Afr. Mat. https://doi.org/10.1007/s13370-018-0546-8
On function spaces with S-transform Baby Kalita1 · Sunil Kumar Singh2
Received: 14 August 2015 / Accepted: 23 January 2018 © African Mathematical Union and Springer-Verlag GmbH Deutschland, ein Teil von Springer Nature 2018
p,q
Abstract For 1 ≤ p, q < ∞, we define the space Wk1 ,k2 (Rn ) by means of the S-transform where k1 and k2 are weight functions on Rn and Rn × Rn0 , respectively. It is a Banach space with respect to a sum norm defined in this paper. Latter we discuss inclusion properties and p,q p,q obtain the dual space of Wk1 ,k2 (Rn ). Furthermore, we define the space Wk1 ,k2 (Rn ) where ξ is fixed and show that it is an essential Banach module over L 1k1 (Rn ).
ξ
Keywords S-transform · Banach module · Multiplier space · Weighted lebesgue spaces Mathematics Subject Classification 65R10
1 Introduction The S-transform was developed by Stockwell et al. [16] in 1996. It has spread in various areas such as mathematics, physics, image processing, signal analysis, geophysics, medicine etc. The n-dimensional continuous S-transform of f with respect to the window function ω is defined as [6,17] (Sω f )(τ, ξ ) = f (t) ω(τ − t, ξ ) e−i2π ξ,t dt; 0 = ξ, τ ∈ Rn , (1.1) Rn
provided the integral exists. In signal analysis, at least in dimension n = 1, R2n is called the time-frequency plane, and in physics R2n is called the phase space.
B
Sunil Kumar Singh
[email protected] Baby Kalita
[email protected]
1
Department of Mathematics, Rajiv Gandhi University, Doimukh, Arunachal Pradesh 791112, India
2
Department of Mathematics, Mahatma Gandhi Central University, Motihari, Bihar 845401, India
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Equation (1.1) can be rewritten as a convolution (Sω f )(τ, ξ ) = f (·)e−i2π ξ,· ∗ ω(·, ξ ) (τ ). Applying the convolution property for the Fourier transform in (1.2), we obtain ˆ ξ ) (τ ), (Sω f )(τ, ξ ) = F −1 fˆ(· + ξ ) ω(·,
(1.2)
(1.3)
where fˆ(η) = (F f )(η) = Rn f (t) e−i2π η,t dt is the Fourier transform of f . We may refer to [7–15] for theoretical properties of the S-transforms. In this paper the window functions ω is assumed to satisfy the condition ω(·, ξ ) 1 = |ω(t, ξ )| dt = M(a constant), (1.4) Rn
:= Without any loss of generality to simplify the calculation we may for all ξ ∈ assume that M = 1. As examples we can consider the following windows Rn0
Rn \{0}.
|ξ | (i) ω1 (t, ξ ) = √ e−t ξ /2κ , κ > 0 (Gaussian window) κ 2π (ii) ω2 (t, ξ ) = δ(t − ξ ) (Dirac delta function). dτ dξ is defined as the set of Definition 1.1 For ξ ∈ Rn0 , the space L 2 Rn × Rn0 , (1+|ξ n+1 |) functions F(τ, ξ ) on Rn × Rn0 such that dτ dξ |F(τ, ξ )|2 < ∞. (1.5) n (1 + |ξ |)n+1 Rn R0 2 2
2
This is a Hilbert space with inner product F, GL 2 = Rn
where dμ(τ, ξ ) =
Rn0
F(τ, ξ ) G(τ, ξ )dμ(τ, ξ ),
(1.6)
dτ dξ . (1+|ξ |)n+1
Theorem 1.2 If Cc (Rn × Rn0 , dτ dξ ) denotes the space of complex-valued continuous functions defined on Rn × Rn0 with compact support, then dτ dξ . Cc (Rn × Rn0 , dτ dξ ) ⊂ L 2 Rn × Rn0 , (1 + |ξ |)n+1
|g(τ,ξ )|2 . Then f Proof For any g ∈ Cc Rn × Rn0 , dτ dξ , let supp g = and f (τ, ξ ) = (1+|ξ |)n+1 is continuous and supp f = . Suppose max f (τ, ξ ) = m, then we have |g(τ, ξ )|2 = g dτ dξ dτ dξ n+1 n 2 n L R ×R0 , Rn ×Rn0 (1 + |ξ |) (1+|ξ |)n+1 dτ dξ ≤m
= mμ( ) < ∞.
dτ dξ . This completes the proof. Therefore, g ∈ L 2 Rn × Rn0 , (1+|ξ n+1 |) We recall the Parseval’s and inversion formula for the S-transform which is given in [15].
