**1. Introduction**

Let *s* ∈ (0, 1) and Δ = *<sup>∂</sup>*2/*∂x*21 + ··· + *<sup>∂</sup>*2/*∂x*2*n*. The fractional Laplacian of a function *u*: *Rn* → *R* is defined as:

$$(-\Delta)^{s}u(\mathbf{x}) = \mathbb{C}\_{n,s} \text{P.V.} \int\_{\mathbb{R}^n} \frac{u(\mathbf{x}) - u(\underline{\zeta})}{|\mathbf{x} - \underline{\zeta}|^{n + 2s}} d\underline{\zeta}\_{\nu} \tag{1}$$

where P.V. stands for the Cauchy principal value, and the constant *Cn*,*<sup>s</sup>* is given by:

$$\mathbb{C}\_{\mathfrak{n},\mathfrak{s}} = \left( \int\_{\mathbb{R}^n} \frac{1 - \cos y\_1}{|y|^{n + 2s} dy} \right)^{-1} = \pi^{-n/2} 2^{2s} \frac{\Gamma(\frac{n + 2s}{2})}{\Gamma(1 - s)} s\_{\mathfrak{s}}$$

and *y* = (*y*1, *y*2, ··· , *yn*) ∈ *Rn*. On the other hand, the fractional Laplacian in *Rn* can be written by the Fourier transform:

$$(-\Delta)^s u(\mathbf{x}) = \frac{1}{(2\pi)^n} \int\_{\mathbb{R}^n} |\zeta|^{2s} (\mathbf{u}, e^{-i\tilde{\zeta}\cdot\mathbf{x}})\_{L^2} e^{i\tilde{\zeta}\cdot\mathbf{x}} d\zeta = \mathcal{F}^{-1}\{ |\zeta|^{2s} \mathcal{F}(\mathbf{u})(\zeta) \}(\mathbf{x})$$

where:

$$\begin{split} & \left( u, \, e^{-i\tilde{\zeta} \cdot x} \right)\_{L^{2}} = \int\_{\mathbb{R}^{n}} u(\mathbf{x}) e^{-i\tilde{\zeta} \cdot x} d\mathbf{x}, \quad \hat{u}(\zeta) = \mathcal{F}\{ u \}(\zeta) = \frac{1}{(2\pi)^{n/2}} \int\_{\mathbb{R}^{n}} u(\mathbf{x}) e^{-i\tilde{\zeta} \cdot x} d\mathbf{x}, \\ & \mathcal{F}^{-1}\{\hat{u}\}(\mathbf{x}) = \frac{1}{(2\pi)^{n/2}} \int\_{\mathbb{R}^{n}} \hat{u}(\zeta) e^{i\tilde{\zeta} \cdot x} d\tilde{\zeta}. \end{split}$$

Hence, the fractional Laplacian is really a pseudo-differential operator with symbol |*ζ*|<sup>2</sup>*s*.

Let *Lp* be the Lebesgue space with *p* ∈ [1, <sup>∞</sup>), *B*0 be the space of continuous functions vanishing at infinity, and *Bbu* be the space of bounded uniformly-continuous functions. M.Kwa´snicki recently presented ten equivalent definitions in [1] for defining (−<sup>Δ</sup>)*<sup>s</sup>* over these three spaces, including the above Fourier definition.

There have been many studies, including numerical analysis approaches, on the fractional Laplacian with applications in solving certain differential equations on bounded domains and in the theory of stochastic processes and anomalous diffusion [2–10]. For example, the work in [11] used the fractional Laplacian for linear and nonlinear lossy media, as well as studying a linear integro-differential equation wave model. The work in [12] studied a finite difference method of solving parabolic partial integro-differential equations, with possibly singular kernels. These arise in option pricing theory when the random evolution of the underlying asset is driven by a Lévy process related to the fractional Laplacian or, more generally, a time-inhomogeneous jump-diffusion process. Using the fractional Laplacian operator, Araci et al. [13] investigated the following *q*-difference boundary value problem:

$$D\_q^\gamma(\phi\_p(D\_q^\delta y(t))) + f(t, y(t)) = 0, \quad (0 < t < 1; \ 3 < \delta < 4)$$

with the boundary conditions:

$$\begin{aligned} y(0) &= (D\_q y)(0) = (D\_q^2 y)(0), \\ a\_1 (D\_q y)(1) + a\_2 (D\_q^2 y)(1) &= 0, \quad \text{and} \quad D\_{0+}^\gamma y(t)|\_{t=0} = 0. \end{aligned}$$

They proved the existence and uniqueness of a positive and nondecreasing solution for this problem by means of a fixed point theorem involving partially-ordered sets.

