Sufficient and necessary condition in order for a homographic function be absolutely monotonic

Question

Let be $I$ a non-empty and open interval of $\mathbb{R}$.

A function $f : I \longrightarrow \mathbb{R}$ is said absolutely monotonic on $I$ if :

1. $f$ is $\mathcal{C}^{\infty}$ on $I$ (differentiable as much as you want, continuous for all derivatives).
2. $\forall n \in \mathbb{N}, \forall x \in I, f^{(n)}(x) \geq 0$ where $f^{(n)}$ is the n-th derivative.

Let be $a, b, \alpha \in \mathbb{R}$, we call an homographic function, a function of this form :

\begin{equation*} f(x) = \dfrac{ax + b}{x - \alpha} \end{equation*}

What would be a sufficient and necessary condition in order for $f(x)$ to be absolutely monotonic?

Here is what I have attempted:

• $f$ is $C^{\infty}$ by algebraic operations (quotient).
• I wanted to find a closed form of the $n$-th derivative of $f$. I have thought of using Leibniz formula, obviously, it didn't work because the numerator becomes too complex to handle.
• I remarked that the denominator is always positive due to being always squared when derivating again.
• Knowing that denominator is always positive, I eventually need to do is to consider when $x < \alpha$ and $x > \alpha$, and prove that all derivatives' numerators have the same sign. So I can deduce the overall sign of the fraction. That's where I don't how to proceed further.

Answer 1

Let be $a, b, \gamma \in \mathbb{R}$ such that $a < b$ and $I = ]a, b[$ so that $\gamma \not \in I$.

Let $\alpha, \beta \in \mathbb{R}$ such that:

$\forall x \in I, f(x) = \dfrac{\alpha x + \beta}{x - \gamma}$, an homographic function.

First, for all $n \geq 1$, for all $a \in \mathbb{C}$, for all $x \in \mathbb{R} \setminus \{ a \}$ :

$\begin{equation*} \dfrac{d^n}{dx^n}\left(\dfrac{1}{x - a}\right) = (-1)^n \dfrac{n!}{(x - a)^{n + 1}} \end{equation*}$

Then, for all $x \in I$ :

$\begin{align*} \dfrac{d^n}{dx^n}f(x) & = \alpha\dfrac{d^n}{dx^n}\left(\dfrac{x}{x - \gamma}\right) + \beta\dfrac{d^n}{dx^n}\left(\dfrac{1}{x - \gamma}\right) \\ & = \alpha \dfrac{d^{n - 1}}{dx^{n - 1}}\left(\dfrac{(x - \gamma) - x}{(x - \gamma)^2}\right) + \beta (-1)^n \dfrac{n!}{(x - \gamma)^{n + 1}} \\ & = -\alpha \gamma \dfrac{d^{n - 1}}{dx^{n - 1}}\left(\dfrac{1}{(x - \gamma)^2}\right) + \beta (-1)^n \dfrac{n!}{(x - \gamma)^{n + 1}} \end{align*}$

By Leibniz's formula :

$\begin{align*} \dfrac{d^{n - 1}}{dx^{n - 1}} \dfrac{1}{(x - \gamma)^2} & = \sum_{k=0}^{n - 1} \binom{n - 1}{k} \dfrac{(-1)^k k!}{(x - \gamma)^{k + 1}} \dfrac{(-1)^{n - k - 1} (n - k - 1)!}{(x - \gamma)^{n - k}} \\ & = (-1)^{n - 1}(n - 1)! \sum_{k=0}^{n - 1} \dfrac{1}{(x - \gamma)^{n + 1}} \\ & = (-1)^{n - 1} \dfrac{n!}{(x - \gamma)^{n + 1}} \end{align*}$

At the end :

$\begin{align*} \dfrac{d^n}{dx^n}f(x) & = -\alpha \gamma \dfrac{d^{n - 1}}{dx^{n - 1}}\left(\dfrac{1}{(x - \gamma)^2}\right) + \beta (-1)^n \dfrac{n!}{(x - \gamma)^{n + 1}} \\ & = \alpha \gamma (-1)^n \dfrac{n!}{(x - \gamma)^{n + 1}} + \beta (-1)^n \dfrac{n!}{(x - \gamma)^{n + 1}} \\ & = \dfrac{(-1)^n n!}{(x - \gamma)^{n + 1}}(\alpha \gamma + \beta) \end{align*}$

Then $f$ is absolutely monotonic if and only if for all $n \in \mathbb{N}$, for all $x \in I$, $f^{(n)}(x) \geq 0$.

