For $u \in L^\infty(\mathbb R)$ and $\eta\_\epsilon$ [mollifier](https://math.stackexchange.com/questions/1366268/standard-mollifier-comparing-the-definition-in-evans-and-wiki), 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?



Fractional Laplacian and convolution $(-\Delta)^\alpha (u \ast \eta_\epsilon) = (-\Delta)^\alpha u \ast \eta_\epsilon$?
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.




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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.




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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.




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References:


* [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](https://doi.org/10.1002/cpa.20153)
* [2] S. Samko, *Hypersingular Integrals and Their Applications*, CRC Press, London–New York, 2001, [DOI:10.1201/9781482264968](https://doi.org/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](https://doi.org/10.1515/9783110571622-007)


