The Browder and Kirk Fixed Point Theorem

Wed, 15 Jul 2026 08:57:07 -0400

At its roots, mathematical analysis is a field about existence of things that may or may not be. Fundamentally this is rooted in our belief that "if we can come arbitrarily close to capturing a particular object, then such an object should exist". It is with this belief that the fundamental concept of completeness arose in analysis: the rational numbers if not complete, as we can take a sequence of rational numbers that seem like it ought to "converge", yet has no rational limit. A significant amount of mathematical analysis resolves around proving theorems that take advantage of two combined assumptions

The completeness of an underlying structure.
The "smallness" of a particular (collection of) object(s).

to derive the existence of a "limiting object".

Two basic examples of "smallness" in the context of complete metric spaces:

A sequence/net of points is "small" if it is Cauchy: this guarantees that it has a limit.
A subset is "small" if it is compact: this guarantees that every sequence/net in it as limit points.

A related technique in analysis is that of a "fixed point construction". Suppose we are trying to find an object $\hat{x}$ in some metric space $M$, based on the satisfaction of a certain criterion $P$. Now $\hat{x}$ is in general hard to find, but supposed that the criterion $P$ is formulated in such a way that given any $x\in M$, we can find a new $y\in M$ that "better meets the requirement than $x$". Then we have hope to generate a "sequence of improvements" \[ x_0 \mapsto x_1 \mapsto x_2 \mapsto x_3 \mapsto \ldots \] that, in principle, should yield $\hat{x}$ in the limit. Let now $\mathcal{F}:M\to M$ be the procedure that generates the improvement $y = \mathcal{F}(x)$ from the initial guess. Then our desired $\hat{x}$ would be an element that cannot be improved further, which would require $\hat{x} = \mathcal{F}(\hat{x})$. The mathematical jargon for this final relation is "$\hat{x}$ is a fixed point of $\mathcal{F}$."

In summary, a lot of constructions in mathematical analysis can be performed by

identifying some metric space $M$
identifying some "improvement map" $\mathcal{F}:M\to M$
looking for fixed point(s) of $\mathcal{F}$

So it is of interest to ask: under what assumptions on the metric space $M$ and improvement map $\mathcal{F}$ can we expect there to be a fixed point? In particular, given that the fixed point construction is motivated by the convergence of some iterative construction, we should expect the "smallness" notions of Cauchy-ness and Compactness to perhaps play a role: the issue is then how do we translate these to "smallness" of $\mathcal{F}$? Results of these type are known as "Fixed Point Theorems".

The two basic fixed point theorems

The two most well-known in these family are those of of Banach and Brouwer, which conveniently reflect the two notion of "smallness" mentioned above.

Theorem [Banach Fixed Point Theorem]

Let $M$ be a complete (non-empty) metric space, and assume that $\mathcal{F}:M\to M$ is a contraction mapping (i.e. there is a $\lambda < 1$ such that $d(\mathcal{F}(x), \mathcal{F}(x')) \leq \lambda d(x,x')$ for every $x,x'\in M$), then $\mathcal{F}$ has a unique fixed point.

Note that contraction mappings shrink distances between points by at least a fixed factor $\lambda$. This allows us to ensure that the sequence \[ x \mapsto \mathcal{F}(x) \mapsto \mathcal{F}(\mathcal{F}(x)) \mapsto \cdots \] is a Cauchy sequence, for any starting $x\in M$.

Theorem [Brouwer/Schauder Fixed Point Theorem]

Let $B$ be a Banach space, and $M\subseteq B$ a compact and convex non-empty subset. Assume that $\mathcal{F}:M\to M$ is a continuous mapping. Then $\mathcal{F}$ has at least one fixed point.

Here the "smallness" is applied on the domain side (that $M$ is compact), but to leverage this smallness we need a compatibility condition on the mapping ($\mathcal{F}$ is continuous; recall that continuous mappings send compact sets to compact sets) as well as a technical and topological condition that $M$ is convex. The significance of convexity is illustrated by the case of the annulus:

Let $B = \mathbb{R}^2$ the plane, and let $M$ be the set of all points $\{ x: |x| \in [1,2]\}$. Then $M$ is a compact set that is not convex. If we let $\mathcal{F}:B\to B$ be the function that rotates the plane by 90 degrees, then we see that its restriction to $M$ has no fixed points.

