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I try to identify conditions for the Fourier-transformation $\mathcal{F}(f)$ of some function $f \in L^1(\mathbb{R}^n)$ to be real-analytic. Namely I want to show that one of the following two conditions is sufficient:

  1. $\exists K \text{ compact}: \text{supp}(f) \subset K$
  2. $\exists C\in (0,\infty): |f|\leq\exp(-C|x|)$

A few notes:

Certainly, what we are looking for is a good bound on the derivatives, in particular $|D^\alpha\mathcal{F}(f)|\leq C_n/r^{|\alpha|}$ where $\alpha$ is a multi-index would be some inequality of the form I am interested in.

I tried to find some resources and found the Paley-Wiener theorems, but I could not find any resource that shows what I am looking for. Rudin shows in "Functional Analysis" (Theorem 7.22+) that if $f$ is smooth and has compact support $\mathcal{F}(f)$ is holomorphic and the restriction of it to the reals is real-analytic then. However, I want to find a proof that doesn't rely on complex-analysis and I would need to remove the restriction that $f$ is smooth.

I currently work with the following definition of the Fourier-transformation:

$\mathcal{F}(f)(t) = \int_{\mathbb{R}^n} f(x)\, e^{-2\pi \mathrm{i} t \cdot x} \,\mathrm{d} x.$

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    By the way, Condition 2 implies that the Fourier transform is the restriction to $\mathbb R$ of a function analytic in the strip |Im(z)| for all B.2012-04-24

1 Answers 1

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In the two cases, we can apply the dominated convergence theorem to show by induction on $|\alpha|$ that $\partial^{\alpha}\mathcal F(f)(t)=\int_{\mathbb R^n}f(x)(-2\pi i)^{\alpha}x^{\alpha}e^{-2\pi it\cdot x}dx,$ hence $|\partial^{\alpha}\mathcal F(f)(t)|=(2\pi)^{\alpha}\int_{\mathbb R^n}|f(x)|\cdot |x|^{\alpha}dx.$ The last function is integrable in the two cases, so the last inequality ensures us that under 1. or 2. the Fourier transform of $f$ is real-analytic.

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    @davide giraudo, This is establishes $\mathscr{F}(f)$ is smooth, but where does the analytic part come from? Is it that trivial?2016-04-20