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At a current approximation $p$, Taylor's theorem replaces $\Phi (p+h)=0$ by the affine equation $\Phi (p)+D\Phi (p)h=0$. Solving this equation gives the Newton iteration in a Banach spaceThe inverse is a bounded linear inverse by the bounded inverse theorem. The displayed operator is defined where the derivative is invertible, in particular on the specified open domain; the hypotheses do not supply a globally defined Newton operator on all of $\mathcal{X}$. If $\Phi (\bar{p})=0$ with $\bar{p}$ in that domain, then $N(\bar{p})=\bar{p}$.

For quadratic convergence, choose a sufficiently small neighbourhood of $\bar{p}$. Continuity of the inverse derivative and of $D^2\Phi$ gives constants $M,L$ there such thatPut $e=p-\bar{p}$ and $B(p)=[D\Phi (p){]}^{-1}$. The quadratic Newton error bound in a Banach space follows from the integral remainder:soTake a closed radius-$r$ ball lying in the neighbourhood with $Cr<1$. This inequality maps the ball into itself and makes errors tend to zero. More explicitly, for $e_n=\Vert p_n-\bar{p}\Vert$, induction gives $C{e}_{n+j}\le (C{e}_n{)}^{2^j}$. Hence every sufficiently close initial point converges to $\bar{p}$ at least quadratically; special maps can converge still faster.

For a fixed-point equation, apply the same method to $F(p)=\Phi (p)-p$. Wherever $D\Phi (p)-I$ is invertible, the exact adapted map isEvery fixed point of $\Phi$ is fixed by $A$, and conversely on this domain. This additional invertibility condition is distinct from invertibility of $D\Phi$; for example, $D\Phi =I$ is invertible but $D\Phi -I=0$ is not.

For certification, a frozen Newton correction for a fixed-point equation is often more useful. Choose a bounded, injective approximate inverse $J$ of $D\Phi (p_0)-I$ and setAgain $A_J(p)=p$ if and only if $\Phi (p)=p$. A well-chosen $J$ makes the corrected operator contract even if the original operator has an expanding direction. The subsequent contraction mapping argument requires the stated derivative and residual bounds; it does not assume that arbitrary inverse approximations provide them.

Let $B=\{p:\Vert p-p_0\Vert \le r\}$ and write $\epsilon =\Vert A(p_0)-p_0\Vert$. Integrating the Fréchet derivative along the line segment between two points in the convex set givesand the operator norm bound impliesIn particular, for $p\in B$,Thus the a posteriori contraction ball is mapped into itself. The ball is a closed subset of a Banach space, hence is a complete metric space. Applying the contraction mapping theorem to $A:B\to B$ gives existence and uniqueness:The estimate also gives $\Vert p_{\ast }-p_0\Vert \le \epsilon /(1-\kappa )$, by applying the same triangle inequality to $A(p_{\ast })=p_{\ast }$. Equality in the residual bound is allowed; strict contraction mapping comes from $\kappa <1$, and a fixed point on the closed ball's boundary is included.

The contraction mapping theorem states that a contraction mapping $f$ of a nonempty complete metric space into itself has a unique fixed point. More explicitly, if $d(f(x),f(y))\le qd(x,y)$ with $0\le q<1$, iteration from any $x_0$ converges to that fixed point.

To prove this, put $x_{n+1}=f(x_n)$. Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for $m>n$ the triangle inequality givesThus $(x_n)$ is a Cauchy sequence, and completeness supplies a limit $x_{\ast }$. A contraction mapping is continuous, so $f(x_{\ast })=\lim f(x_n)=\lim x_{n+1}=x_{\ast }$. Two fixed points satisfy $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing equality of the points. Passing to the limit in the bound also gives the useful error estimate $d(x_n,x_{\ast })\le q^{n}d(x_1,x_0)/(1-q)$.

For the function space, take $X$ nonempty; otherwise the supremum in the printed definition is undefined without a convention. Boundedness of $X$ makes the uniform metric $\rho (f,g)$ finite. Nonnegativity and symmetry follow from $d$; $\rho (f,g)=0$ means $f(x)=g(x)$ at every $x$. Taking suprema in $d(f(x),h(x))\le d(f(x),g(x))+d(g(x),h(x))$ proves the triangle inequality. Hence $\rho$ is a metric.

