# 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:

\(\displaystyle \map {D_x} g\) | \(=\) | \(\displaystyle \map {D_x} f\) | Derivative of Function plus Constant | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map d f\) | \(=\) | \(\displaystyle \map d g\) | 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 $1$