Let $\Omega \subseteq \mathbb{R}^n$ be open. For any compactly supported distribution $u \in \mathcal{E}'(\Omega)$, the distributional Fourier transform $\hat{u}$ is in fact a $C^\infty$ function on $\mathbb{R}^n$ with formula given by $$\hat{u}(\xi) = \langle u(x), \chi(x) e^{i x \cdot \xi} \rangle, $$
where $\chi$ is any element of $C_0^\infty(\mathbb{R}^n)$ such that $\chi \equiv 1$ on the support of $u$.
Suppose we have a sequence $\{u_j\}_{j = 1}^\infty \subseteq \mathcal{E}'(\Omega)$ along with $u \in \mathcal{E}'(\Omega)$ such that the supports of the $u_j$ and of $u$ are contained in a fixed compact set $K \subseteq \Omega$. Moreover, suppose that $$ \langle u_j, \varphi \rangle \to \langle u, \varphi \rangle \qquad \text{as } j \to \infty $$ for all $\varphi$ in the Schwartz Class $\mathcal{S}(\mathbb{R}^n)$.
I would like to know, given any multi-index $\alpha$ and any compact $K' \subseteq \mathbb{R}^n$, must it be the case that the $D^\alpha_\xi \hat{u}_j$ converges uniformly to $D^\alpha_\xi \hat{u}_j$ on $K'$?
Note: it can be shown that $D_\xi^\alpha \hat{u}(\xi) = \langle u, (ix)^\alpha \chi(x) e^{ix\cdot \xi} \rangle$ for all multi-indices $\alpha$.
Edit: Here is my incomplete attempt to show that the statement is true. I guess that they best way to go is to try and use the uniform boundedness principle, but I may be wrong.
Every element of $\mathcal{E}'(\Omega)$ is a continuous linear map from the Fréchet space $C^\infty(\Omega)$ to $\mathbb{C}$. Note that the family of seminorms $\rho_m$, $m \in \mathbb{N}$, that defines the Fréchet space topology on $C^\infty(\Omega)$ is
$$\rho_m(\varphi) = \sup \{|\partial^\beta \varphi(x)| : x \in K_m, \text{ } \beta \le m \}, \qquad \varphi \in C^\infty(\Omega), $$ where $\{K_m\}_{m= 1}^\infty$ is a compact exhaustion of $\Omega$.
The convergence criterion supposed above guarantees that the continuous linear operators $T_j \equiv u_j-u : C^\infty(\Omega) \to \mathbb{C}$ have the property
$$\sup_j |T_j \varphi| < \infty, \qquad \text{for all $\varphi \in C^\infty(\Omega)$.}$$
Therefore, the uniform boundedness principle for Fréchet spaces implies that the family $T_j$ is equicontinuous. This means that for every $\varepsilon > 0$ there exists $\delta_\varepsilon > 0$ and an $m_\varepsilon \in \mathbb{N}$ so that
$$\rho_{m_\varepsilon}(\varphi) < \delta_\varepsilon \implies |T_j(\varphi)| < \varepsilon, \qquad \text{for all $j$.}$$
So, for given $\varepsilon > 0$, we want to show that there exists $J \in \mathbb{N}$ so that
$$|D^\alpha_\xi (\hat{u}_j - u)(\xi) | = |T_j((ix)^\alpha \chi(x) e^{i x \cdot \xi})| < \varepsilon \qquad \text{all } \xi \in K' \text{ and all } j \ge J. $$
At this stage, my intuition is that the convergence in $\mathcal{E}'(\Omega)$ which I suppose is not strong enough to guarantee the convergence in $C^\infty(\mathbb{R}^n)$ I am seeking, because I do not have an effective way to make $(ix)^\alpha \chi(x) e^{i x \cdot \xi}$ small (uniformly in $\xi \in K'$) in the $\rho_\varepsilon$ seminorm.
Although I have yet to come up a with a counterexample.
Hints or solutions are greatly appreciated!
