Complete Spaces and Fixed Points

Fixed Points and the Completion of a Space

The counterexamples in the previous tutorial showed that a contraction on an incomplete space can fail to have a fixed point. Completeness guarantees a fixed point for every contraction, but an individual contraction on an incomplete space may still have one. To understand this distinction, we can place the space inside its completion and extend the contraction there. The extended map always has a unique fixed point; the question is whether that point belongs to the original space.

This viewpoint does not replace the Contraction Mapping Theorem, established earlier in the course. It clarifies what happens when the original space is incomplete: the iteration can converge in the completion even when its limit is missing from the original space. It also explains how a contraction on an incomplete space can have a fixed point when its fixed point remains in that space.

Extending a Contraction to the Completion

Let $(X,d)$ be a nonempty metric space, and let $(\widehat X,\widehat d)$ be its completion. We identify $X$ with its isometric image in $\widehat X$, so $X$ is dense in $\widehat X$. If $f:X\to X$ is a contraction with constant $q$, where $0\leq q<1$, we want to define its value at a point $z\in\widehat X$ by approximating $z$ with points of $X$.

Theorem (Extension of a Contraction to the Completion): Let $(X,d)$ be a nonempty metric space, let $(\widehat X,\widehat d)$ be its completion, and let $f:X\to X$ be a contraction with constant $q<1$. There is a unique map $\widehat f:\widehat X\to\widehat X$ that agrees with $f$ on $X$ and satisfies $$ \widehat d(\widehat f(u),\widehat f(v))\leq q\widehat d(u,v) $$ for all $u,v\in\widehat X$.

Proof. Fix $z\in\widehat X$. Since $X$ is dense in $\widehat X$, there is a sequence $(x_n)$ in $X$ such that $x_n\to z$. This sequence is Cauchy. The contraction inequality gives, for all $m,n$,

d(f(x_n),f(x_m))\leq qd(x_n,x_m).

Thus $(f(x_n))$ is Cauchy in $X$, and hence it converges to a point of $\widehat X$, because $\widehat X$ is complete. Define $\widehat f(z)$ to be this limit.

We must check that this definition does not depend on the approximating sequence. Suppose $x_n\to z$ and $y_n\to z$, with both sequences in $X$. Then

d(f(x_n),f(y_n))\leq qd(x_n,y_n) \leq q\bigl(d(x_n,z)+d(z,y_n)\bigr)\longrightarrow 0.

So the two image sequences have the same limit in $\widehat X$. The definition is therefore well-defined. If $z\in X$, we may use the constant sequence $x_n=z$, which shows that $\widehat f(z)=f(z)$.

Now let $u,v\in\widehat X$. Choose sequences $x_n,y_n\in X$ such that $x_n\to u$ and $y_n\to v$. By the definition of the extension and continuity of the metric,

\widehat d(\widehat f(u),\widehat f(v)) =\lim_{n\to\infty}d(f(x_n),f(y_n)) \leq q\lim_{n\to\infty}d(x_n,y_n) =q\widehat d(u,v).

This proves the contraction inequality. Finally, any continuous map from $\widehat X$ to itself that agrees with $f$ on the dense subset $X$ must agree with $\widehat f$ everywhere: for $z\in\widehat X$, take $x_n\in X$ with $x_n\to z$, and continuity forces the value at $z$ to be the limit of the values at $x_n$. A contraction is continuous, so this applies to any other contraction extension. The extension is unique. $\square$

The proof uses more than a formal extension of the formula for $f$. A point of the completion may not have any expression in terms of operations available inside $X$. Approximating sequences provide a definition that works in any metric space, and the contraction inequality ensures both that the image sequence converges and that its limit is independent of the approximation.

Where the Fixed Point Lies

Because $\widehat X$ is complete, the Contraction Mapping Theorem applies to $\widehat f$. It has a unique fixed point in $\widehat X$. The following consequence identifies exactly when that point also solves the fixed-point problem in $X$.

Theorem (Fixed-Point Location in the Completion): Under the hypotheses of the extension theorem, let $\widehat p$ be the unique fixed point of $\widehat f$ in $\widehat X$. The original map $f:X\to X$ has a fixed point if and only if $\widehat p\in X$. In that case, $\widehat p$ is the unique fixed point of $f$.

Proof. If $\widehat p\in X$, then $\widehat f$ agrees with $f$ on $X$. Hence $\displaystyle f(\widehat p)=\widehat f(\widehat p)=\widehat p$, so $f$ has a fixed point. Conversely, suppose $p\in X$ and $f(p)=p$. Since the extension agrees with $f$ on $X$, we also have $\widehat f(p)=p$. The uniqueness of the fixed point of $\widehat f$ implies $p=\widehat p$. Therefore $\widehat p\in X$. The same argument shows that any fixed point of $f$ must equal $\widehat p$, so it is unique. $\square$

When $X$ itself is complete, its completion adds no points, and this location issue disappears. For an incomplete space, the theorem separates two claims that are easy to conflate: the contraction has a fixed point in the completion, but it has a fixed point in the original space only when that particular limit lies in $X$. Completeness is a guarantee for all contractions, not a requirement for every individual contraction to have a fixed point.

Worked Examples

Worked Example: A Rational Contraction Whose Fixed Point Is Missing

Let $X=\mathbb{Q}\cap[0,1]$ with the usual metric, and define $\displaystyle f(x)=\frac{x^2+1}{3}$. If $x\in X$, then $x^2\in\mathbb{Q}$ and $0\leq x^2\leq1$. Consequently, $\displaystyle \frac13\leq f(x)\leq\frac23$, and $f(x)$ is rational. Thus $f$ maps $X$ into $X$.

