For a group $G$, a positive definite function $f:G \rightarrow \mathbb{C}$ is one that satisfies $$\sum_{i,j=1}^mf(g_ig_j^{-1})c_i\overline{c_j}\geq 0$$
For any $g_1,...,g_m \in G, c_1,...,c_m \in \mathbb{C}$. How can I show that such a function satisfies $f(e_G)\geq |f(g)|$ for all $g\in G$?
Taking $m=1$, $g_1=e_G$, $c_1=1$ shows $f(e)$ is real and nonnegative. Taking $m=2$, $g_1= e,g_2= g, c_1=a, c_2=b $ gives
$$f(e)(|a|^2+|b|^2) + f(g)a\overline{b} +f(g^{-1})\overline{a}b\geq 0.$$ The book I’m reading (Wolf, Harmonic Analysis on Commutative Spaces, $\S$ 8.4 ) claims that taking $a=b=1$ proves that $f(e_G)\geq |f(g)|$ and $f(g^{-1})= \overline{f(g)}$. Indeed that choice does seem to show that the imaginary parts of $f(g^{-1}), f(g)$ differ by a minus sign, but one needs to take $a=1, b=i$ to show their real parts are the same. And I have no idea how that choice would imply $f(e_G)\geq |f(g)|$.
Not "semipositive"? Anyway, once you have $f(g)=\overline{f(g^{-1})},$ set $a=1$ and $b=-f(g)/|f(g)|$ to get
$$0\leq 2f(e)-f(g)\overline{f(g)}/|f(g)| - f(g^{-1}){f(g)}/|f(g)|=2(f(e)-|f(g)|).$$
A different way to think about it is that a $2\times 2$ matrix with constant diagonal $$\begin{pmatrix}f(e)&f(g)\\f(g^{-1})&f(e)\end{pmatrix}$$ is (semi)positive definite (in the sense of $x^*Mx$ real and positive/non-negative) if and only if it is Hermitian and (weakly) diagonally dominant.