When doing nonlinear analysis in function spaces, especially in the context of solving differential or integral equations, we want to work within some Banach space $B$, and the problem we are trying to solve usually dictates what the function $\mathcal{F}:B\to B$ is. Frequently the application of fixed points theorems boils down to finding an appropriate subset $M\subseteq B$: this is the case when using Picard iteration to construct solution to differential equations. For Banach's fixed point theorem we need to find some closed subset $M$ that is mapped into itself by $\mathcal{F}$, and such that on it $\mathcal{F}$ restricts to a contraction mapping. For Brouwer's fixed point theorem we need to find some compact and convex set $M$ which $\mathcal{F}$ maps into itself.

Sometimes, however, neither is possible. Typically finding $M$ that is mapped into itself by $\mathcal{F}$ is not too difficult, as in situations where one can expect a fixed point argument to work one can usually find some nice closed balls in $B$ that satisfies this property. The difficulty is usually in guaranteeing the "smallness" criterion: either $\mathcal{F}$ is a contraction mapping on $M$, or that $M$ can be further shrunk to be compact (remember: on infinite dimensional Banach spaces closed balls of non-zero radius are not compact).

The Browder-Kirk Theorem

The Browder-Kirk Theorem has an in-between notion of "smallness", at the expense of a technical assumption on the space itself.

Theorem [Browder-Kirk Fixed Point Theorem]

Let $B$ be a uniformly convex Banach space, and $M\subseteq B$ a closed, bounded, and convex non-empty subset. Assume that $\mathcal{F}:M\to M$ is a non-expansive mapping (i.e. $\|\mathcal{F}(x) - \mathcal{F}(x')\| \leq \|x - x'\|$). Then $\mathcal{F}$ has at least one fixed point.

Before talking about the proof, let's contrast this theorem against the Banach and Brouwer theorems, focusing on the assumptions. Firstly, in general, recall that the "magic" that begets the existence of a certain element in analytic settings is usually "completeness". So this assumption has to be present in our theorem hypotheses. That in all three theorems the set $M$ is closed reflects this (recall that compact sets are always closed, and that closed subsets of complete metric spaces are themselves complete). Hence we will focus on the other hypotheses. Convexity is easily seen to be required, using the annulus example from the previous section, so we will not speak more of it.

The non-expansive assumption is a relaxation of the contraction mapping requirement of the Banach theorem.
- We see that boundedness is absolutely a price we must pay here (boundedness is not required in Banach theorem). Example: let $B = \mathbb{R}$ and $\mathcal{F}(x) = x + 1$. Then the mapping is non-expansive and has no fixed points. Note that it does not map any bounded non-empty subset of $\mathbb{R}$ to itself.
- Uniform convexity is also required, due to the following example. Let $B = \ell_1$, the space of absolutely summable sequences of real numbers. Let $M$ be the subset $\{\vec{x} = (x_n): x_n \geq 0, \sum x_n = 1\}$. This set is clearly convex and bounded. It is closed as if $\vec{x}_k$ is a converging sequence of elements of $M$, we have that each coordinate much converge separately. Thus as each coordinate is assumed non-negative for each $\vec{x}_k$, so must it be for the limit. Denoting the limit by $\vec{y} = (y_n)$, since $y_n \geq 0$ we have $\sum y_n = \|\vec{y}\| = \|\lim \vec{x}_k\| = \lim \|\vec{x}_k\| = 1$. This shows that the limit $\vec{y}\in M$ also. But now the right-shift function $\mathcal{F}$ defined by $(x_1, x_2, \ldots) \mapsto (0,x_1, x_2, \ldots)$ is clearly an isometry of $\ell_1$, and hence is non-expansive. But $\mathcal{F}$ has no fixed points in $M$. (We will return to the notion of uniform convexity and this counterexample later.)
The assumption of closed and bounded is a relaxation of the compactness requirement of the Brouwer/Schauder theorem. Note that when $B$ is finite dimensional, thanks to the Bolzano-Weierstrass theorem compactness is equivalent to closed and bounded. So this is only a relaxation in the infinite dimensional case.
- Non-expansive is required. (Note that non-expansive maps are automatically Lipschitz continuous, so is a strengthening of the continuity assumption of Brouwer's theorem.) Let $B = \ell_2$, the space of square-summable sequences, and which is a Hilbert space (so is uniformly convex). Let $M$ be the closed unit ball (so closed, bounded, and covex). Let $\mathcal{F}$ be the map that sends $\vec{x} = (x_1, x_2, x_3, \ldots) \mapsto (\sqrt{1 - \|\vec{x}\|}, x_1, x_2, \ldots)$. If $\vec{x} \in M$, then $\|\vec{x}\| \leq 1$ and so $\sqrt{1-\|\vec{x}\|}$ is well-defined. The resulting element has norm 1. Given $\vec{x},\vec{y}\in M$, we have that $\| \mathcal{F}(\vec{x}) - \mathcal{F}(\vec{y})\| \leq |\sqrt{1 - \|\vec{x}\|} - \sqrt{1 - \|\vec{y}\|}| + \|\vec{x} - \vec{y}\|$, which implies that $\mathcal{F}$ is continuous (in particular, there exists a continuous map from the closed unit ball $M$ to its boundary!) This map $\mathcal{F}$ has no fixed points: for $\vec{x} = \mathcal{F}(\vec{x})$, it is necessary that all coordinates are equal. But there are no element of $M$ with norm $1$ and equal coordinates. (We remark that $\mathcal{F}$ is not non-expansive. The origin is mapped to $(1,0,0,\ldots)$, which in turn is mapped to $(0,1,0,0,\ldots)$. The distance between the first two is $1$ and the distance between the latter two is $\sqrt{2} > 1$.)

