The reason for this question is to make a list of the known proofs (or proof ideas) of Fourier inversion formula for functions $f\in L^1(\mathbb{R})$ (obviously adding appropriate hypothesis to get a meaningful result) in order to better grasp the nuances of Fourier transform since, after all, different techniques (could) shed light on different features.

Here the list I know:

  • Proof: via Bochner theorem (see e.g. Rudin - Fourier analysis on groups);
  • Proof: via a summability kernel whose transform is known (see e.g. Rudin - Real & complex analysis);
  • Proof: via Dirichlet kernel and Riemann-Lebesgue's lemma (see e.g. Zemanian - distribution theory and transform analysis);
  • Proof: periodizing $f$ with period $L$, using Fourier inversion formula for $L$-periodic functions and letting $L\rightarrow\infty$ (see e.g. the answer by David Ullrich to this question);
  • Proof idea: via a Riemann series and the Fourier inversion formula for periodic functions (see e.g. this question, and feel free to answer it :) );

Now it's your turn... Let the games begin :)

  • 1
    $\begingroup$ If you permit a tempered distributions, then using $$\int_{-\infty}^\infty e^{i\omega (t-t')}\,d\omega =2\pi \delta(t-t')$$makes it trivial. $\endgroup$
    – Mark Viola
    Commented Aug 5, 2018 at 2:10
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    $\begingroup$ @mark viola I will admit it, if you manage to show that the argument is not circular: for how I understand distributional fourier transform, we first have to prove fourier inversion formula for schwartz test functions to get a sensible definition $\endgroup$
    – Bob
    Commented Aug 5, 2018 at 3:17
  • $\begingroup$ In This Answer, that $\lim_{t\to\infty}\text{PV}\left(\frac{\sin(tx)}{x}\right)=\pi \delta(x)$ in the sense of distributions. Then, note that in the sense of distributions $$\begin{align} \lim_{L\to\infty}\int_{-L}^L e^{i\omega (t-t')}\,d\omega&=\lim_{L\to\infty}\left(\frac{2\sin(L(t-t'))}{t-t'}\right)\\\\ &=2\pi \delta(t-t') \end{align}$$ $\endgroup$
    – Mark Viola
    Commented Aug 5, 2018 at 3:37
  • 1
    $\begingroup$ @MarkViola : the linked answer isn't the inversion formula via an integration by parts argument? :) If it is, I suggest to write an answer to provide a link to that answer $\endgroup$
    – Bob
    Commented Aug 5, 2018 at 3:47
  • $\begingroup$ @MarkViola and basically it seems the same argument in Zemanian, i.e. to use Dirichlet kernel, but you used IBP instead of Riemann-Lebesgue lemma to conclude $\endgroup$
    – Bob
    Commented Aug 5, 2018 at 3:50

4 Answers 4


That answer of mine that you link to is not an actual proof of the Inversion Theorem - it only works for "suitable" $f$, where "suitable" is left undefined. Here's an actual proof.

Just to establish where we're putting the $\pi$'s, we define $$\hat f(\xi)=\int f(t)e^{-it\xi}\,dt.$$

$L^1$ Inversion Theorem. If $f\in L^1(\Bbb R)$ and $\hat f\in L^1(\Bbb R)$ then $f(t)=\frac1{2\pi}\int\hat f(\xi)e^{i\xi t}\,d\xi$ almost everywhere.

We use that periodization argument to establish the theorem under stronger hypotheses:

Partial Inversion Theorem. If $f,f',f''\in L^1(\Bbb R)$ then $\hat f\in L^1$ and $f(t)=\frac1{2\pi}\int\hat f(\xi)e^{it\xi}\,d\xi$.

To be explicit, we're assuming that $f$ is differentiable, $f'$ is absolutely continuous, and $f',f''\in L^1$.