For $x,y\in X$,

|f(x)-f(y)| =\frac{|x^2-y^2|}{3} =\frac{|x-y||x+y|}{3} \leq\frac23|x-y|,

because $0\leq x+y\leq2$. So $f$ is a contraction with constant $2/3$. Its extension to the completion $[0,1]$ is given by the same formula. A fixed point must satisfy

x=\frac{x^2+1}{3} \quad\Longleftrightarrow\quad x^2-3x+1=0,

whose roots are $\displaystyle (3-\sqrt5)/2$ and $\displaystyle (3+\sqrt5)/2$. The first root lies in $[0,1]$, since $2<\sqrt5<3$ implies $0<(3-\sqrt5)/2<1/2$. The second is greater than $1$. The first root is irrational: if it were rational, then $\sqrt5$ would be rational, contrary to the irrationality of $\sqrt5$. Therefore the extension has its fixed point in $[0,1]\setminus X$, and $f$ has no fixed point in $X$.

This example exhibits the completion argument directly. The iterates in the rational space approach the unique fixed point in the real completion, but the limit is not rational, so it is not an available value of $x$ in the original fixed-point equation.

Worked Example: A Contraction on an Incomplete Space with a Fixed Point

Let $X=\mathbb{R}\setminus\{0\}$, with the usual metric, and define $\displaystyle g(x)=\frac{1}{2+|x|}$. For every $x\in X$, $g(x)>0$, so $g(x)\in X$. For $x,y\in X$,

|g(x)-g(y)| =\frac{\bigl||x|-|y|\bigr|}{(2+|x|)(2+|y|)} \leq\frac14|x-y|.

Here $\bigl||x|-|y|\bigr|\leq|x-y|$, and each denominator factor is at least $2$. Thus $g$ is a contraction with constant $1/4$. To solve $g(x)=x$, note that $g(x)>0$, so a fixed point must have $x>0$. The equation becomes $\displaystyle x(2+x)=1$, or $x^2+2x-1=0$. Its roots are $-1+\sqrt2$ and $-1-\sqrt2$. The positive root $\sqrt2-1$ belongs to $X$, so it is the fixed point of $g$.

The space $X$ is incomplete because the sequence $1/n$ is Cauchy in $X$ and converges in $\mathbb{R}$ to the missing point $0$. Nevertheless, this particular contraction has a fixed point. Its extension to $\mathbb{R}$ has the same formula, and its fixed point $\sqrt2-1$ is already in $X$. Incompleteness permits failure of the general guarantee; it does not rule out fixed points.

Worked Example: A Fixed Point in a Complete Subspace

Let $X=[0,1]$ with the usual metric and define $\displaystyle h(x)=\frac{x+2}{4}$. For $x\in[0,1]$, we have $\displaystyle \frac12\leq h(x)\leq\frac34$, so $h$ maps $X$ into itself. Also,

|h(x)-h(y)| =\left|\frac{x+2}{4}-\frac{y+2}{4}\right| =\frac14|x-y|.

Thus $h$ is a contraction. Solving $h(x)=x$ gives $\displaystyle (x+2)/4=x$, so $x=2/3$. This point belongs to $[0,1]$. The completion of $X$ is $X$ itself, so the fixed point of the extended map is the same point. Here completeness guarantees existence by the Contraction Mapping Theorem, while the equation also verifies the fixed point explicitly.

What Completeness Does—and Does Not—Say

The completion viewpoint helps avoid two opposite mistakes. The first is to assume that an incomplete space cannot have fixed points: the punctured real line example disproves that. The second is to assume that a contraction on an incomplete space must have one because its iterates seem to settle down. Those iterates may converge only in the completion, as in the rational example.

For an iterative sequence $x_{n+1}=f(x_n)$, the contraction estimate gives the Cauchy control established earlier in the course. If the sequence is considered in the completion, its limit exists there. The extension theorem ensures that the limit is a fixed point of $\widehat f$: since $x_n\to\widehat p$ and $f(x_n)=x_{n+1}\to\widehat p$, continuity of $\widehat f$ gives $\widehat f(\widehat p)=\widehat p$. What completeness of $X$ adds is that this limit belongs to $X$ itself. Without completeness, the location of the limit must be checked.

In applications, one can sometimes work in a complete space larger than the original domain and then determine whether the resulting fixed point satisfies the desired constraint. Conversely, if the intended solution must lie in an incomplete domain, it is not enough to establish contraction and convergence in a completion; one must also show that the fixed point belongs to the domain. The next tutorial develops more abstract fixed-point arguments, where identifying the right space and the location of the limit remains central.

Check Your Understanding

Use the completion and fixed-point location results to answer each question.

Why does a sequence in $X$ converging to $z\in\widehat X$ produce a convergent image sequence under a contraction?
What ensures that the definition of $\widehat f(z)$ is independent of the chosen approximating sequence?
For $f(x)=(x^2+1)/3$ on $\mathbb{Q}\cap[0,1]$, why does the fixed point in the completion not belong to the original space?
How can an incomplete space support a contraction with a fixed point without contradicting the Contraction Mapping Theorem?
In the fixed-point location theorem, why must every fixed point of $f$ equal the fixed point of its extension?