My advisor made the comment that if $u\in \mathcal{E}'$ is a compactly supported distribution, then $\hat{u}(\xi)\in C^{\infty}(\mathbb{R}^n)$ is *actually* a smooth function (not merely a distribution or generalized function).

I'm trying to prove this, but I'm realizing I'm not even sure how we should properly define the F.T. of a distribution in $\mathcal{E}'$. When $u\in \mathcal{S}'$, then the standard definition $$\left<\hat{u},\phi \right> := \left<u,\hat{\phi} \right>$$ makes perfect sense, since $\phi\in\mathcal{S}\Rightarrow \hat{\phi}\in\mathcal{S}$, and so pairing $\hat{\phi}$ with an element of $\mathcal{S}'$ is well defined.

The problem I'm having with applying this same definition to $u\in\mathcal{E}'$ is that while $\phi\in C^{\infty}$, we don't necessarily have $\hat{\phi}\in C^{\infty}$; and so the statement $\left<u,\hat{\phi} \right>$ might not make sense, as $u$ should only be acting on test functions from $C^{\infty}$.

I've been toying around with the idea of a smooth cutoff function $\rho \equiv 1$ one on $\text{supp}(u)$, and taking $$\left<\hat{u},\phi \right> := \left<u,\widehat{\rho\phi} \right>,$$ to ensure that $\widehat{\rho\phi}$ is at least well-defined. But I'm not sure if that is even close to the right idea, though. So some clarification on this matter would be much appreciated.

That aside, considering that $e^{-ix\cdot\xi}\in C^{\infty}(\mathbb{R}^n_x)$, I'm at least able to show that $\left<u,e^{-ix\cdot\xi} \right>$ is a function of $\xi$ (and not a generalized function). This of course heuristically matches up with the idea that $$\left<u,e^{-ix\cdot\xi} \right> = \int u(x)e^{-ix\cdot\xi}\;dx = \hat{u}(\xi),$$ if $u(x)$ had in fact been a function. Since $\exists\{u_k(x)\}\subset C^{\infty}_0(\mathbb{R}^{n})$ that converge weakly to $u\in\mathcal{E}'$, then we have

\begin{align*} \mathbb{C} \ni \left<u,\phi \right> &= \lim_{k\to\infty}\left<u_k,\phi \right>\\ &= \lim_{k\to\infty}\int u_k(x)e^{-ix\cdot\xi}\;dx\\ &= \lim_{k\to\infty}\hat{u_k}(\xi)\\ &= U(\xi), \end{align*} some function of $\xi$. But when it comes to showing that $U = \hat{u}$ as distributions, we would need to prove that $$\left<U(\xi),\psi(\xi) \right> = \int U(\xi)\psi(\xi) \;d\xi= \left<u,\hat{\psi} \right>,$$ for all $\psi$ in some appropriate space of test functions, which leads to the same issues with $\hat{\psi}$ I was having before.

I also haven't been able to figure out how I would prove that $U(\xi)$ is smooth. So any help with that would also be welcomed.