Uniqueness of the Fixed Point

When the Map Itself Is Not a Contraction

The uniqueness theorem for contractions gives a direct conclusion: a contraction has at most one fixed point. But a map can fail to be a contraction even when repeated application of it contracts distances. In that situation, uniqueness can still be established by studying an iterate of the map. The key is to keep two steps distinct: first obtain a fixed point of the iterate, then use commutation to show that this point is fixed by the original map.

For a self-map \(f:X\to X\), define \(f^1=f\) and \(f^{m+1}=f\circ f^m\) for each positive integer \(m\). Thus \(f^m\) means \(m\) successive applications of \(f\), not a pointwise power. A point fixed by \(f\) is automatically fixed by every iterate \(f^m\). The converse need not hold for an arbitrary map, so it requires an additional argument.

Theorem (Uniqueness from a Contracting Iterate): Let \((X,d)\) be a nonempty complete metric space, let \(f:X\to X\), and suppose that \(f^m\) is a contraction for some positive integer \(m\). Then \(f\) has exactly one fixed point.

Proof. Since \(X\) is nonempty and complete and \(f^m:X\to X\) is a contraction, the Contraction Mapping Theorem guarantees a fixed point \(p\in X\) of \(f^m\). In particular, \(f^m(p)=p\). This first conclusion is about \(f^m\), not yet about \(f\).

The map \(f\) commutes with its own iterate: \(f\circ f^m=f^m\circ f\), by associativity of composition. Therefore

$$ f^m(f(p))=f(f^m(p))=f(p). $$

So \(f(p)\) is also a fixed point of \(f^m\). A contraction has at most one fixed point, by the Uniqueness of a Fixed Point theorem established in “Contractions.” The two fixed points \(f(p)\) and \(p\) of \(f^m\) must therefore be equal. Hence \(f(p)=p\): the point supplied by the Contraction Mapping Theorem for \(f^m\) is a fixed point of \(f\).

Finally, suppose \(r\in X\) is any fixed point of \(f\). Repeated application of \(f(r)=r\) gives \(f^m(r)=r\). Thus \(r\) is a fixed point of \(f^m\), whose fixed point is unique. It follows that \(r=p\). This proves both existence and uniqueness of the fixed point of \(f\). \(\square\)

The proof depends on the order of these arguments. The Contraction Mapping Theorem first supplies a fixed point of \(f^m\); commutation then shows that this point is fixed by \(f\). Only after that do we use the fact that every fixed point of \(f\) is fixed by \(f^m\) to rule out any other fixed points of \(f\). Commutation alone does not establish that \(p\) is fixed by \(f\); it establishes that \(f(p)\) is another fixed point of \(f^m\), where uniqueness can be applied.

Worked Examples: Applying the Iterate Argument

Worked Example: A Map That Is Not a Contraction but Has a Contracting Square

Let \(X=\mathbb{R}^2\), equipped with the maximum metric \(d((x_1,x_2),(y_1,y_2))=\max\{|x_1-y_1|,|x_2-y_2|\}\). Define \(f(x,y)=(2y,x/8)\). This map sends \(X\) into itself. It is not a contraction: for \(u=(0,1)\) and \(v=(0,0)\), \(d(u,v)=1\), while

f(u)=(2,0),\qquad f(v)=(0,0),\qquad d(f(u),f(v))=2.

Thus no contraction constant \(q<1\) can satisfy the contraction inequality for all pairs of points. However, direct calculation gives

$$ f^2(x,y)=f(2y,x/8)=(2(x/8),(2y)/8)=(x/4,y/4). $$

For any \((x,y),(a,b)\in X\),

d(f^2(x,y),f^2(a,b)) =\max\{|x-a|/4,|y-b|/4\} =\frac14d((x,y),(a,b)).

So \(f^2\) is a contraction with constant \(1/4\). The theorem applies, and \(f\) has exactly one fixed point. To identify it directly, solve \(f(x,y)=(x,y)\): the equations are \(x=2y\) and \(y=x/8\). Substituting the second into the first gives \(x=x/4\), so \(x=0\), and then \(y=0\). The unique fixed point is \((0,0)\). This example shows why testing only the one-step distance change can miss contraction behavior that appears after several steps.

Worked Example: Checking Each Part of the Transfer

Use the same map \(f(x,y)=(2y,x/8)\). Its square has the unique fixed point \((0,0)\): solving \(f^2(x,y)=(x,y)\) gives \(x/4=x\) and \(y/4=y\), hence \(x=y=0\). The iterate argument transfers this point to \(f\). Indeed, \(f(0,0)=(0,0)\).

The uniqueness step can also be seen explicitly. If \(r=(a,b)\) is any fixed point of \(f\), then

$$ f^2(r)=f(f(r))=f(r)=r. $$

Since the only fixed point of \(f^2\) is \((0,0)\), we must have \(r=(0,0)\). Notice that the contraction theorem is applied to \(f^2\), not to \(f\). The fixed-point equations confirm the conclusion here, but the theorem provides the general reasoning that works even when those equations cannot be solved explicitly.