Suppose $(f_n)$ is a Cauchy sequence for this uniform metric. At each $x$, $d(f_n(x),f_m(x))\le \rho (f_n,f_m)$, so completeness of $X$ supplies a value $f(x)={\lim }_n{f}_n(x)$. Given $\epsilon >0$, choose $N$ with $\rho (f_n,f_m)<\epsilon /2$ for all $n,m\ge N$. Fix $n\ge N$ and let $m\to \infty$ pointwise; then $d(f_n(x),f(x))\le \epsilon /2$ for every $x$. Thus $\rho (f_n,f)\le \epsilon /2<\epsilon$. The uniform limit theorem makes this limit continuous, so $f\in F$. This proves the completeness of the continuous-map space in the uniform metric.

Finally fix $f\in C$ with contraction constant $q_f<1$, and write $x_f=\theta (f)$ and $x_g=\theta (g)$. Using the fixed point equations,Therefore the fixed-point stability in the uniform metric estimate isChoosing $\rho (f,g)<(1-q_f)\epsilon$ proves continuity of $\theta$ at every $f$. The estimate uses the constant of the fixed map $f$ only; it does not require a common contraction constant for all nearby $g$.

We count periodic orbits whose trajectories remain near the two specified homoclinic orbits; the local hypotheses impose no bound on unrelated distant cycles. Choose the return domain small enough that $A\delta |x{|}^{\delta -1}$ and $B\delta |x{|}^{\delta -1}$ are less than one. Each admissible return branch is then a contraction mapping.

A right-lobe periodic orbit is a fixed point $(p,+1)$ with $p>0$ and $p=-\mu +A{p}^{\delta }$. For small amplitudes it exists exactly when $\mu <0$, and $p=-\mu +o(|\mu |)$. Similarly, a left-lobe periodic orbit has $(x,y)=(-q,-1)$, $q>0$, $q=-\nu +B{q}^{\delta }$, and exists for $\nu <0$. Their Floquet multipliers tend to zero, so both are attracting.

A period-two orbit of the section map corresponds to one flow periodic orbit traversing both lobes. Write its states as $(p,-1)$ and $(-q,+1)$. Its positive amplitudes satisfyThe two-step derivative is $AB{\delta }^2{p}^{\delta -1}{q}^{\delta -1}$, again tending to zero. The asymmetric planar gluing return map therefore gives at most one such attracting two-lobe cycle in the small domain. Its boundaries are obtained by setting one amplitude to zero:For the original flow these are leading equations, with corrections $o({\nu }^{\delta })$ or $o({\mu }^{\delta })$. On $C_R$, the negative unstable branch makes both excursions before returning to the positive stable branch; on $C_L$ the positive unstable branch returns to the negative stable branch. Each is a compound homoclinic orbit, with one extra passage near the saddle equilibrium. They are distinct from the primary loop curves $\mu =0$ and $\nu =0$.

The six open regions and their local cycles are:

There is no additional small alternating cycle when both parameters are negative. Indeed the two positive-amplitude equations would imply $q<A{p}^{\delta }$ and $p<B{q}^{\delta }$, giving $p<B{A}^{\delta }{p}^{{\delta }^2}$, impossible for sufficiently small positive $p$. On each mixed-sign quadrant the contraction equations give the two-lobe cycle precisely on the side of the compound curve toward the positive quadrant.

On $\mu =0$, there is the primary right homoclinic orbit, together with a left cycle if $\nu <0$ or a two-lobe cycle if $\nu >0$. On $\nu =0$ the analogous statement holds with left and right interchanged. On $C_R$ there is a compound homoclinic orbit and the right single-lobe cycle; on $C_L$ there is a compound homoclinic orbit and the left single-lobe cycle. At $(0,0)$ the two primary loops meet at the same saddle equilibrium. The approach to each homoclinic orbit has a diverging period; it is not a saddle-node collision of two finite-period cycles.

