Representative functions form a self-adjoint translation-invariant algebra

The trivial representation $1_K$ is a finite-dimensional continuous unitary representation whose matrix coefficient at a unit vector is the constant function $1$, so every constant function lies in $R(K)$ by [F1]; $R(K)$ is closed under pointwise addition and scalar multiplication because it is by definition a linear span of coefficient functions, and pointwise addition of representatives computed at each $k$ agrees with the sum in the function space of [F3].

If $f=c_{v,w}^{\pi }$ and $f^{\mathrm{\prime }}=c_{v^{\mathrm{\prime }},w^{\mathrm{\prime }}}^{\sigma }$ are matrix coefficients of finite-dimensional continuous unitary representations, then [F2] gives $f(k)f^{\mathrm{\prime }}(k)=c_{v,w}^{\pi }(k)c_{v^{\mathrm{\prime }},w^{\mathrm{\prime }}}^{\sigma }(k)=c_{v\otimes v^{\mathrm{\prime }},w\otimes w^{\mathrm{\prime }}}^{\pi \otimes \sigma }(k)$ for every $k$, a matrix coefficient of the finite-dimensional continuous unitary representation $\pi \otimes \sigma$, hence an element of $R(K)$; general products in $R(K)$ follow by bilinear expansion of finite linear combinations of coefficients, so $R(K)$ is closed under pointwise multiplication.

If $f=c_{v,w}^{\pi }$, then [F2] exhibits $\bar{f}$ as a matrix coefficient of a finite-dimensional continuous unitary representation of $K$, hence $\bar{f}\in R(K)$; since conjugation is additive and conjugate-linear, $\bar{f}\in R(K)$ for every finite linear combination $f$ of matrix coefficients, so $R(K)$ is closed under complex conjugation.

If $f=c_{v,w}^{\pi }$ and $g\in K$, then for every $k\in K$ the unitarity [F4] and the homomorphism property give $f(g^{-1}k)=⟨\pi (g{)}^{-1}\pi (k)v,w⟩=⟨\pi (k)v,\pi (g)w⟩=c_{v,\pi (g)w}^{\pi }(k)$ and $f(kg)=⟨\pi (k)\pi (g)v,w⟩=c_{\pi (g)v,w}^{\pi }(k)$, so the left translate $k\mapsto f(g^{-1}k)$ and the right translate $k\mapsto f(kg)$ are again matrix coefficients of $\pi$; by linearity of translation on functions the same holds for every $f\in R(K)$, so $R(K)$ is closed under left and right translation.

Collecting steps 1.1, 1.2, 1.3 and 1.4: $R(K)$ is a complex linear subspace of the continuous complex functions on $K$ containing the constants and closed under pointwise multiplication, complex conjugation and translation, so it is a self-adjoint unital complex function algebra on $K$ by [F3].