$\DeclareMathOperator\erf{erf}\DeclareMathOperator\sech{sech}\DeclareMathOperator\sgn{sgn}\DeclareMathOperator\sinc{sinc}$This is a cross-post of a question I posted on MSE a couple of weeks ago which has not yet received any comments or answers.
In this question I use the term "nested" Fourier series representation to refer to an infinite series of one or more Fourier series versus a single Fourier series. Whereas a single Fourier series is periodic, a nested Fourier series representation is not necessarily periodic. When it is periodic it can be expressed as a single Fourier series, and therefore doesn't provide as much increased utility. The nested Fourier series representations for $f(x)$ defined in this question are of the non-periodic nature and even seem to converge globally for $x\in\mathbb{C}$ when $f(x)$ is a complex analytic function for which the Fourier inversion theorem holds.
A long time ago I noticed many series representations of functions derived from Mellin convolutions such as
$$f(s)=\left[\delta(x-1)\,*_\mathcal{M}\,f(x)\right](s)=\int_0^\infty\delta(x-1)\, f\left(\frac{s}{x}\right)\,\frac{dx}{x}\tag{1}\label{462653_1}$$
seem to converge for $\Re(s)>0$ (and in some cases for $\Re(s)\ge 0$) when formula \eqref{462653_1} above is evaluated term-wise using nested Fourier series representations such as
$$\delta(x-1)=\lim_{\substack{N, f\to\infty \\ M(N)=0}}\left(\sum\limits_{n=1}^N \frac{\mu(n)}{n} \left(1+2 \sum\limits_{k=1}^{f n} \cos\left(\frac{2 \pi k x}{n}\right)\right)\right)\tag{2}\label{462653_2}$$
and
$$\delta(x-1)=\lim_{N, f\to\infty}\left(\sum\limits_{n=1}^N \frac{\mu(2 n-1)}{2 n-1} \left(\frac{1}{2}+\sum\limits_{k=1}^{2 f (2 n-1)} (-1)^k\, \cos\left(\frac{\pi k x}{2 n-1}\right)\right)\right)\tag{3}\label{462653_3}$$
where the evaluation frequency $f$ in the inner sum over $k$ in formulas \eqref{462653_2} and \eqref{462653_3} above is assumed to be a positive integer.
In Formula \eqref{462653_2} above, the upper evaluation limit $N$ in the outer sum over $n$ must be selected such that $M(N)=0$ where
$$M(x)=\sum\limits_{n=1}^x \mu(n)\tag{4}\label{462653_4}$$
is the Mertens function (which has an infinite number of zeros), whereas formula \eqref{462653_3} above eliminates this conditional convergence requirement.
Formulas \eqref{462653_2} and \eqref{462653_3} above actually converge (in a distributional sense) to $\delta(x+1)+\delta(x-1)$, but the $\delta(x+1)$ term doesn't affect the Mellin convolution defined in formula \eqref{462653_1} above since the integral is over $0\le x\le\infty$.
In this answer to one of my own questions on MSE, I posted several examples of formulas derived from Mellin convolutions such as formula \eqref{462653_1} above using the analytic representation for $\delta(x-1)$ defined in formula \eqref{462653_2} above (see formulas (52) to (70) at my linked answer). As I mentioned above, many of these formulas converge for $\Re(s)>0$ which seems consistent with the fact that many Mellin transforms seem to converge for some subset of $s\in\mathbb{C}$ (e.g. $\Re(s)>\alpha$ for some $\alpha\in\mathbb{R}$).
More recently I noticed many times it seems the function $f(x)$ can be approximated by the nested Fourier series representation
$$f(x)\approx \lim_{N, f\to\infty}\left(\sum\limits_{n=1}^N \frac{\mu(2 n-1) }{2 n-1} \left(\frac{3 F(0)}{8}\\+\frac{1}{2} \sum\limits_{k=1}^{2 f (2 n-1)} (-1)^k\, \cos\left(\frac{\pi k}{2 n-1}\right) \left(F\left(\frac{k}{4 n-2}\right) e^{\frac{i \pi k x}{2 n-1}}+F\left(-\frac{k}{4 n-2}\right) e^{-\frac{i \pi k x}{2 n-1}}\right)\\-\frac{1}{8} \sum\limits_{k=1}^{4 f (2 n-1)} (-1)^k\, \left(F\left(\frac{k}{8 n-4}\right) e^{\frac{i \pi k x}{4 n-2}}+F\left(-\frac{k}{8 n-4}\right) e^{-\frac{i \pi k x}{4 n-2}}\right)\right)\right)\tag{5}\label{462653_5}$$
where
$$F(\omega)=\mathcal{F}_x[f(x)](\omega)=\int\limits_{-\infty}^\infty f(x)\, e^{-2 \pi i \omega x}\, dx\tag{6}\label{462653_6}$$
is the Fourier transform of $f(x)$ and
$$f(x)=\mathcal{F}_{\omega}^{-1}[F(\omega)](x)=\int\limits_{-\infty}^\infty F(\omega)\, e^{2 \pi i x \omega}\, d\omega\tag{7}\label{462653_7}$$
is the related inverse Fourier transform.