The contraction mapping theorem states that a contraction mapping $T:X\to X$ on a nonempty complete metric space has a unique fixed point. Here contraction means that $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one constant $0\le q<1$. Choose $x_0\in X$ and define $x_{n+1}=T{x}_n$. Successive distances satisfy $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for every $m\ge 1$,Thus $(x_n)$ is a Cauchy sequence. Completeness gives $x_n\to x_{\ast }$. Since a contraction mapping is a continuous function, $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ were another fixed point, $d(x_{\ast },y_{\ast })=d(T{x}_{\ast },T{y}_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing distance zero. This proves both existence and uniqueness.

Now suppose $T=f^k$ is a contraction mapping, where $k$ is a positive integer. Let $y$ be its unique fixed point. Since $f$ commutes with its own iterate, $T(fy)=f(Ty)=fy$. Thus $fy$ is also a fixed point of $T$, so $fy=y$. Any fixed point of $f$ is a fixed point of $T$, and must equal $y$. A contractive iterate therefore gives a unique fixed point of $f$, without assuming continuity of $f$ itself.

The space $X=C([0,1],\mathbb{R})$ is complete for the uniform norm. Indeed, a uniformly Cauchy sequence converges pointwise by completeness of $\mathbb{R}$; passing to the limit in its uniform Cauchy bound shows uniform convergence, and the uniform limit of continuous functions is a continuous function. The operator $F$ maps this space to itself: $s\mapsto \varphi (h(s),s)$ is a continuous function, and its integral plus the continuous $g$ is a continuous function.

For $n=0$ the desired bound is the definition of $\Vert h-k{\Vert }_{\infty }$. Suppose it holds for $n$. The Lipschitz condition and the fact that the same $g$ cancels giveThis proves the estimate by induction. Taking the supremum over $0\le t\le 1$, the Lipschitz constant of $F^n$ is at most $M^n/n!$. These constants tend to zero, since the ratio of successive terms is $M/(n+1)$. Some iterate is consequently a contraction. The preceding result gives a unique fixed point of $F$, which is exactly $\boxed{\text{the unique continuous solution of the integral equation on }[0,1]}$. No assumption $M<1$ is needed.

Choose $0<R<R_1$, $0<S<S_1$ and $0<T<T_1$. On this smaller closed polydisc, the holomorphic function $A$ and its $w$-derivative are bounded; write $|A|\le M$ and $|{\partial }_{w}A|\le L$. Choose $0<r<R$ with $rM<S/2$ and $rL<1$, replacing a zero bound by any positive upper bound if necessary, and choose $0<\epsilon <T$.

Consider functions $u(z,\alpha )$ continuous on $\overline{D(0,r)}\times \overline{D(0,\epsilon )}$, jointly holomorphic in the interior, with $u(0,\alpha )=0$ and $\Vert u{\Vert }_{\infty }\le S/2$. This is a closed subset of a Banach space in the supremum norm. The integral operatoris well defined on the straight segment and is jointly holomorphic. Indeed, it equals $z{\int }_0^{1}A(tz,u(tz,\alpha ),\alpha )dt$, an integral of jointly holomorphic functions locally uniformly bounded in both variables. Convexity of the $w$-disc and the derivative bound giveThus the Banach fixed-point theorem gives a fixed point $f(z,\alpha )$. More explicitly, the successive integral iterates starting from zero converge uniformly, and locally uniform convergence of holomorphic functions makes their limit jointly holomorphic. Differentiating the integral equation gives the required complex ordinary differential equation and initial value. This proves local existence, uniqueness and joint holomorphic dependence on $(z,\alpha )$. Uniqueness among arbitrary local solutions follows first on a small disc where their values stay within $|w|<S$, by the same contraction mapping estimate, and then throughout their common connected domain by the identity theorem.

For holomorphic dependence of ordinary differential equations on parameters in the initial value, fix an admissible initial value $w_{\ast }$ and put $u=f-w_0$. The equation becomesIts right-hand side is jointly holomorphic in $(z,u,w_0)$ near $(0,0,w_{\ast })$. The preceding construction supplies a common disc and a jointly holomorphic solution for $w_0$ near $w_{\ast }$. Consequently $f=u+w_0$ is jointly holomorphic in $z$ and its initial value $w_0$.

