Is it true that, $x\rho(x/t)\in H^{s}$ for $\rho\in \mathcal{D}(\mathbb R)$ and $s>3/2$? We put, $\mathcal{D}(\mathbb R)=$ The space of $C^{\infty}-$ functions on $\mathbb R$ with compact support; and the Sobolev space
$$H^{s}(\mathbb R)=\{f\in L^{2}(\mathbb R):[\int_{\mathbb R} |\hat{f}(\xi)|^{2}(1+|\xi|^{2})^{s}d\xi]^{1/2}<\infty \}.$$
We choose smooth positive radial function $\rho \in \mathcal{D}(\mathbb R)$ so that 
$\rho(x)=1$ if $|x|\leq 1$ and $\rho(x)=0$ for $|x|\geq\frac{3}{2}.$
We put, $\rho_{t}(x)= \rho(x/t);$ for $t>0.$

My Question is:  Can we expect, $x\rho_{t}(x) \in H^{s}(\mathbb R)$ for $s>\frac{3}{2}$ ? If yes, how to prove it ?


Yes. $x \rho_t(x)$ is smooth and compactly supported, and any such function $f$ is in $H^s$ for every $s$.  The easiest way to see this may be this:


*

*Let $k$ be any integer larger than $s$.  Verify from the definition that $H^k \subset H^s$.  (Note that $(1+|\xi|^2)^s \le (1+|\xi|^2)^k$).

*Using the fact that the Fourier transform takes differentiation to multiplication by $i\xi$, write $\|f\|_{H^k}$ in terms of $f$ and its first $k$ derivatives.

*Since all the derivatives are bounded and compactly supported, show that the resulting integrals are all finite.  Hence $f \in H^k$.
