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Normalized Haar measure makes this averaging a norm-one projection onto continuous class functions. Averaging a trace coefficient replaces its coefficient endomorphism by $\int \theta (h{)}^{-1}\alpha \theta (h)dh$. On an irreducible summand this average is $(\mathrm{tr}\alpha /\dim V)I$ by Schur lemma. It follows that averaging finite-dimensional matrix coefficients yields finite linear combinations of irreducible characters. Together with the Peter-Weyl theorem, this proves uniform density of those characters among continuous class functions.

A continuous symmetric approximate identity on a compact group gives a compact self-adjoint convolution operator $T$ on $L^2(G)$ which maps into $C(G)$ and commutes with left translations. Its nonzero eigenspaces are finite-dimensional invariant spaces of continuous functions, hence consist of matrix coefficients of finite-dimensional unitary representations. If $S_N$ projects onto increasing finite sums of these eigenspaces, $S_{N}Tf\to Tf$ in $L^2$, and the bound $\Vert Th{\Vert }_{\infty }\le \Vert k{\Vert }_2\Vert h{\Vert }_2$ gives $T{S}_{N}Tf\to T^{2}f$ uniformly. Choosing the approximate identity narrow makes $T^{2}f$ uniformly close to $f$. This supplies uniform approximation without confusing an $L^2$ limit with a uniform limit.

For an Abelian group, on finite formal sums of group points, define $⟨[x],[y]⟩=p(x{y}^{-1})$, with the first variable linear. Quotient by the null space of this positive semidefinite form and complete to a Hilbert space. Left translation is unitary, and $v=[e]$ generates a dense span of translated vectors. For continuous $p$, the equality $\Vert U_x[y]-[y]{\Vert }^2=2(p(e)-\mathrm{Re}p(x))$ gives strong continuity. This realizes $p$ as a matrix coefficient without assuming its integral representation.

By part (i), finite-dimensional continuous unitary representations separate points of the compact group. More explicitly, if every representation sent some $g\ne e$ to the identity, all their matrix coefficients would take the same values at $g$ and $e$. Uniform density would force every continuous scalar function to do so, contradicting separation of points on a compact Hausdorff space. HenceEach representation kernel is a closed normal subgroup. We claim that the intersection can be reduced to finitely many kernels. Begin with $K_0=G$. If $K_j\ne \{e\}$, choose $g_j\in K_j\setminus \{e\}$ and a representation ${\theta }_{j+1}$ with ${\theta }_{j+1}(g_j)\ne I$. ThenAn indefinite continuation would violate the descending chain condition for closed subgroups. Thus after finitely many steps, $K_N=\{e\}$.

The direct sum $\rho ={\theta }_1\oplus \cdots \oplus {\theta }_N$ is consequently a finite-dimensional faithful representation, with continuous injective homomorphismCompactness of $G$ makes this a homeomorphism onto a compact, hence closed, image. The closed-subgroup theorem makes that image an embedded Lie subgroup of the finite-dimensional unitary group. Transporting its smooth structure to $G$ provesThis is the mechanism behind the assertion that compact groups with a descending chain condition are Lie groups: the chain condition turns an arbitrarily large separating family into one finite-dimensional faithful representation.

Interpret $\theta$ as a continuous finite-dimensional complex group representation, so its values lie in ${\mathrm{GL}}_{\mathbb{C}}(V)$, and interpret approximation in the uniform norm. Assume the usual Hausdorff convention for the compact topological group. The following argument proves the required form of the Peter-Weyl theorem without first assuming that $G$ is a Lie group.

Start with normalized Haar measure. It is both left- and right-invariant on a compact group. Average any Hermitian inner product to make a finite-dimensional representation unitary. The left regular action $L_{a}h(x)=h(a^{-1}x)$ is already unitary on $L^2(G)$.

Choose a continuous nonnegative function $k$ supported in a small symmetric identity neighbourhood, with $k(z^{-1})=k(z)$ and $\int k=1$. Such functions are obtained from a nonzero continuous bump near the identity, symmetrizing and normalizing. DefineThe convolution operator $T_k$ commutes with left translations. Its continuous kernel is Hermitian, hence it is self-adjoint. It is compact on $L^2(G)$: approximate the continuous kernel uniformly by finite sums of separated functions of $x$ and $y$, using the Stone-Weierstrass theorem on the compact product. The corresponding operators have finite rank and converge in operator norm. It also maps into $C(G)$ and satisfiesby the Cauchy-Schwarz inequality; continuity follows from the uniform continuity of the translated kernel.

The spectral theorem for compact self-adjoint operators decomposes the orthogonal complement of its kernel into finite-dimensional eigenspaces with nonzero eigenvalues. Every such eigenspace is left-translation invariant, and all its elements are continuous, since $h=T_{k}h/\lambda$ there. If $W\subset C(G)$ is one of these invariant spaces, write $\rho (a)=L_a{|}_W$. Evaluation at the identity givesThus $h$ is a matrix coefficient of the dual representation. This identifies each finite spectral sum with finite-dimensional representation data.

