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In probability theory, the Fourier Inversion theorem for a Borel probability measure $\mu$ on $\mathbb{R}$ reads:

$$ \mu((a,b)) + \frac{1}{2} \mu(\{ a,b \}) = \frac{1}{2\pi} \lim_{T\to\infty} \int^T_{-T}\int^b_a \hat{\mu}(t) e^{-ixt} dx dt $$

On the other hand, in distribution theory, the Fourier Inversion theorem for a tempered distribution $u$ on $\mathbb{R}$ is exceptionally simple:

$$ u = \frac{1}{2\pi} \mathcal{R}(\hat{\hat{u}}) $$

where $\mathcal{R}$ is the reflection operator. Of course, this identity has to be understood in the distributional sense.

My question is: how to reconcile these two inversion theorems?

My try:

First, note that any Borel probability measure $\mu$ is a tempered distribution with Fourier transform given by:

$$ \hat{\mu}(t) = \int_\mathbb{R} e^{ixt} d\mu(x) $$

By the Fourier Inversion theorem for a tempered distribution, for any Schwartz function $\psi \in \mathcal{S}(\mathbb{R}) $:

$$ \begin{align*} \int_\mathbb{R} \psi d\mu &= \frac{1}{2\pi} \langle \mathcal{R}(\hat{\hat{\mu}}), \psi \rangle = \frac{1}{2\pi} \langle \hat{\mu}, \widehat{\mathcal{R}\psi} \rangle \\ &= \frac{1}{2\pi} \int_\mathbb{R} \hat{\mu}(t) \mathcal{R}(\hat{\psi})(t) dt \\ &= \frac{1}{2\pi} \int_\mathbb{R} \hat{\mu}(t) \int_\mathbb{R} \psi(x) e^{-ixt} dx dt \end{align*} $$

Next I want to "substitute" $\psi = \chi_{(a, b)}$, but here are two problems:

  1. The function $\chi_{(a, b)}$ is not a Schwartz function. I guess this is where we cook up the limit $T \to \infty$.
  2. I don't know how the term $\frac{1}{2} \mu(\{ a,b \})$ pops up.

Remark: To be consistent with the notation in probability theory, I define the Fourier transform of an integrable function $f$ on $\mathbb{R}$ to be

$$ \hat{f}(t) = \int_{-\infty}^\infty f(x) e^{ixt} dx $$

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  • $\begingroup$ Instead we can go the other way and prove the Fourier inversion theorem (for L^1 functions, maybe it’s more subtle for distributions) from the probabilistic version. $\endgroup$
    – FShrike
    Apr 10, 2023 at 14:52
  • $\begingroup$ Have you seen a proof of the probabilistic inversion formula? (one that doesn’t use Fourier analysis?) Because it’s not so bad and the “$1/2$” term comes out nicely from the proof that I know. I sketch the proof in my linked answer and can reference a full proof or write one up if you wish $\endgroup$
    – FShrike
    Apr 10, 2023 at 14:53
  • $\begingroup$ @FShrike Yes, I have read the proof in your linked answer. It seems to prove by directly computing the double integral in the inversion formula. Unfortunately I can't see how such direct computation is related to the distributional Fourier inversion. $\endgroup$
    – user141240
    Apr 10, 2023 at 15:37

1 Answer 1

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I am the questioner. I think I might have an answer.

Let's pretend we don't know the identity $ u = \frac{1}{2\pi} \mathcal{R}(\hat{\hat{u}}) $ and try to compute directly. For any Schwartz function $\psi \in \mathcal{S}(\mathbb{R})$, we have

$$ \begin{align*} \frac{1}{2\pi} \langle \mathcal{R}(\hat{\hat{u}}), \psi \rangle &= \frac{1}{2\pi} \int_\mathbb{R} \hat{\mu}(t) \mathcal{R}(\hat{\psi})(t) dt \\ &= \frac{1}{2\pi} \int_\mathbb{R} \int_\mathbb{R} e^{ixt} d\mu(x) \mathcal{R}(\hat{\psi})(t) dt \\ &= \int_\mathbb{R} \frac{1}{2\pi} \int_\mathbb{R} e^{ixt} \mathcal{R}(\hat{\psi})(t) dt d\mu(x) \end{align*} $$

The last step is valid for $\psi \in \mathcal{S}(\mathbb{R})$ because $ \int \int |e^{ixt} \mathcal{R}(\hat{\psi})(t)| dt d\mu = ||\mu|| ||\hat{\psi}||_1 < \infty $

Now I want to substitute $\psi = \chi_{(a, b)}$, the problem is that $\chi_{(a, b)}$ is not a Schwartz function and $ \widehat{\chi_{(a, b)}} $ is not integrable. This is where we need the limit $ T \to \infty $ to permit the interchange of the order of integration.

$$ \begin{align*} \frac{1}{2\pi} \int_\mathbb{R} \hat{\mu}(t) \mathcal{R}(\widehat{\chi_{(a, b)}})(t) dt &= \frac{1}{2\pi} \lim_{T \to \infty} \int_{-T}^T \int_\mathbb{R} e^{ixt} d\mu(x) \mathcal{R}(\widehat{\chi_{(a, b)}})(t) dt \\ &= \lim_{T \to \infty} \int_\mathbb{R} \frac{1}{2\pi} \int_{-T}^T e^{ixt} \mathcal{R}(\widehat{\chi_{(a, b)}})(t) dt d\mu(x) \end{align*} $$

Since $\chi_{(a, b)}$ is piecewise smooth, classical Fourier analysis tells us that we have the pointwise convergence for each $x \in \mathbb{R}$:

$$ \Psi_T(x) := \frac{1}{2\pi} \int_{-T}^T e^{ixt} \mathcal{R}(\widehat{\chi_{(a, b)}})(t) dt \\ \to \frac{\chi_{(a, b)}(x+) + \chi_{(a, b)}(x-)}{2} = \chi_{(a, b)} + \frac{\chi_{\{a, b\}}}{2} $$

By a routine calculation, one can express $\Psi_T$ in terms of sinc functions and show that it is uniformly bounded. Thus, by dominated convergence, we have

$$ \lim_{T \to \infty} \int_\mathbb{R} \Psi_T d\mu =\int_\mathbb{R} \chi_{(a, b)} + \frac{\chi_{\{a, b\}}}{2} d\mu $$

The result follows.

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