We define fractional transforms \(\mathscr{R}_\mu\) and \(\mathscr{H}_\mu\), \(\mu>0\) on the space \(\mathbb{R}\times\mathbb{R}^n\). First, we study these transforms on regular function spaces and we establish that these operators are topological isomorphisms and we give the inverse operators as integro differential operators. Next, we study the \(L^p\)-boundedness of these operators. Namely, we give necessary and sufficient condition on the parameter \(\mu\) for which the transforms \(\mathscr{R}_\mu\) and \(\mathscr{H}_\mu\) are bounded on the weighted spaces \(L^p([0,+\infty[\times\mathbb{R}^n,r^{2a}dr\otimes dx)\) and we give their norms.
Let \(D_j,\ 1\leq j \leq n\), and \(\Xi_\mu,\ \mu>0\), be the singular partial differential operators defined by \begin{eqnarray*}\left \{ \begin{array}{ll} D_j=\displaystyle\frac{\partial}{\partial x_j}\\ \Xi_\mu=\displaystyle(\frac{\partial}{\partial r})^2+\frac{2\mu}{r}\frac{\partial}{\partial r}+ \sum_{j=1}^n(\frac{\partial}{\partial x_j})^2 ; (r,x)\in ]0,+ \infty[\times\mathbb{R}^n, \mu>0. \end{array} \right. \end{eqnarray*} \(\Xi_\mu\) is a Bessel-Laplace operator.
When \(\mu=\frac{n-1}{2}\); \(n\in\mathbb{N}^\ast\), \(\Xi_{\frac{n-1}{2}}\) is the Laplacien operator on \(\mathbb{R}^n\times\mathbb{R}^n\) when acting on the functions \(f:\mathbb{R}^n\times\mathbb{R}^n\longrightarrow\mathbb{C}\), that are radial with respect to the first variable.
For every \((\lambda_0,\lambda)\in \mathbb{C}\times\mathbb{C}^n\), the system \begin{eqnarray*}\left \{ \begin{array}{lll}D_ju(r,x)=\displaystyle-i\lambda_ju(r,x), 1\leqslant j\leqslant n\\ \Xi_\mu u(r,x)=\displaystyle-(\lambda^2_0+\lambda^2)u(r,x)\\ \displaystyle u(0,0)=1, \frac{\partial}{\partial r}u(0,x)=0, \forall x\in\mathbb{R}^ n \end{array} \right. \end{eqnarray*} admits a unique solution given by
The eigenfunction \(\psi_{\lambda_0,\lambda}\) allows us to define the Fourier transform \(\widetilde{\mathscr{F}}_{\mu -\frac{1}{2}}\) connected with the operators \(D_j,\ 1\leqslant j\leqslant n\) and \(\Xi_\mu\) by
Also, many uncertainty principles have been cheked for this transform [11,12,13, 14].
On the other hand, the eigenfunction \(\psi_{\lambda_0,\lambda}\) admits the Poisson integral representation
On the other hand, we shall prove in the next section that for every integrable function \(f\) on \([0, +\infty[\times\mathbb{R}^n\) with respect to the measure \(d\nu_\mu(r,x)\) and for every bounded function \(g\) on \(\mathbb{R}\times\mathbb{R}^n\), even with respect to the first variable, we have the duality relation
where \(dm\) is the Lebesgue measure on \(]0, +\infty[\times\mathbb{R}^n\),
where \(\Lambda\) is the usual Fourier transform defined by \begin{eqnarray*} \Lambda(f)(\lambda_0, \lambda) &=& \int_0^\infty\int_{\mathbb{R}^n}f(r, x)\cos(\lambda_0 r)e^{-i\langle\lambda|x\rangle}dm(r, x), \end{eqnarray*} \(*\) is the convolution product associated with the Fourier transform \(\widetilde{\mathscr{F}}_{\mu-\frac{1}{2}},\)
\(*_o\) is the usual convolution product defined by \begin{eqnarray*} f*_o g(r, x) &=& \int_0^\infty\int_{\mathbb{R}^n}f(s, y)\sigma_{r,x}(g)(s,- y)dm(s, y) \end{eqnarray*} and \(\sigma_{r, x}\) is the usual translation operator given by
Next, we show that the fractional transform \(\mathscr{H}_\mu\) can be extended to \(\mu\in\ \mathbb{R}\) and that for every \(\mu\in\ \mathbb{R}\ ,\ \mathscr{H}_\mu\) is a topological isomorphism from the Schwartz’s space \(\mathscr{S}_e\big(\mathbb{R}\times\mathbb{R}^n\big)\) (the subspace of \(\mathscr{E}_e\big(\mathbb{R}\times\mathbb{R}^n\big)\) consisting of rapidly decreasing functions together with all their derivatives) onto itself whose inverse operator is \(\mathscr{H}_\mu^{-1}\ =\ \mathscr{H}_{- \mu} .\)
The precedent results imply in particular that \(\mathscr{R}_\mu\) and \(\mathscr{H}_\mu\) are transmutation operators of \(D_j,\ 1\leq j\leq n\), and \(\Xi_\mu\) to \(D_j,\ 1\leq j\leq n\) and \(\Delta\), where \begin{eqnarray*} \Delta &=&( \frac{\partial}{\partial r})^2+ \sum_{j=1}^n(\frac{\partial}{\partial x_j})^2. \end{eqnarray*} That is, for every \(f\in \mathscr{E}_e\big(\mathbb{R}\times\mathbb{R}^n\big)\) \begin{eqnarray*} D_j\mathscr{R}_\mu(f)&=&\mathscr{R}_\mu D_j(f),\ 1\leqslant j\leqslant n \\ \Xi_\mu \mathscr{R}_\mu(f) &=& \mathscr{R}_ \mu\ \Delta( f), \end{eqnarray*} and for every \(f\in \mathscr{S}_e \big(\mathbb{R}\times\mathbb{R}^n\big)\) \begin{eqnarray*} D_j\mathscr{H}_\mu(f)&=&\mathscr{H}_\mu D_j(f),\ 1\leqslant j\leqslant n \\ \Delta \mathscr{H}_\mu(f) &=& \mathscr{H}_\mu\ \Xi_\mu(f). \end{eqnarray*}
The third section contains the main results of this paper. In fact, we study the \(L^p-\) boundedness of the operators \(\mathscr{R}_\mu\) and \(\mathscr{H}_\mu\) on the weighted spaces \(L^p\big([0, +\infty[\times\mathbb{R}^n, r^{2a}dr \otimes dx\big),\ p\in\ [1, +\infty].\) We recall in this context, that studing the \(L^p-\) boundedness of integral transforms connected with differential systems is an interesting subject because knowing the range of parameters \(\mu,\ p\) for which an operator is bounded on Lebesgue space gives quantitative information about the rate of growth of the transformed functions [15,16, 17].
In this work, we give necessary and sufficient conditions on the parameters \(\mu,\ a,\ p\) for which the operator \(\mathscr{R}_\mu\) (respectively \(\mathscr{H}_\mu\)) satisfies Lemma 2.1.
i.. For every \(\mu>0 \), the transform \(\mathscr{R}_\mu\) is continuous from \(\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) into itself.
ii. The operator \(\displaystyle \frac{\partial}{\partial r^2}=\frac{1}{r}\frac{\partial}{\partial r}\) is continuous from
\(\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) into itself.
