Lévy-Type Dirichlet Problems on the Half-Line: Probabilistic Mild Solutions and Weighted Energy Estimates

Abstract

This paper studies Dirichlet problems for one-dimensional Lévy-type nonlocal elliptic equations on the half-line. The equation $L^{\mu }\nu (x)=f(x),x>0,\nu (x)=0,x\le 0$ is transformed into a weighted nonlocal equation associated with a multiplicative jump process. Under basic structural assumptions on the Lévy measure, the transformed generator is realized through a martingale problem, and the associated exponential killing representation gives a probabilistic mild solution with an immediate $L^{\infty }$-estimate. For the one-dimensional fractional Laplacian, the transformed process is exactly multiplicative. This yields a new approach in which solution estimates are derived from the stochastic equation of the transformed process; smooth-data resolvent solutions are estimated in weighted $L_p$-spaces and extended to general data by approximation. For more general Lévy measures, a smooth weighted energy estimate is proved. The key analytic input is a weighted adjoint integral inequality for the transformed generator, verified for subordinate Brownian motions associated with Bernstein functions and for non-unimodal logarithmically perturbed stable-type operators.

1. Introduction

Lévy-type nonlocal operators arise as generators of stochastic jump processes and appear widely in analysis, probability, finance, and physics. These operators provide a flexible framework for modeling phenomena with long-range interactions or discontinuous paths, ranging from anomalous diffusion in material science [1] to asset price dynamics in financial markets [2]. Consequently, boundary value problems for Lévy-type operators are of particular importance, as they correspond to the killing or absorption of the underlying stochastic process upon exiting the domain. In this context, the half-line (and, by extension, the half-space in higher dimensions) serves as the fundamental geometric setting. It is the simplest domain that retains all the essential difficulties associated with boundary behavior, making it an ideal proving ground for new analytical techniques. Furthermore, a rigorous understanding of such problems on the half-line for a wide variety of Lévy measures is not merely a mathematical exercise; it is directly motivated by applications in financial mathematics, where these operators model asset price dynamics with absorption at a barrier [2].

For a Lévy measure $\mu$ with $\sigma =\inf \left\{\alpha <2:{\int }_{\left|y\right|\le 1}{\left|y\right|}^{\alpha }\mu \left(dy\right)\right\}\in \left(0,2\right),$ we consider the operator

$$L^{\mu }\nu \left(x\right)=\int \left[\nu \left(x+y\right)-\nu \left(x\right)-{\chi }_{\sigma }\left(y\right)y{\nu }^{\prime }\left(x\right)\right]\mu \left(dy\right),$$

where the truncation function is given by ${\chi }_{\sigma }\left(y\right)=1$ if $\sigma >1,$${\chi }_{\sigma }\left(y\right)={\mathbf{1}}_{\left\{\left|y\right|\le 1\right\}}$ if $\sigma =1$ and ${\chi }_{\sigma }\left(y\right)=0$ if $\sigma <1.$

In this paper, we investigate the weighted Sobolev regularity of the half-line Dirichlet problem

$$\begin{array}{cc}\hfill L^{\mu }\nu \left(x\right)& =f\left(x\right),x>0,\hfill \\ \hfill \nu \left(x\right)& =0,x\le 0.\hfill \end{array}$$

(1)

For the special case of the fractional Laplacian ${\left(-\mathsf{\Delta }\right)}^{\alpha /2}$ with $\mu \left(dy\right)=c_{\alpha }{\left|y\right|}^{-d-\alpha }dy$, weighted Sobolev regularity has been studied in [3,4] with weights consisting of some appropriate powers of the distance to the boundary denoted by $\rho \left(x\right)$. In particular, the following regularity of solutions were obtained for the domain D and $\gamma \in \mathbf{R}$,

$${∥\nu ∥}_{H_{p,\eta -\alpha p/2}^{\gamma +\alpha }\left(D\right)}\le C{∥f∥}_{H_{p,\eta +\alpha p/2}^{\gamma }\left(D\right)}$$

Here, ${\parallel ·\parallel }_{H_{p,\eta }^{\gamma }\left(D\right)}$ denotes the weighted Sobolev norm defined in [3] (Definition 2.16) with $\eta$ in the range of $\left(d-1,d-1+p\right)$. If $\gamma =0$, the above weighted inequality becomes

$${\int }_D\left({\left|{\rho }^{-\alpha /2}\nu \right|}^p+{\left|{\rho }^{\alpha /2}{\mathsf{\Delta }}^{\alpha /2}\nu \right|}^p\right){\rho }^{\eta -d}dx\le C{\int }_D{\left|{\rho }^{\alpha /2}f\right|}^p{\rho }^{\eta -d}dx.$$

(2)

In fact, the $C^{1,\alpha }$-domains, non-zero exterior conditions, and parabolic equations were also studied in [3,4,5]. In approaches based on weighted function spaces, $C^{1,\alpha }$-domains can be reduced to the full-space setting via boundary flattening, albeit with some technical effort. By contrast, the half-space does not admit such a reduction within this framework and must be treated as a genuinely distinct configuration. This highlights a structural difference between the two settings.

Despite its simple form, (1) is a well-known challenging problem, as reflected in technical tools required in previous work. In [3], the authors used the probability density estimates of killed stable processes from [6] to derive the zeroth-order estimates which are tailored specifically to the case of $\mu \left(dy\right)={\left|y\right|}^{-d-\alpha }dy.$ This approach becomes computationally involved and appears difficult to extend beyond this case. In addition, little is known about transition probability density if $\mu$ is not a unimodal measure. In [4], a more analytic approach was developed to obtain zeroth-order estimates. Their method relies on direct technical computations involving the distance to the boundary and the Lévy measure, thereby avoiding the use of transition density estimates. Nevertheless, it remains unclear whether this approach can be extended to cover Lévy measures with more general radial profiles.

This paper develops a probabilistic and weighted framework for Dirichlet problems associated with Lévy-type operators on the half-line. The contribution is threefold. First, we introduce a probabilistic mild solution concept through the martingale problem for a transformed multiplicative jump process and an exponential killing representation. This construction requires only the basic structural assumptions imposed on the Lévy measure and therefore applies to a broad class of examples.

Second, for the one-dimensional fractional Laplacian, exact multiplicativity of the transformed process gives the representation $Y_t^x=x{\mathcal{E}}_t$, where the SDE for ${\mathcal{E}}_t$ can be written explicitly. This structure provides a new route to weighted estimates: the probabilistic mild solutions are estimated directly from the stochastic representation and the multiplicative identity. Starting with smooth data, the corresponding probabilistic mild solutions are estimated in $L_{\theta ,p}\left((0,\infty )\right)$, and the construction is then extended to general data by approximation. The resulting limit solution is well defined in $L_{\theta ,p}\left((0,\infty )\right)$, and the associated transformed operator values also converge in $L_{\theta ,p}\left((0,\infty )\right)$. This argument follows a different route from the Green-kernel and analytic methods used in [3,4]. After the parameter identification

$$\eta =\theta p+{\displaystyle \frac{\alpha p}{2}},$$

the admissible range obtained here agrees with the range in [3]. When $\lambda =0$, the constructed limit solution agrees with the corresponding unique weak solution of [3] (Theorem 2.3(ii)).

Third, for more general Lévy measures, we identify the weighted energy mechanism behind the $L_p$-theory. Theorem 3 proves that, at the smooth level, the weighted $L_p$-norm of u is controlled by the weighted $L_p$-norm of $w{L}^{\mu }u-\lambda u$, provided a weighted adjoint integral inequality holds. Moreover, whenever a graph-convergent approximation

$$u_n\to u,w{L}^{\mu }{u}_n\to w{L}^{\mu }u\mathrm{in}{L}_{\theta ,p}\left((0,\infty )\right)$$

can be constructed, the same estimate passes to the limiting function u. Thus, the theorem separates the analytic weighted estimate from the additional domain question of constructing such approximations. This domain question is left for further future investigation for general Lévy measures. Nevertheless, we view the smooth weighted energy estimate and the verification of the weighted adjoint inequality as significant stepping stones toward a broader weighted $L_p$-theory for nonlocal Dirichlet problems, which remains largely open beyond the stable setting. The required weighted adjoint inequality is verified in Theorem 4 for subordinate Brownian motions associated with Bernstein functions, and the same argument also covers certain non-unimodal logarithmically perturbed stable-type kernels.

2. Notations and Assumptions

Let $\mu$ be a Lévy measure on $\mathbf{R}\setminus \left\{0\right\}$, i.e., ${\int }_{\mathbf{R}}\left(1\wedge {\left|y\right|}^2\right)\mu \left(dy\right)<\infty .$ The Lévy-type operator acting on $\nu \in C_c^{\infty }\left(\mathbf{R}\right)$ is

$$L^{\mu }\nu \left(x\right)=\int \left[\nu \left(x+y\right)-\nu \left(x\right)-{\chi }_{\sigma }\left(y\right)y{\nu }^{\prime }\left(x\right)\right]\mu \left(dy\right),$$

where the truncation function is given by ${\chi }_{\sigma }\left(y\right)=1$ if $\sigma >1,$${\chi }_{\sigma }\left(y\right)={\mathbf{1}}_{\left\{\left|y\right|\le 1\right\}}$ if $\sigma =1$ and ${\chi }_{\sigma }\left(y\right)=0$ if $\sigma <1.$

We consider the following equation

$$\begin{array}{cc}\hfill L^{\mu }\nu \left(x\right)& =f\left(x\right),x>0\hfill \\ \hfill \nu \left(x\right)& =0,x\le 0.\hfill \end{array}$$

(3)

We adopt the following convention in this paper.

• For any function f on $\left(0,\infty \right),$ we write $f_{\ast }={\inf }_{x>0}f\left(x\right)$ and $f^{\ast }={sup}_{x>0}f\left(x\right).$
• We generally use C possibly with subscriptions such as $C_1,C_2,\dots$ to denote generic bounding constant which may vary from line to line.
• We write $f\left(x\right)\asymp g\left(x\right)$ if there exists a constant $C>0$, independent of $x,$ such that

  $$C^{-1}g\left(x\right)\le f\left(x\right)\le Cg\left(x\right),x>0.$$
  We write $f\left(x\right)\lesssim g\left(x\right)$ and $f\left(x\right)\gtrsim g\left(x\right)$ if only one side of the inequalities holds.
• We write $B_b\left((0,\infty )\right)$ for the space of all bounded Borel measurable functions on $(0,\infty )$.

2.1. Auxiliary Functions

For fixed $x>0$, define the tail function $\delta \left(x\right)=\mu \left(\left\{\left|y\right|\ge x\right\}\right),$ and the associated weight function $w\left(x\right)=\delta {\left(x\right)}^{-1}.$ As an example, if $\mu \left(dy\right)=c_{\alpha }{\left|y\right|}^{-1-\alpha }dy,$ then $\delta \left(x\right)\asymp x^{-\alpha },w\left(x\right)\asymp x^{\alpha }.$ Fix $x>0,$ and define the rescaled measures ${\tilde{\mu }}_x\left(dr\right)=w\left(x\right){\mu }_x\left(dr\right).$

We introduce the boundary related coefficients

$$\rho \left(x\right)={\int }_{\left(-\infty ,-1\right]}{\tilde{\mu }}_x\left(dr\right)$$

and for $\sigma \in \left(1,2\right)$

$$\kappa \left(x\right)=-{\int }_{\left(-\infty ,-1\right]}r{\tilde{\mu }}_x\left(dr\right).$$

In fact, if $\mu$ is symmetric, then

$$\rho \left(x\right)=w\left(x\right)\mu \left(\left(-\infty ,-x\right]\right)=w\left(x\right)\mu \left(\left[x,\infty \right)\right)={\displaystyle \frac{1}{2}}w\left(x\right)\delta \left(x\right)={\displaystyle \frac{1}{2}}.$$

2.2. Assumptions

We impose the following standard assumption throughout the paper.

Assumption 1

(Centering Condition). If $\sigma =1$, then ${\int }_{R<\left|y\right|\le R^{\prime }}y\mu \left(dy\right)=0$ for all $0<R<R^{\prime }<\infty .$

Assumption 2

(Uniform Integrability). There exists $C>0$ such that for all $x>0$

$${\int }_{\left|r\right|\le 1}\left|r\right|{\tilde{\mu }}_x\left(dr\right)+{\int }_{\left|r\right|>1}{\tilde{\mu }}_x\left(dr\right)\le C,\mathit{if}\sigma \in \left(0,1\right),$$

$${\int }_{\left|r\right|\le 1}{\left|r\right|}^2{\tilde{\mu }}_x\left(dr\right)+{\int }_{\left|r\right|>1}{\tilde{\mu }}_x\left(dr\right)\le C,\mathit{if}\sigma =1,$$

$${\int }_{\left|r\right|\le 1}{\left|r\right|}^2{\tilde{\mu }}_x\left(dr\right)+{\int }_{\left|r\right|>1}\left|r\right|{\tilde{\mu }}_x\left(dr\right)\le C,\mathit{if}\sigma \in \left(1,2\right).$$

Assumption 3

(Measurability and Regularity).

(i) 

For every Borel set $B\subset \left(-1,\infty \right)$, the map $x\mapsto {\tilde{\mu }}_x\left(B\right)$ is Borel measurable.

(ii) 

The tail function $\delta \left(x\right)=\mu \left(\left|y\right|\ge x\right)$ is continuous on $\left(0,\infty \right)$ with $\delta \left(x\right)>0$ for all $x>0$ and ${lim}_{x\to 0^{+}}\delta \left(x\right)=\infty .$

(iii) 

μ has no atoms on $\mathbf{R}\setminus \left\{0\right\}.$

Numerous examples of $\mu$ that are covered under Assumptions 1–3 that go beyond the classical $\alpha -$stable type were provided in [7] and references therein. Additionally, our main results require the weighted integral inequality (see Theorem 3). Here, we briefly discuss concrete examples for which our main results apply.

Example 1.

