Let $0<\alpha<d$ and let $I_{\alpha}(f)$ be the Riesz potential of a function $f$ on $\mathbb{R}^{d}$, $$ I_{\alpha}(f)(x)=\int_{\mathbb{R}^{d}}\frac{f(y)}{|x-y|^{d-\alpha}}dy. $$ Assuming $f$ is in the Schwartz space (or possibly in a more general function space), it is known that the fractional laplacian is the inverse of the Riesz potential, namely $$ (-\Delta)^{\alpha/2}I_{\alpha}f=f. $$ Does the above formula hold, in a distributional sense, with a positive measure $\mu$ instead of the function $f$ (assuming that $\mu$ integrates $|y|^{\alpha-d}$ at infinity) ? Does such a formula appear in the literature ?
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1$\begingroup$ I must be missing something: "In the sense of distributions" means, I suppose, that $\mu * (I_\alpha (-\Delta)^{\alpha/2} \phi) = \mu * \phi$ for all $\phi$ in, say, $C_c^\infty(\mathbb R^d)$. But this is clearly true, as $I_\alpha (-\Delta)^{\alpha/2} \phi = \phi$. $\endgroup$– Mateusz KwaśnickiCommented Feb 12, 2021 at 22:23
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$\begingroup$ Thanks for the comment. If I understand correctly, you consider both $(-\Delta)^{\alpha/2}$ and $I_\alpha$ as distributions. I wanted to consider $(-\Delta)^{\alpha/2}$ as a distribution but $I_\alpha\mu$ just as a function. Does that make sense ? Maybe I am not understanding something. $\endgroup$– user111Commented Feb 13, 2021 at 7:02
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$\begingroup$ I think I understand now, but, if correct, one has to justify that Fubini applies to get $\int(I_\alpha\mu)(-\Delta)^{\alpha/2}\phi d\lambda=\int I_\alpha(-\Delta)^{\alpha/2}\phi d\mu$. I haven't see the formula $(-\Delta)^{\alpha/2}I_\alpha\mu=\mu$ written anywhere... $\endgroup$– user111Commented Feb 13, 2021 at 9:34
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$\begingroup$ If $\mu$ is compactly supported, justifying Fubini is easy. Regarding a direct reference: similar expressions for more general distributions (say, in the Lizorkin space) are readily available, for example in Samko's book on hypersingular integrals. You can try searching in this book, or in other references on "inversion of Riesz potentials". $\endgroup$– Mateusz KwaśnickiCommented Feb 13, 2021 at 10:40
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$\begingroup$ Thanks, I will check. $\endgroup$– user111Commented Feb 13, 2021 at 11:20
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