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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 : $\limsup_{x \to \pm \infty} \frac{-\ln |\hat{f} (x)|}{|x|^{1-\varepsilon}} = 1$.

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Nonconstant real analytic functions have isolated zeroes, as do polynomials. Compare $e^{-1/x^2} \sin (1/x)$ –  Will Jagy Feb 2 '12 at 21:44

1 Answer 1

up vote 3 down vote accepted

You may be interested in the Paley-Wiener Theorem and its various generalisations. A good reference is Hormander's Analysis of Linear Partial Differential Operators.

Among the various results are:

Theorem Let $f\in L^2(\mathbb{R})$. Then $e^{b|x|}f\in L^2$ for all $0 < a < b$ iff $\hat{f}$ has analytic continuation to the strip $|\Im(z)| < b$ in $\mathbb{C}$ and that for all $0 < a < b$, $$ \sup_{|\eta| < a} \| \hat{f}(\cdot + i\eta)\|_{L^2(\mathbb{R})} < \infty$$

So in particular, you have that an exponential type decay $~e^{-C|x|}$ is sufficient to guarantee that the Fourier transform is real analytic.

This theorem also implies that if there were an analytic function whose Fourier transform decays not better than $\exp(-|x|^{1-\epsilon})$, it cannot be analytically extended to any strip. That is, the radius of convergence of its Taylor expansion around basepoint $x_0$ must go to $0$ as $|x_0|\to \infty$.

Another good reference is Krantz and Parks, A Primer of Real Analytic Functions, chapter 5.

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The sharpest criterion for purely local real analyticity is via the FBI transform, with is a spatially-localised, nonlinear version of the Fourier transform. –  Willie Wong Feb 2 '12 at 22:49
(a bit late) thank you for your answer. Even if it does not solve exactly my problem, I feel like it is "better founded" (i.e. it answer what would be a "good" version of my question), and it made me discover some nice little pieces of maths :) –  D. Thomine Feb 12 '12 at 0:25

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