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On function spaces with S-transform
Theorem 1.3 (Parseval’s formula) Suppose ω1 , ω2 are window functions such that ω1 (·, ξ ) ω2 (·, ξ ) 0 < Cω1 ,ω2 := dξ < ∞. n (1 + |ξ |)n+1 R0
(1.7)
Let f, g ∈ L 2 (Rn ) and (Sω1 f ), (Sω2 g) be the S-transforms of f and g, respectively. Then dτ dξ (Sω1 f )(τ, ξ )(Sω2 g)(τ, ξ ) = C f (t)g(t) dt. (1.8) ω1 ,ω2 (1 + |ξ |)n+1 Rn Rn0 Rn This immediately implies the Plancherel formula Sω f L 2 (Rn ×Rn ) = (Cω )1/2 f L 2 (Rn ) . 0
(1.9)
Theorem 1.4 (Inversion formula) If f ∈ L 2 (Rn ) and window function ω satisfies the condition | ω(·, ξ )|2 dξ < ∞, (1.10) 0 < Cω := n+1 Rn0 (1 + |ξ |) then, f (t) =
1 Cω
Rn
Rn0
(Sω f )(τ, ξ ) ω(τ − t, ξ ) ei2π t,ξ
dτ dξ . (1 + |ξ |)n+1
(1.11)
In this paper Tc is the translation operator defined by Tc f (t) = f (t − c), for any complex valued function f defined on Rn . We also use the Beurling weights such that k(x) ≥ 1, which are positive, real valued. A weight function defined by k(x, ξ ) = (1 + |x| + |ξ |)a , a ≥ 0 on Rn × Rn0 is called a weight of polynomial type. For x, z ∈ Rn and ξ, t ∈ Rn0 , we have k(x + z, ξ ) = (1 + |x + z| + |ξ |)a ≤ (1 + |x + z| + |ξ + t|)a ≤ (1 + |x| + |ξ |)a (1 + |z| + |t|)a = k(x, ξ )k(z, t). For any two weight functions k1 and k2 , we write k1 ≺ k2 if there exists C > 0 such that k1 (x) ≤ C k2 (x) for all x ∈ Rn . Two weight functions k1 and k2 are called equivalent, i.e., k1 ≈ k2 , if and only if k1 ≺ k2 and k2 ≺ k1 . p
n ) is defined as the space of all functions f Definition 1.5 For 1 ≤ p < ∞, the space L k (R p on Rn such that Rn | f k| p < ∞. The space L k (Rn ) is a Banach space with respect to the norm
1 p p p || f || p,k := || f k|| p = | f (x)| k (x)d x . (1.12) Rn
p,q
2 The space Wk1 ,k2 (Rn ) Definition 2.1 Suppose ω(·, ξ ) ∈ S (Rn ), the space of all C ∞ -functions on (Rn ) rapidly decreasing at infinity. Let k1 be a Beurling weight function on Rn and k2 be a weight function p,q of polynomial type on Rn × Rn0 . For 1 ≤ p, q < ∞, we define Wk1 ,k2 (Rn ) as p,q p q (2.1) Wk1 ,k2 (Rn ) = f ∈ L k1 (Rn ) : (Sω f ) ∈ Lk2 (Rn × Rn0 )
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B. Kalita, S. K. Singh
and it is a normed space with respect to || f ||W p,q := || f || p,k1 + ||Sω f ||q,k2 .
(2.2)
k1 ,k2
Theorem 2.2
p,q
Wk1 ,k2 (Rn ), ||.||W p,q
k1 ,k2
is a Banach space. p,q
Proof Let { f n }n∈N be a Cauchy sequence in Wk1 ,k2 (Rn ). It is easy to see that { f n }n∈N and p q p {Sω f n }n∈N are Cauchy sequences in L k1 (Rn ) and Lk2 (Rn ×Rn0 ), respectively. Since L k1 (Rn ) q p q and Lk2 (Rn × Rn0 ) are Banach spaces, there exist f ∈ L k1 (Rn ) and h ∈ Lk2 (Rn × Rn0 ) such that || f n − f || p,k1 → 0 and ||Sω f n − h||q,k2 → 0. Hence || f n − f || p → 0 and ||Sω f n − h||q,k2 =
1/q
|(Sω f n )(τ, ξ ) − h(τ, ξ )|
q
Rn
Rn0
q k2 (τ, ξ )dτ dξ
→0
implies ||Sω f n − h||q → 0, as k1 , k2 ≥ 1. Therefore, for 1 ≤ q < ∞, {Sω f n }n∈N converges to h (a.e.) in L q (Rn × Rn0 ) and hence every subsequence {Sω f nr }nr ∈N converges to h (a.e.). Thus we have, |Sω f − h| = Sω f − Sω f nr + Sω f nr − h ≤ Sω f − Sω f nr + Sω f nr − h −i2π
= e ( f − f nr )(t) ω(τ − t, ξ )dt + Sω f nr − h n R = ω(·, ξ ) ∗ ( f − f n )(·)e−i2π ξ,· + Sω f nr − h ≤ ω(·, ξ ) p f − f nr p + Sω f nr − h → 0 as nr → ∞, where
1 1 + = 1. p p
Therefore, (Sω f ) = h (a.e.) and hence || f n − f ||W p,q = || f n − f || p,k1 + ||Sω f n − Sω f ||q,k2 → 0. k1 ,k2
p,q
p,q
Thus f ∈ Wk1 ,k2 (Rn ) and this proves that Wk1 ,k2 (Rn ) is a Banach space.
