Two representations of the fractional Laplacian operator

There are a few equivalent definitions for the fractional Laplacian operator {(-\Delta)^{\alpha}} for a constant {\alpha \in (0, 2)}. One is given by

\displaystyle - (- \Delta)^{\alpha} \phi(x) = C' \int_{\mathbb R^{d}} (\phi(x+y) - \phi(x) - D \phi(x) \cdot y I_{B_{1}}(y)) \nu(y) dy,

and the other one is given by

\displaystyle - (- \Delta)^{\alpha} \phi(x) = C \int_{\mathbb R^{d}} (\phi(x+y) + \phi(x-y) - 2\phi(x)) \nu(y) dy,

where {C} and {C'} are normalizing constants and {\nu(y) = \frac{1}{|y|^{d + \alpha}}} is a symmetric Levy measure on {\mathbb R^{d}}. In this below, we will show they are actually given equivalent with {2 C = C'}.

For simplicity, we first use the following notions. We denote

\displaystyle I_{1} = \int_{B_{1}^{c}} (\phi(x+y) - \phi(x)) \nu(y)dy,

\displaystyle I_{2} = \int_{B_{1}} (\phi(x+y) - \phi(x) - D \phi(x) \cdot y) \nu(y)dy,


\displaystyle I_{3} = \int_{B_{1}^{c}} (\phi(x+y) + \phi(x-y) - 2\phi(x))\nu(y)dy,

\displaystyle I_{4} = \int_{B_{1}} (\phi(x+y) + \phi(x-y) - 2\phi(x)) \nu(y)dy.

Let’s try to show the fact in a little bit more general case for {\nu}.

Proposition 1

If {\nu} is an even function, then {2 I_{1} = I_{3}} and {2I_{2} = I_{4}} for any function {\phi} to make {I_{1}} and {I_{2}} well defined.

Proof: Indeed, we have the symmetry of {\nu}, i.e. {\nu(y) = \nu (-y)}. Therefore, we obtain

\displaystyle I_{1} = \int_{B_{1}^{c}} (\phi(x - y) - \phi(x)) \nu(y) dy = \frac 1 2 I_{3}.

The last equality in the above holds by adding two representations of {I_{1}} given above. Similarly, we write {I_{2}} by

\displaystyle I_{2} = \int_{B_{1}} (\phi(x - y) - \phi(x) + D \phi(x) \cdot y) \nu(y)dy,

and adding up two representations of {I_{2}} to get {I_{2} = \frac 1 2 I_{4}}. \Box

Example 1

Let {\alpha \in (0, 2)} and {\sigma >0} be two constants. If {W} is a {\alpha}-stable process with {W \sim - (- \Delta)^{-\alpha/2}}, then {X = \sigma W} is {X\sim - \sigma^{\alpha} (- \Delta)^{-\alpha/2}}. {\Box}.

Proof: We denote {L = - (- \Delta)^{-\alpha/2}} and {\hat L \sim X} and try to show {\hat L = \sigma^{\alpha} L}. For any {\phi \in C^{\infty}}, we have, with {\hat \phi(x) = \phi(\sigma x)}

\displaystyle  \begin{array} {ll} \hat L \phi(0) & \displaystyle = \lim_{h\rightarrow 0} \frac{\mathbb E[\phi(\sigma W_{h}) - \phi(0)]}{h} \\ & \displaystyle = \lim_{h\rightarrow 0} \frac{\mathbb E[\hat \phi( W_{h}) - \hat \phi(0)]}{h} = L \hat \phi(0). \end{array}

The proof is accomplished by noting that

\displaystyle  \begin{array} {ll} L \hat \phi(0) & \displaystyle = \int (\phi(\sigma y) + \phi(- \sigma y) - 2 \phi(0) ) \nu(y) dy \\ & \displaystyle = \int (\phi(z) + \phi(- z) - 2 \phi(0) ) \cdot \sigma^{d + \alpha} \nu(z) \cdot \sigma^{-d}dz \\ & \displaystyle = \sigma^{\alpha} L\phi(0). \end{array}

In the above, {d} is the dimension of {X}, and we used change of variable by {z = \sigma y} with the facts { dz = \sigma^{d} dy} and {\sigma^{d + \alpha} \nu(z) = \nu(y)}. \Box

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