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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 $g(x)<x$ 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.

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  • $\begingroup$ At first glance it feels like a contraction mapping theorem type thing, though I could be wrong. $\endgroup$
    – Alan
    May 17, 2021 at 16:05
  • $\begingroup$ Which fixed-point theorems have you investigated? Can you go into more detail how they might relate with your fixed-point result? $\endgroup$
    – supinf
    May 17, 2021 at 17:12
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    $\begingroup$ 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. $\endgroup$
    – Hölderlin
    May 18, 2021 at 9:36

2 Answers 2

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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.

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    $\begingroup$ 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. $\endgroup$
    – Hölderlin
    Jul 8, 2021 at 15:20
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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 $$\|x_{n+1}-x_n\|=\|F(x_n)-F(x_{n-1})\|<g(\|x_n-x_{n-1}\|)<\|x_n-x_{n-1}\|$$ then the sequence is Cauchy's.

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    $\begingroup$ 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. $\endgroup$
    – supinf
    May 17, 2021 at 17:33
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    $\begingroup$ . . neither do I $\endgroup$
    – janmarqz
    May 17, 2021 at 17:34
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    $\begingroup$ 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. $\endgroup$
    – Hölderlin
    May 18, 2021 at 10:58
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    $\begingroup$ In a little while, I will expand the answer $\endgroup$
    – janmarqz
    May 18, 2021 at 13:47
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    $\begingroup$ I'd still be interested in an expanded answer. :) $\endgroup$
    – Hölderlin
    May 24, 2021 at 19:45

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