# On the Fourier-Laplace transform of compactly supported distributions

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 $n\rightarrow\infty$?

• Initial thoughts: Pointwise convergence is for free. By the mean value property, it's sufficient to get $L^1$ convergence on compact subsets, so we might try to show there's a dominating function, or uniform integrability. And if I'm not mistaken, $\mathcal{E}(\mathbb{R})$ is a Frechet space, so we have the uniform boundedness principle available, which seems like it ought to help. – Nate Eldredge Sep 18 '18 at 0:15
• @NateEldredge How is "X-wise convergence is for free" defined? If one knows (x) that elements in a bounded set in $\mathscr E'(\mathbb R)$ have "uniformly bounded" supports and orders, then the assertion follows trivially from formula (2) $\ |f(z)|\le\gamma\,(1+|z|)^N{\rm e}^{\,r|\rm{Im}\,z|}$ in the statement of the Paley−Wiener theorem in Rudin's Functional Analysis (Th. 7.23, p. 183 in my edition). Is (x) "for free"? – TaQ Sep 19 '18 at 8:11
• @TaQ: All I meant by "pointwise convergence is for free" was the trivial observation that we have $\widehat{f_n}(x) \to \widehat{f}(x)$ for each $x$, simply because $\widehat{f_n}(x) = \langle f_n, e^{i x \cdot}\rangle$. I did not think as far ahead as you got. It'd be great if you would post it as an answer. – Nate Eldredge Sep 19 '18 at 13:16

$\mathcal{E}'$ 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.
• It seems your argument proves something stronger because you are using the weak-$\ast$ topology on $\mathcal{E}'$ instead of the strong topology as in Jochen's answer. Although, I don't remember if the two topologies are equivalent from the point of view of convergence of sequence because of more Bairology. – Abdelmalek Abdesselam Sep 19 '18 at 18:42
• They are equivalent for convergence of sequences because of compactology ($\mathcal E'$ is a Montel space, i.e., the closed bounded sets are compact, and weakly convergent sequences are bounded). – Jochen Wengenroth Sep 20 '18 at 6:50
Besides the direct approach to such continuity questions as in Nate Eldredge's comment one can try to use a closed graph argument. The closed graph theorem holds for ultrabornological domain spaces (like the space $\mathcal E'(\mathbb R)$ of compactly supported distributions endowed with the topology of uniform convergence on bounded subsets of $\mathcal E(\mathbb R)$) and webbed range spaces -- a very large class of locally convex spaces containing all Frechet spaces and thus, in particular, the space of entire functions. By Nate's remark that pointwise convergence is for free the Fourier transform $f\mapsto \hat f$ has closed graph and is thus continuous.