Let $\mathcal{E}'(\mathbb{R})$ be the space of all compactly supported distributions on $\mathbb{R}$. 


For $f\in \mathcal{E}'(\mathbb{R})$, let $\widehat{f}$ denote the entire extension of the Fourier transform of $f$. 


**Question:** If $f\_n\stackrel{n\rightarrow\infty}{\longrightarrow}f$ in $\mathcal{E}'(\mathbb{R})$, then does $\widehat{f}\_n$ converge to $\widehat{f}$ uniformly on compact subsets of $\mathbb{C}$ as $n\rightarrow\infty$? 



On the Fourier-Laplace transform of compactly supported distributions
Let me expand Nate Eldredge's comments. A more elementary proof than the one provided by Jochen Wengenroth, still requiring no computations, is as follows:


$\mathcal{E}'$ is the dual space of $C^\infty$, which is endowed with the Fréchet space topology given e.g. by the seminorms $\|f\|\_k:=\|f\|\_{C^k(B\_k(0))}$. The fact that $f\_n\to f$ in $\mathcal{E}'$ means that $\langle f\_n,\phi\rangle\to\langle f,\phi\rangle$ for any $\phi\in C^\infty$. 


Now the uniform boundedness principle holds also for linear maps from a Fréchet space to (say) a normed vector space. The proof is the same as that for the Banach space version; for details, see Theorem 2.6 in Rudin's Functional Analysis (2nd edition). Hence the $f\_n$'s are equicontinuous, i.e. there exists $k\in\mathbb{N}$ and $C>0$ such that
$$ |\langle f\_n,\phi\rangle|\le C\|\phi\|\_{C^k(B\_k(0))}. $$
This gives $|\widehat{f\_n}(\xi)|\le C\|e^{-2\pi i\langle\xi,\cdot\rangle}\|\_{C^k(B\_k(0))}\le C'(1+|\xi|^k)e^{2\pi k|\xi|}$ for $\xi\in\mathbb{C}$. Also, $\widehat{f\_n}(\xi)\to\widehat{f}(\xi)$ for every $\xi\in\mathbb{C}$ (since it corresponds to evaluation of $f\_n$ on the test function $e^{-2\pi i\langle\xi,\cdot\rangle}$). So you have a locally equibounded sequence of pointwise converging holomorphic functions and this implies your claim. This clearly works in any dimension.


