# Proof of a fixed-point lemma

I'm trying to prove the following fixed-point lemma.

Let $$\mathcal X$$ be a Banach space and $$A \neq \emptyset$$ a closed, bounded and convex subset of $$\mathcal X$$. Further let $$g: \mathbb R^+ \to \mathbb R^+$$ be continuous with $$g(x) for all $$x>0$$ and let $$F: A \to A$$ be a function such that $$\left\Vert F(x)-F(y) \right\Vert \leq g(\left\Vert x-y \right\Vert)$$ for all $$x,y\in A$$. Then there exists a fixed-point for $$F$$.

This doesn't seem to be a special case of any fixed-point theorem I know. I've combed the proofs of some of them to find a starting point without success. Can anyone help me with this?

One route I've been exploring is to learn more about $$F$$. It is clearly continuous. If $$\overline{F(M)}$$ were compact for every subset $$M$$ of $$A$$, then the claim would directly follow from Schauder's fixed-point theorem which says that each compact $$F$$ in this setting has a fixed-point. However, I don't think that $$F$$ is compact here, at least I see no way of proving it.

Since there exists an "near-fixed-point", that is a sequence $$(x_n)$$ such that $$\|F(x_n)-x_n\|\to 0$$ as $$n\to\infty$$ (this, I know), it would even suffice to show compactness for $$\overline{F(C)}$$ to yield the claim which is a bit weaker. But again I can't show it and don't know whether this is the case.

• At first glance it feels like a contraction mapping theorem type thing, though I could be wrong.
– Alan
May 17, 2021 at 16:05
• Which fixed-point theorems have you investigated? Can you go into more detail how they might relate with your fixed-point result? May 17, 2021 at 17:12
• There is, e.g., the fixed-point theorem of Browder-Göhde-Kirk which in this setting would yield the existence of a fixpoint due to $F$ being a nonexpansive function, but it requires $X$ to be a uniformly convex Banach space. Then there is a version of Schauder's fixed-point theorem that holds in general Banch spaces for compact functions. This lemma seems to fall somewhere inbetween those. May 18, 2021 at 9:36

Elaboration on the hint given by janmarqz:

First, we can assume that $$g$$ is monotonically non-decreasing: Otherwise we can use $$\hat g(s):=\sup_{0\leq t \leq s} g(t)$$ instead to make $$g$$ non-decreasing.

We also define $$\bar s:=\sup\{\|x-y\| \mid x,y\in A\}<\infty$$. Then we have $$\| F(x)-F(y)\| \leq g(\|x-y\|) \leq g(\bar s) \qquad\forall x,y\in A.$$ A repeated application yields $$\|F^n(x)-F^n(Y)\| \leq g^n(\bar s) \qquad\forall x,y\in A,$$ where $$F^n$$, $$g^n$$ denote the $$n$$-fold composition $$F\circ F\circ\cdots\circ F$$ and $$g\circ g\circ\cdots\circ g$$.

Clearly, $$\{g^n(\bar s)\}_{n\in\Bbb N}$$ is a decreasing sequence. If we denote its limit by $$t$$, then continuity of $$g$$ yields $$g(t)=t$$. But this is only possible if $$t=0$$, i.e. $$g^n(\bar s)\to0$$.

Let us return to the given hint: Let $$x_0\in A$$ be given and let us define $$x_{n+1}:=F(x_n) = F^n(x_0)$$ as suggested by janmarqz. Then we have $$\| x_{n}-x_{n+m}\| = \| F^n(x_0)-F^n(x_m)\| \leq g^n(\bar s) \to0 \quad\text{as }n\to\infty.$$ Thus the sequence $$\{x_n\}_{n\in\Bbb N}$$ is a Cauchy sequence. Let $$\bar x\in A$$ denote its limit. Since $$F$$ is continuous, it follows that $$\bar x$$ is a fixed point for $$F$$.

Remarks on generalization:

This also holds true in complete metric spaces (which is maybe a bit surprising). Therefore this result is a strict generalization of the Banach fixed-point theorem (which is just the case $$g(s)=q s$$ for some $$q<1$$).

A proof of this result can be found in Theorem 1 in this article by Boyd and Wong from the 1960s.

Note that my proof above only works for bounded $$A$$, but the result also holds for unbounded complete metric spaces.

• Thank you very much! Btw, I later found out that the this particular problem where $A$ is in fact convex can be solved much simpler by showing that $F$ is a condensing map and applying the Darbo-Sadovskii fixed-point theorem which says that a fixed point exists precisely in this scenario. Jul 8, 2021 at 15:20

Hint: Take $$x_0\in A$$ and iterate $$x_1=F(x_0)$$, $$x_2=F(x_1)$$, . . . , $$x_{n+1}=F(x_n)$$,... so $$\|x_{n+1}-x_n\|=\|F(x_n)-F(x_{n-1})\| then the sequence is Cauchy's.

• Just out of curiosity: do we even need the convexity of $A$ at all? It seems to me that it is not required in your approach. May 17, 2021 at 17:33
• . . neither do I May 17, 2021 at 17:34
• Why is $(x_n)$ Cauchy? In general, $\|x_{n+1}-x_n\|<\|x_n-x_{n-1}\|$ is not a sufficient condition. Does boundedness make it true? I think not, although I'm still thinking about a suitable counterexample. May 18, 2021 at 10:58
• In a little while, I will expand the answer May 18, 2021 at 13:47
• I'd still be interested in an expanded answer. :) May 24, 2021 at 19:45