$f \in L^1(G) \cap B(G) \implies \hat f \in L^1(\hat G) $ $G$ be a locally compact abelian group. $\hat G$ denotes the group of characters on $G$.
$M(\hat G)$ be the space of regular complex Borel measure on $\hat G$.
And $B(G):=\{f:G \to \mathbb C|\exists \mu \in M(\hat G) \text{ such that } f(x)=\int_{\hat G} \chi(x)\mathrm d\mu(\chi)\}$.
Let $f\in L^1(G) \cap B(G).$
$\hat f:\hat G \to \mathbb C$ is the Fourier transform of $f$ .
Where $$\hat f (\chi)=\int_G f(x)\chi (x^{-1})\mathrm dx$$
$\mathrm dx$ is a haar measure on $G$.
How do I show that $\hat f \in L^1(G)$?
I tried taking modulus but that might have worked if I had $G $ to be a compact group.
 A: *

*I hope this will help you*

Consider the mapping of the Fourier transform F : C
∗
0
(G) → C
∗
0
(Gb) that
takes f → ˆf. We show that this is an isometric isomorphism between the spaces.
Define the inverse map to be Fb : C
∗
0
(Gb) → C
∗
0
(G) where Fb(ψ)(x) = ψˆ(δx−1 ).
Let
B be the space of f ∈ C
∗
0
(G) such that ˆf ∈ C
∗
0
(Gb).
proof, we have that Fb ◦ F(f) = f. Since kfkG = k
ˆˆfk
G
bb
, we have by our definition
of norm on C
∗
0
(Gb) that kfk
∗
0 = kF(f)k
∗
0
. But Cc(Gb) = F(G) is dense in C
∗
0
(Gb) as
proven in the last section, so that F is a surjective isometry from the closure B to
C
∗
0
(Gb). Similarly, the Fourier transform on C
∗
0
(Gb) is dense in C
∗
0
(
bGb), so that by
applying the inverse transform and Pontryagin duality, we have that Fb(C
∗
0
(Gb)) is
dense in C
∗
0
(G). Then F is indeed an isometric isomorphism.
For f that meet the criteria of the formula, ˆf is vanishing and thus ˆf ∈ C
∗
0
(G).
From above, we can find a g ∈ C
∗
0
(Gb) such that ˆg = ˆf and g(x) = ˆf(δx−1 ) for all
x ∈ G. But the Fourier transform is injective, so that f(x) = g(x) = ˆˆf(δx−1 ).