A fixed singularity of a complex differential equation has a location determined by the coefficients, independently of the initial value. A movable singularity of a complex differential equation belongs to an individual continued solution and changes location when the initial value changes. For example, $w^{\prime }=w^2$ has the movable pole $z=c$ in $w=-1/(z-c)$. A fixed exceptional location need not be singular for every solution.

The precise first-order form of the Painlevé determinateness theorem needed here is as follows. Let $w^{\prime }=A(z,w)$ be rational in $w$, with coefficients meromorphic in $z$ on the chosen independent-variable surface. Algebraic coefficients can be made single valued on their own surface. Exclude the fixed coefficient singularities and exceptional fibres where a reduced numerator and denominator vanish together, working also in the dependent coordinate $v=1/w$. If a solution is continued along an arc $\gamma :[0,1)\to U$ with endpoint $z_{\ast }={\lim }_{t\to 1}\gamma (t)$ outside that exceptional set, then $w(\gamma (t))$ has a definite finite or infinite limit. Any singularity there is a pole or an algebraic branch point; there is no movable essential or logarithmic singularity. The standard rectifiable-arc formulation is included in this statement.

Here is a proof. First suppose a sequence along the arc has $(z_n,w_n)\to (z_{\ast },w_{\ast })$ at a point where the vector field is holomorphic. Choose a smaller closed bidisc about this point. Uniform bounds on the vector field and its dependent-variable derivative give, by the integral contraction mapping proof, one radius $r>0$ on which every solution starting sufficiently close to $(z_{\ast },w_{\ast })$ exists and remains bounded. For large $n$, the solution through $(z_n,w_n)$ is therefore defined on a fixed disc about $z_{\ast }$. The entire sufficiently late tail of $\gamma$ lies in that disc. Uniqueness identifies the original continued solution with this local solution throughout that tail, so the solution extends through $z_{\ast }$ and has a limit. The same argument applies to an ordinary point in the reciprocal coordinate $v=1/w$.

If the limit did not exist, consider the cluster set $K$ in the dependent-variable Riemann sphere:Each closed tail is compact and connected, and these sets are nested; hence $K$ is nonempty, compact and connected. The preceding paragraph excludes every ordinary point from $K$. At $z_{\ast }$ the rational vector field has only finitely many nonordinary dependent values, in both coordinate charts: an entire fibre of singular points was excluded among the fixed exceptions. Thus $K$ is a connected subset of a finite set, so it is a singleton. Compactness then forces convergence to its unique point, contradicting the assumed lack of a limit. This proves determinateness.

It remains to determine the local type rather than merely the limit. At a finite limiting value $w_{\ast }$ which is a pole of the vector field, write the reduced equation as $w^{\prime }=P/Q$. Since $z_{\ast }$ is not exceptional, $P(z_{\ast },w_{\ast })\ne 0$, and the inverse equationis holomorphic near $(z_{\ast },w_{\ast })$. The holomorphic dependence of ordinary differential equations on parameters makes its local solution curves jointly holomorphic in $w$ and their initial $z$ values. In particular, the value at the transverse line $w=w_{\ast }$ is a holomorphic first integral labelling these curves. The original graph has constant label; taking its endpoint limit shows that it lies on the inverse solution through $(z_{\ast },w_{\ast })$. That inverse solution cannot be the vertical curve $z\equiv z_{\ast }$: this would require $Q(z_{\ast },w)$ to vanish identically near $w_{\ast }$, an excluded fixed singular fibre. Consequently its power series has a first nonzero term,Here $k\ge 2$ at a vector-field pole. Factor the right side as $(w-w_{\ast }{)}^k$ times a nonvanishing holomorphic function and take its local $k$th root. This gives a new dependent coordinate with nonzero derivative, and the holomorphic inverse function theorem expresses $w$ as a convergent power series in $(z-z_{\ast }{)}^{1/k}$. Thus the singularity is an algebraic branch point. At an infinite limit, repeat the argument for $v=1/w$. An ordinary zero of $v$ gives a pole of $w$, while a finite ramification of $v$ gives an algebraic pole branch. This proves the asserted local types and the theorem.

