Let $G$ be a locally compact abelian (LCA) group.
A complex function $\gamma$ on a LCA group $G$ is called a character of $G$ if $|\gamma(x)| = 1$ for all $x \in G$ and satisfies $$\gamma(x+y) = \gamma(x) \gamma(y).$$ $\Gamma$ is the set of all continuous characters of $G$, called the dual group of $G,$ with operation $+$ given by $$(\gamma_1 + \gamma_2)(x) = \gamma_1(x) \gamma_2(x)$$
Denote $(x,\gamma) = \gamma(x).$
Recall that convolution of $f$ and $g$ is defined as $$(f*g)(x) = \int_G f(x-y)g(y) dy.$$
The following is extracted from 'Fourier Analysis on Groups' by Rudin, page $7:$
(Theorem $1.2.2$) If $\gamma \in \Gamma$ and if
$$\hat{f}(\gamma) = \int_G f(x) (-x,\gamma) dx \,\,\, (f \in L^1(G)).$$ Then the map $f \mapsto \hat{f}(\gamma)$ is a complex homomorphism of $L^1(G),$ and is not identically $0.$ Conversely, every non-zero complex homomorphism of $L^1(G)$ is obtained in this way, and distinct character induce distinct homomorphisms.
Proof (partial):
Suppose $f,g\in L^1(G)$ and $k = f*g.$ Then $$\hat{k}(\gamma) = \int_G (f*g)(x) (-x,\gamma) dx = \int_G (-x,\gamma)dx \int_G f(x-y)g(y)dy $$ $$= \int_G g(y) (-y,\gamma)dy \int_G f(x-y)(-x+y,\gamma)dx = \hat{g}(\gamma)\hat{f}(\gamma).$$
Question: How to obtain second and third equalities, that is, $$\int_G (f*g)(x) (-x,\gamma) dx \stackrel{1}{=} \int_G (-x,\gamma)dx \int_G f(x-y)g(y)dy $$ $$\stackrel{2}{=} \int_G g(y) (-y,\gamma)dy \int_G f(x-y)(-x+y,\gamma)dx?$$
I have no idea how to obtain $1$ and $2.$ Any hint would be appreciated.