Proof.
i.. For every \(f\in \mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\), we have
\begin{eqnarray*}
\mathscr{R}_\mu(f)(r,x)&=&\frac{2\Gamma(\mu+\frac{1}{2})}{\sqrt{\pi}\ \Gamma(\mu)}\int_0^1(1-t^2)^{\mu-1}f(tr,x)dt,
\end{eqnarray*}
this shows that the function \(\mathscr{R}_\mu(f)\) belongs to the
space \(\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\). Moreover, for
every \((\alpha_0,\alpha)\in \mathbb{N}\times\mathbb{N}^n\)
\begin{eqnarray*}
D^{(\alpha_0,\alpha)}(\mathscr{R}_\mu(f))(r,x)&=&\frac{2\Gamma(\mu+\frac{1}{2})}{\sqrt{\pi}\ \Gamma(\mu)}\int_0^1(1-t^2)^{\mu-1}t^{\alpha_0}
D^{(\alpha_0,\alpha)}(f)(tr,x)dt,
\end{eqnarray*}
thus, for every \((m,k) \in \mathbb{N}^2, P_{m,k}(\mathscr{R}_\mu(f))\leqslant P_{m,k}(f).\)
ii.. For every \(f\in \mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\)
\begin{eqnarray*}
\displaystyle\frac{\partial}{\partial
r^2}(f)(r,x)&=&\int_0^1\frac{\partial^2 f}{\partial t^2}(rt,x)dt.
\end{eqnarray*}
Hence, the function \(\displaystyle \frac{\partial}{\partial
r^{2}}(f)\) belongs to the space
\(\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) and for every
\((\alpha_0,\alpha)\in \mathbb{N}\times\mathbb{N}^n\)
\begin{eqnarray*}
\displaystyle D^{(\alpha_{0},\alpha)}(\frac{\partial}{\partial r^{2}}f)(r,x)&=&\int_{0}^{1}t^{\alpha_0}D^{(\alpha_0+2,\alpha)}(f)(rt,x)dt,
\end{eqnarray*}
so, for every \((m,k) \in \mathbb{N}^{2},
P_{m,k}\big(\frac{\partial}{\partial r^{2}}(f)\big)\leqslant
P_{m,k+2}(f).\)
In the following, we shall prove that \(\mathscr{R}_\mu\) is a topological isomorphism from \(\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) onto itself and we give the inverse operator. For this we need the notations $$r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)$$ is the space defined by $$r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)=\big\{ f:\mathbb{R}\backslash\{0\}\times\mathbb{R}^n\longrightarrow\mathbb{C},f$$ is even with respect to the first variable and \(f(r,x)=r^{2a}g(r,x),\ g\in\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\big\} \) $$r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)$$ is equipped by the family of semi-norms $$\widetilde{P}_{m,k,a}(f)=P_{m,k}(r^{-2a}f).$$ \(\widetilde{\mathscr{R}_\mu}\) is the transform defined on \(r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n),\ a>-\frac{1}{2},\) by \begin{eqnarray*} \widetilde{\mathscr{R}_\mu}(f)(r,x)&=& \frac{2r}{2^\mu\ \Gamma(\mu)}\int_0^r(r^2-t^2)^{\mu-1}f(t,x)dt,\ r>0. \end{eqnarray*}
Proposition 2.2.
i. For every \(a>-\frac{1}{2}\), the operator \(\Box\) defined by
\begin{eqnarray*}
\Box(f)(r,x)&=&\frac{\partial}{\partial
r}\big(\frac{f(r,x)}{r}\big)
\end{eqnarray*}
is continuous from \(r^{2(a+1)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) into
\(r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\).
ii. The transform \(\widetilde{\mathscr{R}_\mu}\)
is continuous from
\(r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) into
\(r^{2(a+\mu)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\).
Proof.
i. Let \(f \in r^{2(a+1)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n);
f(r,x)=r^{2a+2}g(r,x), g\in\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n) \)
$$\Box f(r,x)=r^{2a}\big((2a+1)g(r,x)+r\frac{\partial g}{\partial r}(r,x)\big).$$
Since, the map \( :g\longrightarrow(2a+1)g+\displaystyle
r\frac{\partial g}{\partial r}\) is continuous from
\(\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\)
into itself, then, the function \(\Box (f)\) belongs to \(r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n).\)
Moreover, for every \((m,k)\in\mathbb{N}^2\)
\begin{eqnarray*}
\widetilde{P}_{m,k,a}(\Box(f)) &=& P_{m,k}\big((2a+1)g+r\frac{\partial g}{\partial r}\big) \\
&\leqslant & C P_{m’,k’}(g)=C \widetilde{P}_{m’,k’,a+1}(f),
\end{eqnarray*}
where \(C\) is a constant.
ii. For every \(f\in r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n), f=r^{2a}g,\ g\in \mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\ \mbox{and}\ a>-\frac{1}{2}\),
the function
\begin{eqnarray*}
\widetilde{\mathscr{R}_\mu}(f)(r,x) &=& \frac{2r}{2^\mu\ \Gamma(\mu)}\int_0^r(r^2-t^2)^{\mu-1}t^{2a}g(t,x)dt \\
&= &\frac{2r^{2a+2\mu}}{2^\mu\ \Gamma(\mu)}\int_0^1(1-t^2)^{\mu-1}t^{2a}g(tr,x)dt
\end{eqnarray*}
belongs to the space \(r^{2(a+\mu)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\), and for every \((m,k)\in \mathbb{N}^2\)
\begin{eqnarray*}
\widetilde{P}_{m,k,a+\mu}(\widetilde{\mathscr{R}}_\mu(f)) &=&P_{m,k}\big(\frac{2}{2^\mu\ \Gamma(\mu)} \int_0^1(1-t^{2})^{a-1}t^{2a}g(tr,x)dt \big)\\
&\leqslant &\frac{\Gamma(a+\frac{1}{2})}{2^\mu\ \Gamma(\mu+a+\frac{1}{2})}P_{m,k}(g)\\
&= &\frac{\Gamma(a+\frac{1}{2})}{2^\mu\ \Gamma(\mu+a+\frac{1}{2})} \widetilde{P}_{m,k,a}(f).
\end{eqnarray*}
Proposition 2.3. For all \(\mu,\nu > 0\) and \(f\in r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n),\ a>-\frac{1}{2}\), we have \begin{eqnarray*} \displaystyle \widetilde{\mathscr{R}_\mu}\circ\widetilde{\mathscr{R}_\nu}(f) &=& \widetilde{\mathscr{R}}_{\mu+\nu}(f). \end{eqnarray*}
Proof.
For all \(\mu\ ,\nu > 0\) and \(f\in
r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n),\
a>-\frac{1}{2},\)
\begin{eqnarray*}
\widetilde{\mathscr{R}_\mu}\circ\widetilde{\mathscr{R}_\nu}(f)(r,x)= \frac{2r}{2^{\mu+\nu}\ \Gamma(\mu)\ \Gamma(\nu)} \int_0^r(r^2-t^2)^{\mu-1}2t
\Big(\int_0^t(t^2-s^2)^{\nu-1}f(s,x)ds\Big)dt.
\end{eqnarray*}
Applying Fubini’s theorem we get
\begin{eqnarray*}
\widetilde{\mathscr{R}_\mu}\circ\widetilde{\mathscr{R}_\nu}(f)(r,x)= \frac{2r}{2^{\mu+\nu}\ \Gamma(\mu)\ \Gamma(\nu)} \int_0^rf(s,x)
\Big(\int_s^r(r^2-t^2)^{\mu-1}(t^2-s^2)^{\nu-1}2tdt\Big)ds,
\end{eqnarray*}
however, \(\displaystyle\int_s^r(r^2-t^2)^{\mu-1}(t^2-s^2)^{\nu-1}2tdt=\displaystyle\frac{\Gamma(\mu)\ \Gamma(\nu)}{\Gamma(\mu+\nu)}(r^2-s^2)^{\mu+\nu-1}.\)
This completes the proof.
