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Nonlinear operations and factorizations on a class of affine modulation spaces

  • Md Hasan Ali Biswas EMAIL logo and Ramesh Manna
Published/Copyright: November 30, 2024

Abstract

In the first part of the paper, we obtain a necessary and sufficient condition on a complex-valued function 𝐹 on R 2 such that F ( f ) M θ r , s ( R n ) for all f M θ r , 1 ( R n ) , 1 r < , 1 s < 2 , where M θ r , s ( R n ) is the affine modulation space associated with the fractional Fourier transform. In this context, we obtain that if 𝐹 acts on M θ r , 1 ( R n ) by composition, then 𝐹 is necessarily real analytic and F ( 0 ) = 0 . As M θ r , 1 ( R n ) is neither a Banach algebra nor closed under complex conjugation, we prove sufficiency of the condition for a subclass of M θ r , 1 ( R n ) under some additional hypothesis on 𝐹. We provide an example to show that we cannot drop the additional hypothesis. Thus we characterize the functions that operate in the affine modulation space setting. In the second part, we study the factorization problem in the context of M θ r , s ( R n ) . We obtain factorization of M θ r , s ( R 2 n ) , viewing it as an L 1 ( R 2 n ) module with respect to convolution and 𝜆-twisted convolution associated with the fractional Fourier transform. The problems mentioned above originated in the work of Helson, Kahane, Katznelson, Rudin, Cohen, and Hewitt respectively.

MSC 2020: 42B35; 46H25; 42B10; 42A85

Award Identifier / Grant number: DST/INSPIRE/04/2019/001914

Funding statement: Ramesh Manna is thankful for the research grants (DST/INSPIRE/04/2019/001914).

A Appendix

In Example 3.4, we outlined the key steps. Although the verification is elementary, here we provide a detailed justification for the convenience of the reader.

Example A.1

Let 1 < r < 2 and define

F ( x , ω ) = 1 | x | 1 / r ( 1 + ( ln | x | ) 2 ) 1 1 + ω 2 .

Then F L r , 1 ( R 2 ) . Let g ( x ) = e π x 2 and h = V g F . Then, using Proposition 2.13, we get that h M r , 1 ( R ) . Set f = C θ h . Then f M θ r , 1 ( R ) , appealing to Proposition 2.20. Further, using (2.4) and (2.6), we obtain

V C θ g f ( u , η ) = V C θ g C θ h ( u , η ) = e π i u 2 cot θ V g h ( u , η + u cot θ ) = e π i u 2 cot θ R 2 F ( x , ω ) V g g ( u x , η + u cot θ ω ) e 2 π i x ( η + u cot θ ω ) d x d ω = e π i u 2 cot θ R 2 F ( u x , η + u cot θ ω ) V g g ( x , ω ) e 2 π i ( u x ) ω d x d ω = e π i u 2 cot θ R 2 F ( u + x , η + u cot θ ω ) V g g ( ω , x ) e 2 π i u ω d x d ω ,

applying a change of variables. Similarly,

V g C θ f ( u , η ) = V g h ( u , η ) = R 2 F ( u + x , η ω ) V g g ( ω , x ) e 2 π i u ω d x d ω .

Thus, using Remark 2.8 (i), we obtain

V C θ g 2 C θ f 2 ( u , η ) = ( C θ f 2 T u ( C θ g 2 ̄ ) ) ̂ ( η ) = ( C θ f T u ( g ̄ ) f T u ( C θ g ̄ ) ) ̂ ( η ) = ( ( C θ f T u ( g ̄ ) ) ̂ ( f T u ( C θ g ̄ ) ) ̂ ) ( η ) = R ( C θ f T u ( g ̄ ) ) ̂ ( y ) ( f T u ( C θ g ̄ ) ) ̂ ( η y ) d y = R V g C θ f ( u , y ) V C θ g f ( u , η y ) d y .