Let Ω ⊂ *Rn* denote a bounded, open domain. For *u*(*x*): Ω → *R*, D'Elia and Gunzburger [14] investigated the action of the nonlocal diffusion operator L on the function *u*(*x*) as:

$$\mathcal{L}u(\mathbf{x}) = 2 \int\_{\mathbb{R}^n} (\mu(y) - \mu(\mathbf{x})) \gamma(\mathbf{x}, y) dy \quad \forall \mathbf{x} \in \Omega \subseteq \mathbb{R}^n.$$

where the volume of Ω is non-zero and the kernel *<sup>γ</sup>*(*<sup>x</sup>*, *y*): Ω × Ω → *R* is a non-negative symmetric mapping, as well as the nonlocal, steady-state diffusion equation:

$$\begin{aligned} -\mathcal{L}u &= f \quad \text{on } \Omega, \\ u &= 0 \quad \text{on } R^\pi \backslash \Omega, \end{aligned}$$

An example of *<sup>γ</sup>*(*<sup>x</sup>*, *y*) is given by:

$$\gamma(x, y) = \frac{\sigma(x, y)}{|y - x|^{n + 2s}}$$

with *<sup>σ</sup>*(*<sup>x</sup>*, *y*) bounded from above and below by positive constants. This nonlocal diffusion operator has, as special cases, the fractional Laplacian and fractional differential operators that arise in several applications. The corresponding evolution model was further studied in [15]. Recently, Hu et al. [16] studied the following high-dimensional Caputo-type parabolic equation with the fractional Laplacian by using the finite difference method:

$$\begin{aligned} \, \_C D\_{0,t}^{\alpha} u &= -(-\Delta)^s u + f, \quad x \in \Omega, \; t > 0, \\ u(x,0) &= u\_0(x), \\ u(x,t) &= 0 \quad \text{on} \quad x \in R^n \; \bigvee \Omega, \end{aligned}$$

where *α* ∈ (0, <sup>1</sup>),*<sup>s</sup>* ∈ (0, 1) and Ω ⊂ *Rn*. In particular, this involves the half-order Laplacian operator (−<sup>Δ</sup>) 12 when *s* = 1/2. The convergence and error estimate of the established finite difference scheme are shown with several examples.

On the other hand, the Riesz fractional derivative is generally given as:

$$\,\_{\mathbb{R}\mathbb{Z}}D\_{\mathbf{x}}^{a}\mu(\mathbf{x}) = -\frac{(\_{RL}D\_{-\infty,\mathbf{x}}^{a} + \_{RL}D\_{\cos,\mathbf{x}}^{a})\mu(\mathbf{x})}{2\cos(a\pi/2)}\tag{2}$$

where 0 < *α* < 2 and *α* = 1, and:

$$\begin{aligned} \,\_{RL}D\_{-\infty,x}^{\alpha}u(x) &= \frac{1}{\Gamma(n-\alpha)} \frac{d^n}{dx^n} \int\_{-\infty}^{x} (x-\zeta)^{n-\alpha-1} u(\zeta) d\zeta, \\\,\_{RL}D\_{\infty,x}^{\alpha}u(x) &= \frac{1}{\Gamma(n-\alpha)} \frac{d^n}{dx^n} \int\_{\infty}^{x} (x-\zeta)^{n-\alpha-1} u(\zeta) d\zeta \end{aligned}$$

for *n* − 1 < *α* < *n* ∈ *Z*+.

Note that in space fractional quantum mechanics, the *α* = 2 case corresponds to the Schrödinger equation for a massive non-relativistic particle, while the *α* = 1 case needs to be examined carefully, both on physical and mathematical grounds, since Equation (2) is undefined for *α* = 1.

It follows from [10,17–22] that:

$$\,\_{RL}D\_{-\\\\\infty,x}^{\alpha}u(x) = \frac{d^2}{dx^2}[I\_{-\\\infty,x}^{2-\\\alpha}u(x)], \quad 1 < \alpha < 2$$

and:

$$\begin{aligned} \, \_{RL}D\_{-\infty,\mathbf{x}}^{\mu}\mu(\mathbf{x}) &= \, \_{-\infty,\mathbf{x}}\mu(\mathbf{x}) = \frac{1}{\Gamma(2-\alpha)}\frac{d^2}{d\mathbf{x}^2}\int\_{-\infty}^{\infty} \frac{u(t)}{(\mathbf{x}-t)^{\alpha-1}}dt \\ &= \, \_{\Gamma(2-\alpha)}\frac{d^2}{d\mathbf{x}^2}\int\_{0}^{\infty} \zeta^{-\alpha+1}u(\mathbf{x}-\zeta)d\zeta \end{aligned}$$

by making the variable change *ζ* = *x* − *t*.