Note first that $(1+\xi^2)\hat f(\xi)$ is the Fourier transform of $f-f''$ (see Details below), so it's bounded: $$|\hat f(\xi)|\le\frac c{1+\xi^2}.\tag{*}$$

For $L>0$ define $$f_L(t)=\sum_{k\in\Bbb Z}f(t+kL).$$Then $f_L$ is a function with period $L$, and as such it has Fourier coefficients $$c_{L,n}=\frac1L\int_0^Lf_L(t)e^{-2\pi i n t/L}\,dt.$$

Inserting the definition of $f_L$ and using the periodicity of the exponential shows that in fact $$c_{L,n}=\frac1L\hat f\left(\frac{2\pi n}L\right).$$So ($*$) above shows that $\sum_n|c_{L,n}|<\infty$; hence $f_L$ is equal to its Fourier series: $$f_L(t)=\frac1L\sum_n\hat f\left(\frac{2\pi n}L\right)e^{2\pi i nt/L}.$$That's a Riemann sum for a certain integral; we establish convergence by noting that $$\frac1L\sum_n\hat f\left(\frac{2\pi n}L\right)e^{2\pi i nt/L}=\frac1{2\pi}\int g_L(\xi)\,d\xi,$$where $$g_L(\xi)=\hat f\left(\frac{2\pi n}L\right)e^{2\pi i nt/L}\quad(\xi\in[2\pi n/L,2\pi(n+1)/L)).$$Since $\hat f$ is continuous, DCT (using ($*$) for the D) shows that $$\lim_{L\to\infty}\int g_L=\int\hat f(\xi)e^{i\xi t}\,d\xi.$$

So we're done if we can show that $f_L\to f$ almost everywhere as $L\to\infty$. In fact we don't have to worry about whether/how this follows from the hypotheses: It's clear that $f_L\to f$ in $L^1_{loc}$ for every $f\in L^1$, hence some subsequence tends to $f$ almost everywhere.

Deriving IT from PIT is very simple. Say $(\phi_n)$ is an approximate identity; in particular $\phi_n\in C^\infty_c$, the support of $\phi_n$ shrinks to the origin, $||\phi_n||_1=1$ and $\hat\phi_n\to1$ pointwise. Let $f_n=f*\phi_n$. Then $f_n'=f*\phi_n'$, so $f'\in L^1$. Similarly for $f_n''$, so PIT applies to $f_n$. But $f_n\to f$ almost everywhere and DCT shows that $||\hat f_n-\hat f||_1\to0$.

Details, in answer to a comment. Note that here when I say $f,f'\in L^1$ I mean that $f$ is absolutely continuous and $f'\in L^1$.

Proposition. If $f,f'\in L^1(\Bbb R)$ then $\widehat{f'}(\xi)=-i\xi\hat f(\xi)$.

(Unless it's $i\xi\hat f(\xi)$; I never remember - here it doesn't matter since $(-1)^2=1$.)

Of course the proposition is just an integration by parts. Then we have to justify integration by parts in this context and worry about the boundary terms. Seems more instructive to show that

Given $f\in L^1$, the following are equivalent: (i) $f'\in L^1$, (ii) $f$ is "differentiable in $L^1$".

Regarding what (ii) means, see Lemma 2 below. I like to go this way because first, it's cute: "$f'\in L^1$ if and only if $f$ is differentiable in $L^1$", and second it seems to me to say something about what absolute continuity "really means". Anyway:

Exercise. If $f\in L^1$ then $\lim_{t\to0}\int|f(x)-f(x+t)|\,dx=0$.

(Hint: Wlog $f\in C_c(\Bbb R)$.)

Lemma 1. If $f\in L^1$ then $\lim_{h\to0}\int\left|f(x)-\frac1h\int_x^{x+h}f(t)\,dt\right|\,dx=0$.

Proof: $$\begin{align}\int\left|f(x)-\frac1h\int_x^{x+h}f(t)\,dt\right|\,dx &=\int\left|\frac1h\int_0^h(f(x)-f(x+t))\,dt\right|\,dx \\&\le\frac1h\int_0^h\int|f(x)-f(x+t)|\,dxdt.\end{align}$$ Apply the previous exercise and note that $\frac1h\int_0^h\epsilon=\epsilon$.

Lemma 2. If $f,f'\in L^1$ then $\lim_{h\to0}\int\left|f'(x)-\frac{f(x+h)-f(x)}{h}\right|\,dx=0$.

That is, if $f,f'\in L^1$ then $f$ is "differentiable in $L^1$". (We won't use the other implication...)

