Contraction Mapping Proof Workshop

Why Coupled Contraction Proofs Need Care

In the previous tutorial, the unknown was a function and the fixed-point operator encoded a differential equation. A related challenge arises when the unknown has several components: one component of a map may depend on all the others, and a useful estimate for each output can involve several input distances. The ordinary maximum metric treats all components equally. That can make a map appear not to be a contraction even when the cross-dependence is small enough to control by giving the components different weights.

This tutorial develops a proof technique for such situations. We first turn componentwise estimates into one contraction estimate using a weighted maximum metric. We then prove an exact criterion for when suitable weights exist in the two-component case. Throughout, the Contraction Mapping Theorem is the final step: the new work is establishing the right complete metric and checking that the map is a contraction in it.

Weighted Metrics and a Two-Component Criterion

Let \((X_1,d_1)\) and \((X_2,d_2)\) be metric spaces. Choose positive weights \(w_1,w_2\). On \(X_1\times X_2\), define

This is a metric: nonnegativity, symmetry, and separation follow from the corresponding properties of \(d_1,d_2\), and the triangle inequality follows by applying each component triangle inequality before taking the maximum. The weights determine how much a difference in each component counts. In particular, if \(D(u,v)\leq r\), then the component distances satisfy \(d_1(u_1,v_1)\leq w_1r\) and \(d_2(u_2,v_2)\leq w_2r\).

Proof. Equip \(X_1\times X_2\) with the metric \(D\) defined above. This product metric is complete. Indeed, if \((z_n)\) is Cauchy in \(D\), each coordinate sequence is Cauchy in its respective metric, because \(d_i((z_n)_i,(z_m)_i)\leq w_iD(z_n,z_m)\). Completeness of each \(X_i\) gives coordinate limits, say \(z_1\) and \(z_2\). Convergence of both coordinates implies convergence to \((z_1,z_2)\) in \(D\), since the maximum of the two weighted coordinate distances tends to zero.

Now take \(u,v\in X_1\times X_2\). By the definition of \(D\), \(d_1(u_1,v_1)\leq w_1D(u,v)\) and \(d_2(u_2,v_2)\leq w_2D(u,v)\). The first component estimate therefore gives

The second gives

Let \(q\) be the larger of the two parenthesized constants. The hypotheses imply \(0\leq q<1\). Taking the maximum of the two inequalities yields \(D(F(u),F(v))\leq qD(u,v)\), so \(F\) is a contraction on the nonempty complete metric space \((X_1\times X_2,D)\). The Contraction Mapping Theorem gives a unique fixed point. \(\square\)

The conditions on the weights can be checked without guessing them indefinitely. They have a simple exact form for two components.

Proof. Put \(r=w_2/w_1\), so \(r>0\). The required inequalities become \(a+br<1\) and \(d+c/r<1\). If such an \(r\) exists, then \(a<1\) and \(d<1\). Multiplying \(br<1-a\) and \(c/r<1-d\) gives \(bc<(1-a)(1-d)\).

Conversely, suppose the three stated conditions hold. Write \(\alpha=1-a>0\) and \(\delta=1-d>0\). If \(b>0\) and \(c>0\), the product condition says \(c/\delta<\alpha/b\), so choose \(r\) strictly between these two numbers. Then \(br<\alpha\) and \(c/r<\delta\). If \(b=0\) and \(c>0\), choose any \(r>c/\delta\); the first inequality holds because \(a<1\), and the second holds by the choice of \(r\). If \(b>0\) and \(c=0\), choose \(0<r<\alpha/b\). If \(b=c=0\), any \(r>0\) works. In every case, take \(w_1=1\) and \(w_2=r\). This proves the equivalence. \(\square\)

Worked Examples: Choosing and Using the Weights

A Proof Checklist and a Common Pitfall

A coupled-map proof is easiest to audit when its obligations are kept separate. The component estimates do not by themselves give a fixed point. One must also establish that the map takes the proposed space into itself and that the space is complete in the metric actually used. The weighted criterion supplies the contraction estimate, but it cannot repair a failure of either of those other requirements.

A common error is to obtain a bound such as \(d_1(F_1(u),F_1(v))\leq a d_1(u_1,v_1)+b d_2(u_2,v_2)\) and call \(a+b\) the contraction constant without explaining why the two input distances are controlled by the same metric scale. That conclusion is justified for the unweighted maximum metric, because each coordinate distance is at most that maximum. For a weighted metric, the conversion factors \(w_1,w_2\) must appear; they produce the ratios in the theorem. Another error is to find a weighted contraction estimate on a set that is not invariant under the map. A contraction theorem applies to a self-map, not merely to a formula whose outputs have been estimated.

The weighted method is useful beyond two coordinates. For a finite product, one can seek positive weights that make each weighted sum of component influences less than one. In applications, the estimates may come from derivatives, integral bounds, or direct algebra. The proof pattern remains the same: first control each component, then choose a metric that converts those controls into one strict contraction estimate.