I have a question about the convolution of matrix coefficients as follows:
Let $G$ be a compact Lie group. A Map $f:G\rightarrow \mathbb{C}$ is called a matrix coefficient if there is a finite dimensional, (unitary) representation $(\Phi,V)$ and $v_1,~v_2 \in V$ such that $f(g)=(\Phi v_1, v_2)$ for all $g\in G$. Let $f,g$ be two matrix coefficients. Then the convolution $f*g$ is also a matrix coefficient.
Is this true? But I have no any counterexample for this. Any hint and comment are appreciated. Thanks very much!
I'm very sorry for the late reply but yes, it is true and it follows from the definition. Indeed, if $f(\cdot) = \left\langle \phi(\cdot)v, w\right\rangle$ is a matrix coefficient of $G$, and $g$ is any (continuous) function on $G$, then:
$$(f\ast g)(x)$$
$$=\int_{G} f(xh^{-1})g(h) dh$$
$$=\int_{G} \left\langle \phi(xh^{-1})v,w\right\rangle g(h)dh$$
$$=\int_{G} \left\langle \phi(x)\phi(h^{-1})v,w\right\rangle g(h)dh$$
$$=\int_{G} \left\langle \phi(x) g(h) \phi(h^{-1})v,w\right\rangle dh$$
$$=\left\langle \phi(x)\left(\int_{G} g(h)\phi(h^{-1})dh\right)v,w \right\rangle$$
which is also a matrix coefficient. A similar calculation can be done for the case where $f$ is any continuous function and $g$ is a matrix coefficient. Thus, the matrix coefficients (of a fixed representation) constitute a two-sided ideal in the ring of continuous functions (with multiplication given by convolution). Of course, you can also replace continuous by integrable everywhere in my answer and the result is still true.
Hope this helps!