Uniform convexity and where the "smallness" comes from

The relaxation of contraction mapping to non-expansive mapping seemingly removes the use of "Cauchyness" as a smallness condition, while the relaxation of compactness assumption also removes its use as a smallness condition. So where does the requisite smallness come from in the Browder-Kirk theorem? And more precisely, why does the notion of "uniform convexity" help? At this point, it may be useful to recall the definition of uniform convexity:

Definition

A Banach space is uniformly convex if there exists an increasing function $\delta: (0,2]\to (0,1]$ such that for every $\epsilon \in (0,2]$ we have \[ \|x\| \leq 1, \|x'\| \leq 1, \|x - x'\| \geq \epsilon \implies \|(x+x')/2\| \leq 1 - \delta(\epsilon). \] Note that by taking $x' \to x$ we see that $\lim_{\epsilon\to 0}\delta(\epsilon) =0$.

Note that one of the consequences of uniform convexity is strict convexity: if $x$ and $x'$ are distinct unit vectors, then $x + x'$ has norm strictly less than 2. This also means that given two vectors $y$ and $z$, the only element $w$ satisfying $\|w - y\| = \|w - z\| = \frac12 \|y-z\|$ is the midpoint $w = \frac12(y+z)$.

Non-expansive maps are therefore much more rigid on uniformly convex spaces than in general: suppose we know that \[ \| \mathcal{F}(y) - \mathcal{F}(z) \| = \|y - z\| \] Let $x$ be a convex combination $x = \alpha y + (1-\alpha) z$ for $\alpha \in (0,1)$. Then the non-expansive property requires \[ \| \mathcal{F}(x) - \mathcal{F}(y) \| \leq \| x - y\| = (1-\alpha) \|y - z\|, \qquad \| \mathcal{F}(x) - \mathcal{F}(z) \| \leq \|x - z\| = \alpha \|y - z\| \] and so by triangle inequality we have that the two inequalities must be equalities. But then from strict convexity we conclude that in fact \[ \mathcal{F}(x) = \alpha \mathcal{F}(y) + (1-\alpha) \mathcal{F}(z)\] In other words, $\mathcal{F}$ restricts to a linear function on the line segment joining $y$ and $z$.

The $\ell_1$ space used earlier in the example with the shift operator is not uniformly convex: Letting $x = (1,0,0,\ldots)$ and $x' = (0,1,0,0,\ldots)$, we see that $\|x\| = \|x'\| = 1$ and $\|x - x'\| = 2 \geq \epsilon$ for any $\epsilon \in (0,2]$. But the midpoint $(1/2,1/2,0,0,\ldots)$ has norm $1$. Clarkson's inequalities on the other hand guarantees that $\ell_p$ for $p \in (1,\infty)$ are uniformly convex. So let's see why the counterexample doesn't work in $\ell_p$.

