I have a question regarding the half Laplacian $ (-\Delta )^\frac{1}{2}$ on some smooth bounded domain $ \Omega$ in $R^N$. I am attempting to clarify some confusion with the various definitions. Let $ \phi_k(x)$ denote the $L^2$ normalized eigenfunctions of $ -\Delta $ on $ H_0^1(\Omega)$ with eigenvalues $ \lambda_k$ and assume $ \phi_1 >0$.

Example 1. Note that $ (-\Delta)^\frac{1}{2} \phi_1(x) = \lambda_1^\frac{1}{2} \phi_1(x) $ in $ \Omega$ with zero Dirichlet boundary condition. Also note that the right hand side is positive and that $ \phi_1$ is smooth on the closure of $ \Omega$.

Example 2. Let $B$ denote the unit ball in $ R^N$ centered at the origin. Let $ 0 \le f(x) $ be smooth and compactly supported in say $B_\frac{1}{2}$ and not identically zero. Now consider the Green function for this operator (with zero Dirichlet boundary conditions) (see for instance " COMPARISON AND REGULARITY RESULTS FOR THE FRACTIONAL LAPLACIAN VIA SYMMETRIZATION METHODS " page 10, on the archive). If one writes out $u(x)=\int_{B_\frac{1}{2}} G(x,y) f(y) dy$ and using the explicit form of the Green function they see that $ u$ should behave roughly like square root of the distance to the boundary function , when close to the boundary. So this says that $ u$ can't have a gradient on the boundary.

Questions. The above examples seem to be at odds with eachother (one can surely remove this support condition on example 2). So I guess my question is are the two definitions of for the half laplacian on bounded domains equivalent? In addition there appears to be a number of papers that give boundary regularity results regarding the fractional laplacian which appear to be at odds with example 2.

Sorry about being very vague.
thanks for any responses.

  • $\begingroup$ I think you would normally define $(-\Delta)^{1/2}$ via the spectral theorem, with $-\Delta$ being the Dirichlet Laplacian. Then $T=(-\Delta)^{1/2}$ has the same eigenvectors $\phi_n$ as $-\Delta$, and $T^{-1}\phi_1=\lambda_1^{-1/2}\phi_1$. If you can't reproduce this result from a calculation with the integral kernel of $T^{-1}$, then this just means that you miscalculated somewhere. $\endgroup$ Sep 10, 2014 at 18:01
  • $\begingroup$ @Christian. Ya, that is one way to define the half Laplacian. But there is also the other method where you add an extra dimension and consider a local problem. $\endgroup$
    – Craig
    Sep 10, 2014 at 18:04


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