\\
&& \quad\times \left(\int_{\mathbb{R}}|z|^{(\beta-\alpha) q} \tilde h(z)^{q/2} {\mathrm{d}}{z}\right)^{p/q} {\mathrm{d}}{x}
\\
&=& \int_{\mathbb{R}}\left(\int_{\mathbb{R}}\left| u(x+z) - u(x) - z \frac{\partial u}{\partial x}(x)\right|^p {\mathrm{d}}{x}\right) |z|^{-\beta p} \tilde h(z)^{p/2} {\mathrm{d}}{z} 
\\
&& \quad\times \left(\int_{\mathbb{R}}|z|^{(\beta-\alpha) q} \tilde h(z)^{q/2} {\mathrm{d}}{z}\right)^{p/q} 
\\
&\le & 
C_1^p  \Vert \partial_x u\Vert_{X^{\gamma-1/2}}^p
\int_{\mathbb{R}}|z|^{(2\gamma-\beta) p} \tilde h(z)^{p/2} {\mathrm{d}}{z} \left(\int_{\mathbb{R}}|z|^{(\beta-\alpha) q} \tilde h(z)^{q/2} {\mathrm{d}}{z}\right)^{p/q}.\end{aligned}$$ The integrals $C_2=\int_{\mathbb{R}}|z|^{(2\gamma-\beta) p} \tilde h(z)^{p/2} {\mathrm{d}}{z}$ and $C_3=\int_{\mathbb{R}}|z|^{(\beta-\alpha) q} \tilde h(z)^{q/2} {\mathrm{d}}{z}$ are finite provided that $$(2\gamma-\beta) p > -1, \qquad (\beta-\alpha) q = (\beta-\alpha) \frac{p}{p-1} >-1,$$ and $\mu>0, D^\pm\in\mathbb{R}$, or $\mu=0$ and $D^-<0<D^+$. The later inequalities are satisfied if there exists a parameter $\beta$ such that $$\alpha-1+1/p < \beta < 2\gamma+1/p.$$ Such a choice of $\beta$ is possible because we assumed $\gamma>(\alpha-1)/2$. Hence there exists a constant $C>0$ such that $\Vert \tilde f[u]\Vert_{L^p} \le C \Vert \partial_x u\Vert_{X^{\gamma-1/2}}$ for any $u$ satisfying $\partial_x u\in X^{\gamma-1/2}$, as claimed. Due to the continuity of the embedding $X^{\gamma-1/2} \hookrightarrow X$ we have $\Vert f[u]\Vert_{L^p} = \Vert \tilde f[u] + \tilde\omega \partial_x u \Vert_{L^p} \le C \Vert \partial_x u\Vert_{X^{\gamma-1/2}} = C \Vert u\Vert_{X^\gamma}$ for any $u\in X^\gamma$ and $f$ is a bounded linear operator from $X^\gamma$ into $X=L^p$. $\diamondsuit$

Let us denote by $C([0,T],X^{\gamma})$ the Banach space of all continuous functions 