Fourier Transform of a compactly supported distribution : how to write the definition in a "better way"?

Question

I am really troubled by the following Theorem :

*Theorem : If $u\in \mathcal{E}'(\mathbb R^d)$ (space of compactly supported distributions), then its Fourier transform is a function, and : $\hat{u}(\xi)=\langle u,e^{-ix\cdot\xi}\rangle$

For a distribution of $u\in \mathcal{S}'(\mathbb R),$ we define the Fourier transform of $u$ as a distribution, defined for all $\phi \in \mathcal{S}(\mathbb R^d)$ by $\langle\hat u,\phi\rangle=\langle u, \hat\phi \rangle$

Moreover $\mathcal{E}'(\mathbb R^d) \subset \mathcal{S}'(\mathbb R^d)$.

So.. If I follow the definition that is given to me I'm supposed to have, for $u \in \mathcal{E}'(\mathbb R^d),$ $\hat{u}$ as a distribution.

Edit : Evaluating $\hat{u}$ in a real or complex number then doesn't make any sense to me. A distribution of this type has a Schwartz function as argument , and not a complex number to my opinion. So I don't understand how can the symbolism "$\hat{u}(\xi)"$ have a sense in the theorem ? It can also simply be a misunderstanding on my part of the rôle of $\hat{u}$,

A bit stuck with this. Also, if it's just a notation problem, I'd also be interested on how you can easily understand/read this theorem, as this makes a nonsense to me.

Answer 1

The distributional Fourier transform of $u \in \mathcal{E}'(\mathbb{R}^d)$ is by definition a distribution. But there is a function $f \in L^1_{\text{loc}}(\mathbb{R}^d)$ such that $\hat{u}$ is given by integration against $f,$ i.e. $$\langle \hat{u}, \varphi \rangle = \int f(\xi)\,\varphi(\xi)\,d\xi.$$ What the theorem says is that $f(\xi)=\langle u(x), e^{-ix\cdot\xi} \rangle.$

Answer 2

Another way of saying it : (To be confirmed)

Meaning of the theorem : There exists a $L_1^\text{Loc}(R^d)$ function , such as $\hat{u}$=$T_f$ and :

$$\forall \varphi \in D(R^d),\langle\hat{u},\varphi\rangle=\langle T_f, \varphi \rangle, \text{ where } f:\xi \rightarrow \langle u,e^{-ix\cdot\xi}\rangle$$

So for example , if $u$ is a compactly supported distribution , Let's say $u=T_g$ then :

\begin{align} & \forall \varphi \in D(R^d),\,\langle\hat{u},\varphi\rangle=\langle T_f,\varphi\rangle \\[6pt] = {} & \int \langle u,e^{-ix\cdot\xi}\rangle \varphi(\xi) \, d\xi \\[6pt] = {} & \int \varphi(\xi)\int g(x) e^{-ix\cdot\xi} \, dx \,d\xi \end{align}