Let ${\left(S_t\right)}_{t\ge 0}$ be a subordinator (increasing Lévy process starting at 0) with Laplace exponent ϕ and $W_t$ an independent one-dimensional Brownian motion. The process $X_t=W_{S_t}$ is called a subordinate Brownian motion. Its characteristic exponent is given by $\varphi \left({\left|\xi \right|}^2\right),\xi \in \mathbf{R}.$ If ϕ is a complete Bernstein function, then the Lévy measure μ of $X_t$ admits a symmetric density

$$\mu \left(dr\right)=j\left(\left|r\right|\right)dr,$$

where

$$j\left(\left|r\right|\right)={\int }_0^{\infty }{\left(4\pi t\right)}^{-1/2}{e}^{-r^2/\left(4t\right)}{\mu }_S\left(dt\right).$$

Moreover, if ϕ satisfies weak lower and upper scaling conditions (see assumptions (H1) and (H2) of [8]), then some asymptotic estimates of j can be given in terms of $\varphi .$

We recall the following assumptions (H1) and (H2) from [8] for convenience.

• (H1) There exist constants $0<{\delta }_1\le {\delta }_2<1$ and $a_1,a_2>0$ such that

  $$a_1{\left({\displaystyle \frac{R}{r}}\right)}^{{\delta }_1}\le {\displaystyle \frac{\varphi \left(R\right)}{\varphi \left(r\right)}}\le a_2{\left({\displaystyle \frac{R}{r}}\right)}^{{\delta }_2},1\le r\le R.$$
• (H2) There exist constants $0<{\delta }_3\le {\delta }_4<1$ and $a_3,a_4>0$ such that

  $$a_3{\left({\displaystyle \frac{R}{r}}\right)}^{{\delta }_3}\le {\displaystyle \frac{\varphi \left(R\right)}{\varphi \left(r\right)}}\le a_4{\left({\displaystyle \frac{R}{r}}\right)}^{{\delta }_4},0<r\le R\le 1.$$

Under (H1), ϕ is an O-RV function at ∞ with the lower index and the upper index ${\delta }_1\le p_{\infty }\le q_{\infty }\le {\delta }_2.$ Similarly, under (H2), ϕ is an O-RV function at 0 with the lower index and upper index ${\delta }_3\le p_0\le q_0\le {\delta }_4.$

Here, we provide the list of concrete ϕ which are complete Bernstein functions and have weak scaling properties:

(i) 

$\varphi \left(\lambda \right)={\left(\lambda +{\lambda }^{\alpha }\right)}^{\beta },\alpha ,\beta \in \left(0,1\right)$

(ii) 

$\varphi \left(\lambda \right)={\lambda }^{\alpha }{\left(\log \left(1+\lambda \right)\right)}^{\beta },\alpha \in \left(0,1\right),\beta \in \left(0,1-\alpha \right)$

(iii) 

$\varphi \left(\lambda \right)={\left(\log \left(\cosh \left(\sqrt{\lambda }\right)\right)\right)}^{\alpha },\alpha \in \left(0,1\right)$

(iv) 

$\varphi \left(\lambda \right)={\left(\log \left(\sinh \left(\sqrt{\lambda }\right)\right)-\log \sqrt{\lambda }\right)}^{\alpha },\alpha \in \left(0,1\right).$

Assumption 2 can be justified by [7] (Lemma 4) subject to appropriate restrictions on parameters. We omit the details for each choice of $\varphi .$ Assumption 3 is easily verified because j is continuous. Moreover, Theorem 4 establishes that the weighted integral inequality required in Theorem 3.

Example 2.

More generally, we let ϕ satisfy assumptions (H1) and (H2). Suppose $j\left(r\right)=j\left(\left|r\right|\right)$ and

$$j\left(r\right)\asymp r^{-1}\varphi \left(r^{-2}\right),\left|j^{\prime }\left(r\right)\right|\le C{r}^{-2}\varphi \left(r^{-2}\right),r>0.$$

Then, the assumptions of our framework remain valid. In particular, Assumption 2 can again be justified by [7] (Lemma 4) and while the weighted integral inequality required in Theorem 3 can be verified by the same argument as in Theorem 4.

This allows oscillatory and non-unimodal examples which do not arise from Bernstein functions. For instance, we take

$$\varphi \left(\lambda \right)={\lambda }^{\alpha }\left(1+ϵ\sin \left(\log \lambda \right)\right),\alpha \in \left(0,1\right),ϵ\in \left({\displaystyle \frac{\alpha }{\sqrt{1+{\alpha }^2}}},1\right).$$

Since $\left(1-ϵ\right){\lambda }^{\alpha }\le \varphi \left(\lambda \right)\le \left(1+ϵ\right){\lambda }^{\alpha }$, ϕ satisfies assumptions (H1) and (H2) with ${\delta }_1={\delta }_2={\delta }_3={\delta }_4=\alpha .$

• Define

$$j\left(r\right)=r^{-1}\varphi \left(r^{-2}\right)=r^{-1-2\alpha }\left(1-ϵ\sin \left(2\log r\right)\right),r>0.$$

Then, $j\left(r\right)>0$ and $\mu \left(dr\right)=j\left(\left|r\right|\right)dr$ is a symmetric Lévy measure.

• Also,

$$j^{\prime }\left(r\right)=\left(-1-2\alpha \right)r^{-2-2\alpha }\left(1-ϵ\sin \left(2\log r\right)\right)-2ϵr^{-2-2\alpha }\cos \left(2\log r\right).$$

Therefore, $\left|j^{\prime }\left(r\right)\right|\le C{r}^{-2-2\alpha }\le C{r}^{-2}\varphi \left(r^{-2}\right).$

• Moreover,

$$\begin{array}{cc}\hfill {\varphi }^{\prime }\left(\lambda \right)& ={\lambda }^{\alpha -1}ϵ\cos \left(\log \lambda \right)+\alpha {\lambda }^{\alpha -1}\left(1+ϵ\sin \left(\log \lambda \right)\right)\hfill \\ \hfill & ={\lambda }^{\alpha -1}\left(\alpha +ϵ\left(\cos \left(\log \lambda \right)+\alpha \sin \left(\log \lambda \right)\right)\right)\hfill \\ \hfill & ={\lambda }^{\alpha -1}\left(\alpha +ϵH\left(\log \lambda \right)\right).\hfill \end{array}$$

where $H\left(t\right)=\cos \left(t\right)+\alpha \sin \left(t\right)=\sqrt{1+{\alpha }^2}\sin \left(t+\theta \right)$ with $\theta =\arctan \left({\displaystyle \frac{1}{\alpha }}\right).$ Therefore, ${\inf }_{\lambda >0}\left(\alpha +ϵH\left(\log \lambda \right)\right)=\alpha -ϵ\sqrt{1+{\alpha }^2}$, which is negative if ${\displaystyle \frac{\alpha }{\sqrt{1+{\alpha }^2}}}<ϵ<1.$

In particular, ϕ cannot be a Bernstein function because ${\varphi }^{\prime }$ becomes negative for some λ. Thus, the admissible class strictly extends beyond Bernstein-function construction in Example 1.

2.3. Weighted Function Spaces

Let $1\le p<\infty$ and $\theta \in \mathbf{R}$. We define $L_{\theta ,p}\left((0,\infty )\right)$ as the space of all measurable functions u on $(0,\infty )$ such that

$${\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}={\left({\int }_0^{\infty }{\left|u\left(x\right)\right|}^p{x}^{\theta p-1}dx\right)}^{1/p}<\infty .$$

3. Transformed Equation

We now transform (3). First, we investigate operator $L.$

Lemma 1.

If $\nu \in C_c^{\infty }\left(\mathbf{R}\right)$ with supp $\nu \subset \left(0,\infty \right)$ then $w\left(x\right)L^{\mu }\nu \left(x\right)=A\nu \left(x\right)-\rho \left(x\right)\nu \left(x\right),x>0$ where the operator A is given by

$$A\nu \left(x\right)={\int }_{\left(-1,\infty \right)}\left[\nu \left(x+xr\right)-\nu \left(x\right)-{\chi }_{\sigma }\left(r\right)xr{\nu }^{\prime }\left(x\right)\right]{\tilde{\mu }}_x\left(dr\right)+{\mathbf{1}}_{\left\{\sigma >1\right\}}x\kappa \left(x\right){\nu }^{\prime }\left(x\right),x>0,$$

and κ is defined in Section 2.1.

Proof. 

By Assumption 1, we have

$$L^{\mu }\nu \left(x\right)=\int \left[\nu \left(x+y\right)-\nu \left(x\right)-{\mathbf{1}}_{\left\{\sigma >1\right\}}y{\nu }^{\prime }\left(x\right)-{\mathbf{1}}_{\left\{\sigma =1\right\}}{\mathbf{1}}_{\left\{\left|y\right|\le x\right\}}y{\nu }^{\prime }\left(x\right)\right]\mu \left(dy\right).$$

We split the Lévy operator at the boundary point $y=-x$:

$$\begin{array}{cc}\hfill L^{\mu }\nu \left(x\right)& ={\int }_{\left(-x,\infty \right)}\left[\nu \left(x+y\right)-\nu \left(x\right)-{\mathbf{1}}_{\left\{\sigma >1\right\}}y{\nu }^{\prime }\left(x\right)-{\mathbf{1}}_{\left\{\sigma =1\right\}}{\mathbf{1}}_{\left\{\left|y\right|\le x\right\}}y{\nu }^{\prime }\left(x\right)\right]\mu \left(dy\right)\hfill \\ \hfill & +{\int }_{\left(-\infty ,-x\right]}\left[\nu \left(x+y\right)-\nu \left(x\right)-{\mathbf{1}}_{\left\{\sigma >1\right\}}y{\nu }^{\prime }\left(x\right)-{\mathbf{1}}_{\left\{\sigma =1\right\}}{\mathbf{1}}_{\left\{\left|y\right|\le x\right\}}y{\nu }^{\prime }\left(x\right)\right]\mu \left(dy\right)\hfill \\ \hfill & :=L^{\mu ,1}\nu \left(x\right)+L^{\mu ,2}\nu \left(x\right).\hfill \end{array}$$

In the first integral, we substitute $y=xr,$

$$\begin{array}{cc}\hfill & w\left(x\right)L^{\mu ,1}\nu \left(x\right)\hfill \\ \hfill =& w\left(x\right){\int }_{\left(-x,\infty \right)}\left[\nu \left(x+y\right)-\nu \left(x\right)-{\mathbf{1}}_{\left\{\sigma >1\right\}}y{\nu }^{\prime }\left(x\right)-{\mathbf{1}}_{\sigma =1}{\mathbf{1}}_{\left\{\left|y\right|\le x\right\}}y{\nu }^{\prime }\left(x\right)\right]\mu \left(dy\right)\hfill \\ \hfill =& w\left(x\right){\int }_{\left(-1,\infty \right)}\left[\nu \left(x+xr\right)-\nu \left(x\right)-{\mathbf{1}}_{\left\{\sigma >1\right\}}xr{\nu }^{\prime }\left(x\right)-{\mathbf{1}}_{\sigma =1}{\mathbf{1}}_{\left\{x\left|r\right|\le x\right\}}xr{\nu }^{\prime }\left(x\right)\right]\mu \left(xdr\right)\hfill \\ \hfill =& {\int }_{\left(-1,\infty \right)}\left[\nu \left(x+xr\right)-\nu \left(x\right)-{\chi }_{\sigma }\left(r\right)xr{\nu }^{\prime }\left(x\right)\right]{\tilde{\mu }}_x\left(dr\right).\hfill \end{array}$$

Now we deal with $L^{\mu ,2}\nu \left(x\right).$ Note that $y\le -x$ implies $x+y\le 0$ so $\nu \left(x+y\right)=0.$ Thus,

$$w\left(x\right)L^{\mu ,2}\nu \left(x\right)=\left\{\begin{array}{cc}-\nu \left(x\right)\rho \left(x\right),\hfill & \sigma \in \left(0,1\right],\hfill \\ -\nu \left(x\right)\rho \left(x\right)+{\nu }^{\prime }\left(x\right)x\kappa \left(x\right),\hfill & \sigma \in \left(1,2\right).\hfill \end{array}\right.$$

where $\rho \left(x\right)={\int }_{\left(-\infty ,-1\right]}{\tilde{\mu }}_x\left(dr\right)$ and $\kappa \left(x\right)=-{\int }_{\left(-\infty ,-1\right]}r{\tilde{\mu }}_x\left(dr\right).$ Combining both estimates leads to the conclusion. □

Multiplying (3) by $w\left(x\right)$ and subtracting $\lambda \nu \left(x\right)$ for some parameter $\lambda \ge 0$ leads to the transformed equation

$$\begin{array}{cc}\hfill A\nu \left(x\right)-\left(\lambda +\rho \left(x\right)\right)\nu \left(x\right)& =g\left(x\right),g\left(x\right):=w\left(x\right)f\left(x\right),x>0.\hfill \end{array}$$

(4)

4. Probabilistic Mild Solution

In this section, we introduce a probabilistic mild solution concept for the transformed Equation (4). Under Assumptions 1–3, the transformed operator A is first realized through a martingale problem. After fixing a Markov process associated with this martingale problem, the exponential killing factor generated by $\lambda +\rho$ gives a probabilistic representation of a bounded mild solution for bounded Borel data.

Recall for $\nu \in C_c^{\infty }\left(\left(0,\infty \right)\right)$ and $x>0$,

$$A\nu \left(x\right)={\int }_{\left(-1,\infty \right)}\left[\nu \left(x+xr\right)-\nu \left(x\right)-{\chi }_{\sigma }\left(r\right)xr{\nu }^{\prime }\left(x\right)\right]{\tilde{\mu }}_x\left(dr\right)+{\mathbf{1}}_{\left\{\sigma >1\right\}}x\kappa \left(x\right){\nu }^{\prime }\left(x\right).$$

(5)

Testing the operator with $\nu \left(x\right)=e^{i\xi x},\xi \in \mathbf{R}$ then $A{e}^{i\xi x}=-q_A\left(x,\xi \right)e^{i\xi x}$ with the symbol

$$q_A\left(x,\xi \right)={\int }_{\left(-1,\infty \right)}\left(1-e^{i\xi xr}+i\xi xr{\chi }_{\sigma }\left(r\right)\right){\tilde{\mu }}_x\left(dr\right)-{\mathbf{1}}_{\left\{\sigma >1\right\}}i\xi x\kappa \left(x\right).$$

In particular, $q_A\left(x,0\right)=0.$ Now, using ${\tilde{\mu }}_x\left(dr\right)=w\left(x\right)\mu \left(xdr\right)$, we obtain

$$\begin{array}{cc}\hfill q_A\left(x,\xi \right)& =w\left(x\right){\int }_{\left(-x,\infty \right)}\left(1-e^{i\xi r}+i\xi r{\chi }_{\sigma }\left(r{x}^{-1}\right)\right)\mu \left(dr\right)-{\mathbf{1}}_{\left\{\sigma >1\right\}}i\xi x\kappa \left(x\right).\hfill \end{array}$$

We now extend $q_A\left(x,\xi \right)$ and simply define

$${\overline{q}}_A\left(x,\xi \right):=\left\{\begin{array}{cc}{q}_A\left(x,\xi \right)\hfill & ,x>0,\xi \in \mathbf{R},\hfill \\ 0\hfill & ,x\le 0,\xi \in \mathbf{R}.\hfill \end{array}\right.$$

(6)

In the next Lemma, we prove the crucial estimate needed for existence of the Markov process associated with A based on [9] (Corollary 3.2).