Formula \eqref{462653_5} above is based on the Fourier convolution
$$f(x)=[\delta(y)\,*\,f(y)](x)=\int\limits_{-\infty}^\infty \delta(y)\, f(y-x)\, dy\tag{8}\label{462653_8}$$
evaluated term-wise using the nested Fourier series representation
$$\delta(x)=\lim_{N, f\to\infty}\left(\sum\limits_{n=1}^N \frac{\mu(2 n-1)}{2 n-1} \left(\frac{3}{8}+\sum\limits_{k=1}^{2 f (2 n-1)} (-1)^k\, \cos\left(\frac{\pi k}{2 n-1}\right) \cos\left(\frac{\pi k x}{2 n-1}\right)\\-\frac{1}{4} \sum\limits_{k=1}^{4 f (2 n-1)} (-1)^k\, \cos\left(\frac{\pi k x}{2 (2 n-1)}\right)\right)\right)\tag{9}\label{462653_9}$$
which is derived from formula \eqref{462653_3} above (see this answer I posted to one of my own questions on MathOverflow).
I believe formula \eqref{462653_9} above evaluated at $N=\infty$ is equivalent to the limit representation
$$\delta(x)=\lim_{f\to\infty}\int\limits_{-f}^f e^{2 i \pi x t} \, dt=\lim_{f\to\infty}2 f\, \sinc(2 \pi f x).\tag{10}\label{462653_10}$$
Formula \eqref{462653_5} above is derived from formulas \eqref{462653_8} and \eqref{462653_9} above using the convolution theorem.
When $f(x)$ is an even function formula \eqref{462653_5} above simplifies to
$$f(x)\approx \lim_{N, f\to\infty}\left(\sum\limits_{n=1}^N \frac{\mu(2 n-1)}{2 n-1} \left(\frac{3 F(0)}{8}\\+\sum _{k=1}^{2 f (2 n-1)} (-1)^k\, \cos\left(\frac{\pi k}{2 n-1}\right)\, F\left(\frac{k}{4 n-2}\right)\, \cos\left(\frac{\pi k x}{2 n-1}\right)\\-\frac{1}{4} \sum\limits_{k=1}^{4 f (2 n-1)} (-1)^k\, F\left(\frac{k}{8 n-4}\right)\, \cos\left(\frac{\pi k x}{4 n-2}\right)\right)\right),\quad f(-x)=f(x)\tag{11}.\label{462653_11}$$
When $f(x)$ is an odd function formula \eqref{462653_5} above simplifies to
$$f(x)\approx \lim_{N, f\to\infty}\left(i \sum\limits_{n=1}^N \frac{\mu(2 n-1)}{2 n-1} \\\left(\sum\limits_{k=1}^{2 f (2 n-1)} (-1)^k\, \cos\left(\frac{\pi k}{2 n-1}\right)\, F\left(\frac{k}{4 n-2}\right)\, \sin\left(\frac{\pi k x}{2 n-1}\right)\\-\frac{1}{4} \sum\limits_{k=1}^{4 f (2 n-1)} (-1)^k\, F\left(\frac{k}{8 n-4}\right)\, \sin\left(\frac{\pi k x}{4 n-2}\right)\right)\right),\quad f(-x)=-f(x).\tag{12}\label{462653_12}$$
Question: Can the Fourier inversion theorem be stated as follows? If the statement below is too broad, is there a way to qualify it with a set of conditions on $f(x)$ and/or $F(\omega)$ where it holds?
A function $f(x)$ can be recovered from its Fourier transform $F(\omega)$ if-and-only-if the nested Fourier series representation of $f(x)$ converges for $x\in\mathbb{R}$.
The statement above is not intended to preclude convergence over a wider range such as $\Im(x)<\alpha$ when $f(x)$ is a meromorphic function with poles at $x=\pm i \alpha$ for example, or even $x\in\mathbb{C}$ in the case of complex analytic functions. In general I believe when $f(x)$ is recoverable from $F(\omega)$, the nested Fourier series representation for $f(x)$ converges under the same conditions for which the inverse Fourier transform integral defined in formula \eqref{462653_7} above converges even though the Fourier transform $F(\omega)=\mathcal{F}_x[f(x)](\omega)$ and inverse Fourier transform $f(x)=\mathcal{F}_{\omega}^{-1}[F(\omega)](x)$ are generally only considered to be valid for $\omega, x\in\mathbb{R}$.
Table (1) below contains a list of example functions whose nested Fourier series representations seem to converge consistent with the conditions described in the previous paragraph. $F(x)$ in row 8 of Table (1) below is the Dawson function.
