Set\[\caption {(2.1)} {{\Phi }_{\alpha }}=\frac{1}{{{\text{c}}_{\alpha }}}{{\varphi }_{\alpha }}=\frac{\Gamma \left( \frac{\text{n}+\alpha }{2} \right)\Gamma \left( \frac{\text{n}-\alpha }{2} \right)}{{{2}^{2-2\text{n}}}{{\pi}^{\text{n}+1}}}{{\varphi }_{\alpha }},\]where\[\caption {(2.2)} {{\varphi }_{\alpha }}=\frac{1}{{{\left( |\text{z}{{|}^{2}}-\text{it} \right)}^{\frac{\text{n}+\alpha }{2}}}{{\left( |\text{z}{{|}^{2}}+\text{it} \right)}^{\frac{\text{n}-\alpha }{2}}}}\]
In (2. 1) we assume$\frac{\text{n}\pm \alpha }{2}\ne 0, -1, -2, \ldots$. According to Proposition 7.1 of [ 9 ] the operator Kα, defined by\[{{\text{K}}_{\alpha }}\text{f=f }*\text{ }{{\Phi }_{\alpha }}\]\[=\text{c}_{\alpha }^{-1}\int_{{{\text{H}}_{\text{n}}}}{\text{f(y)}{{\varphi }_{\alpha }}({{\text{y}}^{-1}}\text{x})}\text{d}{{\text{H}}_{\text{n}}}(\text{y}),\]$\text{f}\in \text{C}_{0}^{\infty }({{\text{H}}_{\text{n}}})$is inverse to${\cal{L}}_{\alpha }$, as long as α is admissible, i.e.,$\frac{\text{n}\pm \alpha }{2}\ne 0, -1, -2, \ldots$.
2.4 Theorem. Let α be admissible. Then\[\sigma ({{\text{K}}_{\alpha }})(\text{x}, \xi)=\sigma ({{\text{G}}_{\alpha }})(\text{x}, \xi),\]ξ ≠ 0, where$\sigma({{\text{G}}_{\alpha }})(\text{x},\xi)$is given by (1.22). In other words,\[\caption {(2.5)} ({{\text{K}}_{\alpha }}\text{f})(\text{x})={{(2\pi )}^{-2\text{n}-1}}\int_{{{\mathbb{R}}^{2\text{n}+1}}}{{{\text{e}}^{\text{i}<\text{x},\xi >}}\sigma ({{\text{G}}_{\alpha }})(\text{x},\xi )\hat{\text{f}}(\xi )\text{d}\xi }\]whenever$\text{f}\in \cal{S}$,
Proof. Starting with\[\caption {(2.6)} $({{\text{K}}_{\alpha }}\text{f})(\text{x})=\text{c}_{\alpha }^{-1}{{(2\pi )}^{-2\text{n}-1}}\int_{{{\text{H}}_{\text{n}}}}{{{\varphi }_{\alpha }}({{\text{y}}^{-1}}\text{x})\text{d}{{\text{H}}_{\text{n}}}(\text{y})\int_{{{\mathbb{R}}^{2\text{n}+1}}}{{{\text{e}}^{\text{i}<\text{y},\xi >}}}\hat{\text{f}}(\xi )\text{d}\xi }$\]we shall compute\[\caption {(2.7)} \sigma ({{\text{K}}_{\alpha }})(\text{x},\xi )=\text{c}_{\alpha }^{-1}{{\text{e}}^{-\text{i}<\text{x},\xi >}}\int_{{{\mathbb{R}}^{2\text{n}+1}}}{{{\varphi }_{\alpha }}({{\text{y}}^{-1}}\text{x}){{\text{e}}^{\text{i}<\text{y},\xi >}}\text{dy}{{2}^{-\text{n}}}}.\]
We shall do the computation only if${{\xi}_{\text{0}}}>0$. It is similar if${{\xi}_{\text{0}}}<0$. We use the notation\[{{\text{z}}_{\text{j}}}={{\text{x}}_{2\text{j}-1}}+\text{i}{{\text{x}}_{2\text{j}}},\ {{\text{w}}_{\text{j}}}={{\text{y}}_{2\text{j}-1}}+\text{i}{{\text{y}}_{2\text{j}}},\ \text{j=1,}\ldots\text{,n,}\]and recall that\[\text{d}{{\text{H}}_{\text{n}}}(\text{x})={{2}^{-\text{n}}}\text{dx}.\]
First we compute...