To get uniform rather than just $L^2$ approximation, let $S_N$ project onto increasing finite sums of the nonzero eigenspaces. For $f\in C(G)$, $T_{k}f$ is orthogonal to the kernel, so $S_N{T}_{k}f\to T_{k}f$ in $L^2$. Applying the displayed smoothing bound givesMeanwhile, choosing the support of $k$ sufficiently small makes $T_{k}f$ uniformly close to $f$, by uniform continuity under small translations. Since $\Vert T_k{\Vert }_{C\to C}\le 1$, it also makes $T_k^{2}f$ uniformly close to $f$. Every $T_k{S}_N{T}_{k}f$ belongs to a finite sum of invariant eigenspaces, hence is a finite sum of matrix coefficients. This completes the convolution proof of uniform Peter-Weyl approximation.

A coefficient $\ell (\theta (g)v)$ equals $\mathrm{tr}(\alpha \theta (g))$ for the rank-one endomorphism $\alpha =v\otimes \ell$. Finite sums of coefficients can be combined using a direct-sum representation and a block-diagonal $\alpha$. Therefore the conclusion has exactly the requested form:

For a class function $f$, apply conjugation averaging on a compact group,It fixes $f$ and is a contraction in the uniform norm, so averaging an approximation preserves its error bound. A trace coefficient averages to $\mathrm{tr}(\overline{\alpha }\theta (g))$, whereThis endomorphism commutes with the representation. A finite-dimensional unitary representation splits into irreducibles by repeatedly taking invariant orthogonal complements. On each irreducible diagonal block, Schur lemma makes its average a scalar multiple of the identity, namely its trace divided by the block dimension. Off-diagonal blocks do not contribute to the total trace. Thus the averaged coefficient is a finite linear combination of irreducible characters. Consequently irreducible complex characters span a uniformly dense subspace of the continuous class functions.

Use the following normalization for Fourier analysis on a finite group. Choose one unitary irreducible representation $\rho :G\to U(d_{\rho })$ from each equivalence class, including the trivial representation. For a scalar function $f:G\to \mathbb{C}$, putThis convention uses $\rho (x)$, rather than $\rho (x{)}^{\ast }$, in the Fourier transform on a finite group; it makes the normalized convolution on a finite group preserve multiplication order.

The needed representation theory consists of Maschke's theorem and unitarization of a finite-group representation, together with the Schur orthogonality relations:The regular representation contains $d_{\rho }$ copies of each $\rho$, so ${\sum }_{\rho }{d}_{\rho }^2=|G|$. Thus the scaled matrix coefficients $\sqrt{d_{\rho }}\rho (x{)}_{ij}$, and also their complex conjugates, form an orthonormal basis of all scalar functions on $G$. These facts imply Fourier inversion on a finite group and the Parseval identity on a finite group in the formsand henceIn particular, the transform is an isomorphism onto the direct sum of the matrix algebras $M_{d_{\rho }}(\mathbb{C})$, with the displayed weighted Hilbert-Schmidt inner product.

Define the normalized convolution on a finite group bySubstituting $x=yz$ and using the group representation identity yields the convolution theorem on a finite groupUnlike normalized convolution on a finite group on an abelian group, this product need not commute. If $\tilde{f}(x)=\overline{f(x^{-1})}$, then $\hat{\tilde{f}}(\rho )=\hat{f}(\rho {)}^{\ast }$. For left translation of a group function and right translation of a group function $L_{a}f(x)=f(a^{-1}x)$ and $R_{a}f(x)=f(xa)$,For an abelian group, every irreducible representation is one-dimensional; this reduces to Fourier analysis on a finite abelian group with characters relabelled by their inverses. These formulas establish the basic scalar theory, with all normalizations and multiplication orders fixed.

Now suppose every nontrivial irreducible representation has $d_{\rho }\ge m$. If $f$ is a mean-zero function, its component at the trivial representation is zero. The Parseval identity on a finite group gives, for each other $\rho$,Using the convolution theorem on a finite group, the Hilbert-Schmidt norm inequality $\Vert AB{\Vert }_{\mathrm{HS}}\le \Vert A{\Vert }_{\mathrm{op}}\Vert B{\Vert }_{\mathrm{HS}}$, and the Parseval identity on a finite group once more gives the product mixing in a quasirandom group estimateWrite $a,b,c$ for the subset density values of $A,B,C$, respectively, and let $f=1_A-a$, $g=1_B-b$ be balanced subset indicators. Since both are mean-zero functions, $1_A\ast 1_B=ab+f\ast g$. Their squared norms are $a(1-a)$ and $b(1-b)$. Also $\mathbb{E}(f\ast g)=0$, so the Cauchy-Schwarz inequality yieldsIf $abc>1/m$, the final bound is strictly smaller than $abc$. Thus the normalized number of pairs $(x,y)\in A\times B$ with $xy\in C$ is positive. Equivalently,This is the desired conclusion for a quasirandom group; the strict inequality ensures positivity rather than merely a nonnegative lower bound.