For the given equation, the finite-coordinate rational field isThe whole fibre $z=-1$ is singular in the displayed coefficient, and at $(z,w)=(0,0)$ the reduced numerator and denominator vanish together. Hence the finite fixed exceptional set is $\{-1,0\}$. Away from these two locations, the only possible finite dependent limits at a singularity are $w_{\ast }=\sigma z_{\ast }$, with $\sigma =\pm 1$.

At such a point the inverse equation isIt is holomorphic when $z_{\ast }\ne 0,-1$ and $w_{\ast }=\sigma z_{\ast }$. Its inverse solution satisfies $z_w(w_{\ast })=0$, and differentiation givesThereforeandThese are genuine square-root algebraic branch points. The inverse solutions can be based at arbitrary $z_{\ast }\notin \{0,-1\}$ and then continued to nearby ordinary initial points. Their distinct branch locations require different initial data, so these branch points are movable. The signs specify the two inverse branches near each of the two values $w_{\ast }=\pm z_{\ast }$.

There are no movable poles. Indeed the reciprocal equation isFor every finite $z\ne -1$ it is holomorphic at $v=0$, and $v\equiv 0$ is its unique solution with zero initial value. A nonzero reciprocal solution therefore cannot reach zero there. This rules out both an ordinary pole and an infinite algebraic branch at a movable finite location. Together with the theorem, it exhausts the finite movable singularities.

The fixed point zero must not be mistaken for a generic movable collision. It is ordinary for solutions with $w(0)\ne 0$. But solutions reaching $w=0$ can have a fixed square-root singularity. To construct them rather than just balance powers, set $z(w)=w^{2}h(w)$ in the inverse equation. The equation for $h$ isEquivalently,On a small disc this is a contraction mapping of the closed ball $\Vert h-1/2{\Vert }_{\infty }\le 1/4$: both its departure from $1/2$ and its Lipschitz constant are $O(r^2)$. Its unique holomorphic fixed point gives $z=w^2/2+O(w^4)$. Thusis a genuine branch at the fixed location zero. The zero solution itself extends through zero as a function, although the original displayed coefficient is undefined at $(0,0)$.

The point $-1$ can produce a fixed nonalgebraic singularity. For an explicit example take $z=-1+x$ real, $0<x<x_0<1$, and $w=iy$ with initial $y>2$. Put $q=y^2$ and $L=-\log x$. ThenAs $L$ increases, $q$ increases, its derivative is bounded above by $2$ and bounded below by a positive constant, and therefore the solution exists for all large $L$ and $q\to \infty$. The derivative then tends to $2$, so $q/L\to 2$. ConsequentlyThis is unbounded but grows more slowly than every negative power of $z+1$. It cannot be a meromorphic pole even after a finite ramified substitution, demonstrating an actual fixed singularity with logarithmic growth. The determinateness theorem does not exclude such behavior at fixed exceptions.

Finally, on the independent-variable Riemann sphere use $t=1/z$. For finite $w$ the transformed equation iswhich is ordinary at $t=0$. At simultaneous $z=\infty$, $w=\infty$, however, the reciprocal field iswith an indeterminate point at $(t,v)=(0,0)$. Thus infinity is an additional fixed exceptional location on the sphere, rather than an unavoidable singularity of solutions with finite limiting value. There are solutions escaping to infinity there: for real initial data $w(z_0)>z_0>0$, the difference $d=w-z$ stays positive, since its derivative becomes positive as $d\downarrow 0$. Moreover $d^{\prime }\le 0$ when $d\ge 1$, so $0<d\le \max (1,d(z_0))$. Such a solution continues for all positive $z$, with $w=z+O(1)\to \infty$.

In the finite plane, the fixed exceptional locations are $0,-1$; all other solution singularities are movable square-root branches at $w=\pm z$, and there are no movable poles. On the sphere, infinity is also a fixed exception at the infinite dependent value. Being fixed or movable describes the base location, not merely whether the coefficient contains the factors $w-z$ and $w+z$.

