G_\alpha(\xi)  = (1+|\xi|^2)^{-\alpha/2}$, and $G_\alpha$ is given as in Lemma \[A-sectorial\]. For further details we refer the reader to Section 1.6 of [@Henry1981] and [@Stein1970 Chapter 5].

\[lemma-f\] Assume $\nu$ is an admissible activity Lévy measure with shape parameters $\alpha, D^\pm$, and $\mu$ where $\alpha<3$ and either $\mu>0, D^\pm\in\mathbb{R}$, or $\mu=0, D^- + 1<0<D^+$. Suppose that $\gamma\ge 1/2$ and $\gamma>(\alpha-1)/2$. Then, for the mapping $f$ defined by (\[functional\_f\_def\]), there exists a constant $C>0$ such that, for any $u$ satisfying $\partial_x u \in X^{\gamma-1/2}$, the following estimate holds: $$\Vert f[u]\Vert_{L^p} \le C \Vert \partial_x u\Vert_{X^{\gamma-1/2}}.$$ In particular, if $u\in X^\gamma$ we have $\Vert f[u]\Vert_{L^p} \le C \Vert u\Vert_{X^\gamma}$ and the mapping $f$ is a bounded linear operator from the fractional power space $X^\gamma$ into $X=L^p({\mathbb{R}})$.

Proof. The mapping $f$ can be split as follows: $f[u]=\tilde f[u] + \tilde\omega \partial_x u$ where $$\begin{aligned}
\tilde f[u](x) &=& \int_{\mathbb{R}}\left(u(x+z) - u(x) - z \frac{\partial u}{\partial x}(x) \right) \nu({\mathrm{d}}{z}),\end{aligned}$$ and $\tilde\omega=\int_{\mathbb{R}}\left(z - e^z+1 \right) \nu({\mathrm{d}}{z})$. Since $z - e^z+1=O(z^2)$ as $z\to0$, and $$0\le \nu({\mathrm{d}}{z}) = h(z) {\mathrm{d}}{z} \le  |z|^{-\alpha} \tilde h(z) {\mathrm{d}}{z},
\ \ 
\text{where}
\ \ 
\tilde h(z) = C_0 e^{-\mu z^2} \left(e^{D^-z} 1_{z\ge0} + e^{D^+z} 1_{z<0} \right),$$ we have $\tilde\omega\in{\mathbb{R}}$ provided that $0\le 