Let $A,B$ be self-adjoint operators and $A\leq B$. Is $e^A \leq e^B$ true? 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 $n$.
 A: 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 $n\geq 1$.
A: 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.

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 $d=1$).
