Contraction mappings and fixed points

Try

To solve x=(x+3)/2, start anywhere and repeatedly apply T(x)=(x+3)/2. The fixed point is 3. Each application halves the distance between any two inputs, and the error from 3 also halves. The contraction theorem generalizes this predictable iteration to complete metric spaces, including spaces of functions where fixed points can represent solutions of equations.

Predict

Predict which assumptions make the iteration meaningful. The map must send the chosen space into itself, its contraction factor q must be strictly below one, and Cauchy iterates must have a limit in the space. On (0,1), T(x)=x/2 is a contraction self-map, but its limiting fixed point zero is missing. The formula’s algebra alone cannot repair an incomplete domain.

Experiment

Trace the affine example starting at zero.

n	x_n	Distance to fixed point 3
0	0	3
1	3/2	3/2
2	9/4	3/4

The exact error shrinks geometrically. In a general problem the fixed point is unknown, so the theorem provides a bound using the first step and q instead. That bound can certify a requested tolerance without knowing the solution beforehand.

Understand

Let X be a nonempty complete metric space and T:X→X satisfy d(Tx,Ty)≤q d(x,y) for 0≤q<1. Define x_(n+1)=T(x_n). Induction gives d(x_(n+1),x_n)≤q^n d(x₁,x₀). For m>n, the triangle inequality and a geometric sum give d(x_m,x_n)≤q^n d(x₁,x₀)/(1−q). Hence the sequence is Cauchy and converges to p∈X.

A contraction is Lipschitz and therefore continuous. Thus Tx_n→Tp, while Tx_n=x_(n+1)→p. Uniqueness of limits gives Tp=p. If p and r were both fixed, d(p,r)≤q d(p,r), forcing distance zero. Finally, letting m tend to infinity in the tail estimate gives d(x_n,p)≤q^n d(x₁,x₀)/(1−q). This proves existence, uniqueness and an error estimate from the same mechanism.

Apply

An a posteriori estimate uses the latest step: d(x_n,p)≤q d(x_n,x_(n−1))/(1−q) for n≥1. It follows by bounding future step lengths geometrically from the current one. Such a bound depends on a proved global contraction constant on the invariant space; observing a few shrinking steps is not enough.

Repair “a Lipschitz constant one is almost as good.” On complete R, T(x)=x+1 is 1-Lipschitz and has no fixed point, while the identity has every point fixed. Strict contraction is what drives the geometric tail to zero and forces uniqueness.

Transfer: certify an approximate solution and a changed model

A computed iterate is rarely an exact fixed point. Suppose T is a contraction with constant q＜1 on the same nonempty complete metric space X, and p is its fixed point. For any x∈X, the triangle inequality gives

$$d(x,p)\le d(x,Tx)+d(Tx,Tp)\le d(x,Tx)+q\,d(x,p).$$

Rearrange to obtain the residual certificate d(x,p)≤d(x,Tx)/(1−q). Unlike a trace of earlier steps, this certificate uses only the current point and a proved contraction constant. The denominator explains why a tiny residual may still give a weak error bound when q is close to one.

Worked case. For T(x)=x/2+1 on R, use x=1.9. The residual is 0.05 and the error bound is 0.1. Solving the affine equation gives p=2, so this bound is attained. If a rounded evaluation of Tx has error at most ε, the true residual is at most the reported residual plus ε by the triangle inequality; use that larger numerator. A displayed decimal should not be silently treated as exact arithmetic.

Changed model. Suppose S also has a fixed point s in X and d(Tx,Sx)≤η for every x∈X. Then

$$d(p,s)=d(Tp,Ss)\le d(Tp,Ts)+d(Ts,Ss)\le q\,d(p,s)+\eta.$$

Consequently d(p,s)≤η/(1−q). Only T’s contraction is used in this comparison once s exists. To guarantee s exists, one can separately prove that S is a contraction self-map of X. Do not assume existence merely from closeness to T.

Predict, then inspect the certificate

Changed case Residual certificatePerturbed map

T(x)=x/2+1, approximate x=1.9. What bound follows from the residual?

Your prediction Choose a prediction0.050.1No finite bound Check reasoning Reset prediction

Two authored cases are practice, not a proof of the general theorem. Explain which hypothesis changed. Responses are temporary and do not change saved progress.

Changed task. Let Tₐ(x)=3x/4+a on R. Compare a=1 with a=1.02. Prove the fixed-point difference bound, then find both fixed points and check whether the bound is sharp. What happens to the guarantee if the same maps are restricted to (0,1)?

Feedback

The uniform map difference is 0.02, so the bound is 0.02/(1−3/4)=0.08. The fixed points are 4 and 4.08, attaining the bound. The restricted maps are not self-maps of (0,1), and that space is incomplete; no fixed-point conclusion there follows. Check the space before treating the numerical constant as a certificate.

Continue

Optional depth applies the theorem to integral operators on complete continuous-function spaces, a route to local existence for differential equations. One must identify a closed invariant set and a time interval small enough to make the operator a contraction. A formal integral equation alone does not establish these properties.

Save the self-map check, completeness check and contraction inequality before computing iterates. Then derive the number of steps needed for a chosen tolerance from the error estimate. The next lessons use this theorem again in a local proof of the inverse function theorem, after introducing derivative maps and their operator bounds. A contraction may exist only on a smaller closed ball, even when the original formula is defined everywhere. Choosing that ball is part of the proof: every iterate must remain inside it, and the same contraction constant must apply to every pair there.