Algebra of matrix coefficient over a compact group is isomorphic to its dual. Let $K$ be a compact group. Then we have the following definition of matrix coefficient:

Definition: $f: K \rightarrow \mathbb{C}$ is called a matrix coefficient if there is a finite dimensional semisimple $K$-module $V$ with $\alpha \in V^*, v\in V$ such that $f(k) = \alpha(kv)$ for all $k\in K$.

Let $R(K)$ be the set of matrix coefficients of $K$. It is a subalgebra of the algebra $$C(K):=\{\text{Continuous functions } K\to \mathbb{C}\},$$ and isomorphic to $\oplus (F_{\gamma}^*\otimes F_{\gamma})$ as spaces (where $F_{\gamma}$ take over all finite dimensional simple modules). Set $R^v(K)$ to be the dual of $R(K)$. 
Let $\zeta \in R^v(K)$ be defined by $\zeta(\alpha\otimes v) = \alpha(v)$ if $\gamma $ is trivial and $0$ otherwise.
My Question is: Can we show that $R(K) \cong R^v(K)$ via $f \mapsto\zeta(f\cdot)$?
Thanks very much!
 A: (1) Notation: 
$K^{\vee}$ denotes the set of isomorphism classes of simple, finite-dimensional $K$-modules, and $\mu$ denotes the Haar measure on $K$, normalized to $\mu(K)=1$.
(2) Alternative description of matrix coefficients: 
By definition, $R(K)$ is the image of the homomorphism $$\tau: \bigoplus\limits_{E\in K^{\vee}} E\otimes_{\mathbb C} E^{\ast}\to C(K),\quad \varphi\otimes v\mapsto (k\mapsto \varphi(k.v))$$ which in terms of the isomorphism $E\otimes_{\mathbb C} E^{\ast}\to\text{End}_{\mathbb C}(E)$ can also be described as $$\text{End}_{\mathbb C}(E)\ni\varphi\mapsto (k\mapsto \text{tr}_{\mathbb C}(\rho_E(k)\circ\varphi))\in C(K).$$
(3) The canonical form on $R(K)$:
The homomorphism $$\xi: \bigoplus\limits_{E\in K^{\vee}} E\otimes_{\mathbb C} E^{\ast}\twoheadrightarrow {\mathbb C}_{\text{triv}}\otimes_{\mathbb C}{\mathbb C}_{\text{triv}}^{\ast}\cong{\mathbb C}$$
factors through $\tau$ by means of the homomorphism $$C(K)\longrightarrow{\mathbb C},\quad f\mapsto\int_K f\ \text{d}\mu:$$
namely, for $E\in K^{\vee}$ and $\varphi\in\text{End}_{\mathbb C}(E)$ we have $$\int_K \text{tr}_{\mathbb C}(\rho_E(k)\circ\varphi)\ \text{d}\mu(k) = \text{tr}_{\mathbb C}\left[\left(\int_K \rho_E(k)\ \text{d}\mu(k)\right)\circ \varphi\right]=\begin{cases} 0 & \text{ if } E\not\cong {\mathbb C}_{\text{triv}}\\ \text{tr}_{\mathbb C}(\varphi) & \text{ if } E \cong {\mathbb C}_{\text{triv}}\end{cases},$$
where the latter equality holds since $\int_K \rho_E(k)\ \text{d}\mu(k)$ is the projection onto the $K$-invariants. 
Hence, the bilinear form you propose is the restriction to $R(K)$ of the standard pairing on $L^2(K)$, $$(f,g)\mapsto\xi(fg)=\int_K fg\ \text{d}\mu.$$ 
(4) Non-degeneracy of canonical form:
With the above description of $\xi$, we can now check that $\xi$ induces the trace-pairing between between the summands $E\otimes_{\mathbb C} E^{\ast}\cong\text{End}_{\mathbb C}(E)$ and $E^{\ast}\otimes_{\mathbb C} E=\text{End}_{\mathbb C}(E^{\ast})\cong\text{End}_{\mathbb C}(E)$ of $R(K)$, and vanishes otherwise:
First, for $E,F\in K^{\vee}$ and $\varphi\in\text{End}_{\mathbb C}(E)$, $\psi\in\text{End}_{\mathbb C}(F)$ we have $$\text{tr}_{\mathbb C}(\rho_E(k)\circ\varphi)\cdot\text{tr}_{\mathbb C}(\rho_F(k)\circ\psi)=\text{tr}_{\mathbb C}(\rho_{E\otimes_{\mathbb C}F}(k)\circ (\varphi\otimes\psi))$$ (note that the action of $K$ on $E\otimes_{\mathbb C} F$ is componentwise), hence $$(\ddagger)\quad\xi(\tau(\varphi),\tau(\psi))=\int_K \text{tr}_{\mathbb C}(\rho_{E\otimes_{\mathbb C} F}(k)\circ(\varphi\otimes\psi))\ \text{d}\mu(k) = \text{tr}_{\mathbb C}\left[\left(\int_K \rho_{E\otimes_{\mathbb C} F}(k)\ \text{d}\mu(k)\right)\circ (\varphi\otimes\psi)\right].$$ Again, $\int_K \rho_{E\otimes_{\mathbb C} F}(k)\ \text{d}\mu(k)$ is the projection onto the $K$-invariants $(E\otimes_{\mathbb C} F)^K$ in $E\otimes_{\mathbb C} F$, and since $E\otimes_{\mathbb C} F\cong\text{Hom}_{\mathbb C}(E^{\ast},F)$ as $K$-modules, hence $(E\otimes_{\mathbb C} F)^K\cong\text{Hom}_{\mathbb C}(E^{\ast},F)^K=\text{Hom}_K(E^{\ast},F)$, this is  - by Schur's Lemma - zero if $E^{\ast}\not\cong F$ and $1$-dimensional otherwise. This already shows that the summands $E\otimes_{\mathbb C} E^{\ast}$ and $F\otimes_{\mathbb C} F^{\ast}$ of $R(K)$ are orthogonal with respect to $\xi$ if $F\not\cong E^{\ast}$, and it remains to check that the right hand side in $(\ddagger)$ equals $\text{tr}_{\mathbb C}(\varphi\circ\psi^{\ast})$ in case $F=E^{\ast}$. 
For this, note that under the isomorphism $E\otimes_{\mathbb C} E^{\ast}\cong\text{Hom}_{\mathbb C}(E,E)$, the endomorphism $\varphi\otimes\psi$ of $E\otimes_{\mathbb C} E^{\ast}$ corresponds to the endomorphism $f\mapsto \psi^{\ast}\circ f\circ\varphi$ of $\text{Hom}_{\mathbb C}(E,E)$ and $\rho_{E\otimes_{\mathbb C} E^{\ast}}(k)$ corresponds to $f\mapsto \rho_E(k)\circ f\circ\rho_E(k^{-1})$. Hence, since $\text{Hom}_K(E,E)$ is spanned by $\text{id}_E$, we have to show that $\int_K \rho_E(k)\circ (\psi^{\ast}\circ\varphi)\circ\rho_E(k^{-1})\ \text{d}\mu(k) = \text{tr}_{\mathbb C}(\psi^{\ast}\circ\varphi)\cdot\text{id}_E$, which follows by noting firstly that the left hand side is $K$-linear and hence a multiple of $\text{id}_E$, and that secondly the traces of both sides coincide.
