Fractional Laplacian and convolution $(-\Delta)^\alpha (u \ast \eta_\epsilon) = (-\Delta)^\alpha u \ast \eta_\epsilon$? For $u \in L^\infty(\mathbb R)$ and $\eta_\epsilon$ mollifier, it is well-known that for the (distributional) derivative it holds that $(u \ast \eta_\epsilon)' = u'\ast \eta_\epsilon$.
Is it also true for the fractional Laplacian that
$$(-\Delta)^\alpha (u \ast \eta_\epsilon) =  (-\Delta)^\alpha u \ast \eta_\epsilon$$
holds? Where can I find a proof of this?
 A: Answering the question asked in a comment recently: you can read more about distributional definition of the fractional Laplacian in the paper by Luis Silvestre [1] (or in his PhD thesis, if I remember correctly), in an excellent book by Stefan Samko [2], or in Section 5 of my survey [3]. I do not think any of these references has the statement that you are looking for written explicitly, though.

As for the proof, it is in fact a relatively simple consequence of the definition: the distribution $f = (-\Delta)^s u$ is given by
$$ \langle f, w \rangle = \langle u, (-\Delta)^s w \rangle $$
for all infinitely smooth $w$ such that all derivatives of $w$ are absolutely integrable (and $(-\Delta)^s w$ is defined by any of the usual definitions). Now simply apply this to $w(y) = \eta_\epsilon(x - y)$ for a fixed $x$ to get
$$ ((-\Delta)^s f) * \eta_\epsilon(x) = \langle f, w \rangle = \langle u, (-\Delta)^s w \rangle = u * (-\Delta)^s \eta_\epsilon(x) = \ldots $$
Now the right-hand side can be written as
$$ \ldots = \int_{\mathbb R} \int_{\mathbb R} u(x - y) (\eta_\epsilon(y + z) - \eta_\epsilon(y) - z \cdot \nabla \eta_\epsilon(y) \mathbb 1_B(z)) dz dy = \ldots $$
(with $B$ denoting the unit ball $(-1, 1)$), and it is not very difficult to see that we may apply Fubini's theorem to get
$$ \begin{aligned} \ldots & = \int_{\mathbb R} \int_{\mathbb R} u(x - y) (\eta_\epsilon(y + z) - \eta_\epsilon(y) - z \cdot \nabla \eta_\epsilon(y) \mathbb 1_B(z)) dy dz \\
& = \int_{\mathbb R} (u * \eta_\epsilon(x + z) - u * \eta_\epsilon(x) - z \cdot (u * \nabla \eta_\epsilon)(y) \mathbb 1_B(z)) dz = \ldots \end{aligned} $$
We already know that $u * \nabla \eta_\epsilon = \nabla (u * \eta_\epsilon)$, so we eventually get
$$ \ldots = (-\Delta)^s (u * \eta_\epsilon)(x) ,$$
as desired.

If you are familiar with distributional convolution, you can alternatively write $(-\Delta)^s u$ as $L * u$ for an appropriate distribution $L$, and use associativity of convolution to immediately get
$$ (-\Delta)^s u * \eta_\epsilon = (L * u) * \eta_\epsilon = L * (u * \eta_\epsilon) = (-\Delta)^s (u * \eta_\epsilon) .$$
Convolution is indeed associative, because $L$ and $\eta_\epsilon$ are integrable distributions, and $u$ is a bounded distribution.

References:

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*[1] L. Silvestre, Regularity of the obstacle problem for a fractional power of the Laplace operator, Commun. Pure Appl. Math., 60 (2007), 67–112, DOI:10.1002/cpa.20153


*[2] S. Samko, Hypersingular Integrals and Their Applications, CRC Press, London–New York, 2001, DOI:10.1201/9781482264968


*[3] M. Kwaśnicki, Fractional Laplace Operator and its Properties, in: A. Kochubei, Y. Luchko, Handbook of Fractional Calculus with Applications. Volume 1: Basic Theory, De Gruyter Reference, De Gruyter, Berlin, 2019, DOI:10.1515/9783110571622-007
