Reading Stein’s “Singular integrals and differentiability properties of functions” I came across the following statement (this is in the proof of Lemma 3.2, pages 133-134):
We now invoke the $n$-dimensional version of Wiener’s theorem, to wit: If $\Phi_1 \in L^1(\mathbb{R}^n)$ and $\widehat{\Phi_1} + 1$ is nowhere $0$, then there exists a $\Phi_2 \in L^1(\mathbb{R}^n)$ so that $(\widehat{\Phi_1}(x)+1)^{-1} = \widehat{\Phi_2}(x) + 1.$
Stein, however, doesn’t give any reference other than just stating that this is Wiener’s theorem. Is there any reference where I could learn the proof of this theorem? I have searched and I know there is a similar Wiener theorem, but this works only on the torus. I guess the difficulty in extending Wiener’s theorem for the torus to the whole $\mathbb{R}^n$ is the lack of an identity with respect to the convolution, and maybe that’s why we have to add the $+1$s, but I am not sure how to make this precise.
Also, I can prove this if I assume that $\|\Phi_1\|_{L^1(\mathbb{R}^n)} < 1$ by expanding $(1+ \widehat{\Phi_1})^{-1}$ in a series, but I don’t see how this could be bootstrapped to just the requirement of $1 + \widehat{\Phi_1} \neq 0$ (note that this is immediately verified if $\|\Phi_1\|_{L^1(\mathbb{R}^n)} < 1$).
The proof below is adapted from the proof of a `division lemma’
in this paper of Coquand and Stolzenberg:
Given $f=\hat\Phi_1\in \mathcal B$ such that $f+1$ does not vanish,
we wish to show that with
$$
g = \frac{1}{1+f}-1 = \frac{-f}{1+f}.
$$
then $g\in\mathcal B$, meaning $g=\hat\Phi_2$ for some $\Phi_2\in L^1$.
The idea is to approximate $f$
in the denominator by a nice function $h$,
complexify and construct a Cauchy integral
of the form
$$
G = \frac{1}{2\pi i} \int_{\Gamma}
\frac{-f}{1+z+h} \, \frac{1}{z+h-f} \,dz
$$
that converges in the space $\mathcal B$,
where the contour $\Gamma$ loops once around a small circle $|z|=r$.
Since convergence
in $\mathcal B$ implies uniform convergence, evaluation at any point
$x\in\mathbb R^n$ will yield (by the Cauchy integral formula)
the value of $f(x)/(1+z+h(x))$ at the point $z$ where $z+h(x)=f(x)$,
which means $G(x)=g(x)$.
We are given $f=\hat F$ for $F$ in $L^1$, and $r=\inf |1+f|/4>0$.
We can find a smooth function $H$, of compact support say,
such that $\|F-H\|_{L^1}< r/4$. With $h=\hat H$, we have
$\|f-h\|_{\mathcal B}<r/4$, so
$$
\inf_{x\in R^n} |1+z+h(x)| \ge
\inf |1+f| – |z| – \|h-f\|_{\infty} \ge r
$$
whenever $z$ is a complex number with $|z|\le 2r$. Considering the map
$$
z\mapsto \frac{h}{1+z+h},
$$
for fixed $z$ one gets a function that
is smooth and rapidly decaying, and hence in $\mathcal B$, and indeed
the map is continuous in $z$ for $|z|\le r$
with values in $\mathcal B$.
Therefore the same is true for the map
$$
z\mapsto \frac{f}{1+z+h} = \frac1{1+z}\left( f – \frac{fh}{1+z+h}\right).
$$
The function
$$
z\mapsto \frac1{z+h-f} = \frac1z \sum_{k=0}^\infty \left(\frac{f-h}z\right)^k
$$
is represented by a series that converges in $\mathcal B$ for $|z|>r/2$ since
$\|f-h\|_{\mathcal B}<r/4$.
Since $\mathcal B$ is a Banach algebra, the integrand of
$G$ above is continuous in $z$ with values in $\mathcal B$, so the
integral $G$ converges in $\mathcal B$.
Evaluating $G$ at any fixed $x$ in $R^n$, the integral for $G(x)$
converges, and the factor
$$
\frac{-f(x)}{1+z+h(x)}
$$
is analytic in $z$ for $|z|\le r$. By the Cauchy integral formula, it follows
$G(x)=g(x)$, as claimed. Hence $g$ is in $\mathcal B$.