Let $A$ be $C^*$- Algebra and $X$ be a locally compact Hausdorff space and $C_{0}(X,A)$ be the set of all continuous functions from $X$ to $A$ vanishing at infinity. Define $f^{\ast}(t)={f(t)}^{\ast}$ (for $t\in X$). It is well known that $C_0(X,A)$ is $C^{\ast}-$ Algebra. Let $x\in X$ and $\pi$ be a representation of $A$ then the map $\pi_x$ defined as $\pi_x(f)=\pi(f(x))$ is a representation of $C_0(X,A)$
Is is true that all representations of $C_0(X,A)$ are of the form $\pi_x$ for some $x \in X$ and $\pi$ representation of $A$
My guess is that it is true because of identification of $C_0(X,A)$ using min tensor product. Can someone give some reference or a precise argument.
This is not true, unless $X$ is just a point. Recall how the minimal tensor product is defined: Take any faithful representations $(\pi,\mathcal H)$ of $C_0(X)$ and $(\varphi,\mathcal K)$ of $A$, and form the product map $\pi\otimes\varphi:C_0(X)\odot A\to\mathbb B(\mathcal H\otimes\mathcal K)$. Then the closure of $\pi\otimes\varphi(C_0(X)\odot A)$ is isomorphic to $C_0(X)\otimes A\cong C_0(X,A)$, hence yields a representation of $C_0(X,A)$ on $\mathcal H\otimes\mathcal K$, which can't be of the form $\pi_x$, as it's faithful.