*This question (and a second part) have been [asked at MSE](https://math.stackexchange.com/questions/4718733/an-infinitely-conditioned-state-in-a-c-algebra) and gone through two bounties without an answer. I have been beating my head at it for a while without success*.


Let $\mathcal{A}$ be a unital $\mathrm{C}^\*$-algebra. Let $f\in \mathcal{A}$ self-adjoint with spectrum $\sigma(f)$. The bidual $\mathcal{A}^{\*\*}$ contains the spectral projections of $f$, denoted $\mathbf{1}\_{E}(f)$, for Borel $E\subseteq \sigma(f)$. Let $\lambda\in\sigma(f)$, $\varepsilon>0$ and define
$$p\_\varepsilon:=\mathbf{1}\_{(\lambda-\varepsilon,\lambda+\varepsilon)}(f).$$


Suppose that $\varphi$ is a state on $\mathcal{A}$, with an extension to a state on $\mathcal{A}^{\*\*}$ denoted $\omega\_\varphi$. Suppose further, and this is to be a key assumption, that for all $\varepsilon>0$, $$\omega\_\varphi(p\_\varepsilon)>0\qquad(\star).$$


If the following is defined, it defines a state. But is it well-defined?


**Question: Let $g\in\mathcal{A}$. Under the assumption $(\star)$, does the following limit exist:**


$$\lim\_{\varepsilon\to 0^+}\frac{\omega\_\varphi(p\_\varepsilon\,g\,p\_\varepsilon)}{\omega\_\varphi(p\_\varepsilon)}.$$



Conditioning a $\mathrm{C}^*$-algebra state with infinite precision
No, this limit doesn't exist in general. Here's a commutative counterexample.


Let $A = L^\infty[0,1]$, $f(x) = x$, $\phi =$ integration against Lebesgue measure. Let $g$ be the indicator function of the set $\bigcup [10^{-(2n + 1)}, 10^{-2n}] = [.1, 1] \cup [.001, .01] \cup [.00001, .001] \cup \cdots$.


Taking $\lambda = 0$, we get that $p\_\epsilon$ is the indicator function of $[0,\epsilon]$. Then $\frac{\phi(p\_\epsilon g)}{\phi(p\_\epsilon)}$ takes the same value on $\epsilon =$ any even negative power of 10, and the same value on $\epsilon =$ any odd negative power of 10, and those two values aren't the same.


