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Suppose finite-dimensional continuous group representations separate points of a compact Lie group. Choose an identity neighborhood $U$ containing no nontrivial subgroup, using Lie groups have no small subgroups. For each $g\notin U$, choose a representation whose matrix at $g$ is not the identity. The corresponding open nonidentity sets cover the compact complement of $U$, so finitely many representations suffice there. Their direct sum has kernel inside $U$ and therefore trivial kernel. Unitarization of a compact-group representation makes it a finite-dimensional faithful unitary representation.

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 operatorIts 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 $(L_{g}h)(x)=h(g^{-1}x)$ by a change of variable $y=gz$.

By Cauchy-Schwarz inequality and invariance of Haar measure,The same integral formula and continuity of the kernel show $Th$ is continuous for every $h\in L^2(G)$. Moreover, rewriting $y=x{z}^{-1}$ gives $(Tf)(x)=\int f(x{z}^{-1})k(z)dz$. Uniform continuity on the compact group therefore makes $\Vert Tf-f{\Vert }_{\infty }$ arbitrarily small when the support of $k$ is small. Also $\Vert T{\Vert }_{\infty \to \infty }\le 1$ because $k\ge 0$ and has integral one.

Fix $\epsilon >0$ and take $k$ so $\Vert Tf-f{\Vert }_{\infty }<\epsilon /3$. ThenLet $P_N$ 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 $P_{N}Tf\to Tf$ in $L^2$. Applying the smoothing estimate givesThus some $h=T{P}_{N}Tf$ satisfies $\Vert h-f{\Vert }_{\infty }<\epsilon$.

Each nonzero eigenspace is finite dimensional, invariant under $L_g$, and consists of continuous functions: if $Tv=\lambda v$ with $\lambda \ne 0$, 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 $M^{\ast }$ be $\theta (g)\ell =\ell \circ \rho (g^{-1})$. For the linear functional $L_h$ on $M^{\ast }$ defined by $L_h(u)=u(h)$,The rank-one endomorphism $\alpha =\ell \otimes L_h$ of $M^{\ast }$ satisfies $\mathrm{tr}(\alpha \theta (g))=L_h(\theta (g)\ell )$. ThereforeThis 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 $m\Vert X\Vert \ge \delta$. Then $\delta \le m\Vert X\Vert <2\delta <r$, 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 yieldsUsing 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.