(2.3)
Example 2.3 Let k2 be any weight function on R×R0 . Take the weight function k1 (t) = 1+|t| on R. Let ω be the Gaussian window, then −i2π <α,t> | ω(α, ξ )| = e−i2π <α,t> ω(t, ξ ) dt ≤ ω(t, ξ ) dt e R R0 0 |ξ | −t 2 ξ 2 /2κ 2 = dt = 1, √ e R0 κ 2π and hence
Cω =
R0
| ω(·, ξ )|2 dξ ≤ (1 + |ξ |)2
R0
1 dξ < ∞, (1 + |ξ |)2
that is, the Gaussian window satisfies the admissibility condition (1.10). Consider the space 2,q dτ dξ Wk1 ,k2 (R) for 1 ≤ q < ∞. Take any F ∈ Cc (R × R0 , dτ dξ ) ⊂ L 2 R × R0 , (1+|ξ . By |)2
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On function spaces with S-transform
inversion formula (1.11), we have 1 i2π <ξ,t> dτ dξ F(τ, ξ ) ω(τ − t, ξ ) e || f ||2,k1 = C 2 (1 + |ξ |) 2,k1 ω R R0 1 2 2 1 2 dτ dξ i2π <ξ,t> k (t)dt = F(τ, ξ ) ω(τ − t, ξ ) e (1 + |ξ |)2 1 R C ω R R0 1 2 |F(τ, ξ )| 1 2 2 |ω(τ − t, ξ )| k1 (t)dt ≤ dτ dξ Cω R R0 (1 + |ξ |)2 R 1 2 |ξ | −ξ 2 (τ −t)2 2 2 |F(τ, ξ )| 1 2 √ e 2κ k (t)dt dτ dξ ≤ 1 Cω R R0 (1 + |ξ |)2 R κ 2π 21 2 −ξ 2 (τ −t)2 |F(τ, ξ )| ξ 1 2 2 κ e k1 (t)dt dτ dξ = Cω R R0 (1 + |ξ |)2 2πκ 2 R 2 −ξ 2 x 2 21 |F(τ, ξ )| 1 ξ 2 2 ≤ k1 (τ ) e κ k1 (x) d x dτ dξ Cω R R0 (1 + |ξ |)2 2πκ 2 R 2 −ξ 2 x 2 21 |F(τ, ξ )| 1 ξ 2 2 = k1 (τ ) e κ (1 + |x|) d x dτ dξ Cω R R0 (1 + |ξ |)2 2πκ 2 R 2 ∞ −ξ 2 x 2 21 |F(τ, ξ )| 1 ξ 2 2 = k1 (τ ) e κ (1 + x) d x dτ dξ Cω R R0 (1 + |ξ |)2 πκ 2 0 √ 21 √ 2 2 |F(τ, ξ )| ξ π 1 κ π ξ κ + ≤ 1+ (1 + |τ |) dτ dξ Cω R R0 (1 + |ξ |)2 πκ 2 ξ 2 4ξ 2κ 1 2 |F(τ, ξ )| (1 + |τ |) 1 1 ξ κ = dτ dξ. + √ + √ 2 Cω R R0 (1 + |ξ |) π 4ξ π 2κ π If we take a function F and ξ ∈ R0 , then
|F(τ, ξ )| (1 + |τ |) (1 + |ξ |)2
1
1 ξ κ + √ + √ π 4ξ π 2κ π
2
is also continuous. If supp F = , then
|F(τ, ξ )| (1 + |τ |) supp (1 + |ξ |)2
1 κ ξ + √ + √ π 4ξ π 2κ π
1 2
= .
