The normalized matrix coefficients form an orthonormal basis of L2(K)

For classes $\pi ,\sigma$ and indices, [F1] and [F2] give ${\int }_K{u}_{ij}^{\pi }\overline{u_{kl}^{\sigma }}d\mu =\sqrt{d_{\pi }{d}_{\sigma }}{\int }_K⟨\pi (k)e_i^{\pi },e_j^{\pi }⟩\overline{⟨\sigma (k)e_k^{\sigma },e_l^{\sigma }⟩}d\mu (k)$, which is $0$ when $\pi \mathrm{\ne }\sigma$ (the classes are inequivalent) and equals $\sqrt{d_{\pi }^2}\cdot d_{\pi }^{-1}⟨e_i^{\pi },e_k^{\pi }⟩\overline{⟨e_j^{\pi },e_l^{\pi }⟩}={\delta }_{ik}{\delta }_{jl}$ when $\pi =\sigma$, by orthonormality of the fixed bases; hence $\mathcal{B}$ is an orthonormal family in $L^2(K)$.

Let $f=c_{v,w}^{\sigma }$ be a coefficient of a finite-dimensional continuous unitary representation on $V$. By [F3] take an orthogonal decomposition $V={⨁}_{m=1}^r{V}_m$ into irreducible subrepresentations ${\sigma }_m$, with classes ${\pi }_m\in \widehat{K}$. Choose unitary intertwiners $T_m:H_{{\pi }_m}\to V_m$ and write $v={\sum }_m{v}_m$, $w={\sum }_m{w}_m$, $a_m=T_m^{-1}{v}_m$, $b_m=T_m^{-1}{w}_m$. Unitarity and intertwining give $⟨{\sigma }_m(k)v_m,w_m⟩=⟨{\pi }_m(k)a_m,b_m⟩$. Expanding $a_m,b_m$ in the fixed basis of $H_{{\pi }_m}$ and using [F4] gives $f(k)={\sum }_{m,i,j}⟨a_m,e_i^{{\pi }_m}⟩\overline{⟨b_m,e_j^{{\pi }_m}⟩}{u}_{ij}^{{\pi }_m}(k)\mathrm{/}\sqrt{d_{{\pi }_m}}$. Hence $R(K)\subseteq \mathrm{span}\mathcal{B}$. Since $\mu (K)=1$, uniform approximation implies $L^2$ approximation. The uniform density of $R(K)$ in $C(K)$ and the $L^2$ density of $C(K)$ [F5,F6] therefore show that the closed span of $\mathcal{B}$ is $L^2(K)$.

By the Parseval equivalences [F7] applied to the orthonormal family $\mathcal{B}$, completeness is equivalent to the identity ${\sum }_{\pi ,i,j}\mathrm{\mid }⟨f,u_{ij}^{\pi }⟩{\mathrm{\mid }}^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$ for every $f\in L^2(K)$ and to $\mathcal{B}$ being a Hilbert basis of $L^2(K)$; step 1.2 supplies completeness, so both conclusions hold. The Axiom of Choice is inherited through the fixed representatives and bases and the cited suppliers; this proof adds no further choice.

If $K$ is abelian, every irreducible $\pi$ is one dimensional: for each $g$, $\pi (g)$ commutes with every $\pi (h)$ and is a bounded self-intertwiner, hence scalar by [F8]. Every line would therefore be invariant, so irreducibility forces dimension one. The scalar ${\chi }_{\pi }(g)$ is a continuous unit-circle-valued homomorphism. Conversely each such character is an irreducible one-dimensional unitary representation, and two of these representations are equivalent exactly when their characters agree. By [F1] its sole normalized matrix coefficient is ${\chi }_{\pi }$. Finite complete reducibility and coefficient splitting [F3,F4] therefore identify $R(K)$ with the finite linear span of characters. This span is uniformly dense in $C(K)$ by [F5]. The characters separate points: if all had equal values at two points, every finite linear combination, and hence every finite-dimensional matrix coefficient, would also have equal values there, contradicting [F9]. Finally steps 1.1–2.1 identify the character family as an orthonormal Hilbert basis of $L^2(K)$, with Parseval. These are the three compact-abelian Fourier conclusions, derived without general LCA separation or biduality.