y ∈ Rd. Hence we have ∥∥∥TyMγg∥∥∥

1,w ≤ w (y) ‖g‖1,w. (2.52)

By using Proposition 3 in [20], we get

F α

(TyMγg

)(u) = e

d∑j=1

i2 y2

j sinα j cosα j

ed∑

j=1−iu jy j sinα j

F α

(Mγg

)(u − b) (2.53)

and

F α

(Mγg

)(u − b) = e

d∑j=1− i

2 γ2j sinα j cosα j

ed∑

j=1i(u j−b j)γ j cosα j

F αg(u − b − c) (2.54)

= ed∑

j=1− i

2 y2j

cos3α jsinα j e

d∑j=1−iu jy j

cos2α jsinα j

+iy2j

cos3α jsinα jF αg(u)

= ed∑

j=1

i2 y2

jcos3α jsinα j e

d∑j=1−iu jy j

cos2α jsinα jF αg(u)

such that b = (y1 cosα1, ..., yd cosαd) and

c = (γ1 sinα1, ..., γd sinαd) = (−y1 cotα1 sinα1, ...,−yd cotαd sinα1) = −b.

Combining (2.53) and (2.54) we write

F α

(TyMγg

)(u) = e

d∑j=1

i2 y2

j cosα j

(sinα j+

cos2α jsinα j

)e

d∑j=1−iu jy j

(sinα j+

cos2α jsinα j

)F αg(u)

= ed∑

j=1

i2 y2

j cotα j

ed∑

j=1−iu jy j cscα j

F αg(u).

Let τ = (−y1 cscα1, ...,−yd cscαd). Therefore we have

F α

(TyMγg

)(u) = e

d∑j=1

i2 y2

j cotα j

MτF αg(u). (2.55)

18


Since S Θw(Rd) is a solid space, then it is strongly character invariant by Lemma 2.4 in [7]. Thus we get

ed∑

j=1

i2 y2

j cotα j

MτF αg ∈ S Θw(Rd).

By using (2.55), we obtain MτFα f ∈ S Θw(Rd) and∥∥∥∥F α

(TyMγg

)∥∥∥∥S Θ

w=

∥∥∥∥∥∥∥∥ed∑

j=1

i2 y2

j cotα j

MτF αg

∥∥∥∥∥∥∥∥S Θ

w

=

∣∣∣∣∣∣∣∣ed∑

j=1

i2 y2

j cotα j

∣∣∣∣∣∣∣∣ ‖MτF αg‖S Θw

(2.56)

= ‖MτF αg‖S Θw

= ‖F αg‖S Θw.

Finally, by (2.52) and (2.56), we have ∥∥∥TyMγg∥∥∥

1,w ≤ w (y) ‖g‖1,w.

2. First of all we will show continuity at 0. It is easy to see that mapping y → γ from Rd into Rd is continuous.Also mapping y→ Myg is from Rd into S α

w(Rd) continuous for all g ∈ S αw(Rd) by Theorem 2.12 in [21]. Hence

the composition mapping y→ Mγg from Rd into S αw(Rd) is continuous. Let ε > 0 be given. There exists δ1 > 0

such that ∥∥∥Mγg − g∥∥∥

S αw<ε

2(2.57)

when ‖y‖ < δ1 and there exist δ2 > 0 such that∥∥∥∥Ty

(Mγg

)− Mγg

∥∥∥∥S α

w<ε

2(2.58)

when ‖y‖ < δ2 by Theorem 2.10 in [21]. Let δ3 = min {δ1, δ2}. Thus, combining (2.57) and (2.58) we get∥∥∥∥Ty

(Mγg

)− g

∥∥∥∥S α

w=

∥∥∥∥Ty

(Mγg

)− Mγg + Mγg − g

∥∥∥∥S α

w

≤

∥∥∥∥Ty

(Mγg

)− Mγg

∥∥∥∥S α

w+

∥∥∥Mγg − g∥∥∥

S αw

<ε

2+ε

2= ε

when ‖y‖ < δ3 and this proves that the mapping y→ TyMγg is continuous at 0. Now, let us take any fixed pointy∗ =

(y∗1, ..., y

∗d

)∈ Rd. Therefore, we have

Ty−y∗Mγ−γ∗

(Ty∗Mγ∗g

)(x) = Mγ−γ∗

(Ty∗Mγ∗g

)(x − y + y∗)

= e(i(γ−γ∗)(x−y+y∗))(Ty∗Mγ∗g

)(x − y + y∗)

= e(i(γ−γ∗)(x−y+y∗))e(iγ∗(x−y))g(x − y)= e(iy∗γ−iy∗γ∗)e(iγ(x−y))g(x − y)= e(iy∗γ−iy∗γ∗)TyMγg(x)

such that γ∗ =(−y∗1 cotα1, ...,−y∗d cotαd

). Then we write∥∥∥TyMγg − Ty∗Mγ∗g

∥∥∥S α

w=

∥∥∥∥e(iy∗γ∗−iy∗γ)Ty−y∗Mγ−γ∗

(Ty∗Mγ∗g

)− Ty∗Mγ∗g

∥∥∥∥S α

w.

