Contractions isolate maps that pull points uniformly closer together. In a [metric space](/page/Metric%20Space), continuity only says that nearby points remain nearby after applying a map; a contraction imposes a quantitative improvement on every distance at once. This stronger condition is the engine behind fixed point arguments, iterative algorithms, and stability estimates in [complete metric spaces](/page/Complete%20Metric%20Space), [Banach spaces](/page/Banach%20Space), and differential equations. The word also appears in other areas, such as contraction of a loop, contraction of a matroid, contraction under localization, and contraction as a structural rule in [proof theory](/page/Proof%20Theory). Those are separate constructions with their own ambient categories. On this page, the unqualified analytic meaning is the metric-space notion: a self-map whose Lipschitz constant is strictly less than $1$. ## Definition The basic problem is to formalise the idea that applying a map makes all configurations smaller by a uniform factor. A map may decrease some distances and enlarge others; it may even bring a single orbit closer to a point while stretching nearby points. The contraction condition rules out such local accidents by requiring one constant that works for every pair of points. [definition: Contraction] Let $(X,d)$ be a metric space. A map $f: X \to X$ is a contraction if there exists a constant $c \in [0,1)$ such that \begin{align*} d(f(x),f(y)) &\le c\,d(x,y) \end{align*} for all $x,y \in X$. [/definition] A first test case is a map that simply scales all distances by the same factor. This gives the reader a concrete model before the general fixed point machinery enters. [example: Basic Scaling Contraction] Let $X=\mathbb{R}$ with the usual metric $d(x,y)=|x-y|$, and define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=x/2$. For arbitrary $x,y\in\mathbb{R}$, we have \begin{align*} d(f(x),f(y)) &= |f(x)-f(y)| \end{align*} and substituting the formula for $f$ gives \begin{align*} |f(x)-f(y)| &= \left|\frac{x}{2}-\frac{y}{2}\right|. \end{align*} Factoring out $1/2$ inside the absolute value, \begin{align*} \left|\frac{x}{2}-\frac{y}{2}\right| &= \left|\frac{x-y}{2}\right|=\frac{1}{2}|x-y|. \end{align*} Since $1/2\in[0,1)$, the inequality required in the definition holds with equality: \begin{align*} d(f(x),f(y)) &= \frac{1}{2}d(x,y). \end{align*} Thus $f$ is a contraction with contraction constant $1/2$; in this model case every distance is exactly halved. [/example] The constant $c$ is not part of the map's data unless it is explicitly specified. What matters is the existence of some uniform factor strictly below $1$. To compare maps quantitatively, it is useful to record the smallest global distance factor allowed by the metric estimate. [definition: Lipschitz Constant] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, and let $f: X \to Y$ be a map. The Lipschitz constant of $f$ is \begin{align*} \operatorname{Lip}(f) &= \inf\{L \ge 0 : d_Y(f(x),f(y)) \le L d_X(x,y) \text{ for all } x,y \in X\}. \end{align*} [/definition] With this language, a contraction is a self-map admitting a Lipschitz bound with a constant strictly below $1$. A nearby boundary case is important enough to name: maps that never increase distance, but may preserve it, occur naturally in geometry and dynamics. [definition: Nonexpansive Map] Let $(X,d)$ be a metric space. A map $f: X \to X$ is nonexpansive if \begin{align*} d(f(x),f(y)) &\le d(x,y) \end{align*} for all $x,y \in X$. [/definition] Every contraction is nonexpansive, but the converse fails. The identity map on any metric space with at least two points is nonexpansive and is not a contraction. The main payoff of the strict inequality is that iteration is forced toward a single possible endpoint, so the target endpoint needs its own term. [definition: Fixed Point] Let $X$ be a set and let $f: X \to X$ be a map. A point $x^* \in