Now, taking

F 1 ( x ) = 1 | x | 1 / r ( 1 + ( ln | x | ) 2 ) and F 2 ( ω ) = 1 1 + ω 2 ,

we obtain

V C θ g C θ f 2 ( u , η ) = R 2 R 2 V g g ( ω 1 , x 1 ) V g g ( ω 2 , x 2 ) e 2 π i u ( ω 1 + ω 2 ) F 1 ( u + x 1 ) F 1 ( u + x 2 ) × R F 2 ( η y + u cot θ ω 1 ) F 2 ( y ω 2 ) d y d x 1 d ω 1 d x 2 d ω 2 = R 2 R 2 V g g ( ω 1 , x 1 ) V g g ( ω 2 , x 2 ) e 2 π i u ( ω 1 + ω 2 ) F 1 ( u + x 1 ) F 1 ( u + x 2 ) × F 2 F 2 ( η + u cot θ ω 1 ω 2 ) d x 1 d ω 1 d x 2 d ω 2 .

Hence, for s > r , appealing to Minkowski’s integral inequality, we have

I = R ( R | V C θ g C θ f 2 ( u , η ) | s d u ) 1 / s d η = R ( R | R 2 R 2 V g g ( ω 1 , x 1 ) V g g ( ω 2 , x 2 ) e 2 π i u ( ω 1 + ω 2 ) F 1 ( u + x 1 ) F 1 ( u + x 2 ) × F 2 F 2 ( η + u cot θ ω 1 ω 2 ) d x 1 d ω 1 d x 2 d ω 2 | s d u ) 1 / s d η = R ( R | R 2 R 2 V g g ( ω 1 , x 1 ) V g g ( ω 2 , x 2 ) e 2 π i ( ( u η ) ( ω 1 + ω 2 ) tan θ ) F 1 ( ( u η ) tan θ + x 1 ) × F 1 ( ( u η ) tan θ + x 2 ) F 2 F 2 ( u ω 1 ω 2 ) d x 1 d ω 1 d x 2 d ω 2 | s d u ) 1 / s d η ( R ( R | R 2 R 2 V g g ( ω 1 , x 1 ) V g g ( ω 2 , x 2 ) e 2 π i ( ( u η ) ( ω 1 + ω 2 ) tan θ ) F 1 ( ( u η ) tan θ + x 1 ) × F 1 ( ( u η ) tan θ + x 2 ) F 2 F 2 ( u ω 1 ω 2 ) d x 1 d ω 1 d x 2 d ω 2 | d η ) s d u ) 1 / s ( R | R R 2 R 2 V g g ( ω 1 , x 1 ) V g g ( ω 2 , x 2 ) e 2 π i ( ( u η ) ( ω 1 + ω 2 ) tan θ ) F 1 ( ( u η ) tan θ + x 1 ) × F 1 ( ( u η ) tan θ + x 2 ) F 2 F 2 ( u ω 1 ω 2 ) d x 1 d ω 1 d x 2 d ω 2 d η | s d u ) 1 / s ( R ( R R 2 R 2 Re ( V g g ( ω 1 , x 1 ) V g g ( ω 2 , x 2 ) e 2 π i ( ( u η ) ( ω 1 + ω 2 ) tan θ ) ) F 1 ( ( u η ) tan θ + x 1 ) × F 1 ( ( u η ) tan θ + x 2 ) F 2 F 2 ( u ω 1 ω 2 ) d x 1 d ω 1 d x 2 d ω 2 d η ) s d u ) 1 / s = 1 2 ( R ( R R 2 R 2 e π 2 ( x 1 2 + ω 1 2 + x 2 2 + ω 2 2 ) Re ( e 2 π i ( ω 1 x 1 + ω 2 x 2 ( u η ) ( ω 1 + ω 2 ) tan θ ) ) × F 1 ( ( u η ) tan θ + x 1 ) F 1 ( ( u η ) tan θ + x 2 ) F 2 F 2 ( u ω 1 ω 2 ) d x 1 d ω 1 d x 2 d ω 2 d η ) s d u ) 1 / s = 1 2 ( R ( R R 2 R 2 e π 2 ( x 1 2 + ω 1 2 + x 2 2 + ω 2 2 ) cos ( 2 π ( ω 1 x 1 + ω 2 x 2 ( u η ) ( ω 1 + ω 2 ) tan θ ) ) × F 1 ( ( u η ) tan θ + x 1 ) F 1 ( ( u η ) tan θ + x 2 ) F 2 F 2 ( u ω 1 ω 2 ) d x 1 d ω 1 d x 2 d ω 2 d η ) s d u ) 1 / s 1 2 ( R ( R 2 R 2 R e π 2 ( x 1 2 + ω 1 2 + x 2 2 + ω 2 2 ) cos ( 2 π ( ω 1 x 1 + ω 2 x 2 η ( ω 1 + ω 2 ) ) ) F 1 ( η + x 1 ) × F 1 ( η + x 2 ) F 2 F 2 ( u ω 1 ω 2 ) d η d x 1 d ω 1 d x 2 d ω 2 ) s d u ) 1 / s = 1 2 ( R ( R 2 R 2 R e π 2 ( x 1 2 + ω 1 2 + x 2 2 + ω 2 2 ) cos ( 2 π ( ω 1 x 1 + ω 2 x 2 ( η x 2 ) ( ω 1 + ω 2 ) ) ) × F 1 ( η + x 1 x 2 ) F 1 ( η ) F 2 F 2 ( u ω 1 ω 2 ) d η d x 1 d ω 1 d x 2 d ω 2 ) s d u ) 1 / s ,

