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.
On
This answer is actually using a stronger version of Peter Weyl I, which state that the matrix coefficients are even dense in $C(G)$, the space of continuous functions. (So that might not answer your question)
First we show that $G$ must have a faithful representation.
To show that, let $g\in G\setminus \{e\}$ and $f$ be continuous on $G$ so that $f(g) \neq f(e)$. Then by Peter Weyl I, there is a matrix coefficient $u$ so that $u(g) \neq u(e)$. If this $u$ comes from the representation $\phi_1 : G \to GL(V)$, then we see that $g\notin \ker(\phi)$. Let $K_1$ be this kernel. $K_1$ is a closed subgroup of $G$ which is strictly smaller then $G$.
We are done if $K_1 = \{e\}$. If not, then we find $g_2\in K_1\setminus \{e\}$ and similarly construct a representation $\phi_2$ on $G$ so that the kernel $\tilde K_2$ has the property that $K_2:= \cap K_1\neq K_1$.
Then inductively we find a descending sequence of closed subgroups
$$G \supsetneqq K_1 \supsetneqq K_2 \supsetneqq K_3 \supsetneqq \cdots $$
But this chain must stability by dimension consideration and the fact that $G$ is compact. Thus $K_m = \{e\}$ for some $m$. Then the representation
$$\Phi = \phi_1 \oplus \cdots \oplus \phi_m$$
is injective: if $g\in G$ and $\Phi(g) = 0$, then $g\in K_i$ for all $i=1, 2\cdots, m$. In particular $g = e$.
Now the claimed results (Peter Weyl II) follows from the fact that every representation of a compact Lie group admits a $G$-invariant metric. You may check more in the book representation of compact Lie groups, where I learnt the above argument from this book.
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.