Let $A$ be a non-separable $C^*$-algebra. Is it possible that there is a faithful representation $\pi:A\to L(H)$ on a separable hilbert space $H$?
I know that if $A$ is separable, one can choose $H$ separable.
I only have an argument that the answer is "no" for finite dimensional $H$. If $H$ is separable, $H$ is isomorphic to $l^2(\mathbb{N})$ if $H$ is infinite dimensional, and otherwise $H$ is isomorphic to $\mathbb{C}^k$, $k\in\mathbb{N}$. Furthermore, $\pi(A)\cong A$ and I know that $\pi$ is isometric. But now I'm not sure how "big" is $L(l^2(\mathbb{N}))$. For finite dimensional vector spaces $U$ and $V$ with $\dim U=n$, $\dim V=m$ the dimension of $\{f:U\to V; \;\text{f is linear}\}$ is m*n. It follows (linear maps on finite dimensional vector spaces are automatically continuous): If $H$ is finite dimensional, then $A\hookrightarrow L( \mathbb{C}^k)$ contradicts that $A$ is non-separable.
I guess that a similar argument holds if $H\cong l^2(\mathbb{N})$ and that the answer of my question in this case is "no" too, but I'm not sure. Here I'm stuck.
Best.
It is possible, as $L(\ell^2)$ is itself non-separable (and has the identity as representation). To see this, let $A \subseteq \mathbf N$ be any subset and define $T_A \colon \ell^2 \to \ell^2$ by $$ T_A\left(\sum_{n \ge 0} x_n e_n\right) = \sum_{n \in A} x_n e_n $$ Then $T_A \in L(\ell^2)$. Now let $A, B \subseteq \mathbf N$ with $A \ne B$. Let $n \in A \Delta B$, we have $$ \| T_A - T_B \| \ge \| T_A e_n - T_B e_n\| = \|e_n\| = 1 $$ As $\mathscr P(\mathbf N)$ is uncountable, $L(\ell^2)$ is inseparable.