[proofplan] We prove the algebra properties directly from the definition of matrix coefficients. Constants come from the one-dimensional identity representation, products come from [tensor product](/page/Tensor%20Product) representations, and complex conjugates come from conjugate representations. Finally, point separation follows by applying Peter-Weyl density to a [continuous function](/page/Continuous%20Function) that separates two prescribed points of the compact [Hausdorff space](/page/Hausdorff%20Space) $G$. [/proofplan]

[step:Realize constant functions as coefficients of the identity one-dimensional representation] Let $\iota:G\to GL(\mathbb C)$ be the identity one-dimensional representation, defined by $\iota(g)z=z$ for every $g\in G$ and $z\in\mathbb C$. Let $\varepsilon:\mathbb C\to\mathbb C$ be the identity linear functional, $\varepsilon(z)=z$, and let $u:=1\in\mathbb C$. Then \begin{align*} c_{\varepsilon,u}^{\iota}(g)=\varepsilon(\iota(g)u)=1 \end{align*} for every $g\in G$. Hence the constant function $1$ belongs to $\mathcal R(G)$. Since $\mathcal R(G)$ is a complex [vector space](/page/Vector%20Space) by definition as a complex span, every constant complex-valued function belongs to $\mathcal R(G)$, and $\mathcal R(G)$ contains the multiplicative identity of $C(G)$. [/step]

[step:Multiply two coefficients using the tensor product representation]Let $\rho:G\to GL(V)$ and $\sigma:G\to GL(W)$ be continuous finite-dimensional complex representations. Let $\lambda\in V^*$, $\mu\in W^*$, $v\in V$, and $w\in W$. Define the tensor product representation \begin{align*} \rho\otimes\sigma:G&\to GL(V\otimes_{\mathbb C} W) \end{align*} by \begin{align*} (\rho\otimes\sigma)(g)(x\otimes y)=\rho(g)x\otimes\sigma(g)y \end{align*} for every $g\in G$, $x\in V$, and $y\in W$, extended linearly to $V\otimes_{\mathbb C}W$. This is a continuous finite-dimensional complex representation because $\rho$ and $\sigma$ are continuous representations and the tensor product operation is polynomial in matrix coordinates after choosing bases. Let $\lambda\otimes\mu\in (V\otimes_{\mathbb C}W)^*$ denote the linear functional determined by \begin{align*} (\lambda\otimes\mu)(x\otimes y)=\lambda(x)\mu(y). \end{align*} For every $g\in G$, \begin{align*} c_{\lambda\otimes\mu,v\otimes w}^{\rho\otimes\sigma}(g) = (\lambda\otimes\mu)((\rho\otimes\sigma)(g)(v\otimes w)). \end{align*} By the definition of $\rho\otimes\sigma$, this equals \begin{align*} (\lambda\otimes\mu)(\rho(g)v\otimes\sigma(g)w) = \lambda(\rho(g)v)\mu(\sigma(g)w). \end{align*} Therefore \begin{align*} c_{\lambda\otimes\mu,v\otimes w}^{\rho\otimes\sigma}(g) = c_{\lambda,v}^{\rho}(g)c_{\mu,w}^{\sigma}(g). \end{align*} Thus the product of two matrix coefficients is again a matrix coefficient. By bilinearity of pointwise multiplication and because $\mathcal R(G)$ is the complex span of matrix coefficients, $\mathcal R(G)$ is closed under pointwise multiplication.[/step]

[guided]The goal is to show that multiplication does not take us outside the class of representative functions. It is enough to multiply two basic matrix coefficients, because every element of $\mathcal R(G)$ is a finite complex linear combination of such coefficients. Let \begin{align*} \rho:G\to GL(V) \end{align*} and \begin{align*} \sigma:G\to GL(W) \end{align*} be continuous finite-dimensional complex representations. Choose $\lambda\in V^*$, $\mu\in W^*$, $v\in V$, and $w\in W$. We form the tensor product representation \begin{align*} \rho\otimes\sigma:G\to GL(V\otimes_{\mathbb C} W) \end{align*} by requiring \begin{align*} (\rho\otimes\sigma)(g)(x\otimes y)=\rho(g)x\otimes\sigma(g)y \end{align*} for every $g\in G$, $x\in V$, and $y\in W$, and then extending linearly. This is the right construction because the scalar obtained from a tensor product functional on a tensor product vector factors into the product of the two scalar coefficients. The vector space $V\otimes_{\mathbb C}W$ is finite-dimensional because $V$ and $W$ are finite-dimensional. After choosing bases of $V$ and $W$, the entries of $(\rho\otimes\sigma)(g)$ are products of entries of $\rho(g)$ and $\sigma(g)$, so the map $g\mapsto (\rho\otimes\sigma)(g)$ is continuous. Thus $\rho\otimes\sigma$ is again a continuous finite-dimensional complex representation of $G$. Now define $\lambda\otimes\mu\in (V\otimes_{\mathbb C} W)^*$ by \begin{align*} (\lambda\otimes\mu)(x\otimes y)=\lambda(x)\mu(y). \end{align*} For every $g\in G$, the coefficient of the tensor product representation associated to $\lambda\otimes\mu$ and $v\otimes w$ is \begin{align*} c_{\lambda\otimes\mu,v\otimes w}^{\rho\otimes\sigma}(g) = (\lambda\otimes\mu)((\rho\otimes\sigma)(g)(v\otimes w)). \end{align*} Using the defining formula for $\rho\otimes\sigma$, we get \begin{align*} (\rho\otimes\sigma)(g)(v\otimes w)=\rho(g)v\otimes\sigma(g)w. \end{align*} Therefore \begin{align*} c_{\lambda\otimes\mu,v\otimes w}^{\rho\otimes\sigma}(g) = (\lambda\otimes\mu)(\rho(g)v\otimes\sigma(g)w). \end{align*} By the defining property of $\lambda\otimes\mu$, this becomes \begin{align*} c_{\lambda\otimes\mu,v\otimes w}^{\rho\otimes\sigma}(g) = \lambda(\rho(g)v)\mu(\sigma(g)w) = c_{\lambda,v}^{\rho}(g)c_{\mu,w}^{\sigma}(g). \end{align*} Thus the pointwise product of the two original coefficients is exactly one coefficient of the finite-dimensional continuous representation $\rho\otimes\sigma$. Since pointwise multiplication distributes over finite sums, this proves that $\mathcal R(G)$ is closed under multiplication.[/guided]

