Identifying the dual group of a locally compact abelian group with the spectrum of $ {L^{1}}(G) $. Folland stated the following theorem in his book A Course in Abstract Harmonic Analysis on Page 88.

The dual group of a locally compact abelian group can be identified with the spectrum of $ {L^{1}}(G) $.

It came without a proof, so could anyone kindly provide me with some comments on my attempt below?
Proof: Let $ G $ be a locally compact abelian group and $ \widehat{G} $ the dual group of $ G $, i.e., $ \widehat{G} \stackrel{\text{def}}{=} \text{Hom}(G,\mathbb{T}) $, where $ \mathbb{T} \stackrel{\text{def}}{=} \{ z \in \mathbb{C}: |z| = 1 \} $. The spectrum of $ {L^{1}}(G) $, denoted by $ \text{Spec}({L^{1}}(G)) $, is defined as
$$
\text{Spec}({L^{1}}(G))
\stackrel{\text{def}}{=}
\{
F \in {L^{1}}(G)^{\star} \mid
F \not\equiv 0 ~ \text{and} ~ F(f * g) = F(f) \cdot F(g)
\},
$$
where $ * $ denotes convolution. In other words, $ \text{Spec}({L^{1}}(G)) $ is the set of all non-zero multiplicative linear functionals on $ {L^{1}}(G) $.
To show that $ \widehat{G} $ can be identified with $ \text{Spec}({L^{1}}(G)) $, we need to construct a $ 1 $-$ 1 $ correspondence $ \widehat{G} \to \text{Spec}({L^{1}}(G)) $. Define such a correspondence as follows. Every $ \beta \in \widehat{G} $ defines a multiplicative linear functional $ \hat{\beta} $ on $ {L^{1}}(G) $ by
$$
\forall f \in {L^{1}}(G): \quad
\hat{\beta}(f)
\stackrel{\text{def}}{=}
\int_{G} f(g) \beta(g) ~ \mathrm{d}{g}. \quad \blacksquare
$$
Question: Did I define the $ 1 $-$ 1 $ correspondence correctly?
 A: By $-g$ I am assuming you mean $g^{-1}$. Let $G$ is a locally compact group, $\mathcal{H}$ is a Hilbert space and $\rho:G\rightarrow \mathcal{U}(\mathcal{H})$ is a (unitary) representation. Traditionally when one lifts $\rho$ to a representation $\tilde{\rho}:L^1(G)\rightarrow B(H)$, the standard definition is
$$\tilde{\rho}(f) = \int_G f(g)\rho(g)\,dg.$$
This definition is understood weakly. Moreover can be motivated somewhat easily from the setting of finite groups. In your case, $\rho$ is an element of the dual group $\widehat{G}$, $\mathcal{H} = \mathbb{C}$ and $\mathcal{U}(\mathcal{H}) = \mathbb{T}$ since the only unitary elements on $\mathbb{C}$ are toroidal elements. It is not incorrect to consider $\rho(g^{-1})$ instead of $\rho(g)$ in the integral but traditionally we do not. Philosophically the reason is probably to keep expressions as simple as possible without complicating them unnecessarily.
This is definitely a homomorphism on $L^1(G)$ since
\begin{eqnarray}
\tilde{\rho}(f_1\ast f_2) &=& \int_G (f_1\ast f_2)(g)\rho(g)\,dg \\
&=& \int_G \left(\int_G f_1(g')f_2(g'^{-1}g)\,dg'\right)\rho(g)\,dg \\
&=& \int_G \int_G f_1(g')f_2(g'^{-1}g)\,dg'\rho(g')\rho(g'^{-1}g)\,dg \\
&\stackrel{\text{Fubini}}{=}& \int_G f_1(g')\rho(g')\int_G f_2(g'^{-1}g)\rho(g'^{-1}g)\,dg\,dg' \\
&\stackrel{\text{left invariance}}{=}& \int_G f_1(g')\rho(g')\,dg'\int_Gf_2(g)\rho(g)\,dg\\
&=& \tilde{\rho}(f_1)\tilde{\rho}(f_2).
