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A Banach limit is a continuous linear functional $\Lambda$ on $\ell^{\infty}(\mathbb{N})$ satisfying: $\|\Lambda\|=\Lambda(1,1,1,\cdots)=1$; and $\lambda(a_1,a_2,a_3,\cdots)=\Lambda(a_2,a_3,a_4,\cdots)$ for all sequences.

The existence of Banach limits is an exercise, and at the end of 3.1 of A Short Course on Spectral Theory one constructs the noncommutative counterpart of $\Lambda$. Note that for $A\in B(\mathcal{H})$, where $\mathcal{H}$ is a Hilbert space with countable basis $\{e_n\}_{n\in\mathbb{N}}$, $a=(\langle Ae_n,e_n\rangle)\in\ell^{\infty}(\mathbb{N})$ thus one can define \begin{equation} \rho(A)=\Lambda(a). \end{equation}

In the book it is also proved that $\rho$ vanishes on compact operators, and that $\rho(SAS^*)=\rho(A)$ where $S$ is the shift. The second justifies that $\rho$ is the noncommutative counterpart of Banach limits, and the first justifies that compact operators are the noncommutative counterpart of $c_0$, and the study of Calkin algebra is the study of asymptotics of operators.

But I wonder whether there are explicit examples of this $\rho$ (with explicit $\Lambda$), and what do they teach us about $A\in B(\mathcal{H})$.

Thanks!

As pointed out in the comments, the existence of Banach limits is proved using the weak$^*$-compactness of the unit ball in dual of $\ell^{\infty}(\mathbb{N})$, so maybe there are not examples. So let's focus on what information they tell us.

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As far as I know, the construction of Banach Limits invokes the Axiom of Choice (or at least those so far constructed do). This post at MO may be of interest. –  David Mitra Dec 18 '12 at 16:04
    
@DavidMitra Yes, the existence is proved by Axiom of Choice, so maybe my first question is not useful. But assuming there existence, what can they teach us? –  Hui Yu Dec 18 '12 at 16:06
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