Lemma 2.

Let the symbol ${\overline{q}}_A:\mathbf{R}\times \mathbf{R}\to \mathbf{C}$ be given by (6). Then:

(i) 

For every $\xi \in \mathbf{R},$$x\mapsto {\overline{q}}_A\left(x,\xi \right)$ is continuous on $\mathbf{R}.$

(ii) 

${\overline{q}}_A$ is locally bounded. More precisely, for any compact set $K\subset \mathbf{R},$ there exists a constant $C_K>0$ such that

$$\left|{\overline{q}}_A\left(x,\xi \right)\right|\le C_K\left(1+{\left|\xi \right|}^2\right),x\in K,\xi \in \mathbf{R}.$$

(iii) 

${lim\; sup}_{\left|x\right|\to \infty }{sup}_{\left|\xi \right|\le {\left|x\right|}^{-1}}\left|{\overline{q}}_A\left(x,\xi \right)\right|<\infty .$

Proof. 

Let $x>0$ and $\xi \ne 0.$ Using $1-\cos x\le 1\wedge {\left|x\right|}^2$ and Assumption 2,

$$\begin{array}{c}w\left(x\right){\int }_{\left(-x,\infty \right)}\left(1-\cos \left(\xi r\right)\right)\mu \left(dr\right)\le w\left(x\right)\left({\int }_{\left|\xi r\right|\le 1}{\left|\xi r\right|}^2\mu \left(dr\right)+{\int }_{\left|\xi r\right|>1}\mu \left(dr\right)\right)\hfill \\ =w\left(x\right)w{\left({\left|\xi \right|}^{-1}\right)}^{-1}\left({\int }_{\left|r\right|\le 1}{\left|r\right|}^2{\tilde{\mu }}_{{\left|\xi \right|}^{-1}}\left(dr\right)+{\int }_{\left|r\right|>1}{\tilde{\mu }}_{{\left|\xi \right|}^{-1}}\left(dr\right)\right)\le Cw\left(x\right)w{\left({\left|\xi \right|}^{-1}\right)}^{-1}.\hfill \end{array}$$

If $\sigma >1$, using $\left|\sin x-x\right|\le C\left(\left|x\right|\wedge {\left|x\right|}^3\right)$ and Assumption 2,

$$\begin{array}{c}w\left(x\right){\int }_{\left(-x,\infty \right)}\left|-i\sin \left(\xi r\right)+i\xi r\right|\mu \left(dr\right)\le w\left(x\right)\left({\int }_{\left|\xi r\right|\le 1}{\left|\xi r\right|}^3\mu \left(dr\right)+{\int }_{\left|\xi r\right|>1}\left|\xi r\right|\mu \left(dr\right)\right)\hfill \\ \le w\left(x\right)w{\left({\left|\xi \right|}^{-1}\right)}^{-1}\left({\int }_{\left|r\right|\le 1}{\left|r\right|}^3{\tilde{\mu }}_{{\left|\xi \right|}^{-1}}\left(dr\right)+{\int }_{\left|r\right|>1}\left|r\right|{\tilde{\mu }}_{{\left|\xi \right|}^{-1}}\left(dr\right)\right)\le Cw\left(x\right)w{\left({\left|\xi \right|}^{-1}\right)}^{-1}.\hfill \end{array}$$

If $\sigma <1$, using the identity $\left|\sin x\right|\le C\left(1\wedge \left|x\right|\right)$ and Assumption 2,

$$\begin{array}{c}w\left(x\right){\int }_{\left(-x,\infty \right)}\left|-i\sin \left(\xi r\right)\right|\mu \left(dr\right)\hfill \\ \le w\left(x\right)w{\left({\left|\xi \right|}^{-1}\right)}^{-1}\left({\int }_{\left|r\right|\le 1}\left|r\right|{\tilde{\mu }}_{{\left|\xi \right|}^{-1}}\left(dr\right)+{\int }_{\left|r\right|>1}{\tilde{\mu }}_{{\left|\xi \right|}^{-1}}\left(dr\right)\right)\le Cw\left(x\right)w{\left({\left|\xi \right|}^{-1}\right)}^{-1}.\hfill \end{array}$$

If $\sigma =1$ and $\left|\xi \right|\le x^{-1},$ then $\left|\xi x\right|\le 1$. Hence,

$$\begin{array}{c}w\left(x\right){\int }_{\left(-x,\infty \right)}\left|-i\sin \left(\xi r\right)+i\xi r{\mathbf{1}}_{\left\{\left|r\right|\le x\right\}}\right|\mu \left(dr\right)\hfill \\ ={\int }_{\left(-1,\infty \right)}\left|-i\sin \left(\xi xr\right)+i\xi xr{\mathbf{1}}_{\left\{\left|r\right|\le 1\right\}}\right|{\tilde{\mu }}_x\left(dr\right)\hfill \\ \le {\left|\xi x\right|}^3{\int }_{\left|r\right|\le 1}{\left|r\right|}^3{\tilde{\mu }}_x\left(dr\right)+{\int }_{\left|r\right|>1}{\tilde{\mu }}_x\left(dr\right)\le C.\hfill \end{array}$$

We now treat the drift term for $\sigma >1$. Since $\kappa$ is bounded by Assumption 2, $\left|\xi x\kappa \left(x\right)\right|\le C\left|\xi x\right|.$ Combining the preceding estimates yields the desired bound (iii). The local boundedness in (ii) follows with similar arguments with obvious modifications.

Next, we prove the continuity. Fix $x_0>0$ and choose a compact interval $K\subset (0,\infty )$ containing $x_0$. For $x\in K$, both x and $x^{-1}$ are bounded uniformly. Set

$${\mathsf{\Psi }}_x(r,\xi )=1-e^{i\xi r}+i\xi r{\chi }_{\sigma }\left(r{x}^{-1}\right).$$

If $x_n\to x_0$, then

$${\mathsf{\Psi }}_{x_n}(r,\xi )\to {\mathsf{\Psi }}_{x_0}(r,\xi )$$

for $\mu$-almost every r, since $\mu$ has no atoms. Moreover, the standard bounds

$$|e^{iz}-1|\le C(1\wedge \left|z\right|),|e^{iz}{-1-iz|\le C(1\wedge \left|z\right|}^2)$$

show that the integrand is controlled by the small-jump moment and the large-jump mass required in Assumption 2. Since x ranges only over the compact interval K, these bounds are uniform in x. Hence the dominated convergence theorem applies, and the integral term in $q_A(x,\xi )$ is continuous at $x_0$. Together with the continuity of w and of $x\kappa \left(x\right)$ when $\sigma >1$, we obtain the continuity of $x\mapsto q_A\left(x,\xi \right)$ on $\left(0,\infty \right).$

Finally, we discuss the continuity at 0. Trivially, ${\overline{q}}_A\left(0,\xi \right)=0.$ If $\sigma \ne 1$, the preceding estimates gives

$$\left|{\overline{q}}_A\left(x,\xi \right)\right|\le Cw\left(x\right)w{\left({\left|\xi \right|}^{-1}\right)}^{-1}+{\mathbf{1}}_{\left\{\sigma >1\right\}}\left|\xi x\kappa \left(x\right)\right|.$$

Since both $w\left(x\right)\to 0$ and $x\kappa \left(x\right)\to 0$ as $x\downarrow 0$, we have ${\overline{q}}_A\left(x,\xi \right)\to 0$ as $x\downarrow 0.$

• For $\sigma =1,$

$${\overline{q}}_A\left(x,\xi \right)=w\left(x\right){\int }_{\left(-x,\infty \right)}\left(1-e^{i\xi r}+i\xi r{\mathbf{1}}_{\left\{\left|r\right|\le x\right\}}\right)\mu \left(dr\right),x>0.$$

We fix $ϵ>0$ and consider $0<x<ϵ$. Then $\left(-x,\infty \right)=\left(-x,x\right)\cup \left[x,ϵ\right)\cup \left[ϵ,\infty \right).$

• On $\left(-x,x\right),$

$$\begin{array}{cc}\hfill w\left(x\right){\int }_{\left(-x,x\right)}\left|1-e^{i\xi r}+i\xi r\right|\mu \left(dr\right)& \le C_{\xi }w\left(x\right){\int }_{\left|r\right|<x}{r}^2\mu \left(dr\right)\hfill \\ \hfill & =C_{\xi }{x}^2{\int }_{\left|r\right|<1}{r}^2{\tilde{\mu }}_x\left(dr\right)\le C_{\xi }{x}^2.\hfill \end{array}$$

On $\left[x,ϵ\right),$$\left|1-e^{i\xi r}\right|\le C_{\xi }r\le C_{\xi }ϵ.$ Hence,

$$w\left(x\right){\int }_{\left[x,ϵ\right)}\left|1-e^{i\xi r}\right|\mu \left(dr\right)\le C_{\xi }ϵw\left(x\right)\mu \left(\left(x,\infty \right)\right)\le C_{\xi }ϵ.$$

On $\left[ϵ,\infty \right),$

$$w\left(x\right){\int }_{\left[ϵ,\infty \right)}\left|1-e^{i\xi r}\right|\mu \left(dr\right)\le 2w\left(x\right)\mu \left(\left(ϵ,\infty \right)\right).$$

Hence, ${lim\; sup}_{x\downarrow 0}\left|{\overline{q}}_A\left(x,\xi \right)\right|\le C_{\xi }ϵ$. Because $ϵ$ is arbitrary, ${\overline{q}}_A\left(x,\xi \right)\to 0$ as $x\downarrow 0.$ □

The continuity and growth properties established in Lemma 2 allow us to verify the hypotheses of [9] (Corollary 3.2), which we use in the next result to construct the Markov process associated with the operator A.

Lemma 3.

There exists a conservative Markov process Y with state space $\left(0,\infty \right)\cup \left\{\mathsf{\Delta }\right\}$ (with a cemetery point Δ) such that for every $f\in C_c^{\infty }\left(\left(0,\infty \right)\right),$

$$M_t^f=f\left(Y_t\right)-f\left(Y_0\right)-{\int }_0^{t}Af\left(Y_s\right)ds$$

is a martingale. In particular, Y has generator A on $\left(0,\infty \right)$ in the sense of the martingale problem.

Proof. 

By Lemma 2 together with [9] (Lemma 2.1), the symbol ${\overline{q}}_A$ satisfies the hypotheses of [9] (Corollary 3.2). Hence, there exists a conservative solution to the $\left(\overline{A},C_c^{\infty }\left(\mathbf{R}\right)\right)$-martingale problem with symbol ${\overline{q}}_A.$ Restricting this process to $\left(0,\infty \right)$ and adjoining the cemetery point $\mathsf{\Delta }$ yields the desired process with generator A on $\left(0,\infty \right)$. □

We next use the martingale problem associated with A to motivate (formally) the probabilistic formulation of mild solutions.

Define

$$Z_t=\exp \left(-{\int }_0^t\left(\lambda +\rho \left(Y_s\right)\right)ds\right).$$

Formally applying the product rule to the process $Z_t\nu \left(Y_t\right)$ together with the martingale problem for A (Lemma 3), leads to

$$d\left(Z_t\nu \left(Y_t\right)\right)=Z_t\left(A\nu -\left(\lambda +\rho \right)\nu \right)\left(Y_t\right)dt+d{M}_t.$$

where M is a local martingale.

If $\nu$ satisfies $A\nu -\left(\lambda +\rho \right)\nu =g$, then

$$d\left(Z_t\nu \left(Y_t\right)\right)=Z_{t}g\left(Y_t\right)dt+d{M}_t.$$

Integrating up to the killing time ${\tau }_0=\inf \left\{t>0:Y_t=\mathsf{\Delta }\right\}$ and taking expectations yields

$${\mathbf{E}}_x\left[Z_{t\wedge {\tau }_0}\nu \left(Y_{t\wedge {\tau }_0}\right)\right]=\nu \left(x\right)+{\mathbf{E}}_x\left[{\int }_0^{t\wedge {\tau }_0}{Z}_{s}g\left(Y_s\right)ds\right].$$

(7)

Definition 1.

Fix a Markov process Y solving the martingale problem for A in Lemma 3. Let

$${\tau }_0=\inf \{t>0:Y_t=\mathsf{\Delta }\},Z_t=\exp \left(-{\int }_0^{t\wedge {\tau }_0}(\lambda +\rho \left(Y_s\right))ds\right).$$

Let $g\in B_b\left((0,\infty )\right)$. We call ν a probabilistic mild solution of (4) associated with Y if $\nu \in B_b\left((0,\infty )\right)$, $\nu \left(\mathsf{\Delta }\right)=0$, and

$$\nu \left(x\right)={\mathbf{E}}_x\left[Z_{t\wedge {\tau }_0}\nu \left(Y_{t\wedge {\tau }_0}\right)\right]-{\mathbf{E}}_x\left[{\int }_0^{t\wedge {\tau }_0}{Z}_{s}g\left(Y_s\right)ds\right],x>0,t>0.$$

Motivated by the preceding heuristic argument, we now establish the existence of a mild solution via a probabilistic representation formula.

Theorem 1.

