Derivative Function is not Invertible

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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.

$\blacksquare$


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