Fourier transform acts continuously on Schwartz space

The interval FTC for $e^{it}$ gives $\mathrm{\mid }{e}^{it}-1\mathrm{\mid }\le \mathrm{\mid }t\mathrm{\mid }$. Thus the difference quotient in frequency coordinate $j$ is dominated in absolute value by $2\pi \mathrm{\mid }{x}_{j}f(x)\mathrm{\mid }$, integrable by [F2]. By [F4] its limit is $\mathcal{F}(-2\pi i{x}_{j}f)$. The convergence for arbitrary real increments follows either directly from the dominated estimate by truncating the majorant to a finite box and using uniform convergence there, or by the sequential criterion under the stated countable choice. The derivative is continuous by [F3]. Repeating for each weighted function, which remains Schwartz by [F1], proves all ordered derivatives and the second formula.

Fix the other coordinates and integrate in coordinate $j$. For $u=f$ restricted to this line and $v=e^{-2\pi ix\cdot \xi }$, both $u^{\mathrm{\prime }}v$ and $u{v}^{\mathrm{\prime }}$ are integrable on the line: multiply $u,u^{\mathrm{\prime }}$ by $1+x_j^2$ and use their bounded Schwartz seminorms. Also $uv\to 0$ at both ends, since $x_{j}u$ is bounded. [F6] therefore gives the derivative identity in that coordinate. Integrating over the other coordinates is legitimate by [F2] and [F5], since the full integrals of $\mathrm{\mid }{\mathrm{\partial }}_{j}f\mathrm{\mid }$ and $\mathrm{\mid }f\mathrm{\mid }$ are finite. Iteration using [F1] proves the first formula, including zero components of $\xi$ without division by them.

Combine the two identities to obtain ${\xi }^{\alpha }{\mathrm{\partial }}^{\beta }\widehat{f}=(2\pi i{)}^{-\mathrm{\mid }\alpha \mathrm{\mid }}\mathcal{F}({\mathrm{\partial }}^{\alpha }((-2\pi ix{)}^{\beta }f))\mathrm{.}$ By [F3], its supremum is at most $(2\pi {)}^{-\mathrm{\mid }\alpha \mathrm{\mid }}\mathrm{\parallel }{\mathrm{\partial }}^{\alpha }((-2\pi ix{)}^{\beta }f){\mathrm{\parallel }}_1$. By [F2] and the explicit operation bounds in [F1], this is a finite linear combination of input Schwartz seminorms. Thus every output seminorm is finite, and the finite-neighbourhood definition proves continuity.