Proof: Write $\frac{f(x+h)-f(x)}{h}=\frac1h\int_x^{x+h}f'(t)\,dt$ and apply Lemma 1.

Another interesting/instructive version of "differentiable in $L^1$" that we won't use below:

Exercise. Suppose $f\in L^1$, and define $F:\Bbb R\to L^1(\Bbb R)$ by $F(t)(x)=f(x+t)$. Then (i) $f'\in L^1$ if and only if (ii) $F$ is differentiable.

Proof of the proposition: Work out the Fourier transform of the function $x\mapsto\frac{f(x+h)-f(x)}{h}$. Let $h\to0$ (apply Lemma 2).

  • $\begingroup$ can we prove that $(1+\xi ^{2}) \hat {f} (\xi)$ is the F.t. of $f-f''$ by elementary means, i.e., without using generalized functions? $\endgroup$ Commented Aug 6, 2018 at 6:01
  • $\begingroup$ Since $f'$ is absolutely continuous, $|f'(x)-f'(y)|\le\int_x^y|f''(t)|dt\le\|f''\|_1|x-y|$ so $f'$, being Lipschitz and integrable, is such that $f'(M)\rightarrow0, |M|\rightarrow\infty$. Then: $$\mathcal{F}(f'')(\xi) = \int_\mathbb{R}f''(t)e^{-i\xi t}dt = \lim_{M\rightarrow\infty}\int_{-M}^{M}f''(t)e^{-i\xi t}dt = \lim_{M\rightarrow\infty}\left(-f'(-M)e^{i\xi M}+f'(M)e^{-i\xi M}+i\xi\int_{-M}^{M}f'(t)e^{-i\xi t}dt\right) = i \xi \int_{\mathbb{R}}f'(t)e^{-i\xi t}dt = i \xi \mathcal{F}(f')(\xi).$$ Iterating the same argument one more time, we get $\mathcal{F}(f'')(\xi) =-\xi ^2\hat{f}(\xi)$. $\endgroup$
    – Bob
    Commented Aug 6, 2018 at 7:00
  • $\begingroup$ @Bob Thanks. I don't know why I didn't work out the proof myself! $\endgroup$ Commented Aug 6, 2018 at 7:54
  • 1
    $\begingroup$ This is a discussion on the "$L^1$ differentiability" and other modes of differentiability (distributional, etc...) $\endgroup$ Commented Sep 21, 2018 at 9:06
  • 1
    $\begingroup$ @Filippo I said I suspect that, yes. $\endgroup$ Commented Jan 25, 2022 at 12:06

An interesting proof focuses on the Complex Analysis of the resolvent $$ (\lambda I - A)^{-1},\;\;\; A=\frac{1}{i}\frac{d}{dx}. $$

Such analysis can be used to show the completeness of exponentials $\{e^{2\pi inx}\}_{n=-\infty}^{\infty}$ on $[-\pi,\pi]$, and can be used to prove the Plancherel theorem on $L^2(\mathbb{R})$, as well as to derive the Fourier transform and its inverse. Classical pointwise results can also be derived through analysis of the resolvent of differentiation.

In this case, consider $A$ on $\mathcal{D}(A)\subset L^2(\mathbb{R})$ consisting of absolutely continuous $f\in L^1(\mathbb{R})$ with $f'\in L^2(\mathbb{R})$. For $\lambda\notin\mathbb{R}$, solving the resolvent requires solving for $f$ such that $$ \lambda f+if'=g \\ f'-i\lambda f =-ig \\ (e^{-i\lambda t}f)'=-ie^{-i\lambda t}g. $$ Assuming $g\in L^1$ and $\Im\lambda > 0$, then $e^{-i\lambda t}$ decays as $t\rightarrow\infty$, which leads to $$ e^{-i\lambda t}f(t)=i\int_{t}^{\infty}e^{-i\lambda x}g(x)dx \\ f(t) = i\int_{t}^{\infty}e^{-i\lambda(x-t)}g(x)dx,\;\;\Im\lambda > 0. $$


$$ f(t)=-i\int_{-\infty}^{t}e^{-i\lambda(x-t)}g(x)dx,\;\;\Im\lambda < 0. $$ There is a jump discontinuity in the resolvent as $\lambda$ passes through the real axis: $$ \frac{1}{2\pi i}\{((s-i\epsilon)I-A)^{-1}f-((s+i\epsilon)I-A)^{-1}f\} \\ = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-is(x-t)}e^{-\epsilon|x-t|}f(x)dx. $$