Taking $B$ to be the sequence space $\ell_p$, which has norm $\|\vec{x}\| = \Big( \sum |x_n|^p \Big)^{1/p}$, we can let $\mathcal{F}$ again be the shift operator $(x_1, x_2, \ldots) \mapsto (0, x_1, x_2, \ldots)$. The shift operator is again an isometry of $\ell_p$, and so is non-expansive. We continue to let $M$ be the subset of $B$ of those points $\vec{x}$ such that every coordinate is non-negative, and $\sum x_n = 1$. The exact same argument as before shows that the shift operator has no fixed points in this set: a fixed point of $\mathcal{F}$ would have, using the formula, $x_1 = 0$, $x_2 = x_1$, and so on, which would require it to be the $0$ vector, which is not an element of $M$. So why couldn't we apply the Browder-Kirk theorem in this case?

This set $M$ is still clearly convex. And when $p\in (1,\infty)$, this set is bounded: noting that for a sum of non-negative numbers to equal $1$, each coordinate must be no larger than $1$, and hence we have $|x_n|^p \leq x_n$. Thus we find \[ \vec{x} \in M \implies \|\vec{x}\| \leq \Big( \sum x_n \Big)^{1/p} = 1 .\] However, $M$ is no longer closed: let $\vec{x}_k$ be the vector whose first $k$ coordinates equal $1/k$ and the remaining ones are $0$. Each $\vec{x}_k\in M$. For $p \in (1,\infty)$, we can check that $\|\vec{x}_k\| = \Big( k \cdot k^{-p}\Big)^{1/p} = k^{1/p - 1}$, which has a negative exponent and so tend to zero as $k\to\infty$. This shows that $\vec{x}_k$ converges to the 0 vector, so that the 0 vector is in the closure of $M$. Now this $\vec{x}_k$ can be seen to be the average of the first $k$ coordinate unit vectors. The fact that $\vec{x}_k$ norm-converges to $0$ captures the fact that, in uniformly convex spaces, averages of vectors tend to have quantitatively smaller norm than the starting vectors.

Another way to understand this difference is to think about the coordinate unit vectors. In $\ell_1$, the set $M$ is the closed convex hull of the set of all coordinate unit vectors. But in $\ell_p$, taking the closed convex hull yields something larger: the origin must also be included.

Having illustrated the differences, let's talk about how uniform convexity can actually help in this theorem.

Recall that our goal is to find a fixed point. The "compactness" style of argument demonstrates the existence of a limit point by making sure that the domain is not too large. The "Cauchy" style of argument demonstrates the existence of a limit point by ensuring that it converges to a specific point. Now, let's see what can happen in the setting of the Browder-Kirk Theorem.

Consider the situation that $B$ is some Banach space, and $M$ the closed unit ball (so is convex). Consider the mapping that acts as the identity on the ball of radius $1/2$, and outside of this ball it sends $x \mapsto x / (2 \|x\|)$. It is not too hard to see that this mapping in non-expansive. Now, the set of all fixed points of this mapping is the closed ball of radius $1/2$, which cannot be contained in a compact set (in the infinite dimensional case), and so we cannot use the compactness style of argument.

So the only hope is to use some "Cauchy" style of argument. But this requires us to identify a specific point to converge to, but given the set of all possible fixed points, how do we choose one?

This is the genius part of the proof and where uniform convexity is used. The two key observations are:

In a uniformly convex space, the set of all fixed points of a non-expansive map is a closed and convex set.
In a uniformly convex space, given a closed and convex set $K$ and a point $p\not\in K$, there is a unique point $q\in K$ that minimizes the distance to $p$.

The strategy employed in Goebel's proof is to identify such a point using very geometric arguments.

Before talking about Goebel's proof, let's first discuss the first claim about the convexity of the set of fixed points. (As non-expansive maps are continuous, the set of fixed points is necessarily closed.) Observe that as we discussed above, for non-expansive maps on a uniformly convex space, if equality $\|\mathcal{F}(x) - \mathcal{F}(x')\| = \|x - x'\|$, then $\mathcal{F}$ must restrict to a linear map on the line segment joining $x$ and $x'$. Thus if $x$ and $x'$ are fixed by $\mathcal{F}$, so must the line segment joining them.

