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