Absolutely continuous function
Undergraduate (real functions on an interval $[a, b]$)
Consider a function $f : [a, b] \to \reals$ and finite collections of disjoint subintervals $(x_1, y_1), \dots, (x_n, y_n)$ of $[a, b]$. The function $f$ is absolutely continuous if for every $\varepsilon > 0$, there exists $\delta > 0$ such that whenever such a collection has total length
\[\sum_{k=1}^n (y_k - x_k) < \delta,\]the total change of $f$ satisfies
\[\sum_{k=1}^n \abs{f(y_k) - f(x_k)} < \varepsilon.\]
Graduate (Lebesgue integration)
Consider a function $f : [a, b] \to \reals$. The function $f$ is absolutely continuous if there is a Lebesgue integrable function $g$ such that
\[f(x) = f(a) + \int_a^x g(t) \mathrm{d}t\]for every $x \in [a, b]$.
Examples
Examples
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$x \mapsto \abs{x}$ on $[-1, 1]$
High school
$x \mapsto \abs{x}$ on $[-1, 1]$ is absolutely continuous, because it is Lipschitz continuous.
-
$x \mapsto x^2$ on $[0, 1]$
High school
$x \mapsto x^2$ on $[0, 1]$ is absolutely continuous, because it is Lipschitz continuous.
- $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, on $[-1, 1]$ Undergraduate
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$x \mapsto \sqrt{x}$ on $[0, 1]$
Undergraduate
$x \mapsto \sqrt{x}$ is absolutely continuous because it is the integral of its derivative $\frac{1}{2 \sqrt{t}}$, which is integrable on $[0, 1]$.
Non-examples
Things that may be mistaken for examples.
-
The Cantor function on $[0, 1]$
Undergraduate
The Cantor function is not absolutely continuous: it rises from $0$ to $1$, but its derivative is $0$ almost everywhere, so it is not the integral of its derivative.
Properties
- An absolutely continuous function $f$ is differentiable almost everywhere, and \[f(x) = f(a) + \int_a^x f^{\prime}(t) \mathrm{d}t.\]
- Every absolutely continuous function is uniformly continuous and has bounded variation.
Distinctions involving this term
- Types of continuity: There are many ways to describe the continuity of a function. On a compact interval, each is a special case of the next, from continuously differentiable to continuous.