We put, $M(\mathbb R)=$ The space of complex bounded Borel measure on $\mathbb R$
[With each complex Borel measure $\mu$ on $\mathbb R$ there is associated a set function $|\mu|,$ the total variation of $\mu,$ defined by, $$|\mu|(E)= \sup \sum |\mu(E_{i})|,$$ the supreme being taken over all finite collections of pairwise disjoint Borel sets $E_{i}$ whose union is $E.$ Then $|\mu|$ is also a measure on $\mathbb R.$ If $\|\mu\|= |\mu|(\mathbb R)< \infty,$ we say $\mu$ is a bounded complex Borel measure on $\mathbb R.$]
Fact. If $f\in L^{p}(\mathbb R), (1\leq p \leq \infty)$ and $\mu \in M(\mathbb R),$ then the integral $$f\ast \mu (x)= \int_{\mathbb R} f(x-y) d\mu(y)$$ exists for all most all $x,$ $f\ast \mu \in L^{p},$ and $\|f\ast \mu\|_{L^{p}}\leq \|f\|_{L^{p}} \|\mu\|.$
Let $h\in \mathcal{S}(\mathbb R)$(Schwartz space) and $\mu \in M(\mathbb R).$
My Question is: Let $s>\frac{1}{2}.$ Can we show that $ \int_{\mathbb R} |(h\ast \mu)(x)|^{2}(1+|x|^{2})^{s} dx< \infty$ ? (that is, can we expect, $h\ast \mu \in L^{2}(\mathbb R, (1+|x|^{2})^{s})$; in other words, can we expect, $h\ast \mu $ in $L^{2}$ with respect to weight $(1+|x|^{2})^{s}$ ?)
My attempt:
$ (\int_{\mathbb R} |(h\ast \mu)(x)|^{2}(1+|x|^{2})^{s} dx)^{1/2}= (\int_{\mathbb R} |\int_{\mathbb R} h(x-y) d\mu(y)|^{2}(1+|x|^{2})^{s} dx )^{1/2};$ now by Minkowski inequality for the integrals, gives us,
$$\|h\ast\mu\|_{L^{2}(\mathbb R, (1+|x|^{2})^{s})}\leq \int_{\mathbb R} (\int_{\mathbb R} |h(x-y)|^{2} (1+|x|^{2})^{2s}dx)^{1/2} d\mu(y);$$
from here, I don't know how to proceed..
Proof of the above fact. If $f$ and $\mu$ are non-negative, then $f\ast \mu (x)$ exists(possibly being equal to $\infty$) for every $x,$ and by Minkowski's inequality for integrals, $$\|f\ast \mu\|_{L^{p}}\leq \int_{\mathbb R}\|f(\cdot-y)\|_{L^{p}}d\mu(y)=\|f\|_{p}\|\mu\|.$$ In particular, $f\ast \mu(x)<\infty$ for all most all $x.$ In the general case, this argument applies to $|f|$ and $|\mu|,$ and the result follows.