Positive definite functions on an abelian group

Let $G$ be an abelian group written additively (Group and abelian group) and let $\varphi :G\to \mathbb{C}$ be a function (The complex numbers as $\mathbb{R}[x]\mathrm{/}(x^2+1)$, with the real embedding and imaginary unit $i$). Then $\varphi$ is positive definite when for every integer $n\ge 0$, every finite family $x_1,\dots ,x_n\in G$ and all coefficients $c_1,\dots ,c_n\in \mathbb{C}$ one has ${\sum }_{j=1}^n{\sum }_{k=1}^n{c}_j\overline{c_k}\varphi (x_j-x_k)\ge 0.$ The sum for $n=0$ is the empty sum $0$, so the convention covers it; repeated points $x_j=x_k$ are allowed, so the finite matrices tested are the Hermitian matrices $[\varphi (x_j-x_k){]}_{j,k}$. No continuity, boundedness or measurability is part of the definition.

Elementary consequences. The claims below are immediate from the defining inequality and are recorded here for later use. Taking $n=1$, $x_1=0$ and $c_1=1$ gives $\varphi (0)\ge 0$. Taking $n=2$, $x_1=0$, $x_2=x$ and $c_1=1$, $c_2=t$ gives $(1+\mathrm{\mid }t{\mathrm{\mid }}^2)\varphi (0)+\bar{t}\varphi (-x)+t\varphi (x)\ge 0\text{for every }t\in \mathbb{C}\mathrm{.}$ The left side is real and equal to its own conjugate for every $t$, so comparing coefficients at $t=1$ and $t=i$ gives $\varphi (-x)=\overline{\varphi (x)}$ (Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive); hence every tested matrix is Hermitian. If $\varphi (x)\mathrm{\ne }0$, choose $t=-\overline{\varphi (x)}\mathrm{/}\mathrm{\mid }\varphi (x)\mathrm{\mid }$, so $\mathrm{\mid }t\mathrm{\mid }=1$ and $t\varphi (x)=-\mathrm{\mid }\varphi (x)\mathrm{\mid }$. The inequality becomes $2\varphi (0)-2\mathrm{\mid }\varphi (x)\mathrm{\mid }\ge 0$, hence $\mathrm{\mid }\varphi (x)\mathrm{\mid }\le \varphi (0)$; if $\varphi (x)=0$, the same bound follows from $\varphi (0)\ge 0$. Thus a positive definite $\varphi$ satisfies $\varphi (0)\ge 0$, $\varphi (x)=\overline{\varphi (-x)}$ and $\mathrm{\mid }\varphi (x)\mathrm{\mid }\le \varphi (0)$ for all $x\in G$.