Let $M$ be some bounded positive operator with bound $\lVert M\rVert<1$ and let $\rho$ be a positive trace class, self adjoint operator with norm $\lVert\rho\rVert=1$. Assume that these operators act on an infinite dimensional Hilbert space (I don't think that the dimension matters to be honest). Is the following true?
$$ \rho\geq M\rho M $$ in other words is the the operator $\rho- M\rho M$ positive? Yet again in other words, is the following true?
$$ \langle v, \rho v\rangle \geq \langle v, M\rho M v\rangle $$ for any $v$ in the Hibert space that $\rho$ and $M\rho M$ are acting on.
The inequality does not hold even for two dimensional space. For $0<\delta<1$ and $ 0<a<b<{1\over 2}$ let $${\rho} =\begin{pmatrix} 1&0\\ 0&\delta\end{pmatrix},\quad M =\begin{pmatrix} b&a\\ a&b\end{pmatrix}$$ Then $\|\rho\|=1,$ $M>0$ and $\|M\|=a+b<1$ and $$ M{\rho} M= \begin{pmatrix} b^2+\delta a^2&(1+\delta) ab\\ (1+\delta)ab & \delta b^2+a^2\end{pmatrix}$$ Hence $$\rho -M\rho M= \begin{pmatrix} 1-b^2-\delta a^2&-(1+\delta)ab\\ -(1+\delta)ab&\delta-\delta b^2-a^2\end{pmatrix}$$ For $\delta={1\over 25},$ $ a={1\over 5}$ and $b={1\over 4}$ we get $$\delta-\delta b^2-a^2=-\delta b^2=-\textstyle {1\over 400 }<0$$ Hence the matrix $\rho -M\rho M$ is not positive semidefinite.