# Almost completely monotone, but not quite

A function $f\colon (0,\infty)\to\mathbb R$ is called completely monotone if for every integer $n\ge 0$ and every $t>0$ we have $(-1)^n f^{(n)}(t)\ge 0$. (In particular, the derivatives must exist). For example, $f(t)=1/t$ has this property, and so does $f(t)=e^{-t}$. A completely monotone function is positive, decreasing, convex (all in the non-strict sense), and so are its derivatives of even orders. An innocent-looking calculus-type condition. Yet, it has some surprising consequences.

1. Every completely monotone function is real-analytic on $(0,\infty)$: that is, it is locally represented by its Taylor series
2. Moreover, the Taylor series centered at $a>0$ has the radius of convergence $R\ge a$: in other words, $1/t$ is the worst that can happen.
3. Moreover, $f$ extends to a complex-analytic function on the right halfplane $\mathrm{Re}\, z>0$. Of course, this implies the items 1 and 3.

All of this follows at once from Bernstein’s theorem (1928): $f$ is completely monotone if and only if it is the Laplace transform of a positive measure: explicitly, if $\displaystyle f(t)=\int_0^{\infty}e^{-tx}\,d\mu(x)$ for some positive Borel measure $\mu$ on $[0,\infty)$. The measure may be infinite, but the integral is required to converge for all $t>0$. The case when $\mu$ is finite precisely corresponds to $f$ being bounded, and hence continuous on $[0,\infty)$ with $f(0)=\mu([0,\infty))$. Some sources, including Wikipedia, deal only with the finite/bounded case. The complete treatment of Bernstein’s theorem, which also explains what kind of monotonicity it deals with, can be found in Barry Simon’s book Convexity: an analytic viewpoint which I heartily recommend.

This being a Calculus-oriented blog, let’s do an exercise: given $N>0$, construct an infinitely differentiable function on $(0,\infty)$ which satisfies the condition $(-1)^n f^{(n)}(t)\ge 0$ for $n but not for $n\ge N$. It's intuitively clear that such functions exist, but can we find a simple example?

A slight perturbation of the prototypical completely monotone function $e^{-t}$ does the job: take $\displaystyle f(t)=e^{-t}(1+\varepsilon\sin t)$

Let’s see: $f'(t)=-e^{-t}(1+\varepsilon \sin t-\varepsilon\cos t)$, $f''(t)=e^{-t}(1-2\varepsilon \cos t)$
Well, we could survive the computations in this form, but it’s better to use the identity $\sin t-\cos t=\sqrt{2}\sin(x-\pi/4)$ and get everything done at once: $\displaystyle (-1)^nf^{(n)}(t)=e^{-t}\big(1+2^{n/2}\varepsilon\sin (t-\pi n/4)\big),\qquad n=0,1,2,\dots$

Now choose $\varepsilon$ in the range $2^{-N/2}<\varepsilon\le 2^{(1-N)/2}$.

A trigonometric aside. Compared to $\sin^2 t+\cos^2 t=1$, the identity $\sin t+\cos t=\sqrt{2}\sin (t+\pi/4)$ has very few fans. I remember seeing it in high school and shuddering from fear and disgust. What right had $\sqrt{2}$ and $\pi/4$ to be there? Only much later did it dawn on me that $\sin t +\cos t = y+x$ can be taken as a new coordinate $y'$, and that the transformation $x'=x-y$, $y'=x+y$ consists of scaling by $\sqrt{2}$ and rotation by $\pi/4$.