Contraction mapping theorem

A contraction mapping $T:X\to X$ on a nonempty complete metric space has exactly one fixed point $x_{\ast }$, and the sequence $x_{n+1}=T(x_n)$ converges to $x_{\ast }$ from every $x_0\in X$. The same constant $0\le L<1$ must satisfy $d(Tx,Ty)\le Ld(x,y)$ for all $x,y\in X$. The statement applies to metric spaces, including closed subsets of a Banach space; neither linearity nor a vector space structure is required.^[ref]

The force of the result is that existence, uniqueness, and a practical approximation procedure arrive together. Repeated application produces corrections bounded by a geometric series, hence a Cauchy sequence; completeness supplies its limit. The Lipschitz continuity of $T$ makes that limit a fixed point. Two different fixed points would have their positive distance multiplied by at most $L<1$ while remaining unchanged, which is impossible.

Suppose all distances lie in $[0,D]$, $\psi$ is continuous, $0<\psi (r)<r$ for positive $r$, and ${\lim }_{r\to 0}\psi (r)/r<1$. The ratio is uniformly bounded below one near zero, and its maximum on the remaining compact scalar interval is also below one. Thus one constant $q<1$ works on the entire bounded set. If the set is complete and the map preserves it, Banach's theorem gives a unique fixed point. Compactness of the set itself is not required.

For a globally Lipschitz nonlinearity of constant $ϵ$, the map taking $v$ to the zero-Dirichlet Poisson solution with load $f(v)$ is a contraction in the gradient norm if the displayed inequality holds. Poincaré gives the factor $C_P$ from the Sobolev input to $L^2$, and testing the linear Poisson equation gives another factor $C_P$ back to the gradient norm. The condition yields a unique weak solution without requiring $f(0)=0$. With the optimal Poincaré constant it reads $ϵ<{\lambda }_1$.

Successive steps satisfy $d(x_{n+1},x_n)\le L^{n}d(x_1,x_0)$. Summing the remaining steps with the triangle inequality gives a bound available before the iteration starts:After computing a new step, the contraction mapping gives the often sharper boundEquivalently, the residual at any $z\in X$ satisfies $d(z,x_{\ast })\le d(z,Tz)/(1-L)$. These estimates convert visible step sizes into guaranteed distances to an unknown fixed point. When $L$ is close to one, the factor $1/(1-L)$ can be large: a small step alone needs interpretation through the contraction constant.

Use radians and take $T(x)=\cos x$ on the closed interval $[0,1]$. This is a complete metric space with distance $|x-y|$, and $T([0,1])=[\cos 1,1]\subset [0,1]$. The mean value theorem and $|T^{\prime }(x)|=\sin x\le \sin 1$ give a contraction mapping with $L=\sin 1\approx 0.841471<1$. Both the interval and its preservation matter; a small derivative by itself would not ensure that all subsequent approximations stay in the chosen space.

Starting at $x_0=0$ gives $x_1=1$, $x_2\approx 0.540302$, $x_3\approx 0.857553$, and $x_4\approx 0.654290$. The cosine is decreasing here, so successive iterates alternate around the unique fixed point $x_{\ast }\approx 0.739085133215161$. The contraction iteration error bound needs no prior knowledge of this numerical value. At step 40, $x_{40}\approx 0.739085078689123$ and the latest-step bound is below $7.20\times 10^{-7}$, certifying error below one millionth for exact iteration. The initial-step bound requires 91 iterations for the same tolerance. The observed step-40 error is only about $5.45\times 10^{-8}$; the guarantees allow less favorable behavior than this example actually exhibits.

This is the beauty of the contraction mapping theorem: a uniform local improvement controls the entire infinite process. Completeness is essential. On the incomplete interval $(0,1)$, the contraction mapping $T(x)=x/2$ keeps every iterate inside but has no fixed point there; its missing limit is zero. Banach's 1922 paper developed abstract methods for different spaces of functions with applications to integral equations, a motivation still visible whenever repeated approximation is used to construct a function.^[ref]

Let $A$ be differentiable on a neighbourhood of a closed radius-$r$ ball in a Banach space, with $\Vert DA\Vert \le \kappa <1$ on that ball. If $\Vert A(p_0)-p_0\Vert \le (1-\kappa )r$, the mean-value estimate gives $A(B)\subseteq B$ and $\Vert A(p)-A(q)\Vert \le \kappa \Vert p-q\Vert$. The contraction mapping theorem supplies exactly one fixed point in $B$, and its distance from $p_0$ is at most $\Vert A(p_0)-p_0\Vert /(1-\kappa )$.