Goebel's proof

Step 1: Almost fixed points

Without loss of generality we can assume $0\in M$ by overall translation. Since $M$ is convex, a rescaling of $M$ by a factor $< 1$ is still contained in $M$. Consider $\lambda = 1-\epsilon$. Consider the mapping $\mathcal{G} = \lambda \mathcal{F}$: since $\mathcal{F}$ is non-expansive, we see that $\mathcal{G}$ is a contraction mapping of $M$. And so by Banach fixed-point theorem $\mathcal{G}$ has a unique fixed point $x_\epsilon$. We can compute \[ \| x_\epsilon - \mathcal{F}(x_\epsilon)\| = \| \mathcal{G}(x_\epsilon) - \mathcal{F}(x_\epsilon) \| = \epsilon \| \mathcal{F}(x_\epsilon)\| \leq \epsilon \cdot \mathrm{diam}(M). \] Since we assumed that $M$ is bounded, this means that for every $\tilde{\epsilon} > 0$ there exists at least one point $x$ in $M$ such that $\| x - \mathcal{F}(x) \| \leq \tilde{\epsilon}$.

We remark that the argument thus far only used that $M$ is closed, convex, and bounded, and that $\mathcal{F}$ is non-expansive. The uniform convexity assumption is not used anywhere. One can see that the conclusion thus far also applies to $\ell_1$ with the shift operator.

The above argument shows that the following family of sets, indexed by $\epsilon > 0$, are all non-empty. \[ C(\epsilon) = \{ x\in M: \| x - \mathcal{F}(x) \| \leq \epsilon. \] As $\mathcal{F}$ ix non-expansive and hence continuous, we see that $C(\epsilon)$ are nested closed sets.

Note also by the non-expansive property of $\mathcal{F}$, we have $\mathcal{F}(C(\epsilon)) \subseteq C(\epsilon)$.

Now, the intersection $\cap_{\epsilon > 0} C(\epsilon)$ would be precisely the fixed point set of $\mathcal{F}$: our goal is to prove that it is non-empty.

Step 2: Key dichotomy

In general, the fixed point set may be large, as described before. So instead of finding the entire intersection, let's try to locate its nearest point to the origin (which we assumed to be in $M$).

If $0 \in C(\epsilon)$ for all $\epsilon$, then $0$ is a fixed point, and we are done. So we can focus on the case where $0$ is not the fixed point.

Since $C(\epsilon)$ are closed, if $0\not\in C(\epsilon)$ for some $\epsilon$, then some open ball around $0$ is also avoided. We can let $a_\epsilon = \inf \{ \|x\|: x\in C(\epsilon)\}$. By the nested property of $C(\epsilon)$ we see that $a_\epsilon$ increases as $\epsilon \to 0$; since $M$ is bounded and each $C(\epsilon)$ is non-empty, we have also that $a_\epsilon$ is bounded above by the diameter of $M$, and so has a limit, which we call $a$.

Since we are interested in finding the closest point to the origin, we only need to look at those points in $C(\epsilon)$ that are approximately distance $a$ away from the origin.

Now consider the set $D(\epsilon) = \{ x \in C(\epsilon): \|x\| \leq a + \epsilon \}$. By definition $D(\epsilon)$ are also nested, and closed.

If we can show that all of the $D(\epsilon)$ has a common intersection, then we are done.

What is the intuition here? Letting $q\in M$ be the fixed point of $\mathcal{F}$ with the smallest norm, by uniform convexity and the convexity of the fixed point set, $q$ is unique. So for all other points $y$ with norm $\|y\| \leq \|q\|$, there is some $\epsilon$ such that $y\not\in C(\epsilon)$. For all other points $y$ with norm $\|y\| > \|q\|$, there is some $\epsilon$ such that $y\not\in D(\epsilon)$. So we do expect that $D(\epsilon)$ to have a single point of intersection. This suggest that we can use a Cauchy-style argument.

We will therefore leverage Cantor's theorem, which states that a nested family of closed sets with diameter tending to zero has a single unique point of common intersection.

