I currently study matrix monotone. I met some trouble with the proof below.
Prove if $f: \mathbb{R}^{+} \rightarrow \mathbb{R}$ is a matrix monotone function, then $-f$ is a completely monotone function.
I can give an example for the proof below. $f(x)=logx$ is matrix monotone, and then it satisfies the condition for the completely monotone function.
And I know the definition completely monotone funciton $a function f:(0,\infty) \rightarrow[0, \infty)$ is said to be completely monotone if $(-1)^{k} f^{(k)} \geq 0$ for $x>0$ and $ k=0, 1, 2, \dots$
I can not understand how to prove it generally. Can anyone give me some concepts or theorems about the question?
There seem to be some problems here. First, $-\log$ is not positive on $(0,\infty)$, and even if you replace it by something like $-\log(1+\cdot)$, the function and its derivative have the same sign.
What is true is that every matrix monotone function $f\colon (0,\infty)\to(0,\infty)$ is a Bernstein function, namely $(-1)^k f^{(k)}\leq 0$ for $k\geq 1$. In other words, $f'$ is completely monotone. This follows from Löwner's theorem:
The function $f$ has an integral representation of the form $$ f(x)=\int_0^1\frac{x}{t+(1-t)x}\,d\mu(t)=\int_0^1 \frac{1}{tx^{-1}+(1-t)}\,d\mu(t) $$ with a positive measure $\mu$ on $[0,1]$.
By induction and the dominated convergence theorem one can show, $$ f^{(k)}(x)=-(-1)^k k!\int_0^1\frac{t(1-t)^{k-1}}{(t+(1-t)x)^{k+1}}\,d\mu(t), $$ which implies $(-1)^kf^{(k)}\leq 0$, so that $f$ is indeed a Bernstein function.