# The first eigenfunction of fractional laplacian

Let $$\Omega$$ be bounded and smooth domain in $$\mathbb{R}^n$$, $$s\in(0,1)$$, $$e_1\in \mathbb{H}^s(\Omega)$$ the first eigenfunction of fractional laplacian $$(-\Delta)^s$$ with eigenvalue $$\lambda_1>0$$, in weak formulation, that is: $$\frac{C(n,s)}{2}\int_{\mathbb{R}^n\times\mathbb{R}^n}\frac{(e_1(x)-e_1(y))(\phi(x)-\phi(y))}{|x-y|^{n+2s}}\,dx\,dy=\lambda_1\int_\Omega e_1(x)\phi(x)\,dx, \quad\forall\phi\in \mathbb{H}^s(\Omega).$$ I know that $$e_1$$ is continuous on whole $$\mathbb{R}^n$$. I want to prove that: $$(-\Delta)^se_1(x)=\lambda_1e_1(x), \quad\forall x\in\Omega,$$ but i have no idea on how to proceed. Any help would be appreciated.

Here: $$\mathbb{H}^s(\Omega)=\{u\in H^s(\mathbb{R}^n): u=0\,\, \text{ q.o. }\in \mathbb{R}^n\setminus\Omega\},$$ and: $$(-\Delta)^su(x):=\frac{C(n,s)}{2}\int_{\mathbb{R}^n}\frac{2u(x)-u(x+y)-u(x-y)}{|y|^{n+2s}}\,dy,\quad\forall x\in\mathbb{R}^n, \forall u\in\mathcal{S}(\mathbb{R}^n).$$ Moreover, how i can define $$(-\Delta)^s$$ for less regular function?

• The eigenfunctions are known to be smooth in $\Omega$ (in fact, with no regularity conditions on the open set $\Omega$), and equation $(-\Delta)^s e_n(x) = \lambda_{n,s} e_n(x)$ indeed holds pointwise. If you just need a reference, I think the article The Cauchy process and the Steklov problem by Rodrigo Bañuelos and Tadeusz Kulczycki (JFA 2004, DOI:10.1016/j.jfa.2004.02.005) is a good source. Commented Oct 29, 2020 at 11:11
• Regarding the other question, you might like to have a look at my Ten equivalent definitions of the fractional Laplace operator, DOI:10.1515/fca-2017-0002. This is about definitions in all of $\mathbb R^n$, but still hopefully related. Commented Oct 29, 2020 at 11:13
• These articles are more than I need, and in my classroom note i have that only $e_1\in C(\mathbb{R}^n)$, can you give a me sketch of proof that $(-\Delta)^se_1(x)=\lambda_1e_1(x), \forall x\in\Omega$ holds? Please.
– inoc
Commented Oct 29, 2020 at 11:30
• If using something more advanced (e.g. some regularity theory) is forbidden, I do not see a simple, direct proof. Even in order to write $(-\Delta)^s e_1(x)$ one needs $e_1$ to be at least, say, $C^{2s+\epsilon}$ near $x$. Commented Oct 30, 2020 at 0:54
• Do you have some reference that proves that the eigenfunction of $(-\Delta)^s$ are $C^{2s+\epsilon}$?
– inoc
Commented Nov 1, 2020 at 7:28

Just an extended comment.

1. Theorem 4.1 in The Cauchy process and the Steklov problem by Rodrigo Bañuelos and Tadeusz Kulczycki (JFA 2004, DOI:10.1016/j.jfa.2004.02.005) shows that the eigenfunctions $$e_n$$ are even real-analytic for $$s = \tfrac12$$, and the authors write that the proof carries over to general $$s$$ at the price of additional technical difficulties.

2. The eigenfunctions are $$C^\infty$$, as can be easily proved directly using potential-theoretic methods: we have $$\lambda_n e_n(x) = \int_B G_B(x, y) e_n(y) dy = I_{2s} e_n(x) - \int_{B^c} I_{2s} e_n(z) P_B(x, z) dz ,$$ where $$B$$ is a ball contained in $$\Omega$$, $$G_B(x,y)$$ is the Green function, $$P_B(x,z)$$ is the harmonic measure (a.k.a. the Poisson kernel), and $$I_{2s}$$ is the Riesz potential operator. Now it is well-known that if $$f$$ is of class $$C^\beta$$ near a point $$x$$, then $$I_{2s} f$$ is of class $$C^{\beta + 2s}$$ near $$x$$ (see, for example, Stein's book). Furthermore $$P_\Omega(\cdot, z)$$ is known explicitly and it is smooth (even real-analytic). Thus the above display is self-improving, and shows that if $$e_n$$ is merely bounded in $$B$$, then it is automatically $$C^\infty$$ in $$B$$. A similar argument is given in my survey Fractional Laplace Operator and its Properties, DOI:10.1515/9783110571622-007.

3. Alternatively, one can use the PDE-flavoured regularity theory, developed in the last decade by Caffarelli, Silvestre, Serra, Ros-Oton and others. In any case, however, this is not a trivial

4. Once we know that $$e_n$$ is smooth enough, all that remains is to use Fubini's theorem to rearrange the integrals, and use a density argument.

• And once we know that $\int (-\Delta)^s e_n(x) \phi(x) dx = \lambda_n \int e_n(x) \phi(x) dx$ for all smooth $\phi$ compactly supported in $\Omega$, we conclude that $(-\Delta)^s e_n = \lambda_n e_n$ almost everywhere in $\Omega$. Commented Nov 28, 2020 at 20:59
• @mathqestion: I think that in the end step 2 is about weak (rather than pointwise) eigenfunctions. The Green operator can be defined in the weak sense by $\mathcal E(G_B u, v) = \langle u, v\rangle$ for all $u, v \in H_0^s(B)$ (with $G_B u$ itself in $H_0^s(B)$), and the harmonic reduction operator $P_B$ satisfies $\mathcal E(P_B u, v) = 0$ for all $u \in H^s$ and $v \in H_0^s(B)$ (with $P_B u$ equal to $u$ in the complement of $B$, and $P_B u$ in $H^s$). And $e_n$ is in $H^s$ and it is a weak eigenfunction in $\Omega$. This should be enough to get the initial identity in step 2, I think. Commented Feb 11, 2021 at 22:53