Applying two identities:

$$\begin{aligned} \zeta^{-\alpha+1} &= (\alpha - 1) \int\_{\overline{\zeta}}^{\infty} \frac{d\eta}{\eta^{\alpha'}} \\ \frac{\partial^2 u(x - \zeta)}{\partial x^2} &= \frac{\partial^2 u(x - \zeta)}{\partial \zeta^2} \end{aligned}$$

we get:

*RLD<sup>α</sup>*−∞, *x<sup>u</sup>*(*x*) = *α* − 1 Γ(2 − *α*) ( ∞0 *<sup>∂</sup>*2*u*(*x* − *ζ*) *∂ζ*2 ,( ∞*ζ dηηα* - *dζ* = 1 − *α* Γ(2 − *α*) ( ∞0 ,( ∞*ζ dηηα* - *d ∂u*(*x* − *ζ*) *∂ζ* = 1 Γ(1 − *α*) *∂u*(*x* − *ζ*) *∂ζ* ( ∞*ζ <sup>d</sup>ηηα* <sup>∞</sup>*ζ*=<sup>0</sup> + 1 Γ(1 − *α*) ( ∞0 *∂u*(*x* − *ζ*) *∂ζ* 1*ζα dζ* = − 1 Γ(1 − *α*) ( ∞0 1*ζα du*(*x* − *ζ*) = − 1 Γ(1 − *α*) *u*(*x* − *ζ*) *ζα* ∞*ζ*=0 − *α* Γ(1 − *α*) ( ∞0 *u*(*x* − *ζ*) *ζα*+<sup>1</sup> *dζ* = *α* Γ(1 − *α*) ( ∞0 *u*(*x*) *ζα*+<sup>1</sup> *dζ* − *α* Γ(1 − *α*) ( ∞0 *u*(*x* − *ζ*) *ζα*+<sup>1</sup> *dζ* = − *α* Γ(1 − *α*) ( ∞0 *u*(*x* − *ζ*) − *u*(*x*) *ζα*+<sup>1</sup> *dζ*, 1 < *α* < 2

by removing the term:

$$\int\_0^\infty \frac{d\eta}{\eta^{\alpha}},$$

due to the meaning of the finite part integral as the integral is divergent and the finite part:

$$\left. \frac{\eta^{-a+1}}{-a+1} \right|\_{\eta=\infty} = 0$$

With the same argument, we come to:

$$\,\_{RL}D\_{\infty,\mathbf{x}}^{\alpha}u(\mathbf{x}) = -\frac{\mathfrak{a}}{\Gamma(1-\mathfrak{a})} \int\_{0}^{\infty} \frac{u(\mathbf{x}+\boldsymbol{\xi})-u(\mathbf{x})}{\boldsymbol{\zeta}^{\alpha+1}}d\boldsymbol{\zeta}, \quad 1 < \mathfrak{a} < 2.$$

Hence, we have another representation of the Riesz fractional derivative from Equation (2):

$$\delta\_{RZ}D\_{\mathbf{x}}^{\mathbf{x}}u(\mathbf{x}) = \frac{\Gamma(1+a)\sin a\pi/2}{\pi} \int\_{0}^{\infty} \frac{u(\mathbf{x}+\boldsymbol{\zeta}) - 2u(\mathbf{x}) + u(\mathbf{x}-\boldsymbol{\zeta})}{\boldsymbol{\zeta}^{1+a}} d\boldsymbol{\zeta}, \quad 1 < a < 2.$$

Similarly, we can claim that this representation still holds for the entire range 0 < *α* ≤ 2 [23]. In particular,

$$\,\_{\mathbb{R}\mathbb{Z}}D\_{\mathbf{x}}^{1}u(\mathbf{x}) = \frac{1}{\pi} \int\_{0}^{\infty} \frac{u(\mathbf{x} + \boldsymbol{\zeta}) - 2u(\mathbf{x}) + u(\mathbf{x} - \boldsymbol{\zeta})}{\boldsymbol{\zeta}^{2}} d\boldsymbol{\zeta} = \frac{1}{\pi} (\text{P.V.} \frac{1}{\boldsymbol{\zeta}^{2}}, \, u(\mathbf{x} + \boldsymbol{\zeta})) $$

based on the formula:

$$(\text{P.V.} \frac{1}{\mathfrak{x}^2}, \phi(\mathfrak{x})) = \int\_0^\infty \frac{\phi(\mathfrak{x}) - 2\phi(0) + \phi(-\mathfrak{x})}{\mathfrak{x}^2} d\mathfrak{x}.$$

Clearly, Equation (1) becomes:

$$\begin{aligned} (-\Delta)^{1/2} u(\mathbf{x}) &= \prescript{}{}{\mathbf{C}}\_{1,1/2} \text{P.V.} \int\_{R} \frac{u(\mathbf{x}) - u(\underline{\zeta})}{|\mathbf{x} - \underline{\zeta}|^{2}} d\underline{\zeta} \\ &= \frac{1}{\pi} \text{P.V.} \int\_{R} \frac{u(\mathbf{x}) - u(\underline{\zeta})}{(\mathbf{x} - \underline{\zeta})^{2}} d\underline{\zeta} \\ &= -\frac{1}{\pi} \int\_{0}^{\infty} \frac{u(\mathbf{x} + \underline{\zeta}) - 2u(\mathbf{x}) + u(\mathbf{x} - \underline{\zeta})}{\underline{\zeta}^{2}} d\underline{\zeta} \\ &= -\prescript{}{R}{}\_{R} D\_{\mathbf{x}}^{1} u(\mathbf{x}) \end{aligned} \tag{3}$$

for *s* = 1/2 and *n* = 1.

