When does Inverse Fourier transform look close to a positive definite function? Let $G$ be a commutative locally compact group, and $\hat{G}$ be its dual group, consisting of all continuous characters (continuous homomorphisms from $G$ to the circle group $\mathbb{T}$) . I can see using Bochner's Theorem that for any function $\phi\in C_c(\hat{G})$, the inverse Fourier transform  $(|\phi|)^\check{}$ of $|\phi|$ is a positive definite function.
My question is: Given any function $\phi\in C_c(\hat{G})$, under what circumstances can we say that $\check{\phi}$ is a constant multiple of a positive definite function?
My main problem is that I am not being able to relate $(|\phi|)^\check{}$ and $| \check{\phi}|$ in a convenient manner.
Thank you. Any comments/discussions are welcome!
 A: Theorem.  For every $\phi $ in $C_c(\hat G)$,  one  has that $\check \phi $ is a positive definite function if and only if
$$
  \phi (x)\geq 0,  \quad\forall x\in \hat G.
  $$
Proof.  I understand that the OP already knows how to prove the implication "$\Leftarrow$" (because then $\phi =|\phi |$), so I
will only bother to prove "$\Rightarrow$".
We therefore pick $\phi $ in $C_c(\hat G)$, and assume that $\check \phi $ is a positive definite function.
Denoting the duality of  $G$ and $\hat G$ by
$$
  (t, x)\in \hat G\times G\mapsto  \langle t, x\rangle \in {\mathbb C},
  $$
we have for all $x_1, x_2, \ldots , x_n\in G$,  and all $a_1, a_2, \ldots , a_n\in {\mathbb C}$, that
$$
  0 \leq
  \sum_{i, j=1}^n \bar a_i a_j \check \phi (x_i^{-1}x_j) =
  \sum_{i, j=1}^n \bar a_i a_j \int_G\langle t, x_i^{-1}x_j\rangle  \phi (t)\, dt = $$$$ =
  \int_G\sum_{i, j=1}^n \bar a_i a_j \overline{\langle t, x_i\rangle }\langle t, x_j\rangle  \phi (t)\, dt =
  \int_G\Big |\sum_{j=1}^n a_j\langle t, x_j\rangle \Big |^2 \phi (t)\, dt =$$$$=
  \int_K|p(t)|^2\phi (t)\, dt,
  $$
where $K$ is the compact support of $\phi $, and $p(t)=\sum_{j=1}^n a_j\langle t, x_j\rangle $.
Observe that the collection of functions $p$ that one obtains by letting   the $x_j$ range in $\hat G$,  and the $a_j$  range
in ${\mathbb C}$,  gives a subalgebra of $C(K)$,  containing the constant functions and separating the points of $K$.   By
Stone-Weierstrass,  this algebra is uniformly dense in $C(K)$,  whence the set of functions of the form $|p(\cdot)|^2$ is
uniformly dense in
the positive cone of $C(K)$.
The computation above then implies that
$$
  \int_K f(t) \phi (t)\, dt\geq 0, \qquad (*)
  $$
for every nonnegative continuous function  on $K$.
We next claim that  $\phi $ is a nonnegative
function.  To see this, assume by contradiction that there exists $t_0$ in $\hat G$ such that $\phi (t_0)$ lies outside the real
interval $[0, +\infty )$.    Since this interval is a closed subset of
${\mathbb C}$, there exists $\varepsilon >0$, such that the open ball $B_\varepsilon (\phi (t_0))$ has empty intersection with $[0, +\infty )$.
Set
$$
  U= \{t\in  \hat G: |\phi (t)-\phi (t_0)|<\varepsilon /2\},
  $$
so $U$ is an open subset of $\hat G$, contained in $K$.
Pick any nonnegative function $f$ in $C(K)$, vanishing on $K\setminus U$, and such $f(t_0)>0$.  Since Haar measure has
full support,  we have that $\int_Kf(t)\, dt >0$, and if we multiply $f$ by a suitable positive number, we may assume
that
$$
  \int_Kf(t)\, dt =1.
  $$
We then have
$$
  \Big |\phi (t_0) - \int_K f(t)\phi (t)\, dt\Big | =
  \Big |\int_K f(t)(\phi (t_0)-\phi (t))\, dt\Big | \leq  $$$$ \leq 
  \int_U f(t)|\phi (t_0)-\phi (t)|\, dt \leq  \varepsilon /2<\varepsilon .
  $$
This shows that
$$
  \int_K f(t)\phi (t)\, dt \in   B_\varepsilon (\phi (t_0))\cap [0, +\infty ),
  $$
a contradiction.
QED
Using the Theorem it is now clear that $\check \phi $ is a constant multiple of a positive definite function if and only
if $\phi $ is a constant multiple of a (pointwise) nonnegative function.