// Target: dynamical-systems.bigb

For two contraction mappings of a nonempty complete metric space, their fixed points obey $d(x_f,x_g)\le q_{f}d(x_f,x_g)+d_{\infty }(f,g)$. Rearranging proves the displayed bound whenever the uniform metric is finite. It establishes continuity of the fixed-point map at each $f$ without assuming a contraction constant shared by all maps near $f$.

For contraction mappings $f,g$, compare their fixed points using $f$ on both arguments, then bound the change from $f$ to $g$ by the uniform metric. The resulting inequality gives the displayed estimate. Continuity at $f$ only needs its own contraction constant $q_f<1$, even if nearby maps have constants approaching one.

A metric space is complete if and only if every contraction mapping of every nonempty closed subset into itself has a fixed point. For the converse, a nonconvergent Cauchy sequence contains a rapidly approaching subsequence whose point set is closed and whose shift is a contraction without a fixed point. Requiring the property only on the whole space is a different, weaker condition.

A contraction mapping $T$ on a metric space satisfies $d(Tx,Ty)\le qd(x,y)$ for every $x,y$ and some constant $0\le q<1$.

On the maximum product metric, the displayed map is nonexpansive when $f$ is a contraction mapping. If the space has two distinct points, changing only the second input coordinate shows that one step preserves some nonzero distance exactly, so it is not itself a strict contraction. Two iterations apply $f$ to both coordinates and therefore contract with the same constant as $f$. This provides a direct example of a map whose iterate is contractive although the map is not.

A contraction constant for a function $T$ on a metric space is a single $q<1$ that bounds the ratio $d(Tx,Ty)/d(x,y)$ for every distinct pair. Pointwise strict inequalities $d(Tx,Ty)<d(x,y)$ do not imply such a common constant. For a differentiable function on an interval, the mean value theorem shows that $|T^{\prime }|\le q<1$ supplies a contraction constant.

The same self-map can be a contraction mapping for one metric but not for a bilipschitz equivalent metric. On three points with distances $d^{\prime }(x,y)=d^{\prime }(x,z)=2$, $d^{\prime }(y,z)=1$, the map $f(x)=y$, $f(y)=f(z)=z$ has Lipschitz constant $1/2$. For the discrete metric, its Lipschitz constant is one. The metrics satisfy $d\le d^{\prime }\le 2d$, so they preserve topology and completeness but not the strict numerical contraction condition. A contraction mapping theorem fixed-point proof therefore needs the specified metric and its contraction constant.

Suppose $T_{\lambda }$ are contractions of one complete metric space with a common constant $k<1$, and $T_{\lambda }x$ depends continuously on $\lambda$ for each fixed $x$. Their fixed points $x_{\lambda }$ depend continuously on $\lambda$, because for a fixed parameter ${\lambda }_0$,

If a self-map $T$ of a nonempty complete metric space has an iterate $T^n$ that is a contraction mapping, the contraction mapping theorem gives a unique fixed point $a$ of $T^n$. The maps commute, so $T^n(Ta)=T(T^{n}a)=Ta$, and uniqueness forces $Ta=a$. Conversely every fixed point of $T$ is fixed by $T^n$. Thus $a$ is the unique fixed point of $T$.

Let $T$ be a self-map of a nonempty closed subset of a Banach space, with the displayed bounds. Then $T$ has a unique fixed point $p$ and every orbit $x_n=T^n{x}_0$ converges to it. Indeed, $\Vert x_{n+1}-x_n\Vert \le {\alpha }_n\Vert x_1-x_0\Vert$ makes the orbit Cauchy by summing successive increments. The first bound makes $T$ continuous, so the limit is fixed. Since ${\alpha }_n\to 0$, the bounds also imply uniqueness. The tail sum gives $\Vert p-x_n\Vert \le {\sum }_{k\ge n}{\alpha }_k\Vert x_1-x_0\Vert$ for $n\ge 1$. A Volterra integral equation supplies factorially decreasing bounds even when the first iterate is not a contraction mapping.

For the Newton map $g(x)=x-f(x)/f^{\prime }(x)$ near a simple root $r$,Bounds $|f^{\prime }|\ge \delta$ and $|f^{\prime \prime }|\le M$ make $|g^{\prime }|<1$ on a sufficiently small closed interval around $r$. The contraction mapping theorem then gives the unique local fixed point.