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Let $A$ be a separable $C^*$-algebra. Is it possible to that the dual space $A^*$ is separable?

This question is initial from my previous question. A kindly answer claims that separable $C^*$-algebra admits a faithful separable representation. I write down the proof here.

Suppose $A$ is separable. Let $\{x_1,\cdots\}$ be the dense subset in $A$. Then we can choose $\{\phi_1,\cdots\}$ from $S\left(A\right)$ such that $\Vert\phi_i\left(x_i\right)\Vert=\Vert x_i\Vert$. Let $\left(\pi_i, H_i\right)=\left(\pi_{\phi_i},H_{\phi_i}\right)$ be the cycle representation corresponding to the state $\phi_i$. Put $\left(\pi,H\right)=\left(\oplus_{i}\phi_i,\oplus_{i}H_i\right)$. We claim that $\left(\pi,H\right)$ is a faithful separable representation for $A$.

Clearly, $\left(\pi,H\right)$ is a separable representation,since $H_i$ is the completion of $A/N_{\phi_i}$ and $A$ is separable. The following we need to verify is the injectivity of $\pi$. Given $x\in A$, for any $\varepsilon>0$ we could choose $x_n$ from the dense subset $\{x_{i},\cdots\}$ such that $\Vert x-x_n\Vert<\frac{\varepsilon}{2}$. Therefore, we have the following estimation, \begin{equation*} \Vert\pi\left(x\right)\Vert\approx_\frac{\varepsilon}{2}\Vert\pi\left(x_n\right)\Vert=\Vert x_n\Vert\approx_\frac{\varepsilon}{2}\Vert x\Vert. \end{equation*} Due to the arbitary of $\varepsilon$, we conclude that $\left(\pi,H\right)$ is a faithful representation of $A$.

According to the previous statement, we see that a separable $Cˆ*$-algebra $A$ could be embedded in $B\left(l^2\right)$ as a normed closed space.

I would like to know whether the dual space of $B\left(l^2\right)$ is separable or not?

Hao Wang
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    $B(l^2)$ is not separable, so it's dual space isn't either. – PhoemueX Nov 11 '24 at 10:43
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    $B(H)$ contains $\ell^\infty$ - consider diagonal operators – Evangelopoulos Foivos Nov 11 '24 at 10:44
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    For the original question consider e.g. $C(K)$ spaces – leoli1 Nov 11 '24 at 22:01
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    As long as $A$ has a normal element whose spectrum is uncountable, $A^\ast$ is non-separable, and this is quite common for separable $C^\ast$-algebras. Infinite-dimensional $C^\ast$-algebras whose duals are separable are actually somewhat rare. Examples include $c_0$-direct sums of (countably many) matrix algebras and $K(\ell^2)$. – David Gao Nov 11 '24 at 23:39
  • @DavidGao Wow, thank you for your information!!! Normal elements play important role in non-abelian $C^$-algebras. According to the functional calculus, we may restrict some questions to the $C^$-subalgebra generated by normal element, which is commutative. Thank you again for your critical comment. – Hao Wang Nov 12 '24 at 07:24

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