Fix a Markov process Y solving the martingale problem for A in Lemma 3. Let

$${\tau }_0=\inf \{t>0:Y_t=\mathsf{\Delta }\},{\rho }_{\ast }=\underset{x>0}{\inf }\rho \left(x\right),$$

and

$$Z_t=\exp \left(-{\int }_0^{t\wedge {\tau }_0}(\lambda +\rho \left(Y_s\right))ds\right),t\ge 0.$$

For $x>0$, define

$$\nu \left(x\right)=-{\mathbf{E}}_x\left[{\int }_0^{{\tau }_0}{Z}_{t}g\left(Y_t\right)dt\right],$$

and set $\nu \left(\mathsf{\Delta }\right)=0$. Assume that $\lambda +{\rho }_{\ast }>0$ and $g\in B_b\left((0,\infty )\right)$. Then, $\nu \in B_b\left((0,\infty )\right)$ and ν is a probabilistic mild solution of (4) associated with Y. In particular,

$$\nu \left(x\right)={\mathbf{E}}_x\left[Z_{t\wedge {\tau }_0}\nu \left(Y_{t\wedge {\tau }_0}\right)\right]-{\mathbf{E}}_x\left[{\int }_0^{t\wedge {\tau }_0}{Z}_{s}g\left(Y_s\right)ds\right],x>0,t>0.$$

Moreover,

$$\left|\nu \left(x\right)\right|\le {\displaystyle \frac{{\left|g\right|}_{\infty }}{\lambda +{\rho }_{\ast }}},x>0.$$

Proof. 

Since $\rho \left(x\right)\ge {\rho }_{\ast }$ for all $x>0$, we have $Z_t=\exp \left(-{\int }_0^t\left(\lambda +\rho \left(Y_s\right)\right)ds\right)\le e^{-\left(\lambda +{\rho }_{\ast }\right)t}.$ Consequently,

$$\left|\nu \left(x\right)\right|\le {\displaystyle \frac{{\left|g\right|}_{\infty }}{\lambda +{\rho }_{\ast }}},x>0.$$

Next we write for $t>0,$

$$\begin{array}{cc}\hfill \nu \left(x\right)& =-{\mathbf{E}}_x\left[{\int }_0^{{\tau }_0}{Z}_{s}g\left(Y_s\right)ds\right]\hfill \\ \hfill & =-{\mathbf{E}}_x\left[{\int }_0^{t\wedge {\tau }_0}{Z}_{s}g\left(Y_s\right)ds\right]-{\mathbf{E}}_x\left[{\mathbf{1}}_{\left\{t<{\tau }_0\right\}}{\int }_t^{{\tau }_0}{Z}_{s}g\left(Y_s\right)ds\right].\hfill \end{array}$$

(8)

For the second term, observe that for $s\ge t$,

$$Z_s=Z_t\exp \left(-{\int }_t^s\left(\lambda +\rho \left(Y_r\right)\right)dr\right).$$

Hence, by changing the variable of integration on $\left\{t<{\tau }_0\right\}$,

$$\begin{array}{cc}\hfill {\int }_t^{{\tau }_0}{Z}_{s}g\left(Y_s\right)ds& =Z_t{\int }_t^{{\tau }_0}\exp \left(-{\int }_t^s\left(\lambda +\rho \left(Y_r\right)\right)dr\right)g\left(Y_s\right)ds\hfill \\ \hfill & =Z_t{\int }_0^{{\tau }_0-t}\exp \left(-{\int }_0^u\left(\lambda +\rho \left(Y_{t+r}\right)\right)dr\right)g\left(Y_{t+u}\right)du.\hfill \end{array}$$

Using the Markov property at the deterministic time $t,$ we obtain

$$\begin{array}{c}{\mathbf{E}}_x\left[{\mathbf{1}}_{\left\{t<{\tau }_0\right\}}{\int }_t^{{\tau }_0}{Z}_{s}g\left(Y_s\right)ds\right]\hfill \\ ={\mathbf{E}}_x\left[{\mathbf{1}}_{\left\{t<{\tau }_0\right\}}{Z}_t{\mathbf{E}}_x\left[{\int }_0^{{\tau }_0-t}\exp \left(-{\int }_0^u\left(\lambda +\rho \left(Y_{t+r}\right)\right)dr\right)g\left(Y_{t+u}\right)du\left|F_t\right.\right]\right]\hfill \\ ={\mathbf{E}}_x\left[{\mathbf{1}}_{\left\{t<{\tau }_0\right\}}{Z}_t{\mathbf{E}}_{Y_t}\left[{\int }_0^{{\tau }_0}{Z}_{s}g\left(Y_s\right)ds\right]\right]=-{\mathbf{E}}_x\left[{\mathbf{1}}_{\left\{t<{\tau }_0\right\}}{Z}_t\nu \left(Y_t\right)\right].\hfill \end{array}$$

Substituting the above expression into (8) completes the proof. □

5. The Fractional Laplacian

In the preceding sections, the process Y was constructed for general Lévy measures through the transformed martingale problem. In general, the transformed jump kernel depends on the present state x, and therefore the process Y does not have an explicit self-similar form. The homogeneous stable case is exceptional. In this case the transformed process is exactly multiplicative, and this structure allows weighted $L_p$ estimates to be derived directly from the moment identity for the multiplicative factor.

Throughout this section, assume

$$\mu \left(dy\right)={\displaystyle \frac{\alpha }{2}}{\left|y\right|}^{-1-\alpha }dy,0<\alpha <2.$$

Then

$$\delta \left(x\right)=\mu \left(\right\{\left|y\right|\ge x\left\}\right)=x^{-\alpha },w\left(x\right)=x^{\alpha }.$$

With the change of variables $y=xr$, we obtain

$${\tilde{\mu }}_x\left(dr\right)=w\left(x\right)\mu \left(xdr\right)=x^{\alpha }{\displaystyle \frac{\alpha }{2}}{\left|xr\right|}^{-1-\alpha }xdr={\displaystyle \frac{\alpha }{2}}{\left|r\right|}^{-1-\alpha }dr.$$

Moreover,

$$\rho ={\int }_{(-\infty ,-1]}\mu \left(dr\right)={\displaystyle \frac{1}{2}},$$

and, when $\alpha >1$,

$$\kappa =-{\int }_{(-\infty ,-1]}r\mu \left(dr\right)={\displaystyle \frac{\alpha }{2(\alpha -1)}}.$$

Hence, both $\rho$ and $\kappa$ are constants.

The operator A is simplified to

$$A\varphi \left(x\right)={\int }_{(-1,\infty )}\left[\varphi (x+xr)-\varphi \left(x\right)-{\chi }_{\alpha }\left(r\right)xr{\varphi }^{\prime }\left(x\right)\right]\mu \left(dr\right)+{\mathbf{1}}_{\{\alpha >1\}}x\kappa {\varphi }^{\prime }\left(x\right),$$

(9)

where

$${\chi }_{\alpha }\left(r\right)=\left\{\begin{array}{cc}0,\hfill & 0<\alpha <1,\hfill \\ {\mathbf{1}}_{\left\{\right|r|\le 1\}},\hfill & \alpha =1,\hfill \\ 1,\hfill & 1<\alpha <2.\hfill \end{array}\right.$$

Since the jump measure in (9) is independent of x, the transformed process is multiplicative. Let $p(dt,dr)$ be a Poisson random measure on $(0,\infty )\times (-1,\infty )$ with compensator $dt\mu \left(dr\right)$, and let

$$\widehat{p}(dt,dr)=p(dt,dr)-dt\mu \left(dr\right)$$

be the compensated Poisson random measure. Define the pure-jump process $E={\left(E_t\right)}_{t\ge 0}$ by

$$E_t={\int }_0^t{\int }_{(-1,\infty )}rp(ds,dr),0<\alpha <1,$$

$$E_t={\int }_0^t{\int }_{(-1,1]}r\widehat{p}(ds,dr)+{\int }_0^t{\int }_{(1,\infty )}rp(ds,dr),\alpha =1,$$

and

$$E_t={\int }_0^t{\int }_{(-1,\infty )}r\widehat{p}(ds,dr),1<\alpha <2.$$

The transformed process $Y^x={\left(Y_t^x\right)}_{t\ge 0}$, starting from $x>0$, is represented by the multiplicative jump equation

$$d{Y}_t^x=Y_{t-}^{x}d{E}_t+{\mathbf{1}}_{\{\alpha >1\}}\kappa Y_t^{x}dt,Y_0^x=x.$$

(10)

A jump of E of size r sends

$$Y_t^x=Y_{t-}^x(1+r).$$

Thus, positivity is preserved because all relative jumps satisfy $r>-1$.

The generator of $Y^x$ is exactly the operator A in (9). Indeed, Itô’s formula applied to $\varphi \left(Y_t^x\right)$, for $\varphi \in C_c^{\infty }\left((0,\infty )\right)$, gives

$$A\varphi \left(x\right)={\int }_{(-1,\infty )}\left[\varphi (x+xr)-\varphi \left(x\right)-{\chi }_{\alpha }\left(r\right)xr{\varphi }^{\prime }\left(x\right)\right]\mu \left(dr\right)+{\mathbf{1}}_{\{\alpha >1\}}x\kappa {\varphi }^{\prime }\left(x\right).$$

Thus, $Y^x$ solves the martingale problem for A.

Define

$${\mathcal{E}}_t=Y_t^1.$$

By the multiplicative jump representation (10),

$$Y_t^x=x{\mathcal{E}}_t,x>0.$$

This is the multiplicative representation used below. Before using the moment exponent, we record its range of finiteness. For $\mathcal{\ell }\in \mathbf{R}$, set

$$\mathsf{\Gamma }(\mathcal{\ell })={\int }_{(-1,\infty )}\left[{(1+r)}^{\mathcal{\ell }}-1-\mathcal{\ell }{\chi }_{\alpha }\left(r\right)r\right]\mu \left(dr\right)+{\mathbf{1}}_{\{\alpha >1\}}\mathcal{\ell }\kappa .$$

Then, $\mathsf{\Gamma }(\mathcal{\ell })$ is finite precisely when $-1<\mathcal{\ell }<\alpha$. This can be verified by elementary analysis: the condition $\mathcal{\ell }>-1$ controls the singularity near $r=-1$, while the condition $\mathcal{\ell }<\alpha$ controls the tail as $r\to \infty$. Near $r=0$, the truncation term gives the usual cancellation, so no additional restriction is needed (see also Chapter 3 of [10].)

Lemma 4.

Let $\mathcal{\ell }\in (-1,\alpha )$, and define

$$\mathsf{\Gamma }(\mathcal{\ell })={\int }_{(-1,\infty )}\left[{(1+r)}^{\mathcal{\ell }}-1-\mathcal{\ell }{\chi }_{\alpha }\left(r\right)r\right]\mu \left(dr\right)+{\mathbf{1}}_{\{\alpha >1\}}\mathcal{\ell }\kappa .$$

Then

$$\mathbf{E}{\mathcal{E}}_t^{\mathcal{\ell }}=e^{\mathsf{\Gamma }(\mathcal{\ell })t},t\ge 0.$$

Proof. 

We compute A on the power function $\varphi \left(x\right)=x^{\mathcal{\ell }}$. Since ${\varphi }^{\prime }\left(x\right)=\mathcal{\ell }{x}^{\mathcal{\ell }-1}$, we have

$$\varphi (x+xr)-\varphi \left(x\right)-{\chi }_{\alpha }\left(r\right)xr{\varphi }^{\prime }\left(x\right)=x^{\mathcal{\ell }}\left({(1+r)}^{\mathcal{\ell }}-1-\mathcal{\ell }{\chi }_{\alpha }\left(r\right)r\right).$$

Together with the drift term in A, this gives

$$A\varphi \left(x\right)=\mathsf{\Gamma }(\mathcal{\ell })x^{\mathcal{\ell }}.$$

To justify Itô’s formula for power function $x^{\mathcal{\ell }}$, we stop the process before it approaches 0 or infinity. Let

$${\tau }_R=\inf \{t>0:{\mathcal{E}}_t\notin (R^{-1},R)\}.$$

On $[0,{\tau }_R]$, the process $\mathcal{E}$ remains in $(R^{-1},R)$, where $x^{\mathcal{\ell }}$ is bounded with bounded derivatives. Applying Itô’s formula to ${\mathcal{E}}_{t\wedge {\tau }_R}^{\mathcal{\ell }}$ gives

$$\mathbf{E}{\mathcal{E}}_{t\wedge {\tau }_R}^{\mathcal{\ell }}=1+\mathsf{\Gamma }(\mathcal{\ell })\mathbf{E}{\int }_0^{t\wedge {\tau }_R}{\mathcal{E}}_s^{\mathcal{\ell }}ds.$$

Letting $R\to \infty$, we obtain

$$\mathbf{E}{\mathcal{E}}_t^{\mathcal{\ell }}=1+\mathsf{\Gamma }(\mathcal{\ell }){\int }_0^t\mathbf{E}{\mathcal{E}}_s^{\mathcal{\ell }}ds.$$

Thus, $h\left(t\right)=\mathbf{E}{\mathcal{E}}_t^{\mathcal{\ell }}$ satisfies

$$h\left(t\right)=1+\mathsf{\Gamma }(\mathcal{\ell }){\int }_0^{t}h\left(s\right)ds.$$

Solving this integral equation yields

$$h\left(t\right)=e^{\mathsf{\Gamma }(\mathcal{\ell })t}.$$

This proves the claim. □

Lemma 5.

Let $g\in C_c^{\infty }\left((0,\infty )\right)$, and define

$$\nu \left(x\right)=-\mathbf{E}{\int }_0^{\infty }{e}^{-(\lambda +\rho )t}g\left(x{\mathcal{E}}_t\right)dt.$$

Then, ν is a pointwise solution of

$$A\nu \left(x\right)-(\lambda +\rho )\nu \left(x\right)=g\left(x\right),x>0.$$

Consequently,

$$w\left(x\right)L^{\mu }\nu \left(x\right)-\lambda \nu \left(x\right)=g\left(x\right),x>0.$$

Proof. 