The contraction mapping theorem states that a contraction mapping $T:X\to X$ on a nonempty complete metric space has a unique fixed point. Here contraction means that some $0\le q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

For a proof, choose $x_0$ and iterate $x_{r+1}=T{x}_r$. Induction gives $d(x_{r+1},x_r)\le q^{r}d(x_1,x_0)$. For $m>n$, the triangle inequality and a geometric series giveThus the iterates form a Cauchy sequence, and completeness supplies a limit $a\in X$. A contraction mapping is continuous, so $Ta=\lim T{x}_n=\lim x_{n+1}=a$. If $b$ is another fixed point, $d(a,b)=d(Ta,Tb)\le qd(a,b)$, forcing $d(a,b)=0$. This proves existence and uniqueness.

For the exponential problem, use $h=f\circ f$ on the complete metric space $\mathbb{R}$. Its derivative isPutting $t=e^{-x}>0$, the function $t{e}^{-t}$ has maximum $1/e$ at $t=1$, as its derivative is $(1-t)e^{-t}$. By the mean value theorem, $|h(x)-h(y)|\le e^{-1}|x-y|$, so $h$ is a contraction mapping on all of $\mathbb{R}$. Let $a$ be its unique fixed point. The compositions commute, so $h(f(a))=f(h(a))=f(a)$; uniqueness of the fixed point of $h$ forces $f(a)=a$. Conversely every fixed point of $f$ is fixed by $h$. Hence there is exactly one real solution of $x=e^{-x}$. It lies in $(0,1)$ because $a=e^{-a}>0$ and then $e^{-a}<1$. This uses the iterated contraction argument without claiming that $f$ itself contracts on all of $\mathbb{R}$.

Convergence of a numerical method on $[0,T]$ means that its grid error obeys ${\max }_{0\le nh\le T}|y_n-y(nh)|\to 0$ as $h\to 0$, with consistent initial values.

First, the implicit step of Backward Euler method is well-defined for $hL<1$: the map $z\mapsto y_n+hf(t_{n+1},z)$ is a contraction mapping on $\mathbb{R}$, so the Banach fixed-point theorem gives a unique solution. Let $M={\max }_{[0,T]}|y^{\prime \prime }|$, finite under the stated smoothness. The exact solution's step defect isso $|d_n|\le M{h}^2/2$. Subtract the exact and numerical step equations. The Lipschitz condition givesWith exact initial data, sum this geometric progression to obtainFor $hL\le 1/2$ and $nh\le T$, $-\log (1-hL)\le 2hL$, so the bracket is at most $e^{2LT}-1$. Thus the global error is $O(h)$, uniformly on $[0,T]$, and Backward Euler method converges with order one. A vanishing initial error adds only $(1-hL{)}^{-n}|e_0|$.

Order one is sharp: for $y^{\prime }=y$, $y(0)=1$ and $nh=T$, $y_n=(1-h{)}^{-n}=e^T(1+Th/2+O(h^2))$.

For the Dahlquist test equation $y^{\prime }=\lambda y$, the stability function is $R(z)=1/(1-z)$, $z=h\lambda$. Therefore its linear stability domain isThe strict inequality gives decay; equality gives bounded amplification. The whole closed left half-plane lies in this region, so Backward Euler method is A-stable.

The contraction mapping theorem states that a map $f$ of a nonempty complete metric space into itself, satisfying $d(fx,fy)\le qd(x,y)$ for some $0\le q<1$, has a unique fixed point. Iteration from any point converges to it; for example $d(f^{n}x,\tau (f))\le q^{n}d(x,fx)/(1-q)$.

Assume $X$ is nonempty. Its boundedness makes $\delta$ finite. Symmetry, positivity and separation follow pointwise from $d$, and taking the supremum of $d(fx,hx)\le d(fx,gx)+d(gx,hx)$ proves the triangle inequality. Hence $\delta$ is a uniform metric.

