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

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

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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 –  Gerhard Paseman Jan 25 '12 at 19:36
Both functions are log-convex, not log-concave. –  Mark Meckes Jan 25 '12 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 –  Gerhard Paseman Jan 25 '12 at 21:07
The poster said (log f)''<0 (second derivative, not first). Although that's apparently the opposite of what was meant. –  Mark Meckes Jan 26 '12 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.

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