Let $f(x)$ be a completely monotone function on $x>0$, i.e. $f(x) \in C^{\infty}(0,+\infty)$ and for any $n \geqslant 0$ we have $$ (-1)^n \frac{\mathsf d^n f(x)}{\mathsf dx^n} \geqslant 0, \tag{1} $$ and let $f(0) = f(+0) = 1$, but $f(x) \not\equiv 1$. Is it possible that $f(x^{\alpha})$ is also completely monotone for some $\alpha>0$, $\alpha \neq 1$?

We can test inequalities $(1)$ for $f(x^{\alpha})$ but in this case we will obtain an infinite system of differential inequalities that seems insolvable. The other way is to use the Bernstein theorem on monotone functions that states that if $f(x)$ is completely monotone then there exists a Borel nonnegative measure $\mu$ on $[0,+\infty)$, $\mu[0,+\infty) = f(+0)$ and such that $$ f(x) = \int\limits_{0}^{\infty} e^{-xy} \, \mathsf d \mu(y). \tag{2} $$ If $f(x^{\alpha})$ is also completely monotone then there exists some measure $\nu(x)$ such that $$ f(x^{\alpha}) = \int\limits_{0}^{\infty} e^{-xy} \, \mathsf d \nu(y). \tag{3} $$ From $(2)$ and $(3)$ it follows that $$ \int\limits_{0}^{\infty} e^{-x^{\alpha} y} \, \mathsf d \mu(y) = \int\limits_{0}^{\infty} e^{-x y} \, \mathsf d \nu(y). $$ and without loss of generality $\alpha > 1$ (otherwise we can introduce $\tilde{\alpha} = \alpha^{-1}$ and $\tilde{x} = x^{\alpha}$). On the right hand side we have a completely monotone function. Then on the left hand side the function is also completely monotone and by virtue of the Bernstein theorem it satisfies $(1)$. Let's take it derivatives up to second order and write inequalities $(1)$: $$ -\alpha \int\limits_{0}^{\infty} x^{\alpha-1} y e^{-x^{\alpha}y} \, \mathsf d\mu(y) \leqslant 0, - \,\text{it is always true} \\ -\alpha x^{\alpha-2} \int\limits_{0}^{\infty} y \left( \alpha-1 - \alpha yx^{\alpha} \right) e^{-x^{\alpha}y} \, \mathsf d\mu(y) \geqslant 0. $$ Let's divide the last inequality by $-\alpha x^{\alpha-2}$ and rewrite as $$ (\alpha-1) \int\limits_{0}^{\infty} y e^{-x^{\alpha} y} \, \mathsf d \mu(y) \leqslant \alpha x^{\alpha} \int\limits_{0}^{\infty} y^2 e^{-x^{\alpha} y} \, \mathsf d \mu(y). $$ If $y^2$ is integrable with respect to $\mu$ then we can pass to the limit $x \to +0$ to obtain the contradiction: on the left hand side we will obtain a positive number whereas on the right hand side it will be zero. For more general classes of measure $\mu$ that represents completely monotone function $f$ this proof of the fact that $f(x)$ and $f(x^{\alpha})$ can not both be completely monotone nonconstant functions doesn't work.


I am confused... is there any completely monotone function $f$ for which $f(x^\alpha)$ with $0<\alpha<1$ is not completely monotone? I could not find any such example. A completely monotone function has a holomorphic extension to right half-plane. I have a very vague (possibly false) memory of a converse statement such as: a holomorphic function in the right half-plane that is positive and decreasing on $(0,\infty)$ is completely monotone. If this is true, then $f(x^\alpha)$ is always completely monotone when $f$ is completely monotone and $0<\alpha<1$. I hope someone reading this will clarify.

Anyway, here is a concrete example: both $e^{-x}$ and $e^{-\sqrt{x}}$ are completely monotone. For the first this is obvious: $e^{-x}$ is the mother of all completely monotone functions (with S. Bernstein being the father, I suppose). The complete monotonicity of $e^{-\sqrt{x}}$ is a consequence of the following fact: $$(-1)^n \frac{d^n}{dx^n}e^{-\sqrt{x}} = P_n(x^{-1/2})\,e^{-\sqrt{x}} \tag1$$ where $P_n$ is a polynomial with nonnegative coefficients.

Proof of (1), by induction: $n=0$ is trivially true. If (1) holds for particular $n$, then differentiating both sides we get $$\begin{align}(-1)^{n+1} \frac{d^{n+1}}{dx^{n+1}}e^{-\sqrt{x}} &= \left(P_n'(x^{-1/2})\cdot \frac{1}{2}x^{-3/2} + P_n(x^{-1/2})\cdot \frac{1}{2}x^{-1/2}\right) \,e^{-\sqrt{x}} \\& = P_{n+1}(x^{-1/2})\,e^{-\sqrt{x}} \end{align} \tag2$$ where $P_{n+1}(y)=\frac12 (y^3P_n'(y)+yP_n(y))$ clearly has nonnegative coefficients. $\Box$

  • $\begingroup$ It is a nice observation. If $f(x)=e^{-\sqrt{x}}$ is represented by measure $\mu$ then $x^2$ is not integrable with respect to $\mu$, because $+\infty = f''(+0) = \int x^2 \mu(dx)$. Great thanks! $\endgroup$ – Appliqué Mar 25 '13 at 18:48

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