Does complete monotonicity of f imply log-concavity of f?

Question

Let f be a completely monotonic function with $f(0)=1$, that is, $ f:[0, \infty) \rightarrow (0,1] $. My question is:

Is f log concave, that is, is $(logf)''<0$ or equivalently $ f f''< f'^2 $. ?

And what hapens if $f(0)=\infty$, that is if the function is: $ f:(0, \infty) \rightarrow (0,\infty) $.

Answer 1

A counterexample in the second case is $f(x) = e^{1/x}$. A counterexample in the first case is then $f(x) = e^{1/(x+1) - 1}$.

Answer 2

Exercise 6 of this book shows that if $f: (0,\infty) \to \mathbb{R}$ is completely monotonic, then it must be log-convex. Hence, your second claim holds, with concavity replaced by convexity.

Answer 3

Actually, I meant log-convex instead of log-concave. I missed a minus sign in my derivations and this led to the conjecture that log(f) should be concave. I have found a more complete answer in this paper:

Inequalities for Real Powers of Completely Monotonic Functions H. van Haeringen JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS Volume 210, Issue 1, 1 June 1997, Pages 102–113

Theorem 1 establishes a series of inequalities for the derivatives of c.m. functions. In particular, by taking n=0 and m=1 in 3.2 we get the log-convexity of f. I suggest to read this paper because of the relrevance of Theorem 1.

Answer 4

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:

1. 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 $I$.
2. 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 $I$.

References

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