If we set max
(τ,ξ )∈
|F(τ, ξ )| (1 + |τ |) (1 + |ξ |)2
we obtain || f ||2,k1 ≤
M Cω
κ 1 ξ + √ + √ π 4ξ π 2κ π
dτ dξ =
1 2
= M,
M μ( ) < ∞, Cω
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B. Kalita, S. K. Singh
where μ( ) is the area of the set . Therefore, f ∈ L 2k1 (R) ⊂ L 2 (R). Also we have dτ dξ (Sω f ) ∈ L 2 R × R0 , (1+|ξ . We know that the S-transform is one-to-one mapping, 2 |) q
q
therefore, Sω f = F. Since Cc (R × R0 ) ⊂ Lk2 (R × R0 ), we have (Sω f ) ∈ Lk2 (R × R0 ). Therefore, || f ||W 2,q = || f ||2,k1 + ||Sω f ||q,k2 < ∞, k1 ,k2
and hence f ∈
2,q Wk1 ,k2 (R).
Theorem 2.4 Let k1 be any weight function and k2 be a weight function of polynomial type. Then p,q
(i) Wk1 ,k2 (Rn ) is invariant under translations. p,q p,q (ii) The mapping f → Tz f is continuous from Wk1 ,k2 (Rn ) into Wk1 ,k2 (Rn ) for any fixed n z∈R . p,q
p
q
Proof (i) Let f ∈ Wk1 ,k2 (Rn ). Then we have f ∈ L k1 (Rn ) and Sω f ∈ Lk2 (Rn × Rn0 ). Since 1 p p | f (t − z)| p k1 (t)dt Tz f p,k1 = Rn
=
Rn
1 | f (x)|
p
p k1 (z
+ x)d x
≤
1 p
Rn
p
p
| f (x)| p k1 (z)k1 (x)d x
p
= k1 (z) f p,k1 , and Sω (Tz f ) q,k2 = n n R
=
Rn
≤
R0
n
R0
Rn
Rn
Rn
e
e
−i2π <ξ,x+z>
1 q q q f (x) ω(τ − x − z, ξ )d x k2 (τ, ξ ) dμ(τ, ξ )
n
1 q q q q e−i2π <ξ,x> f (x) ω(y − x, ξ )d x k2 (z, t)k2 (y, ξ ) dμ(y, ξ ) n
R0
= k2 (z, t)
1 q q q f (t − z) ω(τ − t, ξ )dt k2 (τ, ξ ) dμ(τ, ξ )
−i2π <ξ,t>
R
Rn
1 q
Rn0
q |(Sω f )(y, ξ )|q k2 (y, ξ ) dμ(y, ξ )
= k2 (z, t) Sω f q,k2 , thus Tz f W p,q = Tz f p,k1 + Sω (Tz f ) q,k2 k1 ,k2
≤ k1 (z) f p,k1 + k2 (z, t) Sω f q,k2 < ∞. Therefore, Tz f ∈
123
p,q Wk1 ,k2 (Rn )
p,q
and hence Wk1 ,k2 (Rn ) is invariant under translations.
On function spaces with S-transform p,q
(ii) Let f ∈ Wk1 ,k2 (Rn ). It is easy to show that f → Tz f is linear. For given > 0, choose δ = k1 (z)+k . Thus if f W p,q < δ, then 2 (z,t) k1 ,k2
f p,k1 ≤ f W p,q < δ and Sω f q,k2 ≤ f W p,q < δ. k1 ,k2
k1 ,k2
Since Tz f p,k1 ≤ k1 (z) f p,k1 and Sω (Tz f ) q,k2 ≤ k2 (z, t) Sω f q,k2 . Therefore, Tz f W p,q = Tz f p,k1 + Sω (Tz f ) q,k2 k1 ,k2
≤ k1 (z) f p,k1 + k2 (z, t) Sω f q,k2 ≤ k1 (z)δ + k2 (z, t)δ = [k1 (z) + k2 (z, t)] δ = [k1 (z) + k2 (z, t)] = .
k1 (z) + k2 (z, t) p,q
p,q
Thus the mapping f → Tz f is continuous from Wk1 ,k2 (Rn ) into Wk1 ,k2 (Rn ) for any fixed z ∈ Rn .
p,q
3 Inclusion properties of the space Wk1 ,k2 (Rn ) p,q
Proposition 3.1 For every non zero function f ∈ Wk1 ,1 (Rn ), there exists one C( f ) > 0 such that C( f )k1 (z) ≤ Tz f W p,q ≤ k1 (z) f W p,q . k1 ,1
k1 ,1
p,q
Proof Let f ∈ Wk1 ,1 (Rn ). By [1, Proposition 1.7], there exists one C( f ) > 0 such that C( f )k1 (z) ≤ Tz f p,k1 ≤ k1 (z) f p,k1 .
