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I have two disjoint open intervals $B_1, B_2 \subset \mathbb{R}$, and variables $0 < s < 1$ and $t \in B_1 \cup B_2$. I want to solve:

$$r_{B_1 \cup B_2}(\Delta^{s} f) = \delta_t$$ for $f$. Here, the support of $f$ is contained in $B_1 \cup B_2$, and $r_{B_1 \cup B_2}$ restricts a function to $B_1 \cup B_2$. How would I go about solving this?

(I think the solution is the Green's function of the fractional Laplacian on two intervals, rather than just one interval, but I have been unable to find any literature on this matter).

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    $\begingroup$ I find your question rather unclear. Is $v$ a discrete measure supported in $S$? Do you really mean $(-\Delta)^{-s}$, and not $(-\Delta)^s$? (The former is called the Riesz potential operator rather than fractional Laplacian.) Green's function typically requires some boundary/exterior condition; what you seem to consider could be called 'trace'. Finally, $(-\Delta)^{-s}$ only makes sense if $2s$ is less then the dimension, and if $v$ has an atom at $q$, then $(-\Delta)^{-s} v$ is infinite at $q$. Can you clarify your question a bit? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 24, 2019 at 18:34
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    $\begingroup$ (1/2) "Makes sense" is perhaps too strong a statement, but if $2s$ is not less than the dimension, there is no solution to $(-\Delta)^s u = f$ (and hence $(-\Delta)^{-s} f$ is not well-defined) unless at least $\int f = 0$, even if we assume that, say, $f$ is smooth and compactly supported. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 25, 2019 at 7:57
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    $\begingroup$ (2/2) However, of course $(-\Delta)^{-s}$ is still a perfectly well-defined unbounded operator on $L^2$. If you are interested in $(-\Delta)^{-s}$ for these values of $s$, you may like the books by Boris Rubin (Fractional Integrals and Potentials, 1996) and by Stefan Samko (Hypersingular Integrals and Their Applications, 2001). $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 25, 2019 at 7:57
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    $\begingroup$ I am rather sure a closed-form expression is not known, and I would be very surprised if it existed. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 27, 2019 at 14:10
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    $\begingroup$ I think so. For the ball, one exploits rotational symmetry, Kelvin transformation, and some somewhat surprising integral identities. As far as I understand, for the interval, one can employ the methods of (one-sided) fractional integration and differentiation. All of this breaks for the union of two intervals. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 27, 2019 at 18:33

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