A non-negative function $f(x)$ is said to be completely monotonic on an interval $I$ if $f(x)$ 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 $n\in\mathbb{N}=\{1,2,3,\dotsc\}$.

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

As for completely monotonic functions, there are two kinds of convexities:

- If a function $f(x)$ is completely monotonic on an interval $I$, by the above definition, it is trivial that the function $f(x)$ is surely convex, that is, $f''(x)\ge0$, on the interval $I$.
- If a function $f(x)$ is completely monotonic on the infinite interval $(0,\infty)$, then the derivative sequence $f^{(n)}(x)$ in $n\ge0$ for $x\in(0,\infty)$ is surely logarithmically convex in $n\ge0$, that is,
\begin{equation}
\frac{f^{(i)}(x)}{f^{(i+1)}(x)}\ge\frac{f^{(i+1)}(x)}{f^{(i+2)}(x)}, \quad i=0,1,2,\dotsc, \quad x\in(0,\infty).
\end{equation}

A positive function $f(x)$ is said to be logarithmically completely monotonic on an interval $I$ if its logarithm $\ln f(x)$ satisfies
\begin{equation*}
0\le(-1)^n[\ln f(x)]^{(n)}<\infty
\end{equation*}
for all $n\in\mathbb{N}$ on $I$.

A logarithmically completely function on an interval $I$ must be also completely monotonic on $I$, but not conversely.

The definition of logarithmically completely monotonic functions and the above relation between completely monotonic functions and logarithmically completely monotonic functions demonstrate that, **if a function $f(x)$ is completely monotonic, but not logarithmically completely monotonic, on an interval $I$, then it is possible, but not sure, that the completely monotonic function $f(x)$ is logarithmically concave or logarithmically convex on $I$.**

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