Proposition 2.4. i. For every \(\mu>1\) and \(f\in r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n),\ a>-\frac{1}{2}\), we have \begin{eqnarray*} \displaystyle \Box\widetilde{\mathscr{R}_\mu}(f)&=&\widetilde{\mathscr{R}}_{\mu-1}(f). \end{eqnarray*} In particular, for every \(\mu>0,\ k\in\mathbb{N}\)
Proof.
i. Let \(f\in r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n),\)
\begin{eqnarray*}
\Box\widetilde{\mathscr{R}\mu}(f)(r,x)&=& \frac{\partial}{\partial r}\big(\frac{2}{2^\mu\ \Gamma(\mu)} \int_0^r(r^2-t^2)^{\mu-1}f(t,x)dt\big)\\
&=&\frac{2.2r(\mu-1)}{2^\mu\ \Gamma(\mu)} \int_0^r(r^2-t^2)^{\mu-2}f(t,x)dt\\
&=&\widetilde{\mathscr{R}}_{\mu-1}(f)(r,x),
\end{eqnarray*}
and by induction, we deduce that for all \(\mu>0,\ k\in\mathbb{N}\)
\begin{eqnarray*}
\Box^k\widetilde{\mathscr{R}}_{\mu+k}(f)&=&\widetilde{\mathscr{R}_\mu}(f).
\end{eqnarray*}
ii. Let \(f\in r^{2(a+1)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\),
by Proposition 2.2, the function \(\Box (f)\) belongs to
the space \(r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) and
we have
\begin{eqnarray*}
\widetilde{\mathscr{R}_\mu}(f)(r,x)=\frac{r}{2^\mu\
\Gamma(\mu+1)}\int_0^r-\displaystyle\frac{\partial}{\partial
t}\big((r^2-t^2)^\mu\big)\frac{f(t,x)}{t}dt.
\end{eqnarray*}
Integrating by parts, we get
\begin{eqnarray*}
\widetilde{\mathscr{R}_\mu}(f)(r,x)=\frac{r}{2^\mu\
\Gamma(\mu+1)}\int_0^r(r^2-t^2)^{\mu}\Box f(t,x)dt,
\end{eqnarray*}
so,
\begin{eqnarray*}
\Box\widetilde{\mathscr{R}_\mu}(f)(r,x)&=&\frac{2r}{2^\mu\ \Gamma(\mu)}\int_0^r(r^2-t^2)^{\mu-1}\Box f(t,x)dt\\
&=&\label{2.2}\widetilde{\mathscr{R}_\mu}(\Box f)(r,x).
\end{eqnarray*}
Now, suppose that for every \(f\in
r^{2(a+k)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n),\)
\(\Box^k\widetilde{\mathscr{R}_\mu}(f)=\widetilde{\mathscr{R}_\mu}(\Box^k f)\),
let \(g\in r^{2(a+k+1)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n).\)
Then, the function \(\Box g\) belongs to \(
r^{2(a+k)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\), and by
hypothesis
\begin{eqnarray*}
\Box^k\widetilde{\mathscr{R}_\mu}(\Box
g)(r,x)=\widetilde{\mathscr{R}_\mu}(\Box^{k+1} g),
\end{eqnarray*}
on the other hand, by relation(15) and the fact that
\(\Box g\in
r^{2(a+k)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\subset
r^{2(a+1)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\), we have
\begin{eqnarray*}
\Box^k\widetilde{\mathscr{R}_\mu}(\Box
g)(r,x)=\Box^{k+1}\widetilde{\mathscr{R}_\mu}( g).
\end{eqnarray*}
The proof is complete by induction.
Theorem 2.5.
For every \(k\in\mathbb{N}\backslash\{0\}\),
the operator \(\widetilde{\mathscr{R}_k}\) is an isomorphism from \(r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) onto
\(r^{2(a+k)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n);\ a>-\frac{1}{2}.\)
The inverse operator is given by
\begin{eqnarray*}
\displaystyle \widetilde{\mathscr{R}_k}^{-1}=\Box^k.
\end{eqnarray*}
Proof. Let \(f\in r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n).\) From Proposition 2.2, the function \(\widetilde{\mathscr{R}_k}(f)\) belongs to \(r^{2(a+k)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) and by relation(14), we have \begin{eqnarray*} \Box^k\widetilde{\mathscr{R}_k}(f)&=&\Box\Box^{k-1}\widetilde{\mathscr{R}}_{1+(k-1)}(f)\\ &=&\Box\widetilde{\mathscr{R}_1}(f)\\ &=& f. \end{eqnarray*} Let \(g \in r^{2(a+k)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\subset r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\), by relation(16) \begin{eqnarray*} \widetilde{\mathscr{R}_k}(\Box^k(g))&=&\Box^k\widetilde{\mathscr{R}_k}(g)\\ &=& g. \end{eqnarray*} This achieves the proof.
Theorem 2.6. For every \(\mu\in ]0,1[\), the fractional transform \(\widetilde{\mathscr{R}_\mu}\) is an isomorphism from \(r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) onto \(r^{2(a+\mu)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n), a>-\frac{1}{2}.\) The inverse operator is given by $$\widetilde{\mathscr{R}_\mu}^{-1}=\Box\widetilde{\mathscr{R}}_{1-\mu}.$$
Proof. Let \(g\in r^{2(a+\mu)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\), $$ g(r,x)=r^{2a+2\mu}h(r,x); \ h\in \mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n), $$ \begin{eqnarray*} \Box\widetilde{\mathscr{R}}_{1-\mu}(g)(r,x)&=& \frac{\partial}{\partial r}\Big(\frac{2}{2^{1-\mu}\Gamma(1-\mu)}\int_0^r(r^2-t^2)^{-\mu}t^{2a+2\mu}h(t,x)dt\Big)\\ &=& \frac{\partial}{\partial r}\Big(\frac{2r^{2a+1}}{2^{1-\mu}\Gamma(1-\mu)}\int_0^1(1-t^2)^{-\mu}t^{2a+2\mu}h(tr,x)dt\Big)\\ &=& 2(2a+1)\frac{r^{2a}}{2^{1-\mu}\ \Gamma(1-\mu)}\int_0^1(1-t^{2})^{-\mu}t^{2a+2\mu}h(tr,x)dt \\&+& 2\frac{r^{2a+1}}{2^{1-\mu}\Gamma(1-\mu)}\int_0^1(1-t^2)^{-\mu}t^{2a+2\mu+1}\frac{\partial h}{\partial t}(tr,x)dt\\ &=& 2\frac{(2a+1)}{2^{1-\mu}\Gamma(1-\mu)}\frac{1}{r}\int_0^r(r^2-t^2)^{-\mu}t^{2a+2\mu}h(t,x)dt \\&+& \frac{2}{2^{1-\mu}\ \Gamma(1-\mu)}\frac{1}{r}\int_0^r(r^2-t^2)^{-\mu}t^{2a+2\mu+1}\frac{\partial h}{\partial t}(t,x)dt. \end{eqnarray*} We deduce that \begin{eqnarray*} \widetilde{\mathscr{R}}_\mu\Big(\Box \widetilde{\mathscr{R}}_{1-\mu}(g)\Big)(r,x)&=& \frac{2(2a+1)2r}{2\Gamma(\mu)\ \Gamma(1-\mu)}\int_0^r(r^2-t^2)^{\mu-1}\frac{1}{t}\Big(\int_0^t(t^2-s^2)^{-\mu}s^{2a+2\mu}h(s,x)ds\Big)dt\\ &+& \frac{2.2r}{2\Gamma(\mu)\ \Gamma(1-\mu)}\int_0^r(r^2-t^2)^{\mu-1}\frac{1}{t}\Big(\int_0^t(t^2-s^2)^{-\mu}s^{2a+2\mu+1}\frac{\partial h}{\partial s}(s,x)ds\Big)dt\\&=&I_{1,\mu}(r,x)+I_{2,\mu}(r,x). \end{eqnarray*} From Fubini’s theorem, we have $$I_{1,\mu}(r,x)= \frac{(2a+1)r}{\Gamma(\mu)\ \Gamma(1-\mu)}\int_0^rh(s,x)\Big(\int_s^r(r^2-t^2)^{\mu-1}(t^2-s^2)^{-\mu}\frac{2t}{t^2}dt\Big)s^{2a+2\mu}ds.