Put $a=\lambda +\rho$. Define

$$\overline{g}\left(z\right)=g\left(e^z\right),\overline{\nu }\left(z\right)=\nu \left(e^z\right),z\in \mathbf{R}.$$

Since $g\in C_c^{\infty }\left((0,\infty )\right)$, it follows that $\overline{g}\in C_c^{\infty }\left(\mathbf{R}\right).$ Using the multiplicative representation $Y_t^x=x{\mathcal{E}}_t$, we have

$$\overline{\nu }\left(z\right)=\nu \left(e^z\right)=-\mathbf{E}{\int }_0^{\infty }{e}^{-at}g\left(e^z{\mathcal{E}}_t\right)dt=-\mathbf{E}{\int }_0^{\infty }{e}^{-at}\overline{g}(z+\log {\mathcal{E}}_t)dt.$$

Let ${\left({\overline{P}}_t\right)}_{t\ge 0}$ be the semigroup of the additive Lévy process $\log {\mathcal{E}}_t$, namely

$${\overline{P}}_{t}h\left(z\right)=\mathbf{E}h(z+\log {\mathcal{E}}_t),$$

and let $\overline{A}$ denote its generator. Then,

$$\overline{\nu }=-{\int }_0^{\infty }{e}^{-at}{\overline{P}}_t\overline{g}dt.$$

By the standard resolvent identity for the additive Lévy semigroup,

$$\overline{\nu }\in D\left(\overline{A}\right),\overline{A}\overline{\nu }-a\overline{\nu }=\overline{g}\mathrm{on}\mathbf{R}.$$

We now translate this identity back to the original variable. Fix $x>0$ and write $z=\log x$. Since $Y_t^x=x{\mathcal{E}}_t=e^z{\mathcal{E}}_t$,

$$\overline{\nu }(z+\log {\mathcal{E}}_t)=\nu \left(e^z{\mathcal{E}}_t\right)=\nu \left(Y_t^x\right).$$

Therefore,

$${\overline{P}}_t\overline{\nu }\left(z\right)=\mathbf{E}\overline{\nu }(z+\log {\mathcal{E}}_t)=\mathbf{E}\nu \left(Y_t^x\right).$$

Hence,

$${\displaystyle \frac{{\overline{P}}_t\overline{\nu }\left(z\right)-\overline{\nu }\left(z\right)}{t}}={\displaystyle \frac{\mathbf{E}\nu \left(Y_t^x\right)-\nu \left(x\right)}{t}}.$$

Since $\overline{\nu }\in D\left(\overline{A}\right)$, the left-hand side converges as $t\downarrow 0$ to $\overline{A}\overline{\nu }\left(z\right)$. Hence, the right-hand side also converges. By the definition of the generator A of Y, this means that $\nu \in D\left(A\right)$ at the point x, and

$$A\nu \left(x\right)=\overline{A}\overline{\nu }\left(\log x\right).$$

Using $\overline{A}\overline{\nu }-a\overline{\nu }=\overline{g}$, we obtain

$$A\nu \left(x\right)-a\nu \left(x\right)=\overline{A}\overline{\nu }\left(\log x\right)-a\overline{\nu }\left(\log x\right)=\overline{g}\left(\log x\right)=g\left(x\right).$$

Since $a=\lambda +\rho$, this gives

$$A\nu \left(x\right)-(\lambda +\rho )\nu \left(x\right)=g\left(x\right),x>0.$$

Since the preceding argument shows that $\nu \in D\left(A\right)$ pointwise, the operator identity obtained in Lemma 1 applies to $\nu$. Therefore,

$$w\left(x\right)L^{\mu }\nu \left(x\right)-\lambda \nu \left(x\right)=g\left(x\right),x>0.$$

The proof is complete. □

We now use the explicit multiplicative representation of the fractional Laplacian case to pass from the probabilistic mild formulation to the closed-graph formulation. The key point is that the scaling identity $Y_t^x=x{\mathcal{E}}_t$ gives a direct weighted $L_{\theta ,p}$-estimate for the resolvent. This yields existence in the closed-graph class for general data $g\in L_{\theta ,p}\left((0,\infty )\right)$.

Theorem 2.

Let $1<p<\infty$, assume $-\theta \in (-1,\alpha )$, and suppose

$$\lambda +\rho >\mathsf{\Gamma }(-\theta ).$$

Let $g\in L_{\theta ,p}\left((0,\infty )\right)$ and $g_n\in C_c^{\infty }\left((0,\infty )\right)$ be any sequence such that

$$g_n\to g\mathit{in}{L}_{\theta ,p}\left((0,\infty )\right).$$

For each n, define

$${\nu }_n\left(x\right)=-\mathbf{E}{\int }_0^{\infty }{e}^{-(\lambda +\rho )t}{g}_n\left(x{\mathcal{E}}_t\right)dt.$$

Then, there exist $\nu ,h\in L_{\theta ,p}\left((0,\infty )\right)$ such that

$${\nu }_n\to \nu ,w{L}^{\mu }{\nu }_n\to h\mathit{in}{L}_{\theta ,p}\left((0,\infty )\right),$$

and

$$h-\lambda \nu =g\mathit{in}{L}_{\theta ,p}\left((0,\infty )\right).$$

The limits ν and h are independent of the approximating sequence $\left(g_n\right)$. Moreover,

$${\left|\nu \right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le {\displaystyle \frac{1}{\lambda +\rho -\mathsf{\Gamma }(-\theta )}}{\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}$$

and

$${\left|h\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le \left(1+{\displaystyle \frac{\lambda }{\lambda +\rho -\mathsf{\Gamma }(-\theta )}}\right){\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

If $\lambda =0$, then ν coincides with the unique weak solution in the corresponding weighted class of [3] (Theorem 2.3(ii)).

Proof. 

We first assume that $g\in C_c^{\infty }\left((0,\infty )\right)$. Define

$$\nu \left(x\right)=-\mathbf{E}{\int }_0^{\infty }{e}^{-(\lambda +\rho )t}g\left(x{\mathcal{E}}_t\right)dt.$$

By Lemma 5, $\nu$ satisfies the pointwise identity

$$w\left(x\right)L^{\mu }\nu \left(x\right)-\lambda \nu \left(x\right)=g\left(x\right),x>0.$$

For each fixed value of ${\mathcal{E}}_t$, the change of variables $y=x{\mathcal{E}}_t$ gives

$$\begin{array}{cc}\hfill {\left|g(·{\mathcal{E}}_t)\right|}_{L_{\theta ,p}\left((0,\infty )\right)}^p& ={\int }_0^{\infty }{\left|g\left(x{\mathcal{E}}_t\right)\right|}^p{x}^{\theta p-1}dx\hfill \\ \hfill & ={\mathcal{E}}_t^{-\theta p}{\int }_0^{\infty }{\left|g\left(y\right)\right|}^p{y}^{\theta p-1}dy.\hfill \end{array}$$

Hence

$${\left|g(·{\mathcal{E}}_t)\right|}_{L_{\theta ,p}\left((0,\infty )\right)}={\mathcal{E}}_t^{-\theta }{\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

By Minkowski’s inequality and Lemma 4,

$$\begin{array}{cc}\hfill {\left|\nu \right|}_{L_{\theta ,p}\left((0,\infty )\right)}& \le {\int }_0^{\infty }{e}^{-(\lambda +\rho )t}\mathbf{E}{\left|g(·{\mathcal{E}}_t)\right|}_{L_{\theta ,p}\left((0,\infty )\right)}dt\hfill \\ \hfill & ={\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}{\int }_0^{\infty }{e}^{-(\lambda +\rho )t}\mathbf{E}{\mathcal{E}}_t^{-\theta }dt\hfill \\ \hfill & ={\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}{\int }_0^{\infty }{e}^{-(\lambda +\rho -\mathsf{\Gamma }(-\theta \left)\right)t}dt\hfill \\ \hfill & ={\displaystyle \frac{1}{\lambda +\rho -\mathsf{\Gamma }(-\theta )}}{\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.\hfill \end{array}$$

Since $w{L}^{\mu }\nu =g+\lambda \nu$, we also have

$$\begin{array}{cc}\hfill {\left|w{L}^{\mu }\nu \right|}_{L_{\theta ,p}\left((0,\infty )\right)}& \le {\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}+\lambda {\left|\nu \right|}_{L_{\theta ,p}\left((0,\infty )\right)}\hfill \\ \hfill & \le \left(1+{\displaystyle \frac{\lambda }{\lambda +\rho -\mathsf{\Gamma }(-\theta )}}\right){\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.\hfill \end{array}$$

Now let $g\in L_{\theta ,p}\left((0,\infty )\right)$, and let $g_n\in C_c^{\infty }\left((0,\infty )\right)$ satisfy

$$g_n\to g\mathrm{in}{L}_{\theta ,p}\left((0,\infty )\right).$$

For each n, define

$${\nu }_n\left(x\right)=-\mathbf{E}{\int }_0^{\infty }{e}^{-(\lambda +\rho )t}{g}_n\left(x{\mathcal{E}}_t\right)dt.$$

By the smooth-data case,

$$w{L}^{\mu }{\nu }_n-\lambda {\nu }_n=g_n\mathrm{pointwise}\mathrm{on}(0,\infty ).$$

Applying the smooth-data estimate to $g_n-g_m$, we obtain

$${\left|{\nu }_n-{\nu }_m\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le {\displaystyle \frac{1}{\lambda +\rho -\mathsf{\Gamma }(-\theta )}}{\left|g_n-g_m\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

Moreover,

$$w{L}^{\mu }({\nu }_n-{\nu }_m)=(g_n-g_m)+\lambda ({\nu }_n-{\nu }_m),$$

and therefore

$${\left|w{L}^{\mu }{\nu }_n-w{L}^{\mu }{\nu }_m\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le \left(1+{\displaystyle \frac{\lambda }{\lambda +\rho -\mathsf{\Gamma }(-\theta )}}\right){\left|g_n-g_m\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

Thus, $\left({\nu }_n\right)$ is Cauchy in $L_{\theta ,p}\left((0,\infty )\right)$, and $\left(w{L}^{\mu }{\nu }_n\right)$ is also Cauchy in $L_{\theta ,p}\left((0,\infty )\right)$. Hence, there exist $\nu ,h\in L_{\theta ,p}\left((0,\infty )\right)$ such that

$${\nu }_n\to \nu ,w{L}^{\mu }{\nu }_n\to h\mathrm{in}{L}_{\theta ,p}\left((0,\infty )\right).$$

Passing to the limit in

$$w{L}^{\mu }{\nu }_n-\lambda {\nu }_n=g_n$$

gives

$$h-\lambda \nu =g\mathrm{in}{L}_{\theta ,p}\left((0,\infty )\right).$$

The estimates for ${\nu }_n$ and $w{L}^{\mu }{\nu }_n$ pass to the limit and give

$${\left|\nu \right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le {\displaystyle \frac{1}{\lambda +\rho -\mathsf{\Gamma }(-\theta )}}{\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}$$

and

$${\left|h\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le \left(1+{\displaystyle \frac{\lambda }{\lambda +\rho -\mathsf{\Gamma }(-\theta )}}\right){\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

We now show that the limits $\nu$ and h are independent of the chosen approximation of g. Let ${\overline{g}}_m\in C_c^{\infty }\left((0,\infty )\right)$ be another sequence such that

$${\overline{g}}_m\to g\mathrm{in}{L}_{\theta ,p}\left((0,\infty )\right),$$

and let

$${\overline{\nu }}_m\left(x\right)=-\mathbf{E}{\int }_0^{\infty }{e}^{-(\lambda +\rho )t}{\overline{g}}_m\left(x{\mathcal{E}}_t\right)dt.$$

Applying the smooth-data estimate to $g_n-{\overline{g}}_m$, we get

$${\left|{\nu }_n-{\overline{\nu }}_m\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le {\displaystyle \frac{1}{\lambda +\rho -\mathsf{\Gamma }(-\theta )}}{\left|g_n-{\overline{g}}_m\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

Letting $n,m\to \infty$ shows that both constructions give the same limit $\nu$. Since

$$w{L}^{\mu }{\nu }_n=g_n+\lambda {\nu }_n,w{L}^{\mu }{\overline{\nu }}_m={\overline{g}}_m+\lambda {\overline{\nu }}_m,$$

the corresponding limits of the transformed operator values also agree and are equal to

$$h=g+\lambda \nu .$$

Thus, $\nu$ and h are independent of the approximating sequence.

Finally, assume that $\lambda =0$. Then, $h=g$. Let $\varphi \in C_c^{\infty }\left((0,\infty )\right)$. For each n, the smooth solution ${\nu }_n$ satisfies

$$w{L}^{\mu }{\nu }_n=g_n.$$

Testing this identity against $\varphi /w$, and using the symmetry of $L^{\mu }$, gives

$${\int }_0^{\infty }{\nu }_n\left(x\right)L^{\mu }\varphi \left(x\right)dx={\int }_0^{\infty }{g}_n\left(x\right){\displaystyle \frac{\varphi \left(x\right)}{w\left(x\right)}}dx.$$

Since

$${\nu }_n\to \nu ,g_n\to g\mathrm{in}{L}_{\theta ,p}\left((0,\infty )\right),$$

and

$$L^{\mu }\varphi ,{\displaystyle \frac{\varphi }{w}}\in L_{1-\theta ,p^{\prime }}\left((0,\infty )\right),$$

we may pass to the limit and obtain

$${\int }_0^{\infty }\nu \left(x\right)L^{\mu }\varphi \left(x\right)dx={\int }_0^{\infty }{\displaystyle \frac{g\left(x\right)}{w\left(x\right)}}\varphi \left(x\right)dx.$$

Thus, $\nu$ is a weak solution with right-hand side $f=g/w$. Under the parameter identification

$$\eta =\theta p+{\displaystyle \frac{\alpha p}{2}},$$

the function $\nu$ has the same weighted regularity as the weak solution in [3] (Theorem 2.3(ii)). Hence, by the uniqueness asserted there, $\nu$ agrees with that weak solution. See also Remark 1 below for the parameter comparison. □

Remark 1.

