Does this function grows faster than any exponential function? I need following result in understanding of a result given in a paper.Any ideas to prove this?
Consider the function $\hat {M}: \bf R \to R$ defined as $$\hat{M}(y)=\int_{\bf R}\vert f(x) \vert e^{\vert xy\vert}dx$$.

Let $f \in L^1(\bf R)$, $f$, and $\hat {f}$ does not have compact support then show that $\hat {M}$ grows faster than any exponential function. 


Lemma. Either for every $M$ one has
$$
\int_M^\infty \lvert f(x) \rvert \, dx > 0
$$
or for every $M$ one has
$$
\int_{-\infty}^M \lvert f(x) \rvert \, dx > 0.
$$
Proof. If by contradiction there exist $M_1, M_2$ such that
$$
\int_{M_1}^\infty \lvert f(x) \rvert \, dx = 0
$$
and
$$
\int_{-\infty}^{M_2} \lvert f(x) \rvert \, dx = 0,
$$
then $f$ is zero almost everywhere outside the interval $[M_2,M_1]$, which contradicts the non-compact support hypothesis.
Assume now, wlog, that the first holds. Fix $p>0$, and set $c = \int_{2p}^{\infty} \lvert f(x) \rvert \, dx > 0$; observe that
\begin{align*}
\frac{\hat{M}(y)}{e^{p\lvert y \rvert}} &= \int_{\mathbb{R}} \lvert f(x) \rvert e^{(\lvert x \rvert - p)\lvert y \rvert}\, dx \geq \int_{2p}^{\infty} \lvert f(x) \rvert e^{(x - p)\lvert y \rvert}\, dx \geq e^{p\lvert y \rvert} \int_{2p}^{\infty} \lvert f(x) \rvert \, dx = ce^{p\lvert y \rvert}
\end{align*}
which tends to $\infty$ as $y \to \pm \infty$. Since $p$ was arbitrary we are done.
