Definition:Piecewise Continuous Function

From ProofWiki
Jump to navigation Jump to search

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 $\set {x_0, x_1, \ldots, x_n}$ 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 improper integrals $\ds \int_{ {x_{i - 1} }^+}^{ {x_i}^-} \map f x \rd x$ 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 $\ds \lim_{x \mathop \to {x_i}^-} \map f x$ 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 $\ds \lim_{x \mathop \to {x_{i − 1} }^+} \map f x$ and $\ds \lim_{x \mathop \to {x_i}^-} \map f x$ 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


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

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