I had a question about checking whether $f(x)=(1-x)^2$ is completely monotonic. My argument is that it is not, because:

  1. It is not strictly monotonic on $[0, \infty)$.
  2. It does not satisfy the following necessary and sufficient condition: $$ (-1)^n \frac {d^n}{dt^n}f(t)\ge0$$ for all nonnegative integers $n$ and for all $t > 0$.

Which is: $$ (-1)^0 \frac {d^0}{dx^0}f(x)=(1-x)^2\ge0$$ $$ (-1) \frac {d}{dx}f(x)=2(1-x)\ge0 \quad {\color{red}{ but \;this \; one\; fails\; for\; x>1}}$$ $$ (-1)^2 \frac {d^2}{dx^2}f(x)=2\ge0$$

My other 3 questions are:

  1. do we need all derivatives to alternate in sign or they can be all zero beyond a particular higher order derivative?
  2. is constant $0$ function completely monotonic?
  3. does complete monotonicity imply strict monotonicity?
  • $\begingroup$ How do you define completely monotonic? $\endgroup$ Mar 16, 2019 at 14:32
  • $\begingroup$ @José Carlos Santos, here is the definition link $\endgroup$
    – Alex
    Mar 16, 2019 at 14:34

2 Answers 2


See, for instance, this Wikipedia article.

Your argument that $(1-x)^2$ is not CM is correct.

As to your other 3 questions. 1: Since "be zero from a certain point on" is a special case of the usual conditions, the change in wording would not change the set of functions described. 2: Yes: the constant function $f(t)=c$, for nonnegative $c$, is the Laplace transform of the Borel measure $c\delta_0$ that puts mass $c$ at $\{0\}$ and none anywhere else. 3: No, because of the constant functions. Otherwise, by the characterization of CM functions as Laplace transforms of positive Borel measures, if $f$ is not a constant function and is CM, then it is strictly decreasing.

  • $\begingroup$ many thanks again! Can I just confirm that I understood correctly: complete monotonicity imply monotonicity, but does not imply strict monotonicity because of constant zero function. Otherwise, if we do not consider the constant zero function, then the implication is true (i.e. it fails because of constant zero function). By the way, is then any positive constant a completely monotonic function? $\endgroup$
    – Alex
    Mar 16, 2019 at 14:59
  • $\begingroup$ Yes, and better than I because I had overlooked nonzero constant functions which are clearly CM, too. I've edited my answer accordingly. $\endgroup$ Mar 16, 2019 at 15:21
  • $\begingroup$ Excellent! Many thanks @kimchi lover! $\endgroup$
    – Alex
    Mar 16, 2019 at 15:37

A non-negative function $f$ is said to be completely monotonic on an interval $I$ if $f$ has derivatives of all orders on $I$ and \begin{equation*} 0\le(-1)^{n-1}f^{(n-1)}(x)<\infty \end{equation*} for all $x\in I$ and $n\in\mathbb{N}=\{1,2,3,\dotsc\}$.

A non-negative function $f$ is said to be absolutely monotonic on an interval $I$ if $f$ has derivatives of all orders on $I$ and \begin{equation*} 0\le f^{(n-1)}(x)<\infty \end{equation*} for all $x\in I$ and $n\in\mathbb{N}$.

If a function $f$ is non-identically zero and completely monotonic on $(0,\infty)$, then $f$ and its derivatives $f^{(n)}(x)$ for $n\in\mathbb{N}$ are impossibly equal to $0$ on $(0,\infty)$.

Answer: the function $f(x)=(1-x)^2$ is completely monotonic on $(0,1)$, is absolutely monotonic on $(1,\infty)$, but is not on $(0,\infty)$.


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