After some careful Complex Analysis, it is possible to equate the integral around the singular part of the resolvent on the real axis to the residue at $\infty$, provided that both $f$ and $\hat{f}$ are in $L^1$. This gives

$$ \frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-is(x-t)}f(x)dx ds \\ = \lim_{\epsilon\,\uparrow\,\infty}\frac{1}{2}\int_{-\infty}^{\infty}e^{-is(x-t)}\epsilon e^{-\epsilon|x-t|}f(x)dx = f(t) $$ The last equality holds because $\int_{-\infty}^{\infty}\epsilon e^{-\epsilon |x|}dx = 1$ The $1/2$ is needed because the residue at $i\infty$ is being added to that at $-i\infty$. And, actually you can conclude $f$ is equal a.e. to a continuous function because of the assumption that $f,\hat{f} \in L^1$.

Cauchy first looked at the residues of the resolvent for the discrete case on $[-\pi,\pi]$ or $[0,2\pi]$. The earliest general pointwise convergence results for general Fourier series and transform pairs were proved in the context of Complex Analysis. One of the earliest proofs of the Spectral Theorem also used Complex Analysis in this way. Trading the sum of all resolvent singularities on the real line for a single residue at $\infty$ is a remarkably powerful and clever technique of Complex Analysis.

  • $\begingroup$ Do you have some reference sources where one could learn in more detail about complex analysis of the resolvent and application to famous problems such as these? $\endgroup$
    – FShrike
    Commented Apr 2, 2023 at 9:39
  • $\begingroup$ @FShrike : One of the best early treatments using Complex Analysis was written by E. C. Titchmarsh .Eigenfunction Expansions Associated with Second-order Differential Equations. Part I (1946)[8] 2nd. edition (1962); Eigenfunction Expansions Associated with Second-order Differential Equations. Part II (1958);[9] $\endgroup$ Commented Apr 2, 2023 at 21:01
  • $\begingroup$ Thanks. I can’t seem to get access online unfortunately $\endgroup$
    – FShrike
    Commented Apr 5, 2023 at 14:14
  • $\begingroup$ @FShrike : look for the book on archive.org $\endgroup$ Commented Apr 6, 2023 at 15:21
  • 1
    $\begingroup$ @FShrike : I'm not sure if you found the book on archive.org, but here is the link to borrow it: openlibrary.org/works/OL5264282W/… $\endgroup$ Commented Apr 10, 2023 at 15:32

Gaussian smushing (Probabilistic Argument)

Assuming we know the characteristic function of the normal distribution $ \newcommand{\cN}{\mathcal{N}} \newcommand{\charFct}{\psi} $ $$ \charFct_{\cN(\mu,\sigma^2)} = \int e^{itx}f_{\cN(\mu,\sigma^2)}(x) dx = \exp(i\mu t - \tfrac12 t^2 \sigma^2) $$

We can prove


For all distribution $\mu$ with bounded and continous density $f$, we have

$$ f(x)= \frac{1}{2\pi} \underbrace{ \int e^{-itx} \charFct_\mu(t)dt }_{=:\hat{\charFct}_\mu(x)} $$

if $\hat{\charFct}_\mu$ is well defined.


Simply plugging in the definition of the characteristic function and using Fubini won't work. The result would be this $$ \hat{\charFct}_f(x) = \int e^{-ixt} \charFct_f(t) d t = \int e^{-ixt}\int e^{iyt} f(y) dy dt = \int\underbrace{\int e^{-ixt} e^{iyt} dt}_{\delta_x(y)} f(y) dy $$

And this would require a notion of distribution functions which we can not assume. And a delta function is certainly not measurable so we can not really use Fubini. So the idea is to smush the function a bit with a gaußian kernel to make everything integrable.


