In this section the mapping relations between W- and \(W^{p}\)-types of spaces are discussed.
Theorem 3.1
Let
\(M(x)\), \(\Omega(y)\)
be the pair of functions which are dual in the sense of Young. Then
$$W^{p}_{M,a}= W_{M,a},\quad 1 \leq p <\infty. $$
Proof
Now, for showing the above theorem we shall prove the following lemma. □
Lemma 3.2
Let
\(1 \leq p <\infty\). Then
\(W^{p}_{M,a} \subset W_{M,a}\).
Proof
Let \(e^{-\frac{|x|^{2}\cot\alpha}{2}}\phi(x)\in W^{p}_{M,a}(\mathbb{R}^{n})\) and \(\sigma= w + i\tau\). Then from the arguments of Theorem 2.1, we get
$$ F_{\alpha}\bigl(W^{p}_{M,a}\bigr)\subset W^{\Omega,\frac{1}{a}}. $$
(3.1)
From the inverse property of the fractional Fourier transform, we have
$$ W^{p}_{M,a}\subset F^{-1}_{\alpha} \bigl(W^{\Omega,\frac{1}{a}}\bigr). $$
(3.2)
Now, let \(\hat{\phi}_{\alpha}(\sigma)\in W^{\Omega,\frac{1}{a}}\). Then by the technique of [2, pp.21-22] and (1.6), we have
$$ \phi(x)= C_{\alpha}\int_{\mathbb{R}^{n}} e^{-\frac{i(|x|^{2} + |\sigma |^{2})\cot\alpha}{2}+ i\langle x, \sigma\rangle\csc\alpha} \hat{ \phi }_{\alpha}(\sigma) \,dw. $$
Therefore,
$$\begin{aligned} \phi^{(k)}(x) =& C_{\alpha}\int_{\mathbb{R}^{n}} D^{k}_{x}\bigl(e^{-\frac {i|x|^{2}\cot\alpha}{2}+ i\langle x, \sigma\rangle\csc\alpha}\bigr)e^{-\frac {i|\sigma|^{2}\cot\alpha}{2}}\hat{ \phi}_{\alpha}(\sigma) \,dw \\ =& C_{\alpha}\int_{\mathbb{R}^{n}}\sum _{r\leq k}\binom{k}{r}\bigl(D^{r}_{x} e^{-\frac{i|x|^{2}\cot\alpha}{2}}\bigr) \bigl(D^{k-r}_{x} e^{i\langle x, \sigma \rangle\csc\alpha} \bigr) e^{-\frac{i|\sigma|^{2}\cot\alpha}{2}}\hat{\phi}_{\alpha}(\sigma) \,dw \\ =& C_{\alpha}\sum_{r\leq k}\binom{k}{r} \int_{\mathbb{R}^{n}}\sum_{\eta \leq r}C_{\eta}( \cot\alpha) x^{\eta}e^{-\frac{i|x|^{2}\cot\alpha}{2}} (i\sigma\csc\alpha)^{k-r} \\ &{} \times e^{i\langle x, \sigma\rangle\csc\alpha}e^{-\frac{i|\sigma |^{2}\cot\alpha}{2}}\hat{\phi}_{\alpha}(\sigma) \,dw \\ =& C_{\alpha}\sum_{r\leq k}\binom{k}{r} \sum_{\eta\leq r}C_{\eta}(\cot \alpha)\int _{\mathbb{R}^{n}}e^{-\frac{i|x|^{2}\cot\alpha}{2}}(i\sigma\csc \alpha)^{k-r} \\ &{} \times\bigl(D^{\eta}_{\sigma}e^{i\langle x, \sigma\rangle\csc\alpha } \bigr)e^{-\frac{i|\sigma|^{2}\cot\alpha}{2}}\hat{\phi}_{\alpha}(\sigma) \,dw \\ =&C_{\alpha}\sum_{r\leq k}\binom{k}{r}\sum _{\eta\leq r}C_{\eta}(\cot \alpha) (i\csc \alpha)^{-|\eta|+|k-r|} (-1)^{|\eta|}\int_{\mathbb{R}^{n}}e^{-\frac{i|x|^{2}\cot\alpha}{2}}e^{i\langle x, \sigma\rangle\csc \alpha} \\ &{} \times D^{\eta}_{\sigma}\bigl(\sigma^{k-r}e^{-\frac{i|\sigma|^{2}\cot \alpha}{2}} \hat{\phi}_{\alpha}(\sigma)\bigr) \,dw \\ =& C_{\alpha}\sum_{r\leq k}\binom{k}{r} \sum_{\eta\leq r}C_{\eta}(\cot \alpha) (i\csc \alpha)^{-|\eta|+|k-r|} (-1)^{|\eta|}\int_{\mathbb{R}^{n}}e^{-\frac{i|x|^{2}\cot\alpha}{2}}e^{i\langle