Definition:Piecewise Continuous Function

Definition

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

$f$ is piecewise continuous if and only if:

there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that:

for all $i \in \set {1, 2, \ldots, n}$, $f$ is continuous on $\openint {x_{i − 1} } {x_i}$.

Further conditions can be placed on this definition to add specialisation as necessary:

With Improper Integrals

$f$ is piecewise continuous with improper integrals if and only if:

there exists a finite subdivision $\left\{{x_0, x_1, \ldots, x_n}\right\}$ of $\left[{a \,.\,.\, b}\right]$, where $x_0 = a$ and $x_n = b$, such that for all $i \in \left\{ {1, 2, \ldots, n}\right\}$:

$(1): \quad f$ is continuous on $\left({x_{i − 1} \,.\,.\, x_i}\right)$

$(2): \quad$ the improper integrals $\displaystyle \int_{ {x_{i - 1} }^+}^{ {x_i}^-} f \left({x}\right) \rd x$ all exist.

Bounded

$f$ is a bounded piecewise continuous function if and only if:

there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that:

$(1): \quad$ for all $i \in \set {1, 2, \ldots, n}$, $f$ is continuous on $\openint {x_{i − 1} } {x_i}$

$(2): \quad$ $f$ is bounded on $\closedint a b$.

With One-Sided Limits

$f$ is piecewise continuous with one-sided limits if and only if:

there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that, for all $i \in \set {1, 2, \ldots, n}$:

$(1): \quad f$ is continuous on $\openint {x_{i − 1} } {x_i}$

$(2): \quad$ the one-sided limits $\displaystyle \lim_{x \mathop \to {x_{i − 1} }^+} \map f x$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \map f x$ exist.

Also known as

Some sources hyphenate: piecewise-continuous.

Also defined as

There are other definitions of Piecewise Continuous Function.

For example, the following variants exist:

Variant 1: $f$ is piecewise defined

Let $\closedint a b$ be a closed interval.

Let $\set {x_0, x_1, \ldots, x_n}$ be a finite subdivision of $\closedint a b$, where $x_0 = a$ and $x_n = b$.

Let $f$ be a real function defined on $\closedint a b \setminus \set {x_0, x_1, \ldots, x_n}$.

$f$ is piecewise continuous with one-sided limits if and only if:

for all $i \in \set {1, 2, \ldots, n}$:

$(1): \quad f$ is continuous on $\openint {x_{i − 1} } {x_i}$

$(2): \quad$ the one-sided limits $\displaystyle \lim_{x \mathop \to {x_{i − 1} }^+} \map f x$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \map f x$ exist.

Variant 2: $f$ is complex-valued

Let $f$ be a complex-valued function defined on a closed interval $\closedint a b$.

$f$ is piecewise continuous if and only if:

there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that:

for all $i \in \set {1, 2, \ldots, n}$, $f$ is continuous on $\openint {x_{i − 1} } {x_i}$.

Variant 3: $f$ has unbounded domain

Let $f$ be a real function defined on $\R$.

$f$ is piecewise continuous if and only if:

for any closed interval $\closedint a b$:

there exists a finite subdivision $\set {x_0, x_1, \ldots, x_n}$ of $\closedint a b$, where $x_0 = a$ and $x_n = b$, such that:

for all $i \in \set {1, 2, \ldots, n}$, $f$ is continuous on $\openint {x_{i − 1} } {x_i}$.

Also see

• Bounded Piecewise Continuous Function has Improper Integrals
• Piecewise Continuous Function with One-Sided Limits is Bounded

Comments

Possible properties of piecewise continuous functions:

• It seems obvious that a linear combination, a product, a quotient, and a composite of piecewise continuous functions are piecewise continuous functions.

• Also, a function piecewise continuous on adjacent intervals should be piecewise continuous on the union of these intervals.

Sources

• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Entry: piecewise continuous
• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Entry: piecewise continuous

This article incorporates material from piecewise on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.

• Weisstein, Eric W. "Piecewise Continuous." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/PiecewiseContinuous.html