Fourier transform of tempered distribution $x^{\alpha}$

Question

Let $\alpha \in \mathbb N_{0}^n$ and $\delta$ the dirac delta function. Using the properties of tempered distribution, I showed that $\langle\widehat{\partial^{\alpha} \delta}, \varphi\rangle $ $= $ $ \langle(2\pi)^{-n/2}(i\xi)^{\alpha}, \varphi\rangle$. I am supposed to follow from this that $\widehat{x^{\alpha}} = (2\pi)^{n/2}|i|^{\alpha}\partial^{\alpha}\delta$ but I am not sure how.

Answer 1

$$ \langle\widehat{x^{\alpha}}, \psi\rangle = \langle x^{\alpha}, \widehat{\psi}\rangle =\frac{1}{(2\pi)^{-n/2}(-i)^{\alpha}} \langle(2\pi)^{-n/2}(-ix)^{\alpha}, \widehat{\psi}\rangle\\ =(2\pi)^{n/2}(i)^{\alpha} \langle(2\pi)^{-n/2}(ix)^{\alpha}, \check{\psi}\rangle=(2\pi)^{n/2}(i)^{\alpha} \langle\widehat{\partial^{\alpha} \delta}, \check{\psi}\rangle\\ =(2\pi)^{n/2}(i)^{\alpha} \langle\partial^{\alpha} \delta, \widehat{\check{\psi}}\rangle=(2\pi)^{n/2}(i)^{\alpha} \langle\partial^{\alpha} \delta, \psi\rangle$$

where as usual $$\check{\psi}(x)=\widehat{\psi}(-x)$$