We are using a Gauß kernel $$ g_\epsilon(x) = \frac{1}{\sqrt{2\pi \epsilon^2}} \exp\left(-\tfrac{x^2}{2\epsilon^2}\right) \to \delta_0(x) \quad (\epsilon\to 0) $$ to smudge the equation $$ \begin{aligned} \hat{\charFct}_\mu * g_\epsilon(x) &= \frac{1}{\sqrt{2\pi \epsilon^2}} \int \hat{\charFct}_\mu(y) \exp\left(-\tfrac{(x-y)^2}{2\epsilon^2}\right)dy\\ &= \frac{1}{\sqrt{2\pi \epsilon^2}} \int \int e^{-ity} \charFct_\mu(t) dt \exp\left(-\tfrac{(x-y)^2}{2\epsilon^2}\right)dy\\ &= \int \int \underbrace{ \frac{1}{\sqrt{2\pi \epsilon^2}}\int e^{-ity} \exp\left(-\tfrac{(x-y)^2}{2\epsilon^2}\right)dy }_{ \begin{aligned}[t] &= \frac{1}{\sqrt{2\pi \epsilon^2}}\int e^{ity} \exp\left(-\tfrac{(y-(-x))^2}{2\epsilon^2}\right)dy\\ &= \charFct_{\cN(-x, \epsilon^2)}(t)\\ &= e^{it(-x) -\tfrac12 t^2\epsilon^2 } \end{aligned} } \charFct_\mu(t) dt \\ &= \int \underbrace{ e^{-\tfrac12 t^2 \epsilon^2} }_{\text{regularization}} \underbrace{e^{-itx}\charFct_\mu(t)}_{\text{Fourier, vgl. } \hat{\charFct}_\mu}dt\\ \end{aligned} $$ Here we can see that another interpretation of the smushing is that we are using a regularized version of the fourier transform on $\charFct_\mu$. Now if $\hat{\charFct}_\mu$ is well defined (i.e. if $e^{-itx}\charFct_\mu(t)$ is integrable), then we have $\hat{\charFct}*g_\epsilon(x)\to\hat{\charFct}(x)$ for $\epsilon\to 0$ by the dominated convergence theorem. Continuing on we get $$ \begin{aligned} \hat{\charFct}_\mu * g_\epsilon(x) &= \int e^{-\tfrac12 t^2 \epsilon^2} \int e^{itz} f(z) dz\\ &= \int \underbrace{ \int e^{it(z-x)} e^{-\tfrac12 t^2\epsilon^2}dt }_{ =\charFct_{\cN\left(0,1/\epsilon^2\right)}(z-x) \sqrt{2\pi 1/\epsilon^2} } f(z) dz\\ &= 2\pi \frac{1}{\sqrt{2\pi\epsilon^2}} \int e^{-\tfrac{z-x}{2\epsilon^2}} f(z) dz\\ &= 2\pi f * g_\epsilon(x). \end{aligned} $$

Taking the limit

What is left to do is taking the limit on this side as well. Here are a couple of approaches for that

  1. $f\in C_c$ $$ f * g_\epsilon(x) = \mathbb{E}_{Y\sim\cN(0,\epsilon^2)}[f(x-Y_\epsilon)] \to \mathbb{E}[f(x-0)] = f(x), $$

    as $Y_\epsilon \to 0$ in $L^2$ so also in distribution which is by definition the above as $f(x-\cdot)$ is a continuos bounded function. The same argument for $\hat{\charFct}_\mu$ results in the statement.

  2. $f$ continuuous in $x$

    let us first transform the problem a bit: $$ f*g_\epsilon = \int \underbrace{f(x-y)}_{=:f_x(y)}g_\epsilon(y) dy $$

    We want to use two statements about weakly converging probability measures from the Portmanteau theorem

    • $\liminf_n \mu_n(O)\ge \mu(O)$ for all open $O$
    • $\limsup_n \mu_n(A) \le \mu(A)$ for all closed $A$