By the first part of this theorem, let us take Ty∗Mγ∗g = h ∈ S αw(Rd). Hence, we obtain∥∥∥TyMγg − Ty∗Mγ∗g

∥∥∥S α

w=

∥∥∥e(iy∗γ∗−iy∗γ)Ty−y∗Mγ−γ∗h − h∥∥∥

S αw

(2.59)

≤∥∥∥e(iy∗γ∗−iy∗γ)Ty−y∗Mγ−γ∗h − e(iy∗γ∗−iy∗γ)h

∥∥∥S α

w

+∥∥∥e(iy∗γ∗−iy∗γ)h − h

∥∥∥S α

w

=∥∥∥Ty−y∗Mγ−γ∗h − h

∥∥∥S α

w+ ‖h‖S α

w

∣∣∣eiy∗γ−eiy∗γ∗∣∣∣ .

19


Moreover, since the point y∗ be an arbitrary fixed point, then mapping y → eiy∗y from Rd into C is obviouslycontinuous and also it is easy to see that mapping y → γ from Rd into Rd is continuous. Hence, compositionmapping y→ eiy∗γ is also continuous. Let ε > 0 be given. There exists δ4 > 0 such that∣∣∣eiy∗γ − eiy∗γ∗

∣∣∣ < ε

2‖h‖S αw

(2.60)

when ‖y − y∗‖ < δ4 and also, by continuity at 0, there exists δ5 > 0 such that∥∥∥Ty−y∗Mγ−γ∗h − h∥∥∥

S αw<ε

2(2.61)

when ‖y − y∗‖ < δ5. Let δ6 = min {δ4, δ5}. Thus, combining (2.59), (2.60) and (2.61) we obtain∥∥∥TyMγg − Ty∗Mγ∗g∥∥∥

S αw≤

∥∥∥Ty−y∗Mγ−γ∗h − h∥∥∥

S αw

+ ‖h‖S αw

∣∣∣eiy∗γ − eiy∗γ∗∣∣∣

<ε

2+ε‖h‖S α

w

2‖h‖S αw

= ε

when ‖y − y∗‖ < δ6. This completes the proof.

Proposition 2.18. Let w be a weight function of regular growth on Rd. If S Θw(Rd) is a solid space, Cc

(Rd

)⊂ S Θ

w(Rd)

and Cc

(Rd

)is dense in S Θ

w(Rd), then S αw(Rd) has an approximate identity with compactly supported fractional Fourier

transforms.

Proof. Let w be a weight function of regular growth on Rd and also S be a finite subset of S αw(Rd) such that S =

{g1, ..., gn}. Let g ∈ S αw(Rd) and γ = (−y1 cotα1, ...,−yd cotαd) for all y = (y1, ..., yd) ∈ Rd. Since S Θ

w(Rd) is a solidspace, Cc

(Rd

)⊂ S Θ

w(Rd) and Cc

(Rd

)is dense in S Θ

w(Rd), then the mapping y → TyMγg from Rd into S αw(Rd) is

continuous by Theorem 2.17. Let ε > 0 be given. There exist δi > 0 such that∥∥∥TyMγgi − gi

∥∥∥S α

w<ε

2

when ‖y‖ < δi for all i = 1, 2, ..., n. Let δ = min {δ1, δ2,..., δn}. Therefore we write∥∥∥TyMγgi − gi

∥∥∥S α

w<ε

2(2.62)

when ‖y‖ < δ for all i = 1, 2, ..., n. Let us take a positive function h ∈ Cc

(Rd

)such that supp h ⊂ B (0, δ) and∫

Rd

h (x)dx = 1. Thus we have

(hΘgi) (x) − gi(x) =

∫Rd

h(y)TyMγgi(x)dy − gi(x) =

∫Rd

h(y)(TyMγgi(x) − gi(x)

)dy

for all x ∈ Rd and i = 1, ..., n. Then by using (2.62), we get

‖(hΘgi) − gi‖S αw

=

∥∥∥∥∥∥∥∥∥∫Rd

h(y)(TyMγgi − gi

)dy

∥∥∥∥∥∥∥∥∥S α

w

≤

∫supp h

|h(y)|∥∥∥TyMγgi − gi

∥∥∥S α

wdy (2.63)

