Derivative Function is not Invertible
Theorem
Let $\Bbb I = \closedint a b$ be a closed interval on the set of real numbers $\R$ such that $a < b$.
Let $A$ denote the set of all continuous real functions $f: \Bbb I \to \R$.
Let $B \subseteq A$ denote the set of all functions differentiable on $\Bbb I$ whose derivative is continuous on $\Bbb I$.
Let $d: B \to A$ denote the mapping defined as:
- $\forall \map f x \in B: \map d f = \map {D_x} f$
where $D_x$ denotes the derivative of $f$ with respect to $x$.
Then $d$ is not an invertible mapping.
Proof
By definition, $d$ is invertible if and only if $d$ is a bijection.
It is sufficient to demonstrate that $d$ is not an injection.
Hence a fortiori $d$ is shown to not be a bijection.
Consider a differentiable function $f \in B$.
Then consider the function $g \in B$ defined as:
- $\forall x \in \Bbb I: \map g x = \map f x + C$
where $C \in \R$ is a constant.
Then we have that:
\(\ds \map {D_x} g\) | \(=\) | \(\ds \map {D_x} f\) | Derivative of Function plus Constant | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map d f\) | \(=\) | \(\ds \map d g\) | Definition of $d$ |
demonstrating that $d$ is not an injection.
The result follows.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions: Exercise $1$