Fourier transform of analytic functions and radius of convergence

Question

Let $\phi$ be a fonction in the Schwartz space $\mathscr S(\mathbb R)$ which is also analytic on the real line and such that the radius of convergence on the real line is bounded below by $\rho>0$. Then, with $\hat \phi(\xi)=\int e^{-2iπ x\xi}\phi(x) dx$, do we have $$ \vert\hat \phi(\xi)\vert\le C e^{-2π\rho \vert\xi\vert}\quad? $$ Is it true for $\phi$ real analytic on $\mathbb R$ which happens to be a temperate distribution? Is there a converse?

Answer 1

You need that $\phi$ is analytic on $|\Im(z)|\le r$,

$L^1$ on horizontal lines and $\lim_{|x|\to \infty}\sup_{|y|\le r} |\phi(x+iy)|=0$ to obtain from the Cauchy integral theorem that $$\int_{-\infty}^\infty e^{2i\pi (x+iy)\xi} \phi(x+iy)dx$$ converges and doesn't depend on $|y|\le r$ so that $$|\hat{\phi}(\xi)| \le (\|\phi(.+ir)\|_{L^1}+\|\phi(.-ir)\|_{L^1}) e^{-2\pi r |\xi|}$$

Conversely if both $\hat{\phi},\hat{\phi}'$ are $O(e^{-2\pi r|\xi|})$ then $\phi$ satisfies the above conditions for $|\Im(z)|\le r-\epsilon$.

Try with $\phi(x)=e^{-e^{x^2}}$ which is unbounded on the horizontal lines others than the real axis, if $\hat{\phi}$ was $O(e^{-4\pi |\xi|})$ we'd have that $\|\phi(.+i)\|_{\infty}\le \|\hat{\phi}e^{2\pi \xi}\|_{L^1}<\infty$.