A list of proofs of Fourier inversion formula

Solution 1:

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).

Solution 2:

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 $$ e^{-i\lambda t}f(t)=i\int_{t}^{\infty}e^{-i\lambda x}g(x)dx \\ f(t) = i\int_{t}^{\infty}e^{-i\lambda(x-t)}g(x)dx,\;\;\Im\lambda > 0. $$

Similarly,

$$ f(t)=-i\int_{-\infty}^{t}e^{-i\lambda(x-t)}g(x)dx,\;\;\Im\lambda < 0. $$ 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.