    If $f$ is continuous in $x$, then $f_x$ is continuous in zero. Now select some $\delta$, then we can use

    $$ \begin{aligned} \min_{z\in(-\delta, \delta)} f_x(z) &= \min_{z\in(-\delta, \delta)} f_x(z) \delta((-\delta, \delta)) \le \liminf_\epsilon \min_{z\in(-\delta, \delta)} f_x(z) \int_{(-\delta,\delta)} g_\epsilon(y) dy\\ &\le \liminf_\epsilon f*g_\epsilon(x) \end{aligned} $$

    So due to continuity in $x$ we get $f(x) \le \liminf_\epsilon f*g_\epsilon(x)$ with $\delta\to 0$

    Now while $\min_{z\in(-\delta, \delta)} f_x(z) \mathbf1_{(-\delta,\delta)}$ is a valid lower bound for $f_x$, we need to improve on this trick for an upper bound. So we split the integral on $[-\delta, \delta]$ and its complement. Where we can use the Portmanteu statement about closed sets

    $$ \limsup_\epsilon f*g_\epsilon(x) \le \max_{z\in[-\delta,\delta]} f(z) +\limsup_\epsilon \int_{[-\delta, \delta]^C}f_x(y)g_\epsilon(y) dy $$

    Now $g_\epsilon$ is bounded on $[-\delta,\delta]^C$ by $c(\epsilon)= \frac{1}{\sqrt{2\pi\epsilon^2}}\exp(-\tfrac{\delta^2}{2\epsilon^2})$ and $f_x$ is a density so integrates to one. Therefore the second integral is bounded by $c(\epsilon)$ which converges to zero for $\epsilon\to0$. So we have

    $$ \limsup_\epsilon f*g_\epsilon(x) \le \max_{z\in[-\delta,\delta]}f_x(z) \to f(x) \quad (\delta\to0) $$

  3. $f\in L^1$

    We are going to show that $$ \tag{1} \forall f\in L^1: \quad f*g_\epsilon \to f \quad \text{in }L^1, $$ while the other limit was a pointwise limit! But this is okay due to Fatou's Lemma $$ \begin{aligned} \int |\hat{\varphi}_\mu (x) - 2\pi f(x) | dx &=\int \liminf_{\epsilon\to 0} |\hat{\varphi}_\mu *g_\epsilon (x) - 2\pi f(x) | dx \\ &\le 2\pi \underbrace{\liminf_{\epsilon\to 0} \int |f*g_\epsilon (x) - f(x) | dx}_{=0}, \end{aligned} $$ which implies $f=\frac1{2\pi}\hat{\charFct}_\mu$ Lebesgue almost everywhere. But densities are only unique up to Lebesgue zero sets, so we would be done.

    Okay so let us show (1): Since $g_\epsilon$ is a density, we can write \begin{align*} f*g_\epsilon(x) - f(x) = \int (f(x-y)-f(x))g_\epsilon(y)dy = \int (f(x-\epsilon z)-f(x))g_1(z)dz, \end{align*} where we used the substitution $y=\epsilon z$ in the last equation. Note that the constant of the normal distribution $\frac1{\sqrt{2\pi\epsilon^2}}$ eats up the $\epsilon$ in the substitution $dy = \epsilon dz$. Therefore we have \begin{align*} \| f*g_\epsilon - f\|_{L^1(\lambda)} &=\int \left| \int (f(x-\epsilon z)-f(x))g_1(z)dz \right| dx\\ &=\int \underbrace{\int |f(x-\epsilon z)-f(x)| dx}_{ =\|f(\cdot-\epsilon z)-f\|_{L^1(\lambda)} } g_1(z)dz. \end{align*} Writing $\tau_{t}(f) = f(\cdot - t)$ for the translation by $t$, we have $$ \| \tau_{\epsilon z}(f) - f \|_{L^1} \le \| \tau_{\epsilon z}(f)\|_{L^1}+ \|f \|_{L^1} = 2\|f\|_{L^1}, $$ because a shift does not change the integral over $\mathbb{R}$. With this upper bound integrable against $\cN(0,1)$, we can use the DCT to move the limit into the integral $$ \lim_{\epsilon\to 0}\|f*g_\epsilon - f\|_{L^1} = \int \underbrace{ \lim_{\epsilon\to 0} \| \tau_{\epsilon z}(f) - f \|_{L^1}}_{ = 0 } g_1(z) dz. $$ Where $\lim_{\epsilon\to 0} \| \tau_{\epsilon z}(f) - f \|_{L^1}= 0$, because this holds for $f\in C_c$. And, as $C_c$ is dense in $L^1$ (since we can approximate any measurable function with linear combinations of indicators which can in turn be approximated by continuous indicators), it also holds by triangle inequality for all $f\in L^1$.

