A quantum operation is defined as \begin{equation} \varepsilon(\rho)=\sum_{k}M_k\rho M_k^{\dagger} \end{equation} where $\varepsilon(\rho)$ takes an initial state $\rho$ to some final state $\rho'$ and $M_k^{\dagger}M_k$'s are positive, contractive operators that satisfy \begin{equation} \sum_kM_k^{\dagger}M_k=\mathbb{I}. \end{equation} It is known through Schauder's fixed point theorem that this admits a fixed point. Note that because of the summation in the first equation, it means that the measurement was not recorded.

QUESTION: If I have a single quantum measurement that was recorded, then I would only have $\varepsilon(\rho)=\frac{M_k\rho M_k^{\dagger}}{\text{tr}(M_k\rho M_k^{\dagger})}$. Would this still admit a fixed point? Is there a way to prove it?

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