Fourier transform of delta constants plane waves and polynomials

Evaluate the transform of ${\delta }_a$ on an arbitrary $\phi \in \mathcal{S}$.

Apply $\mathcal{F}$ to $\mathcal{F}{\delta }_0=1$. Since $R{\delta }_0={\delta }_0$, [F3] gives $\mathcal{F}1={\delta }_0$. Similarly step 1.1 with $a=-b$ gives $\mathcal{F}{\delta }_{-b}=e^{2\pi ib\cdot x}$, so applying $\mathcal{F}$ once more gives $\mathcal{F}(e^{2\pi ib\cdot x})=R{\delta }_{-b}={\delta }_b$.

Apply the derivative identity in [F4] to ${\delta }_0$ and use $\mathcal{F}{\delta }_0=1$ to obtain $\mathcal{F}({\mathrm{\partial }}^{\alpha }{\delta }_0)=(2\pi i\xi {)}^{\alpha }$. Apply the multiplication identity to the constant distribution and use $\mathcal{F}1={\delta }_0$ to obtain the formula for $x^{\alpha }$.

Every polynomial is a finite complex linear combination of monomials, so linearity completes the polynomial assertion. The zero multi-index recovers $\mathcal{F}{\delta }_0=1$ and $\mathcal{F}1={\delta }_0$. Countable Choice is used only through the published Fourier suppliers.