Case $n \geq 1$

Sub-case : $n$ is even

Let's write $n = 2k, k \in \mathbb{N}^{*}$.

Then, for all $x \in I$ :

$\begin{align*} f^{(n)}(x) & = f^{(2k)}(x) \\ & = \dfrac{(-1)^{2k} (2k)! (\alpha\gamma + \beta)}{(x - \gamma)^{2k + 1}} \\ & = \dfrac{(2k)! (\alpha \gamma + \beta)}{(x - \gamma)^{2k + 1}} \end{align*}$

Then, $f^{(n)}(x) \geq 0$ if and only if $(\alpha \gamma + \beta)(x - \gamma)^{2k + 1} \geq 0$.

• Either $\alpha \gamma + \beta \geq 0$ and $x - \gamma \geq 0$ for all $x \in I$, i.e. $x \geq \gamma$, i.e. $a \geq \gamma$ (1)
• Either $\alpha \gamma + \beta \leq 0$ and $x - \gamma \leq 0$ for all $x \in I$, i.e. $x \leq \gamma$, i.e. $b \leq \gamma$ (2)

Sub-case : $n$ is odd

Let's write $n = 2k + 1, k \in \mathbb{N}$.

Then, for all $x \in I$ :

$\begin{align*} f^{(n)}(x) & = f^{(2k + 1)}(x) \\ & = \dfrac{(-1)^{2k + 1} (2k + 1)! (\alpha \gamma + \beta)}{(x - \gamma)^{2k + 2}} \\ & = \dfrac{-(2k + 1)! (\alpha \gamma + \beta)}{((x - \gamma)^k)^2} \end{align*}$

Then $-(\alpha \gamma + \beta) \geq 0$.

That is : $\alpha \gamma + \beta \leq 0$.

Then, in the previous subcase, (1) is impossible, so that we only have (2).

Let's summarize so far.

In order for $f$ to be absolutely monotonic, it is necessary (but not sufficient):

• $\alpha \gamma + \beta \leq 0$
• $b \leq \gamma$.

Case $n = 0$

$f(x) \geq 0$ for all $x \in I$ if and only if $(\alpha x + \beta)(x - \gamma) \geq 0$.

That is:

• Either $\alpha x + \beta \geq 0$ and $x - \gamma \geq 0$ (1)
• Either $\alpha x + \beta \leq 0$ and $x - \gamma \leq 0$ (2)

(1) is impossible because $x < \gamma$ due to the previous case. Then we have (2), i.e. $\alpha x + \beta \leq 0$ for all $x \in I$.

Then : $\alpha x \leq -\beta$.

• If $\alpha > 0$, then $x \leq -\dfrac{\beta}{\alpha}$, i.e. $b \leq -\dfrac{\beta}{\alpha}$, i.e. $b \leq \min \left\{ -\dfrac{\beta}{\alpha}, \gamma \right\}$.
• If $\alpha < 0$, then $x \geq -\dfrac{\beta}{\alpha}$, i.e. $a \geq -\dfrac{\beta}{\alpha}$.

In the end:

If $f$ is absolutely monotonic, necessarily, we have:

• $\alpha \gamma + \beta \leq 0$.
• If $\alpha > 0$, $I \subset \left]-\infty, \gamma\right[$ because $\gamma \leq -\dfrac{\beta}{\alpha}$.
• If $\alpha < 0$, $I \subset \left]-\dfrac{\beta}{\alpha}, \gamma\right[$.

By construction, such an homographic function is always absolutely monotonic, so these conditions are sufficient.