<ε

2

∫supp h

h(y)dy =ε

2

20


for all i = 1, ..., n. Let C = max{‖g1‖S α

w, ..., ‖gn‖S α

w

}. There exists a function f ∈ Fα

0,w

(Rd

)such that

‖h − f ‖1,w <ε

2C(2.64)

by Corollary 1. Since Cc

(Rd

)⊂ S Θ

w(Rd), then f ∈ S αw(Rd). Therefore combining (2.63) and (2.64), we obtain

‖( f Θgi) − gi‖S αw≤ ‖( f Θgi) − (hΘgi)‖S α

w+ ‖(hΘgi) − gi‖S α

w

= ‖h − f ‖1,w‖gi‖S αw

+ ‖(hΘgi) − gi‖S αw

≤ ‖h − f ‖1,wC + ‖(hΘgi) − gi‖S αw

<ε

2CC +

ε

2= ε

for all i = 1, ..., n. Hence S αw(Rd) has an approximate identity with compactly supported fractional Fourier transforms

by 1.3. Proposition in [8].

Since the algebra S αw(Rd) which satisfies the conditions in Proposition 2.18 has an approximate identity, then it is

an algebra without order. Now, we will give definition of multipliers for the Banach algebra S αw(Rd).

Definition 2.19. Let w be a weight function of regular growth on Rd. Let S Θw(Rd) be a solid space and let Cc

(Rd

)be

dense in S Θw(Rd) such that Cc

(Rd

)⊂ S Θ

w(Rd). A multiplier T is a continuous linear operator from S αw(Rd) into S α

w(Rd)which commutes with operator TyMγ, where γ = (−y1 cotα1, ...,−yd cotαd) for all y = (y1, ..., yd) ∈ Rd. The set ofmultipliers on S α

w(Rd) is denoted by M(S α

w(Rd)).

Theorem 2.20. Let w be a weight function of regular growth on Rd. Let S Θw(Rd) be a solid space and let Cc

(Rd

)be

dense in S Θw(Rd), where Cc

(Rd

)⊂ S Θ

w(Rd). Suppose T is an operator from S αw(Rd) into S α

w(Rd). Then T ∈ M(S α

w(Rd))

if and only ifT ( f Θg) = T f Θg = f ΘTg (2.65)

for all f , g ∈ S αw(Rd).

Proof. Let w be a weight function of regular growth on Rd. Let S Θw(Rd) be a solid space and let Cc

(Rd

)be dense in

S Θw(Rd), where Cc

(Rd

)⊂ S Θ

w(Rd). Suppose T is an operator from S αw(Rd) into S α

w(Rd). First of all, we assume that

equality (2.65) holds for all f , g ∈ S αw(Rd). Since S α

w(Rd) is an algebra without order, then we get T ∈ M(S α

w(Rd)),

similar to the first part of the proof of Theorem 2.5.Conversely, let T ∈ M

(S α

w(Rd)). Let

(S α

w(Rd))′

be dual space of the space S αw(Rd). For ϕ ∈

(S α

w(Rd))′

, let usconsider the functional l is given by l (g) = 〈Tg, ϕ〉 for all g ∈ S α

w(Rd). Then, by using linearity of the operators T andϕ, we may write

l (ag + bh) = 〈T (ag + bh) , ϕ〉 = a 〈Tg, ϕ〉 + b 〈Th, ϕ〉 = al(g) + bl(h)

for all f , g ∈ S αw(Rd) and a, b ∈ C. Also, let ‖T‖ and ‖ϕ‖ be operator norms of T and ϕ, respectively. Thus, we have

|〈Tg, ϕ〉| ≤ ‖ϕ‖ ‖Tg‖S αw≤ ‖ϕ‖ ‖T‖ ‖g‖S α

w

for all g ∈ S αw(Rd). Hence l is a bounded linear functional on S α

w(Rd). Since(S α

w(Rd))′

is dual space of the space

S αw(Rd), then there exists a function ψ ∈

(S α

w(Rd))′

such that

〈g, ψ〉 = 〈Tg, ϕ〉 (2.66)

for all g ∈ S αw(Rd). Now, let γ = (−y1 cotα1, ...,−yd cotαd) for all y = (y1, ..., yd) ∈ Rd. Then we define a function k

byk (y) = f (y) TyMγg (2.67)

21


for f , g ∈ S αw(Rd). Also, it is known that the mapping y → TyMγg from Rd into S α

w(Rd) is continuous by Theorem2.17. Thus the function k from Rd into S α

w(Rd) is a measurable. Besides, by using (2.51), we have∫Rd

‖k (y)‖S αwdy =

∫Rd

∥∥∥ f (y)TyMγg∥∥∥

S αwdy ≤

∫Rd

| f (y)|w (y) ‖g‖S αwdy = ‖g‖S α

w‖ f ‖1,w < ∞.

