Show that $\hat{\delta_h} = e^{-2\pi i h \cdot \xi}$

Question

Show that $\hat{\delta_h} = e^{-2\pi i h \cdot \xi}$, where $\delta_h(\phi) = \phi(h)$

By definition, we know that $\hat{\delta_h}(\phi)=\delta_h(\hat{\phi})=\hat{\phi}(h).$

but $\hat{\phi}(h) = \frac{1}{(2\pi)^{N/2}}\int_{\mathbb{R}^N}\phi(x)e^{-ix\cdot h}dx.$ This is really all I got; I'm not sure how this is supposed to be equal to the desired expression.

Any thoughts would be greatly appreciated.

Thanks in advance.

Answer 1

There are two points to be noted. They are using a slightly different definition of FT: $\hat f (h)=\int f(\xi)e^{-2\pi ih.\xi} d\xi$. Once you realize this you can use the following: If $g$ is alocally integrable fuction then we an think of $g$ as a distribution which acts on a test fucnction $\phi$ to give the value $\int g(\xi) \phi (\xi) d\xi$. I hope this clear up your confusions.