If $(f_n)$ is a Cauchy sequence in this uniform metric, each $(f_n(x))$ converges by completeness of $X$; denote its limit by $f(x)$. Given $ϵ>0$, choose $N$ so that $\delta (f_n,f_m)<ϵ$ for $n,m\ge N$. Letting $m\to \infty$ gives $d(f_n(x),f(x))\le ϵ$ for all $x$, so $f_n\to f$ uniformly. To prove continuity, fix $x$ and useChoose $N$ so the outside terms are small, then use continuity of $f_N$ for the middle term. Thus $f\in \mathrm{Maps}(X,X)$, proving completeness of the continuous-map space in the uniform metric.

The contraction subspace is not necessarily complete. On $X=[0,1]$, $f_n(x)=(1-1/n)x$, $n\ge 2$, are contraction mappings and converge to the identity by uniform convergence. Their only possible uniform limit is not a contraction, so this Cauchy sequence has no limit in $\mathcal{C}$.

Nevertheless the fixed point assignment is continuous. Fix $f\in \mathcal{C}$ with contraction constant $q_f<1$, and put $u=\tau (f)$, $v=\tau (g)$. ThenConsequently $d(\tau (f),\tau (g))\le \delta (f,g)/(1-q_f)$. This proves continuity of the fixed-point assignment at $f$ without requiring one contraction constant valid for every $g$.

A contraction mapping $T$ on a metric space satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one constant $0\le q<1$. The contraction mapping theorem says that on a nonempty complete metric space such a map has a unique fixed point; successive iterates from any starting point converge to it. More precisely, if $x_n=T^n{x}_0$, then $d(x_n,x_{\ast })\le q^{n}d(x_1,x_0)/(1-q)$.

The Volterra integration operator maps continuous functions to continuous functions. In the uniform norm,This constant cannot be decreased: for $f=1$ and $g=0$, $Af(x)=x$, and both uniform norms are one. Hence $A$ is not a contraction.

Changing the order of integration over the triangular region givesThus$A^2$ is a contraction, with sharp constant $1/2$. This is an example of an iterated contraction.

The characteristic polynomials of a linear multistep method areThe consistency of a numerical method conditions hold for every real $a$:Both roots of a polynomial of $\rho$, namely $1$ and $-1$, are simple and have modulus one. The root condition for a multistep method therefore gives zero-stability for every $a$. The Dahlquist equivalence theorem now gives convergence for every fixed $a\in \mathbb{R}$. As usual, this means convergence of a numerical method on each fixed finite interval for a sufficiently regular ordinary differential equation with a Lipschitz continuous vector field and starting values tending to the exact starting values. When $a\ne 0$, the implicit time-stepping method update is locally uniquely solvable for sufficiently small $h$, for example by a contraction mapping if $|ha|L<1$. Zero-stability does not assert that a large fixed step is suitable for a stiff differential equation.

We argue by contradiction using the previous subsequence construction. If $X$ is not complete, choose a Cauchy sequence with no limit in $X$. None of its $d_m$ can be zero: if $d_m=0$, then $d(x_m,x_n)\to 0$, giving a limit in $X$. The construction therefore produces pairwise distinct $y_j$ withLet $Y=\{y_1,y_2,\dots \}$. This subset is closed in $X$. Indeed, if a point $x\notin Y$ lay in its closure, there would be points of $Y$ tending to $x$. Their indices must be unbounded, because the distance from $x$ to any finite subset of $Y$ is positive. Passing to increasing indices gives a convergent subsequence of the original Cauchy sequence. Part (a) would then make the whole sequence converge to $x$, a contradiction.

Now define $F:Y\to Y$ by $F(y_j)=y_{j+1}$. Distinctness makes this well-defined, and the displayed inequality makes it a contraction mapping with constant at most $1/2$. It has no fixed point, contradicting the assumed property of closed subsets. ConsequentlyThis is the converse direction of contraction property of closed subsets characterizes completeness.

A contraction mapping on a metric space $(X,d)$ is a map $f:X\to X$ for which there is a single constant $q$ with $0\le q<1$ such thatThe same $q$ must work for every pair. Merely decreasing each nonzero distance strictly does not suffice for this definition.