Let $U$ be an holomorphic function on $\mathbb{C}$ such that, there exist $m \in \mathbb{N}, C_m >0$ such that :
$$|U(z)| \leq C_m(1+|z|)^m e^{R|Imz|} $$ for all $z \in \mathbb{C}$ Prove that the restriction of $U$ on $\mathbb{R}$ define an element of $S'(\mathbb{R})$ and that there exist $u \in S'(\mathbb{R}) $ such that $Fu=U_{|_{\mathbb{R}}}$ where $F$ is the fourier transform. Then compare $supp(u)$ and $B_f(0,R)$ ( the f means closed in topology )
$|U(x)|\le C_m (1+|x|)^m$ so it is a tempered distribution.
Then for $a \in \Bbb{R}^*, b_k\in \Bbb{C}^m-0$ consider things of the form $$f_{a,b}(z)=\sum_{k=0}^m b_k U(z+ka)$$
We get that $g(z)=\frac{f_{a,b}(z)}{z^{m+2}}$ is entire and $O(e^{R|\Im(z)|} /(1+|z|)^2)$.
This implies that for any $\omega> R$ $$\hat{g}(\omega)=\int_{-\infty}^\infty g(x) e^{-i\omega x}dx=\lim_{A\to -\infty} \int_{-\infty}^\infty g(x+iA) e^{-i\omega (x+iA)}dx=0$$ and similarly for $\omega < -R,A\to +\infty$.
Whence $\hat{g}$ is supported on $[-R,R]$, so does its $m+2$th distributional derivative so that the tempered distribution $\widehat{f_{a,b}} =\hat{U} \sum_{k=0}^m b_k e^{i ak\omega}$ is supported on $[-R,R]$ as well.
It remains to show that we can choose a second $A,B_k$ as before and such that $\sum_{k=0}^m b_k e^{i ak\omega},\sum_{k=0}^m B_k e^{i A k\omega}$ don't have any common zero to get that $\hat{U}$ is supported on $[-R,R]$.