x, \sigma\rangle\csc \alpha} \\ &{} \times\sum_{\beta\leq\eta}\binom{\eta}{\beta} \bigl(D^{\eta}_{\sigma }e^{-\frac{i|\sigma|^{2}\cot\alpha}{2}}\bigr) \bigl(D^{\eta-\beta}_{\sigma} \sigma ^{k-r}\hat{\phi}_{\alpha}(\sigma)\bigr)\,dw \\ =&C_{\alpha}\sum_{r\leq k}\binom{k}{r}\sum _{\eta\leq r}C_{\eta}(\cot \alpha) (i\csc \alpha)^{-|\eta|+|k-r|} (-1)^{|\eta|}\int_{\mathbb{R}^{n}}e^{-\frac{i|x|^{2}\cot\alpha}{2}}e^{i\langle x, \sigma\rangle\csc \alpha} \\ &{} \times\sum_{\beta\leq\eta}\binom{\eta}{\beta}\biggl(\sum _{\lambda\leq\eta }C_{\lambda}(\cot\alpha) \sigma^{\lambda}\biggr)e^{-\frac{i|\sigma|^{2}\cot \alpha}{2}} \\ &{}\times\sum_{m\leq\eta-\beta} \binom{\eta-\beta}{m} \bigl(D^{m}_{\sigma} \sigma^{k-r} \bigr) \bigl(D^{\eta-\beta-m}_{\sigma}\hat{\phi}_{\alpha}(\sigma) \bigr)\,dw. \end{aligned}$$
Hence,
$$\begin{aligned} \bigl\vert \phi^{(k)}(x)\bigr\vert \leq& |C_{\alpha}|\sum _{r\leq k}\binom{k}{r}\sum _{\eta \leq r}\bigl\vert C_{\eta}(\cot\alpha)\bigr\vert \vert \csc\alpha|^{-|\eta|+|k-r|} \\ &{}\times\sum_{\beta\leq \eta} \binom{\eta}{\beta} \frac{(k-r)!}{(k-r-m)!} \sum_{\lambda\leq\beta }\bigl|C_{\lambda}( \cot\alpha)\bigr\vert \\ &{} \times\sum_{m\leq\eta-\beta}\binom{\eta-\beta}{m}\int _{\mathbb{R}^{n}}\bigl\vert e^{i\langle x, \sigma\rangle\csc\alpha}\bigr\vert \bigl\vert \sigma^{k-r-m+\lambda} D^{\eta-\beta-m}_{\sigma}\hat{\phi}_{\alpha}( \sigma)\bigr\vert \,dw. \end{aligned}$$
Therefore,
$$\begin{aligned} \bigl\vert \phi^{(k)}(x)\bigr\vert \leq&\vert C_{\alpha} \vert \sum_{r\leq k}\binom{k}{r}\sum _{\eta \leq r}\bigl\vert C_{\eta}(\cot\alpha)\bigr\vert \vert \csc\alpha \vert ^{-|\eta|+|k-r|} \\ &{}\times\sum_{\beta\leq \eta} \binom{\eta}{\beta} \frac{(k-r)!}{(k-r-m)!} \sum_{\lambda\leq\beta }\bigl\vert C_{\lambda}(\cot\alpha)\bigr\vert \\ &{} \times\sum_{m\leq\eta-\beta}\binom{\eta-\beta}{m}\int _{\mathbb{R}^{n}} e^{-\langle x, \tau\rangle\csc\alpha} \biggl(\frac{|\sigma |^{|k-r-m+\lambda|+2} +|\sigma|^{|k-r-m+\lambda|}}{(|w|^{2} +1)} \biggr) \\ &{} \times\bigl\vert D^{\eta-\beta-m}_{\sigma}\hat{ \phi}_{\alpha}(\sigma )\bigr\vert \,dw \\ \leq& C'_{\alpha,k} \exp\biggl[-\langle x, \tau\rangle\csc \alpha+ \Omega \biggl[\biggl(\frac{1}{a} + \rho\biggr)\tau\biggr]\biggr]. \end{aligned}$$
(3.3)
Now using the arguments of [4, p.23], we get
$$ \bigl\vert \phi^{(k)}(x)\bigr\vert \leq C'_{\alpha,k} \exp\bigl[-M\bigl[(a-\delta)x\csc\alpha\bigr]\bigr] $$
(3.4)
for arbitrarily small δ together with ρ. Hence the above expression gives
$$ F_{\alpha}^{-1}\bigl(W^{\Omega,\frac{1}{a}}\bigr)\subset W_{M,a}. $$
(3.5)
Thus (3.2) and (3.5) imply that
$$ W^{p}_{M,a}\subset W_{M,a}. $$
(3.6)
□
Lemma 3.3
Let
\(1\leq p< \infty\). Then
\(W_{M,a}\subset W^{p}_{M,a}\).
