Relation between Dirac function and inverse fourier transform of 1

Question

According to my notes, it holds that $\delta=(2 \pi)^{-n} \widehat{1}$.

How do we get the equality?

We have that $\delta=\frac{1}{(2 \pi)^n} \int_{\mathbb{R}^n} \widehat{\delta}(\xi) e^{i x \xi} d{\xi}= \frac{1}{(2 \pi)^n} \int_{\mathbb{R}^n} e^{i x \xi} d{\xi}$.

But $\widehat{1}=\int_{\mathbb{R}^n} e^{-i x \xi} d{x}$.

So how do we deduce that $\delta=(2 \pi)^{-n} \widehat{1}$?

Answer 1

By definition we have

$\displaystyle \langle \varphi, \widehat{\delta} \rangle= \langle \widehat{\varphi}, \delta \rangle = \widehat{\varphi}(0)=(2\pi)^{-n/2} \int_{\mathbb{R}^n} e^{-i x \cdot 0} \varphi(x)dx = (2\pi)^{-n/2} \langle \varphi , 1 \rangle$

by arbitrariness of $\varphi$ get the identity. Similary

$\displaystyle \langle \varphi, \widehat{1} \rangle = \langle \widehat{\varphi} , 1 \rangle = (2\pi)^{-n/2} \int_{\mathbb{R}^n} e^{i 0 \cdot \xi} \widehat{\varphi}(\xi) d \xi = (2\pi)^{n/2} \mathcal{F}^{-1} \mathcal{F} \varphi(0)=(2\pi)^{n/2} \varphi(0)=(2\pi)^{n/2} \langle \varphi, \delta \rangle$

Note that you can also apply the inversion formula in the first identity. In this case there are the following definitions:

$\displaystyle \widehat{\varphi}(\xi):=(2\pi)^{-n/2} \int_{\mathbb{R}^n} e^{-i x \cdot \xi} \varphi(x)dx$

$\displaystyle \mathcal{F}^{-1}\varphi(\xi):=(2\pi)^{-n/2} \int_{\mathbb{R}^n} e^{i x \cdot \xi} \varphi(x)dx$

Answer 2

Using that $\mathcal{F}\circ\mathcal{F}(f)(x)=f(-x)$ and in the comments you mentioned that you know that $\hat\delta =1$, we get $$ \hat 1 = \hat{\hat \delta} =\delta(-\cdot) =\delta $$ with some normalizing coefficient depending on your Fourier transform.