$f(t) = \min\{1,t\}$ not operator monotone I want to show that the function on $\mathbb{R}^+$ $f(t) = \min \{1,t\}$ is not operator monotone on the complex $2\times 2$ matrices. My plan is to find matrices $A$ and $B$, $B\geq A$, such that the spectrum of $A$ is in $[0,\infty]$ and the spectrum of $B$ contains elements greater than 1. Then applying $f$ should 'make $B$ smaller' such that $f(B) \not \geq f(A)$. However, I struggle to find a concrete example. Can someone help me finding such $A$ and $B$?

You are on the right track. Actually any pair of matrices such that


*

*$A$ has spectrum on $[0, 1]$,

*$B$ has an eigenvalue on $(1, \infty)$, and

*$B - A$ is of rank $1$

*$A$ and $B$ do not commute


will work as an example.
Indeed, $f(t) \leq t$ together with 2. imply that $B \neq f(B) \leq B$. Since also $f(A) = A$,
$$
f(B) - f(A) = f(B) - A \leq B - A.
$$
Since the RHS is of rank $1$, and the second inequality is not equality, the only way the LHS can be positive is that $f(B) - f(A) = c(B - A)$ for some $0 \leq c < 1$, or
$$
 A = \frac{1}{1 - c} \left(f(B) - c B\right).
$$
But this implies that $A$ commutes with $B$, contradiction.
As a concrete example one may for instance take
$$
A = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix},\;\;
B = A + \begin{bmatrix}
\frac12 & \frac12 \\
\frac12 & \frac12
\end{bmatrix}
= \begin{bmatrix}
\frac32 & \frac12 \\
\frac12 & \frac12
\end{bmatrix}.
$$
Again $f(A) = A$. One may calculate $f(B)$ using the definitions directly but the task might be slightly easier using the following observation: if $f$ and $g$ agree on the spectrum of $B$, then $f(B) = g(B)$. One may check that the eigenvalues of $B$ are $1 \pm \frac{1}{\sqrt{2}}$, so it suffices to find a polynomial $p$ with
$$
p\left(1 - \frac{1}{\sqrt{2}}\right) = 1 - \frac{1}{\sqrt{2}}\; \text{ and } p\left(1 + \frac{1}{\sqrt{2}}\right) = 1;
$$
then $f(B) = p(B)$. Degree $1$ example $p(x) = \frac{1}{2} x + \frac{1}{2} - \frac{1}{2 \sqrt{2}}$ will do, and thus
$$
f(B) = p(B) =
\frac{1}{2}
\begin{bmatrix}
\frac32 & \frac12 \\
\frac12 & \frac12
\end{bmatrix}
+ \left(\frac{1}{2} - \frac{1}{2 \sqrt{2}} \right)
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix} \\
= \begin{bmatrix}
\frac54 - \frac{1}{2 \sqrt{2}} & \frac{1}{4} \\
\frac{1}{4} & \frac34 - \frac{1}{2 \sqrt{2}}
\end{bmatrix}.
$$
Finally,
$$
f(B) - f(A) = \begin{bmatrix}
\frac14 - \frac{1}{2 \sqrt{2}} & \frac{1}{4} \\
\frac{1}{4} & \frac34 - \frac{1}{2 \sqrt{2}}
\end{bmatrix} \not\geq 0,
$$
as the determinant of the matrix is $\frac{1}{4} \left(1 - \sqrt{2}\right) < 0$.
