Differentiability of Function with Translation Property
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Theorem
Let $f$ be a real function.
Let $f$ have the translation property.
Let $c$ be a real number.
Let $\map {f'} c$ exist.
Then:
- $f'$ exists
- $f'$ is a constant function
Proof
Let $x$ be a real number.
We have:
\(\ds \map {f'} x\) | \(=\) | \(\ds \lim_{y \mathop \to x} \frac {\map f y - \map f x} {y - x}\) | Definition 1 of Differentiable Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{y \mathop \to x} \frac {\map f {y - x + x - c + c} - \map f {x - c + c} } {y - x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{y \mathop \to x} \frac {\map f {y - x + c + x - c} - \map f {c + x - c} } {y - x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{y \mathop \to x} \frac {\map f {y - x + c + t} - \map f {c + t} } {y - x}\) | $t = x - c$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{z \mathop + x \mathop \to x} \frac {\map f {z + c + t} - \map f {c + t} } z\) | $z = y - x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{z \mathop \to 0} \frac {\map f {z + c + t} - \map f {c + t} } z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{z \mathop \to 0} \frac {\map f {z + c} - \map f c} z\) | Definition of Translation Property | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {f'} c\) | Definition 2 of Differentiable Mapping |
We conclude that $\map {f'} x$ exists as $\map {f'} c$ exists.
Also, $\map {f'} x$ exists everywhere as $x$ is arbitrary.
In other words, $f'$ exists.
Also, $f'$ is a constant function as $\map {f'} x$ equals $\map {f'} c$ for every real number $x$.
$\blacksquare$