Equivalence of Definitions of Piecewise Continuously Differentiable Function

== Theorem ==
A function satisfying Piecewise Continuously Differentiable Function With Closed Intervals is equivalent to being continuous and satisfying Piecewise Continuously Differentiable Function With One-Sided Limits.
Let $f$ be a real function defined on a closed interval $\closedint a b$. 
=== Piecewise Continuously Differentiable Function With One-Sided Limits ===
=== Piecewise Continuously Differentiable Function With Closed Intervals ===
== Proof ==
=== Piecewise Continuously Differentiable Function With Closed Intervals implies continuity and Piecewise Continuously Differentiable Function With One-Sided Limits ===
Assume that $f$ satisfies the requirement of Piecewise Continuously Differentiable Function With Closed Intervals.
We need to prove that $f$ satisfies the requirements of Piecewise Continuously Differentiable Function With One-Sided Limits and that $f$ is continuous.
$f$ is continuous by Piecewise Continuously Differentiable Function/Definition 2 is Continuous.
This proves the continuity requirement.
$f$ is continuously differentiable on $\closedint {x_{i - 1} } {x_i}$, the derivatives at $x_{i - 1}$ and $x_i$ understood as one-sided derivatives, for every $i \in \set {1, \ldots, n}$, by definition.
Therefore, $f$ is continuously differentiable at every point of every $\openint {x_{i - 1} } {x_i}$ since $\openint {x_{i - 1} } {x_i}$ is a subset of $\closedint {x_{i - 1} } {x_i}$.
This proves $(1)$ in Piecewise Continuously Differentiable Function With One-Sided Limits.
$f$ has a continuous right-hand derivative at $x_{i - 1}$ for every $i \in \set {1, \ldots, n}$.
This means that:
:$\map {f'_+} {x_{i - 1} } = \ds \lim_{x \mathop \to {x_{i - 1} }^+} \map {f'} x$
This implies that $\ds \lim_{x \mathop \to {x_{i - 1} }^+} \map {f'} x$ exists for every $i \in \set {1, \ldots, n}$.
Also, $f$ has a continuous left-hand derivative at $x_i$ for every $i \in \set {1, \ldots, n}$.
This means that
:$\map {f'_-} {x_i} = \ds \lim_{x \mathop \to {x_i}^-} \map {f'} x$
This implies that $\ds \lim_{x \mathop \to {x_i}^-} \map {f'} x$ exists for every $i \in \set {1, \ldots, n}$.
These two conclusions prove that requirement $(2)$ in Piecewise Continuously Differentiable Function With One-Sided Limits is satisfied.
This finishes the proof that $f$ satisfies the requirements of Piecewise Continuously Differentiable Function With One-Sided Limits and that $f$ is continuous.
=== Continuity and Piecewise Continuously Differentiable Function With One-Sided Limits imply Piecewise Continuously Differentiable Function With Closed Intervals ===
Assume that $f$ satisfies the requirements of Piecewise Continuously Differentiable Function With One-Sided Limits and that $f$ is continuous.
We need to prove that $f$ satisfies the requirement of Piecewise Continuously Differentiable Function With Closed Intervals.
$f$ is continuously differentiable on $\openint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$ by definition.
Also, the one-sided limits $\ds \lim_{x \mathop \to {x_{i - 1} }^+} \map {f'} x$ and $\ds \lim_{x \mathop \to {x_i}^-} \map {f'} x$ exist for every $i \in \set {1, \ldots, n}$.
$f$ is continuously differentiable on $\closedint {x_{i - 1} } {x_i}$, the derivatives at $x_{i - 1}$ and $x_i$ understood as one-sided derivatives, for every $i \in \set {1, \ldots, n}$, by Extendability Theorem for Derivatives Continuous on Open Intervals.
This is exactly the requirement of Piecewise Continuously Differentiable Function With Closed Intervals.
Category:Piecewise Continuously Differentiable Functions