Infinitely differentiable functions

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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:

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,

Applying this rule again,

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

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,

By induction one shows

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,

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

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 $n \in \mathbb{Z}$.