Peter-Weyl theorem versions Let $G$ be a compact group. I learned the version of the Peter-Weyl theorem which says: the matrix coefficients of $G$ are dense in $L^2(G)$. Call this Peter-Weyl I. 
Apparently there is another version which states: For any $g \in G$ there exists a finite dimensional unitary representation $(\pi,V)$ such that $\pi(g) \neq I$ (identity). Call this Peter-Weyl II.
Can one prove Peter-Weyl II using Peter-Weyl I? A short slick proof is what I'm looking for, of course.

Let $H\subseteq G$ be the intersection of the kernels of all the finite-dimensional unitary representations of $G$; we wish to show $H=\{1\}$.  Let $q:G\to G/H$ be the quotient map.  For any integrable function $f:G/H\to\mathbb{C}$, $f\circ q$ is integrable and we have $\int_G f\circ q=\int_{G/H} f$, where both integrals are with respect to the Haar measures (this can be proven for $f$ continuous by uniqueness of the Haar measure on $G/H$, and then extends to all of $L^1(G/H)$ since continuous functions are dense).  In particular, composition with $q$ defines an isometry $q^*:L^2(G/H)\to L^2(G)$.  Furthermore, every matrix coefficient of $G$ is in the image of $q^*$, because if $f$ is a matrix coefficient, then $f(g)=f(hg)$ for all $h\in H$, $g\in G$ by definition of $H$.
Let us now assume Peter-Weyl I.  The image of $q^*$ is a closed subspace containing every matrix coefficient, so it must be all of $L^2(G)$.  But if $H$ is nontrivial, there is a continuous function $f$ on $G$ which is not constant on $H$, and it is easy to see that such a function cannot be in the image of $q^*$.  Thus $H$ must be trivial, proving Peter-Weyl II.