X$ is a fixed point of $f$ if \begin{align*} f(x^*) &= x^*. \end{align*} [/definition] Fixed points convert equations into geometry. Solving $f(x)=x$ can be easier than solving an equation directly, because repeated application of a contraction forces all starting points into the same limiting location when the ambient space is complete. ## Equivalent Characterisations For computations, it is common to verify contraction through a supremum of distance ratios. This turns a universal inequality over all pairs into a single numerical bound. It is useful when estimates are obtained by comparing output distances with input distances directly. [quotetheorem:9874] The supremum condition is not just a convenient test; it is the sharp way to measure the best possible contraction constant. If the supremum is below $1$, one constant works uniformly for every pair of distinct points. If it is equal to or above $1$, the map may still shrink some distances, but the contraction argument has no global ratio strong enough to force convergence from arbitrary starting points. Many analytic settings do not begin with an abstract distance formula; they begin with a norm and estimates on differences of vectors or functions. To use the distance-ratio criterion there, we need the same idea expressed in terms of norms of differences, because those are the quantities that estimates in vector and function spaces naturally control. [quotetheorem:9875] This formulation is the one behind many arguments in functional analysis. It also explains why estimates in spaces such as $C([0,T])$ or $L^p$ are often arranged to produce a small constant. When a map is differentiable, there is an even more local way to obtain the same global estimate. A derivative measures infinitesimal stretching. If the derivative is uniformly bounded by a number below $1$, then along each line segment the accumulated stretching remains below that same number. The domain must allow those line segments to stay inside it, which is why convexity enters the statement. [quotetheorem:8341] The convexity hypothesis matters because the straight line segment between two points must remain inside the domain when integrating derivative estimates. On nonconvex domains, a derivative bound may control local motion without controlling the metric distance between arbitrary points inside the domain. ## Standard Examples A linear scaling on the real line is the model example. It shows the definition in its purest form: every distance is multiplied by the same number. [example: Linear Scaling] Let $X=\mathbb{R}$ with $d(x,y)=|x-y|$. Fix $a,b\in\mathbb{R}$ and define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=ax+b$. For arbitrary $x,y\in\mathbb{R}$, the distance between the images is \begin{align*} d(f(x),f(y))=|f(x)-f(y)|. \end{align*} Substituting the formula for $f$ gives \begin{align*} |f(x)-f(y)|=|(ax+b)-(ay+b)|. \end{align*} The translation term cancels: \begin{align*} (ax+b)-(ay+b)=ax-ay=a(x-y). \end{align*} Therefore \begin{align*} d(f(x),f(y))=|a(x-y)|=|a|\,|x-y|=|a|\,d(x,y). \end{align*} If $|a|<1$, this is exactly the contraction inequality with contraction constant $c=|a|$. Conversely, if $f$ were a contraction with constant $c<1$, then applying the inequality to $x=1$ and $y=0$ would give \begin{align*} |a|=|f(1)-f(0)|\le c|1-0|=c. \end{align*} Hence $|a|\le c<1$, so $|a|<1$. Thus $f$ is a contraction exactly when $|a|<1$. Assume now that $|a|<1$. Since $a\ne 1$, a fixed point must satisfy \begin{align*} x=ax+b. \end{align*} Subtracting $ax$ from both sides gives \begin{align*} (1-a)x=b. \end{align*} Dividing by $1-a$ gives the unique fixed point \begin{align*} x^*=\frac{b}{1-a}. \end{align*} For the iteration $x_{k+1}=f(x_k)$, subtract the fixed point equation $x^*=ax^*+b$ from $x_{k+1}=ax_k+b$: \begin{align*} x_{k+1}-x^*=a(x_k-x^*). \end{align*} Taking absolute values gives \begin{align*} |x_{k+1}-x^*|=|a|\,|x_k-x^*|. \end{align*} Repeating