# A $L^1$-bounded sequence from a $H^m$-bounded sequence

I am trying to show the following: for any $m > 0$ and $\alpha \in \mathbb{N}^n$, assume $(f_j)$ is a sequence of functions which is bounded in $H^m(\mathbb{R}^n).$ Assume moreover that all the $f_j$ are supported in the open unit ball $B$ centered at the origin. Show that the sequence $(x^\alpha f_j)$ is bounded in $L^1(\mathbb{R}^n).$

What I tried: By Hölder's inequality, we have $$\Vert x^af_j \Vert_{L^1(\mathbb{R}^n)} = \Vert x^af_j \Vert_{L^1(B)} \leq \Vert x^a \Vert_{L^2(B)} \Vert f_j \Vert_{L^2(B)} = C \Vert f_j \Vert_{L^2(B)}.$$

At least under the additional assumption that $m \in \mathbb{Z_+},$ looking at the norm on $H^m$ it seems fair to say that $\Vert f_j \Vert_{L^2(B)} \leq \Vert f_j \Vert_{H^m(B)},$ so I would assume it holds in general. However, I'm not quite sure how to see it. We have

$$\Vert f_j \Vert^2_{H^m(B)} = (2\pi)^{-n} \int (1+\vert \xi \vert^2)^m \vert \hat{f}(\xi) \vert^2 d\xi;$$

is this supposed to be greater than $\int \vert f(x) \vert^2 dx$? Is the rest of the argument somewhat solid?