Polynomial growth of Fourier transforms

Question

I am looking for a theorem that guarantees the polynomial growth of a function $f$ defined by a Fourier integral, that is, when $$f(x)=\int_{-\infty}^{\infty}F(y)e^{ixy}dy.$$ I am only interested in one-sided growth, say $x\rightarrow +\infty$. Further, in the case I have in mind, $f$ is almost everywhere continuous, and everywhere defined.

It seems to me that all such functions should be of polynomial growth, but I can't see how to prove it (beyond the cases when the integral is absolutely convergent, which is too strong for my purposes).

EDIT. In light of the answers given here, it is apparent that I have not defined the function space clearly. Unfortunately, I do not know of a definition for these so I can only state the properties which I may assume, and which seem relavent.

Firstly, $F$ is analytic and not $L^1$. $L^p$ results for larger $p$ are definitely of interest to me.

Secondly, by "everywhere defined", I mean that $f$ has no spikes, but continuity is too strong- I want to allow $f$ to have jumps for arbitrarily large values of $x$.

Further, the improper integral should be understood as $\lim_{R\rightarrow\infty}\int_{-R}^{R}$.

Answer 1

Let me quote the Paley-Wiener-Schwartz theorem. Let $F$ be a tempered distribution on $\mathbb R^n$. Then the two following properties are equivalent.

(i) $F$ is compactly supported with $\text{supp} F\subset\{x\in \mathbb R^n,\vert x\vert\le R\}$

(ii) $\hat F$ is an entire function such that there exists $C\ge 0, N\ge 0$ with $$\forall \zeta \in \mathbb C^n,\quad \vert\hat F(\zeta)\vert\le C(1+\vert\zeta\vert)^N e^{R\vert\Im \zeta\vert} $$

As a consequence, with your notations $f$ will have a polynomial growth on the real line when it is compactly supported, whatever is its regularity.

Answer 2

Edited to more accurately address the OPs (restated) concerns:

It appears that one cannot hope for a general result for $L^p$, analytic (even entire) $F$ if $p \geq 2$. Consider, for example $$ f = \sum_{n = 1}^\infty n^n \chi_{[n, n + e^{-n^2}]}. $$ (By mollifying the cutoffs one could also take $f \in C^\infty$, but let me not worry about that for now.) We have $$ ||f||_{L^p}^p = \sum_{n = 1}^\infty n^{pn}e^{-n^2} < \infty $$ for all $1 \leq p < \infty$ (but $f \notin L^\infty$).

For $\xi \in \mathbb{C}$, define $$ F(\xi) = \int_{\mathbb{R}}f(y)e^{-i\xi y}dy. $$ From the definition of $f$, one immediately sees this is well-defined (the integral is absolutely convergent), and the usual differentiation under the integral sign (again, readily justified from the explicit form of $f$) shows that $F$ is an entire function. By the Hausdorff-Young inequality, $F \in L^p(\mathbb{R})$ for all $2 \leq p \leq \infty$, though we can also say with certitude that $F \notin L^1(\mathbb{R})$, since its Fourier transform $f$ is not $L^\infty$. By Carleson's theorem, $f(x) = \lim_{R \to \infty} \int_{-R}^R F(y)e^{ixy}dy$ for $x$-a.e., and of course $f$ is not of polynomial growth.