\theta z) - G_{2\gamma-1}(\cdot)\Vert_{L^1}^p\Vert\varphi\Vert_{L^p}^p 
\le C_1^p |\theta z|^{(2\gamma-1)p} \Vert \partial_x u\Vert_{X^{\gamma-1/2}}^p.\end{aligned}$$ The latter inequality formally holds true also for the case $\gamma=1/2$ because $$\int_{\mathbb{R}}\left|\frac{\partial u}{\partial x}(x+\theta z) -  \frac{\partial u}{\partial x}(x)\right|^p{\mathrm{d}}{x}
\le 2^p \Vert \partial_x u\Vert_{L^p}^p = 2^p \Vert \partial_x u\Vert_{X^0}^p .$$ The rest of the proof of boundedness of the mapping $f$ holds for $\gamma> 1/2$ as well as $\gamma=1/2$. As $u(x+z) - u(x) - z \frac{\partial u}{\partial x}(x)
=z\int_0^1 \frac{\partial u}{\partial x}(x+\theta z)-  \frac{\partial u}{\partial x}(x){\mathrm{d}}{\theta}$, we obtain $$\begin{aligned}
&& \int_{\mathbb{R}}|u(x+z) - u(x) - z \frac{\partial u}{\partial x}(x)|^p{\mathrm{d}}{x}
=|z|^p \int_{\mathbb{R}}\left| \int_0^1 \frac{\partial u}{\partial x}(x+\theta z)-  \frac{\partial u}{\partial x}(x){\mathrm{d}}{\theta} \right|^p {\mathrm{d}}{x}
\\
&& \le   |z|^p \int_0^1 \int_{\mathbb{R}}\left|  \frac{\partial u}{\partial x}(x+\theta z)-  \frac{\partial u}{\partial x}(x)\right|^p {\mathrm{d}}{x} {\mathrm{d}}{\theta} 
\le C_1^p |z|^{2\gamma p} \Vert \partial_x u\Vert_{X^{\gamma-1/2}}^p.\end{aligned}$$ Now, as $0\le \nu({\mathrm{d}}{z}) = h(z){\mathrm{d}}{z} \le  |z|^{-\alpha} \tilde h(z) {\mathrm{d}}{z}=
(|z|^{-\beta} \tilde h(z)^\frac12 )\cdot (|z|^{\beta-\alpha} \tilde h(z)^\frac12 ) {\mathrm{d}}{z}$, using the Hölder inequality with exponents $p,q$ such that $1/p+1/q=1$ we obtain $$\begin{aligned}
\Vert \tilde f[u]\Vert_{L^p}^p 
&=& \int_{\mathbb{R}}\left| \int_{\mathbb{R}}u(x+z) - u(x) - z \frac{\partial u}{\partial x}(x)\nu({\mathrm{d}}{z})\right |^p{\mathrm{d}}{x}
\\
&\le& \int_{\mathbb{R}}\left|\int_{\mathbb{R}}\left| u(x+z) - u(x) - z \frac{\partial u}{\partial x}(x)\right| h(z){\mathrm{d}}{z}  \right|^p{\mathrm{d}}{x}
\\
&\le& \int_{\mathbb{R}}\int_{\mathbb{R}}\left| u(x+z) - u(x) - z \frac{\partial u}{\partial x}(x)\right|^p |z|^{-\beta p} \tilde h(z)^{p/2} {\mathrm{d}}{z} 
