a little problem in the proof of Peter-Weyl theorem I am trying to understand the proof of Peter-Weyl theorem which has the following form.
$Theorem.$ If $G$ is a compact group, then the linear span of all matrix coefficients for all finite-dimensional irreducible unitary representations of $G$ is dense in $L^2(G)$. 
Let $U$ be the closure in $L^2(G)$ of the linear span of all matrix coefficients of all finite-dimensional irreducible unitary representations of $G$.  Arguing by contradiction, suppose $U\neq L^2(G)$. Then $U^\perp\neq 0$. Note that if $h(x)=\langle R(x)u,v\rangle$ is a matrix coefficient in $U$, then the following functions of $x$ are also matrix coefficients for the same representation $R$:
\begin{align*}
        \overline{h(x^{-1})}&=\overline{\langle R(x^{-1})u,v\rangle}=\langle v,R(x^{-1})u\rangle=\langle R(x)v,u\rangle \\
        h(gx)&=\langle R(gx)u,v\rangle=\langle R(x)u,R(g^{-1})v\rangle \\
        h(xg)&=\langle R(xg)u,v\rangle=\langle R(x)R(g)u,v\rangle 
\end{align*}
 How can we conclude that for any $h\in U^\perp$, the function $h'$ defined by $h'(x):=h(y^{-1}x)$ for some $y\in G$ is in $U^\perp$ ? Thanks.

$My \ attempt.$ If $h=0$ then there is nothing to do. So let $h\in U^\perp \setminus \{0\}.$ Suppose $h'\notin U^\perp$. Let $\mathcal{M}$ be a basis  for $U$ consisting of some matrix coefficients in $U$. Then we may write $$h'=g+\sum_{i=1}^n a_if_i$$ for some $f_1,...,f_n\in \mathcal{M}$ and $g\in U^\perp$ and for some scalars $a_1,...,a_n$. Note that at least one of these scalars is nonzero. Now we  have for all $x\in G$
\begin{equation}
h'(x)=h(y^{-1}x)=g(x)+\sum_{i=1}^n a_i f_i(x)\qquad (*)
\end{equation}
For each $i$, put $f_i(x)=\langle R_i(x)u_i,v_i\rangle$  where $R_i$ is the corresponding representation of $G$ for $f_i$ and also $u_i$ and $v_i$ are corresponding vectors for $f_i$. By the unitarity, we have 
$$f_i(x)=\langle R_i(y^{-1}x)u_i,R_i(y^{-1})v_i\rangle$$ 
So for any $t\in G$, the equation ($*$) becomes 
$$h'(yt)=h(t)=g_1(t)+\sum_{i=1}^n a_if'_i(t)$$ where  $g_1(t):=g(yt)$ and $f'_i(t):=\langle R_i(t)u_i,R_i(y^{-1})v_i\rangle$. Notice that the set $\{f'_i:i=1,...n\}$ is also linearly independent. For simplicity, we write 
$$h=g_1+\sum_{i=1}^n a_if'_i$$ where $g_1=r+s$ for some $r\in U\setminus\{0\},s\in U^\perp\setminus\{0\}$. Then we get 
$$h-s=r+\sum_{i=1}^na_if'_i\in U\cap U^\perp =\{0\}$$
I couldn't get any contradiction from here. Could anyone help me? Thanks.
 A: Asserting that $h(y^{-1}x)\in U^\bot$ means that, for each $f\in U$, $\bigl\langle h(y^{-1}x),f(x)\bigr\rangle=0$, that is, that $\int_Gh(y^{-1}x)\overline{f(x)}\,dx=0$. But$$\int_Gh(y^{-1}x)\overline{f(x)}\,dx=\int_Gh(y^{-1}x)\overline{f(y^{-1}yx)}\,dx=\int_Gh(x)\overline{f(yx)}\,dx\text{,}$$since we are dealing with a Haar measure here. In order to prove that this last integral is $0$, all that remains to be proved is that $\overline{f(yx)}\in U$, since $h(x)\in U^\bot$. Of course, in order to prove that, all I need to do is to prove that $\overline{f(x)}\in U$.
For some unitary representation $R$ of $G$ in a finite dimensional complex vector space $V$, endowed with a scalar product $\langle\cdot,\cdot\rangle$, you have $f(x)=\langle R(x)u,v\rangle$. Now consider a new complex structure on $V$; this time, if $\lambda\in\mathbb C$ and if $u\in V$, then the product of $\lambda$ by $u$ will be $\overline\lambda u$. Consider also a new scalar product $(\cdot,\cdot)$ on $V$, defined by $(u,v)=\overline{\langle u,v\rangle}$. Then $\bigl(R(x)u,v\bigr)=\overline{\bigl\langle R(x)u,v\bigr\rangle}=\overline{f(x)}$. Therefore $\overline{f(x)}\in U$.
