Diagonal unitary coefficients have positive type

For any complex-linear isometry $U:H\to H$, expanding the squared norms using [A4] gives $\mathrm{\parallel }x+y{\mathrm{\parallel }}^2-\mathrm{\parallel }x-y{\mathrm{\parallel }}^2=4\mathrm{Re}⟨x,y⟩$ and $\mathrm{\parallel }x+iy{\mathrm{\parallel }}^2-\mathrm{\parallel }x-iy{\mathrm{\parallel }}^2=4\mathrm{Im}⟨x,y⟩$. Since $U$ is linear and norm-preserving, the two differences are unchanged when $x,y$ are replaced by $Ux,Uy$. Their real and imaginary parts therefore agree, so $⟨Ux,Uy⟩=⟨x,y⟩$. In particular each $\pi (g)$ preserves the inner product.

Fix $n\ge 1$, elements $g_1,\dots ,g_n\in G$, and scalars $c_1,\dots ,c_n\in \mathbb{C}$, and put $v_i=\pi (g_i)\xi$. The homomorphism law and step 1.1 yield $\phi (g_i^{-1}{g}_j)=⟨\pi (g_i{)}^{-1}\pi (g_j)\xi ,\xi ⟩=⟨v_j,v_i⟩$. Hence the positive-semidefinite quadratic form from [A1] is ${\sum }_{i,j=1}^n\overline{c_i}{c}_j\phi (g_i^{-1}{g}_j)={\parallel {\sum }_{j=1}^n{c}_j{v}_j\parallel }^2\ge 0$. The calculation includes repeated group elements, zero coefficients, and $\xi =0$; when $\xi =0$ the value is zero.

By [A2], $\phi =c_{\xi ,\xi }$ is continuous. Step 2.1 proves its positive-type matrix test for every allowed finite list, so [A1] gives that $\phi$ is of positive type. The homomorphism law implies $\pi (e)=I$; therefore $\phi (e)=⟨\xi ,\xi ⟩=\mathrm{\parallel }\xi {\mathrm{\parallel }}^2$, including when $\xi =0$.