Wiener's theorem in $\mathbb{R}^n$

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 $\|\Phi_1\|_{L^1(\mathbb{R}^n)} < 1$).

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The proof below is adapted from the proof of a `division lemma’
in this paper of Coquand and Stolzenberg:

  1. Consider the Banach algebra $\mathcal B$ consisting of all Fourier
    transforms of functions in $L^1(\mathbb R^n)$, where the norm of $f=\hat F$ is
    $$
    \| f\|_{\mathcal B} = \|F\|_{L^1}.
    $$
    The space $\mathcal B$ contains all functions $g$ that are
    sufficiently smooth with sufficiently rapid decay, which ensures the
    inverse Fourier transform $\check g$
    is in $L^1$. Note also that the $\mathcal B$ norm dominates the supremum
    norm: $\|f\|_{\mathcal B} \ge \|f\|_{\infty}$.

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 $G(x)=g(x)$.

  1. 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
    $$
    z\mapsto \frac{f}{1+z+h} = \frac1{1+z}\left( f – \frac{fh}{1+z+h}\right).
    $$

  2. 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
    $\|f-h\|_{\mathcal B}<r/4$.

  3. 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 $\mathcal B$.