using | x + i y | x and applying a change of variables. Our aim is to show that the integral 𝐼 is not finite.

In order to prove the claim, let

X = { ( x 1 , ω 1 , x 2 , ω 2 ) R 4 : 0 ω 1 ( x 1 + x 2 ) + 2 ω 2 x 2 < 1 3 , ω 1 + ω 2 > 0 ,  0 x 2 x 1 1 } .

and for fixed N N , ( x 1 , ω 1 , x 2 , ω 2 ) X , set

Y N = { η R : | η ( ω 1 + ω 2 ) ω 1 ( x 1 + x 2 ) 2 ω 2 x 2 | < 1 3 , x 1 x 2 < 1 N } .

Then, for any ( x 1 , ω 1 , x 2 , ω 2 ) X and η Y N , we have

1 ω 1 + ω 2 ( 1 3 + ω 1 ( x 1 + x 2 ) + 2 ω 2 x 2 ) < η < 1 ω 1 + ω 2 ( 1 3 + ω 1 ( x 1 + x 2 ) + 2 ω 2 x 2 )

and

1 ω 1 + ω 2 ( 1 3 + ω 1 ( x 1 + x 2 ) + 2 ω 2 x 2 ) < 0 .

Taking η 0 = 1 ω 1 + ω 2 ( 1 / 3 + ω 1 ( x 1 + x 2 ) + 2 ω 2 x 2 ) , we obtain

Y N F 1 ( η + x 1 x 2 ) F 1 ( η ) cos ( 2 π ( ω 1 x 1 + ω 2 x 2 ( η x 2 ) ( ω 1 + ω 2 ) ) ) d η 1 2 Y N F 1 ( η + x 1 x 2 ) F 1 ( η ) d η 1 2 0 η 0 F 1 ( η + x 1 x 2 ) F 1 ( η ) d η = 1 2 0 η 0 1 ( η + x 1 x 2 ) 1 / r ( 1 + ( ln ( η + x 1 x 2 ) ) 2 ) 1 η 1 / r ( 1 + ( ln η ) 2 ) d η 1 2 0 η 0 1 ( η + x 1 x 2 ) 2 / r ( 1 + ( ln ( η + x 1 x 2 ) ) 2 ) 2 d η = 1 2 ln ( x 1 x 2 ) ln ( η 0 + x 1 x 2 ) e ( 1 2 / r ) y ( 1 + y 2 ) 2 d y ,

applying a change of variables. Since ( 1 2 / r ) < 0 , the integral

Y N F 1 ( η + x 1 x 2 ) F 1 ( η ) cos ( 2 π ( ω 1 x 1 + ω 2 x 2 ( η x 2 ) ( ω 1 + ω 2 ) ) ) d η

diverges to infinity as N .