$$ Let $$J(r,s)=\int_s^r(r^2-t^2)^{\mu-1}(t^2-s^2)^{-\mu}\frac{2t}{t^2}dt.$$ By the change of variables \(\omega=\frac{r^2-t^2}{r^2-s^2},\) we get \begin{eqnarray*} J(r,s)&=&\frac{1}{r^2}\int_0^1\frac{\omega^{\mu-1}(1-\omega)^{-\mu}}{1-\frac{r^2-s^2}{r^2}\omega}d\omega \\ &=&\frac{1}{r^2}\sum_{k=0}^\infty(\frac{r^2-s^2}{r^2})^k\int_0^1\omega^{k+\mu-1}(1-\omega)^{-\mu}d\omega \\&=&\frac{\Gamma(1-\mu)}{r^2}\sum_{k=0}^\infty\frac{\Gamma(k+\mu)}{k!}(\frac{r^2-s^2}{r^2})^k \\&=&\Gamma(\mu)\ \Gamma(1-\mu)r^{2\mu-2}s^{-2\mu}. \end{eqnarray*} So, $$I_{1,\mu}(r,x)= (2a+1)r^{2\mu-1}\int_0^r h(s,x)s^{2a} ds$$ As the same way, \begin{eqnarray*} I_{2,\mu}(r,x)&=& \frac{r}{\Gamma(\mu)\ \Gamma(1-\mu)}\int_0^r\displaystyle\frac{\partial h}{\partial s}(s,x)\big(\int_s^r(r^2-t^2)^{\mu-1}(t^2-s^2)^{-\mu}\frac{2t}{t^2}dt\big)s^{2a+2\mu+1}ds\\ &=&r^{2\mu-1}\int_0^r\frac{\partial h}{\partial s}(s,x)s^{2a+1} ds. \end{eqnarray*} Consequently, \begin{eqnarray*} \widetilde{\mathscr{R}}_\mu\big(\Box \widetilde{\mathscr{R}}_{1-\mu}(g)\big)(r,x)&=& r^{2\mu-1}\int_0^r\Big((2a+1)s^{2a}h(s,x)+s^{2a+1}\frac{\partial h}{\partial s}(s,x)\Big)ds\\ &=&r^{2\mu-1}\int_0^r\frac{\partial }{\partial s}\big(s^{2a+1}h(s,x)\big) ds\\&=&r^{2a+2\mu}h(r,x),\hbox{ because}\ a>-\frac{1}{2}\\&=&g(r,x). \end{eqnarray*} On the other hand, from Proposition 2.3 and for every \(f\in r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\), \begin{eqnarray*} \Box\widetilde{\mathscr{R}}_{1-\mu}\widetilde{\mathscr{R}}_\mu(f)&=& \Box\widetilde{\mathscr{R}}_1(f)\\&=&f. \end{eqnarray*} This completes the proof.
Lemma 2.7. Let \(\mu\in\mathbb{R},\ \mu\geqslant0.\) For every \(k_1,\ k_2 \in \mathbb{N}\backslash \{0\},\ k_1-\mu>0,\ k_2-\mu>0\) and for every \(f\in r^{2(a+\mu)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\), we have $$ \Box^{k_1}\widetilde{\mathscr{R}}_{k_1-\mu}(f)= \Box^{k_2}\widetilde{\mathscr{R}}_{k_2-\mu}(f) .$$
Proof. Let \(k_1, \ k_2 \in \mathbb{N}\backslash \{0\},\ k_1-\mu>0,\ k_2-\mu>0\), and $k_1< k_2 ,$ $$\Box^{k_2}\widetilde{\mathscr{R}}_{k_2-\mu}(f)=\Box^{k_1}\Box^{k_2-k_1}\widetilde{\mathscr{R}}_{k_2-k_1+(k_1-\mu)}(f), $$ applying relation (14), we get $$\Box^{k_2}\widetilde{\mathscr{R}}_{k_2-\mu}(f)= \Box^{k_1}\widetilde{\mathscr{R}}_{k_1-\mu}(f).$$
The previous Lemma allows us to define the fractional transform \(\widetilde{\mathscr{R}_\mu}\) for every \(\mu\in\mathbb{R}.\) Definition 2.8.
For every \(\mu\in\mathbb{R},\ \mu\geqslant0\), the fractional
transform \(\widetilde{\mathscr{R}_{-\mu}}\) is defined on
\(r^{2(a+\mu)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) by
$$\widetilde{\mathscr{R}_{-\mu}}(f)=\Box^k\widetilde{\mathscr{R}}_{k-\mu}(f),$$
where \( k \in \mathbb{N}\backslash \{0\},\ k-\mu>0.\)
In particular, for \(f \in r^{2(a+\mu)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\)
$$\widetilde{\mathscr{R}_{-\mu}}(f)=\Box^{E(\mu)+1}\widetilde{\mathscr{R}}_{E(\mu)+1-\mu}(f),$$
where \(E(\mu)\) is the entire party of \(\mu .\)
Remark 2.9. According to Definition 2.8 and for every \(f \in r^{2a}\mathscr{E}_{e}(\mathbb{R}\times\mathbb{R}^{n}), \ a>-\frac{1}{2}\), we have $$\widetilde{\mathscr{R}_0}(f)=\Box\widetilde{\mathscr{R}_1}(f)=f,$$ that is $$\widetilde{\mathscr{R}_0}=Id_{ r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)}.$$
Theorem 2.10.
For \(\mu>0\), the fractional transform \(\widetilde{\mathscr{R}_\mu}\) is a topological isomorphism from
\(r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) onto \(r^{2(a+\mu)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n),\ a>-\frac{1}{2}.\)
The inverse operator is given by
$$\widetilde{\mathscr{R}_\mu}^{-1}=\widetilde{\mathscr{R}_{-\mu}}.$$
Proof.
For \(\mu\in\mathbb{N}\), the result follows from Theorem 2.5 and Remark 2.9.
Let \(\mu\in]0,+\infty[\backslash\mathbb{N}\),
for every \(f\in r^{2a}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\)
and from Proposition 2.3 and Theorem 2.5, we have
\begin{eqnarray*}
\widetilde{\mathscr{R}_{-\mu}}\big(\widetilde{\mathscr{R}_\mu}(f)\big)&=&
\Box^{E(\mu)+1}\widetilde{\mathscr{R}}_{E(\mu)+1-\mu}\big(\widetilde{\mathscr{R}}_\mu(f)\big)\\
&=& \Box^{E(\mu)+1}\widetilde{\mathscr{R}}_{E(\mu)+1}(f)\\
&=&f.
\end{eqnarray*}
Conversely, for every \(g\in r^{2(a+\mu)}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n),\)
$$\widetilde{\mathscr{R}_\mu}\circ\widetilde{\mathscr{R}_{-\mu}}(g)=\widetilde{\mathscr{R}_\mu}\Box^{E(\mu)+1}\widetilde{\mathscr{R}}_{E(\mu)+1-\mu}(g),$$
let \(\nu=\mu-E(\mu)\), then \(\nu\in]0,1[\), and
$$\widetilde{\mathscr{R}_\mu}\circ\widetilde{\mathscr{R}_{-\mu}}(g)=\widetilde{\mathscr{R}_\nu}\widetilde{\mathscr{R}}_{E(\mu)}
\Box^{E(\mu)}\Box\widetilde{\mathscr{R}}_{1-\nu}(g). $$ Since,
\(\Box\widetilde{\mathscr{R}}_{1-\nu}(g)\) belongs to
\(r^{2(a+E(\mu))}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n),\) then,
Theorem 2.5 involves that
$$\widetilde{\mathscr{R}}_\mu\circ\widetilde{\mathscr{R}}_{-\mu}(g)=\widetilde{\mathscr{R}}_\nu \Box\widetilde{\mathscr{R}}_{1-\nu}(g).$$
The result follows from Theorem 2.6.