We compare Theorem 2 with the known parameter range for the fractional Laplacian on the half-line. For the one-dimensional fractional Laplacian, $w\left(x\right)=x^{\alpha }$, and the transformed datum is $g=wf=x^{\alpha }f$. Hence

$${\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}^p={\int }_0^{\infty }{\left|f\left(x\right)\right|}^p{x}^{\theta p+\alpha p-1}dx.$$

The known one-dimensional weighted stable estimate is written with a parameter η as

$${\int }_0^{\infty }\left(|x^{-\alpha /2}{\nu \left(x\right)|}^p+{\left|x^{\alpha /2}{L}^{\mu }\nu \left(x\right)\right|}^p\right)x^{\eta -1}dx\le C{\int }_0^{\infty }{\left|x^{\alpha /2}f\left(x\right)\right|}^p{x}^{\eta -1}dx.$$

The solution term on the left-hand side is

$${\int }_0^{\infty }{\left|\nu \left(x\right)\right|}^p{x}^{\eta -1-\alpha p/2}dx.$$

Thus, to identify it with ${\left|\nu \right|}_{L_{\theta ,p}\left((0,\infty )\right)}^p$, we set

$$\eta =\theta p+{\displaystyle \frac{\alpha p}{2}}.$$

(11)

With this identification, the right-hand side becomes exactly the norm of the transformed datum:

$${\int }_0^{\infty }|x^{\alpha /2}{f\left(x\right)|}^p{x}^{\eta -1}dx={\int }_0^{\infty }{\left|f\left(x\right)\right|}^p{x}^{\theta p+\alpha p-1}dx={\left|x^{\alpha }f\right|}_{L_{\theta ,p}\left((0,\infty )\right)}^p.$$

The range in [3] is $0<\eta <p$. By (11), this is equivalent to $0<\theta p+\alpha p/2<p$, or

$$-{\displaystyle \frac{\alpha }{2}}<\theta <1-{\displaystyle \frac{\alpha }{2}}.$$

(12)

Thus, (12) is the fractional Laplacian range written in the notation of this paper.

We now show that the condition in Theorem 2 gives the same range. Since $\rho =1/2$, the condition for $\lambda =0$ is $\mathsf{\Gamma }(-\theta )<1/2$. For $\alpha \ne 1$, Γ has the beta-function expression

$$\mathsf{\Gamma }\left(q\right)={\displaystyle \frac{1}{2}}\left[1-{\displaystyle \frac{\alpha }{\sin (\pi \alpha /2)}}\sin \left({\displaystyle \frac{\pi \alpha }{2}}-\pi q\right)B(\alpha -q,q+1)\right],-1<q<\alpha ,$$

where B is the beta function; see, for example, [10] (Chapter 3). Since $\sin (\pi \alpha /2)>0$ for $0<\alpha <2$, and $B(\alpha -q,q+1)>0$ for $-1<q<\alpha$, the inequality $\mathsf{\Gamma }\left(q\right)<1/2$ is equivalent to

$$\sin \left({\displaystyle \frac{\pi \alpha }{2}}-\pi q\right)>0.$$

This holds exactly when

$${\displaystyle \frac{\alpha }{2}}-1<q<{\displaystyle \frac{\alpha }{2}}.$$

The case $\alpha =1$ gives the same conclusion by the corresponding limiting formula.

Taking $q=-\theta$, the condition $\mathsf{\Gamma }(-\theta )<1/2$ is equivalent to

$$-{\displaystyle \frac{\alpha }{2}}<\theta <1-{\displaystyle \frac{\alpha }{2}}.$$

Therefore, when $\lambda =0$, the moment condition in Theorem 2 recovers exactly the range (12).

6. Smooth Energy Estimate and Weighted Adjoint Inequality

For general Lévy measures, the probabilistic mild formulation gives a robust existence result, but it does not automatically provide the weighted $L_p$-operator information needed for the energy estimate below. The purpose of this section is therefore to isolate the part of the argument that is available at the smooth level.

We first prove the weighted energy estimate for $C_c^{\infty }\left((0,\infty )\right)$ functions. The same estimate also applies to any larger class for which one can construct approximations $u_n$ satisfying

$$u_n\to u,w{L}^{\mu }{u}_n\to h\mathrm{in}{L}_{\theta ,p}\left((0,\infty )\right).$$

For general Lévy measures, constructing such approximations from probabilistic mild solutions, or from other candidate solution constructions, is a separate domain question and is left for further future investigation.

We first record the convexity inequality for the transformed operator A.

Lemma 6.

Let $1<p<\infty$, and let $u\in C_c^{\infty }\left((0,\infty )\right)$. Then, for every $x>0$,

$${p\left|u\left(x\right)\right|}^{p-2}u\left(x\right)Au\left(x\right)\le {A\left(\right|u|}^p)\left(x\right).$$

Proof. 

For $\epsilon >0$, define

$${\psi }_{\epsilon }\left(s\right)={(s^2+{\epsilon }^2)}^{p/2}-{\epsilon }^p.$$

Then ${\psi }_{\epsilon }\in C^{\infty }\left(\mathbf{R}\right)$, ${\psi }_{\epsilon }$ is convex, and ${\psi }_{\epsilon }\left(0\right)=0$. Hence, for all $a,b\in \mathbf{R}$,

$${\psi }_{\epsilon }^{\prime }\left(a\right)(b-a)\le {\psi }_{\epsilon }\left(b\right)-{\psi }_{\epsilon }\left(a\right).$$

Applying this with $a=u\left(x\right)$ and $b=u(x+xr)$, and using

$${\left({\psi }_{\epsilon }\left(u\right)\right)}^{\prime }\left(x\right)={\psi }_{\epsilon }^{\prime }\left(u\left(x\right)\right)u^{\prime }\left(x\right),$$

gives

$$\begin{array}{c}{\psi }_{\epsilon }^{\prime }\left(u\left(x\right)\right)\left(u(x+xr)-u\left(x\right)-{\chi }_{\sigma }\left(r\right)xr{u}^{\prime }\left(x\right)\right)\hfill \\ \le {\psi }_{\epsilon }\left(u(x+xr)\right)-{\psi }_{\epsilon }\left(u\left(x\right)\right)-{\chi }_{\sigma }\left(r\right)xr{\left({\psi }_{\epsilon }\left(u\right)\right)}^{\prime }\left(x\right).\hfill \end{array}$$

After integration with respect to ${\tilde{\mu }}_x\left(dr\right)$, we obtain

$$\begin{array}{c}{\int }_{(-1,\infty )}{\psi }_{\epsilon }^{\prime }\left(u\left(x\right)\right)\left(u(x+xr)-u\left(x\right)-{\chi }_{\sigma }\left(r\right)xr{u}^{\prime }\left(x\right)\right){\tilde{\mu }}_x\left(dr\right)\hfill \\ \le {\int }_{(-1,\infty )}\left[{\psi }_{\epsilon }\left(u(x+xr)\right)-{\psi }_{\epsilon }\left(u\left(x\right)\right)-{\chi }_{\sigma }\left(r\right)xr{\left({\psi }_{\epsilon }\left(u\right)\right)}^{\prime }\left(x\right)\right]{\tilde{\mu }}_x\left(dr\right).\hfill \end{array}$$

If $\sigma \le 1$, this is precisely

$${\psi }_{\epsilon }^{\prime }\left(u\left(x\right)\right)Au\left(x\right)\le A\left({\psi }_{\epsilon }\left(u\right)\right)\left(x\right).$$

If $\sigma >1$, the remaining drift terms agree exactly, since

$${\psi }_{\epsilon }^{\prime }\left(u\left(x\right)\right)x\kappa \left(x\right)u^{\prime }\left(x\right)=x\kappa \left(x\right){\left({\psi }_{\epsilon }\left(u\right)\right)}^{\prime }\left(x\right).$$

Adding this identical drift contribution to both sides gives again

$${\psi }_{\epsilon }^{\prime }\left(u\left(x\right)\right)Au\left(x\right)\le A\left({\psi }_{\epsilon }\left(u\right)\right)\left(x\right).$$

Since $u\in C_c^{\infty }\left((0,\infty )\right)$, the dominated convergence theorem allows us to let $\epsilon \downarrow 0$. The verification follows from the compact support and smoothness of u; the details are standard and are left to the reader. Therefore,

$${p\left|u\left(x\right)\right|}^{p-2}u\left(x\right)Au\left(x\right)\le {A\left(\right|u|}^p)\left(x\right).$$

The proof is finished. □

We now combine Lemma 6 with a weighted adjoint integral inequality.

Theorem 3.

Let $1<p<\infty$, $\theta \in \mathbf{R}$, and $\beta =\theta p-1$. Assume that there exists $M_{\beta }<\infty$ such that, for every non-negative $F\in C_c^{\infty }\left((0,\infty )\right)$,

$${\int }_0^{\infty }AF\left(x\right)x^{\beta }dx\le M_{\beta }{\int }_0^{\infty }F\left(x\right)x^{\beta }dx.$$

Let $\lambda \ge 0$, ${\rho }_{\ast }={\inf }_{x>0}\rho \left(x\right)$, and assume

$$\lambda +{\rho }_{\ast }>{\displaystyle \frac{M_{\beta }}{p}}.$$

Then, for every $u\in C_c^{\infty }\left((0,\infty )\right)$,

$$\left(\lambda +{\rho }_{\ast }-{\displaystyle \frac{M_{\beta }}{p}}\right){\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le {\left|w{L}^{\mu }u-\lambda u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

Moreover, the same estimate passes to any pair $u,h\in L_{\theta ,p}\left((0,\infty )\right)$ for which there exists $u_n\in C_c^{\infty }\left((0,\infty )\right)$, such that

$$u_n\to u,w{L}^{\mu }{u}_n\to h\mathit{in}{L}_{\theta ,p}\left((0,\infty )\right).$$

Namely,

$$\left(\lambda +{\rho }_{\ast }-{\displaystyle \frac{M_{\beta }}{p}}\right){\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le {\left|h-\lambda u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

In particular, if

$$h-\lambda u=g\mathit{in}{L}_{\theta ,p}\left((0,\infty )\right),$$

then

$${\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le {\displaystyle \frac{1}{\lambda +{\rho }_{\ast }-M_{\beta }/p}}{\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)},$$

and

$${\left|h\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le \left(1+{\displaystyle \frac{\lambda }{\lambda +{\rho }_{\ast }-M_{\beta }/p}}\right){\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

Proof. 

We first prove the estimate for $u\in C_c^{\infty }\left((0,\infty )\right)$. Recall that, for smooth functions,

$$w{L}^{\mu }u=Au-\rho u.$$

By Lemma 6,

$${p\left|u\right|}^{p-2}uAu\le {A\left(\right|u|}^p).$$

Using this inequality together with the weighted adjoint inequality, we obtain

$$p{\int }_0^{\infty }{\left|u\right|}^{p-2}uAu{x}^{\beta }dx\le {\int }_0^{\infty }{A\left(\right|u|}^p)\left(x\right)x^{\beta }dx\le M_{\beta }{\int }_0^{\infty }{\left|u\right|}^p{x}^{\beta }dx.$$

For $1<p<2$, this line is justified by the same regularization argument used in the proof of Lemma 6; no additional argument is needed here. Hence

$${\int }_0^{\infty }{\left|u\right|}^{p-2}uAu{x}^{\beta }dx\le {\displaystyle \frac{M_{\beta }}{p}}{\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}^p.$$

Since $w{L}^{\mu }u=Au-\rho u$ and $\rho \left(x\right)\ge {\rho }_{\ast }$, we have

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{\left|u\right|}^{p-2}uw{L}^{\mu }u{x}^{\beta }dx& ={\int }_0^{\infty }{\left|u\right|}^{p-2}uAu{x}^{\beta }dx-{\int }_0^{\infty }\rho \left(x\right){\left|u\right|}^p{x}^{\beta }dx\hfill \\ & \le \left({\displaystyle \frac{M_{\beta }}{p}}-{\rho }_{\ast }\right){\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}^p.\hfill \end{array}$$

Set

$$G=w{L}^{\mu }u-\lambda u.$$

Then

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{\left|u\right|}^{p-2}uG{x}^{\beta }dx& ={\int }_0^{\infty }{\left|u\right|}^{p-2}uw{L}^{\mu }u{x}^{\beta }dx-\lambda {\int }_0^{\infty }{\left|u\right|}^p{x}^{\beta }dx\hfill \\ & \le \left({\displaystyle \frac{M_{\beta }}{p}}-{\rho }_{\ast }-\lambda \right){\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}^p.\hfill \end{array}$$

Therefore,

$$\left(\lambda +{\rho }_{\ast }-{\displaystyle \frac{M_{\beta }}{p}}\right){\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}^p\le -{\int }_0^{\infty }{\left|u\right|}^{p-2}uG{x}^{\beta }dx.$$

By Hölder’s inequality with respect to the measure $x^{\beta }dx$,

$$\left|{\int }_0^{\infty }{\left|u\right|}^{p-2}uG{x}^{\beta }dx\right|\le {\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}^{p-1}{\left|G\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

Thus

$$\left(\lambda +{\rho }_{\ast }-{\displaystyle \frac{M_{\beta }}{p}}\right){\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}^p\le {\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}^{p-1}{\left|w{L}^{\mu }u-\lambda u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

If ${\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}=0$, the desired inequality is immediate. Otherwise, dividing by ${\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}^{p-1}$ gives

$$\left(\lambda +{\rho }_{\ast }-{\displaystyle \frac{M_{\beta }}{p}}\right){\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le {\left|w{L}^{\mu }u-\lambda u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

Now suppose that $u,h\in L_{\theta ,p}\left((0,\infty )\right)$ are obtained from an approximating sequence $u_n\in C_c^{\infty }\left((0,\infty )\right)$ such that

$$u_n\to u,w{L}^{\mu }{u}_n\to h\mathrm{in}{L}_{\theta ,p}\left((0,\infty )\right).$$

Applying the smooth estimate to $u_n$, we have

$$\left(\lambda +{\rho }_{\ast }-{\displaystyle \frac{M_{\beta }}{p}}\right){\left|u_n\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le {\left|w{L}^{\mu }{u}_n-\lambda u_n\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

Passing to the limit gives

$$\left(\lambda +{\rho }_{\ast }-{\displaystyle \frac{M_{\beta }}{p}}\right){\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le {\left|h-\lambda u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

If

$$h-\lambda u=g\mathrm{in}{L}_{\theta ,p}\left((0,\infty )\right),$$

then the preceding estimate gives

$${\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le {\displaystyle \frac{1}{\lambda +{\rho }_{\ast }-M_{\beta }/p}}{\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}.$$

Finally, since $h=g+\lambda u$, we have

$${\left|h\right|}_{L_{\theta ,p}\left((0,\infty )\right)}\le {\left|g\right|}_{L_{\theta ,p}\left((0,\infty )\right)}+\lambda {\left|u\right|}_{L_{\theta ,p}\left((0,\infty )\right)},$$

and the estimate for h follows. □

In the remainder of the paper, we verify the weighted generator inequality required in Theorem 3 for the classes of Lévy measures in Examples 1 and 2. The next lemma gives the preliminary estimates needed for that verification.