[step:Represent complex conjugates using conjugate representations] Let $\rho:G\to GL(V)$ be a continuous finite-dimensional complex representation, let $\lambda\in V^*$, and let $v\in V$. Let $\overline V$ denote the conjugate complex vector space of $V$. For $x\in V$, write $\overline x\in\overline V$ for the corresponding element. Define the conjugate representation \begin{align*} \overline\rho:G&\to GL(\overline V) \end{align*} by \begin{align*} \overline\rho(g)\overline x=\overline{\rho(g)x}. \end{align*} This is a continuous finite-dimensional complex representation. Define $\overline\lambda\in(\overline V)^*$ by \begin{align*} \overline\lambda(\overline x)=\overline{\lambda(x)}. \end{align*} The map $\overline\lambda$ is complex-linear on $\overline V$ because scalar multiplication in $\overline V$ is conjugated from scalar multiplication in $V$. For every $g\in G$, \begin{align*} c_{\overline\lambda,\overline v}^{\overline\rho}(g) = \overline\lambda(\overline\rho(g)\overline v). \end{align*} Using the definition of $\overline\rho$, this equals \begin{align*} \overline\lambda(\overline{\rho(g)v}) = \overline{\lambda(\rho(g)v)} = \overline{c_{\lambda,v}^{\rho}(g)}. \end{align*} Hence the pointwise complex conjugate of every matrix coefficient belongs to $\mathcal R(G)$. Since complex conjugation distributes over finite sums with conjugated scalar coefficients, $\mathcal R(G)$ is closed under pointwise complex conjugation. [/step]

[step:Separate distinct group elements using Peter-Weyl density] Let $g,h\in G$ satisfy $g\ne h$. Since $G$ is compact and Hausdorff, it is a [normal topological space](/page/Normal%20Topological%20Space). By [Urysohn's lemma](/theorems/887) applied to the disjoint closed sets $\{g\}$ and $\{h\}$, there exists a continuous function $F:G\to\mathbb C$ such that $F(g)=0$ and $F(h)=1$. By the Peter-Weyl theorem [citetheorem:8833], the representative functions $\mathcal R(G)$ are dense in $C(G)$ with respect to the [uniform norm](/page/Uniform%20Norm) \begin{align*} \|u\|_{C(G)}:=\sup_{x\in G}|u(x)|. \end{align*} Applying this density statement to $F$ and the number $1/3$, there exists $f\in\mathcal R(G)$ such that \begin{align*} \|f-F\|_{C(G)}<1/3. \end{align*} Evaluating at $g$ and $h$ gives \begin{align*} |f(g)|<1/3 \end{align*} and \begin{align*} |f(h)-1|<1/3. \end{align*} If $f(g)=f(h)$, then $|f(h)|=|f(g)|<1/3$, while the [reverse triangle inequality](/theorems/2300) gives \begin{align*} |f(h)|\ge 1-|f(h)-1|>2/3, \end{align*} a contradiction. Therefore $f(g)\ne f(h)$, and $\mathcal R(G)$ separates points of $G$. [/step]

[step:Conclude the representative functions form the required algebra] The first step shows that $\mathcal R(G)$ contains the constant function $1$. The second step shows that $\mathcal R(G)$ is closed under pointwise multiplication, and it is closed under complex linear combinations by its definition as a complex span. Therefore $\mathcal R(G)$ is a unital subalgebra of $C(G)$. The third step proves closure under pointwise complex conjugation, and the fourth step proves point separation. These are exactly the asserted properties of $\mathcal R(G)$. [/step]