Therefore we may write ⟨∫Rd

k (y)dy, ϕ⟩

=

∫Rd

〈k (y), ϕ〉 dy

by the definition of vector-valued integrals, [22]. Hence, by using (2.67) and definition of the Θ convolution, we get

〈 f Θg, ϕ〉 =

∫Rd

⟨f (y)TyMγg, ϕ

⟩dy =

∫Rd

f (y)⟨TyMγg, ϕ

⟩dy. (2.68)

Thus, by using (2.66), (2.68), we obtain

〈 f ΘTg, ϕ〉 =

∫Rd

f (y)⟨TTyMγg, ϕ

⟩dy =

∫Rd

f (y)⟨TyMγg, ψ

⟩dy = 〈 f Θg, ψ〉 = 〈T ( f Θg) , ϕ〉

for all f , g ∈ S αw(Rd). By using Hahn-Banach theorem, we obtain

T ( f Θg) = f ΘTg.

Consequently, by using commutativity of Θ convolution, the equality (2.65) holds.

References

[1] L. B. Almeida, The fractional Fourier transform and time-frequency representations, IEEE Trans. Signal Process. 42 (11), 3084–3091, 1994.[2] L. B. Almeida, Product and convolution theorems for the fractional fourier transform, IEEE Signal Process. Lett. 4 (1), 15–17, 1997.[3] A. Bourouihiya, Beurling weighted spaces, product-convolution operators, and the tensor product of frames, Ph.D. thesis, University of

Maryland,College Park, Maryland, 2006.[4] A. Bultheel and H. Martinez, A shattered survey of the fractional Fourier transform, Report TW 337, 2002.[5] J. T. Burnham, Closed ideals in subalgebras of Banach algebras. I, Proceedings of the American Mathematical Society 32 (2), 551–555,

1972.[6] J. Cigler, Normed ideals in L1(G), Indagationes Mathematicae 72 (3), 273–282, 1969.[7] M. Dogan and A. T. Gurkanlı, On functions with Fourier tansforms in Sω, Bulletin of Calcutta Mathematical Society 92 (2), 111–120, 2000.[8] R. S. Doran and J. Wichmann, Approximate identities and factorization in Banach modules, 768, Springer-Verlag, New York, 1979.[9] H. G. Feichtinger, On a class of convolution algebras of functions, Annales de l’institu Fourier 27 (3), 135–162, 1977.

[10] H. G. Feichtinger, C. Graham and E. Lakien, Nonfactorization in commutative, weakly selfadjoint Banach algebras, Pacific Journal ofMathematics 80 (1), 117–125, 1979.

[11] R. H. Fischer, A. T. Gurkanlı and T. S. Liu, On a family of weighted spaces, Math. Slovaca 46 (1), 71–82, 1996.[12] K. Grochenig, Weight functions in time-frequency analysis, In: L. Rodino, B. W. Schulze, M. W. Wong, (eds.) Pseudodifferential Operators:

Partial Differential Equations and Time-Frequency Analysis, volume 52 of Fields Inst. Commun., pp. 343–366. Amer. Math. Soc., Providence,RI, 2007.

[13] R Larsen, An introduction to the theory of multipliers, Springer-Verlag, Berlin Heidelberg, New York, 1971.[14] V. Namias, The fractional order of Fourier transform and its application in quantum mechanics, J. Inst. Math. Appl. 25, 241–265, 1980.[15] H. M. Ozaktas, M. A. Kutay and Z. Zalevsky, The fractional Fourier transform with applications in optics and signal processing, John Wiley

and Sons, Chichester, 2001.[16] H. Reiter and J. D. Stegeman, Classical harmonic analysis and locally compact group, Clarendon Press, Oxford, 2000.[17] W. Rudin, Real and complex analysis, McGraw-Hill, New York, 1966.[18] W. Rudin, Functional analysis, Mc Graw-Hill, New York, 1973.[19] A. K. Singh and R. Saxena, On convolution and product theorems for FRFT, Wireless Personal Communications 65 (1), 189–201, 2012.[20] E. Toksoy and A. Sandıkcı, On function spaces with fractional Fourier transform in weighted Lebesgue spaces, J. Inequal. Appl. 87, 2015.[21] E. Toksoy and A. Sandıkcı, On some properties of space S α

w, Erzincan Universitesi Fen Bilimleri Enstitusu Dergisi 13 (2), 923–934, 2020.[22] H. C. Wang, Homogeneous Banach algebras, Marcel Dekker Inc., New York, 1977.[23] A. I. Zayed, On the relationship between the Fourier and fractional Fourier transforms, IEEE Signal Proc. Let. 3, 310–311, 1996.

22

on an abstract segal algebra under fractional convolution

Documents