Proof
Let \(e^{-\frac{i|x|^{2}\cot\alpha}{2}}\phi(x)\in W_{M,a}(\mathbb{R}^{n})\) and \(\sigma= w + i\tau\in\mathbb{C}^{n}\). Then from [8, Theorem 2.1], it follows that
$$ F_{\alpha}(W_{M,a})\subset W^{\Omega,\frac{1}{a}}. $$
(3.7)
Now by the inverse property of the fractional Fourier transform we have
$$ W_{M,a}\subset F_{\alpha}^{-1} \bigl(W^{\Omega,\frac{1}{a}} \bigr). $$
(3.8)
Again let \(\hat{\phi}_{\alpha}(\sigma)\in W^{\Omega,\frac{1}{a}}\). Then from (3.3) we have
$$ \bigl\vert \phi^{k}(x)\bigr\vert \leq C'_{\alpha,k} \exp\biggl[-\langle x, \tau\rangle\csc\alpha + \Omega\biggl[\biggl( \frac{1}{a} + \rho\biggr)\tau\biggr]\biggr]. $$
(3.9)
Using (1.21) and [6, p.385] we get
$$ \bigl\vert \phi^{k}(x)\bigr\vert \leq C'_{\alpha,k} \exp\bigl[-M\bigl[(a-\delta)x\csc\alpha \bigr]-M\bigl[a^{3} \rho^{2} x\csc\alpha\bigr]\bigr]. $$
(3.10)
Therefore,
$$ \bigl\Vert \exp\bigl[M\bigl[(a-\delta)x\csc\alpha\bigr]\bigr] \phi^{k}(x)\bigr\Vert _{p}\leq C'_{\alpha, k} \bigl\Vert e^{-M[a^{3}\rho^{2} x\csc\alpha]}\bigr\Vert _{p}. $$
(3.11)
This implies that
$$ F_{\alpha}^{-1}\bigl(W^{\Omega,\frac{1}{a}}\bigr)\subset W^{p}_{M,a}. $$
(3.12)
From (3.8) and (3.12) we find
$$ W_{M,a}\subset W^{p}_{M,a}. $$
(3.13)
Now from (3.6) and (3.13) we get the result
$$ W^{p}_{M,a}= W_{M,a}. $$
(3.14)
□
Theorem 3.4
Let
\(M(x)\)
and
\(\Omega(y)\)
be the same functions as in Theorem
3.1. Then
$$ W^{\Omega,b,p}= W^{\Omega,b}, \quad 1\leq p<\infty. $$
(3.15)
Proof
Let \(e^{-\frac{i|z|^{2}\cot\alpha}{2}}\phi(z)\in W^{\Omega,b,p}\). Then from Theorem 2.2 it follows that
$$ F_{\alpha}\bigl( W^{\Omega,b,p}\bigr)\subset W_{M,\frac{1}{a}}. $$
(3.16)
By the inverse property of the fractional Fourier transform, we have
$$ W^{\Omega,b,p}\subset F_{\alpha}^{-1} (W_{M,\frac{1}{a}}). $$
(3.17)
Now let \(\hat{\phi}_{\alpha}(x)\in W_{M,\frac{1}{b}}\). Then from the technique of [2, pp.20-21], we have
$$\begin{aligned}& {(i\sigma\csc\alpha)^{k} \phi(\sigma)} \\& \quad = C_{\alpha}\int_{\mathbb{R}^{n}} e^{-\frac{i(|x|^{2} + |\sigma|^{2})\cot \alpha}{2}} \bigl( D^{k}_{x}e^{i\langle x, \sigma\rangle\csc\alpha } \bigr) \hat{ \phi}_{\alpha}(x) \,dx \\& \quad = C_{\alpha}(-1)^{|k|}\int_{\mathbb{R}^{n}}e^{i\langle x, \sigma\rangle \csc\alpha} \bigl( D^{k}_{x} e^{-\frac{i|x|^{2} \cot\alpha}{2}}\hat{\phi }_{\alpha}(x) \bigr) e^{-\frac{ i|\sigma|^{2}\cot\alpha}{2}} \,dx \\& \quad = C_{\alpha}(-1)^{|k|}\int_{\mathbb{R}^{n}}e^{i\langle x, \sigma\rangle \csc\alpha} \sum_{r\leq k}\binom{k}{r} \bigl(D^{r}_{x} e^{-\frac{i|x|^{2} \cot\alpha}{2}} \bigr) \bigl(D^{k-r}_{x} \hat{ \phi}_{\alpha}(x) \bigr)e^{-\frac{ i|\sigma |^{2}\cot\alpha}{2}} \,dx \\& \quad = C_{\alpha}(-1)^{|k|}\int_{\mathbb{R}^{n}}e^{i\langle x, \sigma\rangle \csc\alpha} \sum_{r\leq k}\binom{k}{r}\sum _{\beta\leq r}C_{\beta}(\cot \alpha) e^{-\frac{i|x|^{2} \cot\alpha}{2}}e^{-\frac{ i|\sigma|^{2}\cot\alpha }{2}} \bigl(x^{\beta}D^{k-r}_{x}\hat{\phi}_{\alpha}(x) \bigr)\,dx. \end{aligned}$$
Therefore,
$$\begin{aligned} \bigl\vert (\sigma\csc\alpha)^{k}\phi(s)\bigr\vert \leq& |C_{\alpha}|\sum_{r\leq k}\binom {k}{r}\sum _{\beta\leq r}\bigl\vert C_{\beta}(\cot\alpha)\bigr\vert \int_{\mathbb{R}^{n}} \bigl\vert e^{-\langle x, \tau\rangle\csc\alpha}\bigr\vert \bigl\vert x^{\beta}D^{k-r}_{x} \hat{\phi }_{\alpha}(x)\bigr\vert \,dx \\ \leq&|C_{\alpha}|\sum_{r\leq k} \binom{k}{r}\sum_{\beta\leq r}\bigl\vert C_{\beta }(\cot\alpha)\bigr\vert \int_{\mathbb{R}^{n}} e^{[|x||\tau\csc\alpha|-M[(\frac{1}{b}-\delta )x]]} \,dx. \end{aligned}$$
Now using the arguments of [2, p.21], we get
$$ \bigl\vert (\sigma\csc\alpha)^{k}\phi(\sigma)\bigr\vert \leq C_{\alpha,k} \exp\bigl[\Omega \bigl[(b+\rho)\tau\csc\alpha\bigr]\bigr], $$
where ρ is arbitrarily small together with δ. Thus we have
$$ F_{\alpha}^{-1}(W_{M,\frac{1}{b}})\subset W^{\Omega,b}. $$
(3.18)
Therefore (3.17) and (3.18) yield
$$ W^{\Omega,b,p} \subset W^{\Omega,b} . $$
(3.19)
Again we take \(e^{-\frac{i|z|^{2}\cot\alpha}{2}} \phi(z)\in W^{\Omega ,b}(\mathbb{C}^{n})\). Then from [8, Theorem 2.2], we have
$$ F_{\alpha}\bigl(W^{\Omega,b}\bigr)\subset W_{M,\frac{1}{b}}. $$
(3.20)
By the inverse property of the fractional Fourier transform, we have
$$ W^{\Omega,b}\subset F_{\alpha}^{-1}( W_{M,\frac{1}{b}}). $$
(3.21)
Furthermore, we take \(\hat{\phi}(x)\in W_{M,\frac{1}{b}}(\mathbb{R}^{n})\). Then from the arguments of [2, pp.20-21], we have
$$\begin{aligned}& {\bigl\Vert (i\sigma\csc\alpha)^{k}\phi(\sigma)\bigr\Vert _{p}} \\& \quad = \biggl( \int_{\mathbb{R}^{n}}\bigl\vert (i\sigma\csc \alpha)^{k}\phi(\sigma )\bigr\vert ^{p}\,dw \biggr)^{\frac{1}{p}} \\& \quad = \biggl( \int_{\mathbb{R}^{n}} \biggl( \frac{|i\sigma\csc\alpha|^{|k|+2} +|i\sigma\csc\alpha|^{|k|}}{(|w|^{2}+1)} \biggr)^{p} \bigl\vert \phi(\sigma )\bigr\vert ^{p}\,dw \biggr)^{\frac{1}{p}} \\& \quad \leq \biggl( \int_{\mathbb{R}^{n}} \biggl( \frac{|i\sigma\csc\alpha |^{|k|+2} }{(|w|^{2}+1)} \biggr)^{p} \bigl|\phi(\sigma)\bigr|^{p}\,dw \biggr)^{\frac {1}{p}} + \biggl( \int_{\mathbb{R}^{n}} \biggl( \frac{|i\sigma\csc\alpha |^{|k|}}{(|w|^{2}+1)} \biggr)^{p} \bigl\vert \phi(\sigma)\bigr\vert ^{p}\,dw \biggr)^{\frac {1}{p}} \\& \quad \leq \biggl( \int_{\mathbb{R}^{n}}\frac{dw}{(|w|^{2} +1)^{p}} \biggl( C_{\alpha,k}\int_{\mathbb{R}^{n}}\bigl\vert e^{-\langle x, \tau\rangle\csc\alpha} \hat{\phi}_{\alpha}(x)\bigr\vert \,dx \biggr)^{p} \biggr)^{\frac{1}{p}} \\& \qquad {} + \biggl( \int_{\mathbb{R}^{n}}\frac{dw}{(|w|^{2} +1)^{p}} \biggl( C'_{\alpha,k}\int_{\mathbb{R}^{n}}\bigl\vert e^{-\langle x, \tau\rangle\csc\alpha} \hat{\phi}_{\alpha}(x)\bigr\vert \,dx \biggr)^{p} \biggr)^{\frac{1}{p}} \\& \quad \leq \bigl(C_{\alpha,k} C_{k+2,\delta} +C'_{\alpha,k} C_{k,\delta}\bigr) \biggl( \int_{\mathbb{R}^{n}}\frac{dw}{(|w|^{2} +1)^{p}} \\& \qquad {}\times \biggl(\int_{\mathbb{R}^{n}} \biggl\vert \exp\biggl[-\langle x, \tau \rangle\csc\alpha-M\biggl[\biggl(\frac{1}{b}-\delta \biggr)x\biggr]\biggr] \biggr\vert \,dx \biggr)^{p} \biggr)^{\frac{1}{p}} \\& \quad \leq C_{\alpha,k,\delta} \biggl(\int_{\mathbb{R}^{n}}\biggl\vert \exp\biggl[-\langle x, \tau \rangle\csc\alpha-M\biggl[\biggl(\frac{1}{b}- \delta\biggr)x\biggr]\biggr]\biggr\vert \,dx \biggr) \biggl( \int _{\mathbb{R}^{n}}\frac{dw}{(|w|^{2} +1)^{p}} \biggr)^{\frac{1}{p}} \\& \quad \leq C_{\alpha,k,\delta} C_{p} \int_{\mathbb{R}^{n}} \biggl\vert \exp\biggl[-\langle x, \tau\rangle\csc\alpha-M\biggl[\biggl( \frac{1}{b}-\delta\biggr)x\biggr]\biggr]\biggr\vert \,dx. \end{aligned}$$
(3.22)
Now using the Young inequality (1.21) and from the arguments of [2, p.21], we get
$$ \bigl\Vert (i\sigma\csc\alpha)^{k}\phi(\sigma)\bigr\Vert _{p} \leq C_{\alpha ,k,\delta,p} \exp\bigl[\Omega\bigl[(b+ \rho)\tau\bigr] \bigr]. $$
This implies that
$$ F_{\alpha}^{-1}(W_{M,\frac{1}{b}})\subset W^{\Omega,b,p}. $$
(3.23)
Now (3.21) and (3.23) give
$$ W^{\Omega,b}\subset W^{\Omega,b,p}. $$
(3.24)
Finally, (3.19) and (3.24) give
$$ W^{\Omega,b}= W^{\Omega,b,p}. $$
(3.25)
□
Theorem 3.5
Let
\(\Omega_{0}(y)\)
and
\(M_{0}(x)\)
be the functions which are dual in the sense of Young to the functions
\(M(x)\)
and
\(\Omega(y)\), respectively. Then
$$ W^{\Omega,b,p}_{M,a}=W^{\Omega,b}_{M,a}, \quad 1\leq p< \infty. $$
(3.26)
Proof
Let \(e^{-\frac{i|x|^{2}\cot\alpha}{2}}\phi(x)\in W^{\Omega,b,p}_{M,a} \). Then from Theorem 2.3, it follows that
$$ F_{\alpha}\bigl(W^{\Omega,b,p}_{M,a}\bigr)\subset W^{\Omega_{0},\frac {1}{a}}_{M_{0},\frac{1}{b}}. $$
