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

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: $f:(0, \infty) \rightarrow (0,\infty)$.

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

• Log of both examples is like 1/x, which satisfies the negative derivative condition. What am I missing that makes them counterexamples? Gerhard "Ask Me About System Design" Paseman, 2012.01.25 Jan 25, 2012 at 19:36
• Both functions are log-convex, not log-concave. Jan 25, 2012 at 20:01
• So then when the original poster says (log f)' < 0, they mean f is log convex? Gerhard "Sometimes Confuses Up And Down" Paseman, 2012.01.25 Jan 25, 2012 at 21:07
• The poster said (log f)''<0 (second derivative, not first). Although that's apparently the opposite of what was meant. Jan 26, 2012 at 15:10

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.

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.

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, $$$$\frac{f^{(i)}(x)}{f^{(i+1)}(x)}\ge\frac{f^{(i+1)}(x)}{f^{(i+2)}(x)}, \quad i=0,1,2,\dotsc, \quad x\in(0,\infty).$$$$

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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