# Equivalence of Definitions of Piecewise Continuously Differentiable Function

## Theorem

The following definitions of the concept of **Piecewise Continuously Differentiable Function** are equivalent:

Let $f$ be a real function defined on a closed interval $\closedint a b$.

### Definition 1

$f$ is **piecewise continuously differentiable** if and only if:

- $(1): \quad f$ is continuous
- $(2): \quad$ there exists a finite subdivision $\set {x_0, \ldots, x_n}$ of $\closedint a b$, $x_0 = a$ and $x_n = b$, such that:
- $(2.1): \quad f$ is continuously differentiable on $\openint {x_{i - 1}} {x_i}$ for every $i \in \set {1, \ldots, n}$
- $(2.2): \quad$ 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 $i \in \set {1, \ldots, n}$.

### Definition 2

$f$ is **piecewise continuously differentiable** if and only if:

- there exists a finite subdivision $\set {x_0, \ldots, x_n}$ of $\closedint a b$, $x_0 = a$ and $x_n = b$, such that:
- $f$ is continuously differentiable on $\closedint {x_{i − 1} } {x_i}$, where the derivative at $x_{i − 1}$ understood as right-handed and the derivative at $x_i$ understood as left-handed, for every $i \in \set {1, \ldots, n}$.

## Proof

### Definition 2 implies Definition 1

Assume that $f$ satisfies the requirement of Definition 2.

We need to prove that $f$ satisfies the requirements of Definition 1.

$f$ is continuous by Piecewise Continuously Differentiable Function/Definition 2 is Continuous.

This proves $(1)$ in Definition 1.

$f$ 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 $(2.1)$ in Definition 1.

$f$ 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 $(2.2)$ in Definition 1 is satisfied.

This finishes the proof that $f$ satisfies the requirements of Definition 1.

$\Box$

### Definition 1 implies Definition 2

Assume that $f$ satisfies the requirements of Definition 1.

We need to prove that $f$ satisfies the requirement of Definition 2.

$f$ 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 Definition 2.

$\blacksquare$