(3.1)
Thus we get C( f )k1 (z) ≤ Tz f p,k1 + Sω Tz f q = Tz f W p,q . k1 ,1
Also Tz f W p,q = Tz f p,k1 + Sω Tz f q k1 ,1
≤ k1 (z) f p,k1 + Sω Tz f q ≤ k1 (z) f p,k1 + k1 (z) Sω Tz f q ; (since k1 (z) ≥ 1)
= k1 (z) f p,k1 + Sω Tz f q
= k1 (z) f p,k1 + Sω f q ; since Sω Tz f q = Sω f q = k1 (z) f W p,q . k1 ,1
Hence C( f )k1 (z) ≤ Tz f W p,q ≤ k1 (z) f W p,q . k1 ,1
k1 ,1
(3.2)
By using the techniques of Kulak and Gürkanli [3] we can easily prove the following results.
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B. Kalita, S. K. Singh p,q
p,q
Lemma 3.2 If Wk1 ,k3 (Rn ) ⊂ Wk2 ,k4 (Rn ), where k1 , k2 , k3 and k4 be weight functions, then p,q Wk1 ,k3 (Rn ) is a Banach space under the norm f W = f W p,q + f W p,q . k1 ,k3
k2 ,k4
p,q
p,q
Theorem 3.3 Let k1 and k2 be weight functions. Then Wk1 ,1 (Rn ) ⊂ Wk2 ,1 (Rn ) if and only if k2 ≺ k1 . Proposition 3.4 Let C1 and C2 be two constant numbers and k1 , k2 , k3 and k4 be weight p,q p,q functions such that k3 ≈ C1 and k4 ≈ C2 . Then Wk1 ,k3 (Rn ) ⊂ Wk2 ,k4 (Rn ) if and only if k2 ≺ k1 . p,q
p,q
Corollary 3.5 If k3 ≈ C1 and k4 ≈ C2 , then Wk1 ,k3 (Rn ) = Wk2 ,k4 (Rn ) if and only if k1 ≈ k2 . Proposition 3.6 Let k1 , k2 , k3 and k4 be weight functions. If k = max{k1 , k3 } and l = max{k2 , k4 }, then we have p,q
p,q
p,q
Wk1 ,k2 (Rn ) ∩ Wk3 ,k4 (Rn ) = Wk,l (Rn ). Proposition 3.7 Let k1 , k2 , k3 and k4 be weight functions such that k3 ≺ k1 and k4 ≺ k2 , then p,q p,q Wk1 ,k2 (Rn ) ⊂ Wk3 ,k4 (Rn ). Proof If k3 ≺ k1 and k4 ≺ k2 , then there exist C1 , C2 > 0 such that k3 (t) ≤ C1 k1 (t) and p,q k4 (τ, ξ ) ≤ C2 k2 (τ, ξ ), for all t ∈ Rn and (τ, ξ ) ∈ Rn × Rn0 . Suppose f ∈ Wk1 ,k2 (Rn ), and p q n hence f ∈ L k1 (Rn ) and Sω f ∈ Lk2 (Rn × R0 ). Furthermore, f p,k3 ≤ C1 f p,k1 and Sω f q,k4 ≤ C2 Sω f q,k2 . Therefore, f W p,q = f p,k3 + Sω f q,k4 k3 ,k4
≤ C1 f p,k1 + C2 Sω f q,k2 < ∞. p,q
p,q
p,q
It proves that f ∈ Wk3 ,k4 (Rn ) and hence Wk1 ,k2 (Rn ) ⊂ Wk3 ,k4 (Rn ).
p,q
p,q
Proposition 3.8 Let k1 , k2 , k3 and k4 be weight functions . If Wk1 ,k2 (Rn ) ⊂ Wk3 ,k4 (Rn ), then there exists a constant C > 0 such that f W p,q ≤ C f W p,q k3 ,k4
k1 ,k2
p,q
for every f ∈ Wk1 ,k2 (Rn ).
Theorem 3.9 Let k1 be any weight function and k2 be a weight function of polynomial type. Then there exists C( f ) > 0 such that C( f )k1 (z) ≤ Tz f W p,q ≤ [k1 (z) + k2 (z, t)] f W p,q k1 ,k2
for every 0 = f ∈
p,q Wk1 ,k2 (Rn )
and t ∈
p,q
k1 ,k2
Rn0 . p
q
Proof Let f ∈ Wk1 ,k2 (Rn ). Then f ∈ L k1 (Rn ) and Sω f ∈ Lk2 (Rn × Rn0 ). Moreover, there exists C( f ) > 0 such that C( f )k1 (z) ≤ Tz f p,k1 ≤ k1 (z) f p,k1 . Again we have from Theorem 2.4 Sω Tz f q,k2 ≤ k2 (z, t) Sω f q,k2 .
