Lipschitz continuous function
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
Examples
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$x \mapsto \abs{x}$ on $[-1, 1]$
High school
$x \mapsto \abs{x}$ is Lipschitz continuous with constant $1$, because $\abs{\abs{x} - \abs{y}} \leq \abs{x - y}$.
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$x \mapsto x^2$ on $[0, 1]$
High school
$x \mapsto x^2$ on $[0, 1]$ is Lipschitz continuous because that derivative is at most $2$ on $[0, 1]$.
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$f(x) = x^2 \sin(1/x)$ for $x \neq 0$, $f(0) = 0$, on $[-1, 1]$
Undergraduate
The function $f(x) = x^2 \sin(1/x)$ is Lipschitz continuous on $[-1, 1]$ because it is differentiable everywhere with $\abs{f^{\prime}(x)} \leq 3$.
Non-examples
Things that may be mistaken for examples.
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$f(x) = x^2$ on $\reals$
High school
$x \mapsto x^2$ on $\reals$ is not Lipschitz continuous, because its slope $2x$ is unbounded on $\reals$.
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The Cantor function on $[0, 1]$
Undergraduate
The Cantor function is not Lipschitz continuous, because it rises by $2^{-n}$ over intervals of length $3^{-n}$, so its slopes are unbounded.
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$x \mapsto \sin(1/x)$ on $(0, 1]$
Undergraduate
$x \mapsto \sin(1/x)$ 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.
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$x \mapsto \sqrt{x}$ on $[0, 1]$
Undergraduate
$x \mapsto \sqrt{x}$ 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 $t = 1$.
Distinctions involving this term
- Lipschitz continuous vs. continuously differentiable: Lipschitz continuity and continuous differentiability both rule out functions that are too steep or too wild, but in different ways. A Lipschitz continuous function has a bounded slope but may have corners. A continuously differentiable function has no corners, but its slope may be unbounded on an unbounded domain.
- 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.