Let $A,B$ be self-adjoint operators and $A\leq B$. Is $e^A \leq e^B$ true?

Question

If $A,B$ commute, that $e^A\leq e^B$ follows from functional calculus.

Is this still true when $A,B$ do not commute?

Also I wonder if $A^{2n+1}\leq B^{2n+1}$ is true for natural number $n$.

Answer 1

A continuous function $f:\mathbb R\to \mathbb R$ is said to be operator monotone if, for every self-adjoint operators $A$ and $B$ such that $A\leq B$, one has that $f(A)\leq f(B)$.

Operator monotone functions were characterized by Lowner as being the functions which admit a holomorphic extension $\tilde f$ to the upper half plane $$ \mathbb C_+ = \{z\in \mathbb C: \Im (z) > 0\}, $$ such that $\tilde f(\mathbb C_+)\subset \mathbb C_+$. Since the exponential does not satisfy Lowner's condition, it is not operator monotone.

The same goes for $f(x)=x^{2n+1}$, for $n\geq 1$.

Answer 2

Example.

$$ A = \begin{bmatrix} 1/2&1\cr 1&1 \end{bmatrix} \\ B = \begin{bmatrix} 1&0\cr 0&3 \end{bmatrix} $$ Then $A \le B$ is true, but $A^3 \le B^3$ is false and $e^A \le e^B$ is false.

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How did I find it?
For a $2 \times 2$ matrix $$ \begin{bmatrix} a&b\cr c&d \end{bmatrix} $$ the conditions for positive semidefinite are $$ b=c,\quad a \ge 0,\quad ad-bc \ge 0 . $$ I know that a self-adjoint matrix can be diagonalized, and I normalized an entry to $1$. Using $$ A = \begin{bmatrix} a&b\cr c&d \end{bmatrix},\qquad B = \begin{bmatrix} 1&0\cr 0&f \end{bmatrix} $$ the conditions for $A \le B$ are $$ b=c,\quad a\le 1,\quad ad-af-cb - a + f \ge 0 . \tag1$$ The conditions for $A^3 \le B^3$ are $$ -{a}^{2}b-abd-{b}^{2}c-b{d}^{2}=-{a}^{2}c-acd-b{c}^{2}-c{d}^{ 2}, \\ a^3+2abc+bcd \le 1 \\ {a}^{3}{d}^{3}-{a}^{3}{f}^{3}-3\,{a}^{2}b c{d}^{2}+3\,a{b}^{2}{c}^{2}d-2abc{f}^{3}-{b}^{3}{c}^{3}-bcd{f}^{3}-a bc-2bcd-{d}^{3}+{f}^{3} \ge 0 . \tag2$$ Then we have to find values that satisfy $(1)$ but not $(2)$. Mostly I did it with equality for the inequalities, except for one strict inequality forward in $(1)$ (choosing $a=1/2$) and then one backward in $(2)$ (choosing $d=1$).