this identity yields \begin{align*} |x_k-x^*|=|a|^{k-1}|x_1-x^*|. \end{align*} Because $|a|<1$, the factor $|a|^{k-1}$ tends to $0$, so $x_k\to x^*$ for every initial point $x_1\in\mathbb{R}$. The translation $b$ moves the fixed point, but it does not affect the contraction constant because it cancels in differences. [/example] This example also shows why the additive term $b$ is irrelevant to the contraction constant. Contraction measures how differences evolve, and translations cancel in differences. A standard failure example is the identity map. It preserves every distance, so it is as close to being a contraction as a nonexpansive map can be without crossing the strict threshold. [example: Identity Map Is Not a Contraction] Let $(X,d)$ be a metric space with at least two points, and choose distinct points $x,y\in X$. Let $\operatorname{id}_X:X\to X$ be the identity map, so $\operatorname{id}_X(z)=z$ for every $z\in X$. Then \begin{align*} \operatorname{id}_X(x)=x \quad \text{and} \quad \operatorname{id}_X(y)=y. \end{align*} Therefore \begin{align*} d(\operatorname{id}_X(x),\operatorname{id}_X(y))=d(x,y). \end{align*} Since $x\ne y$, the metric definiteness axiom gives $d(x,y)>0$. The equality above also gives \begin{align*} d(\operatorname{id}_X(x),\operatorname{id}_X(y))\le d(x,y), \end{align*} so $\operatorname{id}_X$ is nonexpansive. We now show that it is not a contraction. If $\operatorname{id}_X$ were a contraction, there would be some $c\in[0,1)$ such that \begin{align*} d(\operatorname{id}_X(u),\operatorname{id}_X(v))\le c\,d(u,v) \end{align*} for all $u,v\in X$. Applying this with $u=x$ and $v=y$ gives \begin{align*} d(\operatorname{id}_X(x),\operatorname{id}_X(y))\le c\,d(x,y). \end{align*} Substituting $d(\operatorname{id}_X(x),\operatorname{id}_X(y))=d(x,y)$ gives \begin{align*} d(x,y)\le c\,d(x,y). \end{align*} Because $d(x,y)>0$, dividing both sides by $d(x,y)$ gives \begin{align*} 1\le c. \end{align*} This contradicts $c\in[0,1)$, which implies $c<1$. Hence the identity map is nonexpansive but not a contraction; preserving every distance is exactly the boundary case excluded by the strict contraction condition. [/example] The failure is not cosmetic. The identity map has every point as a fixed point, so uniqueness of fixed points cannot follow from nonexpansiveness alone. Function spaces provide the most important analytic examples. In applications to differential equations, one often transforms an equation into an integral equation and then proves that the associated integral operator is a contraction on a small time interval. [example: Integral Operator on Continuous Functions] Let $T>0$, fix $x_0\in\mathbb{R}$, and let $X=C([0,T])$ with $\|u\|_\infty=\sup_{t\in[0,T]}|u(t)|$. Suppose $g:[0,T]\times\mathbb{R}\to\mathbb{R}$ is continuous and that there is $L>0$ such that \begin{align*} |g(t,r)-g(t,s)|\le L|r-s| \end{align*} for all $t\in[0,T]$ and $r,s\in\mathbb{R}$. For $u\in X$, define \begin{align*} (Au)(t)=x_0+\int_0^t g(\tau,u(\tau))\,d\mathcal{L}^1(\tau). \end{align*} Since $\tau\mapsto g(\tau,u(\tau))$ is continuous on $[0,T]$, the function $Au$ is continuous, so $A:X\to X$ is well-defined. For $u,v\in X$ and $t\in[0,T]$, subtracting the two defining formulas gives \begin{align*} (Au)(t)-(Av)(t)=\int_0^t g(\tau,u(\tau))\,d\mathcal{L}^1(\tau)-\int_0^t g(\tau,v(\tau))\,d\mathcal{L}^1(\tau). \end{align*} By linearity of the integral, \begin{align*} (Au)(t)-(Av)(t)=\int_0^t \bigl(g(\tau,u(\tau))-g(\tau,v(\tau))\bigr)\,d\mathcal{L}^1(\tau). \end{align*} Taking absolute values and using the triangle inequality for integrals, \begin{align*} |(Au)(t)-(Av)(t)|\le \int_0^t |g(\tau,u(\tau))-g(\tau,v(\tau))|\,d\mathcal{L}^1(\tau). \end{align*} The Lipschitz assumption on $g$ gives \begin{align*} |g(\tau,u(\tau))-g(\tau,v(\tau))|\le L|u(\tau)-v(\tau)|. \end{align*} Since $|u(\tau)-v(\tau)|\le \|u-v\|_\infty$ for every $\tau\in[0,T]$, we get \begin{align*} |g(\tau,u(\tau))-g(\tau,v(\tau))|\le L\|u-v\|_\infty. \end{align*} Therefore \begin{align*} |(Au)(t)-(Av)(t)|\le \int_0^t L\|u-v\|_\infty\,d\mathcal{L}^1(\tau). \end{align*} The integrand is constant in $\tau$, so \begin{align*} \int_0^t L\|u-v\|_\infty\,d\mathcal{L}^1(\tau)=Lt\|u-v\|_\infty. \end{align*} Because $0\le t\le T$, \begin{align*} |(Au)(t)-(Av)(t)|\le LT\|u-v\|_\infty. \end{align*} Taking the supremum over $t\in[0,T]$ gives \begin{align*} \|Au-Av\|_\infty\le LT\|u-v\|_\infty. \end{align*} If $LT<1$, this is the contraction inequality with contraction constant $LT$. Thus $A$ is a contraction on $C([0,T])$, and a fixed point $u=Au$ is exactly a [continuous function](/page/Continuous%20Function) satisfying \begin{align*} u(t)=x_0+\int_0^t g(\tau,u(\tau))\,d\mathcal{L}^1(\tau) \end{align*} for every $t\in[0,T]$. [/example] This example explains the recurring phrase "choose the time interval small enough" in local existence theory. The map may not be a contraction on a long interval, but it becomes one when the estimate gains a small factor from integration over $[0,T]$. Completeness cannot be dropped from the fixed point theorem. The next example shows how iteration may converge toward a point outside the space. [example: Missing Fixed Point in an Incomplete Space] Let $X=(0,1)$ with the usual metric $d(x,y)=|x-y|$, and define $f:X\to X$ by $f(x)=x/2$. If $x\in(0,1)$, then $0<x/2<1/2<1$, so $f(x)\in(0,1)$ and the map is well-defined. For arbitrary $x,y\in X$, the distance between their images is \begin{align*} d(f(x),f(y))=\left|\frac{x}{2}-\frac{y}{2}\right|. \end{align*} Combining the fractions gives \begin{align*} \left|\frac{x}{2}-\frac{y}{2}\right|=\left|\frac{x-y}{2}\right|. \end{align*} Since $1/2>0$, homogeneity of absolute value gives \begin{align*} \left|\frac{x-y}{2}\right|=\frac{1}{2}|x-y|. \end{align*} Thus \begin{align*} d(f(x),f(y))=\frac{1}{2}d(x,y). \end{align*} Because $1/2\in[0,1)$, $f$ is a contraction on $X$ with contraction constant $1/2$. A fixed point in $X$ would be a point $x\in(0,1)$ satisfying $f(x)=x$. Substituting the formula for $f$ gives \begin{align*} \frac{x}{2}=x. \end{align*} Subtracting $x/2$ from both sides gives \begin{align*} 0=\frac{x}{2}. \end{align*} Multiplying both sides by $2$ gives \begin{align*} x=0. \end{align*} But $0\notin(0,1)$, so there is no fixed point in $X$. This contraction points toward the missing endpoint $0$, showing exactly where incompleteness breaks the fixed point conclusion. [/example] The example is the standard warning attached to the [Banach fixed point theorem](/theorems/270). Contraction supplies the Cauchy behaviour; completeness supplies the limit inside the space. ## Properties Before proving existence of fixed points, it is worth separating uniqueness from existence. The contraction inequality alone prevents two different points from both being fixed. No compactness or completeness is needed for this part, because it uses only the distance between two hypothetical fixed points. [quotetheorem:8342] Uniqueness does not place a fixed point inside the space. Starting from a point $x_0$, the contraction forces the iterates $x_0,T(x_0),T^2(x_0),\dots$ to crowd together like a [Cauchy sequence](/page/Cauchy%20Sequence), but this only identifies a candidate limit. In an incomplete space that candidate can lie outside the space, so the equation $T(x)=x$ may still have no solution. Completeness is the hypothesis that turns the limiting candidate produced by the iteration into an actual point of the space. The main existence question is therefore not whether two fixed points can coexist, but whether the iterative candidate actually converges inside the given metric space. The next result packages the standard Picard iteration argument: a contraction on a complete metric space has enough limiting room to produce a fixed point, and the uniqueness result then shows that this point is the only