  • $\begingroup$ This is what I meant by "via a summability kernel whose transform is known (see e.g. Rudin - Real & complex analysis)". $\endgroup$
    – Bob
    Commented Apr 27, 2022 at 19:57
  • $\begingroup$ @Bob ah right - okay :) Still nice to have this available here I guess. And I needed it anyway so not too much additional work $\endgroup$
    – Felix B.
    Commented Apr 27, 2022 at 20:09

$\newcommand{\F}{\mathcal{F}}\newcommand{\B}{\mathscr{B}}\newcommand{\d}{\,\mathrm{d}}$Another probabilistic / measure theoretic argument.


A probability distribution on $\Bbb R^n$ is a nonnegative real measure on $(\Bbb R^n,\B(\Bbb R^n))$ that measures the whole space to be $1$.

For any probability distribution $\mu$ we define the characteristic function $\phi_\mu:\Bbb R^n\to\Bbb C$ via: $$x\mapsto\int_{\Bbb R^n}e^{ix^T\xi}\d\mu(\xi)$$

It is clear that such $\phi_\mu$ are always continuous.

For any $\tau\ge0$ we define $B^{(n)}_\tau\subset\Bbb R^n$ to be the set of all $x$, $\max_{1\le k\le n}|x_k|\le\tau$. Define $R_n$ to be the collection of all subsets of $\Bbb R^n$ of form: $x\in R_n\iff a_j\le x_j\le b_j,\,\forall j$ where $a_j<b_j,\,j=1,\cdots,n$ are given constants. These are like hypercubes.

For $h\in L^1(\Bbb R^n)$ we say: $\F(h)\in C^0(\Bbb R^n)$ is the function given by: $$\xi\mapsto\int_{\Bbb R^n}\exp(-i\xi^Tx)h(x)\d x$$And: $\F^{-1}(h)\in C^0(\Bbb R^n)$ is the function given by: $$\xi\mapsto(2\pi)^{-n}\int_{\Bbb R^n}\exp(i\xi^Tx)h(x)\d x$$

Claim (probabilistic inversion formula):

If $\mu$ is a probability distribution on $\Bbb R^n$ and $A\in R_n$ has $\mu(\partial A)=0$ then: $$\large\mu(A)=\lim_{\tau\to\infty}(2\pi)^{-n}\int_{B^{(n)}_\tau}\phi_\mu(t)\left(\prod_{j=1}^n\frac{\exp(-ia_jt_j)-\exp(-ib_jt_j)}{it_j}\right)\d t$$

The proof sketch is this: expand $\phi_\mu$ into its integral definition. By compactness of $B^{(n)}_\tau$ for any $\tau$ and continuity, we are free to "Fubini it" and integrate with respect to $t$ first and with respect to $\xi$ later. "Fubini it" again to break the integral over $B^{(n)}\tau$ into an iterated integral over intervals $[-\tau,\tau]$ which boil down to a product of integrals. Letting the odd components cancel out over the symmetric $B^{(n)}_\tau$ we get the expression: $$(2\pi)^{-n}\int_{\Bbb R^n}\left(\int_{-\tau}^\tau\frac{\sin((\xi_j-a_j)t)-\sin((\xi_j-b_j)t)}{it}\d t\right)^n\d\mu(\xi)$$

Using the famous Dirichlet integral (which implies boundedness of the inner integrand as $\tau>0$ varies, and $\mu(\Bbb R^n)=1$ then permits use of the dominated convergence theorem) and using the fact that $\partial A=\{(\xi_j)_j:\xi_j=a_j\text{ or }b_j\text{ for at least one $j$}\}$ is $\mu$-null we see this tends to, as $\tau\to\infty$: $$(2\pi)^{-n}\int_{\Bbb R^n}\prod_{j=1}^n((2\pi)\chi_{a_j<\xi_j<b_j})\d\mu(\xi)=\mu(A)$$

As required. This allows for a proof that two probability distributions $\mu,\nu$ on $\Bbb R^n$ agree iff. their characteristic functions agree.