(3.27)
By the inverse property of the fractional Fourier transform we get
$$ W^{\Omega,b,p}_{M,a}\subset F_{\alpha}^{-1} \bigl(W^{\Omega_{0},\frac {1}{a}}_{M_{0},\frac{1}{b}}\bigr). $$
(3.28)
Now let \(\hat{\phi}_{\alpha}(z)\in W_{M_{0},\frac{1}{b}}^{\Omega _{0},\frac{1}{a}}\). Then from the arguments of [2, p.24], we get
$$ \phi(\sigma+ i\tau)= C_{\alpha}\int_{\mathbb{R}^{n}} e^{-\frac{i(|z|^{2} + |\sigma|^{2})\cot\alpha}{2}+i\langle\sigma, z\rangle\csc\alpha} \hat {\phi}_{\alpha}(z) \,dx. $$
Therefore,
$$\begin{aligned}& {\bigl\vert \phi(\sigma+ i\tau)\bigr\vert } \\& \quad \leq |C_{\alpha}|\int_{\mathbb{R}^{n}} \bigl\vert \exp \bigl[-\langle w, y\rangle\csc \alpha-\langle\tau, x\rangle\csc\alpha\bigr]\bigr\vert \bigl\vert \hat{\phi}_{\alpha }(z)\bigr\vert \,dx \\& \quad \leq C_{\alpha} C_{\delta,\rho}\int_{\mathbb{R}^{n}} \bigl\vert \exp\bigl[-\langle w, y\rangle\csc\alpha-\langle\tau, x\rangle\csc \alpha\bigr]\bigr\vert \\& \qquad {}\times\biggl\vert \exp\biggl[-M_{0}\biggl[\biggl( \frac{1}{b}-\delta\biggr)x\biggr]+\Omega_{0}\biggl[\biggl( \frac{1}{a}+\rho \biggr)y\biggr]\biggr]\biggr\vert \,dx \\& \quad \leq C_{\delta,\rho,\alpha} \exp\biggl[\Omega_{0}\biggl[\biggl( \frac{1}{a}+\rho \biggr)y\biggr]-\langle w, y\rangle\csc\alpha\biggr] \\& \qquad {}\times\int _{\mathbb{R}^{n}} \exp\biggl[-M_{0}\biggl[\biggl( \frac {1}{b}-\delta\biggr)x\biggr]+\langle\tau, x\rangle\csc\alpha\biggr] \,dx. \end{aligned}$$
Now using (1.21), we have
$$\begin{aligned} \bigl\vert \phi(\sigma+ i\tau)\bigr\vert \leq& C'_{\delta,\rho,\alpha} \exp \biggl[-M \biggl[\frac{w\csc\alpha}{\frac{1}{a}+ \rho} \biggr] + \Omega \biggl[ \frac{\tau \csc\alpha}{(\frac{1}{b}+ 2\delta)} \biggr] \biggr] \\ \leq&C'_{\delta,\rho,\alpha} \exp \biggl[ -M \biggl[\biggl( \frac{1}{a}-\delta _{0}\biggr)w\csc\alpha \biggr] +\Omega \biggl[\biggl(\frac{1}{b}+ \rho_{0}\biggr)\tau\csc \alpha \biggr] \biggr], \end{aligned}$$
where \(\rho_{0}\) and \(\delta_{0}\) are arbitrarily small together with ρ and δ, respectively. This shows that
$$ F_{\alpha}^{-1}\bigl(W^{\Omega_{0},b,}_{M_{0},a}\bigr) \subset W^{\Omega,b}_{M,a}. $$
(3.29)
Thus from (3.28) and (3.29), we get
$$ W^{\Omega,b,p}_{M,a}\subset W^{\Omega,b}_{M,a}. $$
(3.30)
Similarly it is easy to show that
$$ W^{\Omega,b}_{M,a}\subset W^{\Omega,b,p}_{M,a}. $$
(3.31)
Finally, (3.30) and (3.31) imply that
$$ W^{\Omega,b,p}_{M,a}=W^{\Omega,b}_{M,a}. $$
(3.32)
□