Therefore, investigations of the half-order Laplacian operator (− Δ) 1 2 on *R* are equivalent to studies of the first-order Riesz derivative. We can define −(− Δ) 1 2 *u* as the Riesz derivative *RZ D<sup>α</sup> x<sup>u</sup>*(*x*) in the case of *α* = 1, which is undefined in Equation (2) in the classical sense. The aim of this work is to study the operator (− Δ) 1 2 on *R* in distribution explicitly and implicitly, using a particular delta sequence and the generalized convolution. We also present several interesting examples, such as (− Δ) 1 2 *<sup>δ</sup>*(*x*) and (− Δ) 1 2 *<sup>θ</sup>*(*x*), which are undefined in the classical sense. At the end of this work, we describe applications of such studies to solving the differential equations involving the half-order Laplacian operator.

#### **2. The Explicit Approach to (***−* **<sup>Δ</sup>)1/2***u*

In order to extend the fractional Laplacian (− Δ)1/2 distributionally, we briefly introduce the following basic concepts of distributions. Let D(*R*) be the Schwartz space (testing function space) [24,25] of infinitely-differentiable functions with compact support in *R* and D (*R*) the (dual) space of distributions defined on <sup>D</sup>(*R*). A sequence *φ*1, *φ*2, ··· , *φ<sup>n</sup>*, ··· goes to zero in D(*R*) if and only if these functions vanish outside a certain fixed bounded set and converge to zero uniformly together with their derivatives of any order.

The functional *<sup>δ</sup>*(*n*)(*x* − *<sup>x</sup>*0) is defined as:

$$(\delta^{(\mathfrak{n})}(x - x\_0), \phi(\mathfrak{x})) = (-1)^n \phi^{(\mathfrak{n})}(\mathfrak{x}\_0).$$

where *φ* ∈ <sup>D</sup>(*R*). Clearly, *<sup>δ</sup>*(*n*)(*x* − *<sup>x</sup>*0) is a linear and continuous functional on <sup>D</sup>(*R*), and hence, *<sup>δ</sup>*(*n*)(*x* − *<sup>x</sup>*0) ∈ D (*R*).

Define the unit step function *<sup>θ</sup>*(*x*) as:

$$\theta(\mathbf{x}) = \begin{cases} 1 & \text{if } \mathbf{x} > \mathbf{0}, \\ 0 & \text{if } \mathbf{x} < \mathbf{0}. \end{cases}$$

Then,

$$(\theta(\mathbf{x}), \phi(\mathbf{x})) = \int\_0^\infty \phi(\mathbf{x})d\mathbf{x} \quad \text{for} \quad \phi \in \mathcal{D}(R).$$

which implies *<sup>θ</sup>*(*x*) ∈ D (*R*).

> Let *f* ∈ D (*R*). The distributional derivative of *f* , denoted by *f* or *d f* /*dx*, is defined as:

$$(f', \phi) = -(f, \phi')$$

for *φ* ∈ <sup>D</sup>(*R*).

Clearly, *f* ∈ D(*R*) and every distribution has a derivative. As an example, we are going to show that *<sup>θ</sup>*(*x*) = *<sup>δ</sup>*(*x*) distributionally, although *<sup>θ</sup>*(*x*) is not defined at *x* = 0 in the classical sense. Indeed,

$$(\theta'(\mathbf{x}), \phi(\mathbf{x})) = -(\theta(\mathbf{x}), \phi'(\mathbf{x})) = -\int\_0^\infty \phi'(\mathbf{x})d\mathbf{x} = \phi(0) = (\delta(\mathbf{x}), \phi(\mathbf{x}))\_\*$$

which claims:

> *<sup>θ</sup>*(*x*) = *<sup>δ</sup>*(*x*).

It can be shown that the ordinary rules of differentiation also apply to distributions. For instance, the derivative of a sum is the sum of the derivatives, and a constant can be commuted with the derivative operator.

**Definition 1.** *Let f and g be distributions in* D(*R*) *satisfying either of the following conditions:*


Then, the convolution *f* ∗ *g* is defined by the equation:

$$((f \ast g)(\mathfrak{x}), \phi(\mathfrak{x})) = (\mathfrak{g}(\mathfrak{x}), (f(\mathfrak{y}), \phi(\mathfrak{x} + \mathfrak{y})))$$

for *φ* ∈ D. Clearly, we have:

$$f \ast \mathfrak{g} = \mathfrak{g} \ast f$$

from Definition 1.

> It follows from the definition above that:

$$
\delta^{(n)}(\mathbf{x} - \mathbf{x}\_0) \* f(\mathbf{x}) = f^{(n)}(\mathbf{x} - \mathbf{x}\_0),
$$

for any distribution *f* ∈ <sup>D</sup>(*R*).

