Fixed-Point Theorem Counterexamples

What a Counterexample Can Reveal

The Contraction Mapping Theorem combines several hypotheses: the space is nonempty and complete, and the map is a contraction from that space into itself. The weighted product criterion in the previous tutorial showed how to verify the contraction condition in a suitable metric; it did not make the other hypotheses optional. Counterexamples help separate the role of each condition from the conclusion.

In this tutorial, we will see that a strict contraction can lack a fixed point when the space is incomplete, that a map which only preserves distances need not have one, and that a formula that fails to map a proposed domain into itself cannot be treated as a self-map there. We will also see why the closed-interval fixed-point theorem does not extend to every compact-looking domain. These examples do not say that every hypothesis is necessary for every particular map. They show that dropping one can make a general fixed-point conclusion false.

Completeness Cannot Be Dropped in General

Proof. First, \(T\) maps \(X\) into itself. If \(0\leq x<1\), then \(1/2\leq (x+1)/2<1\), so \(T(x)\in X\). For \(x,y\in X\),

Thus \(T\) is a contraction with constant \(1/2\). If \(T(x)=x\), then \((x+1)/2=x\), which implies \(x=1\). But \(1\notin X\), so there is no fixed point.

To verify incompleteness, consider \(x_n=1-1/n\) for \(n\geq1\). Each \(x_n\in X\), and \(x_n\to1\) in \(\mathbb{R}\), so \((x_n)\) is Cauchy. It cannot converge to a point \(x\in X\): if it did, uniqueness of limits in \(\mathbb{R}\) would force \(x=1\), which is not in \(X\). Hence \(X\) is incomplete. This proves all three claims. \(\square\)

The missing limit is not an incidental detail. Iterating \(T\) from any \(x_0\in X\) gives \(x_{n+1}=(x_n+1)/2\), and direct calculation yields \(x_n=1-(1-x_0)/2^n\). Since \(x_0<1\), every iterate remains below \(1\), while the sequence tends to \(1\), outside the space. The iteration approaches the point that would solve the fixed-point equation, but the space does not contain it.

These examples show that completeness is needed for the general guarantee supplied by the Contraction Mapping Theorem. They do not show that an incomplete space can never support a fixed point. For instance, the constant map \(C:(0,1)\to(0,1)\) defined by \(C(x)=1/3\) has the fixed point \(1/3\). The correct lesson is about guarantees: on an incomplete space, the contraction condition alone does not ensure a fixed point.

The Strict Inequality Also Matters

A contraction must shrink every distance by a single factor strictly less than one. A map that merely does not increase distances is called nonexpansive. That weaker condition does not guarantee a fixed point, even on a complete space.

The contrast is about existence, not just proof technique. A strict contraction on a complete space has a fixed point by the Contraction Mapping Theorem. Distance preservation alone allows motion that never settles at any point, as translation demonstrates. The theorem on uniqueness of a fixed point for a contraction, established earlier in the course, does not supply existence when the contraction hypothesis is absent.

A Formula Must Preserve the Proposed Domain

The self-map condition is a separate obligation. To apply a fixed-point theorem on \(A\), the map must send every point of \(A\) back into \(A\). A contraction estimate for a formula does not establish this condition.

A map may still be studied on a larger space or on a different invariant subset, but that is a different application with its own domain and hypotheses. In a proof, check invariance directly before using a theorem: for each \(x\) in the proposed space, verify that \(T(x)\) belongs to that same space. An estimate comparing \(T(x)\) and \(T(y)\) does not by itself show that either output lies in the space.

Domain Shape Can Matter Too

The fixed-point theorem for a closed interval uses more than the fact that the domain is bounded or has no missing limit points. It is a theorem for continuous self-maps of an interval. On a different domain, even a continuous self-map that preserves distances may have no fixed point.

Proof. If \(u=(x,y)\in S^1\), then \((-x)^2+(-y)^2=x^2+y^2=1\), so \(R(u)\in S^1\). Thus \(R\) is a self-map. For \(u,v\in S^1\), the Euclidean distance satisfies

Therefore \(R\) preserves distances and is continuous. If \(R(u)=u\), then \(-u=u\), so \(u=(0,0)\). But \((0,0)\notin S^1\), since \(0^2+0^2\ne1\). Hence \(R\) has no fixed point. \(\square\)

This example prevents an overbroad reading of the closed-interval theorem: continuity and being a self-map do not imply a fixed point on every domain. The geometry and structure of the domain matter. The unit circle is a closed and bounded subset of \(\mathbb{R}^2\), but it is not an interval, and the antipodal map sends each point to a distinct point on the opposite side.

How to Use Counterexamples in a Proof

When a fixed-point argument seems plausible, identify the exact theorem being invoked and check each hypothesis against the map and its domain. The following sequence helps locate the gap:

A useful counterexample usually changes one feature at a time. The map on \([0,1)\) satisfies the self-map and contraction conditions but lacks completeness. Translation on \(\mathbb{R}\) is a self-map of a complete space but lacks strict contraction. The formula on \([1,2]\) has a strict distance estimate but fails to preserve the proposed set. The antipodal map is a continuous self-map, but the domain is not an interval. Keeping these distinctions clear prevents both invalid applications and overly strong conclusions about what a hypothesis means.