Suppose $f\colon\mathbb R^d\rightarrow\mathbb R$ is Schwarz class and $\int f\,dx=0$. Let $\nu>0$ and let $P_\nu$ be the projection operator such that $\widehat{P_{\nu}f}(k)=\mathbb{1}_{\{|k|\leq \nu\}}\widehat f(k)$ on the Fourier side.
I am trying to prove that, for all $p>1$, $$\|\nabla_x(-\Delta_x)^{-1}P_{\nu}f\|_{L^p}\leq C\|\langle x\rangle f\|_{L^1},$$ for some constant $C>0$ depending on $\nu$ and $p$, where $\langle x\rangle:=\sqrt{1+|x|^2}$.
Start of proof? On the Fourier side, $$|F[\nabla_x(-\Delta_x)^{-1}P_{\nu}f]|\leq \Big|\frac{k}{|k|^2}\mathbb{1}_{\{|k|\leq \nu\}}\widehat f(k)\Big|\leq\Big|\frac{1}{|k|}\mathbb{1}_{\{|k|\leq \nu\}}\big(\widehat f(k)-\widehat f(0)\big)\Big|$$since $\widehat f(0)=0$ by assumption.
Then for $\nu$ sufficiently small, $\widehat f(k)-\widehat f(0)\approx \nabla_k\widehat f(k)\cdot k.$
So $$|F[\nabla_x(-\Delta_x)^{-1}P_{\nu}f]|\lesssim \big|\nabla_k\widehat f(k)\big|=\big|F[xf](k)\big|.$$