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On function spaces with S-transform
Also C( f )k1 (z) ≤ Tz f p,k1 + Sω (Tz f ) q,k2 ≤ k1 (z) f p,k1 + k2 (z, t) Sω f q,k2 ≤ k1 (z) f W p,q + k2 (z, t) f W p,q k1 ,k2
k1 ,k2
≤ [k1 (z) + k2 (z, t)] f W p,q
k1 ,k2
for all t ∈ Rn0 .
p,q
4 Dual space of Wk1 ,k2 (Rn ) p,q
p
q
For 1 ≤ p, q < ∞, we consider the mapping : Wk1 ,k2 (Rn ) → L k1 (Rn ) × Lk2 (Rn × Rn0 ) p,q p,q defined as ( f ) = ( f, Sω f ). Let H = (Wk1 ,k2 (Rn )) = ( f, Sω f ) : f ∈ Wk1 ,k2 (Rn ) . Then ||| ( f )||| = |||( f, Sω f )||| = f p,k1 + Sω f q,k2 is a norm on
p L k1 (Rn )
q
× Lk2 (Rn × Rn0 ). Note that ||| ( f )|||W p,q = f W p,q and is a p,q
k1 ,k2
p
q
p,q
p
k1 ,k2
linear isometry from Wk1 ,k2 (Rn ) into L k1 (Rn ) × Lk2 (Rn × Rn0 ). Furthermore, we define a set K by p q K = (g, h) ∈ L k −1 (Rn ) × Lk −1 (Rn × Rn0 ) : f (y)g(y)dy 1 2 Rn (Sω f )(τ, ξ )h(τ, ξ ) dμ(τ, ξ ) = 0 for all ( f, Sω f ) ∈ H , + Rn ×Rn0
where
1 p
+
1 p
= 1 and
1 q
+
1 q
= 1.
Proposition 4.1 The dual space of Wk1 ,k2 (Rn ) is L k 1 p
+
1 p
= 1 and
1 q
+
1 q
= 1.
1
q
−1
(Rn ) × Lk
2
−1
(Rn × Rn0 )/K , where
p,q
p,q
Proof We know that Wk1 ,k2 (Rn ) is a Banach space. Then H = (Wk1 ,k2 (Rn )) is closed. Using the duality theorem in [4, Theorem 1.7] we get p
H ∼ = ∗
H∗
Lk
1
q
−1
(Rn ) × Lk
2
−1
(Rn × Rn0 )
K p,q
p
q
is dual of H. Also the mapping : Wk1 ,k2 (Rn ) → L k1 (Rn ) × Lk2 (Rn × Rn0 ) is ∗ p,q p,q p q isometry, then (Wk1 ,k2 (Rn ))∗ ∼ = H ∗ . Thus we get Wk1 ,k2 (Rn ) ∼ = L k −1 (Rn )×Lk −1 (Rn × 1 2
Rn0 )/K . where
p,q
5 The space (Wk1 ,k2 )ξ (Rn ) p,q
p,q
We take ξ to be fixed in Wk1 ,k2 (Rn ) and we denote a new space by (Wk1 ,k2 )ξ (Rn ). It is a vector p q space of functions f ∈ L k1 (Rn ) such that their S-transform (Sω f )(·, ξ ) are in L k2 (Rn ). By
123
B. Kalita, S. K. Singh p,q
using the techniques of Theorem 2.2, it is easy to see that (Wk1 ,k2 )ξ (Rn ) is a Banach space with sum norm f (W p,q )ξ := f p,k1 + (Sω f )(·, ξ ) q,k2 . k1 ,k2
p,q
Proposition 5.1 Let k2 be a weight function of polynomial type. Then (Wk1 ,k2 )ξ (Rn ) is dense p in L k1 (Rn ). Proof Since k2 is a weight function of polynomial type then any f ∈ Cc (Rn ) is p q also in f ∈ L k1 (Rn ). Since L k2 (Rn ) is a Banach convolution module over L 1k2 (Rn ) [1, Proposition 1.11] then Sω f q,k2 = f (·)e−i2π <ξ,·> ∗ ω(·, ξ ) q,k2 −i2π <ξ,·> ω(·, ξ ) 1,k2 ≤ f (·)e q,k2
= f q,k2 ω(·, ξ ) 1,k2 < ∞. We get f (W p,q
= f p,k1 + (Sω f )(·, ξ ) q,k2 < ∞.
k1 ,k2 )ξ
p,q
p,q
p
Therefore, f ∈ (Wk1 ,k2 )ξ (Rn ) and hence Cc (Rn ) ⊂ (Wk1 ,k2 )ξ (Rn ) ⊂ L k1 (Rn ). Since p p,q p Cc (Rn ) is dense in L k1 (Rn ), (Wk1 ,k2 )ξ (Rn ) is dense in L k1 (Rn ).
Proposition 5.2 Let C be a non-zero constant number such that k2 ≈ C. Then the space q,q q (Wk1 ,k2 )ξ (Rn ) and L k1 (Rn ) are algebraically isomorphic and homeomorphic. Proof We have from the previous proposition q,q
q
(Wk1 ,k2 )ξ (Rn ) ⊂ L k1 (Rn ).