If $\mu$ is a probability distribution on $\Bbb R^n$ with integrable characteristic function then $\mu$ has a continuous density $f:\Bbb R^n\to[0,\infty)$ given by: $$f(x)=(2\pi)^{-n}\int_{\Bbb R^n}\exp(-ix^Tt)\phi_\mu(t)\d t$$

The proof sketch: define $f$ as above (but a priori with codomain $\Bbb C$). It is clearly continuous and well defined. We get an (a priori complex) measure $\nu$ on $\Bbb R^n$ by using $f$ as a density. If $A\in R_n$ and $\mu(\partial A)=0$ then by integrability and Fubini's theorem we can find $\nu(A)$ equals exactly the limit expression found in the inversion formula, thus $\nu(A)=\mu(A)$. Since there are only countably many hyperplanes parallel to the coordinate axes with positive $\mu$-measure, it follows $\mu(\partial A)\neq 0$ for at most countably many $A\in R_n$. Then agreement of $\nu,\mu$ on $R_n$ with at most countably many exceptions allows, by continuity of measure and taking intersections, for $\nu\equiv\mu$ on $R_n$. Since $R_n$ generates $\B(R^n)$, $\nu=\mu$ and $\nu$ is forced to be a probability distribution, and by continuity of $f$ we then get that $f$'s codomain can be taken to be $[0,\infty)$. $\nu=\mu$ means of course $f$ is a density for $\mu$.

Now we can tackle the Fourier inversion problem: say $g:\Bbb R^n\to\Bbb R$ has both of $g,\F^{-1}(g)$ integrable and that $g$ is continuous at $x_0$. Let $g_{\pm}$ be the positive and negative parts of $g$. We assume $g$ is not almost everywhere zero, as then the Fourier inversion statement is trivial ($g(x_0)=0$ is forced and all the integrals involved will vanish).

We get $h_\pm:=g_\pm\cdot\|g_\pm\|_1^{-1}$ and these induce probability distributions $\mu_\pm$ with densities $h_\pm$. Of course: $$\phi_{\mu_\pm}(x)=\int_{\Bbb R^n}\exp(ix^T\xi)h_\pm(\xi)\d\xi=(2\pi)^n\F^{-1}(g_\pm)(x)\cdot\|g_{\pm}\|_1^{-1}$$Is integrable by assumption.

The probabilistic inversion formula then shows that: $$f_\pm(x):=(2\pi)^{-n}\int_{\Bbb R^n}\exp(-ix^Tt)\phi_{\mu_\pm}\d t$$Are densities for $\mu_{\pm}$. Since any two densities must agree almost everywhere, continuity of $f_\pm$ and $h_\pm$ at $x_0$ proves $f_\pm(x_0)=h_\pm(x_0)$. Hence: $$g_\pm(x_0)=\int_{\Bbb R^n}\exp(-ix_0^Tt)\F^{-1}(g_\pm)(t)\d t=\F(\F^{-1}(g_\pm))(x_0)$$Simply taking $g=g_+-g_-$ finds: $$g(x_0)=\F(\F^{-1}(g))(x_0)$$As desired. $\blacksquare$

Remark 1: breaking $g$ into positive and negative parts is not important, nor is normalising $g$ to $h$ to get a probability measure important. It just fits nicely with the probabilistic perspective.

Remark 2: the proof actually shows that, by almost-everywhere equality of densities, any $g:\Bbb R^n\to\Bbb R$ with both $g$ and $\F^{-1}(g)$ being integrable (equivalently both $g$ and $\F(g)$ being integrable) must almost everywhere equal a continuous function, in particular $g$ must be almost everywhere equal to the continuous $f:=\F(\F^{-1}(g))$. So we can state the Fourier inversion theorem as $g=\F(\F^{-1}(g))$ in $L^1$ since there almost everywhere equal functions are identified.


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