Why should this be the case? Suppose we have two points $x, x'$ in $D(\epsilon)$ that are "far apart" (as in $\gg \epsilon$). Then since $\mathcal{F}$ almost fixes $x$ and $x'$ (up to an $\epsilon$) error, we have that $\mathcal{F}$ is almost a linear function when restricted to the line segment joining $x$ and $x'$. But this will in turn imply that $\mathcal{F}$ also send the midpoint of this line segment almost to itself, which means that the midpoint belongs to some $C(\epsilon')$. But now uniform convexity kicks in: by assumption$x,x'$ are both approximately distance $a$ from the origin, and they are far apart, thus uniform convexity tells us that the midpoint should be significantly closer to the origin (compared to $a$). This would lead to a contradiction if this smaller distance is $< a_{\epsilon'}$.

Step 3: Smallness of $D(\epsilon)$.

To implement our plan, given $x, x'\in D(\epsilon)$, and $y$ their mid-point, we need to be able to control $\| \mathcal{F}(y) - y\|$. To this end, it is more convenient to use the contrapositive of the uniform convexity assumption, which reads: \[ \|z\| \leq 1, \|z'\| \leq 1, \|\frac12 (z + z')\| \geq 1 - \delta(\epsilon') \implies \|z - z'\| \leq \epsilon'. \] We shall apply this with $w$ being one of the $x, x'$, and \[ z = y - w, z' = \mathcal{F}(y) - w. \] Let $2\ell = \|x - x'\|$. By definition, $\|z\| = \ell$. We can estimate \[ \| z' \| = \| \mathcal{F}(y) - \mathcal{F}(w) + \mathcal{F}(w) - w \| \leq \|z\| + \|\mathcal{F}(w) - w\| \leq \ell + \epsilon. \] By triangle inequality, for any point $p$ at least one of $\|p - x\|$ and $\|p - x'\|$ is half of $\|x - x'\|$. So it is possible to select $w$ such that \[ \| \frac12(z + z') \| = \| \frac12 (\mathcal{F}(y) + y) - w\| \geq \ell. \] So if we denote by $\eta$ the inverse function of $\delta$, we see that, applying the contrapositive to $z / (\ell + \epsilon)$ and $z' / (\ell + \epsilon)$, we find their mean has norm $\geq \frac{\ell}{\ell + \epsilon} = 1 - \frac{\epsilon}{\ell + \epsilon}$ and hence we can conclude \[ \|\mathcal{F}(y) - y \| = \|z - z'\| \leq (\ell + \epsilon) \eta( \frac{\epsilon}{\ell + \epsilon}). \] Note that this tends to zero as $\epsilon \to 0$. In fact, using that $M$ is bounded, we have that $\ell \leq \frac12 \mathrm{diam}(M)$. And since $\eta$ is also an increasing function (as it is the inverse of an increasing function) we find that as long as $\ell \geq \sqrt{\epsilon}$, we have \[ \|\mathcal{F}(y) - y\| \leq (\frac12 \mathrm{diam}(M) + \epsilon) \eta(\sqrt{\epsilon}) = \phi(\epsilon). \] The right hand side tends to zero as $\epsilon \to 0$ (noting that $\delta(\epsilon) > 0$ for all $\epsilon$).

This indicates that $y \in C(\phi(\epsilon))$, and by our earlier discussion must satisfy $\|y \| \geq a_{\phi(\epsilon)}$. But as $y$ is the midpoint of $x$ and $x'$, we can apply the contrapositive of the uniform convexity assumption again, which feeds back to provide \[ \|x - x'\| \leq (a + \epsilon) \eta( 1 - \frac{a_{\phi(\epsilon)}}{a + \epsilon} ) \] Nothing that this is under the assumption that $\|x - x'\| \geq 2 \sqrt{\epsilon}$, we conclude the following:

Given $x, x'\in D(\epsilon)$, then either $\|x-x'\| \leq 2 \sqrt{\epsilon}$, or $\|x - x'\| \leq (a + \epsilon) \eta( 1- \frac{a_{\phi(\epsilon)}}{a + \epsilon})$.

Or equivalently: \[ \mathrm{diam}(D(\epsilon)) \leq \max\Big( 2 \sqrt{\epsilon}, (a +\epsilon) \eta( 1- \frac{a_{\phi(\epsilon)}}{a+\epsilon}) .\] Using that $a_{\phi(\epsilon)} \to a$ as $\epsilon \to 0$, we thus conclude that $\mathrm{diam}(D(\epsilon)) \to 0$ and the theorem is proved.

fixed point theorems on QNLW