Now, we have the following important result.
Theorem 2.11. For every \(\mu>0\), the fractional transform \(\mathscr{R}_\mu\) defined by relation (5) is a topological isomorphism from \(\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) onto itself.
Proof.
For every \(f\in\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\),
$$\mathscr{R}_\mu(r,x)=\frac{2^\mu\ \Gamma(\mu+\frac{1}{2})}{\sqrt{\pi}}r^{-2\mu}\widetilde{\mathscr{R}_\mu}(f)(r,x).$$
From Theorem 2.10, the transform
\(\widetilde{\mathscr{R}_\mu}\) is a topological isomorphism from
\(\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\)
onto \(r^{2\mu}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\).
On the other hand, the map
$$f\longrightarrow r^{-2\mu}f$$
is a topological isomorphism from \(r^{2\mu}\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) onto \(\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) .
Consequently, \(\mathscr{R}_\mu\) is a topological isomorphism from \(\mathscr{E}_e(\mathbb{R}\times\mathbb{R}^n)\) onto itself.
Moreover,
\begin{eqnarray*}
\mathscr{R}_\mu^{-1}(f)(r,x)&=&
\frac{\sqrt{\pi}}{2^{\mu}\ \Gamma(\mu+\frac{1}{2})}\widetilde{\mathscr{R}_{-\mu}}\big(r^{2\mu}f)(r,x\big)\\
&=& \frac{\sqrt{\pi}}{2^\mu\
\Gamma(\mu+\frac{1}{2})}\Box^{E(\mu)+1}\widetilde{\mathscr{R}}_{E(\mu)+1-\mu}\big(r^{2\mu}f\big)(r,x).
\end{eqnarray*}
Lemma 2.12. For every \(f\in L^1(d\nu_\mu)\) and \(\mu>0\), the function $$ \mathscr{H}_\mu(f)(t,x)=\frac{1}{2^\mu\ \Gamma(\mu)}\int_t^\infty(r^2-t^2)^{\mu-1}f(r,x)2rdr,$$ is defined almost every where, belongs to \(L^1(dm)\), where \(dm\) is the Lebesgue measure given by relation (8), and we have $$ ||\mathscr{H}_\mu(f)||_{1,m}\leqslant||f||_{1,\nu_\mu}.$$
Proof. By Fubini-Tonnelli Theorem’s, we have \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \int_0^\infty\int_{\mathbb{R}^n}|\mathscr{H}_\mu(f)(t,x)|dm(t,x)&\leqslant& \sqrt{\frac{2}{\pi}} \frac{1}{2^\mu\ \Gamma(\mu)(2\pi)^{\frac{n}{2}}}\int_0^\infty\int_{\mathbb{R}^n}\Big(\int_t^\infty(r^2-t^2)^{\mu-1}|f(r,x)|2rdr\Big)dtdx\\&=& \sqrt{\frac{2}{\pi}} \frac{1}{2^\mu\ \Gamma(\mu)(2\pi)^{\frac{n}{2}}}\int_0^\infty\int_{\mathbb{R}^n}|f(r,x)|\Big(\int_0^r(r^2-t^2)^{\mu-1}dt\Big)2rdrdx\\&=& \frac{1}{2^{\mu-\frac{1}{2}}\Gamma(\mu+\frac{1}{2})(2\pi)^{\frac{n}{2}}}\int_0^\infty \int_{\mathbb{R}^n}|f(r,x)|r^{2\mu}drdx\\&=&\|f\|_{1,\nu_\mu}. \end{eqnarray*}
Proposition 2.13. i. For every \(f\in L^1(d\nu_\mu)\) and every bounded measurable function \(g\) on \([0,+\infty[\times\mathbb{R}^n\), we have the duality relation $$ \int_0^\infty\int_{\mathbb{R}^n}f(r,x)\mathscr{R}_\mu(g)(r,x)d\nu_\mu(r,x)= \int_0^\infty\int_{\mathbb{R}^n}\mathscr{H}_\mu(f)(r,x)g(r,x)dm(r,x). $$ ii. For every \(f\in L^1(d\nu_\mu)\)
Proof.
i. It is clear that for every bounded function \(g\) on \([0,+\infty[\times\mathbb{R}^n\), the function \(\mathscr{R}_\mu(g)\) is also bounded
on \([0,+\infty[\times\mathbb{R}^n\).
Consequently, the integral \(\displaystyle\int_0^\infty\int_{\mathbb{R}^n}f(r,x)\mathscr{R}_\mu(g)(r,x)d\nu_\mu(r,x)\) is well defined, and we have
\begin{eqnarray*}
\int_0^\infty\int_{\mathbb{R}^n}f(r,x)\mathscr{R}_\mu(g)(r,x)d\nu_\mu(r,x)&=&
\int_0^\infty\int_{\mathbb{R}^n}f(r,x)\frac{2r}{2^{\mu-\frac{1}{2}}\sqrt{\pi} \ (2\pi)^{\frac{n}{2}}\Gamma(\mu)}
\\&\times &\Big(\int_0^r(r^2-t^2)^{\mu-1}g(t,x)dt\Big)drdx.
\end{eqnarray*}
By Fubini’s Theorem,
\begin{eqnarray*}
\int_0^\infty\int_{\mathbb{R}^n}f(r,x)\mathscr{R}_\mu(g)(r,x)d\nu_\mu(r,x)&=&
\int_0^\infty\int_{\mathbb{R}^n}g(t,x)\big(\frac{1}{2^\mu\ \Gamma(\mu)}\int_t^\infty(r^2-t^2)^{\mu-1}f(r,x)2rdr\big)\\& \times&\sqrt{\frac{2}{\pi}}
dt\frac{dx}{(2\pi)^{\frac{n}{2}}}\\&=&
\int_0^\infty\int_{\mathbb{R}^n}g(t,x)\mathscr{H}_\mu(f)(t,x)dm(t,x).
\end{eqnarray*}
ii.
Let \(f\in L^1(d\nu_\mu)\), we have
$$\widetilde{\mathscr{F}}_{\mu-\frac{1}{2}}(f)(\lambda_0,\lambda)=\int_0^\infty\int_{\mathbb{R}^n}f(r,x)\Psi_{\lambda_0,\lambda}(r,x)d\nu_\mu(r,x)$$
and by the relation (6),
$$\widetilde{\mathscr{F}}_{\mu-\frac{1}{2}}(f)(\lambda_0,\lambda)=\int_0^\infty\int_{\mathbb{R}^n}f(r,x)\mathscr{R}_\mu\big(\cos(\lambda_0.)
e^{-i\langle\lambda|.\rangle}\big)(r,x)d\nu_\mu(r,x),$$
and by the relation of duality, Proposition 2.13, we obtain
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
\widetilde{\mathscr{F}}_{\mu-\frac{1}{2}}(f)(\lambda_0,\lambda)&=&\int_0^\infty\int_{\mathbb{R}^n}\mathscr{H}_\mu(f)(r,x)\cos(\lambda_0r)
e^{-i\langle\lambda|x\rangle} dm(r,x)\\&=&\Lambda\circ\mathscr{H}_\mu(f)(\lambda_0,\lambda).
\end{eqnarray*}
Corollary 2.14. For every \(\mu >0\), the fractional transform \(\mathscr{H}_\mu\) is a topological isomorphism from \(\mathscr{S}_e(\mathbb{R}\times\mathbb{R}^n)\) onto itself.
