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<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}.

Exponential function in red, almost-completely-monotone in blue

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.

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