A character is positive definite

(Rank-one positivity.) For every finite family $x_1,\dots ,x_n\in G$ and coefficients $c_1,\dots ,c_n\in \mathbb{C}$, put $u_j:=\gamma (x_j)$. Since $\gamma$ is a homomorphism into the unit circle, $\gamma (x_j-x_k)=\gamma (x_j)\gamma (-x_k)=\gamma (x_j)\overline{\gamma (x_k)}=u_j\overline{u_k}$; consequently ${\sum }_{j,k}{c}_j\overline{c_k}\gamma (x_j-x_k)={\sum }_{j,k}{c}_j\overline{c_k}{u}_j\overline{u_k}=\mid {\sum }_j{c}_j{u}_j{\mid }^2\ge 0.$ For $n\ge 1$ the vector $u$ is nonzero because every $u_j$ has modulus one, so the matrix $u{u}^{\ast }$ is positive semidefinite of rank one. For $n=0$ its rank and quadratic form are zero. Thus $\gamma$ is positive definite and $\gamma (0)=1$.

(The point mass represents the character.) Assume now that $G$ is locally compact Hausdorff abelian, so that Bochner's theorem applies. The point mass ${\delta }_{\gamma }$ at the point $\gamma \in \widehat{G}$ is a probability measure and, by [F3], a finite positive Radon measure on $\widehat{G}$. Its inverse (Fourier-Stieltjes) transform is ${\int }_{\widehat{G}}\eta (x)d{\delta }_{\gamma }(\eta )=\gamma (x)(x\in G),$ because the function $\eta \mapsto \eta (x)$ agrees with the constant $\gamma (x)$ off the ${\delta }_{\gamma }$-null set $\widehat{G}\setminus \{\gamma \}$ (evaluation formula of [F3]; the coordinate functions are measurable by joint continuity). By [F4] this transform is continuous and positive definite, so the computation identifies $\gamma$ as the Fourier-Stieltjes transform of the finite positive Radon measure ${\delta }_{\gamma }$; by uniqueness in Bochner's theorem [F2] the point mass is the representing measure of $\gamma$, and its total mass is ${\delta }_{\gamma }(\widehat{G})=1=\gamma (0)$.

Step 1.1 proves that a character is positive definite with $\gamma (0)=1$ and exhibits its rank-one nonempty test matrices and rank-zero empty matrix; step 1.2 identifies the representing probability measure as the point mass ${\delta }_{\gamma }$.