Proof. Since the Fourier transforms \(\Lambda\) and \(\widetilde{\mathscr{F}}_{\mu-\frac{1}{2}}\) are topological isomorphisms from \(\mathscr{S}_e(\mathbb{R}\times\mathbb{R}^n)\) onto itself, the result follows from the relation (18).
Next, we will prove that the fractional transform \(\mathscr{H}_\mu\) can be extended to \(\mu\in \mathbb{R}\) and we give the inverse operator \(\mathscr{H}_\mu^{-1}.\)Proposition 2.15. For every \(\mu,\ \nu >0\) and \(f\in\mathscr{S}_e(\mathbb{R}\times\mathbb{R}^n),\) \begin{eqnarray*} \mathscr{H}_\mu\circ\mathscr{H}_\nu(f)=\mathscr{H}_{\mu+\nu}(f). \end{eqnarray*}
Proof. Let \(\mu,\ \nu >0\) and \(f\in\mathscr{S}_e(\mathbb{R}\times\mathbb{R}^n)\) $$\mathscr{H}_\mu\circ\mathscr{H}_\nu(f)(r,x)=\frac{1}{2^{\mu+\nu}\ \Gamma(\mu)\Gamma(\nu)}\int_r^\infty(t^2-r^2)^{\mu-1} \big(\int_t^{+\infty}(s^2-t^2)^{\nu-1}f(s,x)2sds\big)2tdt.$$ Applying Fubini’s Theorem we get $$\mathscr{H}_\mu\circ\mathscr{H}_\nu(f)(r,x)=\frac{1}{2^{\mu+\nu}\ \Gamma(\mu)\Gamma(\nu)}\int_r^\infty f(s,x) \big(\int_r^s(s^2-t^2)^{\nu-1}(t^2-r^2)^{\mu-1}2tdt\big)2sds,$$ however,$$\int_r^s(s^2-t^2)^{\nu-1}(t^2-r^2)^{\mu-1}2tdt=\frac{\Gamma(\mu)\ \Gamma(\nu)}{\Gamma(\mu+\nu)}(s^2-r^2)^{\mu+\nu-1},$$ this completes the proof.
Proposition 2.16. i. For every \(f\in\mathscr{S}_e(\mathbb{R}\times\mathbb{R}^n)\) and \(\mu>0\), we have
Proof. i. Integrating by parts, we get for every \(f\in\mathscr{S}_e(\mathbb{R}\times\mathbb{R}^n)\), $$\mathscr{H}_\mu(f)(t,x)=-\frac{1}{2^\mu\ \Gamma(\mu+1)}\int_t^\infty(r^2-t^2)^\mu\displaystyle\frac{\partial f}{\partial r}(r,x)dr.$$ Hence, \begin{eqnarray*} \displaystyle \frac{\partial}{\partial t^2}\mathscr{H}_\mu(f)(t,x)&=& \frac{1}{2^\mu\ \Gamma(\mu)}\int_t^\infty(r^2-t^2)^{\mu-1}\frac{\partial f}{\partial r^2}(r,x)2rdr\\&=&\mathscr{H}_\mu(\frac{\partial}{\partial r^2}f)(t,x). \end{eqnarray*} ii. For every \(f\in\mathscr{S}_e(\mathbb{R}\times\mathbb{R}^n),\ \mu>0\), and from relation (19), $$\frac{\partial}{\partial t^2}\mathscr{H}_{\mu+1}(f)=\mathscr{H}_{\mu+1}(\frac{\partial}{\partial t^2}f).$$ So, for every \((t,x)\in\mathbb{R}\times\mathbb{R}^n\), \begin{eqnarray*} \mathscr{H}_{\mu+1}(\frac{\partial}{\partial t^2}f)(t,x)&=& \frac{\partial}{\partial t^2}\Big(\frac{1}{2^{\mu+1}\ \Gamma(\mu+1)}\int_t^\infty(r^2-t^2)^\mu f(r,x)2rdr\Big)\\&=&-\mathscr{H}_\mu(f)(t,x). \end{eqnarray*}
Corollary 2.17. Let \(\mu\) be a real number. For all \(k_1,\ k_2\in \mathbb{N},\ k_1+\mu>0,\ k_2+\mu>0\) and for every \(f\in\mathscr{S}_e(\mathbb{R}\times\mathbb{R}^n)\), we have $$\displaystyle (-1)^{k_1}\mathscr{H}_{\mu+k_1}\Big((\frac{\partial}{\partial t^2})^{k_1}f\Big)= (-1)^{k_2}\mathscr{H}_{\mu+k_2}\Big((\frac{\partial}{\partial t^2})^{k_2}f\Big).$$
Proof. Let \(k_1,\ k_2 \in \mathbb{N},\ k_1 0\) and \(k_2+\mu>0\). From Proposition 2.16, it follows that for every \(f\in\mathscr{S}_e(\mathbb{R}\times\mathbb{R}^n)\), \begin{eqnarray*} (-1)^{k_2}\mathscr{H}_{\mu+k_2}((\frac{\partial}{\partial t^2})^{k_2}f)&=& (-1)^{k_1}(-1)^{k_2-k_1}\mathscr{H}_{\mu+k_1+(k_2-k_1)}\Big((\frac{\partial}{\partial t^2})^{k_2-k_1} (\frac{\partial}{\partial t^2})^{k_1}(f)\Big)\\&=&(-1)^{k_1}\mathscr{H}_{\mu+k_1}((\frac{\partial}{\partial t^2})^{k_1}f). \end{eqnarray*}
Definition 2.18. For every \(\mu\in\mathbb{R}\), the fractional transform \(\mathscr{H}_\mu\) is defined on \(\mathscr{S}_e(\mathbb{R}\times\mathbb{R}^n)\) by $$\mathscr{H}_\mu(f)=(-1)^k\mathscr{H}_{\mu+k}((\frac{\partial}{\partial t^2})^kf)=(-1)^k(\frac{\partial}{\partial t^{2}})^k\mathscr{H}_{\mu+k}(f),$$ where \(k\in \mathbb{N},\ k+\mu>0\).
We have the following properties,Proposition 2.19. i. For every \(\mu,\ \nu \in\mathbb{R}\) and \(f\in\mathscr{S}_e(\mathbb{R}\times\mathbb{R}^n).\)
Proof.
i. Let \(\mu,\ \nu \in\mathbb{R},\ k_1,\ k_2\in \mathbb{N},\ k_1+\mu>0,\ k_2+\mu>0\) and \(f\in\mathscr{S}_e(\mathbb{R}\times\mathbb{R}^n),\) we have
\begin{eqnarray*}
\mathscr{H}_\mu \circ\mathscr{H}_\nu(f)&=&\mathscr{H}_\mu \Big((-1)^{k_2}(\frac{\partial}{\partial t^2})^{k_2}\mathscr{H}_{\nu+k_2}(f)\Big)\\&=&
(-1)^{k_1+k_2}\mathscr{H}_{\mu+k_1}\Big((\frac{\partial}{\partial t^2})^{k_1}\mathscr{H}_{\nu+k_2}\big((\frac{\partial}{\partial t^2})^{k_2}(f)\big)\Big)\\&=&(-1)^{k_1+k_2}\mathscr{H}_{\mu+k_1}\circ\mathscr{H}_{\nu+k_2}\big((\frac{\partial}{\partial t^2})^{k_1+k_2}(f) \big).
\end{eqnarray*}
Now, from Proposition 2.15, we deduce that
\begin{eqnarray*}
\mathscr{H}_\mu\circ\mathscr{H}_\nu(f)&=&(-1)^{k_1+k_2}\mathscr{H}_{\mu+\nu+k_2+k_1}\Big((\frac{\partial}{\partial t^2})^{k_1+k_2}(f)\Big)\\&=
&\mathscr{H}_{\mu+\nu}(f),
\end{eqnarray*}
because \(\mu+\nu+k_1+k_2>0.\)
ii.