\end{eqnarray}
Note that you have not shown one-to-one correspondence with this relation. You have merely shown that every (continuous) character gives rise to an element in $\Delta(L^1(G))$ (the spectrum of $L^1(G)$).
To show the other direction takes much more work. Consider a multiplicative linear functional $\Phi$ on $L^1(G)$, i.e. an element of $\Delta(L^1(G))$. Then by the Riesz representation theorem, $\Phi$ is given by an integral against some $\phi\in L^{\infty}(G)$, i.e.
$$\Phi(f) = \int_G \phi(g)f(g)\,dg.$$
However since $\Phi$ is multiplicative and nonzero, there is $f_1\in L^1(G)$ so that $\Phi(f_1)\neq 0$ and for all $f_2\in L^1(G)$,
\begin{eqnarray}
\Phi(f_1\ast f_2) &=& \Phi(f_1)\Phi(f_2) \\
&=& \int_G \Phi(f_1)\phi(g)f_2(g)\,dg. \tag{1}
\end{eqnarray}
Furthermore,
\begin{eqnarray}
\Phi(f_1\ast f_2) &=& \int_G \phi(g')(f_1\ast f_2)(g')\,dg' \\
&=& \int_G \phi(g')\left(\int_G f_1(g^{-1}g')f_2(g)\,dg\right)\,dg' \\
&\stackrel{\text{Fubini}}{=}&\int_G\left(\int_G\phi(g')f_1(g^{-1}g')\,dg'\right)f_2(g)\,dg \\
&=& \int_G \Phi(L_gf_1)f_2(g)\,dg, \tag{2}
\end{eqnarray}
where $L_gf_1(g') = f_1(g^{-1}g')$. Note the appearance of the inverse. (I loathe Folland's choice for convolution as I think it makes things a little bit more complicated than they need to be.) $(1)$ and $(2)$ are equal for all $f_2\in L^1(G)$ and so we must conclude that
$$\phi(g)\Phi(f_1)=\Phi(L_gf_1).\tag{3}$$
$\Phi$ is a continuous linear functional and so if $g_{\lambda}$ is a net converging to $g$, then $\Phi(L_{g_{\lambda}}f_1)\rightarrow \Phi(L_gf_1)$. Hence $\phi$ agrees almost everywhere with a continuous function and so we may as well assume $\phi$ is continuous.
All that is left to see is that $\phi$ is a homomorphism from $G$ to $\mathbb{C}$ and that $|\phi(g)| = 1$ for all $g\in G$. From the above equality, we know that
\begin{eqnarray}
\phi(gg')\Phi(f_1) &=& \Phi(L_{gg'}f_1) \\
&=& \Phi(L_{g'}L_gf_1) \\
&\stackrel{(3)}{=}& \phi(g')\Phi(L_gf_1) \\
&\stackrel{(3)}{=}& \phi(g')\phi(g)\Phi(f_1)
\end{eqnarray}
Since $\Phi(f_1)$ is nonzero by our above hypothesis, we must conclude that $\phi(gg') = \phi(g')\phi(g)$. However $\phi(g)$ is just a number so in fact $\phi(gg') = \phi(g)\phi(g')$ and therefore $\phi$ is a homomorphism onto $\mathbb{C}$.
To see that $|\phi(g)|= 1$, first note that if $|\phi(g)|\le 1$ for all $g$, then $|\phi(g^{-1})|\le 1$ as well. However this is only possible if $|\phi(g)=1$. So if we can show that $|\phi(g)|=1$, we would be done. This is easily achieved by noting that $\phi(g^n)  = \phi(g)^n$. Since $\phi\in L^{\infty}(G)$, $\phi$ must be bounded but the only way for this to happen is if $|\phi(g)|\le 1$ since otherwise $\phi(g^n)$ would diverge to infinity. Thus every multiplicative linear functional on $L^1(G)$, i.e. every element of $\Delta(L^1(G))$, corresponds to a character. Therefore there is a one-to-one correspondence between the dual group and $\Delta(L^1(G))$.
