$(1-x)^2$ function

Question

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?

Answer 1

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.

Answer 2

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 $(0,\infty)$.

References

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