Continuity of Fourier Transform between Schwartz Space

Question

I'm trying to show for $$f \in \mathcal{S}(\mathbb{R}^{n}) \Rightarrow \hat{f} \in \mathcal{S}(\mathbb{R}^{n}) $$ where $$ \mathcal{S}$$ denotes the Schwartz space.

I've obtained the bound $$||\xi^{\alpha}\partial^{\beta}\hat{f}(\xi)||_{L^{\infty}} \leq C||f||_{L^{\infty}}||\frac{x^{\beta}}{(1+|x|)^{N}}||_{L^{1}} $$

However I don't know how to determine when $$||\frac{x^{\beta}}{(1+|x|)^{N}}||_{L^{1}}$$ is finite.

Answer 1

\begin{align*} x^{\alpha}(\partial^{\beta}\widehat{f})(x)&=x^{\alpha}((-2\pi i\xi)^{\beta}f(\xi))^{\wedge}(x)\\ &=(-2\pi i)^{|\beta|}x^{\alpha}(\xi^{\beta}f(\xi))^{\wedge}(x)\\ &=\dfrac{(-2\pi i)^{|\beta|}}{(2\pi i)^{|\alpha|}}(2\pi ix)^{\alpha}(\xi^{\beta}f(\xi))^{\wedge}(x)\\ &=\dfrac{(-2\pi i)^{|\beta|}}{(2\pi i)^{|\alpha|}}(\partial^{\alpha}(\xi^{\beta}f(\xi)))^{\wedge}(x), \end{align*} so \begin{align*} \|x^{\alpha}(\partial^{\beta}\widehat{f})(x)\|_{L^{\infty}({\bf{R}}^{n})}&=\dfrac{(2\pi)^{|\beta|}}{(2\pi)^{|\alpha|}}\|(\partial^{\alpha}(\xi^{\beta}f(\xi)))^{\wedge}\|_{L^{\infty}({\bf{R}}^{n})}\\ &\leq\dfrac{(2\pi)^{|\beta|}}{(2\pi)^{|\alpha|}}\|\partial^{\alpha}(\xi^{\beta}f(\xi))\|_{L^{1}({\bf{R}}^{n})}\\ &<\infty. \end{align*}