Lemma 7.

Let $\mu \left(dr\right)=j\left(\left|r\right|\right)dr$ be the Lévy measure associated with a subordinate Brownian motion where the Laplace exponent ϕ is a complete Bernstein function satisfying the weak scaling conditions (H1) and (H2). Define

$${\beta }_0=max\left(-1-2p_0,-1-2p_{\infty }\right).$$

Then, for any $\beta \in \left({\beta }_0,0\right)$, the following estimates are valid for all $x>0$:

(i) 

${\int }_x^{\infty }w\left(r\right)r^{\beta }\mu \left(dr\right)\asymp x^{\beta }$ and ${\int }_0^{x}w\left(r\right)r^{\beta }dr\asymp \varphi {\left(x^{-2}\right)}^{-1}{x}^{\beta +1},$

(ii) 

${\int }_{\left|r\right|\le x}{r}^2\mu \left(dr\right)\asymp x^2\varphi \left(x^{-2}\right),$

(iii) 

$\left|w^{″}\left(x\right)\right|\le C\varphi {\left(x^{-2}\right)}^{-1}{x}^{-2}.$

• In addition, if $\sigma >1$, then

(iv) 

$\left|{\displaystyle \frac{\partial }{\partial x}}\left(w\left(x\right)x^{\beta }{\int }_x^{\infty }{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}r\mu \left(dr\right)\right)\right|\le C{x}^{\beta },$ where C is independent of $ϵ>0,$

(v) 

$\left|\kappa \left(x\right)\right|\le C$ and $\left|x{\kappa }^{\prime }\left(x\right)\right|\le C.$

Proof. 

Fix $x>0,$ all approximating constants below are independent of $x.$ Under the assumptions of Example 1 and (H1)–(H2), it is known (see [8] (Lemmas 3.2, 3.3)) that the Lévy density satisfies

$$j\left(r\right)\asymp r^{-1}\varphi \left(r^{-2}\right),r>0.$$

(13)

Hence, by changing the variable of integration $u=r^{-2}$, so that $dr=-{\displaystyle \frac{1}{2}}{u}^{-3/2}du$ and applying Karamata’s theorem for O-regularly varying functions [11] (Theorem 3) in its two-sided form, we obtain

$$\begin{array}{cc}\hfill \delta \left(x\right)& \asymp 2{\int }_x^{\infty }{r}^{-1}\varphi \left(r^{-2}\right)dr={\int }_0^{x^{-2}}{u}^{-1}\varphi \left(u\right)du\asymp \varphi \left(x^{-2}\right).\hfill \end{array}$$

(14)

Consequently,

$$w\left(x\right)=\delta {\left(x\right)}^{-1}\asymp \varphi {\left(x^{-2}\right)}^{-1}.$$

(15)

Proof of (i). Using (13) and (15), since $\beta <0,$

$$\begin{array}{cc}\hfill {\int }_x^{\infty }w\left(r\right)r^{\beta }\mu \left(dr\right)& \asymp {\int }_x^{\infty }\varphi {\left(r^{-2}\right)}^{-1}{r}^{\beta }·r^{-1}\varphi \left(r^{-2}\right)dr={\int }_x^{\infty }{r}^{\beta -1}dr\asymp x^{\beta }.\hfill \end{array}$$

Using the substitution $u=r^{-2}$, $dr=-{\displaystyle \frac{1}{2}}{u}^{-3/2}du$ and applying Karamata’s theorem for $\beta >max\left(-1-2p_0,-1-2p_{\infty }\right)$,

$${\int }_0^{x}w\left(r\right)r^{\beta }dr\asymp {\int }_0^x\varphi {\left(r^{-2}\right)}^{-1}{r}^{\beta }dr\asymp {\int }_{x^{-2}}^{\infty }\varphi {\left(u\right)}^{-1}{u}^{\left(-\beta -3\right)/2}du\asymp \varphi {\left(x^{-2}\right)}^{-1}{x}^{\beta +1}.$$

Proof of (ii). By (13), the substitution $u=r^{-2}$, $dr=-{\displaystyle \frac{1}{2}}{u}^{-3/2}du$ and Karamata’s theorem,

$${\int }_{\left|r\right|\le x}{r}^2\mu \left(dr\right)\asymp {\int }_0^{x}r\varphi \left(r^{-2}\right)dr\asymp {\int }_{x^{-2}}^{\infty }{u}^{-2}\varphi \left(u\right)du\asymp x^2\varphi \left(x^{-2}\right).$$

Proof of (iii). We differentiate:

$${\delta }^{\prime }\left(x\right)=-2j\left(x\right),{\delta }^{″}\left(x\right)=-2j^{\prime }\left(x\right).$$

(16)

By differentiating $j\left(r\right)={\int }_0^{\infty }{\left(4\pi t\right)}^{-1/2}{e}^{-r^2/\left(4t\right)}{\mu }_S\left(dt\right)$ and using the same decomposition used in [8] (Lemma 3.1),

$$\left|j^{\prime }\left(r\right)\right|\le C{r}^{-2}\varphi \left(r^{-2}\right).$$

Since $w\left(x\right)=\delta {\left(x\right)}^{-1}$,

$$w^{\prime }\left(x\right)=-\delta {\left(x\right)}^{-2}{\delta }^{\prime }\left(x\right),w^{″}\left(x\right)=2\delta {\left(x\right)}^{-3}{\left({\delta }^{\prime }\left(x\right)\right)}^2-\delta {\left(x\right)}^{-2}{\delta }^{″}\left(x\right).$$

Hence, by (13), (14), and (16),

$$w^{\prime }\left(x\right)\asymp x^{-1}\varphi {\left(x^{-2}\right)}^{-1}.$$

(17)

Using also $\left|{\delta }^{″}\left(x\right)\right|\le x^{-2}\varphi \left(x^{-2}\right),$ we get

$$\begin{array}{cc}\hfill \left|w^{″}\left(x\right)\right|& \le C{x}^{-2}\varphi {\left(x^{-2}\right)}^{-1}.\hfill \end{array}$$

Proof of (iv). If $\sigma >1$, then Assumption 2 entails $p_0\wedge p_{\infty }>{\displaystyle \frac{1}{2}}.$ Hence, using (13) and Karamata’s theorem gives

$$\begin{array}{cc}\hfill {\int }_x^{\infty }rj\left(r\right)dr& \asymp {\int }_x^{\infty }\varphi \left(r^{-2}\right)dr\asymp x\varphi \left(x^{-2}\right).\hfill \end{array}$$

(18)

Differentiating,

$$\begin{array}{cc}\hfill & {\displaystyle \frac{\partial }{\partial x}}\left(w\left(x\right)x^{\beta }{\int }_x^{\infty }{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}r\mu \left(dr\right)\right)\hfill \\ \hfill & =\left(w^{\prime }\left(x\right)x^{\beta }+\beta w\left(x\right)x^{\beta -1}\right)\left({\int }_x^{\infty }{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}rj\left(r\right)dr\right)-w\left(x\right)x^{\beta +1}j\left(x\right){\mathbf{1}}_{\left\{\left|x\right|>ϵ\right\}}.\hfill \end{array}$$

Using (13), (15), (17), and (18) leads to (iv).

• Proof of (v). We recall, upon changing the variable of integration,

$$\kappa \left(x\right)={\displaystyle \frac{w\left(x\right)}{x}}{\int }_x^{\infty }rj\left(r\right)dr.$$

By (15) and (18), $\left|\kappa \left(x\right)\right|\le C.$ Next, we differentiate

$$\begin{array}{cc}\hfill {\kappa }^{\prime }\left(x\right)& =-w\left(x\right)j\left(x\right)+\left({\displaystyle \frac{x{w}^{\prime }\left(x\right)-w\left(x\right)}{x^2}}\right)\left({\int }_x^{\infty }rj\left(r\right)dr\right),\hfill \\ \hfill x{\kappa }^{\prime }\left(x\right)& =-w\left(x\right)j\left(x\right)x+\left(2w\left(x\right)j\left(x\right)x-1\right)\kappa \left(x\right).\hfill \end{array}$$

Using (13) and (15), we conclude $\left|x{\kappa }^{\prime }\left(x\right)\right|\le C.$ □

Remark 2.

Owing to (15), the function w has two-sided scaling property whose scaling indices are twice those of $\varphi .$

Theorem 4.

Let $\mu \left(dr\right)=j\left(\left|r\right|\right)dr$ be the Lévy measure associated with a subordinate Brownian motion where the Laplace exponent ϕ is a complete Bernstein function satisfying the weak scaling conditions (H1) and (H2). Define

$${\beta }_0=max\left(-1-2p_0,-1-2p_{\infty }\right).$$

Then, for any $\beta \in \left({\beta }_0,0\right)$, there exists a constant $M_{\beta }>0$ such that for every non-negative $F\in C_c^{\infty }\left(\left(0,\infty \right)\right),$

$${\int }_0^{\infty }A\left(F\right)\left(x\right)x^{\beta }dx\le M_{\beta }{\int }_0^{\infty }F\left(x\right)x^{\beta }dx.$$

Consequently, Theorem 3 holds for $\beta =\theta p-1\in \left({\beta }_0,0\right)$ whenever $\lambda +{\displaystyle \frac{1}{2}}>{\displaystyle \frac{M_{\beta }}{p}}.$

Proof. 

Fix a non-negative $F\in C_c^{\infty }\left(\left(0,\infty \right)\right),$ and introduce the truncation parameter $0<ϵ<{\displaystyle \frac{1}{2}}.$ All bounding constants below will be independent of F and $ϵ$. We define

$$\begin{array}{c}{A}_{ϵ}\left(F\right)\left(x\right)\hfill \\ ={\int }_{\left(-1,\infty \right)}{\mathbf{1}}_{\left\{\left|r\right|>{\displaystyle \frac{ϵ}{x}}\right\}}\left(F\left(x\left(1+r\right)\right)-F\left(x\right)-{\chi }_{\sigma }\left(r\right)xr{F}^{\prime }\left(x\right)\right){\tilde{\mu }}_x\left(dr\right)+{\mathbf{1}}_{\left\{\sigma >1\right\}}x\kappa \left(x\right)F^{\prime }\left(x\right).\hfill \end{array}$$

Multiplying by $x^{\beta }$ and integrating in x, we first handle the difference term.

$$\begin{array}{cc}\hfill J^{ϵ}=& {\int }_0^{\infty }{\int }_{\left(-1,\infty \right)}{\mathbf{1}}_{\left\{\left|r\right|>{\displaystyle \frac{ϵ}{x}}\right\}}\left(F\left(x\left(1+r\right)\right)-F\left(x\right)\right){\tilde{\mu }}_x\left(dr\right)x^{\beta }dx\hfill \\ \hfill & ={\int }_0^{\infty }{\int }_{\left(-1,\infty \right)}{\mathbf{1}}_{\left\{\left|r\right|>{\displaystyle \frac{ϵ}{x}}\right\}}F\left(x\left(1+r\right)\right){\tilde{\mu }}_x\left(dr\right)x^{\beta }dx\hfill \\ \hfill & -{\int }_0^{\infty }{\int }_{\left(-1,\infty \right)}{\mathbf{1}}_{\left\{\left|r\right|>{\displaystyle \frac{ϵ}{x}}\right\}}F\left(x\right){\tilde{\mu }}_x\left(dr\right)x^{\beta }dx\hfill \\ \hfill & =I_1^{ϵ}-I_2^{ϵ}.\hfill \end{array}$$

Changing the variable of integration and applying Tonelli theorem yields

$$\begin{array}{cc}\hfill I_1^{ϵ}& ={\int }_0^{\infty }{\int }_{\left(-x,\infty \right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}w\left(x\right)F\left(x+r\right)\mu \left(dr\right)x^{\beta }dx\hfill \\ \hfill & ={\int }_{\mathbf{R}}{\int }_{{\left(-r\right)}^{+}}^{\infty }{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}w\left(x\right)F\left(x+r\right)x^{\beta }dx\mu \left(dr\right)\hfill \\ \hfill & ={\int }_{\mathbf{R}}{\int }_{r^{+}}^{\infty }{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}w\left(x-r\right){\left(x-r\right)}^{\beta }F\left(x\right)dx\mu \left(dr\right)\hfill \\ \hfill & ={\int }_0^{\infty }{\int }_{\left(-\infty ,x\right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}w\left(x-r\right){\left(x-r\right)}^{\beta }F\left(x\right)\mu \left(dr\right)dx.\hfill \end{array}$$

By symmetry of $\mu$, we have

$$\begin{array}{cc}\hfill I_2^{ϵ}& ={\int }_0^{\infty }{\int }_{\left(-x,\infty \right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}w\left(x\right)x^{\beta }F\left(x\right)\mu \left(dr\right)dx\hfill \\ \hfill & ={\int }_0^{\infty }{\int }_{\left(-\infty ,x\right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}w\left(x\right)x^{\beta }F\left(x\right)\mu \left(dr\right)dx.\hfill \end{array}$$

Subtracting,

$$\begin{array}{cc}\hfill J^{ϵ}=& {\int }_0^{\infty }{\int }_{\left(-\infty ,x\right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}\left[w\left(x-r\right){\left(x-r\right)}^{\beta }F\left(x\right)-w\left(x\right)x^{\beta }F\left(x\right)\right]\mu \left(dr\right)dx.\hfill \end{array}$$

We split the domain into three regions: $\left(-\infty ,x\right)=\left(-\infty ,-x/2\right)\cup \left[-x/2,x/2\right)\cup \left[x/2,x\right).$ Correspondingly,

$$\begin{array}{cc}\hfill J^{ϵ}& =J_1^{ϵ}+J_2^{ϵ}+J_3^{ϵ},\hfill \end{array}$$

Now we estimate them one by one.