The result follows from relations (21) and
(22).
Proposition 3.1. For every \(a\in\mathbb{R}\) and every \(\mu>0\), the fractional transform \(\mathscr{R}_\mu\) is bounded from \(L^\infty(d\gamma_a)\) into itself and $$||\mathscr{R}_\mu||_{\infty,\gamma_a}=\sup_{||f||_{\infty,a}\leqslant1}||\mathscr{R}_\mu(f)||_{\infty,a}=1.$$
Proof. Let \(f\) be a bounded measurable function on \([0,+\infty[\times\mathbb{R}^n\). For every \((r,x)\in[0,+\infty[\times\mathbb{R}^n,\) \begin{eqnarray*} |\mathscr{R}_\mu(f)(r,x)|&\leqslant& \frac{2\Gamma(\mu+\frac{1}{2})}{\sqrt{\pi}\Gamma(\mu)}\int_0^1(1-t^2)^{\mu-1}|f(tr,x)|dt \\&\leqslant & ||f||_{\infty,a} \frac{2\Gamma(\mu+\frac{1}{2})}{\sqrt{\pi}\Gamma(\mu)}\int_0^1(1-t^2)^{\mu-1}dt\\&=&||f||_{\infty,a}. \end{eqnarray*} This shows that the operator \(\mathscr{R}_\mu\) is bounded from \(L^\infty(d\gamma_a)\) into itself and that $$||\mathscr{R}_\mu||_{\infty,\gamma_a}\leqslant1.$$ However, \(\mathscr{R}_\mu(1)=1\), this shows that $$||\mathscr{R}_\mu||_{\infty,\gamma_a}=1.$$
Theorem 3.2. The operator \(\mathscr{R}_\mu; \mu>0\) is bounded from \(L^1(d\gamma_a)\) into itself if and only if \(a< 0\) and in this case $$||\mathscr{R}_\mu||_{1,\gamma_a}=\frac{\Gamma(\mu+\frac{1}{2})\Gamma(-a)}{\sqrt{\pi}\ \Gamma(\mu-a)}.$$
Proof. Let \(a \in \mathbb{R},\ a< 0 \). By Fubini-Tonnelli Theorem's and for every \(f\in L^1(d\gamma_a)\), \begin{eqnarray*} \int_0^\infty\int_{\mathbb{R}^n}|\mathscr{R}_\mu(f)(r,x)|d\gamma_a(r,x)&\leqslant & \frac{2\Gamma(\mu+\frac{1}{2})}{\sqrt{\pi}\ \Gamma(\mu)}\int_0^\infty\int_{\mathbb{R}^n}\big(\int_0^1(1-t^2)^{\mu-1} |f(tr,x)|dt\big)d\gamma_a(r,x)\\&=&\frac{2\Gamma(\mu+\frac{1}{2})}{\sqrt{\pi} \ \Gamma(\mu)}\int_0^1(1-t^2)^{\mu-1} \big(\int_0^\infty\int_{\mathbb{R}^n}|f(tr,x)|d\gamma_a(r,x)\big)dt \\&= & ||f||_{1,a} \frac{2\Gamma(\mu+\frac{1}{2})}{\sqrt{\pi}\ \Gamma(\mu)}\int_0^1(1-t^2)^{\mu-1}t^{-(2a+1)}dt \\&=&\frac{\Gamma(\mu+\frac{1}{2})\Gamma(-a)}{\sqrt{\pi} \ \Gamma(\mu-a)}||f||_{1,a}. \end{eqnarray*} Consequently for \(a< 0\), the transform \(\mathscr{R}_\mu\) is a bounded operator from \(L^1(d\gamma_a)\) into itself and $$||\mathscr{R}_\mu||_{1,\gamma_a}\leqslant\frac{\Gamma(\mu+\frac{1}{2})\Gamma(-a)}{\sqrt{\pi} \ \Gamma(\mu-a)}. $$ On the other hand, for every nonnegative \(f\in L^1(d\gamma_a)\), we have $$ ||\mathscr{R}_\mu(f)||_{1,a}=\frac{\Gamma(\mu+\frac{1}{2})\Gamma(-a)}{\sqrt{\pi} \ \Gamma(\mu-a)}||f||_{1,a} $$ We conclude that $$ ||\mathscr{R}_\mu||_{1,\gamma_a}=\frac{\Gamma(\mu+\frac{1}{2})\Gamma(-a)}{\sqrt{\pi} \ \Gamma(\mu-a)}.$$ For converse, let \(a \in \mathbb{R},\ a\geqslant0 \) and let \(f\in L^1(d\gamma_a)\) be a nonnegative function such that \(||f||_{1,a} =1\). We have $$ ||\mathscr{R}_\mu(f)||_{1,\gamma_a}=\frac{\Gamma(\mu+\frac{1}{2})\Gamma(-a)}{\sqrt{\pi} \ \Gamma(\mu-a)}=+\infty.$$ This completes the proof.
Theorem 3.3. Let \(p\in ]1,+\infty[\). The operator \(\mathscr{R}_\mu,\ \mu>0\), is bounded from \(L^p(d\gamma_a)\) into itself if and only if \(2a+1< p\) and in this case $$ ||\mathscr{R}_\mu||_{p,\gamma_a}=\frac{\Gamma(\mu+\frac{1}{2})\Gamma(\frac{p-(2a+1)}{2p})}{\sqrt{\pi} \ \Gamma(\mu+\frac{p-(2a+1)}{2p})}. $$
Proof. Let \(p\in \ ]1,+\infty[,\ 2a+1< p\). From Minkowski's inequality [18] and for every \(f\in L^p(d\gamma_a)\), \begin{eqnarray*} ||\mathscr{R}_\mu(f)||_{p,a}&\leqslant& \frac{2\Gamma(\mu+\frac{1}{2})}{\sqrt{\pi} \ \Gamma(\mu)}\int_0^1(1-t^2)^{\mu-1} \Big(\int_0^\infty\int_{\mathbb{R}^n}|f(tr,x)|^{p}d\gamma_a(r,x)\Big)^{\frac{1}{p}}dt \\&= & \frac{2\Gamma(\mu+\frac{1}{2})}{\sqrt{\pi} \ \Gamma(\mu)}\|f\|_{p,a}\int_0^1(1-t^2)^{\mu-1}t^{-\frac{2a+1}{p}}dt \\&=&\frac{\Gamma(\mu+\frac{1}{2})\Gamma(\frac{p-(2a+1)}{2p})}{\sqrt{\pi} \ \Gamma(\mu+\frac{p-(2a+1)}{2p})} ||f||_{p,a}. \end{eqnarray*} This proves that for \(2a+1< p\), the fractional transform \(\mathscr{R}_\mu\) is bounded from \(L^p(d\gamma_a)\) into itself and
$$ ||\mathscr{R}_\mu||_{p,\gamma_a}=\frac{\Gamma(\mu+\frac{1}{2})\Gamma(\frac{p-(2a+1)}{2p})}{\sqrt{\pi} \ \Gamma(\mu+\frac{p-(2a+1)}{2p})}.$$
Now, we prove that, for \(2a+1>p\), \(\mathscr{R}_\mu\) does not map \(L^p(d\gamma_a)\) into itself. To prove this we have following two cases:
Case 1. Suppose that \(2a+1=p\) and let
$$ g_0(r,x)=\frac{1}{r(1-\ln(r))}\textbf{1}_{]0,1[}(r)\Pi_{j=1}^n\textbf{1}_{]0,1[}(x_j),$$
then, \(g_0\) belongs to \(L^p(d\gamma_a)\) and we have
$$ ||g_0||_{p,a}^p=\int_0^1\frac{dr}{r(1-\ln(r))^p}=\int_{-\infty}^0\frac{ds}{(1-s)^p}=\frac{1}{p-1}.$$
However, for every \((r,x)\in ]0,1[\times]0,1[^n,\)
$$ \mathscr{R}_\mu(g_0)(r,x)= \frac{2\Gamma(\mu+\frac{1}{2})}{\sqrt{\pi}\Gamma(\mu)}r^{1-2\mu}\int_0^r(r^2-t^2)^{\mu-1}\frac{dt}{t(1-\ln(t))}=+\infty,$$
in particular \(\mathscr{R}_\mu(g_0)\) does not belong to \(L^p(d\gamma_a).\)
Case 2. Suppose that \(2a+1>p\) and let \(\eta\in\mathbb{R}; -\frac{2a+1}{p}< \eta p\), \(\mathscr{R}_\mu\) does not map \(L^p(d\gamma_a)\) into itself and this completes the proof of theorem.
