= Solution
We prove uniform approximation, not just approximation in $L^2$. The analytic facts used are: a <continuous> square-integrable kernel gives a compact integral operator on $L^2$; the <spectral theorem for compact self-adjoint operators> decomposes the closure of its range into its finite-dimensional nonzero <eigenspaces>; and <continuous> functions are dense in $L^2$ for normalized <Haar measure> on a <compact Lie group>. Normalized <Haar measure> on a compact <group> is both left- and right-invariant and is preserved by inversion.
Choose a <continuous> nonnegative function $k$ of integral one, supported in a sufficiently small symmetric neighborhood of the identity, with $k(g^{-1})=k(g)$. Such functions are obtained from a local bump and its inverted bump followed by normalization. Define the <convolution> operator
$$
(Th)(x)=\int_Gh(y)k(y^{-1}x)\,dy.
$$
Its kernel is <continuous> on $G\times G$, so $T$ is compact. The symmetry of $k$ makes it self-adjoint. It commutes with the <left translations> $(L_gh)(x)=h(g^{-1}x)$ by a change of variable $y=gz$.
By <Cauchy-Schwarz inequality> and invariance of <Haar measure>,
$$
\|Th\|_\infty\leq\|k\|_2\|h\|_2.
$$
The same integral formula and continuity of the kernel show $Th$ is <continuous> for every $h\in L^2(G)$. Moreover, rewriting $y=xz^{-1}$ gives $(Tf)(x)=\int f(xz^{-1})k(z)\,dz$. Uniform continuity on the compact <group> therefore makes $\|Tf-f\|_\infty$ arbitrarily small when the support of $k$ is small. Also $\|T\|_{\infty\to\infty}\leq1$ because $k\geq0$ and has integral one.
Fix $\varepsilon>0$ and take $k$ so $\|Tf-f\|_\infty<\varepsilon/3$. Then
$$
\|T^2f-f\|_\infty
\leq\|T(Tf-f)\|_\infty+\|Tf-f\|_\infty
<2\varepsilon/3.
$$
Let $P_N$ be the <orthogonal projection> onto increasing finite sums of the nonzero <eigenspaces> of $T$. Since $Tf$ lies in the closure of the range, the <spectral theorem> gives $P_NTf\to Tf$ in $L^2$. Applying the smoothing estimate gives
$$
\|TP_NTf-T^2f\|_\infty
\leq\|k\|_2\|P_NTf-Tf\|_2\longrightarrow0.
$$
Thus some $h=TP_NTf$ satisfies $\|h-f\|_\infty<\varepsilon$.
Each nonzero <eigenspace> is finite dimensional, invariant under $L_g$, and consists of <continuous> functions: if $Tv=\lambda v$ with $\lambda\ne0$, then $v=Tv/\lambda$. The finite sum $M$ containing $h$ therefore carries the finite-dimensional <unitary representation> $\rho(g)=L_g|_M$. It is <continuous>, since translating each of its finitely many <continuous> basis functions depends continuously on $g$ in the uniform and hence the $L^2$ norm.
Let $\ell$ be evaluation at the identity on $M$, and let the <dual representation> on $M^*$ be $\theta(g)\ell=\ell\circ\rho(g^{-1})$. For the linear functional $L_h$ on $M^*$ defined by $L_h(u)=u(h)$,
$$
L_h(\theta(g)\ell)=\ell(\rho(g^{-1})h)=h(g).
$$
The rank-one endomorphism $\alpha=\ell\otimes L_h$ of $M^*$ satisfies $\operatorname{tr}(\alpha\theta(g))=L_h(\theta(g)\ell)$. Therefore
$$
\boxed{\|f-\operatorname{tr}(\alpha\theta(\,\cdot\,))\|_\infty<\varepsilon.}
$$
This is the requested <Peter-Weyl theorem>, proved via the <convolution proof of uniform Peter-Weyl approximation>. The extra convolution after the spectral truncation is what upgrades the $L^2$ estimate to a uniform one.
To deduce a faithful finite-dimensional <representation>, first note that these <representations> separate points. For $g\ne e$, choose a <continuous> function taking different values at $g$ and $e$, and approximate it closely enough by a <trace> coefficient to retain that difference. The <representation> appearing in that <trace> must satisfy $\theta(g)\ne I$.
There is an identity neighborhood $U$ containing no nontrivial subgroup. Here is the <Lie groups have no small subgroups> argument: choose an injective exponential chart on a Lie-algebra ball of radius $r$, and take $U=\exp(B_\delta)$ with $0<\delta<r/2$. If a subgroup contained a nonidentity $\exp X$ in $U$, choose the first integer $m$ with $m\|X\|\geq\delta$. Then $\delta\leq m\|X\|<2\delta<r$, so its power $\exp(mX)$ lies outside $U$, a contradiction. For a zero-dimensional <group>, take $U=\{e\}$.
For every $g\notin U$, choose a <representation> nontrivial at $g$. The sets on which these <representations> are not the identity are open and cover the compact set $G\setminus U$. A finite subcover gives finitely many <representations>, whose <direct sum> has kernel contained in $U$. A subgroup contained in $U$ is trivial, so this <direct sum> is faithful. Averaging its Hermitian form makes it unitary. Thus the <finite faithful representation from point-separating representations> yields
$$
\boxed{G\hookrightarrow U_n\text{ for some finite }n.}
$$
Using the no-small-subgroups neighborhood is essential; $G\setminus\{e\}$ alone need not be compact, so a finite-subcover argument on that punctured set would not be valid.
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