Table (1): Examples of functions $f(x)$ recoverable from their Fourier transform
$$\begin{array}{cccccc} \# & f(x) & F(\omega ) & \text{Convergence} & \int_{-\infty }^{\infty } | f(x)| \, dx & \int_{-\infty }^{\infty } | F(\omega )| \, d\omega \\ 1 & e^{-x^2} & \sqrt{\pi } e^{-\pi ^2 \omega ^2} & x\in \mathbb{C} & \sqrt{\pi } & 1 \\ 2 & -2 x e^{-x^2} & 2 i \pi ^{3/2} e^{-\pi ^2 \omega ^2} \omega & x\in \mathbb{C} & 2 & \frac{2}{\sqrt{\pi }} \\ 3 & e^{-\frac{x^2}{2}} \left(4 x^2-2\right) & 2 \sqrt{2 \pi } e^{-2 \pi ^2 \omega ^2} \left(1-8 \pi ^2 \omega ^2\right) & x\in \mathbb{C} & 2 \sqrt{2} \left(\sqrt{\pi } \left(1-2 \erf\left(\frac{1}{2}\right)\right)+\frac{4}{\sqrt[4]{e}}\right) & 2-4 \erf\left(\frac{1}{2}\right)+\frac{8}{\sqrt[4]{e} \sqrt{\pi }} \\ 4 & \sech(x) & \pi \sech\left(\pi ^2 \omega \right) & -\frac{\pi }{2}<\Im(x)<\frac{\pi }{2} & \pi & 1 \\ 5 & \frac{1}{x^2+1} & \pi e^{-2 \pi | \omega | } & -1<\Im(x)<1 & \pi & 1 \\ 6 & e^{-| x| } & \frac{2}{4 \pi ^2 \omega ^2+1} & x\in \mathbb{R} & 2 & 1 \\ 7 & 2 \sinc(2 \pi x) & \frac{\sgn(1-\omega )+\sgn(\omega +1)}{2} & x\in \mathbb{C} & \infty & 2 \\ 8 & F(x) & -\frac{1}{2} i \pi e^{-\pi ^2 \omega ^2} \sgn(\omega ) & x\in \mathbb{C} & \infty & \frac{\sqrt{\pi }}{2} \\ \end{array}$$
The remainder of this question illustrates some of the nested Fourier series representations of $f(x)$ for some of the example functions in Table (1) above.
Figures (1) to (3) below illustrate formula \eqref{462653_11} for $f(x)=e^{-x^2}$ in orange overlaid on the blue reference function $f(x)$ where $F(\omega)=\sqrt{\pi} e^{-\pi^2 \omega^2}$ and formula \eqref{462653_11} is evaluated at $f=4$ and $N=10$.
Figure (1): Illustration of formula \eqref{462653_11} for $f(x)=e^{-x^2}$ in orange overlaid on the blue reference function $f(x)$
Figure (2): Illustration of real part of formula \eqref{462653_11} for $f(1+i t)$ in orange overlaid on the blue reference function $\Re(f(1+i t))$ where $f(x)=e^{-x^2}$
Figure (3): Illustration of imaginary part of formula \eqref{462653_11} for $f(1+i t)$ in orange overlaid on the blue reference function $\Im(f(1+i t))$ where $f(x)=e^{-x^2}$
Figure (4) below illustrates formula \eqref{462653_12} for $f(x)=-2 x e^{-x^2}$ in orange overlaid on the blue reference function $f(x)$ where $F(\omega)=2 i \pi^{3/2} \omega e^{-\pi ^2 \omega ^2}$ and formula \eqref{462653_12} is evaluated at $f=4$ and $N=10$.
Figure (4): Illustration of formula \eqref{462653_12} for $f(x)=-2 x e^{-x^2}$ in orange overlaid on the blue reference function $f(x)$
Figure (5) below illustrates formula \eqref{462653_11} for $f(x)=e^{-|x|}$ in orange overlaid on the blue reference function $f(x)$ where $F(\omega)=\frac{2}{4 \pi^2 \omega^2+1}$ and formula \eqref{462653_11} is evaluated at $f=4$ and $N=10$.
Figure (5): Illustration of formula \eqref{462653_11} for $f(x)=e^{-|x|}$ in orange overlaid on the blue reference function $f(x)$
\underset{\substack{N, f\to\infty \\ M(N)=0}}{\text{lim}}. I'm not sure why you'd want to use this when the presumably more natural construction\lim_{\substack{N, f \to \infty \\ M(N) = 0}}does the same thing. Perhaps you are concerned that the limit might be off to the side? As you know (because you did it with\int), you can insist that the limit appear below, even in text mode, using\lim\limits: $\lim\limits_{\substack{N, f \to \infty \\ M(N) = 0}}$. … $\endgroup$\tags with\labels, which can then be\eqrefd to get a clickable link, like \eqref{462653_1}. Since these links are just to anchors, they even work across pages; for example, if you need to refer to \eqref{462653_1} in another question, you can link to mathoverflow.net/questions/462653#mjx-eqn-462653_1. I edited accordingly. $\endgroup$