Theorem 3.4. For every \(p\in [1,+\infty]\), the fractional operator \(\mathscr{R}_\mu\) is bounded on \(L^p(d\gamma_a)\) if and only if \(2a+1< p\) and in this case $$ ||\mathscr{R}_\mu||_{p,\gamma_a}=\frac{\Gamma(\mu+\frac{1}{2})\Gamma(\frac{p-(2a+1)}{2p})}{\sqrt{\pi}\ \Gamma(\mu+\frac{p-(2a+1)}{2p})}. $$
Remark 3.5. The case \(a=\mu\) in Theorem (3.4) is important because the measure \(d\nu_\mu\) defined by the relation (3) is connected with the operators \(D_j,\ 1\leqslant j\leqslant n\) and \(\Xi\) and the Fourier-Hankel transform \(\widetilde{\mathscr{F}}_{\mu-\frac{1}{2}}\) given by relation (2) and in this occurrence, \(\mathscr{R}_\mu\) is bounded from \(L^p(d\nu_\mu)\) into itself if and only if \(2\mu+1< p\) and we have $$ ||\mathscr{R}_\mu||_{p,\nu_\mu}=\frac{\Gamma(\mu+\frac{1}{2})\Gamma(\frac{p-(2\mu+1)}{2p})}{\sqrt{\pi} \ \Gamma(\mu+\frac{p-(2\mu+1)}{2p})}. $$
Theorem 3.6. The operator \(\mathscr{H}_\mu,\ \mu>0\) is bounded from \(r^{-2\mu}L^1(d\gamma_a)\) into \(L^1(d\gamma_a)\) if and only if \(2a+1>0\) and in this case $$N_{1,\gamma_a}(\mathscr{H}_\mu)= \sup_{||r^{2\mu}f||_{1,a}\leqslant1}||\mathscr{H}_\mu(f)||_{1,a}=\frac{\Gamma(\frac{2a+1}{2})}{2^\mu\ \Gamma(\mu+\frac{2a+1}{2})}.$$
Proof. Suppose that \(a>-\frac{1}{2}\) and let \(f\in r^{-2\mu}L^1(d\gamma_a)\). We have $$ \big|\mathscr{H}_\mu(f)(r,x)\big|\leq\frac{r^{2\mu}}{2^\mu\ \Gamma(\mu)}\int_1^\infty(t^2-1)^{\mu-1}\big|f(rt,x)\big|2tdt. $$ Applying Fubini-Tonnelli Theorem’s, we get \begin{eqnarray*} \int_0^\infty\int_{\mathbb{R}^n}\big|\mathscr{H}_\mu(f)(r,x)\big|d\gamma_a(r,x)&\leqslant & \frac{1}{2^\mu\ \Gamma(\mu) }\int_1^\infty(t^2-1)^{\mu-1}\Big(\int_0^\infty\int_{\mathbb{R}^n}r^{2\mu+2a}|f(tr,x)|drdx\Big)2tdt \\&=&||r^{2\mu}f||_{1,a}\frac{1}{2^\mu\ \Gamma(\mu)}\int_1^\infty(t^2-1)^{\mu-1}t^{-(2\mu+2a+1)}2tdt. \end{eqnarray*} By the change of variable \(s=\frac{1}{t^2}\), we have $$\frac{1}{2^\mu\ \Gamma(\mu)}\int_1^\infty(t^2-1)^{\mu-1}t^{-(2\mu+2a+1)}2tdt=\frac{\Gamma(\frac{2a+1}{2})}{2^\mu\ \Gamma(\mu+\frac{2a+1}{2})}.$$ This shows that for every \(f\in r^{-2\mu}L^1(d\gamma_a)\), the function \(\mathscr{H}_\mu(f)\) belongs to \(L^1(d\gamma_a)\) and $$||\mathscr{H}_\mu(f)||_{1,a}\leqslant\frac{\Gamma(\frac{2a+1}{2})}{2^\mu\ \Gamma(\mu+\frac{2a+1}{2})}||r^{2\mu}f||_{1,a}$$ On the other hand, for every nonnegative function \(f\in r^{-2\mu}L^1(d\gamma_a)\), we have
Theorem 3.7. For every \(p\in]1,+\infty[\), the fractional transform \(\mathscr{H}_\mu\) is bounded from \(r^{-2\mu}L^p(d\gamma_a)\) into \(L^p(d\gamma_a)\) if and only if \(2a+1>0\) and in this case $$N_{p,\gamma_a}(\mathscr{H}_\mu)= \sup_{||r^{2\mu}f||_{p,a}\leqslant1}||\mathscr{H}_\mu(f)||_{p,a}=\frac{\Gamma(\frac{2a+1}{2p})}{2^\mu\ \Gamma(\mu+\frac{2a+1}{2p})}.$$
Proof. Let \(a>-\frac{1}{2}\) and \(f\in r^{-2\mu}L^p(d\gamma_a)\). By Minkouski’s inequality, we have \begin{eqnarray*} ||\mathscr{H}_\mu(f)||_{p,a} &\leqslant & \frac{1}{2^\mu\ \Gamma(\mu) }\int_1^\infty(t^2-1)^{\mu-1}\Big(\int_0^\infty\int_{\mathbb{R}^n}(r^{2\mu}|f(tr,x)|)^p r^{2a} drdx\Big)^{\frac{1}{p}}2tdt \\&=&||r^{2\mu}f||_{p,a}\frac{1}{2^\mu\ \Gamma(\mu)}\int_1^\infty(t^2-1)^{\mu-1}t^{-\frac{2\mu p+2a+1}{p}}2tdt\\&=&\frac{\Gamma(\frac{2a+1}{2p})}{2^\mu\ \Gamma(\mu+\frac{2a+1}{2p})}||r^{2\mu}f||_{p,a}. \end{eqnarray*} Consequently, for \(a>-\frac{1}{2}\), \(\mathscr{H}_\mu\) is a bounded operator from \(r^{-2\mu}L^p(d\gamma_a)\) into \(L^p(d\gamma_a)\) and
Remark 3.8.
For every \(a\in\mathbb{R}\), the fractional transform \(\mathscr{H}_\mu\) does not map the space \(r^{-2\mu}L^\infty(d\gamma_a)\) into itself.
In fact, the function \(f(r,x)=r^{2\mu}\textbf{1}_{[1,+\infty[}(r)\) belongs to \(r^{-2\mu}L^\infty(d\gamma_a)\), but for every \((r,x)\in]0,+\infty[\times\mathbb{R}^n\)
$$\mathscr{H}_\mu(f)(r,x)=\frac{1}{2^\mu\ \Gamma(\mu) }\int_r^\infty(t^2-r^2)^{\mu-1}t^{2\mu}2tdt=+\infty.$$