Let's say a smooth function is a $\mathcal{C}^\infty$ function on $\mathbb{R}$. Some smooth functions are not analytic, the most notorious example being the bump functions. A non-analytic $\mathcal{C}^\infty$ function seems (formally at least) "less smooth" that an analytic function, but I am wondering how one can quantify this. Looking at derivability does not give us anything (since a $\mathcal{C}^\infty$ is, well, $\mathcal{C}^\infty$), and looking at the radius of convergence of the Taylor series does not give us anything we do not already know.
So, what if we look at the Fourier transform? This is a classical strategy: to quantify the smoothness of a function, quantify the integrability or the decay of its Fourier transform. Let's assume that we work in Scwhartz space $\mathcal{S} (\mathbb{R})$. Then:
- For any $f$ in $\mathcal{S} (\mathbb{R})$, if $\hat{f} (x) =_{\pm \infty} O (e^{-|x|^{1+\varepsilon}})$ for some $\varepsilon > 0$, then $f$ is analytic (I am not one hundred percent sure - I have not proved with utter rigour - but it seems rather safe).
- For all $\varepsilon > 0$, there exists a bump function $f$ such that $\hat{f} (x)$ has a magnitude of roughly(*) $e^{-|x|^{1-\varepsilon}}$ for large enough $|x|$ (again, I am not totally sure, but I think this is what is said in http://math.mit.edu/~stevenj/bump-saddle.pdf).
So I have two questions:
Are there analytic functions $f$ in $\mathcal{S} (\mathbb{R})$ such that $\hat{f} (x)$ has a magnitude of roughly $e^{-|x|^{1-\varepsilon}}$ for some $\varepsilon > 0$ ? If it where not the case, and provided what I said before is true, then we would know that functions whose Fourier transform is of order $e^{-|x|^{1-\varepsilon}}$ are $\mathcal{C}^\infty$ but not analytic, and that functions whose Fourier transform is of order $e^{-|x|^{1+\varepsilon}}$ are analytic.
What about intermediate growth rates, for instance when $\hat{f} (x)$ has a magnitude of roughly $e^{-C|x|}$?
Of course, any more precise criterion or any related reference is welcome.
(*) To be more precise : $\limsup_{x \to \pm \infty} \frac{-\ln |\hat{f} (x)|}{|x|^{1-\varepsilon}} = 1$.