Worked Example: Why a Fixed Point of an Iterate Need Not Be Fixed by the Map

Define \(g:[0,1]\to[0,1]\) by \(g(x)=1-x\). Applying the map twice gives

$$ g^2(x)=g(1-x)=1-(1-x)=x. $$

Consequently every point of \([0,1]\) is fixed by \(g^2\). But \(g(x)=x\) requires \(1-x=x\), so the only fixed point of \(g\) is \(x=1/2\). The transfer theorem does not apply: \(g^2\) is the identity map and is not a contraction on \([0,1]\). For instance, it preserves the distance between \(0\) and \(1\), which is \(1\), and therefore cannot reduce all distances by a factor less than \(1\). The example isolates the necessary ingredient: uniqueness of the fixed point of the iterate is what lets commutation force \(f(p)=p\).

Commuting Maps Preserve the Fixed Point

The same commutation argument is useful even when the map that commutes with a contraction is not itself a contraction. If a map \(S\) commutes with a contraction \(T\), it must send the unique fixed point of \(T\) to another fixed point of \(T\). Uniqueness then forces that image to be the original point.

Theorem (Commuting Maps Fix the Contraction’s Fixed Point): Let \((X,d)\) be a nonempty complete metric space. Suppose \(T:X\to X\) is a contraction and \(S:X\to X\) satisfies \(S\circ T=T\circ S\). If \(p\) is the fixed point of \(T\), then \(S(p)=p\). No continuity or contraction assumption on \(S\) is needed.

Proof. Since \(T(p)=p\) and \(S\) commutes with \(T\),

$$ T(S(p))=S(T(p))=S(p). $$

Thus \(S(p)\) is a fixed point of \(T\). The Contraction Mapping Theorem gives existence of \(p\), and the Uniqueness of a Fixed Point theorem gives that \(T\) has at most one fixed point. Therefore \(S(p)=p\). \(\square\)

Worked Example: A Reflection Commuting with a Contraction

On \(X=\mathbb{R}\) with its usual metric, let \(T(x)=x/4+3/4\) and \(S(x)=2-x\). For all \(x,y\in\mathbb{R}\),

|T(x)-T(y)| =\left|\frac{x-y}{4}\right| =\frac14|x-y|.

Thus \(T\) is a contraction. Its fixed point satisfies \(p=p/4+3/4\), so \(3p/4=3/4\) and \(p=1\). To check commutation, calculate both compositions:

T(S(x))=T(2-x)=\frac{2-x}{4}+\frac34 =\frac54-\frac{x}{4},

S(T(x))=2-\left(\frac{x}{4}+\frac34\right) =\frac54-\frac{x}{4}.

Hence \(T\circ S=S\circ T\). The commuting-map theorem implies that \(S\) fixes \(p\), and indeed \(S(1)=2-1=1\). The map \(S\) is not a contraction because \(|S(x)-S(y)|=|x-y|\), but it still must preserve the contraction’s unique fixed point.

Periodic Points and a Useful Pitfall

A fixed point is a point with period one. More generally, a point \(x\) is periodic if \(f^r(x)=x\) for some positive integer \(r\). If some iterate of \(f\) is a contraction, there can be no periodic points other than the unique fixed point. This rules out cycles such as two points that alternate under repeated application of the map.

Theorem (No Nontrivial Periodic Orbits): Suppose \(X\) is a nonempty complete metric space and \(f:X\to X\) has a contracting iterate \(f^m\). Then every periodic point of \(f\) is the unique fixed point of \(f\).

Proof. By the Uniqueness from a Contracting Iterate theorem, \(f\) has a unique fixed point \(p\). Let \(x\in X\) satisfy \(f^r(x)=x\) for some positive integer \(r\). Set \(L=mr\). Then \(f^L=(f^m)^r\). Since \(f^m\) is a contraction, the Contraction Estimate for Iterates shows that \((f^m)^r\) is a contraction: if \(q<1\) is a contraction constant for \(f^m\), then

d(f^L(u),f^L(v))\leq q^r d(u,v)

for all \(u,v\in X\), and \(q^r<1\). Also, \(f^L(x)=x\), because \(L\) is a multiple of \(r\). Since \(f(p)=p\), we also have \(f^L(p)=p\). The contraction \(f^L\) has at most one fixed point, so \(x=p\). Therefore every periodic point is the same point \(p\), as claimed. \(\square\)

The hypothesis that an iterate is a contraction cannot be replaced by the claim that an iterate merely has fixed points. The map \(g(x)=1-x\) above has every point fixed by its square but has only one fixed point itself. The transfer works because the contracting iterate has a unique fixed point; that uniqueness is precisely what makes commutation decisive.

There are two different uses of completeness in these arguments. It is needed for the Contraction Mapping Theorem to guarantee that a contracting map, such as \(f^m\), has a fixed point in \(X\). Once a fixed point of a contraction is already known, the uniqueness arguments use the contraction inequality and do not need completeness. Keeping these roles separate helps prevent a common logical error: asserting that \(f\) has a fixed point immediately after applying the Contraction Mapping Theorem to \(f^m\), without first showing that the fixed point of \(f^m\) is fixed by \(f\).

Check Your Understanding

Use the iterate and commutation arguments to answer the following questions.

If \(f^m\) is a contraction on a nonempty complete metric space, which map does the Contraction Mapping Theorem first guarantee has a fixed point?
In the proof of uniqueness from a contracting iterate, what does commutation show about \(f(p)\), where \(p\) is fixed by \(f^m\)?
Why is every fixed point of \(f\) also a fixed point of \(f^m\)?
Does the commuting-map theorem require the map \(S\) to be continuous or a contraction? Explain which hypothesis on \(S\) is used.
Why does the map \(g(x)=1-x\) on \([0,1]\) not contradict uniqueness from a contracting iterate?

What You'll Learn