How to verify ihe Interpolation inequality for the weighted Bessel potential spaces?

Question

I am trying to prove the following: Let $w$ be an admissible weight, $p_1,p_2\in[1,\infty)$, $\alpha_1,\alpha_2\in\mathbb{R}$, $\theta\in(0,1)$ and \begin{equation} \alpha=\theta\,\alpha_1+(1-\theta)\,\alpha_2, ~~~ \frac{1}{p} = \frac{\theta}{p_1} + \frac{1-\theta}{p_2}. \end{equation} There exists $C\in(0,\infty)$ such that for all $f\in L^{{\alpha}_1}_{p_1}(\mathbb{R}^d,w^{{1/p}_1})\cap L^{{\alpha}_2}_{p_2}(\mathbb{R}^d,w^{{1/p}_2})$ it holds \begin{equation}\label{1} \|f\|_{L^\alpha_p(\mathbb{R}^d,w^{1/p})} \leq C\,\|f\|^\theta_{L^{\alpha_1}_{p_1}(\mathbb{R}^d,w^{1/p_1})}\, \|f\|^{1-\theta}_{L^{\alpha_2}_{p_2}(\mathbb{R}^d,w^{1/p_2})}\,, \end{equation} where $\|f\|_{L_p^\alpha(\mathbb{R}^d,w)}$ is the weighted Bessel potential of order $\alpha$ with the following Definition: The weighted Bessel potential space $L^\alpha_p(\mathbb{R}^d,w)$ is the Banach space with the norm \begin{equation} \|f\|_{L^\alpha_p(\mathbb{R}^d,w)} := \|(1-\Delta)^{\alpha/2}f\|_{L_p(\mathbb{R}^d,w)}\, \end{equation} such that the norms $\|\cdot\|_{L^\alpha_p(\mathbb{R}^d,w)}$ and $\|w\cdot\|_{L^\alpha_p(\mathbb{R}^d)}$ are equivalent. Moreover, we define for $s\in \mathbb{R}$, the Bessel potential of order $s$ to be the (sequentially) continuous bijective linear operator $\mathbb{J}: S\to S$ by \begin{equation} (1-\Delta)^{s/2}:=\mathbb{J}^{s} u = F^{-1}(1+|\cdot|^2)^{s/2}F u\,. \end{equation} Using the above one has \begin{equation} \mathbb{J}^{s+t} = \mathbb{J}^{s}\mathbb{J}^{t}\,. \end{equation} The above interpolation inequality is known for Besov spaces and Triebel-Lizorkin spaces using section 1.6.7. on page 44 of H. Triebel, Theory of Function Spaces II, (1992).But, I would like to have it for the the weighted bessel potential space $L^{\alpha}_{p}(\mathbb{R}^d,w)$. The following is my idea: \begin{equation} \|f\|^p_{L^\alpha_p(\mathbb{R}^d,w^{1/p})} = \|w^{1/p}(1-\Delta)^{{\alpha}/2} f\|^p_{L_p(\mathbb{R}^d)} = \int_{\mathbb{R}^d} |w(x)^{1/p} (1-\Delta)^{\frac{\alpha}{2}} f(x)|^p dx \end{equation} Using $p = \theta p + (1-\theta) p$, $ \alpha=\theta\,\alpha_1+(1-\theta)\,\alpha_2$ yields \begin{equation} \|f\|^p_{L^\alpha_p(\mathbb{R}^d,w^{1/p})} = \int_{\mathbb{R}^d} |w(x) (1-\Delta)^{\frac{{\alpha}_1}{2} \theta p} f^{\theta p}(x) (1-\Delta)^{\frac{{\alpha}_2}{2} (1-\theta)p} f^{(1-\theta)p}(x) |dx\,. \end{equation} By assumption $1 = \frac{\theta p}{p_1} + \frac{1-\theta p}{p_2}$. Hence, \begin{equation} \|f\|^p_{L^\alpha_p(\mathbb{R}^d,w^{1/p})} \leq C \int_{\mathbb{R}^d}|w(x)^{\frac{\theta p}{p_1}} (1-\Delta)^{\frac{{\alpha}_1}{2} \theta p} f^{\theta p}(x)|\,dx \int_{\mathbb{R}^d} | w(x)^{\frac{1-\theta}{p_2}p}(1-\Delta)^{\frac{{\alpha}_2}{2} (1-\theta)p} f^{(1-\theta)p}(x)| dx, \end{equation} where $C$ only depends on the weight $w(x)$ (In particular, I used the fact that for every $b\in\mathbb{N}^d$ there exists $C\in(0,\infty)$ such that $|\partial^b w (x)|\leq C\, w(x)$ for all $x\in\mathbb{R}^d$). Now, by the Holder inequality one has \begin{equation} \|f\|^p_{L^\alpha_p(\mathbb{R}^d,w^{1/p})} \leq C \Big[ \int_{\mathbb{R}^d}|w(x)(1-\Delta)^{\frac{{\alpha}_1}{2} p_1} f^{p_1}(x)|\Big]^{\frac{\theta p}{p_1}} \int_{\mathbb{R}^d} | w(x)(1-\Delta)^{\frac{{\alpha}_2}{2}p_2} f^{p_2}(x)|^{\frac{1-\theta}{p_2}p} dx \end{equation} It holds, \begin{equation} \|f\|_{L^\alpha_p(\mathbb{R}^d,w^{1/p})} \leq C\,\|f\|^\theta_{L^{\alpha_1}_{p_1}(\mathbb{R}^d,w^{1/p_1})}\, \|f\|^{1-\theta}_{L^{\alpha_2}_{p_2}(\mathbb{R}^d,w^{1/p_2})}\,. \end{equation} It seems ok as far as $\alpha$ is integer but for non-integer values I think there might be a problem. Any idea is welcome and thank you in advance.