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I am currently trying to piece together a proof of the Fourier inversion theorem which is appropriate for a probability theory context. The approach I am taking is smushing with a gaussian kernel $$ g_\epsilon(x) = \frac{1}{\sqrt{2\pi\epsilon^2}} \exp\left(-\frac{x^2}{2\epsilon^2}\right) $$

and then using the fact $$ h * g_\epsilon \to h \quad \text{in } L^1 $$

from this answer. Unfortunately this requires that $h$ is in $L^1$ so to have a complete proof, I need that

$$ \hat{\varphi}_f: x\mapsto \int e^{-itx} \underbrace{\int e^{ity} f(y)dy}_{=\varphi_f(t)}dt \quad \in L^1 $$

for all $f\in L^1$. Because then I have $f*g_\epsilon \to f$ and $\hat{\varphi}_f * g_\epsilon \to \hat{\varphi}_f$. So I get $2\pi f=\hat{\varphi}_f$ by uniqueness of limits.

I can not think of a proof though. Obviously I can not use Fourier inversion already.

EDIT: Simpler Equivalent Question

If I have for all $x\in\mathbb{R}$

$$ \lim_{\epsilon\to 0}\int e^{-\tfrac12 t^2 \epsilon^2} e^{itx}\hat{f}(t)dt = f(x) $$

does that imply that

$$ \int e^{itx} \hat{f}(t)dt =f(x) $$

In other words can I swap limits and integration, if I know that the limits of integrals converges (in this case)?

jjagmath
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Felix Benning
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  • You said that $f\in L^1$. The question is what is $g_\epsilon$. If $g\in L^1,\int g = 1$ and $g_\epsilon(t) = g(t/ \epsilon) / \epsilon$ then $f\ast g_\epsilon \to f$ in $L^1$. The simplest choice of $g$ is $\frac1\pi\frac{\sin(t)^2}{t^2}$ because $g(t)=\frac2\pi \int_{-2}^2 (1-|y/2|) e^{iyt} dy$ so that $f \ast g_\epsilon = ...$ – reuns Apr 28 '22 at 13:16
  • @reuns ah sorry, a gaussian kernel (clarified in the question) – Felix Benning Apr 28 '22 at 13:29
  • You misunderstood the point. Without using any theorem, if $g_\epsilon = \int_{-\infty}^\infty G_\epsilon(y) e^{iyt}dy$ and $\hat{f}(y)=\int_{-\infty}^\infty f(t) e^{-iyt}dt$ (and $f,G_\epsilon$ are $L^1$) then

    $f \ast g_\epsilon (t) = \int_{-\infty}^\infty \hat{f}(t) G_\epsilon(y) e^{iyt}dy$ which is a regularized version of the inverse Fourier transform of $\hat{f}$. With a right choice of $g_\epsilon$ this regularized version converges to $f$ as $\epsilon \to 0$ (in $L^1$), so we have a regularized version of the Fourier inversion theorem.

     – reuns Apr 28 '22 at 13:38
  • @reuns but I want to remove that regularization again to get the unregularized version of the inverse fourier transform theorem. Does that make sense? Or have I misunderstood what you are trying to say? – Felix Benning Apr 28 '22 at 13:43
  • You need some assumptions such as $\hat{f} \in L^1$ to look at the unregularized version. – reuns Apr 28 '22 at 13:44
  • I don't want $\hat{f}\in L^1$, I want $\hat{\hat{f}}\in L^1$ for $f\in L^1$, or $\check{\hat{f}}$ to be more precise – Felix Benning Apr 28 '22 at 13:50
  • The regularized version of $\hat{\hat{f}}$ is equal to $f$ (that's the regularized Fourier inversion theorem, that I'm saying is the main point here). We don't have any definition of $\hat{\hat{f}}$ if $\hat{f}$ is not in $L^1$. – reuns Apr 28 '22 at 13:51
  • @reuns Okay let's put in another way: I am assuming that $f$ is a density, and therefore $\varphi_f$ the characteristic function. The characteristic function always exists. And if I started out with a density, the fourier transform of the characteristic function should be the density again right? Are there really some corner-cases where that blows up in our face? – Felix Benning Apr 28 '22 at 13:54
  • When $\varphi_f$ is not $L^1$, you don't know how to define its Fourier transform. $\int_{-\infty}^\infty \varphi_f(y)e^{iyt}dy$ doesn't have to converge. That's why you regularize it as $\int_{-\infty}^\infty \varphi_f(y) G_\epsilon(y)e^{iyt}dy$ and let $\epsilon \to 0$. With the right choice of $G_\epsilon$ this converges to $f$ in $L^1$. – reuns Apr 28 '22 at 13:59
  • Let us continue this discussion in chat. – Felix Benning Apr 28 '22 at 14:02
  • @reuns Thinking about your comments I phrased and (I think) equivalent question – Felix Benning Apr 28 '22 at 15:20

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Have you tried to write $e^{-t^2\epsilon^2/2}$ as the Fourier transform of a Guassian? If so, then you could write your integral as the convolution of a Guassian and $f$, and then you could take the limit.

Disintegrating By Parts
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