Analysis · Siblings

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.

On a compact interval $[a, b]$:

continuously differentiable ⊂ Lipschitz continuous ⊂ absolutely continuous ⊂ uniformly continuous = continuous

Definitions

continuous

Elementary

Think of a function $y = f(x)$. The graph of $f(x)$ is a plot that shows the value of $y$ for each values of $x$ by plotting the value of $x$ on a horizontal axis and the value $y = f(x)$ on the vertical axis. The function is continuous if you can draw its graph without lifting your pencil.

Caveat. This only works when the domain is one unbroken interval. The function $x \mapsto 1/x$ is continuous, but its graph comes in two pieces because $0$ is not in its domain.
High school

Consider a function $f$. The function $f$ is continuous if

\[\lim_{x \to a} f(x) = f(a)\]

for every $a$ in the domain of $f$.

Undergraduate (metric spaces)

Consider a function $f : X \to Y$ between metric spaces. The function $f$ is continuous if for every $x \in X$ and every $\varepsilon > 0$, there exists $\delta > 0$ such that

\[d(f(x), f(y)) < \varepsilon \quad \text{for every } y \in X \text{ with } d(x, y) < \delta.\]

The choice of $\delta$ may depend on both $\varepsilon$ and $x$.

Graduate (topological spaces)

Consider a function $f : X \to Y$ between topological spaces. The function $f$ is continuous if the preimage $f^{-1}(V)$ is open in $X$ for every open set $V$ in $Y$.

Examples
Non-examples
Properties
  • Sums, products, and compositions of continuous functions are continuous.
  • Continuous images of compact sets are compact, and continuous images of connected sets are connected.
  • A continuous function on $[a, b]$ takes every value between $f(a)$ and $f(b)$ (the intermediate value theorem).

uniformly continuous

High school

A function $f : \reals \to \reals$ is uniformly continuous if for every $\varepsilon > 0$, there exists $\delta > 0$ such that

\[\abs{f(x) - f(y)} < \varepsilon \quad \text{for all } \abs{x - y} < \delta.\]

The choice of $\delta$ depends only on $\varepsilon$, so the same $\delta$ works at every point.

Undergraduate (metric spaces)

Consider a function $f : X \to Y$ between metric spaces. The function $f$ is uniformly continuous if for every $\varepsilon > 0$, there exists $\delta > 0$ such that

\[d(f(x), f(y)) < \varepsilon \quad \text{for all } x, y \in X \text{ with } d(x, y) < \delta.\]

The choice of $\delta$ depends only on $\varepsilon$, so the same $\delta$ works at every point.

Examples
Non-examples
  • $f(x) = x^2$ on $\reals$
    $f(x) = x^2$ is not uniformly continuous on $\reals$, because its graph gets steeper and steeper, so no single closeness of inputs works everywhere.
  • $x \mapsto \sin(1/x)$ on $(0, 1]$
    $x \mapsto \sin(1/x)$ on $(0, 1]$ is not uniformly continuous, because points that are arbitrarily close together have values that differ by $2$. For each natural number $n$, let \[x_n = \frac{1}{\frac{\pi}{2} + 2 \pi n} \quad \text{and} \quad y_n = \frac{1}{\frac{3\pi}{2} + 2 \pi n}.\] Then $\sin(1/x_n) = 1$ and $\sin(1/y_n) = -1$, so $\abs{\sin(1/x_n) - \sin(1/y_n)} = 2$ for every $n$, but $\abs{x_n - y_n} \to 0$ as $n \to \infty$. Therefore, for $\varepsilon = 1$, no $\delta > 0$ satisfies the definition of uniform continuity: some pair $x_n, y_n$ is closer than $\delta$, but the values of the pair differ by more than $\varepsilon$.
Properties
  • Every uniformly continuous function is continuous. On a compact domain, every continuous function is uniformly continuous (the Heine–Cantor theorem).
  • A uniformly continuous function maps Cauchy sequences to Cauchy sequences.

Lipschitz continuous

High school

Consider a function $f$ and the lines through pairs of points on its graph. The function $f$ is Lipschitz continuous if there is a number $L$ such that every one of those lines has a slope between $-L$ and $L$.

Undergraduate (metric spaces)

Consider a function $f : X \to Y$ between metric spaces. The function $f$ is Lipschitz continuous if there exists $L \geq 0$ such that

\[d(f(x), f(y)) \leq L d(x, y) \quad \text{for all } x, y \in X.\]

The smallest such $L$ is the Lipschitz constant of $f$.

