Identity for all uniformly bdd functions follows "by basic Fourier (transform) analysis"? I found an interesting identity in a paper in a paper I am reading (page $9$). The statement is as follows:
$\mathcal F$ is the Fourier transform, $P$ the law of a random variable $X$ and $\varphi$ is the characteristic function of $X$
$$
\int_{\mathbb R}f\ast\mathcal F^{-1}[\frac 1{\varphi}(-\bullet)](x)P(dx)=f(0)
$$ 
for any uniformly bounded function $f$. The statement should follow by basic Fourier analysis, I suspect maybe some kind of Plancherel. Somehow I feel we should utilize (the transform is this way defined to be identical to the characteristic function) $$
\mathcal F[P]=\varphi
$$ 
so that the characteristic functions can cancel out. however so far I couldn't make it work. Since it is supposed to hold for all 
uniformly bounded functions there could also a dirac point measure be involved, not sure. 
Has anyone got an idea?
An attempt (more of an heuristic), I can't justify each step:
Since we have $\mathcal F[P]=\varphi$ we have $P=\mathcal F^{-1}\varphi$ so we write
$$
\int_{\mathbb R}f\ast\mathcal F^{-1}[\frac 1{\varphi}(-\bullet)](x)P(dx)=\int_{\mathbb R}f\ast\mathcal F^{-1}[\frac 1{\varphi}(-\bullet)](x)\mathcal F^{-1}[\varphi](dx)
$$
we now apply Plancherel (the sign vanishes due to complex conjugation)
$$
\int_{\mathbb R}f\ast\mathcal F^{-1}[\frac 1{\varphi}(-\bullet)](x)\mathcal F^{-1}[\varphi](dx)=\frac 1{2\pi}\int_{\mathbb R}\mathcal F[f](u)\frac1{\varphi}(u)\varphi(u)du=\frac 1{2\pi}\int_{\mathbb R}\mathcal F[f](u)du
$$
Now this transfers into 
$$
\frac 1{2\pi}\int_{\mathbb R}\mathcal F[f](u)du=\frac 1{2\pi}\int_{\mathbb R}\Big(\int_{\mathbb R}e^{iux}f(x)dx\Big)du
$$
which becomes using Fubini (or anything which allows us to exchange the order of integration) and $\frac 1{2\pi}\int_{\mathbb R}e^{iux}du=\delta(x)$
$$
\int_{\mathbb R}\Big(\frac 1{2\pi}\int_{\mathbb R}e^{iux}du\Big )f(x))dx=\int_{\mathbb R}\delta(x)f(x)dx=f(0)
$$
 A: Rephrasing with the vocabulary of distributions (how do you define $\mathcal{F}^ {-1}[\frac{1}{\varphi}]$ without the distributions ?)


*

*If $T,S$ are two distributions such that $\mathcal{F}[T] \mathcal{F}[S]$ is a tempered 
distribution then we can define $T \ast S$ as the tempered distribution $\mathcal{F}^{-1}[\mathcal{F}[T] \mathcal{F}[S]]$. 

*If everything is well-defined in the sense of distributions then $\langle f \ast g, h \rangle = \langle f , g \ast h\rangle$

*If $f \in C^\infty_c$ and since $P(dx) = P'(x)dx$ so that $\varphi = \mathcal{F}[P']$ then 
$$f(0) =\langle f, \delta \rangle= \langle f, \mathcal{F}^{-1}[1]\rangle=\langle f, \mathcal{F}^{-1}[\frac{\mathcal{F}[P']}{\varphi}]\rangle = 
 \langle f , P'\ast \mathcal{F}[\frac{1}{\varphi}]\rangle=  \langle f \ast \mathcal{F}^{-1}[\frac{1}{\varphi}], P'\rangle\\ = \int_{-\infty}^\infty f \ast \mathcal{F}^{-1}[\frac{1}{\varphi}] P(dx)$$
Clearly $f \ast \mathcal{F}^{-1}[\frac{1}{\varphi}]$ has little chances to be well-defined if $f \not \in C^\infty_c$ so your statement is not correct as it is.
A: I do not think that you can apply Plancheral's theorem to functions, so it seems that the first equality does not really check out since writing $P(x) = \mathcal{F}^{-1}(dx)$ then applying Plancheral's does not really check out... but it seems that you have the right idea (which is more than I can say for myself not finding even the right idea).
You do, though, have that the second equation is true since we can just apply (1) in the proof of theorem 1 in http://math.mit.edu/~jerison/103/handouts/fourierint2.13.pdf , which says that if $f$ is a Schwartz function and $\mu$ is a finite measure, then
$$\int \hat{f}(x)d\mu(x) = \int f(x) \hat\mu(x) dx,$$
 which is basically what you are doing... I do not know if the theorem is still true though in the general case that you are looking at.
The last equation $$\frac{1}{2\pi}\int_{\mathbb{R}}\mathcal{F}f(\xi)d\xi = f(0)$$ since it is secretly just the Fourier inversion formula:
$$\frac{1}{2\pi}\int_{\mathbb{R}}\mathcal{F}f(\xi)e^{0\xi}d\xi = \mathcal {F}^{-1}\circ \mathcal{F}f(0),$$
which is why you got the dirac delta function going on since people sometimes (try to?) justify the theorem with dirac deltas.
