Constructing a CP map with some decaying property Given some observable $\mathcal O \in \mathcal H$ it is simple to construct a CP (completely positive) map $\Phi:\mathcal{H}\mapsto \mathcal{H}$ that conserves this quantity. All one has to observe is that
$$ \text{Tr}(\mathcal O \, \Phi[\rho]) = \text{Tr}(\Phi^*[\mathcal O] \rho).$$
Therefore, if we impose $\Phi^*[\mathcal O] = \mathcal O$, then $\text{Tr}(\mathcal O \, \Phi[\rho])=\text{Tr}(\mathcal O \rho), \; \forall \rho\in \mathcal H$. That amounts to impose that the Kraus operators of $\Phi^*$ should commute with $\mathcal O$.
I'd like, however, to construct a trace-preserving CP map for which the expectation value of $\mathcal O$ does not increase for any $\rho \in \mathcal H$. More explicitly, given $\mathcal O\in \mathcal H$, I want to construct $\Gamma:\mathcal H \mapsto \mathcal H$ such that 
$$ \text{Tr}(\mathcal O\, \Gamma[\rho]) \le \text{Tr}(\mathcal O \rho), \; \forall \rho \in \mathcal H .$$
How would you go about that? Any ideas?
 A: I'll restrict myself to trace-preserving CP-maps. 
One can rewrite $\mathcal O=\sum_{k,l}o_k|k,l\rangle\langle k,l|$, where the $o_k$ are in decreasing order. The non-increasing condition $\langle\mathcal O\rangle$ corresponds then to  an non-increasing condition on $k$.
Writing $\Gamma$ in terms of Kraus operators, one has
$\Gamma(\rho)=\sum_i B_i\rho B_i^*$ with $\sum_iB_iB_i^*=\mathbb1$.
The condition on $k$ given above is then translated into the following writing of the Kraus operators:
$$B_i=\sum_{\substack{k,l,k',l'\\k\le k'}}B_i^{klk'l'}|k,l\rangle\langle k',l'|.$$
Another way to say the same thing is the condition $B_i^{klk'l'}=0$ if $k>k'$.
Then, of course, the normalization condition imposes
$$ \sum_{\substack{i,k',l'//k\le k'}}\left|B_i^{klk'l'}\right|^2=1, \forall k,l.$$
If you apply the same reasoning with a non-increasing and non-decreasing condition, you find that $k$ has to be conserved, and this is then equivalent to the commutativity condition you give in your question. In the same way, this answer is not general: you have  operations which preserve $\langle\mathcal O\rangle$ without commuting with $\mathcal O$.
A: This is probably not exactly what you had in mind, but how about the channel that discards its input and always outputs the state corresponding the the minimum eigenvalue of $\mathcal{O}$?
A: The formal condition---the correspondent of $\Phi^*[\mathcal{O}]=\mathcal{O}$ in the other case---is $\Gamma^*[\mathcal{O}]\leq \mathcal{O}$. For $\mathcal{O}>0$ (all eigenvalues strictly positive), if one multiplies on the right and on the left by $\mathcal{O}^{-1/2}$, one can see that this condition is equivalent to the map with Kraus operators $\mathcal{O}^{1/2}K_i\mathcal{O}^{-1/2}$ being trace non-increasing.