Let *<sup>δ</sup>n*(*x*) = *nρ*(*nx*) be Temple's *δ*-sequence for *n* = 1, 2, ··· , where *ρ*(*x*) is a fixed, infinitely-differentiable function on *R*, having the following properties [26,27]:

(i) *ρ*(*x*) ≥ 0,

> −∞

(ii) *ρ*(*x*) = 0 for |*x*| ≥ 1,


An example of such a *ρ*(*x*) function is given as:

$$\rho(x) = \begin{cases} \ c e^{-\frac{1}{1-x^2}} & \text{if } |x| < 1, \\ 0 & \text{otherwise} \end{cases}$$

where:

$$c^{-1} = \int\_{-1}^{1} e^{-\frac{1}{1-x^2}} dx.$$

This delta-sequence plays an important role in defining products of distributions [28,29]. Let *f* be a continuous function on *R*. Then,

$$(f \* \delta\_n)(\mathbf{x}) = \int\_{-\infty}^{\infty} f(\mathbf{x} - t)\delta\_n(t)dt = \int\_{-\infty}^{\infty} f(t)\delta\_n(\mathbf{x} - t)dt$$

uniformly converges to *f* on any compact subset of *R*. Indeed, we assume that *f* is continuous on *R*, and *L* is any compact subset of *R*. Then, *f* is uniformly continuous on *L*, and for all > 0, there exists *δ*1 > 0 such that:

$$|f(\mathbf{x} - \mathbf{t}) - f(\mathbf{x})| < \epsilon$$

for all *x* ∈ *L* and |*t*| < *δ*1. This implies that:

$$|(f \* \delta\_n)(\mathbf{x}) - f(\mathbf{x})| \le \int\_{-\infty}^{\infty} |f(\mathbf{x} - t) - f(\mathbf{x})| \delta\_n(t) dt < \epsilon^2$$

for all *x* ∈ *L* and 1/*n* < *δ*1 by noting that:

$$\int\_{-\infty}^{\infty} \delta\_n(t)dt = 1.$$

Furthermore, if *f* ∈ <sup>D</sup>(*R*), then (*f* ∗ *<sup>δ</sup>n*)(*x*) converges to *f* in <sup>D</sup>(*R*). Indeed,

$$\lim\_{n \to \infty} \left( (f \ast \delta\_n)(\mathbf{x}), \phi(\mathbf{x}) \right) = \lim\_{n \to \infty} (f(\mathbf{x}), (\delta\_n(y), \phi(\mathbf{x} + y))) = (f(\mathbf{x}), \phi(\mathbf{x})) $$

by noting that:

$$\lim\_{n \to \infty} (\delta\_n(y), \phi(x+y)) = \phi(x)$$

in <sup>D</sup>(*R*).

In order to study the half-order Laplacian operator in the distribution, we introduce an infinitely-differentiable function *τ*(*x*) satisfying the following conditions:

(i) *τ*(*x*) = *<sup>τ</sup>*(−*<sup>x</sup>*), (ii) 0 ≤ *τ*(*x*) ≤ 1, (iii) *τ*(*x*) = 1 if |*x*| ≤ 1/2,(iv)*τ*(*x*)=0if|*x*|≥1.

Define:

$$\phi\_m(\mathbf{x}) = \begin{cases} 1 & \text{if } |\mathbf{x}| \le m\_\prime \\ \pi(m^m \mathbf{x} - m^{m+1}) & \text{if } \mathbf{x} > m\_\prime \\ \pi(m^m \mathbf{x} + m^{m+1}) & \text{if } \mathbf{x} < -m\_\prime \end{cases}$$

for *m* = 1, 2, ··· . Then, *φm*(*x*) ∈ D(*R*) with the support [−*<sup>m</sup>* − *<sup>m</sup>*<sup>−</sup>*m*, *m* + *<sup>m</sup>*<sup>−</sup>*<sup>m</sup>*]. From Equation (3), we have:

$$(-\Delta)^{1/2}u(\mathbf{x}) = -\frac{1}{\pi}(\text{P.V.}\frac{1}{t^{2\prime}}, u(\mathbf{x}+t)) = -\frac{1}{\pi}(\text{P.V.}\frac{1}{t^{\prime}}, u'(\mathbf{x}+t))$$

if *u* ∈ D(*R*) as:

$$\begin{aligned} \text{(P.V.} \frac{1}{t^2}, \phi(t)) &= \int\_0^\infty \frac{\phi(t) - 2\phi(0) + \phi(-t)}{t^2} dt, \\\frac{d}{dt} \text{P.V.} \frac{1}{t} &= -\text{P.V.} \frac{1}{t^2}. \end{aligned}$$

This suggests the following explicit definition for defining (−<sup>Δ</sup>) 12 *<sup>u</sup>*(*x*). This explicit definition directly evaluates the half-order fractional Laplacian of *u*(*x*) as a function of *x*, without relating to any testing function in the Schwartz space.

