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 :)

  • $\begingroup$ 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 Aug 5 '18 at 2:10
  • 1
    $\begingroup$ @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 Aug 5 '18 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 Aug 5 '18 at 3:37
  • $\begingroup$ @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 Aug 5 '18 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 Aug 5 '18 at 3:50

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? $\endgroup$ – Kavi Rama Murthy Aug 6 '18 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 Aug 6 '18 at 7:00
  • $\begingroup$ @Bob Thanks. I don't know why I didn't work out the proof myself! $\endgroup$ – Kavi Rama Murthy Aug 6 '18 at 7:54
  • $\begingroup$ @Bob The inequality $\int_x^y|f''|\le||f''||_1|x-y|$ is a little overenthusiastic. Fix: The fact that $f'(M)=f'(0)+\int_0^M f''$ and $f''\in L^1$ shows that $\lim_{M\to\infty}f'(M)$ exists; since $f'\in L^1$ the limit is $0$. I think that's all you used the bogus inequality for... $\endgroup$ – David C. Ullrich Aug 6 '18 at 15:11
  • $\begingroup$ @KaviRamaMurthy There's a non-fatal error in what Bob said. See the Details I added for another approach, that for my money explains why it's really true. $\endgroup$ – David C. Ullrich Aug 6 '18 at 15:13

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.