Infinitely differentiable functions

01.06.2012

Let $f,g : X \to \mathbb{R}$ be infinitely differentiable functions, where $X \subset \mathbb{R}$ . In this post I will show some closure properties of this class of functions, such that it becomes easy to see whether a given composite function is infinitely differentiable (but not necessarily that it isn’t). This is of interest when studying the theory of generalized functions.

Derivatives

The derivative of $f$ is infinitely differentiable:

$\begin{equation} \begin{split} D^k(D f) = D^{k + 1} f. \end{split} \end{equation}$

By induction, this also shows that any finite derivative $D^k f$ of $f$ is infinitely differentiable.

Products

The product $fg$ is infinitely differentiable. By the product rule,

$\begin{equation} \begin{split} D(fg) = (Df) g + f(Dg). \end{split} \end{equation}$

Applying this rule again,

$\begin{equation} \begin{split} D^2(fg) = (D^2 f)g + 2(Df)(Dg) + f(D^2 g). \end{split} \end{equation}$

A pattern emerges. By induction one shows the general Leibniz rule:

$\begin{equation} \begin{split} D^k(fg) = \sum_{i = 0}^k \binom{k}{i} (D^i f)(D^{k-i} g), \end{split} \end{equation}$

for $k \in \mathbb{N}$ .

By induction, this also shows that $f^n$ , where $n \in \mathbb{N}$ , is infinitely differentiable (the cases $n = 0$ and $n = 1$ are trivial).

Linear combinations

The linear combination $\alpha f + \beta g$ , where $\alpha, \beta \in \mathbb{R}$ , is infinitely differentiable. By linearity of differentiation,

$\begin{equation} \begin{split} D(\alpha f + \beta g) = \alpha D(f) + \beta D(g). \end{split} \end{equation}$

By induction one shows

$\begin{equation} \begin{split} D^k(\alpha f + \beta g) = \alpha D^k(f) + \beta D^k(g), \end{split} \end{equation}$

for $k \in \mathbb{N}$ .

By induction, this also shows that any finite linear combination is infinitely differentiable.

Compositions

The composition $f \circ g$ , where the range of $f$ is contained in the domain of $g$ , is infinitely differentiable. By writing down the first few formulas, observing a pattern (not easy), and again using induction, one shows the Faà di Bruno’s formula for the $k$ :th derivative of $f \circ g$ .

By induction, this also shows that any finite composition of infinitely differentiable functions is infinitely differentiable.

Multiplicative inverses

The inverse $1/f$ is infinitely differentiable, where $\forall x \in X: f(x) \neq 0$ . Assume $1/f$ is $k$ -times differentiable, with $k \in \mathbb{N}^{>0}$ . Using the general Leibniz rule,

$\begin{equation} \begin{split} 0 & = D^k(1) \\ {} & = D^k(f/f) \\ {} & = \sum_{i = 0}^k \binom{k}{i} (D^i f)(D^{k-i} (1/f)). \end{split} \end{equation}$

It follows that if $1/f$ is $k$ -times differentiable, then the $k$ :th derivative must be given by

$\begin{equation} \begin{split} D^k(1/f) & = -(1/f) \sum_{i = 1}^k \binom{k}{i} (D^i f)(D^{k - i} (1/f)). \end{split} \end{equation}$

Since the $k$ :th derivative is given in terms of lower derivatives, this formula can be proved to hold by induction.

In particular, this shows that $f^n$ is infinitely differentiable for $n \in \mathbb{Z}$ .