**Definition 2.** *Let u* ∈ D(*R*) *and un* = *u* ∗ *δn* = (*u*(*t*), *<sup>δ</sup>n*(*<sup>x</sup>* − *t*)) *for n* = 1, 2, ··· *. We define the half-order Laplacian operator* (−<sup>Δ</sup>) 12 *on* D(*R*) *as:*

$$\begin{aligned} \left( (-\Delta)^{\frac{1}{2}} u(\mathbf{x}) \right) &= -\lim\_{\mathcal{T} \to \infty} \lim\_{n \to \infty} \left( P.V.\frac{1}{t'}, \phi\_m(t) u'\_n(\mathbf{x} + t) \right) \\ &= -\frac{1}{\mathcal{T}} \lim\_{m \to \infty} \lim\_{n \to \infty} \int\_0^\infty \phi\_m(t) \frac{u'\_n(\mathbf{x} + t) - u'\_n(\mathbf{x} - t)}{t} dt \end{aligned} \tag{5}$$

*if it exists.*

> Clearly, the integral:

$$\int\_0^\infty \phi\_m(t) \frac{u\_n'(x+t) - u\_n'(x-t)}{t} dt$$

is well defined as *φm*(*t*) has a bounded support and:

$$\lim\_{t \to 0^+} \frac{u\_n'(\mathbf{x} + \mathbf{t}) - u\_n'(\mathbf{x} - \mathbf{t})}{t} = u\_n''(\mathbf{x}).$$

**Theorem 1.**

$$(-\Delta)^{\frac{1}{2}}\delta(\mathfrak{x}) = -\frac{1}{\pi}P.V.\frac{1}{\mathfrak{x}^2}.$$

**Proof.** Assuming *x* > 0, we choose a large *n* such that 1/*n* < *x*. This infers that *<sup>δ</sup>n*(*<sup>x</sup>* + *t*) = 0, and:

$$\begin{aligned} \left( ( - \Delta )^{\frac{1}{2}} \delta (\mathbf{x}) \right) &= \lim\_{m \to \infty} \lim\_{n \to \infty} -\frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{\delta\_n'(\mathbf{x} + t) - \delta\_n'(\mathbf{x} - t)}{t} dt \\ &= \lim\_{m \to \infty} \lim\_{n \to \infty} \frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{\delta\_n'(\mathbf{x} - t)}{t} dt \end{aligned}$$

from Definition 2. Making the substitution *w* = *x* − *t*, we get:

$$\begin{aligned} (-\Delta)^{\frac{1}{2}}\delta(x) &= \lim\_{m \to \infty} \lim\_{n \to \infty} \frac{1}{\pi} \int\_{-\infty}^{x} \delta\_n'(w) \frac{\phi\_m(x - w)}{x - w} dw \\ &= \lim\_{m \to \infty} \lim\_{n \to \infty} \frac{1}{\pi} \int\_{-\infty}^{\infty} \delta\_n'(w) \frac{\phi\_m(x - w)}{x - w} dw \\ &= \lim\_{m \to \infty} -\frac{1}{\pi} \left. \frac{\partial}{\partial w} \left[ \frac{\phi\_m(x - w)}{x - w} \right] \right|\_{w = 0} \\ &= \lim\_{m \to \infty} \frac{\phi\_m'(x)x - \phi\_m(x)}{\pi x^2} \end{aligned}$$

by noting that *<sup>δ</sup>n*(*w*) is a delta sequence, and:

$$\frac{\phi\_m(x-w)}{x-w}$$

is a testing function of *w* if *w* < *x*, and:

$$
\delta\_n'(w) \frac{\phi\_m(x - w)}{x - w}
$$

is identical to zero if *w* ≥ *x* as *<sup>δ</sup>n*(*w*) = 0. Choosing *m* such that *x* < *m*, we derive that:

$$
\phi'\_m(\mathbf{x}) = 0 \quad \text{and} \quad \phi\_m(\mathbf{x}) = 1
$$

as *φm*(*x*) = 1 if |*x*| < *m*. Hence,

$$(-\Delta)^{\frac{1}{2}}\delta(x) = -\frac{1}{\pi} \text{P.V.} \frac{1}{x^2}.$$

If *x* < 0, we set *y* = −*x* and:

$$\begin{split} -\frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{\delta\_n'(\mathbf{x}+t) - \delta\_n'(\mathbf{x}-t)}{t} dt &= -\frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{\delta\_n'(-y+t) - \delta\_n'(-y-t)}{t} dt \\ &= -\frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{\delta\_n'(y+t) - \delta\_n'(y-t)}{t} dt, \end{split}$$

which implies that:

$$(-\Delta)^{\frac{1}{2}}\delta(x) = \lim\_{m \to \infty} \frac{\phi\_m'(y)y - \phi\_m(y)}{\pi y^2} = -\frac{1}{\pi} \text{P.V.} \frac{1}{x^2}.$$

