Which Hypothesis Has Failed?
The Banach fixed-point theorem gives a powerful guarantee: a contraction from a nonempty complete metric space to itself has exactly one fixed point, and iteration from any starting point converges to it. When an iteration does not produce a fixed point in a particular problem, it matters which hypothesis is missing. Incompleteness removes the existence guarantee, but does not rule out fixed points. Failure of the contraction condition removes the theorem’s guarantee of uniqueness and convergence, but does not by itself imply that there are no fixed points.
Throughout, a metric space \((X,d)\) is complete if every Cauchy sequence in \(X\) converges to a point of \(X\). A map \(T:X\to X\) is a contraction if there is one number \(q\) with \(0\le q<1\) such that \(d(Tx,Ty)\le qd(x,y)\) for all \(x,y\in X\). The requirement \(T:X\to X\) is essential: every iterate must remain in the space. Given \(x_0\in X\), its Picard iterates are \(x_{n+1}=T(x_n)\) for \(n\ge0\).
This tutorial examines what can and cannot be concluded when one of these assumptions is absent. Two facts help organize the examples: contraction still guarantees uniqueness if a fixed point exists, and any limit of Picard iterates that lies in the space must be a fixed point.
What Contraction Still Guarantees
Proof. Let \(p,r\in X\) both be fixed points, so \(T(p)=p\) and \(T(r)=r\). If \(T\) has contraction constant \(q\), then \(d(p,r)=d(T(p),T(r))\le qd(p,r)\). Since \(q<1\), this inequality implies \((1-q)d(p,r)\le0\). The distance is nonnegative and \(1-q>0\), so \(d(p,r)=0\). The defining property of a metric gives \(p=r\). This proves that there cannot be two distinct fixed points, whether or not \(X\) is complete. \(\square\)
Proof. Write \(x_n\to p\). The contraction inequality gives \(d(T(x_n),T(p))\le qd(x_n,p)\), so \(d(T(x_n),T(p))\to0\). But \(T(x_n)=x_{n+1}\), and the shifted sequence \(x_{n+1}\) also converges to \(p\). Limits in a metric space are unique: by the triangle inequality, \(d(p,T(p))\le d(p,x_{n+1})+d(T(x_n),T(p))\), and the right side tends to zero. Thus \(d(p,T(p))=0\), hence \(p=T(p)\). \(\square\)
Consequently, a contraction on an incomplete space may still have a fixed point. What incompleteness can disrupt is the guarantee that the iterates converge to a point in that space. The next example shows this distinction explicitly.
Worked Example: A Contraction Whose Fixed Point Is Missing
Take \(X=(0,1)\) with the usual metric \(d(x,y)=|x-y|\), and define \(T:X\to X\) by \(T(x)=x/2\). For \(0<x<1\), we have \(0<x/2<1/2<1\), so this is indeed a self-map. For any \(x,y\in X\), \(d(T(x),T(y))=|x/2-y/2|=\tfrac12|x-y|\). Thus \(T\) is a contraction with \(q=1/2\).
The space is incomplete: the sequence \(x_n=1/n\), for \(n\in\mathbb N\), is Cauchy in the usual metric because it converges to \(0\) in \(\mathbb R\), but \(0\notin X\). To find fixed points of \(T\), solve \(x=T(x)\): \(x=x/2\) implies \(x=0\), which is not in \(X\). So there is no fixed point in \(X\).
For a specific iteration, start at \(x_0=3/4\). Then \(x_1=3/8\), \(x_2=3/16\), and in general \(x_n=(3/4)/2^n\). Each term is positive and less than \(1\), so each iterate belongs to \(X\); the terms converge in \(\mathbb R\) to \(0\), not to a point of \(X\).
Check: The only possible fixed-point value is \(0\), since \(x=x/2\) forces \(x=0\), and substitution gives \(T(0)=0\) in the larger space \(\mathbb R\). But \(0\notin X\). The example fails the completeness hypothesis, not the self-map or contraction hypotheses.
The completion makes the missing limit visible. The same formula \(T(x)=x/2\) defines a contraction on \(\mathbb R\), where its fixed point is \(0\). In the incomplete space \((0,1)\), the iterates approach that point without reaching any point of the space as their limit. Convergence in a completion does not amount to convergence to a point in the original space.
Worked Example: An Incomplete Space Can Still Have a Fixed Point
Again use \(X=(0,1)\) with its usual metric, but now set \(T(x)=x/2+1/4\). For \(0<x<1\), \(1/4<T(x)<3/4\), so \(T\) maps \(X\) into itself. Moreover, \(d(T(x),T(y))=|(x-y)/2|=\tfrac12d(x,y)\), so it is a contraction.
The fixed-point equation is \(x=x/2+1/4\). Subtracting \(x/2\) from both sides gives \(x/2=1/4\), hence \(x=1/2\), which belongs to \(X\). Substitution checks the answer: \(T(1/2)=1/4+1/4=1/2\).
