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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 $\lim_{R\rightarrow\infty}\int_{-R}^{R}$.

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Could you clarify the distinction between jump discontinuities and 'spikes'? – Jens Apr 15 '13 at 2:40
Well, roughly speaking, a jump discontinuity at $x_0$ is given by $c_0H(x-x_0)$, where $H$ is Heaviside's step function, and a spike/$\delta$-function at $x_0$ is its derivative. Perhaps more formally you'd say that a jump discontinuity is defined as half the sum of the left and right limits. – Kevin Smith Apr 15 '13 at 10:54
I see. And are you assuming the limit $\lim_{R \to \infty} \int_{-R}^R F(y)e^{ixy}dy$ exists for every $x$, or just almost every $x$? – Jens Apr 15 '13 at 17:36
I assume that the limit exists for almost every $x$, and that it does not diverge for any $x$. – Kevin Smith Apr 16 '13 at 6:56
Do you have an example of such an $F$ which is not in $L^1$? It seems like a fairly stringent condition (to ask that the partial integrals be bounded for each $x$). How do you know this is satisfied in the cases you are interested in? – Jens Apr 16 '13 at 18:45

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.

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Ah yes, thank you Bazin- I remember the Paley Wiener theorems now you mention them. Unfortunately though, I require $F$ not compactly supported and $f$ not entire. I should have mentioned this. – Kevin Smith Apr 14 '13 at 11:48

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.

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Thanks for this Jens. In the case in which I am interested, I cannot assume $F$ is $L^1$ but it is analytic on the line of integration and, indeed, I need to define the integral as the limit you suggest. Moreover, $f$ is everywhere defined, ie, it has no spikes. I think perhaps I do not have a good a definition of the space of functions with which I am working here, but these are the properties I know. – Kevin Smith Apr 14 '13 at 12:02

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