1
$\begingroup$

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?

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ 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. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Oct 29, 2020 at 11:11
  • 1
    $\begingroup$ 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. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Oct 29, 2020 at 11:13
  • $\begingroup$ 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. $\endgroup$
    – inoc
    Commented Oct 29, 2020 at 11:30
  • $\begingroup$ 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 $x$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Oct 30, 2020 at 0:54
  • $\begingroup$ Do you have some reference that proves that the eigenfunction of $(-\Delta)^s$ are $C^{2s+\epsilon}$? $\endgroup$
    – inoc
    Commented Nov 1, 2020 at 7:28

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
6
  • $\begingroup$ can you give me the details of the point 4 in your answer please? $\endgroup$
    – inoc
    Commented Nov 24, 2020 at 18:22
  • $\begingroup$ Regarding using Fubini's theorem to rearrange the integrals: I think I already did, in a greater generality, in an answer to a different question. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Nov 28, 2020 at 20:57
  • $\begingroup$ 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 $\Omega$. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Nov 28, 2020 at 20:59
  • $\begingroup$ May I ask a clarifying question? It looks like Steps 2 and 4 justify that any "pointwise-sense" eigenfunciton is a "weak-sense" eigenfunction. But how to show the converse (the exact OP's question)? In particular, can the displayed Green's formula be directly applied to a "weak-sense" eigenfunction? $\endgroup$
    – mathqestion
    Commented Feb 11, 2021 at 13:56
  • $\begingroup$ @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. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Feb 11, 2021 at 22:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .