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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

A Banach space is a normed vector space over the real numbers or complex numbers that is complete for the metric $d(x,y)=\Vert x-y\Vert$. Explicitly, whenever a sequence $(x_n)$ satisfiesthere is an element $x$ of the same space with $\Vert x_n-x\Vert \to 0$. The hypothesis says that the approximations eventually agree to any requested accuracy; completeness says that they describe an actual object in the space. The Cauchy sequence condition can be checked before anyone knows what that object is.

This makes a Banach space a natural setting for constructing solutions by successive improvements. If corrections $v_k$ have ${\sum }_k\Vert v_k\Vert <\infty$, their partial sums form a Cauchy sequence by the triangle inequality, and completeness supplies their sum. In fact, convergence of every such absolutely summable vector series characterizes a Banach space. The contraction mapping theorem uses precisely this mechanism: each correction is smaller than the preceding one by a fixed factor.^[ref]

Every normed vector space of finite dimension is a Banach space. Infinite-dimensional examples include the absolutely summable sequence space and the l-infinity sequence space; the choice of norm remains part of the specification. A collection of increasingly accurate finite descriptions can require an infinite description for its limit, as the finitely supported sequence example below demonstrates. Completion of a normed space adds these missing limits while preserving all original distances.

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.

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]

For finitely many contracting Euclidean similarities, the map $\mathcal{T}$ on nonempty compact subsets of ${\mathbb{R}}^n$ is a contraction in the Hausdorff metric, with contraction constant ${\max }_i{c}_i<1$. The contraction mapping theorem gives a unique nonempty compact invariant set. If the open set condition holds, its Hausdorff dimension is the unique $s$ satisfying ${\sum }_i{c}_i^s=1$.

Let $X$ be a nonempty compact metric space and let $f:X\to X$ satisfy $d(f(x),f(y))<d(x,y)$ whenever $x\ne y$. Then $f$ has exactly one fixed point. The distance condition implies continuity, and $x\mapsto d(f(x),x)$ attains a minimum by compactness. If a minimizer $p$ were not a fixed point, then $d(f^2(p),f(p))<d(f(p),p)$ would contradict minimality. Two distinct fixed points would contradict the same strict distance inequality directly. A uniform Lipschitz constant below $1$ is unnecessary; compactness replaces the quantitative hypothesis of the contraction mapping theorem.

Consider the Volterra integral equationwhere $h$ is continuous, $|h|\le 1$, $h(0)=1$, and the continuous kernel satisfies $0\le K(t,s)\le e^{-c(t-s)}$ for $0\le s\le t$, with $c>0$. If $0<|b|<c/4$, there is exactly one continuous solution on the whole half-line. It is bounded, and Picard iteration beginning at $y_0=bh$ converges in supremum norm with contraction ratio at most $\gamma =4|b|/c<1$.

For a proof, the integral of the kernel is at most $1/c$. Thus $T$ maps the closed ball of radius $2|b|$ in the Banach space of bounded continuous functions into itself, since $\Vert Ty{\Vert }_{\infty }\le |b|+4b^2/c<2|b|$. Factoring $y^2-v^2$ gives $\Vert Ty-Tv{\Vert }_{\infty }\le \gamma \Vert y-v{\Vert }_{\infty }$ on this ball. The contraction mapping theorem gives its unique fixed point. A first-exit argument shows that every continuous solution, even one not initially assumed bounded, stays strictly inside the same ball: a first time with $|y|=2|b|$ would contradict $|y|\le |b|+4b^2/c$. Reapplying the estimate gives the strict global bound $\Vert y{\Vert }_{\infty }\le |b|(1+\gamma )<2|b|$, so uniqueness is unrestricted within the continuous solution class. If $b=0$, zero is the only solution by the Gronwall inequality on each finite interval.

For a uniformly Lipschitz continuous state field $f$ with constant $L$, an implicit midpoint rule step is uniquely solvable when $hL/2<1$, by the contraction mapping theorem. Write its step map as ${\Psi }_h(u)=u+h{\Phi }_h(u)$, where ${\Phi }_h(u)=f(t+h/2,u+h{\Phi }_h(u)/2)$. Subtracting the increment equations givesTherefore $\mathrm{Lip}({\Phi }_h)\le L/(1-hL/2)$, and the triangle inequality gives the bound for ${\Psi }_h$. Combining it with the implicit midpoint consistency bound yieldsfor $L>0$. This follows by summing the geometric error recurrence with forcing $K{h}^3/(1-hL/2)$. For $hL\le 1$ and $nh\le T$, $R_h^n\le e^{2LT}$, proving second-order global error with exact starting data. If $L=0$, the direct bound is $\Vert e_n\Vert \le \Vert e_0\Vert +KT{h}^2$.

An initial value problem combines an ordinary differential equation with an initial condition, for example $y^{\prime }=f(t,y)$ and $y(t_0)=y_0$. The task is to find a solution of a differential equation on an interval containing $t_0$ which has that starting value. For a continuous linear differential equation $u^{\prime }=A(t)u$, every prescribed finite vector $u_0$ determines a unique solution on the coefficient interval. A boundary value problem can instead prescribe values at distinct endpoints.

For the cooling initial value problem $y^{\prime }=-k(y-T_a)$, $y(t_0)=y_0$, where $k>0$ and $T_a$ are constant, the unique solution isTake $T_a=20$ degrees Celsius, $y_0=80$ degrees Celsius, $t_0=0$, and $k=0.1$ per minute. Then $y(10)=20+60e^{-1}\approx 42.07$ degrees Celsius. Reaching $30$ degrees Celsius requires $60e^{-0.1t}=10$, so $t=10\log 6\approx 17.92$ minutes. The time constant $1/k=10$ minutes reduces the excess temperature by a factor $e^{-1}$; it does not reduce the absolute temperature by that factor.

Uniqueness needs assumptions. If $f$ is a continuous function on an open neighborhood of $(t_0,y_0)$ and satisfies a local Lipschitz condition in $y$, uniformly in $t$, the Picard-Lindelöf theorem gives a unique local solution. Concretely, a finite $L$ bounds $|f(t,u)-f(t,v)|\le L|u-v|$ in a sufficiently small rectangle. Cooling satisfies this with $L=k$. Local existence does not promise existence for all future time: $y^{\prime }=y^2$, $y(0)=1$, has $y(t)=1/(1-t)$ on $(-\infty ,1)$ and finite-time blowup at $t=1$.

Continuity alone is insufficient for uniqueness. Consider the dimensionless initial value problem involving a square root and an absolute value:Besides $y(t)\equiv 0$, each delay $a\ge 0$ gives a solution for $t\ge 0$:At $t=4$, delays $a=0,1,2$ give $y=4,9/4,1$, despite identical starting data. These are valid solutions of a differential equation: for $a>0$, both one-sided derivatives at the join are zero; for $a=0$, the right derivative at the starting time is zero. After the join, $y_a^{\prime }=(t-a)/2=\sqrt{|y_a|}$ because $t>a$. Squaring alone would also suggest a parabola before the join, but its negative derivative there would violate the equation. Since $y^{\prime }\ge 0$, a forward solution starting from zero stays nonnegative.

The right-hand side is a continuous function but fails the Lipschitz condition at zero: for $h>0$, $|\sqrt{|h|}-0|/|h|=1/\sqrt{h}$ is unbounded. Thus the Peano existence theorem still supplies local existence, while the Picard-Lindelöf theorem does not supply uniqueness. Related waiting-time examples and a sufficient uniqueness condition appear in MIT 18.034, Lecture 2: Fundamental Principles.

A brief proof outline explains the mechanism: rewrite the initial value problem as the integral equation $y(t)=y_0+{\int }_{t_0}^{t}f(s,y(s))ds$. On a short enough interval, the Lipschitz condition makes the right-hand side a contraction mapping on a closed set of continuous functions near $y_0$. The contraction mapping theorem gives a unique fixed point; the fundamental theorem of calculus turns it back into a solution.

The inverse function theorem answers a practical question: can nearby outputs be used to recover nearby inputs uniquely and smoothly? Its useful surprise is that a nonlinear function can be tested using its linear derivative. If the first-order changes preserve all input directions, then, on a sufficiently small region, the nonlinear map can also be reversed. This is the basis for treating a nonlinear map as a coordinate transformation.

Precisely, let $f:U\to {\mathbb{R}}^n$ be continuously differentiable on an open subset $U$ of Euclidean space, let $a\in U$, and put $b=f(a)$. If $Df(a)$ is an invertible linear map, there are open neighbourhoods $U_0$ of $a$ and $V_0$ of $b$ such that $f:U_0\to V_0$ is a bijection with a continuously differentiable inverse function $g$. Thus this restriction is a diffeomorphism. In coordinates the test is $\det Df(a)\ne 0$, using the Jacobian determinant. The derivative of a local inverse map is obtained by inverting the Jacobian matrix, at the corresponding input. ^[ref]

For a one-dimensional example, $f(x)=x^2$ has $f^{\prime }(1)=2$. Near $x=1$, the inverse function is $g(y)=\sqrt{y}$, and the inverse function rule gives $g^{\prime }(1)=1/2$. But $x^2$ has no inverse function on the whole real line, since $x$ and $-x$ give the same output. The theorem supplies a local restriction, not global injectivity. At zero, the derivative test fails; the nonnegative branch still has a square root inverse, but that inverse has no finite derivative at zero. ^[ref]

Here is a two-dimensional example with an invertible derivative everywhere and infinitely repeated outputs:The exponential function turns $s$ into a positive radius, while $\theta$ is the angle in polar coordinates. Restricting, for example, to $|s|<0.35$ and $|\theta |<0.6$ produces the reversible patch pictured below. On its image, with Cartesian coordinates $(u,v)$ and $u>0$, the inverse function isThe arctangent selects the angle near zero. Without an angle restriction, $f(s,\theta +2\pi k)=f(s,\theta )$ for every integer $k$, so there is no single global inverse function. This map is a local diffeomorphism but is not injective.

The short proof idea is to correct a guessed input using the inverse linear response. With $A=Df(a)$, solving $f(x)=y$ is equivalent to finding a fixed point ofAt $a$, the derivative $D{T}_y(a)=I-A^{-1}Df(a)$ vanishes. Continuity makes $T_y$ a contraction mapping on a sufficiently small closed ball, which it maps into itself when $y$ is sufficiently close to $b$. The contraction mapping theorem then gives existence and uniqueness there. Establishing the smooth dependence on $y$ completes the inverse theorem. This explains why the nonlinearity can be tolerated: its correction has a small local slope.^[ref]

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 $T$.

A different interpretation can treat cyclic spreadsheet formulas as simultaneous equations and seek a fixed point by repeated numerical updates. This requires an explicit update rule, initial values and stopping condition. It is not automatically the meaning of a spreadsheet formula. In particular, a directed cycle alone tells us neither whether a fixed point exists nor whether a chosen update rule approaches it.

Two original toy examples make the distinction exact. The cyclic pair X1=Y1+1, Y1=X1+1 has no solution over the real numbers: substitution gives $X1=X1+2$. For X1=Y1, Y1=X1, every equal pair is a fixed point, so there is no unique answer. Simultaneously updating both spreadsheet cells from the previous pair gives $(x_{n+1},y_{n+1})=(y_n,x_n)$. Starting at $(0,1)$ produces the periodic sequence $(0,1),(1,0),(0,1),\dots$, which does not approach a fixed point. Updating X1 first and immediately using its new value for Y1 instead gives $(1,1)$ in one sweep. That difference follows from the specified rules, not from a claim about any current spreadsheet product. Numerical iteration can be useful for a deliberately chosen model with a justification such as the contraction mapping theorem; ordinary acyclic spreadsheet dependency evaluation already has its answer from prerequisite order.

Suppose the inhomogeneous linear wave equation has an energy estimate in a Banach solution space $X_T$, and its nonlinearity obeystogether with the corresponding locally Lipschitz estimate. For a sufficiently large closed ball and sufficiently small $T$, the map that solves the linear equation with source $F(w)$ is a contraction. The contraction mapping theorem then gives a local weak solution.

Work with the L1 norm on ${\mathbb{R}}^2$. The scalar Lipschitz hypothesis impliesWe may take $L\ge 0$. The compact set $S$ and continuity give a finite bound $M={\max }_S\Vert \mathbf{f}{\Vert }_1$. ChooseLet $I=[a-\eta ,a+\eta ]$. The space $C_{\eta }$ of continuous maps $I\to B_k(\mathbf{b})$ is complete in the supremum norm: it is a closed subset of the Banach space $C(I,{\mathbb{R}}^2)$. Define the Picard operator on a closed ball byFor any $\mathbf{g}\in C_{\eta }$, the integral triangle inequality givesThe integrand is continuous, so the result is continuous and remains in the prescribed closed ball. Thus $\mathcal{P}$ is a self-map of the complete space. Moreover,It is a contraction mapping. The contraction mapping theorem gives a unique fixed point $\mathbf{y}$. By the fundamental theorem of calculus, the fixed-point identity makes $\mathbf{y}$ continuously differentiable and gives ${\mathbf{y}}^{\prime }=\mathbf{f}(x,\mathbf{y})$ and $\mathbf{y}(a)=\mathbf{b}$. Its first component is twice continuously differentiable, because its derivative is the second component.

Conversely, integrating any solution that remains in the ball makes it a fixed point, so it equals $\mathbf{y}$. A solution starting at $\mathbf{b}$ cannot first reach the boundary of this ball during $I$: before such an exit the same estimate gives distance at most $\eta M<k$, contradicting distance $k$ at the first exit. This proves existence and uniqueness on $|x-a|\le \eta$, not merely formal convergence of an iteration. It is the local Picard-Lindelöf theorem construction.

For $0<\eta \le h$, work on the closed subsetIt is a nonempty complete metric space because it is closed in the Banach space of continuous functions with the supremum norm. Define the Picard iteration operatorChoose $\eta$ so that $M\eta \le k$ and $L\eta <1$, omitting a bound whose constant is zero. Such a positive $\eta$ exists. The first inequality makes $T$ a self-map, since $|(Ty)(x)-b|\le M|x-a|\le k$. The Lipschitz condition givesso the contraction mapping theorem yields a unique fixed point. The fundamental theorem of calculus makes it a differentiable solution of $y^{\prime }=f(x,y)$ with $y(a)=b$; conversely every solution staying in the rectangle satisfies the integral equation and is a fixed point. This is the local Picard-Lindelöf theorem construction. The interval-restricted space $C_{\eta }$ avoids treating functions defined on a shorter interval as elements of $C_h$.

The contraction mapping theorem states that a contraction mapping $T:S\to S$ on a nonempty complete metric space has a unique fixed point. Explicitly, if $d(Tu,Tv)\le qd(u,v)$ for all $u,v\in S$ and one fixed $0\le q<1$, then the iterates $u_{n+1}=T{u}_n$ converge to that fixed point from every $u_0\in S$. Both self-mapping and completeness are needed; the set need not be a vector space.

For the direct contraction mapping theorem proof with the usual supremum norm, take the complete metric spaceand the Picard iteration map $Tz(x)=b+{\int }_a^{x}f(s,z(s))ds$. The estimates areThus this proof requiresThe first inequality makes $T$ preserve $\mathcal{B}$; the strict second inequality makes it a contraction mapping. Its unique fixed point is the desired solution of a differential equation.

This is more restrictive than the preceding Weierstrass M-test argument: a single application of $T$ must shrink every difference in the supremum norm, whereas repeated integration in the other proof produces the factorial $(n+1)!$ and convergence even when $Lh\ge 1$. If $L=0$, the additional condition is automatic. These are conditions for the two stated direct proofs; they are not necessary bounds on the actual existence interval.

The rectangle avoids the singular line $y=0$ exactly when $0<k<1$. There $y>0$, and the largest $|x|$ on $[1-h,1+h]$ is $1+h$. Since $1+1/y$ decreases with $y$, the smallest feasible uniform bound isIt is attained at $x=1+h$, $y=1-k$. For the Lipschitz constant, the mean value theorem givesThe quotients approach this upper bound as $x=1+h$ and $y,z\downarrow 1-k$, so the smallest feasible constant isSubstitution into part (a) givesThe contraction mapping restriction is automatic for every M-test-admissible $h$ precisely when $A(k)<B(k)$. Because $0<k<1$,Thereforeis exactly the range where the two proofs guarantee all the same closed symmetric intervals, including the largest M-test interval.

For completeness, set $h_M=(\sqrt{1+4A(k)}-1)/2$ and $h_C=(\sqrt{1+4B(k)}-1)/2$. The M-test allows $0<h\le h_M$; the contraction mapping theorem allows $0<h\le h_M$ with $h<h_C$. At $k=2-\sqrt{2}$, these have the same supremum $h_M=h_C$, but contraction excludes the endpoint $h=h_M$. For $2-\sqrt{2}<k<1$, it gives the strictly shorter range $0<h<h_C<h_M$. For a particular $h$, both proofs work exactly when both displayed inequalities hold, even for these larger $k$.

The contraction mapping theorem states that a self-map $T$ of a nonempty complete metric space with a uniform contraction constant $q<1$ has a unique fixed point $p$. For every starting point, the iteration $x_{n+1}=T{x}_n$ converges to $p$, with $d(x_n,p)\le q^{n}d(x_1,x_0)/(1-q)$.

For the cosine fixed-point iteration, any real $x_0$ gives $x_1\in [-1,1]$. This closed interval is complete, and cosine maps it into $[\cos 1,1]\subseteq [-1,1]$. The mean value theorem giveswith $\sin (1)<1$. Apply the contraction mapping theorem to the tail starting at $x_1$. Therefore $\boxed{x_n\to p,p=\cos p}$ for every real initial value. Numerically $p\approx 0.739085$. The global real-line derivative bound is only one; entering the invariant interval supplies the strict uniform bound needed by the theorem.

Work in the Banach space $C([0,1])$ with its supremum norm, and take the nonempty closed bounded convex set $M=\{u:100\le u(t)\le 121\text{ for all }t\}$. DefineFor $u\in M$ the absolute value is unnecessary, and $\boxed{101\le Tu(t)\le 101+12=113\le 121}$. Thus $T:M\to M$. Also, for $a,b\ge 100$,so $\Vert Tu-Tv{\Vert }_{\infty }\le \frac{1}{20}\Vert u-v{\Vert }_{\infty }$ and in particular $T$ is continuous.

The image is uniformly bounded, and $|Tu(t)-Tu(s)|\le 12|t-s|$, so it is equicontinuous. The Arzelà-Ascoli theorem says that a uniformly bounded equicontinuous family in $C([0,1])$ has relatively compact closure in the uniform norm. Hence $T$ is compact, and the Schauder fixed-point theorem supplies $\boxed{u=Tu}$, the required solution. In fact the displayed contraction estimate also lets the contraction mapping theorem give uniqueness within $M$ and convergence of the corresponding Picard iteration. The initial constant here is the printed integer $101$.