Start instead at \(x_0=1/4\). The iterates are \(x_1=3/8\), \(x_2=7/16\), and \(x_3=15/32\). More generally, \(x_n=\tfrac12-\tfrac{1}{2^{n+2}}\). Indeed, this formula gives \(x_0=1/2-1/4=1/4\); applying \(T\) to the expression at index \(n\) gives \(T(x_n)=\tfrac14+\tfrac12(\tfrac12-\tfrac{1}{2^{n+2}})=\tfrac12-\tfrac{1}{2^{n+3}}=x_{n+1}\). Every term is between \(1/4\) and \(1/2\), hence belongs to \(X\), and the iterates converge to the fixed point \(1/2\).
Check: Substituting \(x=1/2\) into the original formula returns \(1/2\). This contraction has a fixed point despite the incompleteness of its domain. Therefore, incompleteness removes a general guarantee; it does not make fixed points impossible.
When the Map Is Not a Contraction
A map that is not a contraction falls outside the theorem’s hypotheses. That fact alone does not tell us whether it has a fixed point, whether it has several, or whether its iterates converge. The following maps show three different outcomes. Each uses the usual metric on \(\mathbb R\), which is complete.
Worked Example: A Noncontractive Map with No Fixed Point
Let \(T:\mathbb R\to\mathbb R\) be \(T(x)=x+1\). It is a self-map, and for all \(x,y\in\mathbb R\), \(d(T(x),T(y))=|(x+1)-(y+1)|=|x-y|=d(x,y)\). It is not a contraction: for \(x\ne y\), the contraction inequality would require \(1\le q\) after division by \(|x-y|>0\), contradicting \(q<1\).
A fixed point would satisfy \(x=x+1\), which would imply \(0=1\). Thus there is no fixed point. Starting at \(x_0=-2\), the iterates are \(x_1=-1\), \(x_2=0\), \(x_3=1\), and, by induction, \(x_n=-2+n\). Since \(x_n\) grows without bound, this iteration does not converge in \(\mathbb R\).
Check: The formula \(T(x_n)=(-2+n)+1=-2+(n+1)=x_{n+1}\) verifies the stated iterates, while the fixed-point equation has no solution. The failed contraction hypothesis means the Banach theorem does not apply; the direct equation, not the failed hypothesis by itself, proves there is no fixed point.
Worked Example: A Noncontractive Map with Many Fixed Points
Let \(T:\mathbb R\to\mathbb R\) be the identity map \(T(x)=x\). For all \(x,y\in\mathbb R\), \(d(T(x),T(y))=d(x,y)\), so no constant \(q<1\) can satisfy the contraction inequality for distinct \(x\) and \(y\). Nevertheless, every real number is a fixed point: \(T(x)=x\) for each \(x\in\mathbb R\).
For example, \(T(0)=0\), \(T(1)=1\), and \(T(-3/2)=-3/2\). Starting at any \(x_0\), the iteration is \(x_1=x_0\), \(x_2=x_0\), and \(x_n=x_0\) for all \(n\). The iterates converge, but their limit depends on the starting point.
Check: If \(x_0=1\), the limit is \(1\); if \(x_0=0\), the limit is \(0\). Both satisfy the fixed-point equation, confirming that uniqueness fails when the contraction condition is absent.
These examples separate several conclusions that are bundled together in the Banach fixed-point theorem. Completeness supports existence and convergence of the iterates, while strict contraction supplies uniqueness and controls distances between images. Removing a hypothesis means that its conclusion is no longer guaranteed; it does not automatically mean that the conclusion is false.
A Common Pitfall
Common mistake. For \(T(x)=x/2\) on \((0,1)\), one may solve \(x=T(x)\), obtain \(x=0\), and report \(0\) as the fixed point. The algebra is correct, but the conclusion is not: a fixed point of \(T:X\to X\) must belong to \(X\), and \(0\notin(0,1)\). The correction is to check the candidate against the domain after solving the equation. In the completion \(\mathbb R\), \(0\) is a fixed point of the extended formula; in the original space \((0,1)\), there is no fixed point.
A related error is to say that every noncontractive map has no fixed point. The identity map disproves this: it is not a contraction, but every point is fixed. The correct diagnostic is to test each issue separately: is the map a self-map, is there a strict contraction constant, is the metric space complete, and what do the fixed-point equation and actual iterates show?
Check Your Understanding
For each question, distinguish what the hypotheses guarantee from what must be checked directly.
- On \((0,1)\), with the usual metric, determine whether \(T(x)=x/3+1/3\) is a contraction and find any fixed point.
- Why does the sequence \(1/n\) show that \((0,1)\) is incomplete, and where does that sequence converge in the completion?
- Can a contraction on an incomplete metric space have a fixed point? Give an example and verify the fixed point.
- For \(T(x)=x+2\) on \(\mathbb R\), determine whether it is a contraction and solve its fixed-point equation.
- Explain why a contraction has at most one fixed point even when the metric space is incomplete.