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I know that in $\mathbb{R^n}$, the Fourier transform of a fractional laplacian $\widehat{(-\Delta)^s u}(\xi) = |\xi|^{2s} \hat{u}(\xi)$. In particular, when $s=1/2$, $\widehat{(-\Delta)^{\frac{1}{2}} u}(\xi) = |\xi|\hat{u}(\xi)$. I am studying an operator $H_s{v} =- r^{\frac{n-1}{2}}((-\Delta)^{\frac{1}{2}}u)$, where $u$ is radially symmetric and $u(r)=r^{\frac{1-n}{2}}v(r)$. I want to solve $H_s{v} =- v$. Because of nice Fourier invertibility properties for $s=1/2$, I wanted to use Fourier analysis to solve this. I was wondering if there was a simple way to express the Fourier transform of the operator, $\hat{H_s(v)}=\widehat{r^{\frac{n-1}{2}}(-\Delta)^{\frac{1}{2}}(r^{\frac{1-n}{2}}v)}$ in terms of $\hat{v}(\xi)$.

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Let $a = {n-1}/2$. Actually, when $a$ is not a negative even integer, the Fourier transform of $\frac{1}{r^a} = \frac{1}{|x|^a}$ is $\mathcal{F}(\frac{1}{|x|^a}) = \frac{c_{n,a}}{|x|^{n-a}}$ (to be interpreted as the associated Hadamard finite part distribution if $a<0$), so $$ \mathcal{F}(r^{a}(-\Delta)^{1/2} r^{-a} v) = c_{n,a}c_{n,-a}\,\frac{1}{|\xi|^{n+a}} * \left(|\xi| \left(\frac{1}{|x|^{n-a}}* \hat{v}\right)(\xi)\right) $$ You can also rewrite that with fractional laplacians as $$ \mathcal{F}(r^{a}(-\Delta)^{1/2} r^{-a} v) = (-\Delta)^{a/2}\left(|\xi| \left((-\Delta)^{-a/2} \hat{v}\right)(\xi)\right) $$ or in a more integral form $$ \mathcal{F}(r^{a}(-\Delta)^{1/2} r^{-a} v)(\xi) = c_{n,a}\,(-\Delta_{\xi})^{a/2}\int_{\mathbb{R}^n} \frac{|\xi|\, \hat{v}(y)}{|\xi-y|^{n-a}}\,\mathrm{d}y $$

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