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

and $\displaystyle I_{3} = \int_{B_{1}^{c}} (\phi(x+y) + \phi(x-y) - 2\phi(x))\nu(y)dy,$ $\displaystyle I_{4} = \int_{B_{1}} (\phi(x+y) + \phi(x-y) - 2\phi(x)) \nu(y)dy.$

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 ${\sigma^{d + \alpha} \nu(z) = \nu(y)}$. $\Box$