Peter-Weyl theorem as proven in Cartier's Primer I'm reading Pierre Cartier's A primer of Hopf algebras to educate myself. In its subsection 3.3 (which doesn't need any Hopf algebra theory), he sketches a proof why compact Lie groups are algebraic. One step in this proof is the Peter-Weyl theorem. Here is something I don't understand in the proof of this theorem: Why is the space $C_{\lambda, f}$ invariant under left translations $L_g$? It is clear to me that $R_f$ commutes with $L_g$, but I don't see why $R_f^{\ast}$ should also commute with $L_g$, and I would need this assumption to prove the $L_g$-invariance of $C_{\lambda,f}$.
[Disclosure: I am neither an analyst nor a Lie group theorist, so this might be a trivial question.]
 A: (Same as pm's answer, with details.)
Well, if $$ R_f(\varphi)(h) = \int_G \varphi(g) f(g^{-1}h) \ dg $$
then the adjoint satisfies
\begin{align*}
(R_f^*(\varphi)|\psi) &= (\varphi|R_f(\psi))
= \int_{G\times G} \varphi(h) \overline{ \psi(g) f(g^{-1}h) } \ dg \ dh \\
&= \int_{G\times G} \varphi(h) \tilde f(h^{-1}g) \overline{\psi(g)} \ dh \ dg
= (R_{\tilde f}(\varphi) | \psi) \end{align*}
where $\tilde f(g) = \overline{ f(g^{-1}) }$.  So $R_f^* = R_{\tilde f}$ is again a right translation operator.
A: Are you aware of the relations,
$R_{f}^* = R_{f*}$ and $R_f R_h = R_{f \ast h}$? So if $R_f$ commutes with $L_g$ for all $f$, hence so does $R_{f^\ast}$.
So to say $R$ is a $*$ algebra representation by right convolutionn and $L$ are the left convolution. As an intuition why, these are equivalent to right and left translation by elements, which commute for every group. But you can also just check the integrals. The facts are  all purely topological, no smooth structures are required and the proofs work perfectly fine for every compact group.
Actually conversely, every compact group is a projective limit of compact (possibly finite) Lie groups by the Peter-Weyl theorem, but that is another story (see the comments). 
A: More generally, if $A$ is a bounded $G$-invariant operator acting on a unitary representation $(\rho,\mathcal{H})$ of $G$ and $A^*$ is it's adjoint (i.e. $(Ax,y) = (x,A^*y)$ for all $x,y\in \mathcal{H}$), then we have 
$$
(x,A^*\rho(g)y) = (Ax,\rho(g)y) = (\rho(g^{-1})Ax,y)
$$
where in the last step we used unitarity of $\rho$. Now use $G$-invariance of $A$ and retrace the steps back
$$
(\rho(g^{-1})Ax,y) = (A\rho(g^{-1})x,y) = (\rho(g^{-1})x,A^*y) = (x,\rho(g)A^*y).
$$
QED
