A list of proofs of Fourier inversion formula 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 :)
 A: 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).
A: 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.
A: 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
Theorem
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.
Proofsketch
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.
Proof
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

*

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


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


*$f\in L^1$
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 $f\in L^1$.
