Here is a well known statement: Theorem Let $s>(N/2)$: then for any $\rho\in C_c^\infty(\mathbb{R}^N)$ there exists a constant $C_s$ such that \begin{equation} \|\rho \mu\|_{H^{-s}}\leq C_s|\rho \mu|(\mathbb{R}^N) \quad \forall \mu\in \mathcal{M}(\mathbb{R}^N). \end{equation} where $\mathcal{M}(\mathbb{R}^N)$ is a set of all Radon measures.
I have a question. I cannot understand exact meaning of \begin{equation} \|\rho \mu\|_{H^{-s}}\leq C_s|\rho \mu|(\mathbb{R}^N) \quad \forall \mu\in \mathcal{M}(\mathbb{R}^N). \end{equation} More precisely, I guess that this notation implies that \begin{equation} \left(\int_{\mathbb{R}^N}\frac{1}{|1+|\xi|^2|^s}\,|\hat{f}|^2(\xi)\,|\hat{\rho}|^2(\xi) \,\mathrm d\mu\right)^{\frac{1}{2}}\leq C_s\int_{\mathbb{R}^N}|f|\,|\rho| \,\mathrm d\mu. \end{equation}
No, there is no need to add an additional test function here. Since $\mu$ is a measure, it is a tempered distribution and one can take its Fourier transformin the sense of distributions, which is a bounded continuous function. The same is true for $\rho\mu$. Hence your inequality means directly \begin{equation} \left(\int_{\mathbb{R}^N}\frac{|\widehat{\rho\mu}(\xi)|^2}{(1+|\xi|^2)^s}\,\mathrm d\xi\right)^{\frac{1}{2}}\leq C_s\int_{\mathbb{R}^N}\,|\rho| \,\mathrm d|\mu|. \end{equation} In the last integral, you can remove the absolute values on $\mu$ if it is a positive measure.