(5.1) q
Since k2 ≈ C, so there exists N > 0 such that . q,k2 ≤ N . q . For any f ∈ L k1 (Rn ), we have f (W q,q
k1 ,k2 )ξ
= f q,k1 + (Sω f )(·, ξ ) q,k2 ≤ f q,k1 + N (Sω f )(·, ξ ) q ≤ f q,k1 + N f q ω(·, ξ ) 1 ≤ f q,k1 + N f q,k1 ω(·, ξ ) 1 = (1 + N ω(·, ξ ) 1 ) f q,k1 < ∞.
q,q
Therefore, f ∈ (Wk1 ,k2 )ξ (Rn ) and hence q
q,q
L k1 (Rn ) ⊂ (Wk1 ,k2 )ξ (Rn ).
(5.2)
From Eqs. (5.1) and (5.2) we have q,q q Wk1 ,k2 (Rn ) = L k1 (Rn ). ξ
If M = 1 + N ω(·, ξ ) 1 , then f q,k1 ≤ f (W q,q
k1 ,k2 )ξ
q,q
q,q
≤ M f q,k1 , q
for all f ∈ (Wk1 ,k2 )ξ (Rn ). Therefore, (Wk1 ,k2 )ξ (Rn ) and L k1 (Rn ) are algebraically isomorphic and homeomorphic.
123
On function spaces with S-transform
Theorem 5.3 Let k1 be any weight function and k2 be a weight function of polynomial type. Then p,q (i) Wk1 ,k2 (Rn ) is invariant under translations. ξ p,q (ii) The translation mapping z → Tz f is continuous from (Rn ) into Wk1 ,k2 (Rn ). ξ
Proof Let f ∈
p,q (Wk1 ,k2 )ξ (Rn ).
Then by using Theorem 2.4
Tz f (W p,q
= Tz f p,k1 + (Sω Tz f )(·, ξ ) q,k2
k1 ,k2 )ξ
≤ k1 (z) f p,k1 + k2 (z, t) (Sω f )(·, ξ ) q,k2 < ∞. p,q Hence (Wk1 ,k2 )ξ (Rn ) is translation invariant. We know that translation mapping is continp uous from Rn into L k1 (Rn ). Therefore, for any given > 0 there exists δ > 0 such that Tz f − Tu f p,k1 < 2 , whenever z − u < δ for z, u ∈ Rn . Again, since translation mapq ping is continuous from Rn into L k2 (Rn ), so for same > 0 there exists δ > 0 such that Sω (Tz f − Tu f )(·, ξ ) q,k2 < 2 , whenever z − u < δ for z, u ∈ Rn . If δ1 = min(δ, δ ) and z − u < δ1 for z, u ∈ Rn . Then
Tz f − Tu f
p,q 1 ,k2 ξ
Wk
= Tz f − Tu f p,k1 + Sω (Tz f − Tu f )(·, ξ ) q,k2
+ = . 2 2 Therefore, z → Tz f is continuous mapping. <
Theorem 5.4 If k2 = C, a constant number, then module over L 1k1 (Rn ).
p,q (Wk1 ,k2 )ξ (Rn )
is an essential Banach
p,q
p,q
Proof We know that (Wk1 ,k2 )ξ (Rn ) is a Banach space. Let f ∈ (Wk1 ,k2 )ξ (Rn ) and h p ∈ L 1k1 (Rn ). Since L k1 (Rn ) is a Banach convolution module over L 1k1 (Rn ), we can write f ∗ h p,k1 ≤ f p,k1 h 1,k1
(5.3)
and Sω ( f ∗ h)(·, ξ ) q,k2 −i2π <ξ,t> = ( f ∗ h)(t)e ω(τ − t, ξ )dt n R q,k2 −i2π <ξ,t> = f (t − x)h(x)d x e ω(τ − t, ξ )dt Rn
Rn
q,k
2 −i2π <ξ,x> −i2π <ξ,y> = h(x)e f (y)e ω(τ − x − y, ξ )dy dx n R Rn q,k2 (5.4) −i2π <ξ,x> = h(x)e (S f )(τ − x, ξ )d x ω n R q,k2 −i2π <ξ,·> ∗ (Sω f )(·, ξ ) = h(·)e
q,k2
≤ h 1,k2 (Sω f )(·, ξ ) q,k2 = C h 1 (Sω f )(·, ξ ) q,k2 ≤ C h 1,k1 (Sω f )(·, ξ ) q,k2 .