Finally, we have for *x* = 0 by making the variable change *u* = *nt*:

$$\begin{aligned} \int\_0^\infty \phi\_m(t) \frac{\delta\_n'(t) - \delta\_n'(-t)}{t} dt &= \ 2 \int\_0^\infty \phi\_m(t) \frac{\delta\_n'(t)}{t} dt \\ &= \ 2n^2 \int\_0^\infty \phi\_m(t) \frac{\rho'(nt)}{t} dt \\ &= \ 2n^2 \int\_0^1 \phi\_m(u/n) \frac{\rho'(u)}{u} du = 2n^2 \int\_0^1 \frac{\rho'(u)}{u} du \\ &= \ \ \mathcal{O}(n^2) \end{aligned}$$

by noting that the term *φm*(*u*/*n*) = 1, and:

$$\int\_0^1 \frac{\rho'(u)}{u} du = \int\_0^1 \frac{\rho'(u) - \rho'(0)}{u} du$$

is well defined as *ρ*(*u*) is an even function. This completes the proof of Theorem 1.

From [24], we have the distributions P.V.*x*<sup>−</sup>2*<sup>m</sup>* for *m* = 1, 2, ··· and P.V.*x*<sup>−</sup>2*m*−<sup>1</sup> for *m* = 0, 1, ··· given as:

$$\begin{split} \left( \text{P.V.} \, ^{-2m}{\prime}, \phi \right) &= \int\_{0}^{\infty} \text{x}^{-2m} \{ \phi(\mathbf{x}) + \phi(-\mathbf{x}) \\ -2 \left[ \phi(0) + \frac{\mathbf{x}^{2}}{2!} + \dots + \frac{\mathbf{x}^{2m-2}}{(2m-2)!} \phi^{(2m-2)}(0) \right] \mathbf{d}x, \\ \left( \text{P.V.} \, ^{-2m-1}{\prime}, \phi \right) &= \int\_{0}^{\infty} \text{x}^{-2m-1} \{ \phi(\mathbf{x}) - \phi(-\mathbf{x}) \\ -2 \left[ \mathbf{x} \phi'(0) + \frac{\mathbf{x}^{3}}{3!} + \dots + \frac{\mathbf{x}^{2m-1}}{(2m-1)!} \phi^{(2m-1)}(0) \right] \mathbf{d}x. \end{split}$$

**Theorem 2.**

$$(-\Delta)^{\frac{1}{2}}\delta^{(m)}(\mathbf{x}) = \frac{(-1)^{m+1}(m+1)!}{\pi}P.V.\frac{1}{\mathfrak{x}^{m+2}}$$

*for m* = −1, 0, 1, . . . *. In particular, we have for m* = −1 *that:*

$$(-\Delta)^{\frac{1}{2}}\theta(\mathfrak{x}) = \frac{1}{\pi}P.V.\frac{1}{\mathfrak{x}}.$$

**Proof.** We start with the case *m* = −1. Then, by Definition 2,

$$\begin{aligned} (-\Delta)^{\frac{1}{2}}\theta(\mathbf{x}) &= \lim\_{m \to \infty} \lim\_{n \to \infty} -\frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{\delta\_n(\mathbf{x}+t) - \delta\_n(\mathbf{x}-t)}{t} dt \\ &= \lim\_{m \to \infty} \lim\_{n \to \infty} \frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{\delta\_n(\mathbf{x}-t)}{t} dt \end{aligned}$$

for *x* > 0 and a large *n* such that 1/*n* < *x*. Following the proof of Theorem 1, we come to:

$$\begin{split} (-\Delta)^{\frac{1}{2}}\theta(\mathbf{x}) &= \lim\_{m \to \infty} \lim\_{n \to \infty} \frac{1}{\pi} \int\_{-\infty}^{\chi} \delta\_{n}(w) \frac{\phi\_{m}(\mathbf{x} - w)}{\mathbf{x} - w} dw \\ &= \lim\_{m \to \infty} \lim\_{n \to \infty} \frac{1}{\pi} \int\_{-\infty}^{\infty} \delta\_{n}(w) \frac{\phi\_{m}(\mathbf{x} - w)}{\mathbf{x} - w} dw \\ &= \lim\_{m \to \infty} \frac{\phi\_{m}(\mathbf{x})}{\pi \mathbf{x}} \\ &= \frac{1}{\pi} \operatorname{PV} \frac{1}{\chi} .\end{split}$$

