Let $u$ be a Schwartz function on $\mathbb{R}^n$ and for $a\in \mathbb{C}$, let $$f_a(x) = |x|^a u(x).$$ Show that if $\text{Re }a > -\frac{n}{2}$ and $s\in[0, \text{Re }a+\frac{n}{2}),$ then $f_a \in H^s(\mathbb{R}^n)$
I am not sure how to use the condtion for $s$ here. Could you guys take a look.
Recall that $f_a\in H^s(\mathbb{R}^n) $ if and only if $(1+|\xi|^2)^{s/2}\hat{f_a}(\xi) \in L^2(\mathbb{R}^n)$, then we would like to show $$\int_{\mathbb{R}^n} (1+|\xi|^2)^s \left(\int_{\mathbb{R}^n} |x|^a u(x) e^{-i\xi\cdot x} dx\right)^2 d\xi < \infty.\quad\quad (\star)$$ By writing $|u(x)|$ as $\sqrt {|u(x)|}\sqrt{|u(x)}$ and apply Cauchy-Schwarz inequality, we have $$(\star) \leq \left(\int (1+|\xi|^2)^s d\xi\right)\left(\int |x|^{2\text{Re }a} |u(x)|dx\right)\left( \int |u(x)| dx \right) $$ Now, since $2\text{Re }a > -n$, $|x|^{2\text{Re }a}$ is integrable around zero since $$\int_{B(0,1)} |x|^{2\text{Re }a}dx = \int_0^1 \int_{\partial B(0,r)} |x|^{2\text{Re }a}dS dr = c \int_0^1 r^{2\text{Re }a +n -1} dx$$ thus$\int |x|^{2\text{Re }a} |u(x)|dx< \infty$. Third integral is also finite since $u$ is a Schwartz function.