123
B. Kalita, S. K. Singh
Combining Eqs. (5.3) and (5.4), we get f ∗ h (W p,q
= f ∗ h p,k1 + Sω ( f ∗ h)(·, ξ ) q,k2
k1 ,k2 )ξ
≤ f p,k1 h 1,k1 + C h 1,k1 (Sω f )(·, ξ ) q,k2 ≤ C h 1,k1 f (W p,q
k1 ,k2 )ξ
(5.5)
.
p,q
Thus (Wk1 ,k2 )ξ (Rn ) is a Banach convolution module over L 1k1 (Rn ). p,q Now we have to show that (Wk1 ,k2 )ξ (Rn ) is an essential Banach module over L 1k1 (Rn ). p,q p,q First we prove that L 1k1 (Rn ) ∗ (Wk1 ,k2 )ξ (Rn ) is dense in (Wk1 ,k2 )ξ (Rn ). It is known that 1 n L k1 (R ) has a bounded approximate identity [2]. Let N be a neighborhood of unit element of Rn . We can choose an approximate identity (eα )α∈I which is positively bounded and satisfies p,q supp eα ⊂ N , eα = 1 for all α ∈ I . Taking any h ∈ (Wk1 ,k2 )ξ (Rn ) for fixed α0 ∈ I , we get eα ∗ h − h p,q = eα0 (z)Tz h(y)dz − eα0 (z)h(y)dz 0 Wk
1 ,k2 ξ
Rn
= ≤
eα0 (z) Tz h(y) − h(y) dz n
R
Rn
p,q 1 ,k2 ξ
Rn
eα0 (z) Tz h − h
p,q 1 ,k2 ξ
Wk
Wk
p,q 1 ,k2 ξ
Wk
dz.
p,q We know that the translation mapping z → Tz f is continuous from Rn into Wk1 ,k2 (Rn ). ξ
Therefore, for any given > 0, we have Tz h − h (W p,q )ξ < . Thus k1 ,k2 eα ∗ h − h p,q ≤ eα0 (z) dz = . 0 (W ) ξ
k1 ,k2 Rn p,q p,q p,q 1 n n Therefore, L k1 (R ) ∗ (Wk1 ,k2 )ξ (R ) is dense in (Wk1 ,k2 )ξ (Rn ). Consequently (Wk1 ,k2 )ξ (Rn )
is an essential Banach module over L 1k1 (Rn ) by the Module factorization theorem [18].
p,q
6 Space of multipliers of (Wk1 ,k2 )ξ (Rn ) p,q
p
q
For 1 ≤ p, q < ∞, we consider the mapping : (Wk1 ,k2 )ξ (Rn ) → L k1 (Rn ) × L k2 (Rn ) defined as ( f ) = ( f, Sω f ). This mapping is a linear isometric with the norm ||| ( f )||| = |||( f, Sω f )||| = f p,k1 + (Sω f )(·, ξ ) q,k2 . p,q p,q Let H = (Wk1 ,k2 )ξ (Rn ) = ( f, Sω f ) : f ∈ Wk1 ,k2 (Rn ) . Furthermore, we define ξ
a set K by
p L k −1 (Rn ) × 1
q L k −1 (Rn ) 2
: f (y)g(y)dy K = (g, h) ∈ Rn (Sω f )(τ, ξ )h(τ, ξ ) dτ = 0 for all ( f, Sω f ) ∈ H , + Rn
where
1 p
+
123
1 p
= 1 and
1 q
+
1 q
= 1.
On function spaces with S-transform
Proposition 6.1 The dual space ×
q L k −1 (Rn )/K 2
where
1 p
+
1 p
p,q
Wk1 ,k2
= 1 and
1 q
+
ξ
1 q
∗ p (Rn ) is isomorphic to L k −1 (Rn ) 1
= 1. p
q
Theorem 6.2 Let k2 = C, a constant number. Then the spaces L k −1 (Rn ) × L k −1 (Rn )/K 1 2 p,q n ) are algebraically isomorphic and topologi(R and Hom L 1 (Rn ) Wk1 ,k2 (Rn ), L ∞ k −1 ξ
k1
1
cally homeomorphic. p,q Proof Since Wk1 ,k2 (Rn ) is a essential Banach Module over L 1k1 (Rn ). By using corollary ξ
2.13 from [5], we obtain
Hom L 1
k1
(Rn )
p,q
Wk1 ,k2
ξ
n (Rn ), L ∞ (R ) k −1 1
∗ p,q Wk1 ,k2 (Rn ), L 1k1 (Rn ) = Hom L 1 k1 ξ ∗ p,q n 1 n ∼ Wk1 ,k2 (R ) ∗ L k1 (R ) = ξ ∗ p,q n Wk1 ,k2 (R ) =
ξ
∼ =
p L k −1 (Rn ) × 1
K
q
Lk
2
−1
(Rn )
.
Acknowledgements The authors are thankful to the referees for their valuable comments and suggestions.
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