one. [quotetheorem:71] A useful next question is quantitative: after $k$ iterations, how far can $x_k$ be from the fixed point? The first estimate answers this when the initial distance to the fixed point is known or can be bounded by another argument. [quotetheorem:8343] This estimate is a priori because it involves the unknown fixed point. In computations, the distance to the fixed point is usually not known at the start. The next estimate replaces it with the first observed displacement of the iteration. [quotetheorem:8344] The factor $(1-c)^{-1}$ measures the cost of summing the geometric tail. As $c$ approaches $1$, convergence may remain guaranteed but become slow. Structural closure properties help build new contractions from old ones without redoing every estimate from the beginning. Composition is the simplest such closure property. If one map shrinks by $c_g$ and the next shrinks by $c_f$, then the combined operation shrinks by the product. This is useful for iterative schemes assembled from several stages. [quotetheorem:9876] A composition can also be a contraction when only one factor is a contraction and the other is Lipschitz with sufficiently small product of constants. Finally, contraction estimates automatically give ordinary continuity, which locates contractions within the usual hierarchy of regularity. Continuity asks for preservation of closeness near each point. A Lipschitz estimate gives a uniform version of that control over the whole space. Since every contraction is Lipschitz, every contraction inherits this regularity. [quotetheorem:8340] Continuity alone does not imply contraction, even on compact intervals. A continuous map may fold, flatten, or stretch in ways that have no uniform distance factor below $1$. ## Relationship to Other Concepts Contractions sit inside the hierarchy \begin{align*} \text{contraction} \implies \text{nonexpansive map} \implies \text{Lipschitz map} \implies \text{continuous map}. \end{align*} None of the reverse implications holds in general. The strict numerical bound is the extra ingredient that powers fixed point theory. In [ordinary differential equations](/page/Ordinary%20Differential%20Equation), contractions appear after rewriting an [initial value problem](/page/Initial%20Value%20Problem) as an integral equation. The fixed point is the desired solution, and the contraction estimate is obtained from a Lipschitz bound on the vector field and a small time interval. In [Banach spaces](/page/Banach%20Space), contractions are used to solve nonlinear equations of the form $u=Au$ and perturbative equations of the form $Lu=N(u)$ after rearrangement. Many existence arguments in analysis reduce to constructing a complete metric space and proving that a carefully chosen operator maps it into itself as a contraction. In [numerical analysis](/page/Numerical%20Analysis), contraction estimates justify iterative methods. If the update map is a contraction near a solution, then repeated updates converge geometrically. The estimates above quantify how many steps are needed to reach a prescribed tolerance. The graph-theoretic, algebraic, and logical uses of the word "contraction" should not be conflated with the analytic definition. A contraction of a loop, a contraction of a nonloop, a contraction of a matroid, contraction under localization, and contraction as a structural rule in proof theory are operations or inference principles in their own settings. The shared word suggests simplification or reduction, but the metric condition $d(f(x),f(y))\le c d(x,y)$ is the defining feature of the analytic concept. ## References [Metric Space](/page/Metric%20Space). [Complete Metric Space](/page/Complete%20Metric%20Space). [Banach Space](/page/Banach%20Space). Walter Rudin, *Principles of Mathematical Analysis* (1976). Erwin Kreyszig, *Introductory Functional Analysis with Applications* (1978). Lawrence C. Evans, *Partial Differential Equations* (2010).