Let $f \in L^1(\bf R)$ and assume that $\int \int_{\bf R^2}\vert f(x) \hat{f}(y) \vert e^{\vert xy \vert}dxdy< \infty$ where $\hat f$ denotes Fourier transform of $f$. Then $f=0$
Proof: Set $\hat{M}(y)=\int_ {\bf R} \vert f(x) \vert e^{\vert xy \vert}dx$ and $M(x)= \int_{\bf R} \vert \hat{f} (y) e^{\vert xy \vert}dy,$ $\hat{M}(y)$ and $M(x)$ are increasing functions of $\vert y \vert$ and $\vert x \vert$ respectively,and $$ \int_{\bf R} \vert\hat{f}(y)\vert \hat {M}(y) dy=\int_{\bf R}\vert f(x)\vert M(x) dx=\int \int_{\bf R^2}\vert f(x) \hat{f}(y) \vert e^{\vert xy \vert}dxdy < \infty $$
Assume at first that $f$ has compact support.Then $\hat{M}$ must grow exponentially if $f \neq 0$.Hence $f$ is analytic in a band containing the real axis and $f$ must be zero after all.
Can someone please explain me the last paragraph?
By dominated convergence, if $g\in L^1(\mathbb R)$, then $$ \lim_{y\to\infty} e^{-ay}\int_{-a}^a |g(x)|e^{|xy|}\, dx = 0. \quad\quad\quad\quad (1) $$ If now $f\in L^1(\mathbb R)$ is supported by $[-N,N]$ and $f\not\equiv 0$, then pick $0<a<N$ such that $\int_{a<|x|<N} |f|>0$. Since $e^{|xy|}\ge e^{a|y|}$ for these $x$, it now follows from (1) that $\widehat{M}(y)\gtrsim e^{ay}$, as claimed.
Next (and still following the outline you gave), $$ \int |\widehat{f}(y)| e^{ay}\, dy < \infty $$ implies that $f$ has a holomorphic continuation $f(z)$ to $|\textrm{Im}\, z |<a$. We obtain this continuation by simply using the Fourier inversion formula $$ f(z) = \int \widehat{f}(y) e^{izy}\, dy $$ for all such $z$.