Now, for any δ > 0 , we compute

I δ = ( R ( R 2 R 2 | δ | > 0 e π 2 ( x 1 2 + x 2 2 + ω 1 2 + ω 2 2 ) cos ( 2 π ( ω 1 x 1 + ω 2 x 2 ( η x 2 ) ( ω 1 + ω 2 ) ) ) × F 1 ( η + x 1 x 2 ) F 1 ( η ) F 2 F 2 ( u ω 1 ω 2 ) d η d x 1 d ω 1 d x 2 d ω 2 ) s d u ) 1 / s ( R ( R 2 R 2 | δ | > 0 e π 2 ( x 1 2 + x 2 2 + ω 1 2 + ω 2 2 ) F 1 ( η + x 1 x 2 ) F 1 ( η ) × F 2 F 2 ( u ω 1 ω 2 ) d η d x 1 d ω 1 d x 2 d ω 2 ) s d u ) 1 / s R 2 R 2 | δ | > 0 e π 2 ( x 1 2 + x 2 2 + ω 1 2 + ω 2 2 ) F 1 ( η + x 1 x 2 ) F 1 ( η ) × ( R ( F 2 F 2 ( u ω 1 ω 2 ) ) s d u ) 1 / s d η d x 1 d ω 1 d x 2 d ω 2 R 2 R 2 | δ | > 0 e π 2 ( x 1 2 + x 2 2 + ω 1 2 + ω 2 2 ) F 1 ( η + x 1 x 2 ) F 1 ( η ) d η d x 1 d ω 1 d x 2 d ω 2 ,

using Minkowski’s integral inequality. For x 1 , x 2 R , we get

| δ | > 0 F 1 ( η + x 1 x 2 ) F 1 ( η ) d η = δ F 1 ( η + x 1 x 2 ) F 1 ( η ) d η + δ F 1 ( η + x 1 x 2 ) F 1 ( η ) d η = x 2 x 1 F 1 ( η + x 1 x 2 ) F 1 ( η ) d η + x 2 x 1 δ F 1 ( η + x 1 x 2 ) F 1 ( η ) d η + δ F 1 ( η + x 1 x 2 ) F 1 ( η ) d η .

We note that the second integral in the above line is zero if δ > x 1 x 2 . Now, we estimate each of the three integrals separately. Consider

I δ 1 = x 2 x 1 F 1 ( η + x 1 x 2 ) F 1 ( η ) d η = x 2 x 1 1 ( x 2 x 1 η ) 1 / r ( 1 + ( ln ( x 2 x 1 η ) ) 2 ) 1 ( η ) 1 / r ( 1 + ( ln ( η ) ) 2 ) d η
x 1 x 2 1 ( x 2 x 1 + η ) 1 / r η 1 / r d η = x 1 x 2 x 1 x 2 + 1 1 ( x 2 x 1 + η ) 1 / r η 1 / r d η + x 1 x 2 + 1 1 ( x 2 x 1 + η ) 1 / r η 1 / r d η
1 ( x 1 x 2 ) 1 / r x 1 x 2 x 1 x 2 + 1 1 ( x 2 x 1 + η ) 1 / r d η + x 1 x 2 + 1 1 ( x 2 x 1 + η ) 2 / r d η
1 ( x 1 x 2 ) 1 / r 0 1 1 η 1 / r d η + 1 1 η 2 / r d η 1 ( x 1 x 2 ) 1 / r .
Similarly,
I δ 2 = x 2 x 1 δ F 1 ( η + x 1 x 2 ) F 1 ( η ) d η x 2 x 1 δ 1 ( η + x 1 x 2 ) 1 / r ( η ) 1 / r d η 1 δ 1 / r x 2 x 1 δ 1 ( η + x 1 x 2 ) 1 / r d η = 1 δ 1 / r 0 δ + x 1 x 2 1 η 1 / r d η 1 δ 1 / r ( δ + x 1 x 2 ) 1 1 / r ,
I δ 3 = δ F 1 ( η + x 1 x 2 ) F 1 ( η ) d η δ 1 η 2 / r d η δ 1 2 / r .
Since e π 2 ( x 1 2 + ω 1 2 + x 2 2 + ω 2 2 ) S ( R 4 ) , the integral I δ < . This proves that the integral 𝐼 is not finite. In other words, C θ f 2 M s , 1 , which is the same as f 2 M θ s , 1 . Since s > r , M θ r , 1 M θ s , 1 . Thus, in particular, f 2 M θ r , 1 .

Acknowledgements

The authors thank the National Institute of Science Education and Research Bhubaneswar for providing an excellent research facility. We thank the referee for meticulously reading our manuscript and giving us several valuable suggestions.

  1. Communicated by: Anke Pohl

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Received: 2024-04-25
Revised: 2024-10-15
Published Online: 2024-11-30
Published in Print: 2025-06-01

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