Examples
Non-examples
  • $f(x) = x^2$ on $\reals$
    $x \mapsto x^2$ on $\reals$ is not Lipschitz continuous, because its slope $2x$ is unbounded on $\reals$.
  • The Cantor function on $[0, 1]$
    The Cantor function is not Lipschitz continuous, because it rises by $2^{-n}$ over intervals of length $3^{-n}$, so its slopes are unbounded.
  • $x \mapsto \sin(1/x)$ on $(0, 1]$
    $x \mapsto \sin(1/x)$ on $(0, 1]$ is not Lipschitz continuous, because the slopes between pairs of points on its graph are unbounded. With the points $x_n$ and $y_n$ from the uniform continuity explanation, the slope between $(x_n, 1)$ and $(y_n, -1)$ has absolute value \[\frac{2}{\abs{x_n - y_n}} = \frac{2 \left(\frac{\pi}{2} + 2 \pi n\right) \left(\frac{3\pi}{2} + 2 \pi n\right)}{\pi},\] which grows without bound as $n \to \infty$, so no Lipschitz constant $L$ works.
  • $x \mapsto \sqrt{x}$ on $[0, 1]$
    $x \mapsto \sqrt{x}$ is not Lipschitz continuous, because its slope is unbounded near $0$.
Properties
  • Every Lipschitz continuous function is uniformly continuous.
  • A Lipschitz continuous function on an interval is absolutely continuous.
  • A Lipschitz continuous function on $\reals^n$ is differentiable almost everywhere (Rademacher’s theorem).
  • A differentiable function on an interval is Lipschitz continuous with constant $L$ if and only if $\abs{f^{\prime}(x)} \leq L$ for all $x$ (by the mean value theorem).
  • If $f : \reals^n \to \reals^n$ is Lipschitz continuous, then for every initial condition $x_0 \in \reals^n$, the differential equation $\dot{x} = f(x)$ with $x(0) = x_0$ has exactly one solution, and the solution is defined for all $t \in \reals$ (the global Picard–Lindelöf theorem). If $f$ is only locally Lipschitz continuous, then the solution exists and is unique on some time interval around $0$, but the solution can blow up in finite time: the solution of $\dot{x} = x^2$ with $x(0) = 1$ is $x(t) = 1 / (1 - t)$, which blows up at $t = 1$.

continuously differentiable

High school

A function $f$ is called continuously differentiable if the derivative $f^{\prime}$ exists at each point in the domain of $f$ and the derivative is itself a continuous function.

Caveat. Having a derivative everywhere is not enough. The function given by $f(x) = x^2 \sin(1/x)$ for $x \neq 0$ and $f(0) = 0$ is differentiable everywhere, but its derivative is not continuous at $0$.
Undergraduate (functions on an open set $U \subseteq \reals^n$)

Consider a function $f : U \to \reals^m$ on an open set $U \subseteq \reals^n$. The function $f$ is continuously differentiable if every partial derivative of $f$ exists and is continuous on $U$.

Examples
Non-examples
  • $f(x) = \abs{x}$
    $f(x) = \abs{x}$ is not continuously differentiable, because $f$ has no derivative at $0$, so the derivative is not a continuous function on all of $\reals$.
  • $x \mapsto \abs{x}$ on $[-1, 1]$
    $x \mapsto \abs{x}$ is not continuously differentiable, because it is not differentiable at $0$.
  • $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$
    $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, is not continuously differentiable, because $f^{\prime}(x) = 2x \sin(1/x) - \cos(1/x)$ has no limit as $x \to 0$.
  • $f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, on $[-1, 1]$
    The function $f$ is not continuously differentiable, because its derivative is not continuous at $0$. At $0$, we have $f^{\prime}(0) = 0$ because \[\lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} h \sin(1/h) = 0.\] Away from $0$, the product and chain rules give \[f^{\prime}(x) = 2x \sin(1/x) - \cos(1/x),\] The points $x_n = 1 / (2 \pi n)$ tend to $0$, but $f^{\prime}(x_n) = 0 - \cos(2 \pi n) = -1$ for every $n$, so $f^{\prime}(x_n)$ does not tend to $f^{\prime}(0) = 0$. That is, \[f^{\prime}(0) \neq \lim_{x \to 0} f^{\prime}(x),\] because the limit does not exist. The discontinuity of $f^{\prime}$ at $0$ is not a jump discontinuity, at which the limits from the left and from the right exist but differ. Instead, $f^{\prime}$ oscillates: as $x \to 0$, the term $2x \sin(1/x)$ tends to $0$, but the term $-\cos(1/x)$ oscillates between $-1$ and $1$ faster and faster, so $f^{\prime}(x)$ has no limit from either side. This oscillation is the only way that a derivative can fail to be continuous, by Darboux’s theorem: every derivative has the intermediate value property, so a derivative cannot have a jump discontinuity, such as the jump of a step function.
  • The Cantor function on $[0, 1]$
  • $x \mapsto \sqrt{x}$ on $[0, 1]$
    $x \mapsto \sqrt{x}$ on $[0, 1]$ is not Lipschitz continuous or continuously differentiable on $[0, 1]$, because its slope is unbounded near $0$.
Properties

absolutely continuous

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
Non-examples
  • The Cantor function on $[0, 1]$
    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

Examples

Example continuous uniformly continuous Lipschitz continuous continuously differentiable absolutely continuous Why
$x \mapsto x^2$ on $[0, 1]$ ✓✓✓✓✓
$f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, on $[-1, 1]$ ✓✓✓✗✓
$x \mapsto \sqrt{x}$ on $[0, 1]$ ✓✓✗✗✓
The Cantor function on $[0, 1]$ ✓✓✗✗✗
$x \mapsto \sin(1/x)$ on $(0, 1]$ ✓✗✗✓–
$f(x) = x^2$ on $\reals$ ✓✗✗✓– The squaring function, as a function $f : \reals \to \reals$.

Notes

  • Each neighboring pair in the chain has its own page, listed under “See also.”
  • On domains other than compact intervals, the chain changes. Lipschitz continuous ⊂ uniformly continuous ⊂ continuous still holds, but continuous functions need not be uniformly continuous, and continuously differentiable functions need not be Lipschitz continuous (see the last two examples).
  • Hölder continuity lies between Lipschitz continuity and uniform continuity. It is not covered here yet.

See also

All distinctions · All definitions