Estimate of $J_1^{ϵ}$: On the region $r\in \left(-\infty ,-x/2\right)$, we have $\left|r\right|\le x-r\le 3\left|r\right|.$ By the scaling property of w (see Remark 2), there exists $C>0$ such that

$$C^{-1}w\left(\left|r\right|\right)\le w\left(x-r\right)\le Cw\left(\left|r\right|\right).$$

Hence, using symmetry of $\mu$ and recalling $\delta =w^{-1}$,

$$\begin{array}{cc}\hfill & \left|J_1^{ϵ}\right|\hfill \\ \hfill \le & {\int }_0^{\infty }{\int }_{\left(-\infty ,-x/2\right)}\left[w\left(x-r\right){\left(x-r\right)}^{\beta }F\left(x\right)+w\left(x\right)x^{\beta }F\left(x\right)\right]\mu \left(dr\right)dx\hfill \\ \hfill \le & C{\int }_0^{\infty }{\int }_{\left(-\infty ,-x/2\right)}w\left(\left|r\right|\right){\left|r\right|}^{\beta }\mu \left(dr\right)F\left(x\right)dx+{\int }_0^{\infty }{\int }_{\left(-\infty ,-x/2\right)}\mu \left(dr\right)w\left(x\right)x^{\beta }F\left(x\right)dx\hfill \\ \hfill =& C{\int }_0^{\infty }{\int }_{\left(x/2,\infty \right)}w\left(r\right)r^{\beta }\mu \left(dr\right)F\left(x\right)dx+{\displaystyle \frac{1}{2}}{\int }_0^{\infty }w{\left(x/2\right)}^{-1}w\left(x\right)x^{\beta }F\left(x\right)dx.\hfill \end{array}$$

By Lemma 7 (i) and the scaling property of w,

$$\left|J_1^{ϵ}\right|\le C{\int }_0^{\infty }{x}^{\beta }F\left(x\right)dx.$$

Estimate of $J_2^{ϵ}$: We set $G\left(y\right)=w\left(y\right)y^{\beta },y>0.$ Then

$$J_2^{ϵ}={\int }_0^{\infty }{\int }_{\left[-x/2,x/2\right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}\left[G\left(x-r\right)-G\left(x\right)\right]F\left(x\right)\mu \left(dr\right)dx.$$

By Taylor expansion, $G\left(x-r\right)-G\left(x\right)=-r{G}^{\prime }\left(x\right)+{\displaystyle \frac{r^2}{2}}{G}^{″}\left(x-\theta r\right)$ for some $\theta =\theta \left(x,r\right)\in \left(0,1\right).$

Plugging in:

$$J_2^{ϵ}={\int }_0^{\infty }{\int }_{\left[-x/2,x/2\right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}\left[-r{G}^{\prime }\left(x\right)+{\displaystyle \frac{r^2}{2}}{G}^{″}\left(x-\theta r\right)\right]F\left(x\right)\mu \left(dr\right)dx.$$

Due to symmetry of $\mu \left(dr\right)$, the first term vanishes and

$$J_2^{ϵ}={\displaystyle \frac{1}{2}}{\int }_0^{\infty }{\int }_{\left[-x/2,x/2\right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}{r}^2{G}^{″}\left(x-\theta r\right)F\left(x\right)\mu \left(dr\right)dx.$$

By a direct computation,

$$G^{″}\left(y\right)=w^{″}\left(y\right)y^{\beta }+2\beta w^{\prime }\left(y\right)y^{\beta -1}+\beta \left(\beta -1\right)w\left(y\right)y^{\beta -2}.$$

Using estimates of $w^{″},w^{\prime }$ and w in Lemma 7,

$$\left|G^{″}\left(y\right)\right|\le C\varphi {\left(y^{-2}\right)}^{-1}{y}^{\beta -2}.$$

On the region $r\in \left[-x/2,x/2\right)$, we have ${\displaystyle \frac{x}{2}}\le x-\theta r\le {\displaystyle \frac{3x}{2}}.$ Since $\varphi$ satisfies the scaling assumptions, $\varphi \left({\left(x-\theta r\right)}^{-2}\right)\asymp \varphi \left(x^{-2}\right).$ Hence,

$$\left|G^{″}\left(x-\theta r\right)\right|\le C\varphi {\left(x^{-2}\right)}^{-1}{x}^{\beta -2}.$$

Using this bound and Lemma 7 (ii),

$$\begin{array}{cc}\hfill \left|J_2^{ϵ}\right|& \le C{\int }_0^{\infty }\varphi {\left(x^{-2}\right)}^{-1}{x}^{\beta -2}\left({\int }_{\left[-x/2,x/2\right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}{r}^2\mu \left(dr\right)\right)F\left(x\right)dx\hfill \\ \hfill & \le C{\int }_0^{\infty }{x}^{\beta }F\left(x\right)dx.\hfill \end{array}$$

Estimate of $J_3^{ϵ}$: We split further as

$$\begin{array}{cc}\hfill J_3^{ϵ}& ={\int }_0^{\infty }{\int }_{\left[x/2,x\right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}\left[w\left(x-r\right){\left(x-r\right)}^{\beta }F\left(x\right)-w\left(x\right)x^{\beta }F\left(x\right)\right]\mu \left(dr\right)dx\hfill \\ \hfill & ={\int }_0^{\infty }{\int }_{\left[x/2,x\right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}w\left(x-r\right){\left(x-r\right)}^{\beta }F\left(x\right)\mu \left(dr\right)dx\hfill \\ \hfill & -{\int }_0^{\infty }{\int }_{\left[x/2,x\right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}w\left(x\right)x^{\beta }F\left(x\right)\mu \left(dr\right)dx\hfill \\ \hfill & =J_{31}^{ϵ}-J_{32}^{ϵ}.\hfill \end{array}$$

For the first term, we have

$$\begin{array}{cc}\hfill \left|J_{31}^{ϵ}\right|& \le {\int }_0^{\infty }{\int }_{\left[x/2,x\right)}w\left(x-r\right){\left(x-r\right)}^{\beta }F\left(x\right)\mu \left(dr\right)dx\hfill \\ \hfill & ={\int }_0^{\infty }{\int }_{\left[0,x/2\right)}w\left(r\right)r^{\beta }j\left(x-r\right)drF\left(x\right)dx\hfill \\ \hfill & \le C{\int }_0^{\infty }\left({\int }_0^{x/2}w\left(r\right)r^{\beta }dr\right)j\left(x\right)F\left(x\right)dx.\hfill \end{array}$$

where in the last inequality we used $j\left(x-r\right)\asymp j\left(x\right)$ since ${\displaystyle \frac{x}{2}}\le x-r\le x.$

Applying Lemma 7 (i) and the scaling property of $\varphi$,

$$\left|J_{31}^{ϵ}\right|\le C{\int }_0^{\infty }{x}^{\beta }F\left(x\right)dx.$$

Using the scaling property of w,

$$\begin{array}{cc}\hfill \left|J_{32}^{ϵ}\right|& \le {\int }_0^{\infty }\delta \left({\displaystyle \frac{x}{2}}\right)w\left(x\right)x^{\beta }F\left(x\right)dx\hfill \\ \hfill & \le C{\int }_0^{\infty }{x}^{\beta }F\left(x\right)dx.\hfill \end{array}$$

Combining the preceding estimates

$$\left|J^{ϵ}\right|\le C{\int }_0^{\infty }{x}^{\beta }F\left(x\right)dx.$$

We now estimate the derivative term. By symmetry of ${\tilde{\mu }}_x\left(dr\right)$ the contribution from $\left(-1,1\right)$ vanishes.

If $\sigma =1$, this removes the derivative term entirely. If $\sigma >1$, using integration by parts over $\left(1,\infty \right),$

$$\begin{array}{cc}\hfill K^{ϵ}& ={\int }_0^{\infty }{\int }_{\left(1,\infty \right)}{\mathbf{1}}_{\left\{\left|r\right|>{\displaystyle \frac{ϵ}{x}}\right\}}xr{F}^{\prime }\left(x\right){\tilde{\mu }}_x\left(dr\right)x^{\beta }dx\hfill \\ \hfill & ={\int }_0^{\infty }{\int }_{\left(x,\infty \right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}w\left(x\right)x^{\beta }r{F}^{\prime }\left(x\right)\mu \left(dr\right)dx\hfill \\ \hfill & ={\int }_0^{\infty }\left[w\left(x\right)x^{\beta }{\int }_{\left(x,\infty \right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}r\mu \left(dr\right)\right]F^{\prime }\left(x\right)dx\hfill \\ \hfill & =-{\int }_0^{\infty }{\displaystyle \frac{\partial }{\partial x}}\left(w\left(x\right)x^{\beta }{\int }_{\left(x,\infty \right)}{\mathbf{1}}_{\left\{\left|r\right|>ϵ\right\}}r\mu \left(dr\right)\right)F\left(x\right)dx.\hfill \end{array}$$

By Lemma 7 (iv), $\left|K^{ϵ}\right|\le C{\int }_0^{\infty }{x}^{\beta }F\left(x\right)dx.$

The drift term, present only when $\sigma >1$, is also handled with integration by parts. Indeed,

$$\begin{array}{cc}\hfill L={\int }_0^{\infty }x\kappa \left(x\right)F^{\prime }\left(x\right)x^{\beta }dx& =-{\int }_0^{\infty }{\displaystyle \frac{d}{dx}}\left(x^{\beta +1}\kappa \left(x\right)\right)F\left(x\right)dx\hfill \\ \hfill & =-{\int }_0^{\infty }\left(\left(\beta +1\right)\kappa \left(x\right)+x{\kappa }^{\prime }\left(x\right)\right)x^{\beta }F\left(x\right)dx.\hfill \end{array}$$

Therefore, by Lemma 7 (v),

$$\left|L\right|\le C{\int }_0^{\infty }{x}^{\beta }F\left(x\right)dx.$$

Because all estimates are independent of $ϵ$, we let $ϵ\downarrow 0$ and obtain the desired inequality. The last statement follows from the fact that ${\rho }_{\ast }={\displaystyle \frac{1}{2}}$ thanks to symmetry. □

Remark 3.

(i) Examining the proof of Lemma 7 and Theorem 4, we see that all estimates also hold for non-Bernstein Example 2. Consequently, Theorem 3 applies in this setting as well.

(ii) 

The constant $M_{\beta }$ depends on scaling indices through the applications of Karamata’s theorem in Lemma 7.

(iii) 

The symmetry assumption on μ is crucial in the proof of Theorem 4. For example, in the estimate of $J_2^{\epsilon }$, the Taylor expansion

$$G(x-r)-G\left(x\right)=-r{G}^{\prime }\left(x\right)+{\displaystyle \frac{r^2}{2}}{G}^{″}(x-\theta r)$$

produces the first-order contribution

$$-{\int }_0^{\infty }{G}^{\prime }\left(x\right)F\left(x\right){\int }_{(-x/2,x/2)}{\mathbf{1}}_{\left\{\right|r|>\epsilon \}}r\mu \left(dr\right)dx.$$

This term vanishes when μ is symmetric. For a non-symmetric Lévy measure, the inner integral need not vanish and is not controlled by the present argument. A complete treatment of weighted estimates for non-symmetric kernels is left for further future investigation.

7. Conclusions

This paper developed a probabilistic and weighted approach to one-dimensional Lévy-type Dirichlet problems on the half-line. By multiplying the original equation by the tail weight w, the Dirichlet problem was transformed into a weighted nonlocal equation associated with a multiplicative jump mechanism and a boundary killing term. Under the basic structural assumptions on the Lévy measure, the transformed operator was realized through a martingale problem. After fixing a Markov process solving this martingale problem, the exponential killing representation gives a probabilistic mild solution for bounded Borel data and yields the immediate $L^{\infty }$-estimate. This is the general existence result obtained in the paper.

In the one-dimensional fractional Laplacian case, the transformed jump measure is independent of the spatial variable. Consequently, the transformed process has the exact multiplicative representation $Y_t^x=x{\mathcal{E}}_t$, where ${\mathcal{E}}_t$ satisfies an explicit multiplicative jump equation. This special structure allows the probabilistic mild formulation to be connected with weighted $L_{\theta ,p}$-estimates. Starting from smooth data, the corresponding mild solutions are estimated in $L_{\theta ,p}\left((0,\infty )\right)$, and the construction is extended to general data by approximation in the transformed graph norm. The resulting limit solution is therefore an $L_{\theta ,p}$-solution in the closed-graph sense for the transformed operator $w{L}^{\mu }-\lambda$. In the case $\lambda =0$, this solution agrees with the corresponding weak solution from [3] (Theorem 2.3(ii)). The comparison of parameters shows that the range obtained from the present multiplicative method recovers the known weighted range for the fractional Laplacian.

For more general Lévy measures, the paper identifies the analytic weighted energy mechanism behind the $L_p$-estimate. The estimate is proved at the smooth level: for $u\in C_c^{\infty }\left((0,\infty )\right)$, the weighted $L_{\theta ,p}$-norm of u is controlled by the weighted $L_{\theta ,p}$-norm of $w{L}^{\mu }u-\lambda u$, provided the weighted adjoint integral inequality holds. The same estimate also passes to any limiting function for which a graph-convergent approximation

$$u_n\to u,w{L}^{\mu }{u}_n\to w{L}^{\mu }u\mathrm{in}{L}_{\theta ,p}\left((0,\infty )\right)$$

is available. Thus, the result gives a conditional closed-graph estimate, while the construction of such approximating sequences for general Lévy measures is a separate domain problem.

The required weighted adjoint integral inequality was verified for symmetric subordinate Brownian motions associated with Bernstein functions and for non-unimodal logarithmically perturbed stable-type kernels. These examples show that the weighted energy mechanism is not limited to the classical stable kernel and can be established without relying on heat kernel estimates for killed processes. At the same time, the present argument uses symmetry in an essential way; the corresponding theory for non-symmetric kernels remains open.

The main issues left for further future investigation are therefore the domain passage from probabilistic mild solutions to graph-convergent weighted $L_{\theta ,p}$-operator solutions for general Lévy measures, the extension of the weighted adjoint inequality to non-symmetric kernels, and the development of parabolic and higher-dimensional analogues of the present framework.