The case *x* < 0 follows similarly. To compute (−<sup>Δ</sup>) 12 *<sup>δ</sup>*(*m*)(*x*), we note that for a large *n*:

$$(-\Delta)^{\frac{1}{2}}\delta^{(m)}(\mathbf{x}) = \lim\_{m \to \infty} \lim\_{n \to \infty} \frac{1}{\pi} \int\_{-\infty}^{\infty} \delta\_n^{(m+1)}(w) \frac{\phi\_m(\mathbf{x} - w)}{\mathbf{x} - w} dw$$

from the proof of Theorem 1 and:

$$\begin{split} &\lim\_{m\to\infty} \frac{(-1)^{m+1}}{\pi} \left. \frac{\partial^{m+1}}{\partial w^{m+1}} \right| \left. \frac{\phi\_m(\mathbf{x} - \boldsymbol{w})}{\mathbf{x} - \boldsymbol{w}} \right| \bigg|\_{\mathbf{w}=\mathbf{0}} \\ &= \lim\_{m\to\infty} \frac{(-1)^{m+1}}{\pi} \sum\_{k=0}^{m+1} \binom{m+1}{k} (-1)^k \phi\_m^{(k)}(\mathbf{x} - \boldsymbol{w}) \left. \left( \frac{1}{\mathbf{x} - \boldsymbol{w}} \right)^{(m+1-k)} \right|\_{\mathbf{w}=\mathbf{0}} \\ &= \lim\_{m\to\infty} \frac{(-1)^{m+1}}{\pi} \binom{m+1}{0} \phi\_m(\mathbf{x} - \boldsymbol{w}) \left. \left( \frac{1}{\mathbf{x} - \boldsymbol{w}} \right)^{(m+1)} \right|\_{\mathbf{w}=0} \\ &= \frac{(-1)^{m+1} (m+1)!}{\pi} \text{P.V.} \frac{1}{\mathbf{x}^{m+2}} \end{split}$$

since lim *m*→∞ *φm*(*x*) = 1 and:

$$\lim\_{m \to \infty} \frac{(-1)^{m+1}}{\pi} \sum\_{k=1}^{m+1} \binom{m+1}{k} (-1)^k \phi\_m^{(k)}(x-w) \left. \left( \frac{1}{x-w} \right)^{(m+1-k)} \right|\_{w=0} = 0$$

from the definition of *φm*(*x*). This completes the proof of Theorem 2.

We should note that Theorem 2 cannot be derived by the standard definition of the fractional Laplacian via Fourier transform, and:

$$(-\Delta)^{\frac{1}{2}}\theta(\mathbf{x}) = \frac{1}{\pi}\text{P.V.}\frac{1}{\mathbf{x}}$$

holds in the distributional sense and:

$$\frac{d}{dx}\theta(x) = \delta(x),$$

although *<sup>θ</sup>*(*x*) is discontinuous at *x* = 0 in the classical sense. **Theorem 3.**

$$(-\Delta)^{\frac{1}{2}}\sin x = \sin x,\tag{6}$$

$$(-\Delta)^{\frac{1}{2}}\cos x = \cos x,\tag{7}$$

$$(-\Delta)^{\frac{1}{2}}(a\mathbf{x}+b) = 0.\tag{8}$$

*where a and b are arbitrary constants.*

**Proof.** Clearly, sin *x* is an infinitely-differentiable function on *R*. This claims that:

$$(\sin \mathfrak{x})' \ast \delta\_{\mathfrak{n}} = \cos \mathfrak{x} \ast \delta\_{\mathfrak{n}}$$

uniformly converges to cos *x* on any compact subset of *R*. Therefore,

$$\begin{aligned} \left(( - \Delta)^{\frac{1}{2}} \sin x \right) &= \lim\_{m \to \infty} -\frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{\cos(x + t) - \cos(x - t)}{t} dt \\ &= \lim\_{m \to \infty} -\frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{-2 \sin x \sin t}{t} dt \\ &= \frac{2 \sin x}{\pi} \int\_0^\infty \frac{\sin t}{t} dt = \sin x \end{aligned}$$

by using:

$$\int\_0^\infty \frac{\sin t}{t} dt = \pi/2.$$

Similarly,

$$\begin{array}{rcl} ( - \Delta )^{\frac{1}{2}} \cos x &=& \lim\_{m \to \infty} -\frac{1}{\pi} \int\_{0}^{\infty} \phi\_{m}(t) \frac{-\sin(x+t) + \sin(x-t)}{t} dt \\ &=& \lim\_{m \to \infty} -\frac{1}{\pi} \int\_{0}^{\infty} \phi\_{m}(t) \frac{-2 \cos x \sin t}{t} dt \\ &=& \frac{2 \cos x}{\pi} \int\_{0}^{\infty} \frac{\sin t}{t} dt = \cos x. \end{array}$$

Finally,

$$\begin{aligned} (-\Lambda)^{\frac{1}{2}}(ax+b) &= \lim\_{m \to \infty} \lim\_{n \to \infty} -\frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{(b\*\delta\_n)(x+t) - (b\*\delta\_n)(x-t)}{t} dt \\ &= \lim\_{m \to \infty} -\frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{b-b}{t} dt = 0. \end{aligned}$$

This completes the proof of Theorem 3.

We would like to mention that Theorem 3 extends several classical results obtained in [30] to distributions by Definition 2 with a new approach.
